XtraEdge for IIT-JEE 2 FEBRUARY 2012 Volume-7 Issue-8 February, 2012 (Monthly Magazine) NEXT MONTHS...

XtraEdge for IIT-JEE 1 FEBRUARY 2012

Dear Students,

All of us live and work within fixed patterns. These patterns and habits determine the quality of our life and the choices we make in life. There are a few vital things to know about ourselves. We should become aware of how much we influence others, how productive we are and what can help us to achieve our goals. It is important to create an environment which will promote our success. We should consciously create a system that would enable us to achieve our goals. Most of us live in systems which have come our way by an accident, circumstances or people we have met over a period of time. We are surrounded by our colleagues or subordinates who happened to be there by the fact of sheer recruitment earlier or later by the management. Our daily routines and schedules have been formed on the basis of convenience, coincidence, and the expectations of society and sometimes due to superstitions. The trick for success is to have an environment that helps in attaining our goals. Control your life. Make an effort to launch your day with a great start. A law of physics says that an object set in motion tends to remain in motion. It is the same thing with daily routine. To have a good start each morning will keep you upbeat during the day. If you begin the day stressed, you will tend to remain so that way. The best is to create a course of action or conditions where you are not hassled for being late for a meeting, worried about household affairs or distracted by happenings in the world.

Aim to be highly successful. Control the direction of your life. Not only should you start the day on a cheerful note but also continue to do so during the day. Keep yourself stimulated and invigorated during the entire day. Start your day with a purpose. Have a daily direction and trajectory of action. It will keep you on your course all day long. Throughout the day reinforce your positive values and your choices. Anything that helps you in maintaining your highest values and your most important priorities should be welcome. Be in control of your life and work. Create and sustain a wonderful environment filled with beauty, peace, inspiration and hope.

Plan your day in such a way that suits your plans objectives and makes you feel just right with the right amount of encouragement during the entire day. You should give a direction to your day and timing.

Presenting forever positive ideas to your success. Yours truly

Pramod Maheshwari, B.Tech., IIT Delhi

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Worry is a misuse of imagination.

Volume - 7 Issue - 8 February, 2012 (Monthly Magazine)

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Volume-7 Issue-8 February, 2012 (Monthly Magazine)

NEXT MONTHS ATTRACTIONS

Much more IIT-JEE News.

Know IIT-JEE With 15 Best Questions of IIT-JEE

Challenging Problems in Physics,, Chemistry & Maths

Key Concepts & Problem Solving strategy for IIT-JEE.

IIT-JEE Mock Test Paper with Solution

AIEEE & BIT-SAT Mock Test Paper with Solution

S

Success Tips for the Months

• If one asks for success and prepares for failure, he will get the situation he has prepared for.

• Loser's visualize the penalties of failure. Winner's visualize the rewards of success.

• Treat others as if they were what they ought to be and you help them to become what they are capable of being.

• You never achieve real success unless you like what you are doing

• The first step toward success is taken when you refuse to be a captive of the environment in which you first find yourself.

• Believe in yourself ! Have faith in your abilities ! without a humble but reasonable confidence in your own powers you can not be successful or happy.

CONTENTS

INDEX PAGE

NEWS ARTICLE 3 • 25% increase in number of girls from Bombay zone for IIT-JEE • IIT placements : Companies Back on hiring track

IITian ON THE PATH OF SUCCESS 5 Dr. Amitabha Ghosh

KNOW IIT-JEE 6 Previous IIT-JEE Question

XTRAEDGE TEST SERIES 47 Class XII – IIT-JEE 2012 Paper Class XI – IIT-JEE 2013 Paper

Mock Test-3 (CBSE Board Pattern) [Class # XII] 68 Solution of Mock Test-2 & 3 (CBSE Pattern)

Regulars ..........

DYNAMIC PHYSICS 14

8-Challenging Problems [Set # 10] Students’ Forum

Physics Fundamentals Matter Waves, Photo-electric Effect Thermal Expansion, Thermodynamics CATALYSE CHEMISTRY 29

Key Concept Carbonyl Compound Co-ordination Compound & Metallurgy

Understanding : Physical Chemistry

DICEY MATHS 36

Mathematical Challenges Students’ Forum

Key Concept Integration Trigonometrical Equation

Study Time........

Test Time ..........


25% increase in number of girls from Bombay zone for IIT-JEE There are far more girls set to appear for the Indian Institute of Technology Joint Entrance Exam (IIT-JEE) in the Bombay zone in April 2012 than before. The number of applications by girls has increased by nearly 25% compared to last year. The Bombay zone includes Maharashtra, Goa, This year, girls account for nearly 27% of the total candidates applying, with around 20,300 girls from among 76,600 candidates. The IITs have been constantly trying to address the skewed male-female ratio at the IITs. “This is a significant increase,” said AV Mahajan, chairperson for JEE 2012, Bombay zone. “One reason for this could be that it was free for girls to apply online this year as well as cheaper to apply offline as compared to male applicants.”

Two IT-BHU students to start entrepreneurship club for school students VARANASI: When they cracked IIT-JEE two years ago, getting attractive jobs in top IT companies was their dream. Now, they want to promote entrepreneurial skills in young school children (Class X onwards) by establishing entrepreneurship club for schools. Bhanu Swami and Varun Agrawal, B Tech (III) students of electrical engineering of Institute of Technology, Banaras Hindu University (IT-BHU), are on their way to launch country's first entrepre-neurship club targeting students (Class X onwards). The initiative not only aims at tapping and promoting entrepreneurial skills in young students, but also promises to turn these students into entrepreneurs. This way they would become job providers rather than job seekers.

Akash 2 tablet to be launched by April 2012: Govt. The government has expressed confidence it will be successful in bringing out the improved version of the Aakash tablet by April this year. The Aakash 2 is likely to have several improved specifications such as elongated battery life and faster processor. The government has been striving hard to make its Aakash project successful. The device, which is also touted as the world's cheapest tablet, faced harsh criticism from several quarters for its poor quality and dismal features. IIT Rajasthan, which developed the prototype of the device along with Data Wind, found a series of faults in the device, prompting the government to reconsider extension of the LC to the Canadian company. Under pressure to provide better endowed low-cost device, the government has now its turned attention to the Aakash 2, which is likely to come at the same price tag of the original Aakash. HRD Minister Kapil Sibal has said that the government will be looking for more manufacturers to manufacture the Aakash 2 tablets, attributing the massive demand for the move. “We want to make sure that the upgraded product caters to the need of the customers... We have involved ITI in order to upgrade it... We will be able to bring in Aakash-II by April," Kapil Sibal said. Last week media reports had hinted that the government might shelve its Aakash project in view of the harsh criticism from most quarters. The reports were triggered by DataWind's opposition to certain test criteria suggested by the IIT. Read our previous coverage

IIT Techfest 2012 : All about Robotics One of the highlights of the Techfest 2012, which held at IIT Bombay, was its international exhibitions arena. Evidently dominated by robots, the robotic section had covered several aspects - humanoid, surveillance, educational and industrial. The robotic platform at the Techfest 2012 showcased bots expressing emotions, playing intelligent games and serving as a model for military training, among others. With a great potential for applications in future in varying fields, lets look at these robots one by one. Enjoy Moneycontrol.com on iPad and be prepared for a fantastic experience. Get real time stock quotes, interactive charts, market buzz, and watch CNBC-TV18, CNBC Awaaz live on your iPad. Check out the free moneycontrol app

IIT-KGP celebrates the 9th annual Alumni Meet 2012 while its alumni commits to give back to their Alma meter IIT Kharagpur celebrated its 9th Annual Alumni Meet. While it honoured its alumni with the Distinguished Service Award more alumni committed towards the cause of contributing to their Alma mater. A galaxy of alumni from world wide graced the 3-day dynamic event Auto Expo 2012 : Hydrogen-powered 3-wheeler at expo NEW DELHI: The world's first hydrogen-powered three-wheeler, 'HyAlfa', was showcased at theAuto Expo on Monday. Part of a development project dubbed 'DelHy 3w', a fleet of 15 HyAlfa three-wheelers will run on an experimental basis at Pragati Maidan, where a


hydrogen refuelling station has also been set up. India Trade Organisation Promotion ( ITPO) will use the vehicles on an experimental basis. The HyAlfa has been developed under a joint project by the United Nations Industrial Development Organisation ( UNIDO) International Centre for Hydrogen Energy Technologies ( ICHET), Mahindra & Mahindra and IIT-Delhi, with support from the Ministry of New and Renewable Energy. "The aim of this project is to convert vehicles so that they can carry and use hydrogen - a carbon-free fuel - and thus remove all pollutants," Mahindra & Mahindra president (automotive) Pawan Goenka said. He said the vehicle is not yet ready for commercial production and further fine-tuning will be required before moving in that direction. "Moreover, we also have to look at commercial viability of running a hydrogen-powered three-wheeler as the cost of hydrogen will be around Rs 250 per kg, which is not affordable at all," he said. This is a step in the right direction. Finally India Inc. with support from the likes of MNRE etc. is making headway in vehicles driven by alternative energy sources. Jury is still out on whether Hydrogen is the way of the future. Judging by the progress made in the USA and Germany on Hydrogen powered vehicles, we are not quite there yet. There are not only supply chain challenges, but Hydrogen continues to be a costly proposition. You can either produce Hydrogen from hydrocarbon cracking (which in turn has dependence on fossil fuels) or by splitting of water through electrolysis. This itself requires around 5-6 kWh of electricity. So unless the electricity source is a renewable source such as solar or wind, the electricity required to split water itself is most likely to come from thermal power plants, thereby not giving any benefit. However, as everyone knows, human ingenuity knows no bounds and technologies only develop incrementally. I firmly believe that we are one step away from a miracle where both Hydrogen consumption

(by means of fuel cells etc.) and Hydrogen generation (by means of hydrocarbon cracking or electrolyzing/splitting water) become viable for mass production and consumption. In the short term, we can and must focus immediately on "Hydrogen supplementation". By this I mean - Hydrogen CNG (or HCNG in short). New Delhi moved to CNG public transport a while back. In the short term, this did bring down the pollution levels and particulate matter in the atmosphere. After prolonged use, we are now becoming aware of other problems such as NOX emissions due to unclean or inefficient burning of CNG. NOX is highly carcinogenic and now the levels of NOX in New Delhi are far exceeding permissible limits of WHO. One way to solve this problem is to supplement CNG with Hydrogen. A blended product called Hythane (trade name for HCNG) is already under experimentation by Indian Oil. In conclusion, while Hydrogen powered vehicles are all a step in the right direction, the government should put impetus on technologies that are more feasible in solving current big-city problems that Indian cities face.

IITian tops CAT 2011 CHENNAI: Ajinkya Deshmukh, an IIT - Madras graduate, decided to put in 100% effort into preparing for the Common Admission Test this time, and got 100 percentile in return. He is one of nine MBA aspirants in the country to secure the top score this year. The IIMs published the CAT 2011 scores on their website on Wednesday. On his success, Ajinkya, who wrote the test for the third time this year, said, "I always thought I haven't been making a full effort while preparing or writing the test. This time I was determined to give my best. I told myself that I had to do it this year." Indian American sworn in as America's top science official IIT Madras alumnus, Subra Suresh, has been sworn in as the director of America's National Science Foundation (NSF), the top US science body with a $7.4 billion budget to support scientific institutions.

"We are very grateful to have Subra taking this new task," said President Barack Obama at the White House Science after Suresh was sworn in as the 13th NSF director by John Holdren, Obama's science advisor. "He has been at MIT (Massachusetts Institute of Technology) and has been leading one of the top engineering programmes in the country, and for him now to be able to apply that to the National Science Foundation is just going to be outstanding," he said. "So we're very grateful for your service." Suresh, 54, was confirmed by the US Senate for a six-year term. He has served as dean of the engineering school and as Vannevar Bush Professor of Engineering at MIT. A mechanical engineer, who later became interested in materials science and biology, Suresh has done pioneering work studying the biomechanics of blood cells under the influence of diseases such as malaria. From 2000 to 2006, Suresh served as the head of the MIT Department of Materials Science and Engineering. He joined MIT in 1993 as the R.P. Simmons Professor of Materials Science and Engineering and held joint faculty appointments in the departments of mechanical engineering and biological engineering, as well as the division of health sciences and technology. Suresh holds a bachelor's degree from the Indian Institute of Technology in Madras and a master's degree from Iowa State University. Suresh was nominated by President Obama to become the new NSF director in place of Arden L. Bement Jr, who led the agency from 2004 until he resigned in May this yea


Dr Amitabha Ghosh was the only Asian on NASA's Mars Pathfinder mission. At present, he is a member of the Mars Odyssey Mission and the Mars Exploration Rover Mission.

During the Mars Pathfinder Mission, he conducted chemical analysis of rocks and soil on the landing site. The simple and unassuming 34-year-old planetary geologist has won several accolades, which include the NASA Mars Pathfinder Achievement Award in 1997 and the NASA Mars Exploration Rover Achievement Award in 2004.

The journey from India to NASA. It has been an intriguing experience. I was keen on geologic research data interpretation and solar system formation. During my geological research days in India, I had slept in railway stations while traveling to various places.

After my post graduation in applied geology from IIT Kharagpur, I wrote a letter to a professor at NASA expressing a desire to work at the space agency.

I made certain suggestions; in fact, it was a critical letter. In India, you can never imagine criticising your professor.

My suggestions were approved, while I got an opportunity to work at NASA.

I think one requires luck and to put in sincere effort to achieve one's goals. Being in the right place at the right time is also important.

In Mumbai for the Pravasi Bharatiya Divas, he spoke about his work at NASA and his vision for India.

The Vision for India : I feel there India has a great future. We have world-class companies. Today, companies like Infosys can be compared with world leaders like Oracle. Like the Information Technology revolution, we can have a science or space revolution. We have the potential to bring about revolutions in other sectors as well.

How India can we develop science and technology sector : It should be treated as a business. There should be more private participation. We must have an external review to evaluate the system and make changes as science and technology can take the country forward.

We must check brain drain. About 80,000 students migrate to the US for further studies, and settle there. They find the facilities much better abroad. We need to reverse brain drain by enhancing and upgrading institutes in India.

The state of space research in India : I don't want to make controversial statements. All I can say is India is not at the frontier of space research. We have made commendable progress but there is a long way to go. We can do much better. I would be glad to be of help in any way. Investment in research is investment in imagination. It is a matter of national pride and internal recognition. We need to allocate more funds to enhance research and development work.

We need good educational institutes like IITs and IIMs, but IITians don't rule the world. You must remember that Microsoft co-founder (Bill Gates does not have a college degree.

Youngsters must look around for role models and see what it is that they are doing right. Individuals must make use of their inherent strengths to succeed.

How can India become a leading global player : Globalisation will reap huge and long-term benefits and India must make the best use of the opportunities. At the PBD seminar, I found people presenting grandiose plans. Instead, we should look at the realities and immediate solutions.

The private sector has to be actively involved in the development of the country and the government has to respond to the needs of the people. Fifteen years ago, we didn't have an Infosys, today we have many global companies.

There should be drastic reduction in paper work. We need a scenario where one can start any business in a day, like in the US.

Can India have something like NASA: The answer is: Yes, India can. All it requires is the right kind of investment, infrastructure, people and support from the government.

Success Story Success Story This article contains stories/interviews of persons who succeed after graduation from different IITs

Dr. Amitabha Ghosh • Post graduation in applied

geology from IIT Kharagpur,

• Working at NASA


PHYSICS

1. A large open top container of negligible mass and uniform cross-sectional area A has a small holes of cross-sectional area A/100 in its side wall near the bottom. The container is kept on a smooth horizontal floor and contains a liquid of density ρ and mass m0. Assuming that the liquid starts flowing out horizontally through the hole at t = 0, Calculate

[IIT-1997] (i) the acceleration of the container, and (ii) its velocity when 75% of the liquid has drained

out. Sol. (i) Let at any instant of time during the flow, the

height of liquid in the container is x. The velocity of flow of liquid through small hole in the orifice by Toricelli's theorem is

v = gx2 ...(i)

The mass of liquid flowing per second through the orifice

= ρ × volume of liquid flowing per second

dtdm = ρ × gx2 ×

100A ...(iii)

Therefore the rate of charge of momentum of the system in forward direction

= dtdm × v =

100Agx2 ρ×× (From (i) and (ii))

The rate of charge of momentum of the system in the backward direction

= Force on backward direction = m × a where m is mass of liquid in the container at the instant t

m = Vol. × density = A × x × ρ

xρ

v

∴ The rate of charge of momentum of the system in the backward direction

Axρ × a By conservation of linear momentum

Axp × a = 100gxA2 ρ

⇒ a = 50g

(ii) By toricell's theorem v' = )h25.0(g2 × where h is the initial height of the liquid in the container m0 is the initial mass

∴ m0 = Ah × ρ ⇒ h = ρA

m0

∴ v' = ρ

××Am

25.0g2 0 = ρA2

gm0

2. Two parallel plate capacitors A and B have the same separation d = 8.85 × 10–4 m between the plates. The plate area of A and B are 0.04 m2 and 0.02 m2

respectively. A slab of dielectric constant (relative permittivity) K = 9 has dimensions such that it can exactly fill the space between the plates of capacitor B. [IIT-1993]

A

110 V

B

(c)

A

(b)(a)

B

(i) The dielectric slab is placed inside A as shown

in figure (a). A is then charged to a potential difference of 110 V. Calculate the capacitance of A and the energy stored in it.

(ii) The battery is disconnected and then the dielectric slab is moved from A. Find the work done by the external agency in removing the slab from A.

(iii) The same dielectric slab is now placed inside B, filling it completely. The two capacitors A and B are then connected as shown in figure (c). Calculate the energy stored in the system.

KNOW IIT-JEE By Previous Exam Questions


Sol. (i) The capacitor A with dielectric slab can be considered as two capacitors in parallel, one having dielectric slab and one not having dielectric

slab each capacitor has an area of 2A . The

combined capacitance is

A

110 V+

+ + + +

–

– – – +A/2 A/2

C = C1 + C2

= d

)2/A( 0ε +

dr)2/A( 0εε

= d2

A 0ε[1 + εr]

= 4–

12–

1085.821085.84.0

×××× [1 + 9] = 2 × 10–9 F

∴ Energy stored

= 21 CV2 =

21 × 2 × 10–9 × (110)2 = 1.21 × 10–5 J

(ii) Work done in removing the dielectric state = (Energy stored in capacitor without dielectric) – (Energy stored in capacitor with dielectric).

It may be noted that while taking out the dielectric the charge on the capacitor plate remains the same.

∴ W = 'C2

q 2 –

C2q 2

Here, C = 2 × 10–9 F,

C' = d

A 0ε= 4–

14–

1085.81085.804.0

××× = 0.4 × 10–9 F

q = CV = 2 × 10–9 × 110 = 2.2 × 10–7C

∴ W =

×××

9–9–

27–

1021–

104.01

2)102.2(

=

××××

4.024.0–2

102102.22.29–

14–

= 1.21 × 4.06.1 × 10–5 = 4.84 × 10–5 J

(iii) The capacitance of B = dA Br0εε

= 4–

12–

1085.802.091085.8

××××

CB = 1.8 × 10–9 F The charge on A, qA = 2.2 × 10–7 C gets distributed

into two parts. ∴ q1 + q2 = 2.2 × 10–7 C Also the potential

difference across A = p.d. across B

∴ A

1

Cq =

B

2

Cq

⇒ q1 = B

A

CC . q2 = 9–

9–

108.1104.0

×× . q2 = 0.22 q2

∴ 0.22 q2 + q2 = 2.2 × 10–7

⇒ q2 = 22.12.2 × 10–7 = 1.8 × 10–7 C

⇒ q1 = 0.4 × 10–7 C Total energy stored

= A

21

C2q +

B

22

C2q

= 9–

14–

104.02104.04.0

×××× + 8–

14–

108.12108.18.1

××××

= 0.2 × 10–5 + 0.9 × 10–5 = 1.1 × 10–5 J. 3. A particles of mass m = 1.6 × 10–27 kg and charge q =

1.6 × 10–19 C enters a region of uniform magnetic field of strength 1 Tesla along the direction shown in figure. The speed of the particle is 107 m/s. (i) the magnetic field is directed along the inward normal to the plane of the paper. The particle leaves the region of the field at the point F. Find the distance EF and the angle θ. (ii) If the direction of the field is along the outward normal to the plane of the paper, find the time spent by the particle in the region of the magnetic field after entering it at E. [IIT-1984]

F

E

45º

θ × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × ×

Sol. (a) m = 1.6 × 10–27 kg, q = 1.6 × 10–19 C B = 1 T, v = 107 m/s F = q . v B sin α (acting towards O by Fleming's left hand rule)

F = qvB [Q α = 90º] But F = ma ∴ qvB = ma

∴ a = m

qvB

= 27–

719–

106.1110106.1

××××

= 1015 m/s2


× × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × ×

O

F θ

E 45º

45º2VV2 =

2VV2 =

This is the centripetal acceleration

∴ r

v2 = 1015 ∠OEF = 45º

(Q OE, act as a radius)

⇒ r = 15

14

1010 = 0.1 m By symmetry

∠OFE = 45º ∴ ∠EOF = 90º (by Geometry) If the magnetic field is in the outward direction

and the particle enters in the same way at E, then according to Fleming's Left hand rule, the particle will turn towards clockwise direction and cover 3/4th of a circle as shown in the figure.

× × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × ×

× 45º

45º

E

O

45º

F

∴ Time required =

π

×v

r243 = 4.71 × 10–8 sec.

4. A long solenoid of radius 'a' and number of turns per unit length n is enclosed by cylindrical shell of radius R, thickness d(d<<R) and length L. A variable current i = i0 sin ωt flows through the coil. If the resitivity of the material of cylindrical shell is ρ, find the induced current in the shell. [IIT-2005]

Sol. The magnetic field in the solenoid is given by B = µ0ni

a

L

d R

⇒ B = µ0 n i0 sin ωt [Q i = i0 sin ωt given] The magnetic flux linked with the solenoid

φ = →B .

→A = BA cos 90º

= (µ0 n i0 sin ωt) (πa2) ∴ The rate of change of magnetic flux through the

solenoid

dtdφ = π µ0 n a2 i0 ω cos ωt

The same rate of change of flux is linked with the cylindrical shell. By the principle of electromagetic induction, the induced emf produced in the cylinderical shell is

ITOP VIEW

e = –dtdφ = – πµ0 n a2 i0 ω cos ωt ...(i)

The resistance offered by the cylindrical shell to the flow of induced current I will be

R = ρAl

Here, l = 2πR, A = L × d

∴ R = ρLd

R2π ...(ii)

The induced current I will be

I = R

|e| = R2

Ld]tcosinaµ[ 02

0

π×ρ×ωωπ

I = R2

Ldtcosinaµ 02

0

ρ×ωωπ

Ans.

5. A particle of charge equal to that of an electron, –e,

and mass 208 times the mass of the electron (called a nu-meson) moves in a circular orbit around a nucleus of charge + 3e. (Take the mass of the nucleus to be infinite). Assuming that the bohr model of the atom is applicable to this system.

(i) Derive an expression for the radius of the nth Bohr orbit.

(ii) Find the value of n for which the radius of the orbit is approximately the same as that of the first Bohr orbit for the hydrogen atom.

(iii) Find the wavelength of the radiation emitted when the mu-meson jumps from the third orbit of the first orbit. [IIT-1988]


Sol. (i) Let m be the mass of electron. Then the mass of mu-meson is 208 m. According to Bohr's postualte, the angular momentum of mu-meson should be an integral multiple of h/2π.

+3e

e

r

∴ (208 m) vr = π2

nh

∴ v = mr

nh2082 ×π

= mr

nhπ416

…(1) Since mu-meson is moving in a circular path

therefore it needs centripetal force which is provided by the electrostatic force between the nucleus and mu-meson.

∴ r

v)m208( 2 =

041πε

× 23

ree×

∴ r = 2

0

2

mv2084e3

×πε

Substituting the value of v from (1) we get

r = 220

2

20844164163

hnmrmre

×πεπ×π×

⇒ r = 20

22

624 mehnπ

ε

…(2) (ii) The radius of the first orbit of the hydrogen

atom

= 2

20

meh

π

ε

…(3) To find the value of n for which the radius of

the orbit is approximately the same as that of the first Bohr orbit for hydrogen atom, we equate equation (2) and (3)

20

22

624 mehnπ

ε = 2

20

meh

π

ε ⇒ n = 624 ≈ 25

(iii) λ1 = 208 R × z2

22

21

1–1nn

⇒ λ1 = 208 × 1.097 × 107 × 32

22 31–

11

⇒ λ = 5.478 × 10–11 m

CHEMISTRY

6. An organic compound CxH2yOy was burnt with

twice the amount of oxygen needed for complete combustion to CO2 and H2O. The hot gases, when cooled to 0 ºC and 1 atm pressure, measured 2.24 L. The water collected during cooling weighed 0.9 g. The vapour pressure of pure water at 20 ºC is 17.5 mm Hg and is lowered by 0.104 mm Hg when 50 g of the organic compound is dissolved in 1000 g of water. Give the molecular formula of the organic compound. [IIT-1983]

Sol. The combustion reaction is CxH2yOy + x O2 → x CO2 + y H2O To start with, the amount of O2 taken is 2x. Hence,

after the combustion reaction, we will be left with the following amounts.

Amount of oxygen left unreacted = x Amount of carbon dioxide = x Amount of water = y When this mixture is cooled to 0 ºC and 1 atm, we

will be left with oxygen and carbon dioxide. Hence, the amount 2x occupies the given volume of 2.24 L at STP. Hence,

Amount x = 1Lmol4.22L)2/24.2(−

= 0.05 mol

Now, Mass of water collected = 0.9 g Amount of water collected,

y = 1molg18g9.0

−= 0.05 mol

Thus, the empirical formula of the compound is C0.05H2 × 0.05 O0.05, i.e. CH2O.

Now, according to Raoult's law

–*pp∆ = x2

i.e. mmHg5.17mmHg104.0 =

)molg18/g1000()M/g50()M/g50(

1−+

Solving for M, we get M = 150.5 g mol–1 Number of repeating units of CH2O in the

molecular formula = 16212

5.150++

≈ 5

Hence, Molecular formula of the compound is C5H10O5.

7. The oxides of sodium and potassium contained in a 0.5 g sample of feldspar were converted to the respective chlorides. The weight of the chlorides thus obtained was 0.1180 g. Subsequent treatment of the chlorides with silver nitrate gave 0.2451 g of silver chloride. What is the percentage of Na2O and K2O in the mixture ? [IIT-1979]


Sol. Mass of sample of feldspar containing Na2O and K2O = 0.5 g.

According to the question, Na2O + 2HCl → 2NaCl + H2O ..(1) 2 × 23 + 16 = 62g 2(23 + 35.5) = 117 g K2O + 2HCl → 2KCl + H2O ...(2) 2 × 39 + 16 = 94g 2(39 + 35.5) = 149 g Mass of chlorides = 0.1180 g Let, Mass of NaCl = x g ∴ Mass of KCl = (0.1180 – x)g Again, on reaction with silver nitrate, NaCl + AgNO3 → AgCl + NaNO3 ...(3) 23 + 35.5 = 58.5g 108 + 35.5 = 143.5g KCl + AgNO3 → AgCl + KNO3 ...(4) 39 + 35.5 = 74.5g 108 + 35.5 = 143.5g Total mass of AgCl obtained = 0.2451 g Step 1. From eq. (3) 58.5 g of NaCl yields = 143.5 g AgCl

∴ x g of NaCl yields = 5.585.143 x g AgCl

And from eq. (4), 74.5 g of KCl yields = 143.5 g of AgCl ∴ (0.1180 – x)g of KCl yields

= 5.745.143 (0.1180 – x)g AgCl

Total mass of AgCl

5.585.143 x +

5.745.143 (0.1180 – x) = 0.2451

which gives, x = 0.0342 Hence, Mass of NaCl = x = 0.0342 g And Mass of KCl = 0.1180 – 0.0342 = 0.0838g Step 2. From eq.(1), 117 g of NaCl is obtained from = 62 g Na2O ∴ 0.0342 g NaCl is obtained from

= 11762 × 0.032 = 0.018 g Na2O

From eq. (2), 149 g of KCl is obtained from = 94 g K2O ∴ 0.0838 g of KCl is obtained from

= 14994 × 0.0838 = 0.053 g K2O

Step 3. % of Na2O in feldspar = 5.0

018.0 × 100

= 3.6%

% of K2O in feldspar =5.0

053.0 × 100 = 10.6 %

8. 5 ml of 8 N of nitric acid, 4.8 ml of 5 N hydrochloric acid and a certain volume of 17 M sulphuric acid are mixed together and made upto 2 litre. 30 ml of this mixture exactly neutralizes 42.9 ml of sodium carbonate solution containing one gram of Na2CO3.10H2O in 100 ml of water. Calculate the amount in grams of the sulphate ions in solution. [IIT-1985]

Sol. Given that,

3HNON = 8 N

3HNOV = 5 ml

NHCl = 5 N VHCl = 4.8 ml

42SOHM = 17 M

Step 1. Meq. of HNO3 in 2L solution =

3HNON × 3HNOV

= 8 × 5 = 40 ∴ Meq. of HNO3 in 30 ml solution

= 2000

40 × 30 = 0.6

Step 2. Meq. of HCl in 2L solution = NHCl × VHCl = 5 × 4.8 = 24 ∴ Meq. of HCl in 30 ml solution

= 2000

24 × 30 = 0.36

Step 3. Meq. of H2SO4 in 2L solution = Valency factor ×

42SOHM × 42SOHV

= 2 × 17 × 42SOHV

∴ Meq. of H2SO4 in 30 ml solution

= 2000

30V3442SOH ××

= 0.5142SOHV

Step 4. Also given that, Volume of Na2CO3.10H2O = 100 ml Mass of Na2CO3.10H2O = 1 g Equivalent mass of Na2CO3.10H2O

= 2

massMolecular = 2

286 = 143 g equiv–1

We know, Normality

= (ml) Volume solute of mass Equivalent

1000 solute of Mass×

×

= 100143

10001×

× = 0.070 N

Meq. of Na2CO3.10H2O = OH10.CONa 232

N × OH10.CONa 232V

= 0.070 × 42.9 = 3.003


Step 5. 1 gram equivalent acid neutralizes 1 gram equivalent of base.

∴ 0.6 + 0.36 + 0.5142SOHV = 3.003

Solving , 42SOHV = 4 ml

Step 6. 1000 ml of 1 M H2SO4 contains = 96 g SO4

2–ions

∴ 4ml of 17 M H2SO4 contains = 1000

41796 ××

= 6.528g SO42– ions

9. The equilibrium constant Kp of the reaction 2SO2(g) + O2(g) 2SO3(g) is 900 atm–1 at

800 K. A mixture containing SO3 and O2 having initial partial pressures of 1 atm and 2 atm, respectively, is heated at constant volume to equilibrate. Calculate the pressure of each gas at 800 K. [IIT- 1989]

Sol. Since to start with SO2 is not present, it is expected that some of SO3 will decompose to give SO2 and O2 at equilibrium. If 2x is the partial pressure of SO3 that is decreased at equilibrium, we would have

2SO2(g) + O2(g) 2SO3(g) t = 0 0 2 atm 1 atm teq 2x 2 atm + x 1 atm – 2x

Hence, Kp = )p()p(

)p(

22

3

O2

SO

2SO =

)xatm2()x2()x2atm1(

2

2

+−

= 900 atm–1 Assuming x << 2 atm, we get

)atm2()x2(

)x2atm1(2

2− = 900 atm–1

or 2

2

)x2()x2atm1( − = 1800

or x2

atm1 – 1 = 42.43

or x = 43.432

1×

atm = 0.0115 atm

Hence, p(SO2) = 2x = 0.023 atm; p(O2) = 2atm + x = 2.0115 atm and p(SO3) = 1 atm – 2x = 0.977 atm

10. A white substance (A) reacts with dil. H2SO4 to produce a colourless gas (B) and a colourless solution (C). The reaction between (B) and acidified K2Cr2O7 solution produces a green solution and a slightly coloured precipitate (D). The substance (D) burns in air to produce a gas (E) which reacts with (B) to yield (D) and a colourless liquid. Anhydrous copper sulphate is turned blue on addition of this colourless liquid. Addition of aqueous NH3 or NaOH to (C) produces first a

precipitate, which dissolves in the excess of the respective reagent to produce a clear solution in each case. Identify (A), (B), (C), (D) and (E). Write the equations of the reactions involved. [IIT-2001]

Sol. (A) is ZnS,

)A(ZnS + H2SO4 →

)C(4ZnSO +

)B(2SH

)B(2SH3 + K2Cr2O7 + 4H2SO4 → K2SO4 + Cr2(SO4)3

+ 7H2O + )D(greyWhite

S3

)D(

S + Air

2O → )E(2SO

)E(2SO +

)B(2SH2 →

)C(liquid Colourless2OH2 +

)(DS3

5H2O + White

4CuSO → Blue

24 OH5.CuSO

ZnSO4 + 2NaOH → Zn(OH)2 + Na2SO4 Zn(OH)2 + 2NaOH → Na2ZnO2 (soluble) + 2H2O Also in excess of NH4OH it forms soluble

complex [Zn(NH3)4](OH)2.

MATHEMATICS

11. If f (x) is twice differentiable function such that f (a) = 0, f (b) = 2, f (c) = 1, f (d) = 2, f (e) = 0, where a < b < c < d < e, then the minimum number of zero's of g(x) = f ' (x)2 + f '' (x). f (x) in the interval [a, e] is? [IIT-2006]

Sol. Let, g(x) = dxd [f (x) . f '(x)]

to get the zero of g(x) we take function h(x) = f (x). f ' (x)

between any two roots of h(x) there lies at least one root of h' (x) = 0

⇒ g(x) = 0 ⇒ h(x) = 0 ⇒ f (x) = 0 or f ' (x) = 0 If f (x) = 0 has 4 minimum solutions. f ' (x) = 0 has 3 minimum solutions. h(x) = 0 has 7 minimum solutions. ⇒ h' (x) = g(x) = 0 has 6 minimum solutions

12. The position vectors of the vertices A, B and C of a

tetrahedron ABCD are î + ^j + ^k , î and 3 î ,

respectively. The altitude from vertex D to the opposite face ABC meets the median line through A of the triangle ABC at a point E. If the length of the side AD is 4 and the volume of the tetrahedron

is 3

22 , find the position vector of the point E for

all its possible positions. [IIT-1996]


Sol. F is mid-point of BC i.e., F ≡ 23^^ ii+ = 2 î

and AE ⊥ DE (given)

A(i + j + k) D

C(3i) F(2i) B(i)

E 1

λ

Let E divides AF in λ : 1. The position vector of E is given by

1

)(12 ^^^^

+λ+++λ kjii = ^

112 i

+λ+λ + ^

11 j+λ

+ ^

11 k+λ

Now, volume of the tetrahedron

= 31 (area of the base) (height)

⇒ 3

22 = 31 (area of the ∆ABC) (DE)

But area of the ∆ABC = |)(|21 →→

× BABC

= |)(2|21 ^^^

kji +× = |)|^^^^kiji ×+×

= |)–|^^jk = 2

Therefore, 3

22 = 31 )2( (DE)

⇒ DE = 2 Since ∆ADE is a right angle triangle, AD2 = AE2 + DE2 ⇒ (4)2 = AE2 + (2)2 ⇒ AE2 = 12

But →

AE =^

112 i

+λ+λ +

^

11 j+λ

+^

11 k+λ

– )(^^^kji ++

= ^

1i

+λλ –

^

1j

+λλ –

^

1k

+λλ

⇒ 2||→AE = 2)1(

1+λ

[λ2 + λ2 + λ2] = 2

2

)1(3+λλ

Therefore, 12 = 2

2

)1(3+λλ

⇒ 4(λ + 1)2 = λ2 ⇒ 4λ2 + 4 + 8λ = λ2 ⇒ 3λ2 + 8λ + 4 = 0 ⇒ 3λ2 + 6λ + 2λ + 4 = 0

⇒ 3λ(λ + 2) + 2(λ + 2) = 0 ⇒ (3λ + 2) (λ + 2) = 0 ⇒ λ = – 2/3, λ = – 2 Therefore, when λ = – 2/3, position vector of E is

given by

^

112 i

+λ+λ +

^

11 j+λ

+ ^

11 k+λ

= ^

13/2–1)3/2.(–2 i

++ +

^

13/2–1 j

+ +

^

13/2–1 k

+

= ^

332–

13/4– i+

+ + ^

332–

1 j+

+ ^

332–

1 k+

= ^

3/13

34–

i

+

+ ^

3/11 j +

^

3/11 k

= –î +

^3 j +

^3k

and when λ = – 2 Position vector of E is given by,

^

12–1)2(–2 i

++× +

^

12–1 j+

+^

12–1 k+

=^

1–14– i+ –

^j –

^k

= ^

3 i – ^j –

^k

Therefore, –î +

^3 j +

^3k and +

^3 i –

^j –

^k are

the position vector of E.

13. Evaluate ∫π

π−

π

+−

+π3/

3/

3

3||cos2

4

x

x dx [IIT-2004]

Sol. Let,

I = ∫π

π−

π

+−

π3/

3/

3||cos2 x

dx + 4 ∫π

π−

π

+−

3/

3/

3

3||cos2 x

dxx

Using ∫−

a

axf )( dx=

=−

−=−

∫ )()(,)(2

)()(,0

0xfxfdxxf

xfxfa

∴ I = 2 ∫π

π

+−

π3/

0

3||cos2 x

dx + 0

π

+−∫

π

π−

3/

3/

3

3||cos2

oddisx

dxxas

I = 2π ∫π

π+−

3/

0 )3/cos(2 xdx

= 2π ∫π

π −

3/2

3/ cos2 tdt , where x +

3π = t


= 2π ∫π

π +

3/2

3/ 2

2

2tan31

2sec

t

dtt

= 2π ∫ +

3

3/1 2312

udu =

34π . 3

3/11 3tan3 u−

= 3

4π (tan–1 3 – tan–11) = 3

4π tan–1

21

∴ ∫π

π−

π

+−

+π3/

3/

3

3||cos2

4

x

x dx = 3

4π tan–1

21 .

14. An unbiased die, with faces numbered 1, 2, 3, 4, 5, 6, is thrown n times and the list on n numbers showing up is noted. What is the probability that among the numbers 1, 2, 3, 4, 5, 6 only three numbers appear in this list ? [IIT-2001]

Sol. Let us define at onto function F from A : [r1, r2 ... rn] to B : [1, 2, 3] where r1r2 .... rn are the readings of n throws and 1, 2, 3 are the numbers that appear in the n throws.

Number of such functions, M = N – [n(1) – n(2) + n(3)] where N = total number of functions and n(t) = number of function having exactly t elements

in the range. Now, N = 3n, n(1) = 3.2n, n(2) = 3, n(3) = 0 ⇒ M = 3n – 3.2n + 3 Hence the total number of favourable cases = (3n – 3.2n + 3). 6C3

⇒ Required probability = n

nn C6

)32.33( 36×+−

15. Show that the value of tanx/tan3x, wherever defined never lies between 1/3 and 3. [IIT-1992]

Sol. y = xx

3tantan =

xxx

x

2

3

tan3–1tan–tan3

tan

= xxxx

3

2

tan–tan3)tan3–1(tan

= xx

2

2

tan–3tan3–1 [Q tan 3x ≠ 0 ⇒ 3x ≠ 0]

⇒ x ≠ 0 ⇒ tan x ≠ 0

0 ∞

+ – +3 1/3

Let tan x = t

⇒ y = 2

2

–33–1tt

⇒ 3y – t2y = 1 – 3t2 ⇒ 3y – 1 = t2y – 3t2 ⇒ 3y – 1 = t2 (y – 3)

⇒ 3–1–3

yy = t2

⇒ 3–1–3

yy ≥ 0, t2 ≥ 0 ∀ t ∈ R

⇒ y ∈ (– ∞, 1/3) ∪ (3, ∞) Therefore, y is not defined in between (1/3, 3).

Physics Facts

Nuclear Physics 1. Alpha particles are the same as helium nuclei and have the symbol . 1. The atomic number is equal to the number of protons (2 for alpha) 2. Deuterium ( ) is an isotope of hydrogen ( ) 3. The number of nucleons is equal to protons + neutrons (4 for alpha) 4. Only charged particles can be accelerated in a particle accelerator such as a cyclotron or Van Der Graaf

generator. 5. Natural radiation is alpha ( ), beta ( ) and gamma (high energy x-rays) 6. A loss of a beta particle results in an increase in atomic number. 7. All nuclei weigh less than their parts. This mass defect is converted into binding energy. (E=mc2) 8. Isotopes have different neutron numbers and atomic masses but the same number of protons (atomic

numbers). 9. Geiger counters, photographic plates, cloud and bubble chambers are all used to detect or observe radiation.


1. A particle of charge Q and of negligible initial speed

is accelerated through a potential difference of U. The particle reaches a region of uniform magnetic field of induction B, where it undergoes circular motion. If potential difference is doubled & B is also doubled then magnetic moment of the circular current due to circular motion of charge Q will become

(A) double (B) half (C) four times (D) remain same

2. Three long rod AA, AB, CC are moving with a speed v in a uniform magnetic field B perpendicular to the plane of paper as shown in figure. The triangle formed between the three wires is always an equilateral triangle. If resistance per unit length of wire is λ , then the induced current in the triangle is

A B

C A

C B

v

v B0

× × × × × × × × × × × × × × × × × × × × × × × × × × × ×

× × × × ×

B0 v

(A) λ3vB0 (B)

λ3vB2 0

(C) λ3vB0 (D)

λvB0

3. In the diagram shown, the wires P1Q1 and P2Q2 are made to slide on the rails with same speed of 5m/s. In this region a magnetic field of 1T exists. The electric current in Ω9 resistor is

P1 P2

Q2 Q1

2Ω 2Ω 9Ω 4cm

× × × × × × × ×× × × × × × × × ×× × × × × × × × ×× × × × × × × × ×× × × × × × × × ×

(A) zero if both wires slide towards left (B) zero if both wires slide in opposite direction

(C) 0.2mA if both wires move towards left (D) 0.2mA if both wires move in opposite direction

4. A circular current loop is shown in the adjacent figure. The magnetic field in the region is along x-axis and its magnitude in the space is increasing with increasing y-coordinate. The net magnetic force on the loop is

yB

x

(A) along + z-axis (B) along –z-axis (C) along +y-axis (D) None of these

5. A point charge q moves from A to B along a parabolic path (AB is latus rectum). Electric field is along x-axis. The work done by the field is

y2 = 4ax

x E

A B

y

(A) – qEa (B) –2qEa

(C) +qeA (D) 2qEa

6. In a cylindrical zone of radius R, magnetic field →B is

present perpendicular to the plane of the paper into it.

Magnitude of →B is varying with time as B0(p + qt)

where p & q are positive constant. Consider a static triangle loop OAM. Emf induced in the triangular loop

This section is designed to give IIT JEE aspirants a thorough grinding & exposure to variety of possible twists and turns of problems in physics, that would be very helpful in facing IIT JEE. Each and every problem is well thought of in order to strengthen the concepts and we hope that this section would prove a rich resource for practicing challenging problems and enhancing the preparation level of IIT JEE aspirants.

By : Dev Sharma Director Academics, Jodhpur Branch

Physics Challenging Problems

Solut ions wil l be published in next issue

Set # 10


A M

O B R

60º 60º

× × ×

× ×

× × × × × ×

× × × ×

× × × ×

× × ×

(A) 3qRB2 2

0π (B)

3qRB 2

0

(C) 3PRB 2

0 (D) 2RPB 2

0

7. Consider cylindrical region of the magnetic field shown in the figure. Region I and II have fields directed perpendicularly outward and inward respectively. Fields are varying with time as

Region I : B = 3B0t Region II : B = B0t There is no net induced electric field in the region

outside the magnetic field then the ratio of =

2

1

2

rr

Region II

× × ×

× ×

× × × × × ×

× × × ×

×

Region I r1

r2

8. Match the column: Column – I Column – II (A) In a series RLC (P) Z may increase circuit if C decreases

(B) In a series RLC (Q) Z may decrease circuit if L increases (C) In a series RLC (R) resonance circuit if R decreases frequency increases (D) In a series RLC (S) power factor circuit if C increases decrease

• After travelling 2.4 billion miles in just over 6 years to reach Jupiter, Galileo missed its target at the Jovian moon Io by only 67 miles. That's like shooting an arrow from Los Angeles at a bull's-eye in New York and missing by only 6 inches!

• Utopia ia a large, smooth lying area of Mars.

• The biggest star has a diameter of 1800 million miles, making it 2000 times bigger than the Sun.

• 15% of the world's fresh water flows doen the Amazon.

• In 1995, each American used an annual average of 731 pounds of paper, more than double the amount used in the 1980's. Contrary to predictions that computers would displace paper, consumption is growing.

• The term 'black hole' was coined in 1968 when John Wheeler described how an in-falling object 'becomes dimmer millisecond by millisecond...light and particles incident from outside...go down the black hole only to add to its mass and increase its gravitational attraction.'

• The 'Red Planet' isn't really red at all, Nasa photographs indicate that it is more of a tan or butterscotch colour.

• The International Space Station orbits at 248 miles above the Earth.

• The axis of orbit of the planet Uranus is tilted at a 90 degree angle.

• Astronomers have discovered over 10,000 asteroids - but put them together and they would be smaller than the Moon.

• Have you ever seen a ring around the moon? Folklore has it that this means bad weather is coming.


1. For the process BA →

VM1RT

231U ∝⇒

ρ∝

or T ∝ V So AB is isobaric process (pressure in constant)

== A0

0A RT

23U4As

R3U8

T

0

A 2MVρ

=

so M3

U16P 00

Aρ

= and 0

B 4MVρ

=

so WAB = P[VB - VA]

0

00

00

00

4M

M3U16

2M

4M

M3U16

ρ×

ρ−=

ρ

−ρ

ρ⇒

Option [D] is correct

2. The slope of the graph is 0

0

V2P

m −=

The equation of this graph is given by

2P3

V2VP

P 0

0

0 +−=

2P3

V2VP

VnRT 0

0

0 +−

=⇒

VnR2P3

VnRV2

VPT 02

0

0 +−

−=⇒

0nR2P3

nRV2VP2

dVdT 0

0

00 =+−

=

2V3

VnR2P2

nRVVP 00

0

0 =⇒=⇒

At ,2V3

V 0= temp will be maximum [n = 1]

2V3

R2P3

4V9

RV2P

T 0020

0

0max ×+×−=

RVP

49

RVP

89 0000 +−⇒

RVP

89 00⇒

Option [C] is correct

3. TV-3 = k kVnRPV 3 =⇒ −

for this polytrophic process, PV-2 = constant so x = -2

01212 nRT

32

3)TT(nR

x1)TT(nR

w =−

=−

−=

∆U = nCV∆T = n 3/2 R(3T0 – T0) = 3nRT0

00 nRT3

11nRT332WUQ =

+==∆=∆

∆Q = nC∆T R6

11)T2(n

nRT3

11Tn

QC0

0 ==∆

∆=⇒

Option [A,C,D] is correct

4.

y

R

FFx

Fyωt x

F = mω2R Fy = F sin ωt Fy = mω2R sin (ωt)

* In figure as cos ωt = Rx ⇒ tcosRx ω=

* For one complete circle, tω=θ

*

y

R

vx

ωt x vx = – ωR sin ωt

v = ωR

Option [A,C] is correct

5. From 1st law, ∆Q = ∆U + W nCαT + nCvαT + pαV ⇒ nCαT = nCVαT + nRT/V αV

VeTV

nRTVeTCVeTn V0

V0V

V0 α+αα=αα⇒ ααα

⇒ αC = αCV + R/V ⇒ C = CV + R/αV

Option [B] is correct

6. Option [C] is correct 7. Option [B] is correct 8. Option [D] is correct

Solution Physics Challenging Problems

Set # 9

8 Questions were Published in January Issue


1. AB and CD are two ideal springs having force constant K1 and K2 respectively. Lower ends of these springs are attached to the ground so that the springs remain vertical. A light rod or length 3a is attached with upper ends B and C of springs. A particle of mass m is fixed with the rod at a distance a from end B and in equilibrium, the rod is horizontal. Calculate period of small vertical oscillations of the system.

Bm

a 2a C

K2K1

A D Sol. Let, in equilibrium, compressive forces in left and

right springs be F1 and F2, respectively. Considering free body diagram of rod BC, (figure)

B C

F1

a

mg

2a

F2 F1 + F2 = mg ...(i) Taking moments about B, mg × a = F2 × 3a

or F2 = 31 mg

Substituting this value in equation (i),

F1 = 32 mg

Let the particle be pressed from its equilibrium position by applying a force 'F'. Let left and right springs be further compressed through y1 and y2, respectively. Increase in compressive forces in the spring will be K1y1 and K2y2 respectively. In other words, total compressive forces in two springs will be (F1 + K1y1) and (F2 + K2y2) respectively.

Now considering new free body diagram (figure) of the rod BC,

(F1 + K1y1) + (F2 + K2y2) = F + mg ...(ii) Taking moments about B, (figure)

B C

(F1 + K1y1)

a

mg

2a

(F2 + K2y2)

F

(F + mg) × a = (F2 + K2y2) × 3a ...(iii) Solving above equations,

K1y1 = 3F2 and K2y2 =

3F .

or y1 = 1K3

F2 and y2 = 2K3

F

Since, distance of the particle from left spring is 'a' and that from right spring is '2a', therefore, downward displacement of the particle from equilibrium position will be

y = 3y2 1 +

3y2 =

lK9F4 +

2K9F =

21

21

KK9F)K4K( +

or F = )K4K(

KK9

21

21

+ ...(iv)

Now, if the particle be released, it starts accelerating upwards due to excess compressive force in springs. Hence, the restoring force is (K1y1 + K2y2), which is numerically equal to F.

or Restoring force = F = )K4K(

KK9

21

21

+. y

or Restoring acceleration, a =mF =

)K4K(mKK9

21

21

+.y

Since, acceleration is restoring and is directly proportional to displacement y, therefore, the particle performs SHM. Its period of oscillations is

T = ay2π

or T = 21

21

KK)K4K(m

32 +π Ans.

Expert’s Solution for Question asked by IIT-JEE Aspirants

Students ForumPHYSICS


2. A plane spiral coil is made on a thin insulated wire and has N = 100 turns. Radii of inside and outside turns are a = 10 cm and b = 20 cm respectively. A magnetic field normal to the plane of spiral exists in the space. The magnetic field increases at a constant rate α = 0.3 tesla/second. Calculate potential difference between the ends of the spiral.

Sol. Since, magnetic field is increasing, therefore flux linked with the coil is also increasing. Due to increase in flux an emf is induced in it. Potential difference between ends of the spiral coil is equal to magnitude of emf induced in it.

Since there are N turns in a radial width (b – a), therefore number of turns in the spiral coil per unit radial width is

n = )a–b(

N

Consider a concentric circular ring of radius x and radial thickness dx

Number of turns in this ring = n dx. Let at some instant magnetic field strength be B, Flux linked with the ring, φ = πx2B

Emf induced in this ring = (n . dx) dtdφ

= (n . dx)

π

dtdBx 2 = πnα x2dx

∴ Total emf induced in the spiral coil,

e = ∫=

=

απbx

ax

2 )dxxn( = 31

πn α (b3 – a3)

Substituting value of n,

e = 31

πN α(a2 + b2 + ab) = 2.2 volt Ans.

3. A spherical shell of radius R is filled with water. Temperature of atmosphere is (– θ) ºC. The shell is exposed to atmosphere and all water comes down to 0ºC and then it starts freezing from outer surface towards the centre of the shell. Assuming shell to be highly conducting, calculate time for whole mass of water to freeze. Thermal conductivity of ice is K and latent heat of its fusion is L. Density of water is ρ.

Neglect expansion during freezing. Sol. Heat flows from water to atmosphere because

atmospheric temperature is below 0ºC. Water is filled in spherical shell, therefore heat flows in radial direction.

Let at some instant t, thickness of ice layer be x. Then a concentric sphere of radius (R – x) is in liquid form as shown in figure. Heat flows from this sphere to atmosphere through ice layer.

To calculate rate of heat flow, first we have to calculate thermal resistance of this ice layer.

Considering a concentric spherical shell of radius r [(R – x) < r < R] and radial thickness dr as shown in figure.

rR

(R–x)

Its thermal resistance = Kr4

dr2π

∴ Total thermal resistance of ice layer

∫ π

R

)x–R(2Kr4

dr = )x–R(KR4

xπ

Temperature just inside and outside the ice layer is 0ºC and (– θ)ºC, respectively.

∴ Temperature difference = 0 – (– q) = θº C

Hence, rate of heat flow, H =

π

θ

)x–R(KR4x

= x

)x–R(KR4 θπ

rate of freezing of mass = LH

∴ Rate of freezing of volume = ρL

H

But it is equal to 4π(R – x)2 . dtdx

∴ xL

)x–R(KR4ρθπ = 4p (R – x)2 .

dtdx

or θ

ρKR

)x–R(xL . dx = dt

At instant t = 0, thickness of ice layer was equal to x = 0 and we have to calculate time t when whole mass has frozen or when x = R

Substituting these limits,

∫ ∫θρ

=t

0

R

0

dx)x–R(xKRLdt

or t = θ

ρK6

LR 2 Ans.


4. In a Young's experiment, the upper slit is covered by a thin glass plate of refractive index µ1 = 1.4 while the lower slit is covered by another glass plate having the same thickness as the first one but having refractive index µ2 = 1.7. Interference pattern is observed using light of wavelength λ = 5400 Å. It is found that the point P on the screen where the central maxima fell before the glass plates ware inserted, now has 3/4 the original intensity. It is further observed that what used to be the fifth maxima earlier, lies below the point P while the sixth minima lies above the point P. Neglecting absorption of light by glass plates, calculate thickness of the glass plates.

Sol. If glass plates are not inserted in slits, central maximum occurs at perpendicular bisector of slits. Hence, point P lies on perpendicular bisector of S1S2 as shown in figure.

S1

S2

P

Screen

When a plate is inserted in upper slit, the

interference pattern shifts upwards and due to plate inserted in lower slit, it shifts downwards.

Since, distance of nth maxima (nth bright fringe) from central bright fringe is nω, where ω is fringe width and fifth maxima now lies below the point P, therefore, shift of fringe pattern is downwards and it is greater that 5 ω.

Since, sixth minima (sixth dark fringe) lies above point P and distance of mth minima from central

bright fringe is

21–m ω, therefore, shift of fringe

pattern is less than 5.5 ω. Hence, the shift lies between 5 ω and 5.5 ω.

Let intensity of light due to each slit be I0. Then l1 = l2 = l0.

Since, before insertion of glass plates, a bright fringe (central bright fringe) was formed at P, therefore, original intensity at P was equal to Imax.

But Imax = ( 1I + 2I )2 = 4I0 But now intensity at P is 3/4 of the original intensity,

therefore now intensity at P is I = 3/4 Imax = 3I0. But I = I1 + I2 + 21II2 cos φ where φ is phase

difference between two rays reaching P. Substituting I1 = I2 = I0 and I = 3I0 in above

equation,

cos φ = 21 or φ = (2nπ + π/3) where n is an integer

But phase difference, φ = 2πωs where s is shift of

fringe pattern.

Hence, s =

+

61n ω.

But s lies between 5ω and 5.5 ω, therefore,

s =

+

615 ω =

631

ω where ω = dDλ

∴ s = d6D31λ where D is distance of screen from

slits and d is distance between slits. Let thickness of each glass plate be t. ∴ Upward shift due to upper glass plate,

s1 = d

tD)1–( 1µ

and downward shift due to lower glass plate,

s2 = d

tD)1–( 2µ

∴ Resultant downward shift s = s2 – s1. Substituting values of s, s1 and s2,

t = )–(6

31

12 µµλ = 9.3 µm Ans.

5. A point isotropic light source of power P = 12 watt is

located on the axis of a circular mirror plate of radius R = 3 cm. If distance of source from the plate is a = 39 cm and reflection coefficient of mirror plate is α = 0.70, calculate force exerted by light rays on the plate.

Sol. When light rays are incident on the mirror plate, a part of then is reflected and a part is absorbed by the plate. Therefore, momentum of light rays changes. Due to change in the momentum, a force is experienced by the plate. Magnitude of the force is equal to rate of change of momentum. To calculate rate of change of momentum, first we have to consider a circular ring coplanar and co-axial with the plate. Let the radius of that ring be x and radial thickness dx as shown figure

Sourceθ

a Distance of every point of this ring from the source

is

r = 22 xa + . Hence, intensity of light incident on the ring is

I = 2r4Pπ


Direction of incident rays is inclined at angle θ with normal to the plane of ring. Therefore, power incident on the ring, p = (2πxdx)I cos θ

or p = 2r2dxxcosP θ

Since, incident power is in the form of light rays which are incident at angle θ with normal to the plate, therefore, net rate of incidence of momentum on the ring considered

= c

cosP θ = cr2

dxxcosP2

2 θ

Since, 70% of the rays are reflected and 30% are absorbed by the plate, therefore, rate of change of momentum from the ring considered

= c1 [(0.7 p cos θ) × 2 + (0.3 p cos θ)]

But it is equal to force dF on the ring.

∴ dF = 0.85 222

2

)xa(cdxxPa

+

or Total force on the plate,

F = ∫=

=+

Rx

0x222

2

)xa(dxx

cPa85.0

=

+ )Ra(1–

a1

c2Pa85.0

222

2

= c)Ra(2

PR85.022

2

+ = 1 × 10–10 N Ans.

NUCLEAR REACTOR TYPES – ARE THEY REALLY SAFE? The nuclear power industry along with the reactor technology has been constantly developing for more than five decades now. Nuclear reactors can be classified based on their nuclear reaction, the moderator material used, generation of the reactor, fuel phase, fuel type, coolant used, etc. The fission nuclear reactors are mostly dealt with because the fusion reactors are still in the developing stages and the fission reactors are already being used for the past six decades.

• Based on nuclear reaction

This type refers to the thermal (slow) reactors and the fast reactors based on the speed of neutrons. Thermal reactors are the most affordable and common as they use the natural and raw uranium; and the neutrons are decelerated from their natural speed when emitted from the broken atomic nuclei, and uses a moderating material in the process. The Fast reactors are very expensive that require more enriched fuel.

• Based on moderator material Thermal reactors (because of the presence of the moderating material), and Graphite, Normal water and Heavy water are also used as moderators. The moderating materials in the Graphite and the Heavy water reactors thermalize the neutrons and keep the natural uranium intact without any enrichment.

• Based on generation Generation I reactors were the first prototype reactors, Generation II used standard designs till 50s, Generation III were more modern, lightweighted, more efficient and were used till late 90s, the latest i.e. Generation IV reactors targeting on economical and minimal waste, are still in the research and development stage which may officially work until late 2020s.

• Based on fuel phase and fuel type It is Solid, Liquid or Gas reactor where Solid is most typical. The fuel type reactors also come with fuel phase- uranium or thorium, which are available in abundant quantities on the land.


Matter Waves : Planck's quantum theory : Wave-particle duality - Planck gave quantum theory while explaining the

radiation spectrum of a black body. According to Planck's theory, energy is always exchanged in integral multiples of a quanta of light or photon.

Each photon has an energy E that depends only on the frequency ν of electromagnetic radiation and is given by :

E = hν .....(1) where h = 6.6 × 10–34 joule-sec, is Planck's

constant. In any interaction, the photon either gives up all of its energy or none of it.

From Einstein's mass-energy equivalence principle, we have

E = mc2 .....(2) Using equations (1) and (2), we get ;

mc2 = hν or m = 2chν .....(3)

where m represents the mass of a photon in motion. The velocity v of a photon is equal to that of light, i.e., v = c.

According to theory of relativity, the rest mass m0 of a photon is given by :

m0 = 2

2

cv1m −

Here, m = 2chν and v = c

Hence, m0 = 0 ....(4) i.e., rest mass of photon is zero, i.e., energy of

photon is totally kinetic.

The momentum p of each photon is given by :

p = mc = 2chν × c =

chν =

ν/ch =

λh ......(5)

The left hand side of the above equation involves the particle aspect of photons (momentum) while the right hand side involves the wave aspect (wavelength) and the Planck's constant is the bridge between the two sides. This shows that electromagnetic radiation exhibits a wave-

particle duality. In certain circumstances, it behaves like a wave, while in other circumstances it behaves like a particle.

The wave-particle is not the sole monopoly of e.m. waves. Even a material particle in motion according to de Broglie will have a wavelength. The de Broglie wavelength λ of the matter waves is also given by :

λ = mvh =

ph =

mKh

2

where K is the kinetic energy of the particle. If a particle of mass m kg and charge q coulomb

is accelerated from rest through a potential difference of V volt. Then

21 mv2 = qV or mv = mqV2

Hence, λ = mqVh

2 =

V34.12 Å

Photoelectric effect : When light of suitable frequency (electromagnetic

radiation) is allowed to fall on a metal surface, electrons are emitted from the surface. These electrons are known as photoelectrons and the effect is known as photoelectric effect. Photoelectric effect, light energy is converted into electrical energy.

Laws of photolectric effect : The kinetic energy of the emitted electron is

independent of intensity of incident radiation. But the photoelectric current increases with the increase of intensity of incident radiation.

The kinetic energy of the emitted electron depends on the frequency of the incident radiation. It increases with the increase of frequency of incident radiation.

If the frequency of the incident radiation is less than a certain value, then photoelectric emission is not possible. This frequency is known as threshold frequency. This threshold frequency varies from emitter to emitter, i.e., depends on the material.

There is no time lag between the arrival of light and the emission of photoelectrons, i.e., it is an instantaneous phenomenon.

Matter Waves, Photo-electric Effect

PHYSICS FUNDAMENTAL FOR IIT-JEE

KEY CONCEPTS & PROBLEM SOLVING STRATEGY


Failure of wave theory : Wave theory of light could not explain the laws of

photoelectric effect. According to wave theory, the kinetic energy of

the emitted electrons should increase with the increase of intensity of incident radiation.

Kinetic energy of the emitted electron does not depend on the frequency of incident radiation according to wave theory.

Wave theory failed to explain the existence of threshold frequency.

According to wave theory there must be a time lag between the arrival of light and emission of photoelectrons.

Einstein's theory of photoelectric effect : Einstein explained the laws of photoelectric effect

on the basis of Planck's quantum theory of radiation.

Einstein treated photoelectric effect as a collision between a photon and an atom in which photon is absorbed by the atom and an electron is emitted.

According to law of conservation of energy,

hν = hν0 + 21 mv2

where hν is the energy of the incident photon; hv0 is the minimum energy required to detach the electron from the atom (work function or ionisation energy) and (1/2) mv2 is the kinetic energy of the emitted electron.

The above equation is known as Einstein's photoelectric equation. Kinetic energy of the emitted electron,

= 21 mv2 = h(ν – ν0) = hν – W

Explanation of laws of photoelectric effect : (a) The KE of the emitted electron increases with the

increase of frequency of incident radiation since W (work function) is constant for a given emitter. KE is directly proportional to (ν – ν0)

(b) Keeping the frequency of incident radiation constant if the intensity of incident light is increased, more photons collide with more atoms and more photoelectrons are emitted. The KE of the emitted electron remains constant since the same photon collides with the same atom (i.e., the nature of the collision does not change). With the increase in the intensity of incident light photoelectric current increases.

(c) According to Einstein's equation, if the frequency of incident radiation is less than certain minimum value, the photoelectric emission is not possible. This frequency is known as threshold frequency. Hence, the frequency of incident radiation below which photoelectric emission is not possible is known as threshold frequency or cut-off frequency. It is given by :

ν0 = h

mv)2/1(h 2−ν

On the other hand, if the wavelength of the incident radiation is more than certain critical value, then photoelectric emission is not possible. This wavelength is known as threshold wavelength of cut-off wavelength. It is given by :

λ0 = ]mv)2/1(h[

hc2−ν

(d) Since Einstein treated photoelectric effect as a collision between a photon and an atom, he explained the instantaneous nature of photoelectric effect.

Some other important points : Stopping potential : The negative potential

applied to the collector in order to prevent the electron from reaching the collector (i.e., to reduce the photoelectric current to zero) is known as stopping potential.

eV0 = 2.maxmv

21 = hν – W = h(ν – ν0)

Millikan measured K.E. of emitted electrons or stopping potentials for different frequencies of incident radiation for a given emitter. He plotted a graph with the frequency on x-axis and stopping potential on y-axis. The graph so obtained was a straight line as shown in figure.

ν0

Frequency of incident light

V0(s

topp

ing

pote

ntia

l)

Millikan measured the slope of the straight line

(=h/e) and calculated the value of Planck's constant.

I

Full intensity

75% intensity 50% intensity 25% intensity

V0– +

Potential difference


The intercept of V0 versus ν graph on frequency axis is equal to threshold frequency (ν0). From this, the work function (hν0) can be calculated.

Graphs in photoelectric effect : (a) Photoelectric current versus potential difference

graphs for varying intensity (keeping same metal plate and same frequency of incident light) : These graphs indicate that stopping potential is independent of the intensity and saturation current is directly proportional to the intensity of light.

I

(V0)1 – +

Potential difference (V0)2

ν2 ν1

ν2>ν1

(b) Photoelectric current versus potential difference

graphs for varying frequency (keeping same metal plate and same intensity of incident light) : These graphs indicate that the stopping potential is constant for a given frequency. The stopping potential increases with increase of frequency. The KE of the emitted electrons is proportional to the frequency of incident light.

A1 A2 A3

ν0

B1

B2

B3

Frequency

Stop

ping

pot

entia

l

(c) Stopping potential versus frequency graphs for

different metals : These graphs indicate that the stops is same for all metal, since they are parallel straight lines. The slope is a universal constant (=h/e). Further, the threshold frequency varies with emitter since the intercepts on frequency axis are different for different metals.

1. (i) A stopping potential of 0.82 V is required to stop

the emission of photoelectrons from the surface of a metal by light of wavelength 4000 Å. For light of wavelength 3000 Å, the stopping potential is 1.85 V. Find the value of Planck's constant.

(ii) At stopping potential, if the wavelength of the incident light is kept at 4000 Å but the intensity of light is increased two times, will photoelectric current be obtained? Give reasons for your answer.

Sol. (i) We have 1λ

hc = eV1 + W

and 2λ

hc = eV2 + W

⇒

λ

−λ 12

11hc = e(V2 – V1)

or h =

λ

−λ

−

12

12

11)(

e

VVe =

×−

××

−×

−−

−

778

19

1041

1031103

)82.085.1(106.1

= 6.592 × 10–34 Js (ii) No, because the stopping potential depends only

on the wavelength of light and not on its intensity. 2. A small plate of a metal (work function = 1.17 eV) is

plated at a distance of 2m from a monochromatic light source of wavelength 4.8 × 10–7 m and power 1.0 watt. The light falls normally on the plate. Find the number of photons striking the metal plate per square metre per second. If a constant magnetic field of strength 10–4 tesla is parallel to the metal surface, find the radius of the largest circular path followed by the emitted photoelectrons.

Sol. Energy of one photon = λhc = 7

834

108.4103106.6

−

−

×

×××

= 4.125 × 10–19 J Number of photons emitted per second

= 1910125.40.1

−× = 2.424 × 1018

Number of photons striking the plate per square metre per second

= 2

18

)2(14.3410424.2××

× = 4.82 × 1016

Maximum kinetic energy of photoelectrons emitted from the plate

Emax = λhc – W

= 4.125 × 10–19 – 1.17 × 1.6 × 10–19 = 2.253 × 10–19 J

Solved Examples


3. A monochromatic light source of frequency ν illuminates a metallic surface and ejects photoelectrons. The photoelectrons having maximum energy are just able to ionize the hydrogen atom in ground state. When the whole experiment is repeated with an incident radiation of frequency (5/6) ν, the photoelectrons so emitted are able to excite the hydrogen atom beam which then emits a radiation of wavelength 1215 Å. Find the work function of the metal and the frequency ν.

Sol. In the first case, Emax = Ionization energy = 13.6 eV = 21.76 × 10–19 J So, hν = 21.76 × 10–19 J ....(1) In the second case,

E'max = λhc

= 10

834

101215103106.6

−

−

×

×××

=16.3×10–19 J

So, 6

5 hν = 16.3 × 10–19 + W ...(2)

Dividing Eq.(1) by Eq.(2)

56 =

W103.16W1076.21

19

19

+×

+×−

−

Solving, we get W = 11.0 × 10 – 19 J = 6.875 eV

From Eq.(1) ν = 34

1919

106.6100.111076.21

−

−−

×

×+×

= 5 × 1015 Hz 4. The radiation, emitted when an electron jumps from

n = 3 to n = 2 orbit in a hydrogen atom, falls on a metal to produce photoelectrons. The electrons from the metal surface with maximum kinetic energy are made to move perpendicular to a magnetic field of 1/320 T in a radius of 10–3 m. Find (i) the kinetic energy of electrons, (ii) wavelength of radiation and (iii) the work function of metal.

Sol. (i) Speed of an electron in the magnetic field,

v = m

Ber

Kinetic energy of electrons

Emax = 21 mv2 =

mreB

2

222

= 2

3201

× 31

23219

101.92)10()106.1(

−

−−

××

××

= 1.374 × 10–19 J = 0.8588 eV

(ii) Energy of the photon emitted from a hydrogen atom

hν = λhc =

− 22 3

121

= 1.888 eV Wavelength of radiation,

λ = 19

834

106.1888.11031062.6

−

−

××

×××

= 6.572 × 10–7m = 6572 Å (iii) Work function of metal W = hν – Emax = 1.8888 – 0.8588 = 1.03 eV 5. X-rays are produced in an X-ray tube by electrons

accelerated through a potential difference of 50.0 kV. An electron makes three collisions in the target before coming to rest and loses half of its kinetic energy in each of the first two collisions. Determine the wavelengths of the resulting photons. Neglect the recoil of the heavy target atoms.

Sol. Initial kinetic energy of the electron = 50.0 keV Kinetic energy after first collision = 25.0 keV Energy of the photon produced in the first collision, E1 = 50.0 – 25.0 = 25.0 keV Wavelength of this photon

λ1 = 1E

hc = 319

834

100.25106.1103106.6×××

×××−

−

= 0.495 × 10–10 m = 0.495 Å Kinetic energy of the electron after second collision = 12.5 eV Energy of the photon produced in the second

collision, E2 = 25.0 – 12.5 = 12.5 keV Wavelength of this photon

λ2 = 2E

hc = 319

834

105.12106.1103106.6×××

×××−

−

= 0.99 × 10–10 m = 0.99 Å Kinetic energy of the electron after third collision = 0 Energy of the photon produced in the third collision,

E3 = 12.5 – 0 = 12.5 keV This is same as E2. Therefore, wavelength of this photon, λ3 = λ2 = 0.99 Å.


Thermal Expansion :

.(a) When the temperature of a substance is increased, it expands. The heat energy which is supplied to the substance is gained by the constituent particles of the substance as its kinetic energy. Because of this the collisions between the constituents particles are accompanied with greater force which increase the distance between the constituent particles.

∆l = lα∆T ; ∆A = Aβ∆T ; ∆V = Vγ∆T or l' = l (1 + α∆T) ; A' = A(1 + β∆T) ; V' = V(1 + γ∆T) (b) Also ρ = ρ'(1 + γ∆T) where ρ' is the density at

higher temperature clearly ρ' < ρ for substances which have positive value of γ

* β = 2α and γ = 3α Water has negative value of γ for certain temperature

range (0º to 4ºC). This means that for that temperature range the volume decreases with increase in temperature. In other words the density increases with increase in temperature.

0 ml 5 ml 10 ml 15 ml 20 ml 25 ml 30 ml

If a liquid is kept in a container and the temperature

of the system is increased then the volume of the liquid as well as the container increases. The apparent change in volume of the liquid as shown by the scale is

∆Vapp = V(γ – 3α) ∆T Where V is the volume of liquid at lower temperature ∆Vapp is the apparent change in volume γ is the coefficient of cubical expansion of liquid α is the coefficients of linear expansion of the

container. Loss or gain in time by a pendulum clock with

change in temperature is ∆t = 21

α(∆T) × t

Where ∆t is the loss or gain in time in a time interval t ∆T is change in temperature and d is coefficient of

linear expansion. If a rod is heated or cooled but not allowed to expand

or contract then the thermal stresses developed

AF = γα∆T.

If a scale is calibrated at a temperature T1 but used at a temperature T2, then the observed reading will be wrong. In this case the actual reading is given by

R = R0(1 + α∆T) Where R0 is the observed reading, R is the actual

reading. For difference between two rods to the same at all

temperatures l 1α1 = l2α2. Thermodynamics

According to first law of thermodynamics q = ∆U + W

For an isothermal process (for a gaseous system) (a) The pressure volume relationship is ρV = constt. (b) ∆U = 0 (c) q = W (d) W = 2.303 nRT log10

i

f

VV

= 2.303 nRT log10 f

ipp

(e) Graphs T2 > T1

T2T1

P

V

P

T

V

T These lines are called isotherms (parameters at

constant temperature) For an adiabatic process (for a gaseous system)

(a) The pressure-volume relationship is PVγ = constt. (b) The pressure-volume-temperature relationship is

TPV = constt.

(c) From (a) and (b) TVγ–I = constt. (d) q = 0 (e) W = –∆U

Thermal Expansion, Thermodynamics

PHYSICS FUNDAMENTAL FOR IIT-JEE

KEY CONCEPTS & PROBLEM SOLVING STRATEGY


(f) ∆U = ncv∆T where cv = 1−γ

R

(g) W = 1−γ

− ffii VpVp =

1)(

−γ

− fi TTnR

(h) Graphs

V

P

P

T

V

T Please note that P-V graph line (isotherm) is

steeper. For isochoric process

(a) P ∝ T (b) W = 0 (c) q = ∆U

(d) ∆U = nCv∆T where Cv = 1−γ

R

(e) Graphs

P

V

P

T

V

T For isobaric process

(a) V ∝ T (b) W = P∆V = P(Vf – Vi) = nR(Tf – Ti) (c) ∆U = nCv∆T (d) q = nCp∆T (e) Graphs

P

V

P

T

V

T For a cyclic process

(a) ∆U = 0 ⇒ q = W (b) Work done is the area enclosed in p-V graph.

For any process depicted by P-V diagram, area under the graph represents the word done.

Kirchoff's law states that good absorbers are good emitters also.

Problem solving Strategy : Thermal Expansion Step 1: Identify the relevant concepts: Decide

whether the problem involves changes in length (linear thermal expansion) or in volume (volume thermal expansion)

Step 2: Set up the problem using the following steps: Eq. ∆L = αL0∆T for linear expansion and Eq. ∆V = βV0∆T for volume expansion. Identify which quantities in Eq. ∆L = αL0∆T or

∆V = βV0∆T are known and which are the unknown target variables.

Step 3: Execute the solution as follows: Solve for the target variables. Often you will be

given two temperatures and asked to compute ∆T. Or you may be given an initial temperature T0 and asked to find a final temperature corresponding to a given length or volume change. In this case, plan to find ∆T first; then the final temperature is T0 + ∆T.

Unit consistency is crucial, as always. L0 and ∆L (or V0 ∆V) must have the same units, and if you use a value or α or β in K–1 or (Cº)–1, then ∆T must be in kelvins or Celsius degrees (Cº). But you can use K and Cº interchangeably.

Step 4: Evaluate your answer: Check whether your results make sense. Remember that the sizes of holes in a material expand with temperature just as the same way as any other linear dimension, and the volume of a hole (such as the volume of a container) expands the same way as the corresponding solid shape.

Problem solving strategy : Thermodynamics Ist Law Step 1: Identify the relevant concepts : The first law

of thermodynamics is the statement of the law of conservation of energy in its most general form. You can apply it to any situation in which you are concerned with changes in the internal energy of a system, with heat flow into or out of a system, and/or with work done by or on a system.

Step 2: Set up the problem using the following steps Carefully define what the thermodynamics system is. The first law of thermodynamics focuses on

systems that go through thermodynamic processes. Some problems involve processes with more than one step. so make sure that you identify the initial and final state for each step.

Identify the known quantities and the target variables.

Check whether you have enough equations. The first law, ∆U = Q – W, can be applied just once to each step in a thermodynamic process, so you will often need additional equations. These often

include Eq. ∫=2

1

V

V

dVpW for the work done in a

volume change and the equation of state of the material that makes up the thermodynamic system (for an ideal gas, pV = nRT).

Step 3: Execute the solution as follows : You shouldn't be surprised to be told that

consistent units are essential. If p is a Pa and V in m3, then W is in joules. Otherwise, you may want to convert the pressure and volume units into units of Pa and m3. If a heat capacity is given in terms of calories, usually the simplest procedure is to convert it to joules. Be especially careful with moles. When you use n = mtot/M to convert


between total mass and number of moles, remember that if mtot is in kilograms, M must be in kilograms per mole. The usual units for M are grams per mole; be careful !

The internal energy change ∆U in any thermodynamic process or series of processes in independent of the path, whether the substance is an ideal gas or not. This point is of the utmost importance in the problems in this topic. Sometimes you will be given enough information about one path between the given initial and final states to calculate ∆U for that path. Since ∆U is the same for every possible path between the same two states, you can then relate the various energy quantities for other paths.

When a process consists of several distinct steps, it often helps to make a chart showing Q, W, and ∆U for each step. Put these quantities for each step on a different line, and arrange them so the Q's, W's, and ∆U's form columns. Then you can apply the first law to each line ; in addition, you can add each column and apply the first law to the sums. Do you see why ?

Using above steps, solve for the target variables. Step 4: Evaluate your answer : Check your results for

reasonableness. In particular, make sure that each of your answers has the correct algebraic sign. Remember that a positive Q means that heat flows into the system, and that a negative Q means that heat flows into the system, and that a negative Q means that heat flows out of the system. A positive W means that work is done by the system on its environment, while a negative W means that work is done on the system by its environment.

1. A metallic bob weighs 50 g in air. It it is immersed

in a liquid at a temperature of 25ºC, it weighs 45 g. When the temperature of the liquid is raised to 100ºC, it weighs 45.1 g. Calculate the coefficient of cubical expansion of the liquid given that the coefficient of linear expansion of the metal is 2 × 10–6(ºC)–1.

Sol. Loss in weight in liquid at 25ºC = (50 – 45) = 5 gm Weight of liquid displaced at 25ºC = V25ρ25g ∴ 5 = V25ρ25g ...(1) Similarly, V100ρ100g = 50 – 45.1 = 4.9 gm ...(2) From eq.(1) & (2) we get,

9.4

5 = 100

25

100

25 .ρρ

VV

Now, V100 = V25(1 + γmetal × 75)= V25(1 + 3αmetal × 75) = V25(1 + 3 × 12 × 10–6 × 75) or V100 = V25(1 + 0.0027) = V25 × 1.0027 Also, ρ25 = ρ100(1 + γ × 75) where, γ = Required coefficient of expansion of the liquid

9.4

5 = 100

100

25

25 )751(0027.1 ρ

γ+ρ×

×VV =

0027.1751 γ+

or γ = 3.1 × 10–4 (ºC)–1

2. A one litre flask contains some mercury. It is found that at different temperature the volume of air inside the flask remains the same. What is the volume of mercury in flask ? Given that the coefficient of linear expansion of glass = 9 × 10–6(ºC)–1 and coefficient of volume expansion of mercury = 1.8 × 10–4 (ºC–1).

Sol. Let V = Volume of the vessel V' = Volume of mercury For unoccupied volume to remain constant increase

in volume of mercury should be equal to increase in volume of vessel.

∴ V' γm∆T = Vγg∆T or V' = m

gVγ

γ×

∴ V' = 4

6

108.110271000

−

−

×

×× = 150 cm3

3. A clock with a metallic pendulum gains 6 seconds each day when the temperature is 20ºC and loses 12 seconds each day when the temperature is 40ºC. Find the coefficient of linear expansion of the metal.

Sol. Time taken for one oscillation of the pendulum is

T = gL

π2 or T2 = 4π2 × gL .....(1)

Partially differentiating, we get

2T∆t = 4π2 × gL∆ .....(2)

Dividing (2) by (1), we get

TT∆ =

LL

2∆ =

LtL

2∆α = t∆α

21

where ∆t is the change in temperature. Now, One day = 24 hours = 86400 sec Let t be the temperature at which the clock keeps

correct time. At 20ºC, the gain in time is

6 = 21

α × (t – 20) × 86400 ....(3)

At 40ºC, the loss in time is

12 = 21

α× (40 – t) × 86400 ...(4)

Dividing (4) by (3), we have

6

12 = 20

40−

−t

t

which gives t = 3

80 ºC.

Using the value in equation(3), we have

6 = 21 × α ×

− 20

380 × 86400

which gives α = 2.1 × 10–5 perºC

Solved Examples


4. A piston can freely move inside a horizontal cylinder closed from both ends. Initially, the piston separates the inside space of the cylinder into two equal parts each of volume V0, in which an ideal gas is contained under the same pressure p0 and at the same temperature. What work has to be performed in order to increase isothermally the volume of one part of gas η times compared to that of the other by slowly moving the piston ?

Sol. Let volume of chambers changes by ∆V. According to the problem, the final volume of left chamber is η times final volume of right chamber.

∴ V0 + ∆V = η(V0 – ∆V)

or ∆V = 011 V

+η−η

P0,v0,T0 P0,v0,T0

As piston is moved slowly therefore, change in

kinetic energy is zero. By work-energy theorem, we can write

Wgas in right chamber + Wgas in left chamber + extAgentW = ∆KE

extAgentW = (Wgas(R) + Wgas(L))

We know that in isothermal process, work done is given by

W = nRT ln

i

f

VV

∴ Work done by gas in left chamber (WL)

= P0V0 ln

∆+

0

0

VVV = P0V0 ln

+ηη

12

Similarly, work done by gas in right chamber (WR)

= P0V0 ln

∆−

0

0

VVV = P0V0 ln

+ηη

12

extAgentW = –P0V0 ln

+ηη

12 – P0V0 ln

+ηη

12

= P0V0 ln 2

41

η+η

5. A smooth vertical tube having two different sections is open from both ends equipped with two pistons of different areas figure. Each piston slides within a respective tube section. One mole of ideal gas is enclosed between the pistons tied with a non-stretchable thread. The cross-sectional area of the upper piston is ∆S greater than that of the lower one. The combined mass of the two pistons is equal to m. The outside air pressure is P0. By how many kelvins must the gas between the pistons be heated to shift the pistons through l.

P0

P0

Sol. Let A1 = Cross section of upper piston A2 = Cross section of lower piston T = Tension in the string P = Gas pressure m1 = Mass of upper piston m2 = Mass of lower piston Now, consider FBD of upper piston

PA1 m1g

P0 A1

From equilibrium consideration of upper piston we get, P0A1 + T + m1g = PA1 Similarly, consider FBD of lower piston

PA2

m2g P0 A2

T

∴ P0A2 + T = m2g + PA2 Eliminating T, we get

P = P0 + 21

21 )(AA

gmm−

+

According to problem m = m1 + m2 and ∆S = A1 – A2

∴ P = P0 + S

mg∆

Now, PV = RT

or P∆V = R∆T or ∆T = R

VP∆

But ∆V = (A1 – A2)l = ∆S. l

∴ ∆T =

∆+

SmgP0 ∆S.l

ll

l


Reduction of Aldehydes and Ketones by Hydride Transfer :

H3B – H + C = OR

R´ H – C – O

R

R´

– H – C – O – H

R

R´

H – OHδ+ δ–

Hydride transfer Alkoxide ion Alcohol

These steps are repeated until all hydrogen atoms attached to boron have been transferred.

Sodium borohydride is a less powerful reducing agent than lithium aluminum hydride. Lithium aluminum hydride reduces acids, aldehydes, and ketones but sodium borohydride reduces only aldehydes and ketones :

O

C O– R

O

C OR´ R

O

C R´ R

O

C HR

< < <

Reduced by LiAlH4

Reduced by NaBH4

Ease of reduction

Lithium aluminum hydride reacts violently with water, and therefore reductions with lithium aluminum hydride must be carried out in anhydrous solutions, usually in anhydrous ether. (Ethyl acetate is added cautiously after the reaction is over to decompose excess LiAlH4; then water is added to decompose the aluminum complex.) Sodium borohydride reductions, by contrast, can be carried out in water or alcohol solutions.

The Addition of Ylides : The Wittig reaction : Aldehydes and ketones react with phosphorus ylides

to yield alkenes and triphenylphosphine oxide. (An ylide is a neutral molecule having a negative carbon adjacent to a positive heteroatom.) Phosphorus ylides are also called phosphoranes :

C = O + (C6H5)3P – C

R

R

..+ R´´

R´´´

C = C R´´

R´´´

R

R´+ O =P(C6H5)3

Aldehyde orketone

Phosphorus ylide or phosphorane

Alkene [(E) and(Z) isomers]

Triphenyl phosphineoxide

This reaction, known as the Wittig reaction, has proved to be a valuable method for synthesizing alkenes. The Wittig reaction is applicable to a wide variety of compounds, and although a mixture of (E) and (Z) isomers may result, the Wittig reaction offers a great advantage over most other alkene syntheses in that no ambiguity exists as to the location of the double bond in the product. (This is in contrast to E1 eliminations, which may yield multiple alkene products by rearrangement to more stable carbocation intermediates, and both E1 and E2 elimination reactions, which may produce multiple products when different β hydrogens are available for removal.)

Phosphorus ylides are easily prepared from triphenylphosphine and primary or secondary alkyl halides. Their preparation involves two reactions :

General Reaction Reaction 1

(C6H5)3P : + CH – X → (C6H5)3P – CH X–

R´´

R´´´

+

Triphenylphosphine

R´´

R´´´

An alkyltriphenylphosphoniumhalide

Reaction 2

(C6H5)3P – C – H : B– → (C6H5)3P – C :– + H:B

R´´

R´´´

+R´´

R´´´

A phosphorus ylide

+

Specific Example Reaction 1

(C6H5)3P : + CH3Br → (C6H5)3P – CH3Br –C6H6+

Methyltriphenylphosphoniumbromide (89%)

Reaction 2

Organic Chemistry

Fundamentals

CARBONYL COMPOUND

KEY CONCEPT


(C6H5)3P – CH3 + C6H5Li →

(C6H5)3P – CH2 :– + C6H6 + LiBr+

+

Br–

The first reaction is a nucleophilic substitution

reaction. Triphenylphosphine is an excellent nucleophile and a weak base. It reacts readily with 1º and 2º alkyl halide by an SN2 mechanism to displace a halide ion from the alkyl halide to give an alkyltriphenylphosphonium salt. The second reaction is an acid-base reaction. A strong base (usually an alkyllithium or phenyllithium) removes a proton from the carbon that is attached to phosphorus to give the ylide.

Phosphorus ylides can be represented as a hybrid of the two resonance structures shown here. Quantum mechanical calculations indicate that the contribution made by the first structure is relatively unimportant.

(C6H5)3P = C

R´´

R´´´ (C6H5)3P – C :–

R´´

R´´´

+

The mechanism of the Wittig reaction has been the

subject of considerable study. An early mechanistic proposal suggested that the ylide, acting as a carbanion, attacks the carbonyl carbon of the aldehyde or ketone to form an unstable intermediate with separated charges called a betaine. In the next step, the betaine is envisioned as becoming an unstable four-membered cyclic system called an oxaphosphetane, which then spontaneously loses triphenylphosphine oxide to become an alkene. However, studies by E. Vedejs and others suggest that the betaine is not an intermediate and that the oxaphosphetane is formed directly by a cycloaddition reaction. The driving force for the Wittig reaction is the formation of the very strong (∆Hº = 540 kJ mol–1) phosphorus –oxygen bond in triphenylphosphine oxide.

R–C + –:C–R´ R – C – C – R´´´

R ´

–:O:

R ´´

P(C6H5)3..+

R – C – C – R´´´

R ´

:O –

R ´´

P(C6H5)3..

Aldehyde or ketone

R ´ R ´´

P(C6H5)3 +

:O:

Ylide Betaine (may not be formed)

Oxaphosphetane

C = C + O = P(C6H5)3R´´

R´´´

R´

R

Alkene (+diastereomer)

Triphenylphosphineoxide

Specific Example :

O + :CH2 – P(C6H5)3 CH2

O P(C6H5)3+–

CH2

O –P(C6H5)3

CH2 + O=P(C6H5)3

Methylenecyclohexane(86%)

– +

Michael Additions : Conjugate additions of enolate anions to

α-β-unsaturated carbonyl compound are known generally as Michael additions. An example is the addition of cyclohexanone to C6H5CH=CHCOC6H5 :

O O

O–

C6H5CH=CH–CC6H5

O –

O CH

C6H5

CHδ–

C—Oδ–

C6H5

O CH

C6H5

C

C = O

C6H5

H

H

+H3O+

OH–

The sequence that follows illustrates how a conjugate

aldol addition (Michael addition) followed by a simple aldol condensation may be used to build one ring onto another. This procedure is known as the Robinson anulation (ring-forming) reaction (after the English chemist, Sir Robert Robinon, who won the Nobel Prize in chemistry in 1947 for his research on naturally occurring compounds) :

O

+ CH2 = CHCCH3

O CH2

CH3

CH2

CH3C

O

2-Methylcyclo- hexane-1, 3-dione

CH3

O

O

O O

CH3

O (65%)

OH–

CH3OH (conjugateaddition) Methyl vinyl

ketone

aldol condensation

base (–H2O)


Inorganic Chemistry

Fundamentals

Tetragonal distortion of octahedral complexes (Jahn-Teller distortion) : The shape of transition metal complexes are affected

by whether the d orbitals are symmetrically or asymmetrically filled.

Repulsion by six ligands in an octahedral complex splits the d orbitals on the central metal into t2g and eg levels. It follows that there is a corresponding repulsion between the d electrons and the ligands. If the d electrons are symmetrically arranged, they will repel all six ligands equally. Thus the structure will be a completely regular octahedron. The symmetrical arrangements of d electrons are shown in Table.

Symmetrical electronic arrangements :

Electronic configurat

ion

t2g eg

d5

d6

d8

d10

All other arrangements have an asymmetrical

arrangement of d electrons. If the d electrons are asymmetrically arranged, they will repel some ligands in the complex more than others. Thus the structure is distorted because some ligands are prevented from approaching the metal.

as closely as others. The eg orbitals point directly at the ligands. Thus asymmetric filling of the eg orbitals in some ligands being repelled more than others. This causes a significant distortion of the octahedral shape. In contrast the t2g orbitals do not point directly at the ligands, but point in between the ligand directions. Thus asymmetric filling of the t2g orbitals has only a very small effect on the stereochemistry. Distortion caused by asymmetric filling of the t2g orbitals is usually too small to measure. The electronic arrangements which will produce a large distortion are shown in Table.

The two eg orbitals 22 yxd − and 2zd are normally

degenerate. However, if they are asymmetrically filled then this degeneracy is destroyed, and the two

orbitals are no longer equal in energy. If the 2zd orbital contains one.

Asymmetrical electronic arrangements :

Electronic configurati

on

t2g eg

d4

d7

d9

more electron than the 22 yxd − orbital then the ligands

approaching along +z and –z will encounter greater repulsion than the other four ligands. The repulsion and distortion result in elongation of the octahedron along the z axis. This is called tetragonal distortion. Strictly it should be called tetragonal elongation. This form of distortion is commonly obsered.

If the 22 yxd − orbital contains the extra electron, then

elongation will occur along the x and y axes. This means that the ligands approach more closely along the z-axis. Thus there will be four long bonds and two short bonds. This is equivalent to compressing the octahedron along the z axis, and is called tetragonal compression, and it is not possible to predict which will occur.

For example, the crystal structure of CrF2 is a distorted rutile (TiO2) structure. Cr2+ is octahedrally surrounded by six F–, and there are four Cr–F bonds of length 1.98 – 2.01 Å, and two longer bonds of length 2.43 Å. The octahedron is said to be tetragonally distorted. The electronic arrangement in Cr2+ is d4. F– is a weak field ligand, and so the t2g level contains three electrons and the eg level contains one electron. The 22 yxd − orbital has four lobes whilst

the 2zd orbital has only two lobes pointing at the ligands. To minimize repulsion with the ligands, the single eg electron will occupy the 2zd orbital. This is equivalent to splitting the degeneracy of the eg level so that 2zd is of lower energy, i.e. more stable, and

22 yxd − is of higher energy, i.e. less stable. Thus the

CO-ORDINATION COMPOUND

& METALLURGY

KEY CONCEPT


two ligands approaching along the +z and –z directions are subjected to greater repulsion than the four ligands along +x, –x, +y and –y. This causes tetragonal distortion with four short bonds and two long bonds. In the same way MnF3 contains Mn3+ with a d4 configuration, and forms a tetragonally distorted octahedral structure.

Many Cu(+II) salts and complexes also show tetragonally distorted octahedral structures. Cu2+ has a d9 configuration :

t2g

eg

To minimize repulsion with the ligands, two

electrons occupy the 2zd orbital and one electron

occupies the 22 yxd − orbital. Thus the two ligands

along –z and –z are repelled more strongly than are the other four ligands.

The examples above show that whenever the 2zd and

22 yxd − orbitals are unequally occupied, distortion

occurs. This is know as Jahn–Teller distortion. Leaching : It involves the treatment of the ore with a suitable

reagents as to make it soluble while impurities remain insoluble. The ore is recovered from the solution by suitable chemical method. For example, bauxite ore contains ferric oxide, titanium oxide and silica as impurities. When the powdered ore is digested with an aqueous solution of sodium hydroxide at about 150ºC under pressure, the alumina (Al2O3) dissolves forming soluble sodium meta-aluminate while ferric oxide (Fe2O3), TiO2 and silica remain as insoluble part.

Al2O3 + 2NaOH → 2NaAlO2 + H2O Pure alumina is recovered from the filtrate NaAlO2 + 2H2O → Al(OH)3 + NaOH

2Al(OH)3 )autoclave(Ignited → Al2O3 + 3H2O

Gold and silver are also extracted from their native ores by Leaching (Mac-Arthur Forrest cyanide process). Both silver and gold particles dissolve in dilute solution of sodium cyanide in presence of oxygen of the air forming complex cyanides.

4Ag + 8NaCN + 2H2O + O2 → 4NaAg(CN)2 + 4NaOH Sod. argentocyanide 4Au + 8NaCN + 2H2O + O2 → 4NaAu(CN)2 + 4NaOH Sod. aurocyanide Ag or Au is recovered from the solution by the

addition of electropositive metal like zinc.

2NaAg(CN)2 + Zn → Na2Zn(CN)4 + 2Ag ↓ 2NaAu(CN)2 + Zn → Na2Zn(CN)4 + 2Au ↓ Soluble complex Special Methods : Mond's process : Nickel is purified by this method.

Impure nickel is treated with carbon monoxide at 60–80º C when volatile compound, nickel carbonyl, is formed. Nickel carbonyl decomposes at 180ºC to form pure nickel and carbon monoxide which can again be used.

Impure nickel + CO 60–80ºCNI(CO)4

Ni + 4CO

180ºC

Gaseous compound

Zone refining or Fractional crystallisation : Elements such as Si, Ge, Ga, etc., which are used as

semiconductors are refined by this method. Highly pure metals are obtained. The method is based on the difference in solubility of impurities in molten and solid state of the metal. A movable heater is fitted around a rod of the impure metal. The heater is slowly moved across the rod. The metal melts at the point of heating and as the heater moves on from one end of the rod to the other end, the pure metal crystallises while the impurities pass on the adjacent melted zone.

Molten zonecontainingimpurity

Pure metalMoving circular

heater

Impure zone

Different metallurgical processes can be broadly

divided into three main types. Pyrometallurgy : Extraction is done using heat

energy. The metals like Cu, Fe, Zn, Pb, Sn, Ni, Cr, Hg, etc., which are found in nature in the form of oxides, carbonates, sulphides are extracted by this process.

Hydrometallurgy : Extraction of metals involving aqueous solution is known as hydrometallurgy. Silver, gold, etc., are extracted by this process.

Electrometallurgy : Extraction of highly reactive metals such as Na, K, Ca, Mg, Al, etc., by carrying electrolysis of one of the suitable compound in fused or molten state.


1. It is possible to supercool water without freezing. 18

g of water are supercooled to 263.15 K(–10ºC) in a thermostat held at this temperature, and then crystallization takes place.

Calculate ∆rG for this process. Given: Cp(H2O,1) = 75.312 J K–1 mol–1

Cp (H2O,s) = 36.400 J K–1 mol–1 ∆fusH (at 0ºC) = 6.008 kJ mol–1 Sol. The process of crystallization at 0ºC and at 101.325

kPa pressure is an equilibrium process, for which ∆G = 0. The crystallization of supercooled water is a spontaneous phase transformation, for which ∆G must be less than zero. Its value for this process can be calculated as shown below.

The given process H2O(1, – 10ºC) → H2O(s, –10ºC) is replaced by the following reversible steps. (a) H2O(1, – 10ºC) → H2O(1, 0ºC) ...(1)

∆rH1 = ∫K15.273

K15.263m,p )1(C dT

= (75.312 J K–1 mol–1 ) (10 K) = 753.12 J mol–1

∆rS1 = ∫K15.273

K15.263

m,p

R)1(C

dT

= (75.312 J K–1mol–1) × ln

K15.263K15.273

= 2.809 J K–1 mol–1 (b) H2O(1, 0ºC) → H2O(s, 0ºC) ...(2) ∆rH2 = – 6.008 kJ mol–1

∆rS2 = – )K15.273(

)molJ6008( 1– = – 21.995 J K–1 mol–1

(c) H2O(s, 0ºC) → H2O(s, –10ºC) ...(3)

∆rH3 = ∫K15.263

K15.273m,p )s(C dT

= (36.400 J K–1 mol–1)(–10 K) = – 364.0 J mol–1

∆rS3 = ∫K15.263

K15.273

m,p

T)s(C

dT

= (36.400 J K–1 mol–1) ×ln

K15.273K15.263

= – 1.358 J K–1 mol–1 The overall process is obtained by adding Eqs. (1),

(2) and (3), i.e. H2O(1, –10ºC) → H2O(s, –10ºC) The total changes in ∆rH and ∆rS are given by ∆rH = ∆rH1 + ∆rH2 + ∆rH3 =(753.12 – 6008 – 364.0) J mol–1

= – 5618.88 J mol–1 ∆rS = ∆rS1 + ∆rS2 + ∆rS3 = (2.809 – 21.995 – 1.358) J K–1 mol–1 = – 20.544 J K–1 mol–1 Now ∆rG of this process is given by ∆rG = ∆rH – T∆rS = – 5618.88 J mol–1 – (263.15 K)( –20.544 J K–1 mol–1 ) = – 212.726 J mol–1

2. From the standard potentials shown in the following diagram, calculate the potentials º

1E and º2E .

BrO3– 0.54 V BrO– 0.45 V

21 Br2

1.07 V Br–

0.17 V

E2º

E1º

Sol. The reaction corresponding to the potential Eº1 is

BrO3– + 3H2O + 5e– =

21 Br2 + 6OH– ...(1)

This reaction can be obtained by adding the following two reduction reactions:

BrO3– + 2H2O + 4e– = BrO– + 4OH– ...(2)

BrO– + H2O + e– = 21 Br2 + 2OH– ...(3)

Hence the free energy change of reaction (1) will be º

)1(reactionG∆ = º)2(reactionG∆ + º

)3(reactionG∆

Replacing ∆Gºs in terms of potentials, we get – 5FE1º = – 4F(0.54 V) – 1F (0.45 V) = (–2.61 V) F

UNDERSTANDINGPhysical Chemistry


Hence E1º = 5

V61.2 = 0.52 V

Now the reaction corresponding to the potential E2º is BrO3

– + 2H2O + 6e– = Br– + 6OH– ...(4) This reaction can be obtained by adding the

following three reactions. BrO3

– + 2H2O + 4e– = BrO– + 4OH– (Eq.2)

BrO– + H2O + e– = 21 Br2 + 2OH– (Eq.3)

21 Br2 + e– = Br– ...(5)

Hence

º)4(reactionG∆ = º

)2(reactionG∆ + º)3(reactionG∆

+ º)5(reactionG∆

or – 6F(E2º) = – 4F(0.54 V) – 1F(0.45 V) – 1F (1.07 V) = (– 3.68 V) F

or E2º = 668.3 = 0.61 V.

3. What is the solubility of AgCl in 0.20 M NH3 ? Given : Ksp(AgCl) = 1.7 × 10–10 M2 K1 = [Ag(NH3)+] / [Ag+] [NH3] = 2.33 × 103 M–1 and K2 = [Ag(NH3)2

+]/[Ag(NH3)+][NH3] = 7.14 × 103 M–1 Sol. If x be the concentration of AgCl in the solution, then [Cl–] = x From the Ksp for AgCl, we derive

[Ag+] = ]Cl[

Ksp−

= x

M107.1 210−×

If we assume that the majority of the dissolved Ag+ goes into solution as Ag(NH3)2

+ then [Ag(NH3)2+] = x

Since two molecules of NH3 are required for every Ag(NH3)2

+ ion formed, we have [NH3] = 0.20 M – 2x Therefore,

Kinst =])NH(Ag[

]NH][Ag[

23

23+

+

= x

)x2M20.0(x

M107.1 2210

−

× −

= 6.0 × 10–8 M2 From which we derive

2

2

x)x2M20.0( − = 210

28

M107.1M100.6

−

−

×× = 3.5 × 102

which gives x = [Ag(NH3)2+] = 9.6 × 10–3 M, which

is the solubility of AgCl in 0.20 M NH3

4. Potassium alum is KA1(SO4)2.12H2O. As a strong electrolyte, it is considered to be 100% dissociated into K+, Al3+, and SO4

2–. The solution is acidic because of the hydrolysis of Al3+, but not so acidic as might be expected, because the SO4

2– can sponge up some of the H3O+ by forming HSO4

–. Given a solution made by dissolving 11.4 g of KA1(SO4)2.12H2O in enough water to make 0.10 dm3 of solution, calculate its [H3O+] :

(a) Considering the hydrolysis Al3+ + 2H2O Al(OH)2+ + H3O+ with Kh = 1.4 × 10–5 M (b) Allowing also for the equilibrium HSO4

– + H2O H3O+ + SO42–

with K2 = 1.26 × 10–2 M

Sol. (a) Amount of alum = 1molg38.474g4.11

− = 0.024 mol

Molarity of the prepared solution = 3dm1.0mol024.0

= 0.24 M Hydrolysis of Al3+ is Al3+ + 2H2O Al(OH)2+ + H3O+

Kh = ]Al[

]OH][)OH(Al[3

32

+

++

If x is the concentration of Al3+ that has hydrolyzed, we have

Kh = xM24.0

)x)(x(−

= 1.4 × 10–5 M

Solving for x, we get [H3O+] = x = 1.82 × 10–3 M (b) We will have to consider the following equilibria. Al3+ + 2H2O Al(OH)2+ + H3O+ H3O+ + SO4

2– HSO4– + H2O

Let z be the concentration of SO42– that combines

with H3O+ and y be the net concentration of H3O+ that is present in the solution. Since the concentration z of SO4

2– combines with the concentration z of H3O+, it is obvious that the net concentration of H3O+ produced in the hydrolysis reaction of Al3+ is (y + z). Thus, the concentration (y + z) of Al3+ out of 0.24 M hydrolyzes in the solution. With these, the concentrations of various species in the solution are

zyM24.0

3Al−−

+ + 2H2O zy

2)OH(Al+

+ + y

3OH +

y

3OH + + zM48.0

24SO

−

− z

4HSO− + H2O

Thus, Kh = )zyM24.0(

)y)(zy(−−

+ = 1.4 × 10–5 M ...(i)


K2 = )zM48.0(y

z−

= M1026.1

12−×

...(ii)

From Eq. (ii), we get

z = y)M1026.1(

y)M48.0(2 +× −

Substituting this in Eq. (i), we get

+×−−

+×+

−

−

y)M1026.1(y)M48.0(y24.0

yy)M1026.1(

y)M48.0(y

2

2

= 1.4 × 10–5

Making an assumption that y <<1.26 × 10–2 M, and then solving for y, we get

[H3O+] = y = 2.932 × 10–4 M 5. The freezing point of an aqueous solution of KCN

containing 0.189 mol kg–1 was – 0.704 ºC. On adding 0.095 mol of Hg(CN)2, the freezing point of the solution became –0.530ºC. Assuming that the complex is formed according to the equation

Hg(CN)2 + x CN– → –x2x)CN(Hg +

Find the formula of the complex. Sol. Molality of the solution containing only KCN is

m = f

f

K)T(–∆

= )molkgK86.1(

)K704.0(1– = 0.379 mol kg–1

This is just double of the given molality ( = 0.189 mol kg–1) of KCN, indicating complete dissociation of KCN. Molality of the solution after the formation of the complex

m = f

f

K)T(–∆

= )molkgK86.1(

)K530.0(1– = 0.285 mol kg–1

If it be assumed that the whole of Hg(CN)2 is converted into complex, the amounts of various species in 1 kg of solvent after the formation of the complex will be

n(K+) = 0.189 mol, n(CN–) = (0.189 – x) mol ))CN(Hg(n –x

2x+ = 0.095 mol Total amount of species in 1 kg solvent becomes ntotal = [0.189 + (0.189 – x) + 0.095] mol = (0.473 – x) mol Equating this to 0.285 mol, we get (0.473 – x) mol = 0.285 mol i.e. x = (0.473 – 0.285) = 0.188

Number of CN– units combined = mol095.0mol188.0 = 2

Thus, the formula of the complex is –24)CN(Hg .

CHEMISTRY JOKES

Joke 1 : If you succeeding in guessing the answer to the previous joke, then you figure out this one:

Q: What is the chemical name of the following benzene-like molecule?

4 \ C - C 4 / \ / C C \ / C - C A : Metaphor

Joke 2 : Q: What is the name of the molecule bunny-O-

bunny? A: An ether bunny

Joke 3 : Q: If H-two-O is the formula for water, what is the

formula for ice? A: H-two-O-CUBED

Joke 4 : Q: What is the chemical symbol for diarrhea? A: (CO(NH2)2)2

Joke 5 : Q: Why do chemists like nitrates so much? A: They're cheaper than day rates.

Joke 6 : Q: What is the chemical formula for the molecules in

candy? A: Carbon-Holmium-Cobalt-Lanthanum-Tellurium

or CHoCoLaTe

Joke 7 : Here is a historical note: In the 1980's, in an effort to increase public awareness about the importance of chemistry, the American Chemical Society posted billboards with a picture of C6H10 and the title, "It takes alkynes to make a world."


1. Let f (x) = sin x and

g(x) =

π>π≤≤≤≤

xxxforxttf

;2/sin0;0);(max

2

Discuss the continuity and differentiability of g(x) in (0, ∞)

2. Is the inequality sin2 x < x sin(sin x) true for 0 < x < π/2 ? Justify your answer.

3. A shop sells 6 different flavours of ice-cream. In how many ways can a customer choose 4 ice-cream cones if

(i) they are all of different flavours;

(ii) they are not necessarily of different flavours;

(iii) they contain only 3 different flavoures;

(iv) they contain only 2 or 3 different flavoures ?

4. Using vector method, show that the internal (external) bisector of any angle of a triangle divides the opposite side internally (externally) in the ratio of the other two sides containing the triangle.

5. Prove that

(a) cos x + nC1 cos 2x + nC2 cos 3x + ............

...... + nCn cos(n + 1)x = 2n. cosnx/2. cos

+ xn

22

(b) sin x + nC1 sin 2x + nC2 sin 3x + ............... ....... + nCn sin(n + 1)x = 2n . cosn x/2 . sin

+ xn

22

6. In a town with a population of n, a person sands two letters to two sperate people, each of whom is asked to repeat the procedure. Thus, for each letter received, two letters are sent to separate persons chosen at random (irrespective of what happened in the past). What is the probability that in the first k stages, the person who started the chain will not receive a letter ?

7. Prove the identity :

∫ −x zzxe0

2dz = ∫ −x zx ee

0

44 22 dz, deriving for the

function f (x) = ∫ −x zzxe0

2dz a differential equation

and solving it.

8. Prove that ∫ θθsecsin n dθ

= –1

)1cos(2−

θ−n

n – ∫ θθθ dn sec)2–sin( dθ.

Hence or otherwise evaluate

∫π

θθθ2/

0 cos3sin5cos dθ.

9. Find the latus rectum of parabola

9x2 – 24 xy + 16y2 – 18x – 101y + 19 = 0.

10. A circle of radius 1 unit touches positive x-axis and positive y-axis at A and B respectively. A variable line passing through origin intersects the circle in two points in two points D and E. Find the equation of the lines for which area of ∆ DEB is maximum.

`tàxÅtà|vtÄ VtÄÄxÇzxá This section is designed to give IIT JEE aspirants a thorough grinding & exposure to variety of possible twists and turns of problems in mathematics that would be very helpful in facing IIT JEE. Each and every problem is well thought of in order to strengthen the concepts and we hope that this section would prove a rich resource for practicing challenging problems and enhancing the preparation level of IIT JEE aspirants.

By : Shailendra Maheshwari Joint Director Academics, Career Point, KotaSolut ions wil l be published in next issue

10Set

Behavior

• Behavior is a mirror in which everyone displays his image.

• Behavior is what a man does, not what he thinks, feels, or believes.

• Behave the way you'd like to be and soon you'll be the way you behave.


1. as φ (a) = φ (b) = φ (c) so by Rolle’s theorem there must exist at least a point

x = α & x = β each of intervals (a, c) & (c, b) such that φ′(α) = 0 & φ′(β) = 0. Again by Rolle’s theorem, there must exist at least a point x = µ such that α < µ < β where φ′(µ) = 0

so )()(

)(2caba

af−−

+ )()(

)(2abcb

bf−−

+ )()(

)(2bcac

cf−−

– f ′′ (µ) = 0

so )()(

)(caba

af−−

+ )()(

)(abcb

bf−−

+ )()(

)(bcac

cf−−

= 21 f ′′ (µ)

where a < µ < b. 2. Required probability

1 . 65 .

65 .

65 ........

65 .

61 =

2r

65 −

.

61 (r – 2) times

Note : any number in 1st loss same no. does not in 2nd (any other comes). Now 3rd is also diff. (and in same r − 2 times) Now (r − 1)th & r th must be same. 3. 2s = a + b + c ON = − BN + BO Let BN = x 2BN + 2CN + 2AR = 2s x + (a − x) + (b − a + x) = s x = s − b

R

O C

N B

M

A

r

I (h,k)

so h = ON =

2a − (s − b)

= 2

22 bas ++− = 2

cb − & r = k.

so r = k = s∆ =

scsbsass ))()(( −−−

r = k = s

csbsass ))()(( −−−

2sk = ))()(( cbacbaass −++−−

= )2)(2)(( xaxaass +−−

2sk = )4)(( 22 haass −− required locus is 4s2y2 = A(a2 – 4x2)

⇒ s2y2 + Ax2 = 4

2Aa

where A is = s (s – a) here h2 < as so it is an ellipse

4. f (0) = c f (1) = a + b + c & f (−1) = a − b + c solving these,

a = 21 [f (1) + f (−1) − 2 f (0)] ,

b = 21 [f (1) − f (−1)] & c = f (0)

so f (x) = 2

)1( +xx f (1) + (1− x2) f (0) +2

)1( −xx f(−1)

2 | f (x) | < | x | | x + 1 | + 2| 1 − x2 | + | x | | x − 1| ; as | f (1) | , | f (0) |, | f (−1) | ≤ 1. 2 | f (x) | ≤ | x | (x + 1) + 2 (1 − x2) + | x | (1 − x) as

x ∈ [−1, 1]

so 2 | f (x) | ≤ 2 (|x| + 1 − x2) ≤ 2 . 45 so | f (x) | ≤

45

Now as g (x) = x2 f (1/x) = 21 (1 + x) f (1)

+ (x2 − 1) f (0) + 21 (1 − x) f (−1)

so 2 | g (x) | ≤ | x + 1 | + 2 | 1 − x2 | + | 1 − x| ⇒ 2 | g (x) | ≤ x + 1 + 2 (1 − x2) | + 1 − x ; as x ∈ [−1, 1] ⇒ 2 | g (x) | < − 2x2 + 4 ≤ 4. ⇒ |g (x) | ≤ 2.

5. Oil bed is being shown by the plane A′ PQ. θ be the angle between the planes A′ PQ & A′ B′ C′. Let A′ B′ C′ be the x − y plane with x-axis along A′ C′ and origin at A′. The P.V.s of the various points are defined as follows

MATHEMATICAL CHALLENGES SOLUTION FOR JANUARY ISSUE (SET # 9)


B

A C

B´

A´ P

Q

C´

point C′ : b i , point B′ : cos A i + c sin A j ,

point Q : b i – z k , point P : cos A i + c sin A j – y k normal vector to the plane A′ B′ C′ = 1n

r = bc sin A k

normal vector to the plane A'PQ = 2nr

= cz sin A i + (by – cz cos A) j + bc sin A k

so cos θ = ||||

.

11

21nn

nnrr

rr

= 2/12222222 ]sin)cos(sin[sin

AcbAczbyAzcAbc

+−+

cos θ = 2/12222222 )]cos2(sin[sin

AbyczybzcAcbAcb

−++

so tan θ = Abc

Abyczybzcsin

]cos2[ 2/12222 −+

so tan θ . sin A = Abcyz

cy

bz cos2

2

2

2

2−+

6. ∫ +−

xxx

5cos217cos8cos .

xx

5sin25sin2 dx

= ∫ ++−−)10sin5(sin2

2sin12sin3sin13sinxx

xxxx dx

= ∫ +−+

)10sin5(sin212sin–3sin2sin13sin

xxxxxx dx

= ∫−

25cos

215sin.2.2

29cos

215sin2

211cos

215sin2

xx

xxxx

dx

= ∫−

25cos2

29cos

211cos

x

xx

dx

= ∫−

25cos2

2sin5sin2

x

xxdx

= − 2 ∫

2sin

25sin xx dx

= ∫

−

24cos

26cos xx dx

= ∫ − dxxx )2cos3(cos

= 33sin x −

22sin x + C

7. 2

2

dxyd = 2 ∫

x

dttf0

)(

integrate using by parts method

dxdy = 2

− ∫∫

xx

dxxfxdttfx00

)(.)(

= 2

−∫

x

dttftx0

)()(

again integrating,

y = 2

−−− ∫∫∫ dxdttfxdttftxx

xxx

000

0)()()(

=2

+−− ∫∫∫ dxxfxdttfxdttftxx

xxx

0

2

0

2

0

)(2

)(2

)()(

= ∫ −x

dttfxtx0

2 )()(2 − ∫x

dttfx0

2 )( + ∫x

dttft0

2 )(

y = ∫ +−x

dttftxtx0

22 )()2( = ∫ −x

dttftx0

2 )()(

8. To prove that αα

+

/1

1ba <

ββ

+

/1

1ba

Let ba = c > 0

so (cα + 1)1/α < (cβ + 1)1/β. Let f (x) = (cx + 1)1/x ; x > 0

f ′(x) = (cx + 1)1/x ln (cx + 1)

− 2

1x

+x1 (cx + 1) x

1 –1. cx ln c

= 2

11

)1(x

c xx −+ ])1()1([ xxxx cnccnc ll +++− < 0

so f (x) is decreasing function so f (α) < f (β). Hence proved.


9. Point P (x, 1/2) under the given condition are length PB = OB

B (t, 1)

A (t – 1)

C O

P θ

rθ = t ; so θ = t

from ∆PAB : 2

PB = PA sin 2θ

⇒ PB = 2 sin 2t ........(1)

Now ∠ PBC = 2θ =

2t ;

so from ∠ PCB ; 2θ =

2t

so from ∆ PCB ; PB

2/1 = sin 2t ........(2)

from (1) & (2) PB = 1 ; so θ = t = π/3

thus | PB |2 = (t − x)2 + 41 = 1.

| t − x | = 23 ; t − x =

23 ; as t > x

so x = 3π −

23

10. Let xn = 1−n + 1+n be rational, then

nx

1 = 11

1++− nn

is also rational

nx

1 = 2

11 −−+ nn is also rational

1+n − 1−n is also rational

as 1+n + 1−n & 1+n − 1−n are rational

so 1+n + 1−n must be rational i.e. (n + 1) & (n – 1) are perfect squares. This is not possible as any two perfect squares differe

at least by 3. Hence there is not positive integer n for

which 1−n + 1+n is a rational.

Regents Physics You Should Know

Modern Physics : • The particle behavior of light is proven by the

photoelectric effect.

• A photon is a particle of light wave packet.

• Large objects have very short wavelengths when moving and thus can not be observed behaving as a wave. (DeBroglie Waves)

• All electromagnetic waves originate from accelerating charged particles.

• The frequency of a light wave determines its energy (E = hf).

• The lowest energy state of a atom is called the ground state.

• Increasing light frequency increases the kinetic energy of the emitted photo-electrons.

• As the threshold frequency increase for a photo-cell (photo emissive material) the work function also increases.

• Increasing light intensity increases the number of emitted photo-electrons but not their KE.

Mechanics : • Centripetal force and centripetal acceleration

vectors are toward the center of the circle- while the velocity vector is tangent to the circle.

• An unbalanced force (object not in equilibrium) must produce acceleration.

• The slope of the distance-tine graph is velocity.

• The equilibrant force is equal in magnitude but opposite in direction to the resultant vector.

• Momentum is conserved in all collision systems.

• Magnitude is a term use to state how large a vector quantity is.


1. If a

α4sin + b

α4cos = ba +

1 , show that

3

8sina

α + 3

8cosb

α = 3)(1ba +

.

Sol. Here a

ba + sin4α + b

ba + cos4α = 1

or sin4α + cos4α + ab sin4α +

ba cos4α = 1

or (sin2α + cos2α)2 – 2 sin2α . cos2α

+ ab sin4α +

ba cos4α = 1

or 2

2sin

α

ab – 2 .

ab sin2α .

ba cos2α

+ 2

2cos

α

ba = 0

or 2

22 cos–sin

αα

ba

ab = 0

∴ ab sin2α =

ba cos2α

or sin2α = ba cos2α

∴ a

α2sin = b

α2cos = ba +

α+α 22 cossin

∴ sin2α = ba

a+

, cos2α = ba

b+

∴ 3

8sina

α + 3

8cosb

α = 31a

. 4

4

)( baa+

+ 31

b. 4

4

)( bab+

= 4)( baa

+ + 4)( ba

b+

= 4)( baba

++ = 3)(

1ba +

2. Let [x] stands for the greatest integer function find

the derivative of f(x) = xxxx sin3 2])1[( +++ , where it

exists in (1, 1.5). Indicate the point(s) where it does not exist. Give reason(s) for your conclusion.

Sol. The greatest integer [x3 + 1] takes jump from 2 to 3 at 3 2 and again from 3 to 4 at 3 3 in [1, 1.5] and therefore it is discontinuous at these two points. As a result the given function is discontinuous at 3 2 and hence not differentiable.

To find the derivative at other points we write :

in (1, 3 2 ), f(x) = xxx sin2)2( ++

⇒ f ´(x) = 1sin2)2( −++ xxx

x2 + sin x + (x + 2) (2x + cos x) log (x + 2)

in ( 3 2 , 3 3 ), f(x) = xxx sin2)3( ++ ,

f ´(x) = 1sin2)3( −++ xxx x2 + sin x

+ (2x + cos x) (x + 3) × loge (x + 3)

in ( 3 5 , 1.5), f(x) = xxx sin2)4( ++ ,

f ´(x) = 1sin2)4( −++ xxx , x2 + sin x + (2x + cos x)

(x + 4) × loge(x + 4)

3. The decimal parts of the logarithms of two numbers taken at random are found to six places of decimal. What is the chance that the second can be subtracted from the first without "borrowing"?

Sol. For each column of the two numbers, n(S) = number of ways to fill the two places by the

digits 0, 1, 2, ... , 9 = 10 × 10 = 100.

× × × × × ×

y × × × × × ×

x

Let E be the event of subtracting in a column without

borrowing. If the pair of digits be (x, y) in the column where x is in the first number and y is in the second number then

E = (0, 0), (1, 0), (2, 0), .. ,(9, 0), (1, 1), (2, 1), ..., (9, 1), (2, 2), (3, 2), ..., (9, 2), (3, 3), (4, 3), ..., (9, 3), ...... (8, 8), (9, 8), (9, 9)

Expert’s Solution for Question asked by IIT-JEE Aspirants

Students' ForumMATHS


∴ n(E) = 10 + 9 + 8 + ... + 2 + 1 = 211.10 = 55

∴ the probability of subtracting without borrowing

in each column = 10055 .

∴ the required probability = 6

10055

=

6

2011

.

4. Let S be the coefficients of x49 in given expression f(x) and if P be product of roots of the equation

f(x) = 0, then find the value of PS , given that :

f(x) = (x – 1)2

− 2

2x

−

21x

− 3

3x

−

31x ,

.........

− 25

25x

−

251x

Sol. Here we can write f(x) as :

f(x) =

−

−

−− 25

25...3

32

2)1( xxxx

×

−

−

−−

251...

31

21)1( xxxx

Now roots of f(x) = 0 are;

12, 22, 32, ..... , 252 and 1, 21 ,

31 , .....,

251

Now f(x) is the polynomial of degree 50, So coefficient of x49 will be : S = – (sum of roots)

= – (12 + 22 + ... + 252) –

++++

251....

31

211

= –

+

×× K6

512625 where, K = ∑=

25

1

1

n n

⇒ S = –(K + 5525). Product of roots :

12 . 22 . 32 .... 252 . 1 . 21 .

31 ....

251 = 1 . 2 . 3 ...25

∴ P = 25 !

Hence PS =

!25)5525K( +− , where K = ∑

=

25

1

1

n n

5. A traveller starts from a certain place on a certain day and travels 1 km on the first day and on subsequent days, he travels 2 km more than the previous day. After 3 days, a second traveller sets out from the same place and on his first day he travels 12 km and on subsequent days he travels 1 km more than the previous day. On how many days will the second traveller be ahead of the first?

Sol. The first traveller travels on different days as follow (in km) 1, 1 + 2, 1 + 2 + 2, ... .

After 3 days the first traveller is already ahead by (1 + 3 + 5) km, i.e., 9 km.

1 3 5

The second traveller travels on different days as

follows : 0, 0, 0, 12, 13, 14, ... After n days from the day the second traveller starts,

the distance covered by the first = 1 + 3 + 5 + (7 + 9 + ... to n terms) = 1 + 3 + 5 + ... to (n + 3) terms = (n + 3)2 and the distance covered by the second = 12 + 13 + 14 + ... to n terms

= 2n 24 + (n – 1).1 =

2)23( +nn

The second traveller is ahead of the first on the nth day (after the second sets off) if

2

)23( +nn > (n + 3)2

or n2 + 23n > 2(n2 + 6n + 9) or n2 – 11n + 18 < 0 or (n – 2) (n – 9) < 0. So n – 2 > 0 and n – 9 < 0 ...(i) or n – 2 < 0 and n – 9 > 0 ...(ii) (i) ⇒ n > 2 and n < 9 (ii) ⇒ n < 2 and n > 0 (absurd) Thus, from the begining of the 3rd day to the end of

the 9th day the second traveller is ahead of the first. So, the second is ahead of the first on the 3rd, 4th, 5th,

..., 9th days (after the second sets off). Hence, the required number of days = 7.

6. Let P(x) be a polynomial of degree n such that

P(i) = 1+i

i for i = 0, 1, 2 ..... n. If n is odd than find

the value of P(n + 1). Sol. Let Q(x) = (x + 1) P(x) – x clearly Q(x) is polynomial of degree n + 1. Also

Q(i) = (i + 1) 1+i

i – i = 0 for i = 1, 2, 3 .....n

Thus we can assume Q(x) = kx(x – 1) (x – 2) ...... (x – n) where k is a constant. Now Q(–1) = k(–1)(–2)(–3) ...... (–1 – n) 1 = (–1)n + 1 k(n + 1) !

⇒ k = !)1n(

1+

(Q n is odd)

Thus P(x) =

+

+−−−

+x

xnxxxx

x !)1())....(2)(1(

11 ,

where n is odd , ∴ P(n + 1) = 1


Integration :

If dxd f(x) = F(x), then ∫ )(xF dx = f(x) + c, where c

is an arbitrary constant called constant of integration.

1. ∫ dxxn = 1

1

+

+

nxn

(n ≠ –1)

2. ∫ dxx1 = log x

3. ∫ dxex = ex

4. ∫ dxa x = a

a

e

x

log

5. ∫ dxxsin = – cos x

6. ∫ dxxcos = sin x

7. ∫ dxx2sec = tan x

8. ∫ dxxec2cos = – cot x

9. ∫ sec x tan x dx = sec x

10. ∫ cosec x cot x dx = – cosec x

11. ∫ sec x dx = log(sec x + tan x) = log tan

π

+42

x

12. ∫ cosec x dx = – log (cosec x + cot x) = log tan

2x

13. ∫ tan x dx = – log cos x

14. ∫ cot x dx = log sin x

15. ∫− 22 xa

dx = sin–1

ax = – cos–1

ax

16. ∫ + 22 xadx =

a1 tan–1

ax = –

a1 cot–1

ax

17. ∫− 22 axx

dx = a1 sec–1

ax = –

a1 cosec–1

ax

18. ∫ − 221

ax =

a21 log

axax

+− , when x > a

19. ∫ − 221

xadx =

a21 log

xaxa

−+ , when x < a

20. ∫− 22 ax

dx = log

−+ 22 axx = cos h–1

ax

21. ∫+ 22 ax

dx = log

++ 22 axx = sin h–1

ax

22. ∫ − 22 xa dx = 21 x 22 xa − +

21 a2 sin–1

ax

23. ∫ − 22 ax dx = 21 x 22 ax −

– 21 a2log

−+ 22 axx

24. ∫ + 22 ax dx = 21 x 22 ax +

+ 21 a2 log

++ 22 axx

25. ∫ )()´(

xfxf dx = log f(x)

26. ∫ )()´(

xfxf dx = 2 )(xf

Integration by Decomposition into Sum : 1. Trigonometrical transformations : For the

integrations of the trigonometrical products such as sin2x, cos2x, sin3x, cos3x, sin ax cos bx, etc., they are expressed as the sum or difference of the sines and cosines of multiples of angles.

2. Partial fractions : If the given function is in the form of fractions of two polynomials, then for its integration, decompose it into partial fractions (if possible).

Integration of some special integrals :

(i) ∫ ++ cbxaxdx

2

This may be reduced to one of the forms of the above formulae (16), (18) or (19).

INTEGRATION

Mathematics Fundamentals MATHS


(ii) ∫++ cbxax

dx2

This can be reduced to one of the forms of the above formulae (15), (20) or (21).

(iii) ∫ ++ cbxax2 dx

This can be reduced to one of the forms of the above formulae (22), (23) or (24).

(iv) ∫ +++

cbxaxdxqpx

2)( , ∫

++

+

cbxax

dxqpx2

)(

For the evaluation of any of these integrals, put px + q = A differentiation of (ax2 + bx + c) + B

Find A and B by comparing the coefficients of like powers of x on the two sides.

1. If k is a constant, then

∫ dxk = kx and ∫ dxxfk )( = k ∫ dxxf )(

2. ∫ ± dxxfxf )()( 21 = ∫ dxxf )(1 ± ∫ dxxf )(2

Some Proper Substitutions :

1. ∫ f(ax + b) dx, ax + b = t

2. ∫ f(axn + b)xn–1dx, axn + b = t

3. ∫ fφ(x) φ´(x) dx, φ(x) = t

4. ∫ dxxfxf)()´( , f(x) = t

5. ∫ − 22 xa dx, x = a sin θ or a cos θ

6. ∫ + 22 xa dx, x = a tan θ

7. ∫ +−

22

22

xaxa dx, x2 = a2 cos 2θ

8. ∫ ± xa dx, a ± x = t2

9. ∫ +−

xaxa dx, x = a cos 2θ

10. ∫ − 22 xax dx, x = a(1 – cos θ)

11. ∫ − 22 ax dx, x = a sec θ

Substitution for Some irrational Functions :

1. ∫ ++ baxqpxdx)(

, ax + b = t2

2. ∫+++ cbxaxqpx

dx2)(

, px + q = t1

3. ∫ +++ baxrqxpxdx

)( 2 , ax + b = t2

4. ∫++ caxrpx

dx22 )(

, at first x =t1 and then a + ct2 = z2

Some Important Integrals :

1. To evaluate ∫ β−α− ))(( xxdx , ∫

−β

α−x

x dx,

∫ −βα− ))(( xx dx. Put x = α cos2θ + β sin2θ

2. To evaluate ∫ + xbadx

cos, ∫ + xba

dxsin

,

∫ ++ xcxbadx

sincos

Replace sin x =

+

2tan1

2tan2

2 x

x

and cos x =

+

−

2tan1

2tan1

2

2

x

x

Then put tan 2x = t.

3. To evaluate ∫ +++

xcxbaxqxp

sincossincos dx

Put p cos x + q sin x = A(a + b cos x + c sin x) + B. diff. of (a + b cos x + c sin x) + C A, B and C can be calculated by equating the

coefficients of cos x, sin x and the constant terms.

4. To evaluate ∫ ++ xcxxbxadx

22 sincossin2cos,

∫ + bxadx2cos

, ∫ + xbadx

2sin

In the above type of questions divide Nr and Dr by cos2x. The numerator will become sec2x and in the denominator we will have a quadratic equation in tan x (change sec2x into 1 + tan2x).

Putting tan x = t the question will reduce to the form

∫ ++ cbtatdt

2

5. Integration of rational function of the given form

(i) ∫ +++

424

22

akxxax dx, (ii) ∫ ++

−424

22

akxxax dx, where

k is a constant, positive, negative or zero. These integrals can be obtained by dividing

numerator and denominator by x2, then putting

x – x

a2 = t and x +

xa2

= t respectively.

Integration of Product of Two Functions :

1. ∫ f1(x) f2(x) dx = f1(x) ∫ f2(x) dx – [ ]∫ ∫ dxxfxf )()(( 2'

1 dx

Proper choice of the first and second functions : Integration with the help of the above rule is called


integration by parts, In the above rule, there are two terms on R.H.S. and in both the terms integral of the second function is involve. Therefore in the product of two functions if one of the two functions is not directly integrable (e.g. log x, sin–1x, cos–1x, tan–1x etc.) we take it as the first function and the remaining function is taken as the second function. If there is no other function, then unity is taken as the second function. If in the integral both the functions are easily integrable, then the first function is chosen in such a way that the derivative of the function is a simple functions and the function thus obtained under the integral sign is easily integrable than the original function.

2. ∫ + )sin( cbxeax dx

= 22 baeax

+[a sin (bx + c) – b cos (bx + c)]

= 22 ba

eax

+sin

−+ −

abcbx 1tan

3. ∫ + )cos( cbxeax dx

= 22 baeax

+[a cos (bx + c) + b sin(bx + c)]

= 22 ba

eax

+cos

−+ −

abcbx 1tan

4. ∫ ekxkf(x) + f '(x) dx = ekxf(x)

5. ∫ xelog = x(logex – 1) = x loge

ex

Integration of Trigonometric Functions : 1. To evaluate the integrals of the form

I = ∫ sinmx cosnx dx, where m and n are rational

numbers. (i) Substitute sin x = t, if n is odd; (ii) Substitute cos x = t, if m is odd; (iii) Substitute tan x = t, if m + n is a negative even

integer; and

(iv) Substitute cot x = t, if 21 (m + 1) +

21 (n – 1) is an

integer.

2. Integrals of the form ∫ R (sin x, cos x) dx, where R is

a rational function of sin x and cos x, are transformed into integrals of a rational function by the substitution

tan 2x = t, where –π < x < π. This is the so called

universal substitution. Sometimes it is more

convenient to make the substitution cot2x = t for

0 < x < 2π. The above substitution enables us to integrate any

function of the form R (sin x, cos x). However, in practice, it sometimes leads to extremely complex rational functions. In some cases, the integral can be simplified by –

(i) Substituting sin x = t, if the integral is of the form

∫R (sin x) cos x dx.

(ii) Substituting cos x = t, if the integral is of the form

∫R (cos x) sin x dx.

(iii) Substituting tan x = t, i.e. dx = 21 tdt+

, if the

integral is dependent only on tan x. Some Useful Integrals :

1. (When a > b) ∫ + xbadx

cos

= 22

2

ba −tan–1

+−

2tan x

baba + c

2. (When a < b) ∫ + xbadx

cos

= –22

1

ab −log

baaxab

baaxab

++−

+−−

tan

tan

3. (when a = b) ∫ + xbadx

cos=

a1 tan

2x + c

4. (When a > b) ∫ + xbadx

sin

= 22

2

ba − tan–1

−

+

22

2tan

ba

bxa + c

5. (When a < b) ∫ + xbadx

sin

= 22

1

ab − log

22

22

2tan

2tan

abbxa

abbxa

−++

−−+

+ c

6. (When a = b) ∫ + xbadx

sin=

a1 [tan x – sec x] + c


Functions with their Periods :

Function Period

sin (ax + b), cos (ax + b), sec (ax + b), cosec (ax + b)

2π/a

tan(ax + b), cot (ax + b) π/a

|sin (ax + b)|, |cos (ax + b)|, |sec (ax + b)|, |cosec (ax + b)|

π/a

|tan (ax + b)|, |cot (ax + b)| π/2a

Trigonometrical Equations with their General Solution:

Trgonometrical equation General Solution

sin θ = 0 θ = nπ

cos θ = 0 θ = nπ + π/2

tan θ = 0 θ = nπ

sin θ = 1 θ = 2nπ + π/2

cos θ = 1 θ = 2nπ

sin θ = sin α θ = nπ + (–1)n α

cos θ = cos α θ = 2nπ ± α

tan θ = tan α θ = nπ + α

sin2θ = sin2α θ = nπ ± α

tan2θ = tan2α θ = nπ ± α

cos2θ = cos2α θ = nπ ± α

*coscossinsin

α=θα=θ

θ = 2nπ + α

*tantansinsin

α=θα=θ

θ = 2nπ + α

*coscostantan

α=θα=θ

θ = 2nπ + α

* If α be the least positive value of θ which satisfy two given trigonometrical equations, then the general value of θ will be 2nπ + α.

Note : 1. If while solving an equation we have to square it,

then the roots found after squaring must be

checked whether they satisfy the original equation or not. e.g. Let x = 3. Squaring, we get x2 = 9, ∴ x = 3 and – 3 but x = – 3 does not satisfy the original equation x = 3.

2. Any value of x which makes both R.H.S. and L.H.S. equal will be a root but the value of x for which ∞ = ∞ will not be a solution as it is an indeterminate form.

3. If xy = xz, then x(y – z) = 0 ⇒ either x = 0 or

y = z or both. But xy =

xz ⇒ y = z only and not

x = 0, as it will make ∞ = ∞. Similarly, if ay = az, then it will also imply y = z only as a ≠ 0 being a constant.

Similarly, x + y = x + z ⇒ y = z and x – y = x – z ⇒ y = z. Here we do not take x = 0 as in the above because x is an additive factor and not multiplicative factor.

4. When cos θ = 0, then sin θ = 1 or –1. We have to verify which value of sin θ is to be chosen which

satisfies the equation cos θ = 0 ⇒ θ =

+

21n π

If sin θ = 1, then obviously n = even. But if sin θ = –1, then n = odd.

Similarly, when sin θ = 0, then θ = nπ and cos θ = 1 or –1.

If cos θ = 1, then n is even and if cos θ = –1, then n is odd.

5. The equations a cos θ ± b sin θ = c are solved as follows :

Put a = r cos α, b = r sin α so that r = 22 ba + and α = tan–1 b/a.

The given equation becomes

r[cos θ cos α ± sin θ sin α] = c ;

cos (θ ± α) = rc provided

rc ≤ 1.

Relation between the sides and the angle of a triangle: 1. Sine formula :

TRIGONOMETRICAL EQUATION

Mathematics Fundamentals MATHS


a

Asin = b

Bsin = c

Csin = R21

Where R is the radius of circumcircle of triangle ABC.

2. Cosine formulae :

cos A = bc

acb2

222 −+ , cos B = ac

bca2

222 −+ ,

cos C = ab

cba2

222 −+

It should be remembered that, in a triangle ABC If ∠A = 60º, then b2 + c2 – a2 = bc If ∠B = 60º, then a2 + c2 – b2 = ac If ∠C = 60º, then a2 + b2 – c2 = ab 3. Projection formulae : a = b cos C + c cos B, b = c cos A + a cos C c = a cos B + b cos A Trigonometrical Ratios of the Half Angles of a Triangle:

If s = 2

cba ++ in triangle ABC, where a, b and c are

the lengths of sides of ∆ABC, then

(a) cos2A =

bcass )( − , cos

2B =

acbss )( − ,

cos2C =

abcss )( −

(b) sin2A =

bccsbs ))(( −− , sin

2B =

accsas ))(( −− ,

sin2C =

abbsas ))(( −−

(c) tan2A =

)())((

asscsbs

−−− ,

tan2B =

)())((

bsscsas

−−− , tan

2C

)())((

cssbsas

−−−

Napier's Analogy :

tan2

CB − = cbcb

+− cot

2A , tan

2AC − =

acac

+− cot

2B

tan2

BA − = baba

+− cot

2C

Area of Triangle :

∆ = 21 bc sin A=

21 ca sin B =

21 ab sin C

∆ =)sin(

sinsin21 2

CBCBa

+=

)sin(sinsin

21 2

ACACb

+=

)sin(sinsin

21 2

BABAc

+

sin A = bc2 ))()(( csbsass −−− =

bc∆2

Similarly sin B = ca∆2 & sin C =

ab∆2

Some Important Results :

1. tan2A tan

2B =

scs − ∴ cot

2A cot

2B =

css−

2. tan2A + tan

2B =

sc cot

2C =

∆c (s – c)

3. tan2A – tan

2B =

∆− ba (s – c)

4. cot2A + cot

2B =

2tan

2tan

2tan

2tan

BA

BA+

= cs

c−

cot2C

5. Also note the following identities : Σ(p – q) = (p – q) + (q – r) + (r – p) = 0 Σp(q – r) = p(q – r) + q(r – p) + r(p – q) = 0 Σ(p + a)(q – r) = Σp(q – r) + aΣ(q – r) = 0 Solution of Triangles : 1. Introduction : In a triangle, there are six

elements viz. three sides and three angles. In plane geometry we have done that if three of the elements are given, at least one of which must be a side, then the other three elements can be uniquely determined. The procedure of determining unknown elements from the known elements is called solving a triangle.

2. Solution of a right angled triangle : Case I. When two sides are given : Let the

triangle be right angled at C. Then we can determine the remaining elements as given in the following table.

Given Required

(i) a, b tanA =

ba

, B = 90º – A, c = A

asin

(ii) a, c sinA =

ca

, b = c cos A, B = 90º – A

Case II. When a side and an acute angle are given – In this case, we can determine

Given Required

(i) a, A B = 90º – A, b = a cot A, c =

Aa

sin

(ii) c, A B = 90º – A, a = c sin A, b = c cos A


a

PHYSICS

Questions 1 to 9 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. + 3 marks will be given for each correct answer and – 1 mark for each wrong answer.

1. In the circuit shown, the cell is ideal, with e.m.f. = 15 V. Each resistance is of 3Ω. The potential difference across the capacitors in steady state –

R=3Ω C=3µF

R R

R R

15V

+ –

(A) 0 (B) 9 V (C) 12 V (D) 15 V

2. In a young double slit apparatus the screen is rotated by 60º about an axis parallel to the slits. The slits separation is 3mm, slits to screen distance (i.e AB) is 4 m & wavelength of light is 450 nm. The separation between the 3rd dark fringe on the either side of B.

60º

A B

screen

(A) 6 mm (B) 8 mm (C) 4 3 mm (D) 2 3 mm

3. A black body emits radiation at the rate P when its temperature is T. At this temperature the wavelength at which the radiation has maximum intensity is λ0. If at another temperature T' the power radiated is P' &

wavelength at maximum intensity is 20λ

then –

(A) P'T' = 32 PT (B) P'T' = 16 PT (C) P'T' = 8 PT (D) P'T' = 4 PT

4. If two identical string are stretched such that there is fractional increase in their length, the fractional increase in length of first string is f and second string is 2f. Then the ratio of their fundamental frequency is. (Assume both obey the Hook Law i.e. tension ∝ elongation in string) -

(A) 2

1f1f21

++ (B)

f1f212

++

(C) f21

f12

1++ (D)

f21f1

++

5. A infinite line charge of charge density λ lies along the x axis and let the surface of zero potential passes through (0, 5, 12) m. The potential at point (2, 3, –4) is –

z

V = 0(0,5,12)

y

IIT-JEE 2012

XtraEdge Test Series # 10

Based on New Pattern

Time : 3 Hours Syllabus : Physics : Full Syllabus, Chemistry : Full Syllabus, Mathematics : Full syllabus Instructions : Section - I • Question 1 to 9 are multiple choice questions with only one correct answer. +3 marks will be awarded for

correct answer and -1 mark for wrong answer. • Question 10 to 13 are Reason and Assertion type question with one correct answer. +3 marks will be awarded for

correct answer and –1 mark for wrong answer.

• Question 14 to 19 are passage based questions. +4 marks will be awarded for correct answer and –1 mark for wrong answer.

• Question 20 to 22 are Column Matching type questions. +6 marks will be awarded for the complete correctly matched answer (i.e. +1 marks for each correct row) and No Negative marks for wrong answer.


(A) 02πε

λ ln 5

13 (B) 0

2επλ

ln3

13

(C) 04 επ

λ ln5

13 (D) 0επ

λln

313

6. A satellite is put in an orbit just above the earth atmosphere with a velocity 5.1 times the velocity for a circular orbit at that height. The initial velocity imparted is horizontal. What would be the maximum distance of satellite from earth when it is in the orbit-

(A) 3R (B) 4R (C) 2 R (D) 5 R 7. The density of a uniform rod with cross section area

A is ρ, its specific heat capacity is C and the coefficient of its linear expansion is α. Calculate the amount of heat that should be added in order to increase the length of the rod by ∆l.

(A) α

∆ρ lCA2 (B) l∆

αρCA

(C) α

∆ρ lCA (D) α∆ρ

A2C l

8. For the system shown each spring has a stiffness of

175 N/m. The mass of the pulleys may be neglected. The period of vertical oscillation of block – (mass of block is 28 kg)

k k

Block fricationless surface

28 kg

(A) 2 s (B) π 2 s (C) 5π s (D)

5π 2 s

9. A uniform elastic rod of cross section area A, natural length L and young modulus Y is placed on a smooth horizontal surface. Now two horizontal force (of magnitude F and 3F) directed along the length of rod and in opposite direction act at two of its ends as shown. After the rod has acquired steady state (i.e. no further extension take place), the extension of the rod will be –

F 3F

Elastic rod

(A) YA

F2 L (B) YA

F4 L

(C) YAF L (D)

YA2F3 L

This section contains 4 questions numbered 10 to 13, (Reason and Assertion type question). Each question contains Assertion (A) and Reason (R). Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. +3 marks will be given for each correct answer and – 1 mark for each wrong answer. The following questions given below consist of an "Assertion" (A) and "Reason" (R) Type questions. Use the following Key to choose the appropriate answer. (A) If both (A) and (R) are true, and (R) is the correct

explanation of (A). (B) If both (A) and (R) are true but (R) is not the

correct explanation of (A). (C) If (A) is true but (R) is false. (D) If (A) is false but (R) is true.

10. Assertion (A) : As the earth revolves around the sun it has an acceleration which is directed towards centre of the sun.

Reason (R) : Angular momentum of the earth about the sun remains constant.

11. Assertion (A) : In electric circuits , wires carrying currents in opposite direction are often twisted together.

Reason (R) : The magnetic field in the surrounding space of a twisted wire system in not precisely zero.

12. Assertion (A) : A metal ball is floating in mercury. Coefficient of volume expansion of metal is less than that of mercury. If temperature is increased, fraction of volume immersed of metal will increase.

Reason (R) : Effect of heating on density of mercury will be more compared to that of metal.

13. Assertion (A) : Work done by static friction is always zero.

Reason (R) : Static friction acts when there is no relative motion between two bodies in contact.

This section contains 2 paragraphs; each has 3 multiple choice questions. (Questions 14 to 19) Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. +4 marks will be given for each correct answer and – 1 mark for each wrong answer.

Passage # 1 (Ques. 14 to 16)

A conducting rod PQ of mass M rotates without friction on a horizontal plane about Ο on circular rails of diameter 'l'. The centre O and the periphery are connected by resistance R. The system is located in a uniform magnetic field perpendicular to the plane of the loop. At t = 0, PQ starts rotating clockwise with angular velocity ω0. Neglect the resistance of the rails and rod, as well as self inductance.


B

O ω0

Q

R P

⊗

14. Magnitude of current as a function of time

(A) t2

0 eR2

B α−ω l (B) t22

0 eR16

B α−ω l

(C) t2

0 eR8

B α−ω l (D) t22

0 eR8

B α−ω l

Where α = RM8B3 22l

15. Total charge flow through resistance till rod PQ stop rotating .

(A) B8M0ω (B)

B3M0ω (C)

B6M0ω (D)

B9M0ω

16. Heat generated in the circuit by t = ∞

(A) 24

M 20

2ωl (B) 8

M 20

2ωl (C) 3

M 20

2ωl (D) 32

M 20

2ωl

Passage # 2 (Ques. 17 to 19) A cylindrical spool of radius R is rigidly attached to a

pulley of radius 2R. The mass of the combination is m and the radius of gyration about the centroidal axis G is R. A constant horizontal force F is applied at one end of the tape. Assume rolling motion between the pulley and the ground. I is the moment of inertia about centroidal axis.

2R

G R

PF

17. The acceleration of point P on the tape relative to the

ground is -

(A) m2F (B)

m5F (C)

m3F (D)

m4F

18. The length of the tape wound/unwound on the spool in time t is -

(A) m10

Ft2 (B)

m5Ft2

(C) m4

Ft2 (D)

m2Ft2

19. The linear acceleration of centre G is -

(A) I

FR2 2 (B) 2

2

mRIFR2

+

(C) 2

2

mR2IFR2

+ (D) 2

2

mR4IFR2

+

This section contains 3 questions (Questions 20 to 22). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, B-Q, B-R, C-P, C-Q and D-S, then the correctly bubbled 4 × 4 matrix should be as follows :

ABCD

P Q R S

S

P

P P Q R

R R

Q Q

S S

P Q R S

Mark your response in OMR sheet against the question number of that question in section-II. + 6 marks will be given for complete correct answer and No Negative marks for wrong answer. However, 1 mark will be given for a correctly marked answer in any row.

20. A charged particle passes through a region that could have electric field only or magnetic field only or both electric and magnetic field or none of the fields. Match the possibilities

Column-I (A) Kinetic energy of the particle remain constant. (B) Acceleration of the particle is zero (C) Kinetic energy of the particle changes and it also

suffers deflection (D) Kinetic energy of the particle changes but it

suffers no deflection Column-II (P) Under special condition this is possible when

both electric and magnetic fields are present (Q) The region has electric field (R) The region has magnetic field only (S) The region contains no field

21. Match the column : Column-I Phenomena on which machine work (A) Electromagnetic induction (B) Light of suitable frequency falling on a material

result in emissions of electrons from the material (C) Change of orientation of a coil in a magnetic

field results in e.m.f. across the coil (D) Mutual induction Column-II Machine or instrument (P) Photocell (Q) DC motor (R) AC generator (S) Transformer


22. Match the column : Column-I (A) A photon stimulates the emission of another

photon of (B) Photons of electromagnetic wave of different

wavelengths may have (C) Two points on a wavefront must have (D) For constructive interference the waves must

have Column-II (P) Same direction (Q) Same energy (R) Same phase (S) Same frequency

CHEMISTRY


1. The phenomenon of optical activity will be shown by -

(A)

B

A

B

B B

A

M

(B)

A

B

M

A

B

en

(C)

A

A

en en M (D)

B

B

M

en

en

2. Titration curve of Na2CO3 and HCl is as given below. The indicators In1 and In2 respectively, must be –

10 20 30 40 50 60

2

4

6

8

10

12

In1

In2

pH

Volume of HCl (ml)

(A) phenolphthalein and methyl orange (B) methyl orange and phenolphthalein (C) methyl orange and phenol red (D) phenolphthalein and phenol red.

3. Metallic sulphates can be obtained by reacting the metals (above hydrogen in ECS), or its oxide, hydroxide or carbonate with dil.H2SO4. Gp IA metals also form hydrogen sulphates which can be isolated in solid. In general metal sulphates are soluble in water and crystallizes with water of crystallization. Sulphate are thermally more stable than nitrates. Select the stable hydrogen sulphate which can be obtained in solid state -

(A) KHSO4 (B) CaHSO4 (C) FeHSO4 (D) All of these

4. Which of the following represent glyptal -

(A)

––O–CH2–CH2–O–C

O C–––

O

4

(B)

––NH–(CH2)6–NH–C––(CH2)4–

O C–––

O

4

(C)

O O

O O

4

(D) O

4N H

5. In the Cannizzaro's reaction given below :

2Ph – CHO →–OH Ph–CH2OH + PhCOO–-

the slowest step is - (A) the attack of OH– at the carbonyl group (B) the transfer of hydride to the carbonyl group (C) the abstraction of proton from the carboxylic acid (D) the deprotonation of Ph–CH2OH

6. The relative rates of solvolysis in 80% EtOH of the following bromides is in the order –

Br

I

Br

II

Br

III

(A) I > II > III (B) III > II > I (C) II > III > I (D) II > I > III


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7. NaOH can be prepared by two methods each of the method has two steps having 100% yield.

Method I : 2Na + 2H2O → 2NaOH + H2

2H2 + O2 → 2H2O

Method II : 2Na + 21 O2 → Na2O

Na2O + H2O → 2NaOH Which of the above method gives better yield of

NaOH ? (A) Method I (B) Method II (C) Method I & Method II give equal yields. (D) Yield cannot be determined

8. Which of the following value θ is correspond to the maximum dipole moment of the triatomic molecule XY2

X

θ

Y

Y

(A) θ = 90º (B) θ = 120º (C) θ = 150º (D) 180º

9. Photons having energy equivalent to binding energy of 2nd state of Li+ ion are used at metal surface of work function 10.6 eV. If the ejected electrons are further accelerated through the potential difference of 5 V then the minimum value of de-Broglie wavelength associated with the electron is –

(A) 2.45 Å (B) 9.15 Å (C) 5 Å (D) 11 Å




10. Assertion (A) :

N O O N OO

O O

is a macrocyclic ligand

Reason (R) : The ligand in which donor atoms are N and O is known a cryptands.

11. Assertion (A) : For adsorption ∆G, ∆S and ∆H all have negative values.

Reason (R) : Adsorption is spontaneous process accompained by increase in entropy.

12. Assertion (A) : For the concentration cell, Zn(s) | Zn2+ (aq)(C1) | | Zn2+ (aq) (C2)/Zn, for spontaneous cell reaction C1 < C2.

Reason (R) : For concentration cell,Ecell = 1

2

CCln

nFRT

For spontaneous reaction, Ecell +ve ⇒ C2 > C1

13. Assertion (A) : Aryl halides undergo nucleophilic substitution with ease.

Reason (R) : Carbon-halogen bond in aryl halides has partial double bond character.



FeCl3 on reaction with K4 [Fe(CN)6] in aqueous solution gives blue colour, according the

equation 4FeCl3 + 3K4[Fe(CN)6] → Fe4[Fe(CN)6]3 + 12 KCl At 300 K two aqueous solution of K4[Fe(CN)6] &

FeCl3 with equal concentrations 0.1 M. These two solutions are separated by a semipermeable membrane AB as shown in figure.

A

B

0.1M0.1M

FeCl3

side'y' side'x'

K4[Fe(CN)6]

14. What is correct – (A) side ‘x’ is hypotonic (B) side ‘y’ is hypotonic (C) both are isotonic (D) None of these

15. What is true about the solutions – (A) blue colour forms in side ‘x’ (B) blue colour forms in side ‘y’ (C) blue colour forms on both sides (D) No blue colour formation

16. By applying external pressure osmosis can be stopped it should be applied to –

(A) side ‘x’ (B) side ‘y’ (C) equal on both the sides (D) can not be stopped by applying external pressure


Passage # 2 (Ques. 17 to 19) An activating group activates all positions of the

benzene ring; even the positions meta to it are more reactive than any single position in benzene itself. It directs ortho and para simply because it activates the ortho and para positions much more than it does the meta.

A deactivating group deactivates all positions in the ring, even the positions meta to it. It directs meta simply because it deactivates the ortho and para positions even more than it does the meta. Thus, both ortho and para orientation and meta orientation arise in the same way : The effect of any group whether activating or deactivating is strongest at the ortho and para positions.

But certain groups (– NH2 and – OH, and their derivatives) act as powerful activators towards electrophilic aromatic substitution, even though they contain electronegative atoms and can be shown in other ways to have electron-withdrawing inductive effects.

Halogens are unusual in their effect on electrophilic aromatic substitution; they are deactivating yet ortho, para-directing. Deactivation is characteristic of electron withdrawal, whereas ortho-para orientation is characteristic of electron release.

17. Which will undergo Friedel-Crafts alkylation reaction?

(1)

CH3

NO2

(2)

CH2CH3

(3)

COOH

(4)

OH

(A) 1, 2 and 4 (B) 1 and 3 (C) 2 and 4 (D) 1 and 2 18. Which of the following is the strongest acid ?

(A) OH

NO2 (B)

OH

Cl

(C)

OH

NO2

(D)

OH

NO2

19. Reactivity in halogen substituted benzene rings is controlled by :

(A) resonance (B) inductive effect (C) inductive effect dominates resonance effect (D) resonance effect dominates inductive effect


ABCD

P Q R S

S

P

P P Q R

R R

Q Q

S S

P Q R S

Mark your response in OMR sheet against the question number of that question in section-II. + 6 marks will be given for complete correct answer and No Negative marks for wrong answer. However, 1 mark will be given for a correctly marked answer in any row. 20. One mole of an ideal gas is subjected to the following

process : Column-I Column-II (A) ∆W = 0 (P) If the gas undergoes free expansion (B) ∆E = 75Cv (Q) If the gas is cooled reversibly at constant pressure from 373 K to 298 K (C) ∆H = –75 Cp (R) If the gas is heated from 298 K to 373 K reversively at constant pressure (D) ∆W = –75 R (S) If the gas is heated from 298 K to 373 K at constant volume 21. Match the Column : Column-I Column-II (Gases X and Y (Tatio of times taken of diffusion) taken) (A) X = 100 ml of H2 at 1 bar, 25º C (P) 1 : 1225 Y = 200 ml of O2 at 2 bar, 25ºC (B) X = 100 ml of O2 at 1 bar, 25ºC (Q) 1 : 0.7 Y = 200 ml of O3 at 2 bar, 25ºC


(C) X = 100 ml of SO2 at 1 bar, 25ºC (R) 1 : 1.36 Y = 100 ml of O2 at 1 bar, 25ºC (D) X = HCl gas to travel 100 cm (S) 1 : 8 length in a tube Y = NH3 gas to travel 200 cm length using the same tube (P, V, T) = same in both cases)

22. Match the Column : Column-I Column-II (A) Soluble in dil HCl (P) SrS2O3 (B) Yields SO2 with dil (Q) SrSO3

HCl on bonding (C) Soluble in aqueous (R) SrSO4

solution (D) Insoluble in aqueous (S) CdS solution

MATHEMATICS


1. The area enclosed by | y | – | x | = 1 and x2 + y2 =1 is (A) 2 units2 (B) zero units2 (C) infinite units2 (D) none of these

2. Least value of the expression 9sec2θ + 4cosec2θ, is- (A) 6 (B) 1 (C) 36 (D) 25

3. Let f (x) = ∫ −−+−−x

dxxxxx1

223 ))2()1(3)2)(1(2( ,

then- (A) f has exactly 4 critical points

(B) f has maximum at x = 2 (C) x = 7/5 is minima & x = 1 is maxima

(D) none of these

4. The locus of the middle points of chords of a parabola which subtend a right angle at the vertex of the parabola is-

(A) Circle (B) Parabola (C) Ellipse (D) Straight line

5. The probability that a particular day in the month of july is a rainy day is 3/4. Two person whose credibility are 4/5 and 2/3 respectively claim that 15th july was a rainy day. The probability that it was real a rainy day.

(A) 3/4 (B) 24/25 (C) 8/9 (D) none

6. Domain of f (x) =

−−

][][2sin 1

xxx , where [.] denotes

the greatest integer function, is

(A) (–∞, 1) – 0 (B)

− 0,

34 ∪ 0

(C) (–∞, 0) ∪ I+ (D) (–∞, ∞) – [0, 1) 7. The number of different words of three letters which

can be formed from the word "PROPOSAL", if a vowel is always in the middle are-

(A) 53 (B) 52 (C) 63 (D) 32

8. Let a1, a2, a3, ...... be terms of an A.P. If

q

p

aaaaaa

+++

+++

.........

21

21 = 2

2

qp , p ≠ q, then

21

6

aa

equals-

(A) 41/11 (B) 7/2 (C) 2/7 (D) 11/41

9. The curve y = ax3 + bx2 + cx is inclined by 45º to x-axis at origin and it touches x-axis at (1,0). Then-

(A) a = –2, b = 1, c = 1 (B) a = 1, b = 1, c = –2 (C) a = 1, b = –2, c = 1 (D) a = –1, b = 2, c = 1 This section contains 4 questions numbered 10 to 13, (Reason and Assertion type question). Each question contains Assertion (A) and Reason (R). Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. +3 marks will be given for each correct answer and – 1 mark for each wrong answer. The following questions given below consist of an "Assertion" (A) and "Reason" (R) Type questions. Use the following Key to choose the appropriate answer. (A) If both (A) and (R) are true, and (R) is the correct



10. Assertion (A) : Let z be a complex number satisfying |z – 3| ≤ |z – 1|, |z – 3| ≤ |z – 5|, |z – i| ≤ |z + i| and |z – i| ≤ |z – 5i|. Then the area of region in which z lies is 12 sq. units.

Reason (R) : Area of trapezium = 21 (sum of parallel

sides) (Distance between parallel sides)

11. Let f (x) = | 1 – x | and g(x) = sin–1(f | x |) Assertion (A) : Number of values of x, where g(x) is

non differentiable is 3. Reason (R) : Domain of g(x) is [–1, 1]


12. Assertion (A) : If eccentricities of two ellipse are same then their areas are also same.

Reason (R) : Area of the ellipse 1by

ax

2

2

2

2=+

(a < b, a > 0, b > 0) is π ab square units.

13. Consider a circle S : (x – 2)2 + (y – 3)2 = 13 and a line L : y = x – 12.

Assertion (A) : Chord of contact of pair of tangents drawn from every point on L = 0 to S = 0 passes through P(3, 2)

Reason (R) : Pole of polar L = 0 with respect to S = 0 is P(3, 2)

This section contains 2 paragraphs; each has 3 multiple choice questions. (Questions 14 to 19) Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. +4 marks will be given for each correct answer and – 1 mark for each wrong answer. Passage # 1 (Ques. 14 to 16)

If f (xy) = f (x) . f (y) and f is differentiable at x = 1 such that f '(1) = 1 also f (1) ≠ 0, then

14. f (x) is - (A) continuous for all x ∈ R (B) discontinuous at x = –1, 0, 1 (C) differentiable for all x ≠ 0 (D) None of these

15. f '(7) equals- (A) 7 (B) 14 (C) 1 (D) None

16. Area bounded by curve f(x), x axis and ordinate x = 4, is-

(A) 64/3 (B) 8 (C) 16 (D) None


There exists a G.P. with first term A1 and common

ratio A (A > 1). If we add 21 in the sum of first n

terms of the sequence, it equals to the sum of the coefficients of even power of x in the expansion of (1 + x)n. If we interchange the first term & common ratio of given G.P., the sum of new infinitely decreasing G.P. is equal to B, where A, B and n are related by the relation

∫−

−

+2B

2A

n dx)x1( = 3

364

17. The value of Ax

Bn)x1(limA

Ax −−−+

→ is-

(A) 3 (B) 6 (C) e (D) 8

18. Area bounded by f(x) = xA and g(x) = xB is-

(A) n

BA + (B) n

AB −

(C) nB2A

A2++

(D) BAn

B++

19. Number of real roots of the equation (xB – nxA)1/A = 6 are (A) 2 (B) 4 (C) 1 (D) 0


ABCD

P Q R S

S

P

P P Q R

R R

Q Q

S S

P Q R S


20. Match the Column : Column-I Column-II (A) The reflection of the point (P) 5 (t – 1, 2t + 2) in a line is (2t + 1, t) then the line has slope equal to (B) If θ be the angle between (Q) 6 two tangents which are drawn

to the circle x2 + y2 – 6 3 x – 6y + 27 = 0

from the origin, then 2 3 tanθ equals to

(C) The shortest distance between (R) 72 parabolas y2 = 4x and y2 = 2x – 6

is d then d2 =

(D) Distance between foci of the (S) 1 curve represented by the equation x = 1 + 4cosθ, y = 2 + 3sinθ is


21. Column-I Column-II

(A) If y = 2[x] + 9 = 3[x + 2], where (P) – 1 [.] denotes greatest integer function,

then 61 [x + y] is equal to

(B) If x

x x1cos

x1sinlim

+

∞→ = ek/2 then (Q) 0

k is equal to (C) If three successive terms of a G.P. (R) 2 with common ratio r, (r > 1) forms the sides of a triangle then [r] + [–r] is equal to (where [.] denotes greatest integer function) (D) Let f(x) = (x2 – 3x + 2)(x2+3x+2) (S) 3 and α, β, γ are the roots of f '(x) = 0, then [α]+[β] + [γ] is equal to (where [.] denotes greatest integer function) 22. Column-I Column-II (A) The order and degree of the (P) 13 differential equation

0x7dx

yd4dxdy

2

23 =−− are

a and b then a + b is

(B) If k3j2ia ++=r

, kji2b +−=r

(Q) 102

and kj2i3c ++=r

and )cb(arrr

××

is equal to czbyaxrrr

++ , then x + y + z is equal to (C) The number of 4 digit numbers (R) 5 that can be made with the digits 1,2,3,4,3,2

(D) If ∫ ++ )4x)(1x(dx

22 = (S) 7

kdxtan

caxtan

ba 11 +

− −−

where k is constant of integration, then 2a + b + c + d is (where a & b and a & c are co-prime numbers)

Chemistry Facts • At 0 degress Celsius and 1 atmospheric pressure,

one mole of any gas occupies approximately 22.4 liters.

• Atomic weight is the mass of an atom relative to the mass of an atom of carbon-12 which has an atomic weight of exactly 12.00000 amu.

• If the atom were the size of a pixel (or the size of a period), humans would be a thousand miles tall.

• It would require about 100 million (100,000,000) atoms to form a straight line one centimeter long.

• The weight (or mass) of a proton is 1,836.1526675 times heavier than the weight (or mass) of an electron.

• The electron was first discovered before the proton and neutron, in 1897 from English physicist John Joseph Thomson.

• The neutron was discovered after the proton in 1932 from British physicist James Chadwick, which proved an important discovery in the development of nuclear reactors.

• Carbon dioxide was discovered by Scottish chemist Joseph Black.

• When silver nitrate is exposed to light, it results in a blackening effect. (Discovered by Scheele, which became an important discovery for the development of photography).


PHYSICS


1. Ice point on a particular scale measure 22º. The scale agrees with Fahrenheit scale at 72ºF. The temperature in Fahrenheit scale when new scale reads 27º is -

(A) 40ºF (B) 38ºF (C) 36ºF (D) 34ºF

2. A hot body is kept in a chamber maintained at fixed lower temperature. Time taken by body in loosing half the maximum heat it can lose is 5 min. Time taken by body in loosing the maximum heat it can lose is -

(A) 10 min (B) 12 min (C) 20 min (D) Infinite

3. A point source of power 4 W is placed at the centre of spherical shell. The shell is kept in a vacuum chamber maintained at 27ºC. If shell attains a constant temperature 37ºC, emissivity of surface of shell is nearly -

(A) 21 (B)

31 (C)

32 (D)

54

4. In a resonance–column experiment, a long tube, open at top, is clamped vertically. Water level inside tube can be moved up or down. First resonance is occuring when water level is at depth 20 cm below open end. Let second resonance occurs when water level is at a distance ‘x’ below opening, then -

(A) x = 40 cm (B) x = 60 cm (C) x < 60 cm (D) x > 60 cm

5. Figure shows a toy-whistle. It a disk made of plastic having two conical grooves at diametrically opposite point. When it is rotated about its centre with sufficiently high speed air intercepted by groove produce whistling sound. An observer is at a distance 10 m from the centre of toy. Radius of toy is 15 cm. Frequency of sound emitted by toy when it is rotating

with ω = π

1800 rpm is 10 kHz. If velocity of sound

in air 300 m/s, beat frequency heard by observer is –

O

(A) 10 Hz (B) 15 Hz (C) 20 Hz (D) None

6. An ice cube is fixed at the bottom of container containing water. Water in the container will be cooled majorly by -

(A) Convection (B) Conduction (C) Radiation (D) Convection and conduction

7. In a region of space a constant force F newton acts on a particle of mass m, which is released from rest at point A. When the particle reaches B its –

A B m

→F

IIT-JEE 2013

XtraEdge Test Series # 10

Based on New Pattern

Time : 3 Hours Syllabus : Physics : Full Syllabus, Chemistry : Full Syllabus, Mathematics : Full syllabus Instructions :

• Question 1 to 9 are multiple choice questions with only one correct answer. +3 marks will be awarded for correct answer and -1 mark for wrong answer.

• Question 10 to 13 are Reason and Assertion type question with one correct answer. +3 marks will be awarded for correct answer and –1 mark for wrong answer.

• Question 14 to 19 are passage based questions. +4 marks will be awarded for correct answer and –1 mark for wrong answer.

• Question 20 to 22 are Column Matching type questions. +6 marks will be awarded for the complete correctly matched answer (i.e. +1 marks for each correct row) and No Negative marks for wrong answer.


(A) potential energy (PE) increases but kinetic energy (KE) decreases.

(B) PE decreases but KE increases. (C) PE remains constant but KE increases. (D) PE decreases but KE remains constant

8. A L shaped rod whose one rod is horizontal and other is vertical is rotating about a vertical axis as shown with angular speed ω. The sleeve shown in figure has mass m and friction coefficient between rod and sleeve is µ. The minimum angular speed ω for which sleeve cannot sleep on rod is –

m

ω

sleeve

l

(A) lµ

=ωg (B)

l

gµ=ω

(C) gµ

=ωl (D) None of these

9. Four particles of equal mass M move along a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is -

(A) R

GM (B)

RGM22

(C)

+ )122(

RGM (D)

+4

122R

GM




10. Assertion (A) : A string of length 28 cm, fixed at one end vibrates in its 3rd overtone. Wave motion will not be disturbed by holding the string at a distance 8 cm from fixed end.

Reason (R) : Position of nodes in the above string will at distance 4 cm, 8 cm, …. from fixed end.

11. Assertion (A) : Two tube of same length but different diameter vibrating in same harmonic. Frequency of tube having smaller diameter will be more.

Reason (R) : End-correction is less in tube of smaller diameter.

12. Assertion (A) : When an observer moves towards stationary source frequency heard by observer is more than that frequency emitted by source.

Reason (R) : Wavelength of sound wave received by observer becomes smaller.

13. Assertion (A) : At height h from ground and at depth h below ground, where h is approximately equal to 0.62 R, the value of g acceleration due to gravity is same.

Reason (R) : Value of g decreases both sides, in going up and down.



A vertical cylindrical tube has its lower end closed. There is small opening in the bottom of the tube by which level of water in tube can be adjusted. Length of tube is 50 cm. Velocity of sound in air and water is 320 m/s and 1440 m/s respectively. Water is filled in tube upto height 30 cm.

14. Minimum frequency with which tube can resonate is- (A) 300 Hz (B) 320 Hz (C) 400 Hz (D) 450 Hz 15. Minimum frequency so that standing waves can be

formed both in air and liquid column - (A) 400 Hz (B) 800 Hz (C) 640 Hz (D) 900 Hz

16. Air column in tube is resonating with a tuning fork at its lowest possible frequency. Water level is lowered slowly. Minimum distance by which water level has to be lowered so that intensity of sound become maximum is -

(A) 10 cm (B) 20 cm (C) 40 cm (D) 50 cm Passage # 2 (Ques. 17 to 19)

A wire of length 1 m is clamped horizontally between two rigid support.

17. Wire is plucked at a distance 30 cm from one end. Lowest harmonic in which wire can resonate is -

(A) 3rd (B) 4th (C) 5th (D) 10th


18. A light ring which can slip frictionlessly over wire is slipped over wire. When a tuning fork start resonating with wire, ring is found to be at rest at a distance 15 cm from one end. If velocity of wave in wire is 10 m/s, minimum frequency of tuning fork is-

(A) 50 Hz (B) 60 Hz (C) 70 Hz (D) None of these

19. Now two light rings are slipped on wire which can move frictionlessly. When wire is plucked, the two rings are found to be at rest separated by 12 cm. Maximum wavelength of wave in wire is -

(A) 4 cm (B) 8 cm (C) 12 cm (D) 24 cm


A B C D

P Q R S

S

P

P P Q R

R R

Q Q

S S

P Q R S


20. A small ring of mass m passes through a smooth wire bent in form of horizontal circle. The ring is connected to a spring whose other end is fixed at A on the wire as shown. The natural length of spring is

R and spring constant is R

mg where m is mass of the

ring and R is also the radius of circle. Initially the ring is released from rest from position B and it moves towards C as in the figure. (N = Normal reaction between wire and ring, v = speed of ring)

B

wire C

A 60º

Column-I Column-II (A) N = mg (P) at position B (B) N = zero (Q) at position C (C) N = mg2 (R) some where between position B and position C (D) v = gR (S) never

21. A body of mass ‘m’ is acted upon by net force Fr

. vandr be their initial position and velocity. All

quantity are in S.I. unit . Then match the following. Column-I Column-II

(A) ix4F 2−= , 0r = , (P) Motion : S.H.M.

2m,i2v ==

(B) j)y42(F −= , 0r = (Q) Motion : Non-periodic

0v = , m = 1

(C) i)xx(2F 3+−= , (R) Path : Straight line

i10r 2–= 0v = , m = 1

(D) )jyix(4F +−= (S) Time period : π

j2r = , 1m,i6v == 22. The cubical container filled with water is given

acceleration →a = a0 i + a0 j + a0 k , then : (neglect

the effect of gravity)

B

C

E F

D

G

A z

x

y

H

Column-I Column-II (A) Pressure at point A is (P) less than pressure at point G (B) Pressure at point D is (Q) less than pressure at point F (C) Pressure at point E is (R) Greater than pressure at point C (D) Pressure at point H is (S) Greater than pressure at point B

CHEMISTRY


1. Which of the substance has maximum number of hydrogen atoms per gm of the substance ?

At. wts. : Cu – 63.5 ; S – 32 ; O –16 ; H –1 (A) CH4 (B) CuSO4

. 5H2O

(C) H2O2 (D) H2O


2. In the reaction 3 Cu + 8 HNO3 → 3 Cu(NO3)2

+ 2NO + 4 H2O what is the equivalent weight of HNO3? if molecular

weight of HNO3 is M -

(A) M (B) 3M (C) M

43 (D) M

34

3. The order of magnitude of ionic radii of ions Na+, Mg2+, Al3+ and Si4+ is :

(A) Na+ > Si4+ > Al3+ > Mg2+

(B) Mg2+ > Na+ > Al3+ > Si4+ (C) Na+ > Mg2+ > Al3+ > Si4+

(D) Si4+ > Al3+ > Mg2+ > Na+

4. The correct order of tendency of polymerization is - (A) SiO4

4– < PO43– < SO4

2– < ClO4–

(B) PO43– < SiO4

4– < SO42– < ClO4

– (C) ClO4

– < SO42– < SiO4

4– < PO43–

(D) SiO44– > PO4

3– > SO42– > ClO4

–

5. C

P

KK for the gaseous reaction –

(a) 2 A + 3 B 2C (b) 2 A 4B (c) A + B + 2C 4D would be respectively - (A) (RT)–3 , (RT)2, (RT)º (B) (RT)–3 , (RT)–2, (RT)–1 (C) (RT)–3 , (RT)2, (RT) (D) None of the above

6. No. of heteroatoms (other than C) present in the following heterocyclic compound is –

O

N–H

O (A) 3 (B) 2 (C) 1 (D) 0

7. & are – (A) Tautomers (B) Functional (C) Position (D) All the above

8. Which does not exists in solid state - (A) LiHCO3 (B) CaCO3 (C) NaHCO3 (D) Na2CO3 9. Decomposition of H2O2 is retarded by - (A) Acetanilide (B) MnO2 (C) Zinc (D) Finely divided metals




10. Assertion (A) : F atom has a less negative electron gain enthalpy than Cl atom.

Reason (R) : Additional electron are repelled more effectively by 3p electron in Cl atom than by 2p electron in F atom.

11. Assertion (A) : When a non-volatile solute is added to water ice equilibrium, some ice melts.

Reason (R) : Ice melts to dilute the solution in order to increase the vapour pressure in accordance to Le-Chatelier's principle.

12. Assertion (A) : H

CN is called cyclohexancarbonitrile.

Reason (R) : It is an aromatic compound.

13. Assertion (A) : The following compounds are optically inactive.

CH3

H

CH3

ClH Cl

CH3H

CH3

Cl

H Cl

CH3

H

CH3

Cl

H Cl

CH2 CH2

CH2

CH3

H

CH3

Cl

H Cl

CH2

CH2

CH2

Reason (R) : the meso compounds do not have any chiral C atom so have optical rotation is zero.




PCl5 in solid state exists as PCl4+ and PCl6

–. Also in some solvents it undergoes dissociation as

2PCl5 PCl4+ + PCl6

–

14. The geometry and hybridisation of PCl5 is - (A) Trigonal bipyramid, sp3d (B) Tetrahedral, sp3 (C) Octahedral, sp3 d2

(D) none of these

15. The geometry and hybridisation of PCl4+ is -

(A) Tetrahedral, sp3 (B) Octahedral, sp3 d2 (C) Trigonal pyramid, sp3d (D) See-saw, sp3 d

16. The geometry and hybridisation of PCl6– is -

(A) Octahedral, sp3 d2

(B) Tetrahedral, sp3 d2 (C) Square planar bipyramid, sp3 d (D) See-saw, sp3 d Passage # 2 (Ques. 17 to 19) The IE1 and IEII in kJ/mol of a few elements are

given in the following table

Element IEI in kJ/Mol IEII in kJ//mol

P 2372 5251

Q 520 7300

R 900 1760

S 1680 3380 17. Which of the above element is likely to be alkali

metal ? (A) P (B) Q (C) R (D) S

18. Which of the above element is likely to be alkaline earth metal ?

(A) P (B) Q (C) R (D) S

19. Which of the above element is likely to be noble gas? (A) P (B) Q (C) R (D) S


ABCD

P Q R S

S

P

P P Q R

R R

Q Q

S S

P Q R S


20. Column-I Column-II (A) Primary standard base (P) Na2C2O4

(B) Equiv. wt =2

wtMolecular (Q) Na2B4O7

(C) Primary standard reducing (R) Na2CO3 agent (D) Capable of reducing (S) As2O3 hardness of water due to the presence of Ca2+

21. Column-I Column-II (A) Tc/Pc (P) Z (B) Tc/Vc (Q) a/Rb (C) TB (R) 8b/R (D) Ti/TB (S) 8a/81 Rb2

22. Column-I Column-II

(A)

V

PA

B

(P)

(B)

T

P A B(Q) Isotherm

(C)

T

P

A

B(R) Isochoric

Temperature is increasing


(D)

B

A

T

P (S) Pressure is increasing

MATHEMATICS


1. In a ∆ABC, if r = r2 + r3 –r1 and ∠A > π/3 then range

of as is equal to-

(A)

2,

21 (B)

∞,

21 (C)

3,

21 (D) (3, ∞)

2. Vector perpendicular to i – j – k and in the plane of

i + j + k and – i + j + k is

(A) i + k (B) 2 i + j + k

(C) 3 i +2 j + k (D) 4 i – 2 j –2 k

3. If a, b ∈ R, a ≠ 0 and roots of ax2 – bx + 1 = 0 imaginary, then a + b + 1 is

(A) Zero (B) Positive (C) Negative (D) None of these

4. Number of distinct normals that can be drawn to the curve x2 = 4y from point (1, 2), is

(A) 0 (B) 1 (C) 2 (D) 3

5. ABC is a triangle whose medians AD and BE are perpendicular to each other. If AD = p and BE = q then area of ∆ABC is-

(A) pq32 (B) pq

23 (C) pq

34 (D) pq

43

6. If (21.4)a = (0.00214)b =100, then the value of

a1 –

b1 is :

(A) rational but not integral (B) prime (C) irrational (D) composite

7. For 3 ≤ r ≤ n nCr + 3 nCr–1

+ 3 nCr–2 + nCr–3 is-

(A) n+3Cr (B) 2 n+2Cr+2 (C) 3 n+1Cr+1 (D) 3 nCr

8. Point of intersection of straight lines represented by 6x2 + xy – 40y2 – 35x – 83y + 11 = 0 is-

(A) (3, 1) (B) (3, –1) (C) (–3, 1) (D) (–3, –1)

9. Number of points outside the hyperbola 3x2 – y2 = 48 from where two perpendicular tangents can be drawn to the hyperbola is/are -

(A) 1 (B) 2 (C) infinite (D) None




10. Assertion (A) : The length of the shortest intercept made by the family of lines (1 + λ) x + (λ – 1) y + 2 (1– λ) = 0 on the parabola x2 = 4(y – 1) is 5.

Reason (R) : Latus rectum is the shortest focal chord of the parabola.

11. Assertion (A) : If S1 and S2 are non-concentric circles then their radical axis must exist.

Reason (R) : S1,S2, S3 are three circles such that no two are concentric then their radical centre is defined.

12. Assertion (A) : If two straight lines intersect the x-axis at A and B and y-axis at C and D such that OA.OB = OC.OD, O being origin then points A, B, C, D are concyclic.

Reason (R) : If a secant through a point P intersects a circle at Q and R then PQ.PR is independent of the direction of the secant.

13. Assertion (A) : The equation (log x)2 – log x3 + 2 = 0 has only one solution. Reason (R) : log x2 = 2 log x if x > 0 This section contains 2 paragraphs; each has 3 multiple choice questions. (Questions 14 to 19) Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. +4 marks will be given for each correct answer and – 1 mark for each wrong answer.


Passage # 1 (Ques. 14 to 16) It is given that A = (tan–1x)3 + (cot–1x)3, where

x > 0 & B = (cos–1t)2 + (sin–1t)2, where

t∈

21,0 & sin–1x + cos–1x =

2π for –1 ≤ x ≤ 1 &

tan–1x + cot–1 x = 2π ∀ x∈R

14. The interval in which A lie is

(A)

ππ2

,8

33(B)

ππ8

,32

33 (C)

ππ5

,10

33(D) None

15. Maximum value of B is

(A) 8

2π (B) 16

2π (C) 4

2π (D) None

16. If least value of A is λ & max. value of B is µ then

cot–1

µ

µπ−λcot =

(A) 8π (B)

8π− (C)

87π (D)

87π−


Consider lines

L1 : 12x − =

13y − =

k4z

−− , L2 : 2

1x − = 2

4y − = 1

5z −

17. A vector perpendicular to L1 &L2 and of length 3 2 is- (When k = 1)

(A) 3 i –3 j (B) 2(3 i +2 j – k )

(C) –3 i +4 j + 2 k (D) 3 i + 4 k

18. Value of 'k' so that lines L1 and L2 are coplanar, is - (A) –1 (B) –1/2 (C) –2 (D) 2

19. Equation of plane containing these lines is (A) x – y – 2 = 0 (B) 2x – y + 2 = 0 (C) x – y + 7 = 0 (D) None of these


A B C D

P Q R S

S

P

P P Q R

R R

Q Q

S S

P Q R S


20. Match the following Consider a plane P = 0 on whom foot of

perpendicular from point (1, 1, 1) is (2, 3, 4). Column-I Column-II (A) sum of intercepts of P = 0 (P) 52/7 on coordinate axis is (B) perpendicular distance of (Q) 110/3 (0, 0, 0) from plane is λ then λ2 is (C) A line through (0, 0, 0) and (R) 120 perpendicular to plane is

ax =

by =

cz then a + b+ c may be

(D) Radius of circle obtained by (S) 200/7 plane 'P' and sphere x2 + y2 + z2

= 36 is 'r' then r2 is equal to

21. Match the Column : Column-I Column-II (A) Distance of 3x + 4y – 5 = 0 (P)1 from (1, 1) is

(B) Area of ∆ formed by x + y = 2 (Q) 4/3 with co-ordinate axis is (C) Circumcentre of ∆ formed by (R) 2/5 3x + 4y – 7 = 0 and axis is (h, k) then h + k is (D) Two sides of ∆ are x + y = 1 (S) 49/24 and 2x + y + 4 = 0. If circumcentre is (2, 1) then slope of third side is 22. Match the Column : Column-I Column-II (A) Find the number of 6 digit (P) 1 natural numbers, where each digit appears at least twice (B) In how many ways can five (Q) 677 different books be tied up in three bundles (C) How many non-empty (R) 11754 collections are possible by using 5P's and 6 Q's (D) How many students do you (S) 25 need in a school to guarantee that there are atleast 2 students, who have the same 1st two initials in their 1st names


PHYSICS

1. The spectrum of hydrogen atom has many lines although hydrogen atom contain only one electron why ?

2. What happens to the frequency when light passes from one medium to another.

3. What are coherent sources of light

4. If the intensity of the incident radiation on a metal is doubled, what happens to the kinetic energy of the emitted photoelectrons.

5. Prove mathematically that the average power over a complete cycle of alternating current through an ideal inductor is zero.

6. Figure shows a branch of a circuit. Find out the value of potential difference between A and B.

A 1A

2Ω + – 1Ω B

2V 7. Write the following radiations in an ascending order

in respect of their frequencies : X–rays, microwaves, ultraviolet rays, radiowaves

8. In a closed surface incoming flux and outgoing flux is 5 × 105 and 4 × 105 V-m respectively then find out the charge enclosed by surface.

9. A freshly prepared radioactive substance of half life 2 hours emits radiation of intensity which is 64 times the permissible safe level. Find the minimum time after which it would possible to work with this source safely.

10. Why does the conductivity of a semiconductor change with the rise in temperature.

11. In optical fiber refractive index of cladding is less than core. Why ?

12. Where should an object be placed from a convex lens to from an image of the same size ? Can it happen in the case of a concave lens ?

13. A transmitting antenna at the top of tower has a height of 36 m and the height of the receiving antenna is 49 m. What is the maximum distance between them for satisfactory communication in the LOS mode ?

(Radius of earth = 6400 km)

General Instructions : Physics & Chemistry • Time given for each subject paper is 3 hrs and Max. marks 70 for each. • All questions are compulsory. • Marks for each question are indicated against it. • Question numbers 1 to 8 are very short-answer questions and carrying 1 mark each. • Question numbers 9 to 18 are short-answer questions, and carry 2 marks each. • Question numbers 19 to 27 are also short-answer questions, and carry 3 marks each. • Question numbers 28 to 30 are long-answer questions and carry 5 marks each. • Use of calculators is not permitted.

General Instructions : Mathematics • Time given to solve this subject paper is 3 hrs and Max. marks 100. • All questions are compulsory. • The question paper consists of 29 questions divided into three sections A, B and C. Section A comprises of 10 questions of one mark each. Section B comprises of 12 questions of four marks each. Section C comprises of 7 questions of six marks each. • All question in Section A are to be answered in one word, one sentence or as per the exact requirement of the question. • There is no overall choice. However, internal choice has been provided in 4 questions of four marks each and

2 question of six marks each. You have to attempt only one of the alternatives in all such questions. • Use of calculators is not permitted.

MOCK TEST PAPER-3

CBSE BOARD PATTERN

CLASS # XII

SUBJECT : PHYSICS , CHEMISTRY & MATHEMATICS Solut ions wil l be published in same issue


14. Derive the expression for the capacitance of a parallel plate capacitor, having two identical plates each of area A and separated by a distance d, when the space between the plates is filled by a dielectric medium.

15. In a meter bridge shown in circuit diagram, the balance point is found to be at 40 cm from the end A when resistor Y is of value 20 Ω. Calculate the shifting in of balancing point on the same wire if X and Y are interchanged.

X Y

B

C G A

D

E

16. Two parallel wires of infinite length have linear charge densities λ1 and λ2 coulomb/metre. Derive an expression for force per unit length acting between them.

17. A 10 µF capacitor is charged by a 30 V d.c. supply and then connected across an uncharged 50 µF capacitor. Calculate (i) the final potential difference across the combination and (ii) the initial and final energies. How will you account for the difference in energy ?

18. Amplitude of electric field in an electromagnetic wave is 3×106 V/m. Find amplitude of magnetic field.

19. Explain it (i) work function (ii) Threshold frequency (iii) Threshold wave length (iv) Stopping potential

20. Show that the energy of the first excited state of He+ atom is equal to the energy of the ground state of hydrogen atom

21. In a young's double slit experiment, the slits are separated by 0.56 mm and the screen is placed 2.8 m away. The distance between the central bright fringe and the fifth bright fringe is measured to be 1.5 cm. Determine the wavelength of light used in the experiment.

22. A communication system having operating wavelength λ in m can use only x% of its source frequency as its channel bandwidth. The system is to be used for transmitting TV signals requiring bandwidth of F Hz, How many channels can this system transmit simultaneously

23. The given figure shows a network of resistances R1, R2, R3 and R4. Using Kirchhoff's laws, establish the balance condition for the network.

G

R2 R1

R3 R4

D

B

+ –

A C

24. Distinguish between resistance, reactance and

impedance for an a.c. circuit. Draw graph showing variation for reactance of (i) a capacitor (ii) an inductor with frequency of the circuit.

25. A 0.5 m long metal rod PQ completes the circuit as shown. The area of the circuit is perpendicular to the magnetic field of flux density 0.15 T. If the resistance of the total circuit is 3 Ω. Calculate the force needed to move the rod in the direction as indicated with speed 2 ms–1.

× × × × × × × × × ×

× × × × × × × × × × × × × × × × × × × ×

× × × × × × × × × × × × × × × × × × × × v

Q

P

26. Explain the principle and working of cyclotron with the help of a labelled diagram.

27. Using Gauss' theorem, deduce an expression for the electric field intensity at any point due to a thin, infinitely long wire of charge/length λ C/m.

28. What is diffraction ? write down its difference with interference ? If single slit diffraction is obtained by using light of wavelength 5000Å and slit width 0.5 mm. Then calculate angular width of central maxima.

OR What is dispersion ? How the angular dispersion is

defined ? Which colours is deviated maximum by a prism ? Calculate dispersive power of crown glass if refractive indices for red, yellow and violet colours are 1.5140, 1.5170 and 1.5318 respectively.

• 29. What is spectrum ? write down types of spectrum ?

• OR • If electrons trans it from first excited state of

H-atom to ground state and emitted radiations incidents over a metallic surface of threshold wavelength 4000 Å, then what is the value of


maximum kinetic energy of electrons emitted from this surface ?

30. With the help of a neat and labelled diagram, explain the underlying principle and working of a moving coil galvanometer. What is the function of:

(i) uniform radial field (ii) soft iron core in such a device ?

OR Derive a mathematical expression for the force per

unit length experienced by each of the two long current carrying conductors placed parallel to each other in air. Hence define one ampere of current. Explain why two parallel straight conductors carrying current in the opposite direction kept near each other in air repel ?

CHEMISTRY

1. Complete the following reactions : CH3–CH2–CH=CH2 + HCl → ………….

2. Arrange the following compounds in an increasing order of their acid strengths :

(i) Br2 – CH2 – CH2 – COOH (ii) Cl – CH – CH2 – COOH

| Cl

(iii) F – CH2 – CH2 – COOH (iv) CH2 – CH2 – COOH

| I

3. How will you distinguish between propanal & propanone.

4. What are amino acid. Write zwitterionic structure of alanine.

5. What happen when an electric field is applied to a colloidal solution ?

6. Explain the Frenkel defect with suitable example.

7. Why is the elevation in b.p. of water different in the following solutions ? (i) 0.1 molar NaCl solution.

(ii) 0.1 molar sugar solution.

8. Write a IUPAC name of following :

BrCH3 – CH – C – C – OCH3

O O

9. Complete the following reaction sequense :

(i) C6H5N2Cl + Cu2Cl2 →∆ ……………

(ii) HC ≡ CH 42

4SOHSOHg → − ..................

10. For the following conversion reactions write the chemical equations :

(i) Ethyl isocyanide to ethylamine (ii) Aniline to N-Phenylethanamide 11. Write the structural formulae of A, B, C and D in the

following sequence of reaction :

A + CH3MgBr H2O CH3CH2 – CH – CH3

OH

–H2O

H2SO4BBr2 CAlc. KOHD

12. Give the structural formulae and names of the products of the following reactions :

(i) Chloroform is heated with aniline in presence of alc. KOH.

(ii) Phenol is treated with an excess of aqueous bromine.

13. For a Ist order chemical reaction rate constant is 2.303 sec–1. Determine the time required for 90% completion of a chemical reaction

14. The vapour pressure of pure liquid A and B are 450 mm Hg and 700 mm Hg at 350 K respectively. Find the mole fractions of the components if total vapour pressure is 600 mm Hg.

15. Calculate the mass of a nonvolatile solute (molecular mass 40) which should be dissolved in 114 g octane to reduce its vaour pressure to 80%.

16. Describe the following with an example for each : (i) Aldol condensation (ii) Trans-esterification

17. Write reactions starting conditions for the following conversions :

(i) Benzene to Acetophenone (ii) Ethanal to Propanone

18. (a) Compare the magnetic moments of the following Cr+3, V+3, Fe+3 (b) Which of the following is coloured CuSO4(aq), [Sc(H2O)6]+3, ZnCl2

19. How are the following conversions carried out ? (Write reactions only)

(i) 1-bromopropane to 2-bromopropane (ii) Propanone to iodoform (iii) Phenol to salicylic acid

20. Describe in brief the structure of DNA.

21. Write the structure of polymers : (a) PVC (b) PMMA (C) PTFE

22. What are antacids. Explain with example

23. What is adsorption ? How does adsorption of a gas on a solid surface vary with


(a) temperature and (b) Pressure ? Illustrate with the help of appropriate graphs. 24. Explain the following terms with suitable examples : (i) Schottky defect, (ii) Interstitials (iii) F-centres.

25. A first order reaction is 20% complete in 10 minutes. Calculate (i) k of the reaction and (ii) time taken for the reaction to go upto 75% completion.

26. (a) Name the lanthanoide element which forms tetrapositive ions in aqueous solution.

(b) Why do Zr and Hf exhibit similar properties ? (c) Colour of a solution of K2Cr2O7 depends on pH of the solution. Why ?

27. (a) Write preparation of K2Cr2O7. Give reactions. (b) Complete reactions : K2Cr2O7 + H2SO4 + KI → KMnO4 + H2SO4 + H2S → 28. (a) Describe the preparation of acetic acid from

acetylene. (b) How can the following be obtained from acetic

acid (i) Acetone (ii)Acetaldehyde (c) In what way can acetic acid be distinguished

from acetone ? (d) Why do carboxylic acids not give the

characteristic reactions of a carbonyl group ? • OR

• (a) How would you account for the following : • (i) Aldehydes are more reactive than ketones

towards nucleophiles. (ii) The boiling points of aldehydes and ketones

are lower than of the corresponding acids • (iii) The aldehydes and ketones undergo a

number of addition reactions. • (b) Give chemical tests to distinguish between : (i) Acetaldehyde and benzaldehyde • (ii) Propanone and propanol • • 29. (a) Calculate the EMF of the cell Mg (s)/Mg2+

(0.2 M) || Ag+ (1 × 10–3 M) / Ag ; Mg/Mg2E +Θ = –

2.37 V, Ag/AgE +Θ = + 0.80 V. What will be the effect on EMF if concentration of Mg2+ ion is decreased to 0.1 M ?

• (b) (i) Arrange the following metals in the order in which they displace each other from the solution of their salts.

(ii) Given the standard electrode potentials ; K+ / K = –2.93 V, Ag+ / Ag = 0.80 V, Hg2+ / Hg = + 0.79 V, Mg2+ / Mg = – 2.37 V,

Cr3+ / Cr = – 0.74 V

• Arrange these metals in their increasing order of reducing power.

30. Arrange the following in order of the property mentioned and give reason.

(i) F2, Cl2, Br2, I2 (Bond energy) (ii) HCl, HF, HI, HBr (acidic strength) (iii) PH3, NH3, SbH3, AsH3 (basic strength) (iv) H2O, H2S, H2Se, H2Te (boiling point) (v) HOCl, HClO2, HClO3, HClO4 (acidic strength)

MATHEMATICS

Section A 1. From the differential equation representing the

family of curves y = A cos (x + B), where A and B are parameter.

2. Evaluate : ∫ + dxx 2/sin1 .

3. If →a = kji ˆ3ˆ2–ˆ + and

→b = k3–i , find

|→b × 2

→a |.

4. If →a = k3–j2i + ;

→b = kji ˆ2ˆ–ˆ3 + , show that

(→a +

→b ) is perpendicular to (

→a –

→b ).

5. Show that the four points whose position vectors are

ji ˆ7–ˆ6 , kji ˆ4–ˆ29–ˆ16 , kj ˆ6–ˆ3 and

kji ˆ10ˆ5ˆ2 ++ are coplanar. 6. Write the following function in the simplest form

tan–1

22 – xa

x .

7. Find gof : If f (x) = |x| and g(x) = |5x – 2|.

8. If matrix A has order 3 × 4, matrix B has order

4 × 2 and matrix C has order 2 × 3, Find order of (A.B.C).

9. If A =

+03–2

20x

x is skew symmetric,

find x.

10. Find value of x such that 1542

= x

x6

42.


Section B

11. The odds against a certain event are 5 to 2 and the

odds in favour of another independent event are 6 to 5. Find the chance that at least one of the events will happen.

OR In two successive throws of a pair of dice, determine

the probability of getting a total of 8, each time.

12. Solve (1 + x2) dxdy +2xy – 4x2 = 0.

13. Evaluate : ∫

++ dx

xxex

2)2(1 .

OR

Evaluate ∫−−

dxxx

x421

2

14. Evaluate : ∫ + )cossin2)(cos2–(sin xxxxdx .

15. The position vectors of two points A and B are kji ˆ2ˆˆ3 ++ and kji ˆ4–ˆ2–ˆ respectively. Find the

vector equation of the plane passing through B and perpendicular to the vector AB.

16. If →a = kji ˆˆ3–ˆ + ,

→b = kji ˆˆ–ˆ + and

→c = kji ˆˆ–ˆ2 + , verify that

→a × (

→b ×

→c ) =

(→a .

→c )

→b – (

→a .

→b )

→c .

17. Find the equation of tangent to curve y = 2–3x . Which is parallel to line 4x – 2y + 5 = 0.

18. If y = ∝.....

xx

x , then show that

dxdy =

)log–2(

2

xyxy .

OR

Differentiate tan–1

−++

−−+22

22

11

11

xx

xx w.r.t,cos–1 x2

19. If xy = ex–y, then prove that

dxdy = 2)log1(

logx

x+

20. If f (x) = x

x

2cot

–4

tan

π

, x ≠ π/4, find the value which

can be assigned to f (x), at x = π/4 so that f (x) becomes continuous every where in [0, π/2].

21. Find the value of parameter α for which

f (x) = 1 + αx, α ≠ 0 is the inverse of itself. 22. Use properties of determinants proof that

48810445

++++

xxyxxxyxxxyx

= x3.

OR

Show that xyzzxyyzx

111

= 2

2

2

111

zzyyxx

and hence factorize.

Section C

23. One bag contains five white and four black ball. Another bag contains seven white and nine black balls. A ball is transferred from the first bag to the second and then a ball is drawn from the second. Find the probability that the ball drawn is white.

24. Find the area of the region bounded by x2 + y2 = 1 and (x – 1)2 + y2 = 1.

25. Evaluate : dxee

exx

x

∫π

+0

cos–cos

cos.

Or

Evaluate : dxx

x∫

π

π

π+4/

4/–2cos–24/ .

26. Find the shortest distance between the following

lines : →r = (1 – t) i + (t – 2) j + (3 – 2t) k and

→r = (s + 1) i + (2s – 1) j – (2s + 1) k .

27. A farmer has a supply of chemical fertilizer of type A which contains 10% nitrogen and 6% phosphoric acid and of type B which contains 5% of nitrogen and 10% of phosphoric acid. After soil testing it is found that at least 7 kg of nitrogen and the same quantity of phosphoric acid is required for a good crop. the fertilizer of type A costs Rs. 5.00 per kg and the type B costs Rs. 8.00 per kg. Using linear programming find how many kgs of each type of the fertilizer


should be bought to meet the requirement and the cost be minimum. Solve the problem graphically.

OR

A furniture dealer deals only in two items-tables and chairs. He has Rs.10,000 to invest and a space to store at most 60 pieces. A table costs him Rs. 500 and a chair Rs.200. He can sell a table at a profit of Rs.50 and a chair at a profit of Rs.15. Assume that he can sell all items that he buys. Using linear programming formulate the problem for maximum profit and solve it graphically.

28. A wire of length 36 m is to be cut into two pieces. One of the pieces is to be made into a square and other into a circle. What should be lengths of two pieces, so that combined area of the squared and circle is minimum.

29. Solve the system of equation by using matrix method.

x2 +

y3 +

z10 = 4 ;

x4 –

y6 +

z5 = 1 ;

x6 +

y9 –

z20 = 2

Brain Teaser Pears - There are a few trees in a garden. On one of them, a pear tree, there are pears (quite logical). But after a strong wind blew, there were neither pears on the tree nor on the ground. How come? Pears – Solution At first, there were 2 pears on the tree. After the wind blew, one pear fell on the ground. So there where no pears on the tree and there were no pears on the ground.Another possible solutions: The wind blew so hard that the pears fell of the tree and blew along the ground into the water or hovering in the air in a tornado.

1. Emeralds have been produced synthetically in

labs since 1848 and can be virtually indistinguishable from the genuine article.

2. In the last 200 years the use of metals has

increased as scientists have discovered new ones: until the 17th Century only 12 metals were known - there are now 86.

3. The only person to have an element named after

him while still alive was Glenn Seaborg, the most prolific of all the element hunters.

4. Traffic lights with red and green gas lights were

first introduced in London in 1868. Unfortunately, they exploded and killed a policeman. The first successful system was installed in Cleveland, Ohio in 1914.

5. In 1998, design student Damini Kumar at South

Bank University patented a teapot with a special grooved spout, which she claims virtually rules out dribbling.

6. Even though most items in the home today are

technologically up to date, most of us are still using the standard light bulb designed in 1928!

7. A chest x-ray is comprised of 90,000 to 130,000

electron volts. 8. The strength of early lasers was measured in

Gillettes, the number of blue razor blades a given beam could puncture.

9. The first commercial radio station in the United

States, KDKA Pittsburgh, began broadcasting in November 1920.

10. A British rocket attack on US soldiers is

celebrated in the lyrics of the US National Anthem.

11. Until the late 1800s, it was forbidden for women

in the United States to obtain a patent, so if a woman had invented something she would file for a patent under her husband or father's name. For this reason, the number of early female inventors remains a mystery.

12. Milt Garland, a 102 year old engineer, invented a

technology that forms ice on the exterior of a casing instead of inside it, which is used to create indoor ice rinks.


PHYSICS

1. Refractive index is maximum for violet and minimum for red colour

2. Path difference between two waves should be odd multiple of λ/2 or phase difference between two waves should be odd multiple of π.

3. 10

34

105000106.6hP −

−

××

=λ

= = 1.32 × 10–27 kg-m/s

4. R = R0A1/3 ⇒ R ∝ A1/3

∴ 2

1

RR =

31

2

1

AA

⇒

2RR =

3/1

278

=

32

∴ R2 = 2R3

5. It should be very high i.e. in KHz to GHz

6. µ

=cv ; v2 < v3 < v1

7. Resistivity or specific resistance of a material is defined as the resistance offered by a wire made by that material having unit area of cross-section and unit length.

8. V = qW = 7–

5–

102.3106.1

×× = 50 volt

9. Objective lens must have greater focal length than eye-piece because magnifying power of lens

=

e

0P f

fM increases in this case.

10. Conditions of interference (i) waves must travel in the same medium in the same direction. (ii) Their frequencies and wavelengths must be

same (iii) Their plane of polarisation must be same (iv) Their must be maintained a constant phase

difference between the waves.

11. Ionisation energy:- Minimum energy required to remove electron from its ground state.

Excitation energy:- Energy required to transit an electron to one of its excited state from ground state

KE = r2

KZe2; U = –

rKZe2

∴ KE2U −=

12. Emax. = 12V; Emin. = 4V

Ma = 21

412412

EEEE

minmax

minmax =+−

=+−

∴ ma in percentage = 21 × 100 = 50%

13. Modulating signal being of low frequency cannot travel to large distance so it is modulated with high frequency carrier. Frequency modulation is virtually free from noise where as amplitude modulation suffers from noise pollution.

14. Semiconductor atoms are tetravalent. The form covalent bonds by sharing their electrons with the neighboring atoms. When donor impurity of valency S is doped, the donor atoms takes the place of a semiconductor atom in the crystal lattice. Donor atoms forms covalent bond by sharing its four electrons whereas the fifth electron of donor atom is so loosely attached to the atom such that it behaves as a free electron at room temperature.

15. (i) width of depletion layer of zener diode becomes very small due to heavy doping of p and n regions.

(ii) The field across depletion layer becomes very

high

=

dVEQ

16. Let at any instant charge on the capacitor is q, then

p.d. V = Cq

Then work done to put additional charge dq is

dW = (dq) V =

Cq dq

∴ Total work done to give charge Q is

W = ∫

Q

0

dqCq =

C2Q2

= 21 CV2 (Q Q = CV)

MOCK TEST-2 (SOLUTION) MOCK TEST– 2 PUBLISHED IN JANUARY ISSUE


This work done is stored inside the capacitor as P.E.

∴ 2CV21WU ==

17. (i) A galvanometer can be converted into ammeter by connecting a low resistance called shunt parallel to the galvanometer coil.

(ii) A galvanometer can be converted into voltmeter by connecting a large resistance in series with the galvanometer coil.

18. (i) Electromagnetic waves are produced by accelerated charge.

(ii) The velocity of electromagnetic wave in free space is equal to the velocity of light in free space.

(iii) It is transverse in nature. 19. α = 2º β = ? fo = 20 cm fe = 4 cm

MP = αβ and MP =

e

o

ff ∴

e

o

ff

=αβ ⇒ β =

e

o

ff

α

∴ β =

420 × 2 = 10

20. mqV2h

=λ

for proton

meV2h

=λ …(i)

For α-particle

λ = 'eV2m42

h××

…(ii)

By (i) and (ii) 2meV = 2 × 4m × 2eV'

8/V'V =

21.

ab

c d

e Fe 56

a → 2He4

b → 4Be8

c → 6C12

d → 8O16

e → 10Ne20

BE/A

A

22. 2 is NAND and 1 is OR gate A B Y' Y 0 1 1 0 23. R = (AB × 10c ± 5)% Yellow A = 4 Violet B = 7 Brown C = 1 ∴ R = (47 × 10 ± 5)% = (470 ± 5)%

24. (i) To make strong electromagnet (ii) To make high speed computers

25. tanθ = H

V

BB

∴ BV = BH tanθ = 0.4 × 10–4 tan60º = 0.69 × 10–4 T and BH = Be cos θ

⇒ Be = º60cos

104.0cosB 4

H−×

=θ

= 0.8 × 10–4 T

26. Biot–Savart′s law This law is applicable to determine magnetic field due to small current element. Magnetic field due to

small element →ld at point A is given as

→r

A →ld

→ld

i

30

r)rd(

4idB

→→→ ×

πµ

=l

or 20

rsinid

4dB θ

πµ

=l

where µ0 ⇒ Permeability of free space,

θ ⇒ between →ld and

→r .

27. (i) On increasing distance between the coils, flux

linked decreases, hence mutual inductance decreases

(ii) Q M = l

ANN 210µ

∴ on increasing no. of turns, M increases (iii) when iron sheet (µr) is inserted

Then M = l

ANN 21r0µµ

∴ M increased, because for iron µr > > 1


28 Parallel light rays incidenting over a lens converges to a point (convex lens) or seems to diverge from the point (concave lens) after refraction from lens. Then this point is called principle focus of lens. When lens is dipped into a liquid of refractive index greater than lens material refractive index then nature of lens gets changed. Focal length of lens is maximum for red and minimum for violet colour.

µ2 µ1 = 1 µ1 = 1 – +

I II

R1 = + 20; R2 → ∞ ;

f1 = (µ2 – 1)

∞−

+1

201

f1 = (2 –1)

∞−

1201

∴ cm20f +=

OR

i M r

β γ α

u P R v

µ2

I C O

µ1

By snell's law m1 sin i = µ2 sin r

But for small aperture MP;

µ1i = µ2r …(i)

In ∆OCM; i = α + β

and In ∆ICM; β = r + γ ⇒ r = β – γ.

∴ By (i)

µ1 (α + β) = µ2 (β – γ)

µ1

+

PCMP

OPMP = µ2

−

PIMP

PCMP

µ1

++

− R1

u1 = µ2

+−

+ v1

R1

– vRRu2211 µ

−µ

=µ

+µ

Ruv1212 µ−µ

=µ

−µ

29. Principal of van deGraff Generator : (i) Let a small charged conducting shell of radius r be located inside a larger charged conducting shell of radius R. If they are connected with a conductor, then charge q from the small shell will move to the outer surface of bigger shell irrespection of its own charge Q.

Here potential difference

r

Q

q

R

= V(r) – V(R)

= 04

qπε

R1–

r1

In this way, the potential of the outer shell increases considerably.

(ii) Sharp pointed surfaces have larger charge densities, so these can be used to set up discharging action.

Conducting Shell

HvR

Insulating Column

Target

Grounded Steal Tank

C1 , C2 metal Comb

C2

S

Working :Let spray comb C1 be charged to a high +ve potential which spray +ve charge to the belt which in turn becomes positively charged. Since belt is moving up, so it carries this positive charge upward. Opposite charge appears on the teeth of collecting comb C2 by induction from the belt. As a result of this, positive charge appears on the outer surface of shell S. As the belt is moving continuously, so the charge on the shell S increase continuously. Consequently, the potential of the shell (S) rises, to a very high value. Now the charged particles at the top of the tub (T) are very high potential with respect to the lower end of the tube which is earthed. Thus these particles get accelerated downward and hit the target emerging from the tube.


Use : It can be used to accelerate particles which are used in nuclear physics for collision experiments.

Or (i) Intensity of the Electric field at a point on the Axis of a Diople : A

O –q +q

l l

r

B E2 E1P

The intensities E1 and E2 are along the same line in opposite directions. Therefore, the resultant intensity E at the point P will be equal to their difference and in the direction BP (since E1 > E2). That is,

E = E1 – E2

= K4

1

0πε 2)–r(ql

– K4

1

0πε 2)r(ql+

= K4

q

0πε

+ 22 )r(1–

)–r(1

ll

= K4

q

0πε

+222

22

)–r()–r(–)r(

l

ll

= K4

q

0πε

222 )–r(

r4l

l

= K4

1

0πε

222 )–r(r)q2(2

l

l

But 2ql = p (electric dipole moment).

∴ E = K4

1

0πε 222 )–r(

rp2l

If l is very small compared to r (l << r), then l2 may be neglected in comparison to r2. Then the electric intensity at the point P due to the dipole is given by

E = K4

1

0πε 4rrp2 =

K41

0πε 3r

p2 N/C.

For vacuum (or air) K = 1.

∴ 1–3

0NC

rp2

41Eπε

= …..(i)

Intensity of the Electric Field at a point on the Equatorial Line of a Dipole.

A θ

r

→P

BO

l l

–q +q

θ E P

E1

E2

E2

Pθ

E1 sin θ

E2 sin θ

E1 cos θ + E2 cos θ

E2

E1 = K4

1

0πε

)r(q

22 l+

(in the direction BP) and E2 = K41

0πε

)r(q

22 l+

(in the direction PA) The magnitude of E1 and E2 are equal (but directions are different). On resolving E1 and E2 into two components parallel and perpendicular to AB, the components perpendicular to AB (E1 sin θ and E2 sin θ) cancel each other (because they are equal and opposite) while the components parallel to AB (E1 cos θ and E2 cos θ), being in the same direction ,add up. Hence the resultant intensity of electric field at the point P is E = E1 cos θ + E2 cos θ

=K4

1

0πε )r(q

22 l+cos θ +

K41

0πε )r(q

22 l+cos θ

= K4

1

0πε )r(q

22 l+ 2 cos θ

But from fig. cos θ = PBOB = 2/122 )OBOP(

OB+

= 2/122 )r( l

l

+.

∴ E = K4

1

0πε )r(q

22 l+ 2/122 )r(

2l

l

+

= K4

1

0πε 2/322 )r(

q2l

l

+.

But 2ql = p (moment of electric dipole).

E = K4

1

0πε 2/322 )r(pl+

.

If l is very small as compared to r ( l < < r), then l2 can be neglected in comparison to r2. Then the electric intensity at the point P due to the dipole is

E =K4

1

0πε 2/32 )r(

p = K4

1

0πε 3r

p N/C.

For vacuum (or air) K = 1

.NCrp

41 1–

30πε

…..(ii)

The direction of the electric field E is parallel to the axis of the dipole from the positive charge towards the negative charge. Comparing eq. (i) and (ii) we see that for a short dipole the intensity of the electric field on an axial point is twice the intensity at the same distance on the equatorial line. (ii) W = Uf – Ui =PE (1 – cos θ) = PE (1 – cos 180°)

= PE (1 + 1) = 2 PE


30.

A

t

φ

E

Or

Let a source of alternating e.m.f. be connected to a circuit containing a pure inductance only, Fig. Suppose the alternating e.m.f. supplied is represented by

tsinEE 0 ω=

L

If dI/dt is the rate of change of current through L at any instant, then induced e.m.f. in the inductor at the same instant is = – L dI/dt . The negative sign indicated that induced e.m.f. opposes the change of current. To maintain the flow of current, the applied voltage must be equal and opposite to the induced voltage

i.e. E = –

dtdIL– = Eo sin ωt

or dI = LE0 sin ω t dt

Integrating both sides, we get

I = LE0 ∫ ω dttsin

= LE0

ωωtcos– = –

LE0

ωcos ωt

= – L

E0

ωsin

ω

π t–2

or I = L

E0

ωsin (ωt – π/2) …(i)

The current will be maximum i.e. I = I0, when sin (ωt– π/2) = maximum = 1.

From (i), I0 = L

E0

ω× 1 …(ii)

Putting in (i), we get

)2/–tsin(II 0 πω=

This is the form of alternating current developed.

CHEMISTRY

1 acidoictanBuCOOHCHCHCH

altanBuCHOCHCHCH 223

)reagent'Tollens(

AgNOAmmonical223

3 →

2. 3-Amino-2-chloro butanamide

3. These are partly produced in body. Ex. Histidine and arginine 4. yes, because O2 is paramagnetic

5. In B.C.C no. of effective atoms are = 2 In F.C.C no. of effective atoms are = 4

6. In lyophillic colloids affinity is present between dispersed phase and dispersion medium and therefore it is more stable.

7. r = k [NO]2 [H2]

8. 6-Amino 3-chloro 4-hydroxy hexanamide

9. (i) Aldol condensation : Two molecules of an aldehyde or a ketone having at least one α-hydrogen atom, condense in the presence of a dilute alkali to give β-hydroxy aldehyde or β-hydroxy ketone.

CH3–C + HCH2CHO

O

H

Dil. NaOH CH3–C–CH2CHO

OH

HEthanal Ethanal Aldol

(ii) Gabriel phthalimide synthesis : Phthalimide on treatment with ethanoic KOH gives potassium phthalimide which on heating with a suitable alkyl halide gives N-substituted phthalimides. These upon hydrolysis with dil HCl under pressure give primary amines.

CO

–H2OCONH + KOH (alc)

CO

CO

N–H+

Pot. Phthalimide

Phthalimide

C2H5I, ∆–KI

CO

CON–C2H5

N-Ethylphthalimide

H+ / –H2O

COOH

COOHC2H5NH2 +

Phthalic acid Ethylamide


10.

COCH3

Benzene

(i) CH3COCl + Anhyd. AlCl3

F.C.acylation

Acetophenone

CH3 – C = O Ethanal

(ii)

H CH3MgBrDry ether CH3 – C – OMgBr

H

CH3

H+/H2O

CH3 – C – OH

H

CH3

Cu 573 KCH3 – C = O + H2

CH3

Propanone

11. The compound is CH3CH2CHO, Its IUPAC name is propanal. The isomer is CH3COCH3, acetone.

12. (i) Treat the compound with Lucas reagent (conc. HCl + anhy. ZnCl2) 2-propanol gives turbidity in 5 min whereas ethanol gives no turbidity at room temperature

CH3CH2OH + HCl → 2ZnCl No reaction

CH3CHCH3 + HCl OH

CH3–CH–CH3 + H2O

Cl

ZnCl2

Turbidity appears in 5 min

(ii) Acetaldehyde reduces Tollen's reagent to silver mirror but acetone does not.

CH3CHO + 2[Ag(NH3)2]+ + OH– → CH3COO + 2H2O + 2Ag↓ + 4NH3

CH3COCH3 → reagents'Tollen No action Sol.3 (a) Where concentration of reactant become unity or

one litre rate of reaction is called as rate constant. (b) The reaction which has order greater than one

but follows the kinetics of Ist order reaction is called pseudo unimolecular reaction.

CH3 COOC2H5 + H2O(excesss) →HCl CH3COOH + C2H5OH

velocity = K[CH3COOC2H5]

14. (a) When NaCl is added neutralization of charge takes place which results in precipitation of colloidal solution and coagulation takes place.

(b) Due to scattering of blue wavelength of light it makes image in our eye and sky appears blue.

15. (i) The reaction involved is 2H2O → O2 + 4H+ + 4e– 2 mol H2O required ––– 4F charge ∴ 1 mol H2O required ––– 2F charge (ii) 2FeO + H2O ––– Fe2O3 + 2n+ + 2e– 2 mol –––––––––––––2F 1 mol –––––––––––––1F

16. (i) Fluorine (ii) Iodine

17. (i) Transition metals have unpaired e– (ii) Transition metals and compounds have unpaired e–

18.

H – C = O + CH3CH2MgBr (i)| H

Methanal Grignard reagent

Dry ether

H – C – OMgBr

CH2CH3

H

H2O/H+

CH3CH2CH2OH + Mg OH

Br Propan-1-ol

KMnO4/H+ (ii)

CH2OH

Benzyl alcohol

COOH

Benzoic alcohol

19.

2CH3 – CHI + 2Na (i) | CH3

Dry ether

CH3 – CH – CH – CH3 + 2NaI | CH3

| CH3

2,3-Dimethyl butane

(CH3)2CO (ii)LiAIH4 CH3–CH–CH3

| OH

2-Propanol


OH

(iii)COOH

+

CH3CO

CH3CO

O

OCOCH3

COOH

+ CH3COOH

Aspirin 20. (a) ––CH2–CH=CH2––CH––CH––CH2––

C6H5 H

(b)

––CH2–C––

C6H5 H

H

(c) ––CH2–CH–– CN

H

21. Antibiotics are natural chemicals which kill/stop the growth

of bacteria Types – (A) Bactericidal : Ex- penicillin, afloxacin (B) Bacteriostatic: Ex tetracycline, chloremphenicol 22. Vitamins are organic compounds required in small quantity for normal functioning of body vitamin (C) is ascorbic acid. Deficiency- scurvy vitamin B1 is thiamine acid. Deficiency – beri-beri vitamin (B12) is cyanocobalamine. Deficiency anaemia

23 (i) Rate of Reaction Rate constant of Reaction

1. It is the speed at which the reactants are converted into the products at any moment of time

2. It depends upon the concentration of reactants species at that moment of time.

3. It generally decreases with progress of the reaction

1. It is the constant of proportionality in the rate law equation

2. It refers to the rate of the reaction at the specific point when concentration of every reacting species is unity

3. It is constant and does not depend on the progress of the reaction

(ii) Molecularity Order

1. It is number of reacting species

1. It is the sum of powers of

undergoing simultaneous collisions in the reaction.

2. It is a theoretical concept.

3. It can have integral value only.

4. It cannot be zero.

5. It does not tell us anything about the mechanism of the reaction.

concentration terms in the rate law expression

2. It is determined experimentally.

3. It can have even zero value.

4. It can also have fractional values.

5. It tells us about the slowest step in the mechanism and hence gives some clue about mechanism of the reaction.

24. (a) The conductivity of all the ions of the solution

which is kept between electrodes 1 cm apart and area of the electrodes 1 cm2.

Molar conductivity can be defined as conductance of all the ions present in the solution which contain 1 mol electrolyte in certain volume and solution is kept between electrodes 1 cm apart and area of the electrodes such that whole solution is present between them.

(b) In this cell Zn acts as anode and Ag as cathode

°cellE = °

AgE – °ZnE

= 0.344 – (– 0.76) = 1.104 V ∆G° = – n F E°cell = – 2 × 96500 × 1.104 = – 2.13 × 105 J

25. (A) We know d = ANV

zM×

⇒ 7.86 = 2338– 1002.6)1068.2(56Z

××××

z ~– 2 i.e. B.C.C. structure

(B) Total no. of atoms surrounding a particular atom in crystal structure is called coordination number.

(i) In C.C.P → C.No. 12 ; (ii) In B.C.C → C.No. 8

26. (a) 27Co+3 = [Ar] 3d 4s 4p

d2sp3

Octahedral geometry


NH3

NH3

Co+3

NH3 NH3

NH3

NH3

No unpaired e– therefore diamagnetic

(b) 28Ni+2 = [Ar] 3d 4s 4p

dsp2

CN–

Ni+2

CN–

CN– CN–

Square planar No unpaired e– ∴ Diamagnetic in nature

(c) 28Ni = [Ar] 3d 4s 4p

sp3

Ni

CO

CO

CO

CO

Tetrahedral geometry No unpaired e– ∴ Diamagnetic in nature

27. (a) (i) K2Cr2O7 + 7H2SO4 + 6KI → Cr2(SO4)3 + K2SO4 + 7H2O + 3I2 (ii) 6 KMnO4 + 6 KOH + KI → 6K2MnO4 + 3H2O + KIO3 (b) Lanthanoids have almost similar properties.

28. (a) According to Henry's law mole fraction of a gas is directly proportional to the pressure at which gas is disolved

p = kHx App. (1) Dissolution of gases in cold drinks (2) Dissolution of O2 in haemoglobin (b) π = CRT

= 180

8.1 × 10

1000 × 0.0821 × 298 = 2.446 atm

∆Tf = Kfm = 1.86 × 0.1 = 0.186 T = – 0.186ºC

29. (a) Acetylene is first oxidized with 40% H2SO4 in the presence of HgSO4

H–C≡C – H + H2O4

42

HgSO%1

SOH%,40 → CH3–CHO

Acetylene Acetaldehyde Acetaldehyde is finally oxidized to acid with air in

the presence of manganous acetate catalyst

acidAcetic3

Acetate

Manganous

deAcetaldehy3 COOHCH]O[CHOCH →+

(b) (i) CH3COOH → 2)OH(Ca (CH3COO)2 Ca Calcium acetate

∆ →Ca(HCOO)2

3deAcetaldehy

3 CaCO2CHOCH +

(ii) CH3COOH → 2)OH(Ca (CH3COO)2Ca Acetic acid

3Acetone

33 CaCOCOCHCH +→∆

(c) When heated with I2 + NaCO3 Solution, acetone gives yellow crystals of iodoform CH3COCH3 + 3NaOI → CH3I + CH3COONa

Acetone Yellow ppt. (Iodoform)

Acetic acid does not give iodoform test. (d) The carbonyl group in – COOH is inert and does

not show nucleophilic addition reaction like carbonyl compounds. It is due to resonance stabilization of carboxylate ions.

R – C = O

O–

R – C = O–

O Or

(a) (i) Due to smaller + I-effect of one alkyl group in aldehydes as a compared to larger +I-effect of two alkyl groups, the magnitude of positive charge on the carbonyl carbon is more in aldehydes than in ketones. As a result nucleophilic addition reaction occur more readily in aldehyde than in ketones.

(ii) The boiling points of aldehydes and ketones are lower than corresponding acids and alcohols due to absence of intermolecular hydrogen bonding .

(iii) Aldehydes and ketones undergo a number of addition reactions as both possess the carbonyl functional group which reacts a number of nucleophiles such as HNC, NaHSO3, alcohols, ammonia derivatives and Grignard reagents.

(b) (i) Distinction between acetaldehyde and benzaldehyde: Acetaldehyde and benzaldehyde can be distinguish by Fehling solution.

Acetaldehyde give red coloured precipitate with Fehling solution while benzaldehyde does not.

→++ −+44 344 21

SolutionFehling

23 OH52CuCHOCH


OHOCuCOOCH 2.pptred

23 ++−

(ii) Distinction between Propanone and Propanol : Propanone (CH3COCH3) and propanol (CH3CH2CH2OH) can be distinguish by iodoform test. Propanone when warmed with sodium hypoiodide (NaOI) i.e. I2 in NaOH, it gives yellow ppt of idoform while propanol does not respond to iodoform test.

→+ NaOI3COCHCHopanonePr

33

NaOH2COONaCHICH 3pptYellow

3 ++↓

30 (a) Mass of unit cell 2310023.6M2

××

=

Density = 7.2 g cm–3 α = 288 pm = 288 × 10–10 cm

Now density = Volume

Mass

∴ 7.2 = 310–23 )10288(10023.6M2

××××

or M = 2

10)288(10023.62.7 30–323 ××××

= 51.79 g mol–1.

(b) Those compounds containing two or more halogen atoms in their molecules are known as Interhalogen compounds. Properties : (i) They are covalent compounds and diamagnetic in nature.

(ii) They are more reactive than the constituent halogens. It is because A – X bond is relatively weaker than X – X bond.

(iii) They are very good oxidizing agents. Their melting point and boiling point are higher than

halogens and increases with increase in the difference of electronegativity.

MATHEMATICS

Section A

1. Given line is 12

−+x =

73+y =

2/33−z .

∴ Its direction ratios are –1, 7, 3/2 Or –2, 14, 3 Equation of line pasing through (1, 2, 3) is

22

−+x =

142−y =

33−z .

2. →b +

→c = i + 3 j + k + i + k = 2 i + 3 j + 2 k

Projection of →b +

→c on

→a =

||).(

aacb→→→

+ .

=222 )1()2()1(

)ˆˆ2ˆ).(ˆ2ˆ3ˆ2(

++

++++ kjikji =141262

++

++ =6

10

3. The vectors →a ,

→b ,

→c are coplanar if

→a . [

→b ×

→c ] = 0

λ1–8–

341–

2–6–4–

= 0

– 4 + (4λ + 3) + 6 (– λ + 24) – 2 (1 + 32) = 0 –16λ – 12 – 6λ + 144 – 66 = 0 – 22λ + 66 = 0 – 22λ = –66 ⇒ λ = 3 4. order = 2

Q The equation can't be expressed in polynomial from

⇒ degree is not defined 5.

⇒ dxxxxx

xxxxxx∫ −+

−++22222

222244

cossin2)cos(sin)cos(sin)cos(sin)cos(sin

⇒ dxxx

xxx∫ +

+44

44

cossin)2cos(–)1()cos(sin

= – 22sin x + c

6. A =

5432

A′ =

5342

A – A′ =

−0110

(A – A′)′ = ′

−0110

=

− 01

10

⇒ (A – A′) = –

−0110

⇒ (A – A′)′ = – (A – A′)

⇒ (A – A′) is skew symmetric matrix.


7. f(A) = A2 – 5A + 7 I2

=

− 21

13.

− 21

13– 5

− 21

13 + 7.

1001

=

− 35

58 –

− 105

515 +

7007

=

0000

= Null Matrix.

8. Area of ∆ = 0

⇒ 21

1264121122

kkkk

kk

−−−+−

− = 0

k . [2k – 6 + 2k] – (2 – 2k) [–k + 1 + 4 + k] + 1 [(–k + 1) (6 – 2k) –2k (–4 – k)] = 0 k [4k – 6] – (2 – 2k) (5) + [2k2 – 8k + 6 + 8k + 2k2] = 0 4k2 – 6k – 10 + 10k + 4k2 + 6 = 0 8k2 + 4k – 4 = 0 2k2 + k – 1 = 0

(2k – 1) (k + 1) = 0 ⇒ k = 1/2, – 1

9. y = f (ex)

y' = f '(ex). ex

xxxxx eefeeefy ).(".).('" +=

10. f (g(x)) = 1)( 2 −= xxg

g(f (x)) = f (x)2 – 1 = 1)( 2 −x = x – 1

Section B 11. A : 3 cards have the same number

n(A) = 13 C(4, 3) = 13

13

4= 13 (4) = 52

n(S) = C(52, 3) =493

52= 22100

Required probability

= P(A) =)()(

SnAn =

2210052 =

552513 =

4251

OR Let E : Candidate Reaches late A1 = Candidate travels by bus A2 : Candidate travels by scooter A3 : Candidate travels by other modes of transport

P(A1) = 103 , P(A2) =

101 , P(A3) =

53

P(E/A1) = 41 , P(E/A2) =

31 , P(E/A3) = 0

∴ By Baye's Theorem

P(A1/E) = )A/E(P)A(P)A/E(P)A(P)A/E(P)A(P

)A/E(P)A(P

332211

11

++

= 0

301

403

41

103

++

× =

139

12. Here, kjiA ˆ9ˆ2ˆ3 ++=→

; kjiB ˆ3ˆˆ +λ+=→

⇒ )ˆˆˆ()ˆ9ˆ2ˆ3( kjikjiba +++++=+→→

= kji ˆ12ˆ)2(ˆ4 +λ++

)ˆ3ˆˆ()ˆ9ˆ2ˆ3( kjikjiba +λ+−++=−→→

= kji ˆ6ˆ)2(ˆ2 +λ−+

Since

−⊥

+

→→→→baba we have

−⋅

+

→→→→baba = 0

⇒ 0]ˆ6ˆ)2(ˆ2[]ˆ12ˆ)2(ˆ4[ =+λ−++λ++ kjikji

⇒ 4 × + (2 + λ) × (2 – λ) + 12 × 6 = 0 ⇒ 8 + 4 – λ2 + 72 = 0

⇒ λ2 = 84 ⇒ 212±=λ

13. Equation of plane passing through the inter sections of planes

x + 2y + 32 – 4 = 0 and 2x + 4 – z + 5 = 0 (x + 2y + 32 – 4) + λ (2x y – z + 5) = 0 …(i) x + 2y + 32 – 4 + 2λx + λy –λz + 5λ = 0 (1 + 2λ) x + (2 + λ) y + (3 –λ) z – 4 + 5λ = 0 Since the plane (i) is perpendicular to 5x + 3y + 6z + 8 = 0 ∴ (1 + 2λ) ⋅ 5+ (2 + λ) ⋅ 3 + ( 3 – λ) ⋅ 6 = 0 5 + 10λ + 6 + 3λ + 18 – 6 λ = 0

7λ + 29 = 0 ⇒ 729−

=λ

∴ required equation of plane is

(x + 2y + 3z – 4) 729− (2x + y – z + 5) = 0

7x + 14 y + 21z – 28 – 54x – 29y + 29z – 145 = 0 – 47 x – 15y + 50z –173 = 0 47x + 15y – 50z + 173 = 0


OR The equation of a plane passing through the

intersection of the given planes is (4x – y + z –10) + λ(x + y – z –4) = 0 ⇒ x(4 + λ) + y (λ –1) + z (1 –λ) –10 – 4λ = 0 This plane is parallel to the line with direction ratios

proportional to 2, 1,1 ∴ 2(4 + λ) + 1(λ –1) + 1(1 –λ) = 0 ⇒ λ = – 4 Putting λ = – 4 in (i), we obtain 5y – 5z – 6 = 0 This is the equation of the required plane. Now, length of the perpendicular from (1, 1, 1) on (ii)

is given by

d = 22 )5(5

61515

−+

−×−× = 5

23

14. dxax

aax∫ −

+−)sin(

)sin(

dxax

aaxaax∫ −

−+−)sin(

sin)cos(cos)sin(

= cos a. x + sin a. ln sin (x – a) + c

15. sin x = t cos x dx = dt

∫−− 322 tt

dt

∫−− 22 2)1(t

dt

log ctt +−−+− |2)1()1(| 22 replace t

16. take log m log x + n log y = (m + n) log (x + y)

diff.

+++

=+dxdy

yxnm

dxdy

yn

xm 1

)(

⇒ xyx

mynxdxdy

yxymynx

)()( +−

=+

−⇒ xy

dxdy

/=

17. C1 → C1 + C2 + C3

(a + b + c) cbc

cbbcaba

3131

1

+−+−+−+−

R2 → R2 – R1, R3 – R1

(a + b + c) acacababcaba

++−+−++−+−

2121

1

(a + b + c) [(2b + a) (2c + a) – (a – b) (a – c)]

= (a + b + c) [3bc + 3ab + 3ac] = 3 (a + b + c) (ab + bc ca) hence proved.

18. at y-axis

==

byx 0

⇒ pt = (0, b)

Now axeab

dxdy /−−=

ab

dxdy

b−=

,0

equation of tangent y – b = – ab (x – 0)

⇒ 1/ =+ byax

19. centre = (h, 0) rad. = h equation is

(h, 0)

(x – h)2 + (y – 0)2 = h2 ⇒ x2 + y2 – 2hx = 0 … (1)

Deff. 2x + 2y . y1 – 2h = 0 ⇒ 'yyxh += …(2)

From (1), (2)

xyyxyx )'(222 +=+

OR Let P(x, y) be any point on the curve. The equation of

the normal at P (x, y) to the given curve is

Y – y = –dxdy /

1 (X –x) … (i)

It is given that the normal at each point passes through (2, 0). Therefore, (i) also passes through (2, 0). Putting Y = 0 and x = 2 in (i), we get

0 – y = – dxdy /

1 (2 –x)

⇒ ydxdy = 2 – x

⇒ ydy = (2 – x) dx [On integrating both sides]

⇒ 2

2y = – 2

)2( 2x− + C

⇒ y2 = – (2 – x)2 + 2C … (ii) This passes through (2, 3). Therefore,

9 = 0 + 2C ⇒ C = 29

Putting C = 29 in (ii), we get


y2 = – (2 – x)2 + 9 This is the equation of required curve.

20. Q f (x) is conti. at x = 1 ∴ f (1) = f (1–) ⇒ 1 =

0lim→h

a (1 – h)2 – b

⇒ 1 = a – b …(1)

21. Let xx

xxy−

−

+

−=

10101010

1

Componendo – dividendo rule

⇒ )1010()1010()1010()1010(

11

xxxx

xxxx

yy

−−

−−

++−+−−

=+−

⇒ x

x

yy

10.210.2

11 −

−=+−

⇒ y

y−+

11 = 102x

⇒ x = 21

ln10

−+

yy1

1

∴

−+

=−

xxnxf1

121)( 10

1 l

OR

f(1) = 2

11 + = 1

f(2) = 22 = 1

many-one function If n → odd natural number then 2n –1 is also odd

number

f(2n –1) = 2

112 +−n = n

If n → even natural number then 2n is also an even natural number

f(2n) = 2

2n = n

⇒ f is onto function.

22. Let y = sin–1

+ 212

xx

put x = tan θ

y = sin–1

θ+θ2tan1

tan2

y = sin–1 (sin 2θ) y = 2tan–1 x

212xdx

dy+

=

z = tan–1 x

21

1xdx

dz+

=

Now 2

2

1/11/2

xx

dzdy

++

= = 2

Section C

23. Let x denotes the number of spades. Clearly x can take the values 0, 1, 2, or 3.

Probability of getting a spade =5213 =

41

Probability of not getting a spade = 1 – 41 =

43

P(x = 0) = P (No spade) = 43 ×

43 ×

43 =

6427

P(x = 1) = P (One spade and two non spade)

= 3C1 × 41 ×

43 ×

43 =

6427

P(x = 2) = P (Two spades and one non spade)

= 3C2 × 41 ×

41 ×

43 =

649

P(x = 3) = P (Three spade) = 41 ×

41 ×

41 =

641

∴ The probability distribution of the random variable x is

x 0 1 2 3

P(x) 6427

6427

649

641

24. The required line is to pass through

kjia ˆ5ˆ3ˆ2 −−=→

and is ⊥ to the plane given by

02)ˆ5ˆ3ˆ6.( =++−→

kjir ……(i)

Here the vector kjib ˆ5ˆ3ˆ6 ++=→

is normal to the plane (1) ⇒ The required line is along the direction of this

vector. Hence its equation is →→→

+= btar

⇒ )ˆ5ˆ3ˆ6(ˆ5ˆ3ˆ2 kjitkjir +−+−−=→

….(ii) Now line (2) meets the plane (1), when

2

1 1


[ ])ˆ5ˆ3ˆ6()ˆ5ˆ3ˆ2( kjitkji +−+−− ⋅ 02)ˆ5ˆ3ˆ6( =++− kji i.e. when 6 (2 + 6t) + 3 (3 + 3t) + 5 (5t – 5) = –2

i.e. when 351

=t

Substituting this value of t in (2), we get

)ˆ5ˆ3ˆ6(351ˆ5ˆ3ˆ2 kjikjir +−+−−=

→

= )ˆ170ˆ108ˆ76(351 kji −−

∴ The required point of intersection is

−−

35170,

35108,

3576

25. Let the distance covered at the speed 25 km/hour = x km and the distance covered at the speed 40 km hour

= y km maximum distance z = x + y subject to constraints

2x + 5y ≤ 100

14025

≤+yx

x, y ≥ 0 Table for 2x + 5y = 100 …(i)

x 0 50 y 20 0

Table for 14025

=+yx

x 0 25 y 40 0

Plot lines 2x + 5y = 100 and 14025

=+yx on a graph

paper. The shaded region satisfy the given inequalities.

D(25,0)

40 50XB

I302010O

II

C

50

40

30

A (0, 20)20

10

(0, 40)

Y

340,

350P

The coordinates of feasible region is A (0, 20)

340,

350P , D (25, 0)

Corner points z = x + 5

A (0, 20) z = 0 + 20 = 20

P

340,

350 z = 30

340

350

=+

D (25, 0) z = 25 + 0 = 25

∴ Max. distance covered = 30 km 50/3 km at the speed of 25 km/hour and 40/3 km at the speed of 40 km/hour

26. Ist part : Area = dxxxx∫ −−−3

0

2 )]()2[(

= 3

0

32

323

−

xx

area = 9/2

(0, 0)(2, 0)

(3, –3)

2nd part : Both curve intersect at

)2,32(),2,32( −


(4, 0)

(0, 4)

)2,32(

Area = 2

−+ ∫∫

4

2

22

0166 dyydyy

=

×+−+ −

4

2

1220

2/3 4/sin162116

21)(

32.62 yyyy

= 3

163

34 π+

27. I = ∫π 2/

0xdxsinlog …(1)

By property

I = ∫π 2/

0dxxcoslog …(2)

2I = ∫π

2/

0dx

2x2sinlog

2I = ∫π

−2/

0)2(logxdx2sinlog ∫

π 2/

0dx

2I = I1 – log2 2π …(3)

Now In I1 2x = t dx = 1/2 dt

I1 = ∫π

0dttsinlog

21

= ∫π 2/

0dttsinlog2.

21

I1 = I …(4)

From (3), (4), 2I = I – π/2 log 2 I = – π/2 log 2

28. System of equations can be written as :

−−112112

111 .

zyx

=

225

AX = B

X = A–1 . B =

|A|

)A(Adj B

X = 81−

−−−−

−−

134314220

225

⇒

zyx

=

221

x = 1, y = 2, z = 3 OR

System of equations can be written as :

−

−

132211212

.

zyx

=

− 223

A . X = B

X = A–1 . B =

|A|AAd j . B

X =

−−−

−

−141265

077

71 .

− 223

⇒ X =

zyx

=

−11

1

x = –1, y = –1, z = 1 29. Let length, width → x height → y volume v = x2y …(1) are s = x2 + 4xy …(2)

s = x2 + xv4 (from (1), (2))

dxds = 2x – ,4

2xv 082

32

2>+=

xv

dxsd

Now 0=dxds ⇒ 2x – 04

2 =xv

⇒ x = 2y (Q v = x2y)


PHYSICS

1. There are many atoms in an atomic hydrogen gas sample and electrons of different atoms can make different transitions and different wavelength radiations are emitted.

2. Frequency remains constant when light passes from one medium to other.

3. Those sources which have constant phase difference are called coherent sources. These are derived from a single sources.

4. Kinetic energy depends on frequency of incident radiations not on intensity so kinetic energy remains same.

5. Average power over a complete cycle of ac through an ideal inductor (Resistance = 0) is zero.

Pav = (Irms) × (Vrms) cos φ

cos φ = ZR = 0

Pav = 0

6. A B + –

1A 1Ω 2Ω

VA – VB = 1 × 2 + 2 + 1 × 1 = 5 V 7. X–rays –→ 3 × 1019 to 1 × 1016 Hz Microwave –→ 3 × 1011 to 1 × 109 Hz U-V rays –→ 5 × 1017 to 8 × 1014 Hz Radio Waves –→ 3 × 107 to 3 × 104 Hz

8. φ = →→A.E , ∆ φ =

0

qε

φi = EA cos 180° φf = EA cos 0° φi = – EA φf = + EA φi = – 5 × 105 V– m φf = 4 × 105 V– m

– 5 × 105 + 4 × 105 = 0

qε

, q = – ε0 × 105 Cb

9. Intensity of radiations is 64 = 26 times the safe value. So after six half lines it would be safe so t = 6T = 6 × 2 = 12 hours.

10. On temperature rise more valance band electrons become free and more free electrons and hole pair will be produced and conductivity of semiconductor increases.

11. In optical fibres refractive index of cladding is less than core. So that electromagnetic wave traveling in core when incidents other cladding at an angle greater than ic. Total internal reflection will occur which is necessary for transmission of signals from one place to other.

12. Object should be placed at 2F of convex lens so as to form image of same size. It can't happen in case of concave lens because it always forms a diminished image.

13. Using d = RT Rh2Rh2 +

= )hh(R2 RT +

= 20.8 km5 = 46.5 km

14. When a charge +Q is given to the insulated plate, then a charge –Q is induced on the nearer face of the plate.

E

+ Q – Q

Net electric field between the plate

E = 0ε

σ also, E = dV

∴ V = Ed = 0

dεσ =

CQ ⇒ C =

dA0ε

with dielectric. C' = d

Ar0εε

MOCK TEST-3 (SOLUTION) MOCK TEST– 3 PUBLISHED IN SAME ISSUE


15. X = ? Y = 20 Ω

l1 = 40 cm l – l1 = 60 cm X Y B

C G A

D

V1 V2 l1 l – l1

V1 ∝ l1 V2 ∝ (l – l1)

)–(

YX

11 lll=

40X =

6020 or

40X =

31 or X =

340

Ω

When X and Y are interchanged Y X B

CG A

D

l2 l – l2

2

Yl

= )–(

X

2ll

⇒ 2

20l

= )–100(3

40

2l

2

1l

= )–100(3

2

2l

300 – 3l2 = 2l2 or 5l2 = 300 l2 = 60 cm

16. d2

F0

21

πελλ

=

λ1 λ2

d

17. (i) Vcom = 21

2211

CCVCVC

++

(ii) Ui = 222

211 VC

21VC

21

+

Uf = 21 (C1 + C2) 2

comV

18. B0 = c

E0

19. Work-function : Minimum energy needed by a free electron to remove it from metallic surface.

Threshold frequency : Minimum frequency required for photo-electron emission.

Threshold wavelength : Maximum wavelength of incident radiations required for photoelectric emission.

Stopping potential : Minimum potential required across the photo cell to completely stop the photo current.

20. (En) = 2n6.13– eV, for ground state n = 1

∴ (E1) = – 13.6 eV For hydrogen like atom, (En) HLA = Z2(En) H

= eVn

6.13Z2

2 ×

For He+ atom, Z = 2 and n = 2

∴ (E2)He = 4

46.13– = – 13.6 eV

21. m105.1dD5 2−×=

λ

d = 0.56 × 10–3 m D = 2.8 m

∴ 23 105.1

1056.08.25 −

−×=

×λ××

∴ λ = 6 × 10–7 m ∴ λ = 6000 Å

22. Source frequency ν = λc

Band width of system = ν100

x = λ

×c

100x

Number of channels N which can be transmitted simultaneously can be found out by dividing band width by the system with band width of one channel. i.e.,

N = F100

xcF

100xc

λ=

λ×

23. 4

3

2

1

RR

RR

=

24. (1) Resistance = R (2) Reactance are of two types (a) Capacitive Reactance

XC = C

1ω


(b) Inductive Reactance XL = ωL

(3) Impendence = Z = 2CL

2 )X–X(R +

(4)

ν

XC

ν

XL

25. Force = F = Bil

For current (i)

i =

Remf =

RBvl ⇒ F =

RvB 22 l

26. Cyclotron work on the fact that a positively charged particle can be accelerated to a sufficiently high energy with the help of smaller values of oscillating electric field by making it to cross the same electric field time and again with use of strong magnetic field.

North

target

South

B (Magnetic Field)H.F oscillator

Dees D1

Dees D2

Magnet

27.

+ + + + + + + + + + + + + +

λ

l r

0

inqds.E∈

=∫r

E.2πrl = 0

·∈λ l

E = r2 0∈π

λ

28. Bending of light from sharp edges is called diffraction. Difference between interference and diffraction.

Interference Diffraction

(i) Two coherent sources are necessary

(i) One source is Necessary

(ii) All fringes are of same width

(ii) CM has double width than all other fringes

(iii) All bright fringes has equal intensity

(iii) as order of bright fringes increases, intensity goes down

(iv) For bright fringes path difference = nλ

(iv) For bright fringes, Path difference

= (2n – 1) 2λ .

(v) For dark fringes, path difference = (2n –

1) 2λ

(v) For dark fringes, path difference = nλ

Angular width of CM = 3

10

105.01050002

a2

−

−

×××

=λ

= 2 × 10–3 Radian OR

When white light incidents over prism then it divide it into component colours. This phenomenon is called diffraction and angle between end colours is called angular dispersion. Prism deviates violet colour maximum as its refractive index for violet colour is maximum.

ω = 1y

Rv

−µµ−µ

∴ ω = 15170.1

5140.15318.1−

−

∴ ω = .034

29. white

Types of spectrum Spectrum is of 2 types. (i) Emission (ii) Absorption.


(i) Emission : - If radiations emitted from a substance is obtained ones screen after passing through the prism then spectrum is called emission spectrum.

It is line emission spectrum for substances at atomic level in gaseous state.

It is continuous spectrum for substances in liquid or solid state.

(ii) Absorption spectrum : - The substance of which absorption spectrum is to be find out is placed in a transparent tube and white light is passed through it now substance absorb radiation corresponding to some particular wavelength and black lines are obtained for these absorbed radiations. These black lines are called absorption spectrum.

OR Energy of emitted photons from H-atom = 10.2 eV Work – function of metallic surface

= 4000

12400 eV = 3.1 eV

Now According to Einstein's Law of Photo-electric effect '

E = w + K.E.max.

10.2 = 3.1 + K.E.max ∴ K.E.max = 7.1 eV Ans 30. Moving Coil Galvanometer – Principle : When a

coil carrying current is placed in a magnetic field, it experiences a Torque τ = NiAB sin α. This Torque is proportional to the current passed through it.

N S

Torsion Head

Mirror

Frame

Core

Copper coil

Phosphor Bronze

Spring

T1

T2

Working : When a current is passed through the

coil, the two vertical limbs experience a force normal

to it, equal in magnitude and parallel but opposite in sense. No forces act on the horizontal sides as they are parallel to the field .Since the field is radial, the forces acting on the vertical parts of coil remain always perpendicular to the plane of the coil in all its positions so that the perpendicular distance between the forces is always equal to breadth of the coil. Thus the coil is subjected to a torque whose magnitude is given by

τ = NiAB … (i) Under the action of this deflecting torque, the coil

rotates on its axis. Due to elastic forces, a restoring torque τr is produced in the suspension coil, which is perpendicular to the twist θ produced in the wire.

τr = θ or τr = Cθ … (ii) where C is restoring torque per unit twist, and it is

called the constant of twist for the suspension. As soon as restoring torque becomes equal to the deflecting torque, equilibrium is established. The coil does not rotate further. Let φ be the angle through which the coil rotates till it reaches the equilibrium position. τ = τr or NiAB = Cφ … (iii)

i = φNAB

C = K.φ

K = NAB

C is constant for the instrument, called

Galvanometer constant. ∴ i ∝ φ .

Thus deflection of coil is proportional to the current

passing through it. The deflection can be measured

by using a lamp and scale arrangement.

OR Consider two infinitely long thin conductors

carrying currents in opposite directions. Magnetic field B1 due to I1 at P on conductor CD is

given by

B1 = r2I10

πµ

B I1 D

P

B1F2 F1B2

r

A C I2

Q

The magnetic field B1 is perpendicular to plane of

paper and directed inward. This field will produce a force/length F2 on conductor given CD by


F2 = B1I2 = r2II 210

πµ [Q F = BI l, Here l = 1]

By Fleming's left hand rule direction of F2 is away from conductor AB.

Similarly the current I2 will create a field B2 at Q directed inward which in turn will create force/length F1

F1 = B2I1 = r2II 210

πµ

By Fleming's left hand rule, the direction of F1 is away from the conductor CD. Hence the two conductors repel each other.

Definition of Ampere If I1 = I2 = 1A, and r = 1m, then

F = 177

0 Nm1022104

2−−

−

×=π

×π=

πµ

Thus one ampere is that current which on flowing through each of the two parallel uniform linear conductors placed in free space at a distance of one meter from each other produces between them a force of 2 × 10–7 N per metre of their lengths.

CHEMISTRY

1. CH3–CH2–CH=CH2 + HCl →

Cl

CHCHCHCH|

323 −−−

2. (ii) > (iii) > (i) > (iv) 3. By fehling solution, tollen reagent etc. 4. Amino acid are compound containing amino group

and carboxylic group. Structure of alanine is –

H3N–CH–COO– +

CH3 5. Movement of colloidal particles towards opposite

terminal takes place which is called as electrophoresis.

6. It is a stoichiometric defect in which an ion (cation)

get displaced from its position and get arranged as an interstitial particle it is called Frenkel defect in this defect a vacancy or interstitial defect are formed simultaneously. Due to Frenkel defect.

(i) Density remains same (ii) Electrical conductivity improved

7. No of particle are different in sugar and NaCl. As dissociation in NaCl takes place.

8. Methyl 3-bromo 2-oxo butanoate

9. (i) Sandmayer reaction. —Cl

(ii) Kuchrous reaction CH ≡ CH → CH2 = CH

| OH

CH3 – CHO

10. (i) Hydrolysis

H

isocyanideEthyl52 HOH2CNHC →+=−

+

HCOOHNHHCeminaEthyl252 +

(ii) NHH

Aniline

N–C–CH3 + HClH

N-Phenylethanamide

+ Cl–C–CH3

H

Acetyl chloride

∆

Base

O

11. The given reaction will be as

CH3CH2–C–H + CH3MgBr H2O

OH

–H2O

H2SO4

Alc. KOH

(B)

O

(A)CH3CH2–CH–CH3

CH3CH=CH–CH3Br2 CH3–CH–CH–CH3

Br Br

CH3–C≡C–CH3

(C)

(D)

Structural formula of A : HCCHCH||

O

23 −−

Structural formula of B : CH3CH=CH–CH3 Structural formula of C : CH3–CH–CH–CH3

Br Br

Structural formula of D : CH3–C≡C–CH3

12.

NC

Aniline

(i) ∆

Carbylamine

+ CH3Cl + 3 KOH

Chloroform

NH2

+ 3 KCl + 3 H2O


Phenol

(ii) + Br2

OH

H2O

2,4,6-tribromophenol

OH

Br Br

Br

13. Ist Method : Given K = 2.303

∴ t1/2 = K

3.0303.2 × = 303.2

3.0303.2 × = 0.3 sec.

t90 = 3

10 × t1/2 = 3

10 × 0.3 = 1 sec

II method

K = t303.2 ln

− xa

a

0

0

2.303 = 90t303.2 ln

10100

t90 = 1 sec

14. P = B0BA

0A xPxP + PºA = 450 mm

PºB = 700 mm 600 = 450 xA + 700 xB = 450 (1 – xB) + 700 xB = 450 + 250 xB 250 xB = 600 – 450 = 150

xB = 53 & xA =

52

15. vapour pressure of solution pA = 0.8 pA

let mass of solute be Wg moles of solute = 40W

mole of octane n0 = 114114 = 1

xB = 140/W

40/W+

°

∆

A

A

pp =

°°°

A

AA

pp8.0–p = xB =

140/W40/W

+

0.2 = 140/W

40/W+

, W = 8.0402.0 × = 10g

16. (i) Aldol condensation : Two molecules of an aldehyde or a ketone having at least one α-hydrogen atom, condense in the presence of a dilute alkali to give β-hydroxy aldehyde or β-hydroxy ketone.

CH3–C + HCH2CHO

O

H

Dil. NaOH CH3–C–CH2CHO

OH

H

Ethanal Ethanal Aldol

(ii) Trans-Esterification : An ester on reaction with excess of alcohol in the presence of mineral acid forms a new ester.

CH3–C–OCH3 + CH3CH2OH

OH+

CH3–C–OCH2CH3 + CH3OH

Methyl ethanoate

Ethanol

O

Ethyl ethanoate Methanol

17

COCH3

Benzene

(i) CH3COCl + Anhyd. AlCl3

F.C.acylation

Acetophenone

CH3 – C = OEthanal

(ii)H

CH3MgBrDry ether CH3 – C – OMgBr

H

CH3

H+/H2O

CH3 – C – OH

H

CH3

Cu 573 KCH3 – C = O + H2

CH3

Propanone

18. (a) Cr+3, µs = 15 Fe+3 > Cr+3 > V+3

V+3, µs = 8

Fe+3, µs = 35

(b) CuSO4 (aq)

19. Conversions : (i) 1-bromopropane to 2-bromopropane

CH3CH2CHBr alc. KOH CH3CH = CH2HBr

(Peroxide)

CH3CHCH3

Br

(ii) CH3–C–CH3I2

OCH3–C–C.I3

NaOHO

CHI3

Na(iii)

OH

Phenol

CO2

ONa

4–7 atm.

OH

COONa


20. DNA has a hydrogen bonded double helical structure. The two strands are antiparallel. It can be considered a polymer of nucleotide (Base + sugar + phosphoric acid). In a nucleotide base and phosphate are linked to C1′ and C5 of sugar molecule respectively. Two nucleotide are linked by 3 ′ – 5′ phosphodiester bonds. Hydrogen bonds are formed between –

Puric bases

Adenine = Thyonine & Guanine ≡ cytosine

Pyrimidal bases

P

PS

S

5 ′

3 ′ 5 4 3 2 1

H bond

A = TP

S

S G C ≡P

3 ′

5 ′ 21. (a) CH2–CH

Cl H

(b) CH2–C OCOCH3

H

CH3

(c) CF2–CF2 H

22. Antacids are drugs used to relieve acidity these can

act in any of the following ways – (a) neutralize acid in stomach Ex – NaHCO3, milk of magnesia (b) stop acid production in stomach Ex – oneprazole, lasoprazole (c) antihistaminic Ex- Ranitidine (zantac) 23. Absorption at the surface is called adsorption. In

adsorption the conc. of adsorbate is greater at the surface than in the inner bulk.

Adsorption of gases : Temperature : As temperature ↑ physisorption

continuously ↓ where as with increase in temperature chemisorption first increases than decreases.

Phys

isor

ptio

n

T

chem

isor

ptio

n

T Pressure : With increases in pressure adsorption of

gases increases.

24. (i) Schottky defect : It is a stoichiometric defect in which pair of cations and anions get displaced from their position in such a manner a pair of vacancies are created simultaneously

In this defect (1) Density ↓ (2) Electrical conductiriy↑ (ii) Interstials : when particles may be an additional

particle of same material a foreign particle get arranged in the spaces between regular arrangement of particles.

(iii) F-centres : when crystal of NaCl is heated in the vapours of Na Cl– ions get displaced from its position and diffused into the Na vapour. And NaCl is formed which get deposited at the surface and in place of Cl– ion e– get arranged. This results in colour in the NaCl crystal. Other example LiCl in Li vapours become Pink, KCl in K vapours become violet.

25. (i) Let a0 = 100 M, conc. after 10 min a0 – x = 100 – 20

t303.2k = log

x–a

a

0

0

= 10303.2 log

80100 = 0.02231 min–1

(ii) Let the time for 75% completion = t min x = 75 M

∴ t = k303.2 log

x–a

a

0

0

= 02231.0303.2 log

25100 = 62.15 min

26. (a) Cerium (Ce), (b) Due to lanthanoide contraction. (c) In basic medium, K2Cr2O7 converts into K2CrO4

hence its colour changes while in acidic medium K2CrO4 converts into K2Cr2O7

27 (a) 4FeCr2O4+16 NaOH+)air(2O7 →

8 Na2CrO4+2Fe2O3 + 8H2O 2Na2CrO4 + H2SO4 → Na2Cr2O7 + Na2SO4 + H2O Na2Cr2O7 + 2KCl → K2Cr2O7 +2NaCl (b) K2Cr2O7 + 7H2SO4 + 6KI → Cr2(SO4)3 + 4K2SO4 + 3I2 + 7H2O 2KMnO4+ 3H2SO4 + 5H2S →

2MnSO4 + K2SO4 + 8H2O + 5S


28. (a) Acetylene is first oxidized with 40% H2SO4 in the presence of HgSO4

H – C ≡ C – H + H2O 4

42

HgSO%1

SOH%,40 → CH3 – CHO

Acetylene Acetaldehyde Acetaldehyde is finally oxidized to acid with air in the

presence of manganous acetate catalyst

acidAcetic3

Acetate

Manganous

deAcetaldehy3 COOHCH]O[CHOCH →+

(b) (i) CH3COOH → 2)OH(Ca (CH3COO)2 Ca Calcium acetate

∆ →Ca(HCOO)2

3deAcetaldehy

3 CaCO2CHOCH +

(ii) CH3COOH → 2)OH(Ca (CH3COO)2Ca Acetic acid

3Acetone

33 CaCOCOCHCH +→∆

(c) When heated with I2 + NaCO3 Solution, acetone gives yellow crystals of iodoform CH3COCH3 + 3NaOI → CH3I + CH3COONa

Acetone Yellow ppt. (Iodoform)

Acetic acid does not give iodoform test. (d) The carbonyl group in – COOH is inert and does

not show nucleophilic addition reaction like carbonyl compounds. It is due to resonance stabilization of carboxylate ions.

R – C = O

O–

R – C = O–

O

Or

(a) (i) Due to smaller + I-effect of one alkyl group in aldehydes as a compared to larger +I-effect of two alkyl groups, the magnitude of positive charge on the carbonyl carbon is more in aldehydes than in ketones. As a result nucleophilic addition reaction occur more readily in aldehyde than in ketones.

(ii) The boiling points of aldehydes and ketones are lower than corresponding acids and alcohols due to absence of intermolecular hydrogen bonding .

(iii) Aldehydes and ketones undergo a number of addition reactions as both possess the carbonyl functional group which reacts a number of nucleophiles such as HNC, NaHSO3, alcohols, ammonia derivatives and Grignard reagents.

(b) (i) Distinction between acetaldehyde and benzaldehyde: Acetaldehyde and benzaldehyde can be distinguish by Fehling solution.

Acetaldehyde give red coloured precipitate with Fehling solution while benzaldehyde does not.

→++ −+44 344 21

SolutionFehling

23 OH52CuCHOCH

OHOCuCOOCH 2.pptred

23 ++−

(ii) Distinction between Propanone and Propanol : Propanone (CH3COCH3) and propanol (CH3CH2CH2OH) can be distinguish by iodoform test. Propanone when warmed with sodium hypoiodide (NaOI) i.e. I2 in NaOH, it gives yellow ppt of idoform while propanol does not respond to iodoform test.

→+ NaOI3COCHCHopanonePr

33

NaOH2COONaCHICH 3pptYellow

3 ++↓

29. (a) Net cell reaction Mg (s) + 2Ag+ (aq) → 2Ag(s) + Mg+2 (aq) According to nernst equation

Ecell = E°cell – n

0591.0 log [ ][ ]2

2

Ag

Mg+

+

= 0.80 – (– 2.37) –2

0591.0 log 23– )101(2.0

×

= 3.32 V when conc. of Mg+2 is decreased to 0.1, the new emf is

Ecell = 3.17 – 2

0591.0 log 23– )101(1.0

×

= 3.34 V (b) (i) Mg > Αl > Zn > Fe > Cu As we move downwards in the electro chemical

series tendency to displace increases. (ii) K > Mg > Cr > Hg > Ag 30. (i) Cl2 > Br2 > F2 > I2 (due to exceptionally small size

of F, F2 has lower bond energy than expected) (ii) HF < HCl < HBr < HI (according to bond length) (iii) NH3 > PH3 > AsH3 > SbH3 (according to density of lonepair on central atom) (iv) H2O > H2Te > H2Se > H2S (according to strength of intermolecular bonding) (v) HOCl < HClO2 <HClO3 < HClO4 (according to no. of O atoms attached to Cl)


MATHEMATICS

Section A 1. y = A cos (x + B)

dxdy

= – A sin (x + B)

2

2

dxyd

= – A cos (x + B)

2

2

dxyd

= – y ⇒ 2

2

dxyd + y = 0

2. ∫ ++ 4/cos4/sin24/sin4/cos 22 xxxx dx

dxxx∫ + )4/sin4/(cos = 4 sin (x/4) – 4 cos (x/4) + c

3. →a = kji ˆ3ˆ2–ˆ + ;

→b = ki ˆ3–ˆ ,

⇒ 2→a = kji ˆ6ˆ4–ˆ2 +

∴ →b × 2

→a =

64–23–01

ˆˆˆ kji

= – kji ˆ4–ˆ12–ˆ21

= )ˆˆ3ˆ3(4– kji ++

⇒ |→b × 2

→a | = |)ˆˆ3ˆ3(4–| kji ++

= 222 1334 ++ = 194

4. →a = kji ˆ3–ˆ2ˆ + ,

→b = kji ˆ2ˆ–ˆ3 +

⇒ →a +

→b = kji ˆ–ˆˆ4 + ,

→a –

→b = kji ˆ5–ˆ3ˆ2 +

(→a +

→b ) . (

→a –

→b ) = ( kji ˆ–ˆˆ4 + ).( kji ˆ5–ˆ3ˆ2 + )

= 4 × (– 2) + 1 × 3 – 1 × (– 5) = – 8 + 3 + 5 = 0

⇒ →a +

→b is perpendicular to

→a –

→b .

5. Let the position vectors of A, B, C, D be j7–i6 ,

k4–j29–i16 , k6–j3 and k10j5i2 ++

respectively. Then

AB = OB – OA

= ( k4–j29–i16 ) – ( j7–i6 )

= k4–j22–i10

AC = ( k6–j3 ) – ( j7–i6 )

= – k6–j10i6 +

AD = ( k10j5i2 ++ ) – ( j7–i6 )

= k10j12i4– ++

Now ∆ = 10124–

6–106–4–22–10

Operate R1 → R1 + R2 + R3

= 10124–

6–106–000

= 0

⇒ The points A, B, C and D are coplanar.

6. Put x = a sin θ

= tan–1

θθ

cosasina

= tan–1 (tan θ) = θ

= sin–1

ax

7. g (f(x)) = |5 f(x) – 2|

= | 5 | x | – 2| =

<≥

0x|,2–x5–|0x|,2–x5|

8. ( )2443 BA ×× ⋅ . C2×3

= (X)3×2 . C2×3 = (Y)3×3 ∴ final order = 3 × 3 9. x + 2 = – (2 × –3) ⇒ 3x = 1 ⇒ x = 1/3

10. 2 – 20 = 2x2 – 24 ⇒ 2x2 = 6 ⇒ x = ± 3

Section B

11. Let A and B denote the two events respectively

P( A ) = 25

5+

= 75 , P(B) =

566+

= 116

P(A) = 1 – 75 =

72 , P( B ) = 1 –

116 =

115

P(At least one of A and B happens) = 1 – P(none of A and B happens) = 1 – P ( A ∩ B )

= 1 – 75 ×

115 =

7752


OR Let A → total of 8 in first throw B → total of 8 in 2nd throw Number of exhaustive cases when a pair of dice is

thrown = 6 × 6 = 36 Cases favourable to a total of 8 in each throw are

(2, 6), (6, 2), (3, 5), (5, 3), (4, 4) Their number = 5 P(A) = P(B) = 5/36

P(A and B) = P(A). P(B) = 365 ×

365 =

129625

12. dxdy + 2x1

x2+

y = 2

2

x1x4

+

Here P = 2x1x2

+, Q = 2

2

x1x4

+

I.F. = ∫

+dx

x1x2

2e = )x1(n 2e +l = 1 + x2

Solution is : y (1 + x2) = ∫ + 2

2

x1x4 (1 + x2) dx + c

y(1+x2) = 3x4 3

+ c

13. ∫

++ dx

)x2(1–)x2(e 2

x

∫

++

+dx

)x2(1–

x21e 2

x

↓ ↓ f(x) f ′(x)

= cx2

1.ex ++

OR

dxxx

x∫

−− 222 )(1

2

Let x2 = t. Then, d(x2) = dt ⇒ 2x dx = dt ⇒ dx = x

dt2

∴ I = ∫−− 21 tt

dt = ∫−+− 1 2 tt

dt

= ∫−−++− 1

41

41 2 tt

dt

⇒ Ι = ∫

−

+−

45

21 2

t

dt = ∫

+−

2

21

45 t

dt

= ∫

+−

22

21

25 t

dt

⇒ I = sin–1

+

2/52/1t + C

= sin–1

+

512t + C

= sin–1

+

512 2x + C

14. Divide by cos2x in Nr & Dr both

∫ +dx

)1xtan2)(2–x(tanxsec2

Let tan x = t sec2x dx = dt

∫ + )1t2)(2–t(dt

∫ + )2/1t)(2–t(dt

21

∫

+dt

2/1t1–

2–t1

52.

21

c)2/1t(

)2–t(log51

+

+

15. Position vector of A is

k2ji3 ++ ⇒ A (3, 1, 2)

Position vector of B is

k4j2–i − ⇒ B(1, –2, – 4)

d.r.s of the line AB are 3, –1, 1 – (– 2), 2 – (– 4) i.e. 2, 3, 6 ∴ Eq. of the required plane through B and

perpendicular line AB is 2 (x – 1) + 3 (y + 2) + 6 (z + 4) = 0 ⇒ 2x + 3y + 6z + 28 = 0

16. →a = kj3–i + ,

→b = kj–i + and

→c = k–j–i2

→b ×

→c =

1–1–211–1kji

= kj3i2 ++

∴ L.H.S = →a × (

→b ×

→c )


= 13213–1kji

= k9ji6– ++

Next →a .

→c = ( kj3–i + ).( k–j–i2 )

= 2 + 3 – 1 = 4

→a .

→c = ( kj3–i + ). ( kj–i + )

= 1 + 3 + 1 = 5

∴ R.H.S = (→a .

→c )

→b – (

→a .

→b )

→c

= 4( kj–i + ) – 5( k–j–i2 )

= kji6– ++ = L.H.S.

17. dxdy =

2–x323 = 2 (slope of line)

⇒ 43 = 2–x3 ⇒ 9/16 = 2x – 2 ⇒ x =

4841

y = 3/4 eqn of tangent

(y –43 ) = 2

4841–x ⇒ 48 x – 24 y = 23

18. Let y = y

x ⇒ y = xy/2

log y = xlog2y

dxdy

xlog

21–

y1 =

x2y

dxdy =

)xlogy–2(xy2

OR

Let y = tan–1

−++

−−+22

22

11

11

xx

xx , z = cos–1 x2

put x2 = cos 2θ

y = tan–1

θ−+θ+

θ−−θ+

2cos12cos12cos12cos1

y = tan–1

θ+θθ−θ

sincossincos

dxdz =

41

1

x−

− .2x

y = tan–1 [tan (π/4 –θ)]

y = 4π –

21 cos–1 x2

dxdy =

21−

−

− xx

2.1

14

dxdy =

41 x

x

−

∴ dzdy =

dxdzdxdy

// =

4

4

1/2

1/

xx

xx

−−

− = 21−

19 Taking log y log x = x – y

y = xlog1

x+

⇒ dxdy = 2)xlog1(

)x/1(x–1).xlog1(+

+

dxdy = 2)xlog1(

xlog+

20. Q f(π/4) = f(π/4+)

= 0h

lim→

)h4/(2cot

)]h4/(–4/tan[+π

+ππ

= 0h

lim→ h2tan–

htan– = 0h

lim→ h2

h2h2tan

hh

htan

= 1/2

21. Let y = f(x) y = 1 + αx

x = α

1–y ∴ f–1(x) = α

1–x

Now f(x) = f–1(x)

1 + αx = α

1–x

Compose 1 = –1/α, α = 1/α

α = – 1 α = ± 1 ,

α = – 1

22. L.H.S = 4xx8x10

xx4x5xxx

+ +

4xx8y8xx4y4xxy

+

= x2

4x810x45x11

++ y.x

x4x8y8xx4y4x11

+

= x2

+4810045011

x810x45x11

+ 0


= x2

+ 0172131110

x + 0 = x3.(7 – 6) = x3

OR

xyzzxyyzx

111

= xyz1

xyzzzxyzyyxyzxx

2

2

2

= xyzxyz

1zz1yy1xx

2

2

2

C1 ↔ C3

⇒ 2

2

2

111

zzyyyx

R1 → R1 – R2, R2 → R2 – R3

= 2

22

22

100

zzzyzyyxyx

−−−−

= (x – y) (y – z) 21

1010

zzzyyx

++

= (x – y) (y – z) zyyx

++

11

= (x – y) (y – z) [(y + z) – (x + y)] = (x – y) (y – z) (z – x)

Section C

23. The ball transferred from the first bag to the second

bag can either be white or black. Case I : When white ball is transferred from the Ist

bag to the second.

Probability of getting white ball from Ist bag = 45

5+

= 95 .

Now, second bag contains 8 white and 9 black balls. The probability of drawing a white ball from the

second bag = 98

8+

= 178 .

∴ The probability of both these events taking place

together is = 95 ×

1718 =

15340

Case II : When a black ball is transferred from the Ist bag to the second probability of getting black ball

from Ist bag = 45

4+

= 94 .

The probability of drawing a white ball from the

second bag is 107

7+

= 177

Q The probability of both these events taking place

together is = 94 ×

177 =

15328

Hence, the required probability

= 15340 +

15328 =

15368 =

94 .

24.

(0, 0) (1, 0)

A(1/2, 2/3 )

(2, 0)

required are

A =

+ ∫∫ dxx–1dx)1–x(–12

1

2/1

22/1

0

2

A = 2

+

2/1

0

1–2 )1–x(sin21)1–x(–1)1–x(

21

++

1

2/1

1–2 xsin21x–1x

21 =

32π –

23

25. Ist Part

I = ∫π

+0

xcos–xcos

xcos

eee .... (1)

By property

I = ∫π

+0

xcosxcos–

xcos–

eee .... (2)

By property

π=∫∫

ππ

00

dx)x–(fdx)x(f

eqn (1) + (2)

2 I = ∫π

++

0xcos–xcos

xcos–xcos

eeee dx ⇒ I = π/2

2nd Part

I = ∫∫π

π

π

π

π+4/

4/–

4/

4/–x2cos–2

dx4/dxx2cos–2

dx


I1 I2

Since I1 is an odd function therefore I1 = 0

I = ∫π

π

+

π4/

4/–2

2

xtan1xtan–1–2

dx4/

= ∫π

π+

π4/

4/–22

2

)xtan3(1xsec4/

= ∫π

+π

4/

022

2

)xtan3(1xsec4/2 [Q f(x) is even function]

Let 3 tan x = t ⇒ sec2x dx = 3

dt

= ∫ +π

3

02t13/dt

2 = ( ) 3

01– ttan32/π =

36

2π

26. The given lines are

→r = ( k3j2–i + ) + t ( k2–ji– + )

and →r = ( k–j–i ) + s ( k2–j2i + )

Here →

1a = k3j2–i + ; →

1b = k2–ji– +

→

2a = k–j–i ; →

2b = k2–j2i +

∴ →

1b × →

2b = 2–212–11–

kji= k3–j4–i2

The S.D. between the gives lines

= |bb|

)a–a).(bb(

21

2121

××

→→→→

= |k3–j4–i2|

)k4–j).(k3–j4–i2(

= 222 342

124

++

+ = 298

27. Part Ist Let x kg and y kg. of fertilizer A and B be mixed by

the farmer. Then LPP is Minimize : C = 5x + 8y subject to the constraints

x10010 + y

1005 ≥ 7 ⇔ 2x + y ≥ 140

x100

6 + y10010

≥ 7 ⇔ 3x + 5y ≥ 350

x ≥ 0, y ≥ 0

y

Px

A C O

D(0, 140)B

(1)

(50, 40)

0,

3350

(2)

Draw the lines 2x + y = 140 ..... (1) 3x + 5y = 350 ..... (2) The shaded unbounded region is the feasible region. These line meet at P(50, 40) Now value of c = 5x + 8y

at C

0,

3350 , is

31750 = 583

31

at P(50, 40) is 570 at B(0, 140) is 1120 ∴ The cost is minimum at P(50, 40) This occurs when 50 kg of type & fertilizer and 40

kg of type B fertilizer are mixed to the meet the requirement.

Part 2nd Let x and y represent the number of tables and chairs

respectively that the dealer sells and P be the profit of the dealer. Then the required LPP is .

Maximize: P = 50x + 15y Subject to the constraints 500x + 200 y ≤ 10,000 ⇔ 5x + 2y ≤ 100 x + y = 60 ....(1) 5x + 2y = 100 .... (2)

y

xO (1)

(2)

B(0, 60)

A(60, 0)

C (20, 0)

D(0, 50)

A shaded region OCD is the feasible region. Now P = 50x + 15y At O, value of P is O At C, value of P is 50 × 20 = 1,000 At D, value of P is 15 × 50 = 750 ∴ Profit is maximum at C (20, 0)


28.

x x

y

∴ x → side of square y → rad. of circle ∴ 4x + 2πy = 36 ..... (1) Now area A = x2 + πy2

A = x2 + π2x2–18

π [From (1)]

dxdA = 2x –

π4 (18 – 2x)

2

2

dxAd = 2 + 8/π > 0 ⇒ minm area

Form minm areadxdA = 0

⇒ x = 4

36+π

From (1) , y = 4

18+π

Length of one piece = 4x = 4

144+π

Length of other piece = 2πy = 4

36+π

π

29. Let x1 = u ;

y1 = v;

z1 = w

Given equation can be written in form

20–9656–4

1032

wvu

=

214

A ⋅ X = B X = A–1 ⋅ B

X = |A|

)A(Adj . B, |A| = 1200

X = 1200

214

.24–072

30100–1107515075

X =

wvu

=

1200240400600

=

5/13/12/1

u = 1/2 ⇒ x = 2 v = 1/3 ⇒ y = 3 w = 1/5 ⇒ z = 5

At a Glance

Some Important Practical Units 1. Par sec : It is the largest practical unit of

distance.

1 par sec = 3.26 light year

2. X-ray unit : It is the unit of length.

1 X-ray unit = 10–13 m

3. Slug : It is the unit of mass.

1 slug = 14.59 kg

4. Chandra Shekhar limit : It is the largest practical unit of mass.

1 Chandra Shekhar limit = 1.4 × Solar mass

5. Shake : It is the unit of time.

1 Shake = 10–6 second

6. Barn : It is the unit of area.

1 barn = 10–28 m2

7. Cusec : It is the unit of water flow.

1 cusec = 1 cubic foot per second flow

8. Match No. : This unit is used to express velocity of supersonic jets.

1 match no. = velocity of sound

= 332 m/sec.

9. Knot : This unit is used to express velocity of ships in water.

1 knot = 1.852 km/hour

10. Rutherford : It is the unit of radioactivity.

1 rutherford (rd) = 1 × 106 disintegrations/sec

11. Dalton : It is the unit of mass.

1 dalton = 121 mass of C12 = 931 MeV

= 1 a.m.u.

12. Curie : It is the unit of radioactivity.

1 curie = 3.7 × 1010 disintegration / sec


XtraEdge Test Series ANSWER KEY

PHYSICS Ques 1 2 3 4 5 6 7 8 9 10 A n s C A A A A C C D A B Ques 11 12 13 14 15 16 17 18 19 A n s B A D B B A B A D

20 A → P,R,S B → P,R,S C → P,Q D → P,Q 21 A → R,S B → P C → R D → S Column

Matching 22 A → P,Q,R,S B → P,R C → Q,R D → R

CHEMISTRY

Ques 1 2 3 4 5 6 7 8 9 10 A n s B A A C B B C A A B Ques 11 12 13 14 15 16 17 18 19 A n s C A D B D A C C C

20 A → P,S B → S C → Q D →RP 21 A → S B → P C → Q D → R Column

Matching 22 A → P,Q B → P C → P,Q,R D → S

MATHEMATICS

Ques 1 2 3 4 5 6 7 8 9 10 A n s B D C B B C A D C D Ques 11 12 13 14 15 16 17 18 19 A n s C D A A C B B C A

20 A → S B → Q C → P D → R 21 A → S B → R C → P D → P Column

Matching 22 A → R B → S C → Q D → P

PHYSICS

Ques 1 2 3 4 5 6 7 8 9 10 A n s C D A D A B A A D C Ques 11 12 13 14 15 16 17 18 19 A n s A C B C B C C D B

20 A → R B → P,S C → P,Q D → Q 21 A → Q B → P,R,S C → P,R D → S Column

Matching 22 A → P,Q B → Q C → R,S D → S

CHEMISTRY

Ques 1 2 3 4 5 6 7 8 9 10 A n s A D C D A C B A A C Ques 11 12 13 14 15 16 17 18 19 A n s A C C A A A B C A

20 A → Q,R B → P,Q,R,S C → P,S D → P,Q,R 21 A → R B → S C → Q D → P Column

Matching 22 A → P,S B → P C → Q D → P,R,S

MATHEMATICS

Ques 1 2 3 4 5 6 7 8 9 10 A n s A B B B A B A B D D Ques 11 12 13 14 15 16 17 18 19 A n s C D D B C A A B D

20 A → Q B → S C → R D → P 21 A → R B → P C → S D → Q Column

Matching 22 A → R B → S C → P D → Q

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XtraEdge for IIT-JEE 2 FEBRUARY 2012 Volume-7 Issue-8 February, 2012 (Monthly Magazine) NEXT MONTHS...

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