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JEE Main Exam 2016

(Paper & Solution)

Code – E Date: 3-04-2016

Important Instructions:

1. Immediately fill in the particulars on this page of the Test Booklet with only Blue/Black Ball Point Pen

provided by the Board.

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take out the Answer Sheet and fill in the particulars carefully.

3. The test is of 3 hours duration.

4. The Test Booklet consists of 90 questions. The maximum marks are 360.

5. There are three parts in the question paper A, B, C consisting of Physics, Chemistry and Mathematics

having 30 questions in each part of equal weightage. Each question is allotted 4 (four) marks for each

correct response.

6. Candidates will be awarded marks as stated above in instruction No. 5 for correct response of each

question. 1/4(one fourth) marks will be deducted for indicating incorrect response of each question.

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Part A – Physics 1. A student measures the time period of 100 oscillations of a simple pendulum four times. The data set is

90 s, 91 s, 95 s and 92 s. If the minimum division in the measuring clock is 1 s, then the reported mean

time should be:

(a) 92±2 s (b) 92 ± 5.0 s (c) 92 ± 1.8 s (d) 92 ± 3 s

Ans. (a)

Solution:

Time period of 100 oscillations

90 sec, 91 sec, 95 sec, 92 sec

Mean value of time = 90 91 95 92

4

mt 92 sec

Absolute error

1 m 1r t t 2 sec

2 m 2r t t 1 sec

3 m 3r t t 3 sec

4 m 4r t t 0 sec

mean absolute error

mean

2 1 3 0t

4

meant 1.5 sec 2s

mean lime (92 ± 2 sec)

2. A particle of mass m is moving along the side of a square of side ‘a’, with a uniform speed ν in the x-y

plane as shown in the figure:

Which of the following statements is false for the angular momentum L about the origin?

(a) mν

L R k2

when the particle is moving from A to B.

(b) R

L mν a k2

when the particle is moving from C to D.




(c) R

L mν a k2

when the particle is moving from B to C.

(d) mν

L R k2

when the particle is moving from D to A.

Ans. (a, d)

Solution:

When particle move along AB

L mvr k

RL mg k

2

when particle move along C to D

L mvr k

RL mv a k

2

when particle move from B to C

RL mvr k mv a k

2

when particle move from D to A

RL mvr k mv k

2

so (b) and (4) are false

3. A point particle of mass m, moves along the uniformly rough track PQR as shown in the figure. The

coefficient of friction, between the particle and the rough track equals μ . The particle is released, from

rest, from the point P and it comes to rest at a point R. The energies, lost by the ball, over the parts, PQ

and QR, of the track, are equal to each other, and no energy is lost when particle changes direction

from PQ to QR.

The values of the coefficient of friction μ and the distance x (=QR), are, respectively close to:

(a) 0.2 and 6.5 m (b) 0.2 and 3.5 m (c) 0.29 and 3.5 m (d) 0.2 and 6.5 m

Ans. (c)

Solution:




From work-energy theorem

1 2g f fW W W K

mgh 2W 0 1 2f fW W W

mghW

2

and if

mghW

2

0 mghμmg cos 30 4

2

μ 0.29

and 2f

mghW

2

mghμmgx

2

hx 3.5 m

2μ

4. A person trying to lose weight by burning fat lifts a mass of 10kg upto a height of 1m 1000 limes.

Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he

use up considering the work done only when the weight is lifted up? Fat supplies 3.8 107 J of energy

per kg which is converted to mechanical energy with a 20% efficiency rate. Take g = 9.8 ms–2:

(a) 2.45 10–3 kg (b) 6.45 10–3kg (c) 9.89 10–3kg (d) 12.89 10–3 kg

Ans. (d)

Solution:

PE= 1000 mgh

= 1000 10 9.8 1

= 98000 J

mechanical energy per unit mass = 723.8 10

100

= 7.6 105 J/kg

Fat bum = 5

9800kg

7.6 10

= 12.89 10–3 kg

5. A roller is made by joining together two cones at their vertices O. It is kept on two rails AB and CD

which are placed asymmetrically (see figure), with its axis perpendicular to CD and its centre O at the

centre of line joining AB and CD (see figure). It is given a light push so that it starts rolling with its

centre O moving parallel to CD in the direction shown. As it moves, the roller will tend to:




(a) turn left. (b) turn right.

(c) go straight. (d). turn left and right alternately.

Ans. (a)

Solution:

When this cone proceeds then on AB rail the inner point-will touch the rail whose velocity in forward

direction which slip forward and friction acts in backward direction. Due to this friction the system

turn left.

6. A satellite is revolving in a circular orbit at a height ‘h’ from the earth's surface (radius of earth

R; h <<R). The minimum increases in its orbital velocity required, so that the satellite could escape

from the earth's gravitational field, is close to: (Neglect the effect of atmosphere.)

(a) 2gR (b) gR (c) gR /2 (d) gR( 2 1)

Ans. (d)

Solution:

Orbital velocity of a satellite revolving in a circular orbit at a height (h <<R)

0v Rg

and the velocity required to escape is

ev 2Rg

So increase in velocity

e 0v v v 2 1 Rg

7. A pendulum clock loses 12 s a day if the temperature is 400C and gains 4 s a day if the temperature is

200C. The temperature at which the clock will show correct time, and the co-efficient of linear

expansion (α ) of the metal of the pendulum shaft are respectively:

(a) 0 5 o25 C; α 1.85 10 / C (b) 0 4 o60 C; α 1.85 10 / C

(c) 0 3 o30 C; α 1.85 10 / C (d) 0 2 o55 C; α 1.85 10 / C Ans. (a)

Solution:

Let at temperatureθ , clock gives correct time

1T α θ T,

2

T = 1 day = 86400s

1

12 α 40 θ T2

…(i)

1

4 α θ 20 T2

…(ii)




12 40 θ

4 θ 20

3θ 60 40 θ 0θ 25 C

from (ii) 4 = 1α

2 (25 – 20) 86400

5 o8α 1.85 10 / C

5 86400

8. An ideal gas undergoes a quasi-static, reversible process in which its molar heat capacity C remains

constant. If during this process the relation of pressure P and volume V is given by PVn = constant, then

n is given by (Here CP and CV are molar specific heat at constant volume, respectively):

(a) P

V

Cn

C (b) P

V

C Cn

C C

(c)

P

V

C Cn

C C

(d) V

P

C Cn

C C

Ans. (b)

Solution:

V

PdV R RC C

ndT γ 1 1 n

P

V

R RC

C 1 n1

C

on solving

P

V

C Cn

C C

9. 'n' moles of an ideal gas undergoes a process A B as shown in the figure. The maximum

temperature of the gas during the process will be:

(a) 0 09P V

4nR (b) 0 03P V

2nR (c) 0 09P V

2nR (d) 0 09P V

nR

Ans. (a)

Solution:




0 00 0

0 0

P 2PP 2P V V

2V V

00 0

0

PP 2P V P

V

00

0

PP V 3P

V

00

0

PnRTV 3P

V V

200

0

P1T V 3P V

nR V

for max

dTT , 0

dV

00

0

2P1V 3P 0

nR V

0

3V V

2

2

0max 0 0 0

0

P1 3 3T V 3P V

nR V 2 2

= 0 00 0 0 0

9P V1 9 9P V P V

nR 4 2 4nR

10. A particle performs simple harmonic motion with amplitude A. Its speed is trebled at the instant that it

is at a distance 2A

3 from equilibrium position. The new amplitude of the motion is:

(a) A

413

(b) 3A (c) A 3 (d) 7A

3

Ans. (d)

Solution:

Velocity at any position is given by

2 2v ω A x

at 2A

x3




2

2ω

4A 5v ω A v A

9 3

if speed at x = 2A

3 is tripled i.e.

v' 3v 5 Aω

So new amplitude A' is 2

2 4Av' ω A'

3

22 4A

5 Aω ω A'9

7AA'

3

11. A uniform string of length 20m is suspended from a rigid support. A short wave pulse is introduced at

its lowest end. It starts moving up the string. The time taken to reach the support is: (take g = 10 ms–2)

(a) 2π 2 s (b) 2s (c) 2 2 s (d) 2 s Ans. (c)

Solution: Let m mass/length

T = (xm)g

TV

m

= xm g

m

= gx

VdV da gx gx

dx dx

= 1

gx g2 x

ga

2 = constant

21at

2

2 2 20t 2 2 2s

a g /2 10

12. The region between two concentric spheres of radii 'a’ and 'b', respectively (see figure), has volume

charge density A

ρ ,r

where A is a constant and r is the distance from the centre. At the centre of the

spheres is a point charge Q. The value of A such that the electric field in the region between the spheres

will be constant, is




(a) 2

Q

2πa (b)

2 2

Q

2π(b a ) (c)

2 2

2Q

π(a b ) (d)

2

2Q

πan

Ans. (a)

Solution:

in

0

qE. dS

r

2 r a

0

Q ρdVE. 4πr

r2

2 a

0

AQ 4πr dr

rE. 4πr

2 2

20

1 r aE Q 4πA

4π r 2

2 2

20

1E [Q 2πA r a ]

4π r

2 2 2

2 2 2 20 0 0 0 0

A r aQ Q A AE

4π r 2 r 4π r 2 2 r

E = constant

E Independent of r

2

2 20 0

Q Aa

4π r 2 r

2

QA

2πa

13. A combination of capacitors is set up as shown in the figure. The magnitude of the electric field, due to

a point charge Q (having a charge equal to the sum of the charges on the 4μF and 9 μF capacitors), at a

point distant 30 m from it, would equal

(a) 240N/C (b) 360N/C (c) 420N/C (d) 480 N/C




Ans. (c)

Solution:

eq

4 12q C V 8 24μC

4 12

1

9q 24μC 18μC

9 3

1Q q q 24 18 42μC

9 6

22

KQ 9 10 42 10E 420 N /C

r 30

14. The temperature dependence of resistances of Cu and undoped Si in the temperature range 300-400 K,

is best described by

(a) Linear increase for Cu, linear increase for Si

(b) Linear increase for Cu, exponential increase for Si

(c) Linear increase for Cu, exponential decrease for Si

(d) Linear decrease for Cu, linear decrease for Si

Ans. (c)

Solution:

Cu is metal, its resistance increases linearly on increasing temperature, where as Si is semiconductor

which resistance decreases exponentially on increasing temperature.

15. Two identical wires A and B, each of length ‘l’ carry the same current I. Wire A is bent into a circle of

radius Rand wire B is bent to form a square of side 'a'. If BA and BB are the values of magnetic field at

the centres of the circle and square respectively, then the ratio A

B

B

Bis

(a) 2π

8 (b)

2π

16 2 (c)

2π

16 (d)

2π

8 2

Ans. (d)

Solution:

2πR 4a




R2π

a4

0 0 01

μ I μ I μ πIB

2R 22π

0 00 02

μ I μ IB [sin α sin β] 4 [sin 45 sin45 ] 4

a4πr 4π2

= 0 0 0μ I μ I μ I2 2 2 2 8 2

ππa π

4

02

1

02

μ πIB π

μ IB 8 28 2π

16. Hysteresis loops for two magnetic materials A and B are given below :

These materials are used 10 make magnets for electric generators, transformer core and

electromagnet core. Then it is proper to use

(a) A for electric generators and transformers

(b) A for electromagnets and B for electric generators

(c) A for transformers and B for electric generators

(d) B for electromagnets and transformers

Ans. (d)

Solution:

fig. A Hard magnetic material

fig. B soft magnetic material

Transformer core and electromagnets are made by soft magnetic materials, which are easy to

magnetise and demagnetise where as for electric generators we use hard magnetic materials.

17. An arc lamp requires a direct current of 10 A at 80 V to function. If it is connected to a 220 V (rms),

50 Hz AC supply, the series inductor needed for it to work is close to

(a) 80H (b) 0.08 H (c) 0.044 H (d) 0.065 H

Ans. (d)

Solution:




Resistance of lamp R = V 80

8i 10

RV 80

iR = 80

2 2L

220R 80

R X

2 2L

11R X R

4

2 2 2L

121R X R

16

2 2 2L

105 105X R 8

16 16

2LX 420

LX 20.5

2πfL 20.5

2 3.14 50L 20.5

L = 0.065 H

18. Arrange the following electromagnetic radiations per quantum in the order of increasing energy :

A : Blue light B : Yellow light

C : X-ray D : Radiowave

(a) D,B, A, C (b) A, B, D, C (c) C, A, B, D (d) B, A, D, C

Ans. (a)

Solution:

D, B, A, C

19. An observer looks at a distant tree of height 10 m with a telescope of magnifying power of 20. To the

observer the tree appears:

(a) 10 times taller (b) 10 times nearer (c) 20 times taller (d) 20 times nearer

Ans. (c)

Solution:

Magnifying power MP = β

α

(MP) = 20

β20

α




object is of height 10 m which is very faraway and when image also very faraway and then height of

image h.

h 10mβ : α

x x

β h

α 10

h20

10

h = 20 10 m

20 limes taller

20. The box of a pin hole camera, of length L, has a hole of radius a. It is assumed that when the hole is

illuminated by a parallel beam of light of wavelength λ , the spread of the spot (obtained on the

opposite wall of the camera) is the sum of its geometrical spread and the spread due to diffraction.

The spot would then have its minimum size (say bmin) when:

(a) 2 2

min

λ 2λa and b

L L

(b)

2

min

2λa λL and b

L

(c) mina λL and b 4λL (d) 2

min

λa and b 4λL

L

Ans. (c)

Solution: Width of CM = 2Dλ

a

= 2Lλ 2Lλ

a λL

= 4λL

21. Radiation of wavelength λ , is incident on a photocell. The fastest emitted electron has speed v. If the

wavelength is changed to 3λ

4, the speed of the fastest emitted electron will be:

(a)

1

24v

3

(b)

1

24v

3

(c)

1

24v

3

(d)

1

23v

4

Ans. (a)

Solution:

21 hcmv

2 λ …(i)

21 hcm(v')

3λ2

4

21 4 hcm(v')

2 3 λ

2 21 4 1m(v') mv

2 3 2

2 21 4 1

m v' mv2 3 2 3




2 21 4 1

m v' mv2 3 2

4

v' v3

22. Half-lives of two radioactive elements A and B are 20 minutes and 40 minutes, respectively. Initially,

the samples have equal number of nuclei. After 80 minutes, the ratio of decayed numbers of A and B

nuclei will be:

(a) 1 : 16 (b) 4: 1 (c) 1 : 4 (4) 5 :4

Ans. (d)

Solution:

1 1

2 2

A T 20 min B T 40min.

No No

0N

2 0N

2

0N

4 0N

4

No of decayed nuclei

0N

8 = 0 0

0

N 3NN

4 3

0N

16

No of decayed nuclei

= 0 00

N 15NN

16 16

Ratio of decayed Nuclei = 0 015N 3N 5/

16 4 4

23. If a, b, c, d are inputs lo a gale and x is its output, then, as per the following lime graph, the gate is:

(a) NOT (b) AND (c) OR (d) NAND

Ans. (c)




Solution:

If one of the inputs is high then output will be high in OR gate.

24. Choose the correct statement:

(a) In amplitude modulation the amplitude of the high frequency carrier wave is made lo vary in

proportion to the amplitude of the audio signal

(b) In amplitude modulation the frequency of the high frequency carrier wave is made lo vary in


(c) In frequency modulation the amplitude of the high frequency carrier wave is made lo vary in


(d) In frequency modulation the amplitude of the high frequency carrier wave is made to vary in

proportion to the frequency of the audio signal

Ans. (a)

Solution:

(1) is correct but language of this option is not up to the mark.

25. A screw gauge with a pitch of 0.5 mm and a circular scale with 50 divisions is used to measure the

thickness of a thin sheet of Aluminium. Before starting the measurement, it is found that when the two

jaws of the screw gauge are brought in contact, the 45th division coincides with the main scale line and

that the zero of the main scale is barely visible. What is the thickness of the sheet if the main scale

reading is 0.5 mm and the 25th division coincides the main scale line?

(a) 0.75 mm (b) 0.80 mm (c) 0.70 mm (d) 0.50 mm

Ans. (b)

Solution:

pitchLC

no. of div. on circular scale

= 0.5 mm

0.01 mm50

Zero error e = N LC = 5 0.01

= 0.05 mm

True reading = observed reading + e

= LSR + CSR + e

= 0.5 mm+ 25 0.01 +0.05

= 0.75 + 0.05 = 0.80 mm

26. A pipe open at both ends has a fundamental frequency f in air. The pipe is dipped vertically in water so

that half of it is in water. The fundamental frequency of the air column is now :

(a) f

2 (b)

3f

4 (c) 2f (d) f

Ans. (d)




Solution:

27. A galvanometer having a coil resistance of 100 gives a full scale deflection, when a current of 1 mA

is passed through it. The value of the resistance which can convert this galvanometer into ammeter

giving a full scale deflection for a current of 10 A, is:

(a) 0.01 (b) 2 (c) 0.1 (d) 3

Ans. (a)

Solution:

G = 100 3

gi 1 mA 10 A

i 10A

s ?

g

g

i Gs

i i

= 3

3

10 100

10 10

= 210

28. In an experiment for determination of refractive index of glass of a prism by i δ, plot, it was found

that a ray incident at angle 35°, suffers a deviation of 40° and that it emerges at angle 79°. In that case

which of the following is closest 10 the maximum possible value of the refractive index ?

(1)1.5 (2) 1.6 (3) 1.7 (4) 1.8

Ans. (a)

Solution:

i = 350 0δ 40

0e 79

A = ?

δ i e A

A = 350 + 790 – 40°

A = 740




mδ Asin

2μ

Asin

2

0m

0

74 δsin

2μ

sin 37

0 mδ5μ sin 37

3 2

0 05 5μ sin 57 sin 60

3 3

5μ μ 1.44

2 3

29. Identify the semiconductor devices whose characteristics are given below, in the order (a), (b), (c), (d):

(a) Simple diode, Zener diode, Solar cell, Light dependent resistance

(b) Zener diode, Simple diode, Light dependent resistance, Solar cell

(c) Solar cell. Light dependent resistance, Zener diode, Simple diode

(d) Zener diode, Solar cell, Simple diode, Light dependent resistance

Ans. (a)

Solution: By theory

30. For a common emitter configuration, if α and β have their usual meanings, the incorrect relationship

between α and β is:

(a) 1 1

1α β (b)

βα

1 β

(c)

βα

1 β

(d)

2

2

βα

1 β

Ans. [2, 4]

Solution:




1 11

α β

or β

α1 β

Part B-CHEMISTRY

31. At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL air containing 20% O2 by

volume for combustion. After combustion the gaseous occupy 330 mL. Assuming that the water

formed is in liquid form and the volumes were measured at the same temperature and pressure, the

formula of the hydrocarbon is :

(a)C3H6 (b)C3H8 (3)C4H8 (d) C4H10

Ans. (Bonus)

Solution:

x y 2 2 2

y yC H g x O g xCO g H O

4 2

Initially 15 ml 75 ml 0 –

After reaction 0 0 15x –

As per problem:

15 75

y1x

4

So, y

x 54

...(i)

and 15x = 330 – 300

x = 2 ..(ii)

from eq. (i) & (ii)

y = 12

In this problem it seems lobe there is a printing error if volume of gaseous products after combustion

is 345 ml instead of 330 ml.

15x = 345 – 300

x = 3

& x +y

4= 5

y = 8

C3H8 Ans.

32. Two closed bulbs of equal volume (V) containing an ideal gas initially at pressure pi and temperature

T1 are connected through a narrow tube of negligible volume as shown in the figure below. The

temperature of one of the bulbs is then raised to T2. The final pressure pf is:

(a) 1 2i

1 2

T Tp

T T

(b) 1

i

1 2

T2p

T T

(c)

2i

1 2

T2p

T T

(d)

1 2i

1 2

T T2p

T T




Ans. (c)

Solution:

Initial condition

So, initial mole in both the bulb is same, say (n)

So ,n = i

i

p V

R T

Final condition

So, n1 + n2 = 2n [Total mole must be same]

f f i

1 2 1

P V P V PV2

RT RT RT

if

1 2 1

2P1 1P

T T T

2f i

1 2

TP 2P

T T

33. A stream of electrons from a healed filament was passed between two charged plates kept at a

potential difference V esu. If e and in are charge and mass of an electron, respectively, then the value of

h / λ , (where λ , is wavelength associated with electron wave) is given by:

(a) meV (2) 2meV (c) meV (d) 2meV

Ans. (d)

Solution:

hλ

p ..(1)

p = mv …(2)

K. E. = 21mv

2

2KEv

m …(3)

from eq. (1), (2) & (3)

hλ

2KEm

m

hλ

2mKE …(4)

If electron is accelerated by potential difference of V then its K.E. is given as follows

K.E. = eV …(5)




from eq. (4) and (5)

hλ

2meV

h2meV

λ

34. The species in which the N atom is in a stale of sp hybridization is :

(a) NO 2 (b) NO 2

(c) NO 3 (d) NO2

Ans. (a)

Solution:

In 2NO , 'N' has zero lone pair and two bond pairs (ςbond). Hence, hybridization is sp.

35. The heats of combustion of carbon and carbon monoxide are – 393.5 and 283.5 kJ mol–1, respectively.

The heat of formation (in kJ) of carbon monoxide per mole is:

(a) 110.5 (b) 676.5 (c) – 676.5 (d) – 110.5

Ans. (d)

Solution:

Given

2 2 1C s O g CO g , H 393.5 kJ / mol

2 2 2

1CO g O g CO g , H 283.5 kJ / mol

2

Object reaction,

2

1C s O g CO g H ?

2

eq (i) – eq (ii)

1 2H H H

= –393.5 + 283.5

= –110 KJ/mol.

36. 18 g glucose (C6H12O6) is added to 178.2 g water. The vapour pressure of water (in torr) for this

aqueous solution is:

(a) 7.6 (b) 76.0 (c) 752.4 (d) 759.0

Ans. (3)

Solution:

We assume the solution is prepared at boiling point temperature of water

oP = 760 mm Hg o

so

P P n

P n N

s760 P 0.1

760 10

sP1 0.01

760

sP 752.4 mm Hg or torr.




37. The equilibrium constant at 298 K for a reaction A + B C + D is 100. If the initial concentration of

all the four species were 1 M each, then equilibrium concentration of D (in mol L–1) will be :

(a) 0.182 (b) 0.818 (c) 1.818 (d)1.182

Ans. (3)

Solution:

A + B C + D

t = 0 1 1 1 1

eqm: 1 – x 1 – x 1 + x 1 + x

At the given state Q = 1; therefore Q < K

reaction will move in forward direction.

So, 100 =

1 x 1 x

1 x 1 x

1 x10

1 x

10 – 10x = 1 + x

11x = 9

9x

11

So, [D] = 1 + x = 1 + 9

11

= 20

11 = 1.818 M

38. Galvanization is applying a coaling of:

(a) Pb (b)Cr (c)Cu (d) Zn

Ans. (4)

Solution:

In Galvanization Zn coating is applied an iron. 0 0OXZn OXFeE E

39. Decomposition of H2O2 follows a first order read ion. In fifty minutes the concentration of H2O2

decreases from 0.5 to 0.125 M in one such decomposition. When the concentration of H2O2 reaches

0.05 M, the rate of formation of O2 will be:

(a) 6.93 10–2 mol min–1 (b) 6.93 10–4 mol min–1

(c) 2.66 L min–1 at STP (d) 1.34 10–2 mol min–1

Ans. (2)

Solution:

Here solution is given assuming unit of rate is the option to be mol L–1 min–1.

For 1st order reaction,

1 0.5K ln

50 0.125

1ln 4K min

50




Now, 12 2R K[H O ]

1ln 4R 0.05 M min

50

3 1 1R 2 0.693 10 lit min

And, 2 2 2 2

1H O H O O

2

So rate of reaction, R = 2O 2R (Rate of formation of O )

1

2

Or 2

31 1

O

1 2 0.693 10R R mol Lit min

2 2

2

4 1 1OR 6.93 10 mol L min

40. For a linear plot of log (x/m) versus log p in a Freundlich adsorption isotherm, which of the following

statements is correct ? (k and n are constants)

(a) Both k and 1/n appear in the slope term (b) 1/n appears as the intercept

(c) Only 1/n appears as the slope (d) log (1/n) appears as the intercept

Ans. (c)

Solution:

1/nxKP

m

x 1log

m n log P + log K

Slope is 1

n

41. Which of the following atoms has the highest first ionization energy ?

(a) Rb (b)Na (c)K (d) Sc

Ans. (d)

Solution:

Order of first ionization energy : Sc > Na > K > Rb

42. Which one of the following ores is best concentrated by froth floatation method?

(a) Magnetite (b) Siderite (c) Galena (d) Malachite

Ans. (c)

Solution:

Froth floatation method is mainly used for sulphide ores

Magnetite Fe3O4

Siderite FeCO3

Galena PbS

Malachite Cu(OH)2CuCO3

43. Which one of the following statements about water in FALSE ?

(a) Water is oxidized to oxygen during photosynthesis

(b) Water can act both as an acid and as a base

(c) There is extensive intramolecular hydrogen bonding in the condensed phase




(d) Ice formed by heavy water sinks in normal water

Ans. (c)

Solution:

In water, there is inter molecular hydrogen bonding, not intramolecular hydrogen bonding.

44. The main oxides formed on combustion of Li, Na and K. in excess of air are, respectively:

(a) Li2O, Na2O and KO2 (b) LiO2, Na2O2 and K2O

(c) Li2O2, Na2O2 and KO2 (d) Li2O, Na2O2 and KO2

Ans. (4)

Solution:

22 excess

12Li O Li O

2

2 22 excess2Na O Na O

22 excessK O KO

45. The reaction of zinc with dilute and concentrated nitric acid, respectively, produces:

(a) N2O and NO2 (2) NO2 and NO (c) NO and N2O (d) NO2 and N2O

Ans. (a)

Solution:

Zn + HNO3(dil) Zn(NO3)2 + NH4NO3 + H2O

N2O

Zn + HNO3(conc.) Zn(NO3)2 + NO2 + H2O 46. The pair in which phosphorous atoms have a formal oxidation state of+3 is:

(a) Orthophosphorous and pyrophosphorous acids

(b) Pyrophosphorous and hypophosphoric acids

(c) Orthophosphorous and hypophosphoric acids

(d) Pyrophosphorous and pyrophosphoric acids

Ans. (a)

Solution:

Orthophosphorous acid is H3PO5 and Pyrophosphorous acid is H4P2O5. In both compounds, oxidation

state of ‘P' is +3

47. Which of the following compounds is metallic and ferromagnetic?

(a) TiO2 (b) CrO2 (c) VO2 (d) MnO2

Ans. (d)

Solution:

Ferromagnetic materials are Fe, Co, Ni, Gd, CrO2 etc.

48. The pair having the same magnetic moment is:

[At. No, : Cr = 24, Mn = 25, Fe = 26, Co = 27]

(a) 2 22 6 4[Cr(H O) ] and [C0Cl ] (b) 2 2

2 6 2 6[Cr(H O) ] and [Fe(H O) ]

(c) 2 22 6 2 6[Mn(H O) ] and [Cr(H O) ] (d) 2 2

4 2 6[CoCl ] and [Fe(H O) ]

Ans. (b)

Solution:




22 6[Cr(H O) ]

2 0 4Cr [Ar] 4s 3d

It has 4 unpaired electrons

22 6[Fe(H O) ]

2 0 6Fe [Ar] 4s 3d

It has 4 unpaired electrons

49. Which one of the following complexes shows optical isomerism?

(a) [Co(NH3)3Cl3] (b) cis[Co(en)2Cl2]Cl

(c) trans[Co(en)2Cl2]Cl (d) [Co(NH3)4Cl2]Cl

Ans. (2)

Solution:

cis-[Co(en)2Cl2]Cl shows optical isomerism. It's general case is M(aa)2b2

Mirror image is non-super imposable. Hence, this compound is optically active.

50. The concentration of fluoride, lead, nitrate and iron in a water sample from an underground lake was

found to be 1000 ppb, 40 ppb, 100 ppm and 0.2 ppm, respectively. This water is unsuitable for

drinking due to high concentration of:

(a) Fluoride (b) Lead (c) Nitrate (d) Iron

Ans. (c)

Solution:

Substance Upper limit

Fluoride 2 ppm or 2000 ppb

Lead 0.05 ppm or 50 ppb

Nitrate 50 ppm

Fe 0.2 ppm

In water sample, concentration of nitrate is higher than upper limit. Hence, this water is unsuitable for

drinking.

51. The distillation technique most suited for separating glycerol from spent-lye in the soap industry is:

(a) Simple distillation (b) Fractional distillation

(c) Steam distillation (d) Distillation under reduced pressure

Ans. (d)




Solution:

Distillation under reduced pressure is the technique most suited for separation of Glycerol from

Spent-lye (Soap Solution obtain from basic hydrolysis of fats)

(Mentioned in NCERRT in chapter purification methods)

52. The product of the reaction given below is :

Ans. (b)

Solution:

53. The absolute configuration of

(a) (2R,3S) (b)(2S,3R) (c) (2S, 3S) (d) (2R, 3R)

Ans. (b)

Solution:




54. 2-chloro-2-melhylpentane on reaction with sodium methoxide in methanol yields :

(a) All of these (b) (a) and (c) (c) (c) only (d) (a)and(b)

Ans. (a)

Solution:

55. The reaction of propene with HOCl (Cl2 + H2O) proceeds through the intermediate:

(a) CH3–CH+–CH2–OH (b) CH3–CH+–CH2–Cl (c) CH3–CH(OH)–CH 2 (d)CH3–CHCl–CH 2

Ans. (b)

Solution:

56. In the Hofmann bromamide degradation reaction, the number of moles of NaOH and Br2 used per mole

of amine produced are:

(a) One mole of NaOH and one mole of Br2.

(b) Four moles of NaOH and two moles of Br2

(c) Two moles of NaOH and two moles of Br2.

(d) Four moles of NaOH and one mole of Br2.

Ans. (d)

Solution:




Total (d) moles of NaOH and 1 mole of Br2 is required for conversion from benzamide lo per mole of

amine

57. Which of the following statements about low density polythene is FALSE?

(a) Its synthesis requires high pressure.

(b) It is a poor conductor of electricity.

(c) Its synthesis requires dioxygen or a peroxide initiator as a catalyst.

(d) It is used in the manufacture of buckets, dust-bins etc.

Ans. (d)

Solution:

(a) It synthesis requires high pressure : TRUE

(b) Low density polythene is not poor conductor of electricity so it is: TRUE

(c) It synthesis requires dioxygen or a peroxide initiator as a catalyst: TRUE

(d) It is used in the manufacture of buckets, dust-bins etc : FALSE

58. Thiol group is present in:

(a) Cylosine (b) Cysline (c) Cysteine (d) Methionine

Ans. (c)

Solution:

Thiol (–SH) group is present in cysteine Amino Acid

Structure of Cysteine : -

59. Which of the following is an anionic detergent?

(a) Sodium stearate (b) Sodium lauryl sulphate

(c) Cetyltrimethyl ammonium bromide (d) Glyceryl oleate

Ans. (b)




Solution:

Sodium lauryl sulphate is an example of anionic detergent

60. The hottest region of Bunsen flame shown in the figure below is:

(a) regional (b) region 2 (c) region 3 (d) region 4

The very top of the inner cone is Hottest because when you have a definite inner cone, that's when

there is the most air flow through the Bunsen burner. When you have a definite inner cone, that's when

there is the most air flow through the Bunsen burner and the most heat and that is concentrated at the

tip of inner cone.

Part C – Mathematics

61. If 1

f x 2f 3x, x 0, and S {x R : f x f x }x

; then S:

(a) is an empty set (b) contains exactly one element

(c) contains exactly two elements (d) contains more than two elements

Ans. (c)

Solution:

Given that 1

f x 2f 3x, x 0x

…(1)

Replace 1

xx

then 1 3

2f x fx x

…(2)

from (1) & (2)

f(x) – x + 2

x

2

f x xx

f x f x (Given)

2 2

x xx x

22x 4




2x 2

x 2

So numbers of solutions of equal to 2.

62. A value of θ for which 2 3i sin θ

1 2i sin θ

is purely imaginary, is :

(a) π

3 (b)

π

6 (c) 1 3

sin4

(d) 1 1sin

3

Ans. (d)

Solution:

For purely imaginary, real part of the given complex number should be zero.

2 3i sin θ

Re 01 2i sin θ

2 3i sin θ 1 2i sinθR 0

1 2i sin θ 1 2i sinθ

2

2 2

2 6 sin θ i7 sin θ0

1 4 sin θ 1 4 sin θ

2

2

2 6 sin θ

1 4 sin θ

= 0

2 1sin θ

3

1

sin θ3

1 11 1θ sin , sin

3 3

63. The sum of all real values of x satisfying the equation 2x 4x 60

2x 5x 5 1

is:

(a) 3 (b) – 4 (c) 6 (d) 5

Ans. (a)

Solution:

Given equation 2x 4x 60

2x 5x 5 1

Case I : Any finite real number

1 = 1

(x2 – 5x + 5) = 1 and (x2 + 4x – 60) = any finite real number

(x2 – 5x + 4) = 0

(x – 1) (x – 4) = 0

x = 1, 4 x = 1, 4 satisfies given equation

Case II: even int eger

1 1

(x2 – 5x + 5) = – 1 and x2 + 4x – 60 = even integer

x2 – 5x + 6 = 0

x = 2, 3




Check (A) : At x = 2 x2 + 4x – 60 = – 48 = (even integer)

x = 2 is root

Check (B): At x = 3 x2 + 4x – 60 = – 39 even integer

Case III: (Any non zero real number)0= 1

(x2 – 5x + 5) is any non zero real number and x2 + 4x – 60 = 0

(x+ 10)(x – 6) = 0

x = 6, – 10 are the roots

overall roots are 1, 4, 2, – 10, 6

Hence sum of roots =1+4 + 2 + 6 – 10 = 3

64. If A = 5a b

3 2

and A adj A = AAT, then 5a + b is equal to

(a) – 1 (b) 5 (c) 4 (d) 13

Ans. (b)

Solution:

5a bA

3 2

A. Adj A = A. AT (Given)

2

5a b 5a 3A I . ,

3 2 b 2

{using property A. Adj A = A . I2}

2 21 0 25a b 15a 2b

10a 3b .0 1 15a 2b 13

15a – 2b = 0 …(1)

10a + 3b = 13 …(2)

25a2 + b2 = 13 …(3)

By solving eq. (1) and (2)

2a ; b 3

5

Check eq. (3)

Also At 2

a , b 35

2 2 425a b 25 9

25

= 13

satisfying equation (3)

2

5a b 5 3 2 3 55

65. The system of linear equations

x + λ y – z = 0

λ x – y – z = 0

x + y – λ z = 0

has a non-trivial solution for

(a) infinitely many values of λ (b) exactly one value of λ




(c) exactly two values of λ (d) exactly three values of λ

Ans. (d)

Solution:

For non-trivial solution, 0

1 λ 1

λ 1 1 0

1 1 λ

21 λ 1 λ λ 1 1 λ 1 0

3λ λ 0

2λ λ 1 0

λ 0, 1, 1

exactly three values of λ , exist for which the given system of linear equations has non-trivial

solution.

66. If all the words (with or without meaning) having five letters, formed using the letters of the word

SMALL and arranged as in a dictionary: then the position of the word SMALL is:

(a) 46th (b) 59th (c) 52nd (d) 58th

Ans. (d)

Solution:

Method-I: Short Trick

S M A L L

Method-II : Actual Method

(1) Words start with A

(2) Words start with M




(3) Words start with SA

(4) Words start with SA

(5) Words start with SL

(6) Words start with SMALL

67. If the number of terms in the expansion of n

2

2 41 , x 0,

x x

is 28, then the sum of the coefficients

of all the terms in this expansion, is

(a) 64 (b) 2187 (c)243 (d) 729

Ans. (d)

Solution:

Given expression n

2

2 41 , x 0,

x x

n

2

2n

x 2x 4, x 0

x

No. of terms in given polynomial = 2n + 1 = 28

27

n2

which is not possible

If multinomial theorem is used then

n

2n

2 2n

x 2x 42 41

x x x

then no. of terms = n 3 13 1C 28

n 2

2

n 2 n 1C 28 28

2!

n 2 n 1 8 7

n + 2 = 8

n = 6

sum of coefficient (put x = 1) = n 6

62 41 3 729

1 1

Although, application of multinomial theorem for one variable expression is conceptually wrong, hence

bonus must be awarded.




68. If the 2nd, 5th and 9th terms of a non-constant A. P. are in G.P., then the common ratio of this G.P. is

(a) 8

5 (b)

4

3 (c) 1 (d)

7

4

Ans. (b)

Solution:

Let a be the first term and d is the common difference of A. P.

Now 2nd , 5th and 9th terms of A.P. are

T2 = a + d

T5 = a + 4d

T9 = a + 8d

These terms are in G.P.

(a + 4d)2 = (a + d) (a + 8d)

a2 + 16d2 + 8ad = a2 + 9ad + 8d2

8d2 = ad

8d = a (d 0)

Now common ratio r = a 4d 8d 4d 4

a d 8d d 3

69. If the sum of the first ten terms of the series 2 2 2 2

23 2 1 4 161 2 3 4 4 ...., is

5 5 5 5 5

m, then m is

equal to

(a)102 (b) 101 (c) 100 (d) 99

Ans. (b)

Solution:

2 2 2 2

2

1 16[8 12 16 20 .... upto 10 terms] m

5 5

22 3 2

2

4 16[2 3 ... 11 ] m

5 5

2 2 2 21[ 1 2 .... 11 1 ] m

5

1 11 12 231 m

5 6

22 23 1 5m

m = 101

70. Let 1

2 2x

x 0p lim 1 tan x

then log p is equal to

(a) 2 (b) 1 (c) 1

2 (d)

1

4

Ans. (c)

Solution:

Indeterminate form 1 .

So 2

x 0

1lim (tan x )

2xp e




1/2p e

1

log p2

71. For x R, f x |log 2 sin x| and g(x) f(x)), then :

(a) g is not differentiate at x = 0

(b) g'(0) = cos(log2)

(c) g'(0) = – cos(log 2)

(d) g is differentiable at x = 0 and g'(0) = sin(log2)

Ans. (b)

Solution:

f(x) = log 2 – sin x, { when x 0, log 2 – sin x > 0}

g(x) = log 2 – sin (log 2 – sin x)

g’(x) = 0 + cos (log 2 – sin x). cos x

g’(0) = cos (log 2)

72. Consider 1 1 sin x πf x tan , x 0, .

21 sin x

A normal to y = f(x) at x = π

6also passes through the

point:

(a) (0, 0) (b) 2π

0,3

(c) π

,06

(d) π

,04

Ans. (b)

Solution:

1 1 sin x πf x tan , x 0,

1 sin x 2

1

x xcos sin

2 2f x tan ,x x

cos sin2 2

x π

0,2 4

1

x xcos sin

2 2f x tanx x

cos sin2 2

1 x xf x tan tan

4 2

π x

f x4 2

1

f ' x2

Slope of tangent = 1

2

Slope of normal at π π

,6 3

is




π πy 2 x

3 6

2π

y 2x3

passes through 2π

0,3

73. A wire of length 2 units is cut into two parts which are bent respectively to forma square of side = x

units and a circle of radius = r units. If the sum of the areas of the square and the circle so formed is

minimum, then:

(a) 2x π 4 r (b) 4 π x πr (c) x = 2r (d) 2x = r

Ans. (c)

Solution:

Circle Sq. (side = x units)

(Radius = r) units

OQ + QP = 2(given)

4x 2πr 2

2x πr 1 …(1)

1 2xr

π

Sum of areas = A = 2 2x πr

2

2

2

1 2xA x π

π

2 1 2xdA2x 2

dx π

2x – 4r = 0

4r = 2x

2r = x

74. The integral

12 9

35 3

2x 5x

x x 1

dx is equal to -

(a)

5

25 3

x

x x 1

+C (b)

10

25 3

xC

2 x x 1

(c)

5

25 3

xC

2 x x 1

(d)

10

25 3

xC

2 x x 1

Ans. (b)

Solution:

12 9

35 3

2x 5xI dx

x x 1

3 6

3

2 5

2 5

x xI dx1 1

1x x




2 5

3 6

1 1Let 1 t

x x

2 5dx dt

x x

3

dtI

t

3I t dt

2

2

t 1I C C

2 2t

2

2 5

1I C

1 12 1

x x

10

25 3

xI C

2 x x 1

75.

1/n

2nn

n 1 n 2 ....3nlim

n

is equal to -

(a) 4

18

e (b)

2

27

e (c)

2

9

e (d) 3 log 3 – 2

Ans. (b)

Solution: 1/n

n 1 n 2 n 2ny .....

n n n

Taking log both sides 2n

r 1

1 rlog y log 1

n n

2

0

log y log 1 x dx Put 1 x t

dx dt

3

3

11

log y log t dt t log t t

log y 3 log 3 3 [ 1]

log y log 27 2

log 27 2y e

2y 27 e

76. The area (in sq. units) of the region {(x, y) : 2y 2x and 2 2x y 4x, x 0, y 0} is -

(a) 4

π3

(b) 8

π3

(c) 4 2

π3

(d) π 2 2

2 3

Ans. (b)

Solution:




2y 2x ..(1)

2 2x y 4x

2 2x 4x y 0

2 2 2x 2 y 2 ..(2)

for point of intersection

x2 + 2x = 4x

x2 – 2x = 0

x(x – 2) = 0

x = 0, 2

at x = 0, y = 0

and at x = 2, y = 2

P(2, 2) and Q(2, – 2)

Hence required area

= 2

2

0

( 4x x 2x )dx

= 2

2

0

( 4 x 2 2x )dx

=

23/2

2 1

0

x 2 4 x 2 2 x4x x sin

2 2 2 3/2

= 3/2 2 4 π2 2

3 2 2

= 4 2 8

π π3 3

77. If a curve y = f(x) passes through the point (1, –1) and satisfies the differential equation, y(1 + xy)dx = x

dy, then f 1

2

is equal to -

(a) 2

5 (b)

4

5 (c)

2

5 (d)

4

5

Ans. (d)

Solution:

dy

y 1 xy xdx




2dy yy

dx x

2dy yy

dx x

2

1 dy 1 1. 1

y dx x y Let

2

1 1 dy dtt

y y dx dx

dt 1.t 1

dx x

I.F. = 1

dxn xxe e

I.F. = x

Solution of this differential equation is

t.x x dx

2x

tx C2

2x x

Cy 2

this curve passes through (1, –1) so

1

1 C2

1

C2

2x x 1

y 2 2 ..(1)

from at 1

x2

1 1 1

2y 8 2

1 1 41 y

y 4 5

78. Two sides of a rhombus are along the lines, x – y + 1 = 0 and 7x – y –5 = 0. If its diagonals intersect at

(–1, –2), then which one of the following is a vertex of this rhombus?

(a) (–3, – 9) (b) (–3, – 8) (c) 1 8

,3 3

(d)

10 7,

3 3

Ans. (c)

Solution:

Equation of AB x – y +1 = 0 …(1)

Equation of AD 7x – y – 5 =0 ….(2)

Solve (1) & (2) A(1, 2)

Point P(–1, – 2) is the mid-point of the diagonals AC & BD




11

11

x 11 x 3

2AP CP Cy 2

2 y 62

C(–3, –6)

Equation of BC 17x y λ 0

It's passes through the point C, then

17 3 6 λ 0

1λ 15

Equation of BC is 7x – y + 15 = 0 …(3)

Solve (1) & (3) 7 4

B ,3 3

Equation of CD 2x y λ 0

x y 3 0 …(4)

Solve (2) & (4) 1 8

D ,3 3

79. The centres of those circles which touch the circle, x2 + y2 – 8x – 8y – 4 = 0, externally and also touch

the x-axis, lie on:

(a) a circle (b) an ellipse which is not a circle

(c) a hyperbola (d) a parabola

Ans. (d)

Solution:

x2 + y2 – 8x – 8y – 4 = 0 ...(1)

C1 = centre (4,4)

r1 = radius = 16 16 4 6

Locus of the centre C2(h, k) = ?

radius of circle (b) is

2r k

& circle (2) touches the circle (1) externally.

So, C1C2 = r1 + r2

(h + 4)2 +(k – 4)2 = (6 + k )2

h2+ 16 – 8h + k2+ 16 – 8k=36 + k2+ 12 k




Replaced h by x & k by y then

2x 8x 8y 12 y 4 0

y; If y 0

y or

y; If y 0

Case (I): If y 0

2x 8x 20y 4 0

Case (II): If y < 0

x2 – 8x + 4y – 4 = 0

Both of these equations are the equation of parabola.

80. If one of the diameters of the circle, given by the equation, x2 + y2 – 4x + 6y – 12 = 0, is a chord of a

circle S, whose centre is at (– 3,2), then the radius of S is -

(a) 5 2 (b) 5 3 (c) 5 (d) 10

Ans. (b)

Solution:

x2 + y2 – 4x + 6y – 12 = 0 ..(1)

centre C1(2, – 3)

radius r1 = 5

& centre of circle (2) is C2 (–3, 2)

2 2

1 2C C 2 3 3 2 25 25 5 2

radius (r2) of circle (2) is

2

2 2 22 2 1 1 2r C A C A C C 5 5 2

= 25 50 75 5 3

81. Let P be the point on the parabola, y2 = 8x which is at a minimum distance from the centre C of the

circle, x2 + (y + 6)2= 1. Then the equation of the circle, passing through C and having its centre at P is -

(a) x2 + y2 – 4x + 8y + 12 = 0 (b) x2 + y2 – x + 4y – 12 = 0

(c) x2 + y2 – x

4 + 2y – 24 = 0 (d) x2 + y2 – 4x + 9y + 18 = 0

Ans. (a)

Solution:

y2 = 8x

Let, the point P (2t2, 4t) lies on the parabola & centre of the given circle is C (0, –6)

Distance between the points C & P.




2

D 4 4 4t 6

For minimum distance

4

4 4 3 2 4t 6 4dD0

dt 2 4t 4t

4t3 + 8t + 12 = 0

t3 + 2t + 3 = 0

t2(t + 1) – t(t + 1) + 3(t + 1) = 0

(t2 – t + 3)(t + 1) = 0

t = –1

point P (2, – 4)

centre C (0, – 6)

CP = radius of circle = 4 4 2 2

Equation of circle (x – 2)2 + (y + 4)2 = 8

x2 – 4x + 4 + y2 + 8y+ 16 = 8

x2 + y2 – 4x + 8y + 12= 10

82. The eccentricity of the hyperbola whose length of the latus rectum is equal to 8 and the length of its

conjugate axis is equal to half of the distance between its foci, is –

(a) 4

3 (b)

4

3 (c)

2

3 (d) 3

Ans. (c)

Solution:

Let eqn. of hyperbola is 2 2

2 2

x y1

a b

It's length of latus-rectum is 22b

8a

b2 = 4a ...(1)

& length of it's conjugate axis = half of the distance between it's foci

2b = 1

2(2ae)

ae

b2

...(2)

2 2 2b a e 1 …(3)

By neq (2) & (3)

2 2

2 2a ea e 1

4

2

2 ee 1

4




23e

14

2

e3

83. The distance of the point (1, –5, 9) from the plane x – y + z = 5 measured along the line x = y= z is :

(a) 3 10 (b) 10 3 (c) 10

3 (d)

20

3

Ans. (b)

Solution:

Given plane x y + z = 5 …(1)

and line x = y = z …(2)

Given line is parallel 10 line PM

DR's of line PM = DR's of given line

= 1, 1, 1

Eqn. of line PM is x 1 y 5 z 9

λ1 1 1

(Let)

Coordinates of point M is ( λ + 1, λ – 5, λ + 9)

Point M lies on the plane x – y + z = 5

That's why ( λ + 1) – ( λ – 5) + ( λ + 9) = 5

λ + 15 = 5

λ = – 10

M(–9, –15, – )

2 2 2

PM 1 9 5 15 9 1 100 100 100

= 10 3

84. If the line, x 3 y 2 z 4

2 1 3

lies in the plane, lx + my – z = 9, then l2 + m2 is equal to :

(a) 26 (b) 18 (c) 5 (d)2

Ans. (d)

Solution:




x 3 y 2 z 4

2 1 3

..(1) lie on plane

Let point P lies on line (1)

P(3, –2, –4)

Point P also lies on plane lx + my – z = 9 …(2)

3l – 2m + 4 = 9 3l – 2m = 5 …(3)

2l – m – 3 = 0 4l – 2m = 6 …(4)

On solving eq. (3) & (4)

_____________

–l = – 1

l = 1

m = – 1

l2 + m2 = 2

85. Let a, b and c be three unit vectors such that 3

a (b c) (b c).2

If b is not parallel to c , then the

angle between a and b is:

(a) 3π

4 (b)

π

2 (c)

2π

3 (d)

5π

6

Ans. (d)

Solution:

3a (b c) (b c)

2

3 3(a. c) b (a. b) c b c

2 2

Now comparing both sides ( b and c are not parallel)

3

a. c2

& 3

a. b2

a b 1 and if angle between a and b is θ then

3

cos θ2

5π

θ6

86. If the standard deviation of the numbers 2, 3, a and 11 is 3.5, then which of the following is true ?

(a) 3a2 – 26a + 55 = 0 (b) 3a2 – 32a + 84 = 0

(c) 3a2 – 34a + 91 = 0 (d) 3a2 – 23a + 44 = 0

Ans. (b)

Solution: (a)

Observations are 2, 3, a, 11

Mean 222 3 a 11 a

x 4 and ς 3.5 12.254 4




Now, n 2

2 2i

i 1

1ς x x

n

2

21 a12.25 [4 9 a 121] 4

4 4

2 2134 a a

12.25 16 2a4 16

2 2a a

12.25 33.5 16 2a4 16

23a

2a 5.25 016

3a2 – 32a + 84 = 0

87. Let two fair six-faced dice A and B be thrown simultaneously. If E1 is the event that die A shows up

four, E2 is the event that die B shows up two and E3 is the event that the sum of numbers on both dice

is odd, then which of the following statements is NOT true ?

(a) E1 and E2 are independent (b) E2 and E3 are independent

(c) E1 and E3 are independent (d) E1, E2 and E3 are independent

Ans. (d)

Solution:

1

1P E

6 1E shows 4

2

1P E

6 2E shows 2

3

18P E

36 E3 = sum equal to odd or both dice

1 2

1P E E

36 1 3E E { 4, 1 , 4,3 , 4,5 }

2 3

3P E E

36 2 3E E { 2, 1 , 2,3 , 2,5 }

1 2 3E E E {}

Check by options

(a) 1 2 1 2P E E P E E

1 1 1

36 6 6 E1 and E2 are independent True

(b) 1 2 1 2P E E P E P E

1 1 1


(c) 1 3 1 3P E E P E E

3 1 18





(d) 1 2 3 1 2 3P E E E P E P E P E

1 1 18

06 6 36

E1, E2, E3 are not independent False

88. If 0 x 2π, then the number of real values of x, which satisfy the equation cos x + cos 2x + cos 3x

+ cos 4x = 0, is :

(a) 3 (b) 5 (c) 7 (d) 9

Solution: (c)

cos x + cos 2x + cos 3x + cos 4x = 0

4 1 4xcos x x . sin

2 20

xsin

2

[By using series formula]

5x

cos . sin 2x 02

But x

sin 02

sin 2x = 0 5x

cos 02

2x nπ 5x π

2n 12 2

nπ

x , n I2

π

x 2n 1 , n I5

π 3π

x 0, , π, , 2π2 2

π 3π 7π 9π

x , , π, ,5 5 5 5

But x = 0, 2π does not satisfy ( sin x

02 )

Solutions are π 3π π 3π 7π 9π

x , π, , , , ,2 2 5 5 5 5

7 solutions

89. A man is walking towards a vertical pillar in a straight path, at a uniform speed. At a certain point A on

the path, he observes that the angle of elevation of the top of the pillar is 30°. After walking for 10

minutes from A in the same direction, at a point B, he observes that the angle of elevation of the top of

the pillar is 60°. Then the time taken (in minutes) by him, from B to reach the pillar, is :

(a)6 (b) 10 (c) 20 (d) 5

Ans. (d)

Solution:




0 hIn ACD, tan 30

x y

x y

h3

Now in BCD, tan 0 h60

y

3 y h

x y

3 y3

x = 2y

y = x

2

Given that travelling time x = 10 min

travelling lime y= 10

2 = 5 min

90. The Boolean Expression (p ~ q) q (~ p q) is equivalent to :

(a) ~ p q (b) p q (c) p q (d) p ~q

Ans. (c)

Solution:

P Q ~p ~q (p ~ q) (1) (~ p q)

(p ~q) q (p ~ q)

q (~ p q) (2)

p q (3)

p q (4)

(p ~ q)

T T F F F F T T T T T

T F F F T F T F F T T

F T T F F T T F F T T

F F T F F F F F F F T

Pioneer Education The Best Way To Success NTSE | … · having 30 questions in each part of equal...

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