Physics (Code-A) Sample Question Paper for Class XII

(1)

Physics (Code-A) Sample Question Paper for Class XII

Time : 3 Hours Max. Marks : 70

General Instructions :

(i) All questions are compulsory.

(ii) There is no overall choice. However an internal choice has been provided in one question of 2 marks,

one question of 3 marks and all three questions of 5 marks each. You have to attempt only one of the

choices in such questions.

(iii) Questions 1 to 8 are very short answer type and carry 1 mark each.

(iv) Questions 9 to 16 are short answer type and carry 2 marks each.

(v) Questions 17 to 25 are short answer type and carry 3 marks each.

(vi) Question 26 is a value based question and carry 4 marks

(vii) Questions 27 to 29 are long answer type and carry 5 marks each.

(viii) Use of calculator is not permitted.

(ix) You may use the following values of physical constants wherever necessary :

c = 3 × 108

ms–1

h = 6.626 × 10–34

Js

e = 1.602 × 10–19

C

�0

= 4� × 10–7

T mA–1

9 2 2

0

1

9 10 Nm C

4

� ��

Mass of neutron mn

= 1.675 × 10–27

kg

Boltzmann’s constant k = 1.381 × 10–23

JK–1

Avogadro’s number NA

= 6.022 × 1023

/mol–1

SECTION-A

Very Short Answer Type Questions :

1. A wire of resistivity � is stretched to three times its length. What will be its new resistivity? [1]

2. What oscillates in electromagnetic waves? Are these waves transverse or longitudinal? [1]

3. State the essential condition for diffraction of light to occur. [1]

4. The electric current flowing in a wire in direction B to A is decreasing. What is the direction of induced current

in the metallic loop kept above wire as seen from top? [1]

A B

5. What are the two possible extreme values of the magnifying power of a simple microscope made of convex

lens of focal length 5.0 cm? [1]

6. If the maximum kinetic energy of a photoelectron is 3 eV, what is its stopping potential? [1]

(2)

Sample Question Paper for Class XII Physics (Code-A)

7. The frequency of the alternating current applied to a series circuit containing resistance, inductance and

capacitance is doubled. What happens to R, XL

and XC

? [1]

8. An electric dipole of dipole moment 20 × 10–6

cm is enclosed in a closed surface. What is net flux coming

out of the surface? [1]

SECTION-B

Short Answer Type Questions :

9. Calculate the capacitance of the unknown capacitor C if the equivalent capacitance between P and Q is

40 �F. [2]

Q

10 F�

80 F�

C

P

10. Using Gauss law, derive an expression for electric field intensity due to a uniformly charged spherical shell at

a point : [2]

(a) Outside and

(b) Inside the shell

11. Define self inductance and give its SI unit. Derive an expression for self inductance of a long air cored solenoid

of length l, radius r and turns per unit length n. [2]

12. State Ampere’s circuital law. Use it to derive an expression for the magnetic field along the axis of a current

carrying toroidal solenoid of N number of turns having radius r. [2]

13. What do you mean by power of a lens? What is its unit? What is the expression for power of two thin lenses

in contact with each other? [2]

14. Draw a neat and labelled diagram of image formed by a compound microscope. Write an expression for its

magnifying power. [2]

15. On the basis of energy band diagrams, distinguish between [2]

(i) A metal

(ii) An insulator

(iii) A semiconductor.

16. A coil has an inductance of 1 H. [2]

(a) At what frequency will it have a reactance of 3142 �?

(b) What should be the capacitance of a capacitor which has the same reactance at that frequency?

OR

A 15.0 �F capacitor is connected to 110 V, 50 Hz source. Find the capacitive reactance and the rms current.

(3)

Physics (Code-A) Sample Question Paper for Class XII

SECTION-C


17. Using Biot-Savart’s law, derive an expression for the magnetic field intensity at the centre of a current carrying

circular coil. Also find the magnitude of field due to a semicircular coil of radius r and carrying current I. [3]

18. Define capacitance. Derive an expression for the capacitance of a parallel plate capacitor. How will the

capacitance of the capacitor be affected if a conducting slab of some thickness is partially filled in the space

between two plates. [3]

19. Define the term electron mobility. Explain how electron mobility changes for a good conductor when : [3]

(a) The temperature of the conductor is decreased, keeping potential difference constant.

(b) Applied potential difference is doubled keeping temperature constant.

20. In certain cases when an object and screen are separated by a distance d, two positions x of the converging

lens relative to the object will give an image on the screen. Show that these two values are

1/2

4

1 1

2

d f

x

d

⎡ ⎤⎛ ⎞� ⎢ ⎥⎜ ⎟

⎝ ⎠⎣ ⎦

.

Under what conditions will no image be formed? [3]

21. What is electron emission? Discuss various forms of electron emission in brief. Establish Einstein’s photoelectric

equation? [3]

22. A series LCR circuit is connected to an AC source. Using the phasor diagram, derive the expression for the

impedance of the circuit. Plot a graph to show the variation of current with frequency of source, explaining the

nature of its variation. [3]

23. Mention three modes of propagation used in communication system. Explain with the help of a neat and labelled

diagram how long distance communication can be achieved by ionosphere reflection of radio waves. [3]

24. Obtain the binding energy of the nuclei 56

26

Fe in units of MeV from the following data : [3]

1.007825 up

m �

1.008665 un

m �

� �56

26

Fe 55.934939 um � . Also calculate binding energy/nucleon.

25. With the help of an example, explain how the neutron to proton ratio changes during -decay. Also briefly explain

what are nuclear forces. Give their important properties. [3]

OR

A doubly ionised lithium atom is hydrogen like with atomic number 3. Find

(a) The wavelength of radiation required to excite the electron in Li++

from the first to the third Bohr’s orbit.

(b) How many spectral lines are observed in the emission spectrum of the above excited system?

26. Mr. Feynmann and his students went to north pole on a picnic. They found beautiful coloured curtain like

structures are hanging from the polar sky. What is the reason behind this? [4]

(4)

Sample Question Paper for Class XII Physics (Code-A)

SECTION-D

Long Answer Type Questions :

27. (a) In Young’s double slit experiment, derive the condition for (i) Constructive interference and (ii) Destructive

interference at a point on the screen. Also derive expression of fringe width. [3]

(b) Estimate the distance for which ray optics is a good approximation for an aperture of 4 mm and wavelength

400 nm. [2]

OR

State Huygens Principle. Using the geometrical construction of secondary wavelets, explain the refraction of a

plane wavefront incident at a plane surface. Hence verify the Snell’s law. Illustrate with the help of a diagram the

action of (i) Convex lens and (ii) concave mirror on a plane wavefront incident on it.

28. (a) Explain briefly with the help of a circuit diagram, how V-I characteristics of a p-n junction diode are obtained

in

(i) Forward bias

(ii) Reverse bias

Draw the shapes of curves obtained. [3]

(b) A photodiode is fabricated from a semiconductor with a band gap of 2.8 eV. Can it detect a wavelength of

6000 nm? Justify. [2]

OR

(a) With the help of a circuit diagram, explain working of a transistor as an amplifier.

(b) Calculate the base current, collector current and collector to emitter voltage for the circuit given

RC = 2 k�

RB = 300 k�

C

B

VCC

= 9 V

E � = 50

29. (a) Prove that for a short magnetic dipole axial

equatorial

2

B

B � . [3]

(b) The susceptibility of magnesium at 300 K is 1.2 × 10–5

. At what temperature will the susceptibility increase

to 1.8 × 10–5

? [2]

OR

(a) State Biot-Savart’s Law. Use it to obtain the magnetic field at an axial point, distance x from the centre of a

circular coil of radius a, carrying a current I.

(b) A galvanometer having 30 divisions has a current sensitivity of 20 �A/div. It has a resistance of 25 �. How will

you convert it into an ammeter of range 0 - 1 A?

� � �

(5)

Chemistry (Code-A) Sample Question Paper for Class XII




(ii) The question paper consists of four sections A, B, C and D.

Section A contains 8 questions of 1 mark each.

Section B contains 10 questions of 2 marks each.

Section C has 9 questions of 3 marks each, whereas Section D contains 3 questions of 5 marks each.

(iii) There is no overall choice. However, an internal choice has been provided in one question of 2 marks,

one question of 3 marks and all three questions of 5 marks weightage. A student has to attempt only one of the

alternatives in such questions.

(iv) Wherever necessary, the diagrams drawn should be neat and properly labelled.

SECTION-A


1. Give IUPAC name of the compound: [1]

CH — C — CH — C — CH3 2 3

—

F

—

C H2 5

—

C H2 5

—

Br

2. Consider the reaction R �� P. The change in the concentration of R with time is shown in the following

plot. [1]

Time

[R]t

(i) Predict the order of the reaction.

(ii) Write the expression for half life of this reaction.

3. Predict the shape of BrF3

on the basis of VSEPR theory. [1]

4. What are the dispersed phase and dispersion medium in milk? [1]

5. A compound contains two type of atoms A and B. It crystalises in a cubic lattice with atom A at the corners

of the cube and atom B at the face centres. What is the simplest formula of the compound? [1]

6. An ionic compound AB is 50% dissociated in aqueous solution. Determine the value of Van’t Holf factor for

this compound. [1]

7. An ore sample of galena (PbS) is contaminated with zinc blende (ZnS). Name one chemical reagent which

can be used to concentrate selectively by froth floatation method. [1]

8. Write balanced equation for the complete hydrolysis of XeF4

. [1]

(6)

Sample Question Paper for Class XII Chemistry (Code-A)

SECTION-B


9. Write the structure of 4-Methylpent-3-en-2-one. Will it respond positive to iodoform test? If so, Why? [2]

10. Define osmotic pressure. How does it change with [2]

(a) Temperature

(b) Atmospheric pressure

11. Write the equations of the reactions involved in the extraction of copper from its sulphide ore (Cu2

S). [2]

12. Name the two components of starch. How do they differ from each other structurally? [2]

13. Give the equations of the reactions involved when glucose is treated with [2]

(a) HI

(b) (CH3

CO)2

O

14. Describe the mechanism of the formation of ethene from ethanol in the presence of concentrated sulphuric

acid. [2]

15. Give one example each of [2]

(a) Hofmann bromamide reaction

(b) Gabriel phthalimide reaction

16. Distinguish chemically between [2]

(a) Butan-1-ol and 2-Methyl propan-2-ol

(b) Phenol and benzyl alcohol

17. Account for the following: [2]

(a) Aniline does not undergo Friedel Crafts alkylation

(b) Aniline undergoes bromination even in the absence of halogen carrier

OR

(a) Unlike alkylation, acylation of amines stops after first step

(b) Tertiary amines do not undergo acylation

18. Explain the type of deviation shown by a mixture of acetone and chloroform. [2]

SECTION-C


19. An element X with atomic mass 60 g/mole has a density 6.23 g/cm3

. If the edge length of unit cell is 400 pm,

identify the type of cubic unit cell. Calculate the radius of an atom of this element. (NA

= 6 × 1023

mole–1

) [3]

20. Write the names of the monomers of these polymers and classify them as addition or condensation polymers. [3]

(a) Terylene

(b) Teflon

(c) Natural rubber

(7)

Chemistry (Code-A) Sample Question Paper for Class XII

21. (a) Name the isomerism exhibited by the following pair of coordination compounds: [3]

[Co(NH3

)5

Br]SO4

& [Co(NH3

)5

SO4

]Br. Give one chemical test to distinguish between these two

compounds.

(b) Using valence bond theory compare the structures of [FeF6

]4–

and [Fe(CN)6

]4–

(Atomic number of Fe = 26)

22. Give three differences each between [3]

(a) Physisorption and chemisorption

(b) Lyophilic sol and lyophobic sol

23. Account for the following : [3]

(a) ICl is more reactive than l2

(b) NO2

dimerises to form N2

O4

(c) H3

PO3

is a reducing agent but not H3

PO4

24. Give the structures of the following compounds : [3]

(a) XeOF4

(b) PF5

(c) H2

S2

O8

OR

Give the products and balance the following equations:

(a) Ca3

P2

+ H2

O ��

(b) XeF6

+ NaF ��

(c) Cu + H2

SO4

(conc.) ��

25. Activation energy of a reaction at 300 K is 55 kJ/mole. When the same reaction is carried out in the presence

of a catalyst, activation energy is lowered to 45 kJ/mole at 300 K. Determine the extent to which the rate of

the reaction is increased. [3]

26. Account for the following: [3]

(a) Haloalkanes react with KCN to give alkyl cyanide as a major product and not alkyl isocyanide.

(b) In chloro benzene, Cl is electron withdrawing group but it undergoes electrophilic substitution at ortho and

para position.

(c) Allyl chloride is more reactive towards nucleophilic substitution than n-propyl chloride.

27. Why are vitamin A and C essential to us? Also write their chemical names. [3]

SECTION-D


28. (a) Carry out the following conversions: [3]

(i) Ethanal to butane-1, 3-diol

(ii) Acetone to tert. butyl alcohol

(iii) Benzoic acid to m-nitro benzoic acid

(8)

Sample Question Paper for Class XII Chemistry (Code-A)

(b) Give one example each of [2]

(i) Clemmensen reduction

(ii) Cannizzaro reaction

OR

(a) Carry out the following conversions: [3]

(i) Ethanal to butan-1-ol

(ii) Benzaldehyde to 3-phenyl propanol

(iii) Ethyl benzene to benzene

(b) Give one example each of [2]

(i) Aldol condensation

(ii) HVZ reaction

29. (a) At what pH of HCl solution will the standard hydrogen electrode have a potential – 0.118 V at 298 K? [2]

(b) Write the equations of the reactions involved at each electrode in a H2

– O2

fuel cell. [2]

(c) Why does the potential of a mercury cell remain constant? [1]

OR

(a) Calculate the equilibrium constant and work done by the cell [2]

Ni + Cu2+

�� Ni2+

+ Cu, given 2 2

o o

Ni /Ni Cu /Cu

E 0.25 V; E 0.34 V� ��

(b) Write the equations of the reactions involved at each electrode during the electrolysis of [2]

(i) CuSO4

(aq) using inert electrode

(ii) Dilute solution of sulphuric acid

(c) How many coulombs of electricity are needed for the conversion of 9 grams of Al from molten AlCl3

?

(Atomic mass of Al = 27) [1]

30. Police usually disperse the indisciplined mob by using tear gas shell. One of the person in the mob advised

the people either to use water wetted cloth on eyes or to avoid smoke.

(i) Write the chemical formula of tear gas. [1]

(ii) Write the value involved as advised by the person present in the mob. [2]

(iii) Write IUPAC name of tear gas. [2]

OR

Arrange the following in order of property indicated:

(i) F2

, Cl2

, Br2

, I2

– Increasing bond dissociation enthalpy [1]

(ii) HF, HCl, HBr, HI – Increasing acid strength [1]

(iii) NH3

, PH3

, AsH3

, SbH3

– Increasing base strength [1]

(iv) H2

O, H2

S, H2

Se, H2

Te – Increasing boiling point [1]

(v) HF, HCl, HBr, HI – Increasing boiling point [1]

� � �

(9)

Mathematics (Code-A) Sample Question Paper for Class XII




(ii) The question paper consist 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 and

section C comprises of 7 questions of six marks each.

(iii) All questions in Section A are to be answered in one word, one sentence or as per the exact requirement of

the question.

(iv) There will be no overall choice. However, internal choice has been provided in Four Questions of 4 Marks

each and Two Questions of 6 Marks each. You have to attempt only one of the alternatives in all such

questions.

(v) Use of calculator is not permitted. You may ask for logarithmic tables, if required.

SECTION-A


1. Using principal values, write the value of � �1 1 11 3

sin cos tan 3

2 2

⎛ ⎞ � � ⎜ ⎟

⎝ ⎠

[1]

2. Let � be a binary operation on N given by HCF( , )a b a b� � for all a, b � N. Find 12 27.� [1]

3. If

2 4 1

3 2 1

0 0 2

� � , write cofactor of the element a31

. [1]

4. Find value of x for which the matrix

1 2

3 1

x

x

�⎡ ⎤

⎢ ⎥⎣ ⎦

is singular. [1]

5. Evaluate � �� 1 1 1x x x dx � �∫ . [1]

6. If ˆ ˆ ˆˆ 2a i j k� � and ˆ ˆ ˆ ˆ,b i j k� � find | |a b�

��

[1]

7. Write the degree of differential equation [1]

22

2

0

dy d y

y

dx dx

⎛ ⎞⎛ ⎞ � � �⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

8. If

2 1 4 0 9 2

,

3 2 1 2 14 x

⎡ ⎤ ⎡ ⎤ ⎡ ⎤

�⎢ ⎥ ⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦ ⎣ ⎦

then find value of x. [1]

(10)

Sample Question Paper for Class XII Mathematics (Code-A)

9. Write the vector equation of a line given by [1]

1 1

2 3 4

x y z � �

10. Find distance of point (1, 2, 3) from the plane 2x + 3y – 4z = 0. [1]

SECTION-B


11. If x = 2(� – sin�) and y = 2(1 + cos�). Find

dy

dx

at .

2

�� . [4]

12. Find the value of ‘a’ for which the function f defined as [4]

3

( 1), if 0

( ) tan sin

, if 0

a x x

f x x x

x

x

� �⎧

⎪� ⎨ �⎪⎩

is continuous at x = 0

13. Evaluate

1

2

sin

1

x x

dx

x

∫ [4]

OR

Evaluate

sin( )

sin

a x

dx

x

∫

14. Show that 1 1 1

3 8 84

sin sin cos

5 17 85

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ �⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠

[4]

15. Let

3

: {–1}

2

f R R

⎧ ⎫ � ⎨ ⎬

⎩ ⎭

be defined as

2 3

( ) .

3 2

x

f x

x

��

Find the function g(x) such that fog = gof = IR

. Also

find domain of g(x) [4]

OR

Show that f : R � R defined as f(x) = 2x + 7 is one-one onto function. Also find its inverse

16. Solve the differential equation [4]

sec sec tan

dy

y x x x

dx

� �

17. If , ,a b c

��

are three vectors such that | | 5, | | 8a b� ��

�

and | | 3c ��

and ˆ ˆ8 6 .a b c i j� � � �

��

Find the value of

.a b b c c a� � � � ��

� � � �

[4]

(11)

Mathematics (Code-A) Sample Question Paper for Class XII

18. Evaluate

1

11

x

dx

e �∫ [4]

OR

Evaluate

0

(| cos | | sin |)x x dx

�

�∫

19. Find the point on the curve y = x3

– 6x + 3 at which equation of tangent is 3x + y – 1 = 0. [4]

OR

Using differentials approximate 25.2

20. Using matrix method, solve the system of equations [4]

x + 2y + z = 1

x + y – z = 2

2x + 3y + z = 1

21. Write the equation of line in vector form which is perpendicular to the two lines

1 1 3

2 2 4

x y z � � and

1

1

4 3

x y

z

� � � , and passes through point (2, 1, 2). [4]

22. How many times a fair coin must be tossed, so that the probability of getting atleast one tail is more than

90%? [4]

SECTION-C


23. Show that of all the rectangles inscribed in a given fixed circle, the square has maximum area. [6]

24. Write the equation of plane passing through points A(1, 2, 1), B(2, 3, 1) and C(2, 1, 0). Also find its distance

from point (1, 1, 1). [6]

25. Using integration, find the area of bounded region between the curve y = x2

and y = –|x| + 2. [6]

OR

Find the area of region

� �2

( , ) : 0 1, 0 1, 0 2x y y x y x x� � � � � � � �

26. A factory makes nuts and bolts. A nut takes 1.5 hrs of machine time and 3 hrs of worker’s time in its making

while a bolt take 3 hrs of machine time and 1 hr of worker’s time. In a day, the factory has the availability of

not more than 42 hrs of machine time and 24 hrs of worker’s time. If the profit on a nut is Rs. 20 and on a

bolt is Rs. 10, find the number of nuts and bolts the factory should manufacture to earn maximum profit. Make

it as L.P.P and solve it graphically. [6]

(12)

Sample Question Paper for Class XII Mathematics (Code-A)

27. Using elementary transformations, find the inverse of matrix [6]

1 3 2

3 0 1

2 1 0

⎡ ⎤

⎢ ⎥

⎢ ⎥

⎢ ⎥ ⎣ ⎦

28. Find the particular solution of the differential equation log 3 4

dy

x y

dx

⎛ ⎞ � ⎜ ⎟

⎝ ⎠

given that y = 0 when x = 0. [6]

29. In an examination of a school having 60 students in XII A and 40 students in XII B, 15 and 20 students failed

in XII A and XII B, respectively. A student is called by the principal at random, find the probability that he is

from XII A if he passed in the examination. [6]

OR

Let bag A contains 2 red and 3 white balls, another bag B contains 2 white and 3 black balls and another

bag C contains 2 black and 3 red balls. A bag and a ball out of it are selected at random. What is the

probability that selected ball is white assuming selections of bag A, B and C are equally likely?

� � �

(1)

Physics (Code-A) Solutions of Sample Question Paper for Class XII

SECTION-A

1. No change, resistivity is material dependent.

2. Electric and magnetic fields. Transverse waves.

3. Diffraction of light occurs when size of obstacle (or aperture) is comparable to the wavelength of light used.

4. By Lenz law it is in clockwise direction.

5. When image is formed at distinct vision,

25

1 1 6

5

D

M

f

� � � � �

When image is formed at infinity,

25

5

5

D

M

f

� � �

6. KEmax

= eVSP

3 eV

3 volt.SP

V

e

� �

7. R - Remains unchanged

XL

- Becomes doubled

XC

- Becomes halved

8. Since dipole consists of equal and opposite charge hence net charge and thus net flux is zero.

SECTION-B

9. 80 �F and 10 �F are in parallel

Therefore, equivalent is 80 + 10 = 90 �F

This is in series with C

Hence, equivalent

90

40

90

C

C

��

72 FC⇒ � �

SOLUTIONS

(2)

Solutions of Sample Question Paper for Class XII Physics (Code-A)

10. (a) Outside the shell,

enc

0

q

E ds� ��∫

��

�

2

0

4

q

E r⇒ � ��

Rr R>

2

0

4

q

E

r

⇒ ��

(b) Inside the shell, Take Gaussian surface

enc

0

q

E ds� ��∫

��

�

as enc

0q �

r

R

0E⇒ �

11. The property of an electric circuit by virtue of which it opposes any change of flux or current in it by inducing a

current in itself is called self induction. It is numerically equal to the flux linked when unit current flows through it.

�� i

Li� �

L

i

��

SI unit is henry.

Consider, air cored solenoid

I I

BAN� �

� �� 2

0

ni r nl� � � �

2 2

0

L n r l

i

��

12. Ampere’s circuital law : The magnetic circulation around any closed curve is equal to �0

times the electric

current threading the curve.

Amperian curve

0 enclosed

B dl I� � �∫

� �

�

Magnetic field due to a toroid

r

Current

in each

turn is I

0 enclosed

B dl I� � �∫

� �

�

0

2B r NI� � � �

0

2

NI

B

r

��

(3)


13. Power of lens is a measure of its ability to converge or diverge a light beam falling on it. Power of lens is defined

as the tangent of the angle by which the lens converges (or diverges) a beam of light falling on it at a unit distance

from its optical centre.

tan

h

f

� �

when h = 1,

1

tan

f

� �

Power of a lens

1

tanP

f

� � � f

hF

�

Unit of power of lens is dioptre (D) or m–1

.

When lenses are in contact

eq 1 2

P P P� �

14.

Magnification :

(a) For strained eye, 1

o e

L D

M

f f

⎡ ⎤� �⎢ ⎥

⎣ ⎦

(b) For relaxed eye,

o e

LD

M

f f

�

15. (i) Metal : A solid is a conductor if its valence band overlaps its conduction band.

VB

CB

(4)


(ii) Insulator : A solid is an insulator if the valence band and the conduction band do not overlap and are separated

by an energy gap between 3 eV and 6 eV.

VB

CB

Eg (~3 to 6 eV)

(iii) Semiconductor : A solid is a semiconductor if the energy gap is much smaller as compared to that in case

of insulators. Band energy gap is in the range (0.1 - 1.0 eV)

VB

CB

Eg (~0.1 to 1.0 eV)

16. (a) Here, L = 1 H, XL

= 3142 �

XL

= 2�fL

f = 500 Hz

(b)

1

0.10 F

2

C

X C

fC

� ⇒ � ��

OR

15 FC � �

1

212.3

2

C

X

fC

� � ��

rms

rms

0.52 A

C

V

I

X

� �

SECTION-C

17.

0

3

,

4

I dl r

dB B

r

� ��

��

� �

directed outwards

0

2

4

I dl

dB

r

��

r

dl

I

0 0

2

4 2

Idl I

B dB

r r

� �� ∫ ∫

For half coil

0

4

I

B

r

��

(5)


18. Capacitance of a conductor is defined as the ratio of charge on it to its potential

Numerically,

Q

C

V

� .

Numerically, capacitance of a conductor is equal to the amount of charge required to raise potential through unity.

0

E

��

V

E

d

�

–

–

–

–

–

–

–

–

–

+

+

+

+

+

+

+

+

+

E

P

Q –Q

0

d

V

��

0

AQ

C

V d

��

On insertion of a conducting slab, capacitance of the conductor increases as 0

A

C

d t

��

, t is thickness of the

conducting slab inserted between the plates.

19. Mobility is defined as the magnitude of drift velocity per unit electric field.

,

d

e d e

V eE

V

E m

� � � !

e

e

e

m

!⇒ � �

(a) When temperature of the conductor decreases, e

! decreases and consequently e

� decreases.

(b)e

� doesn’t depend upon potential difference and thus it remains unchanged.

20. For a lens

1 1 1

f v u

�

Since the object is real, u = –x

Thus,

vx

f

v x

��

Further, as the distance between the object and the screen is d,

� �v d x�

Thus,

� �� d x x

f

d x x

� �

2

0x dx fd⇒ � �

2

4

2

d d fd

x

⇒ �

For the image to be formed, x should be real,

i.e. 1 4 0

f

d

�

4d f⇒ �

When d < 4f, no image is formed.

(6)


21. The emission of free electron from metal surface is called electron emission. It can take place through any of

following physical processes.

(i) Thermionic Emission : The release of electron from metal as a result of its temperature, i.e. by heating.

(ii) Field emission : It is a kind of electron emission in which a very strong electric field pulls the electron out of

metal surface.

(iii) Photoelectric Emission : It is that kind of electron emission in which light of suitable frequency ejects the

electrons from a metal surface.

Photoelectric effect equation max 0

KE h h� " " .

22. Phasor diagram,

#

VR

I0

E0

VC

V – L

VC

VL

0

sinE E t#�

L R C

0 0

,

R L L

V I R V I X� � and 0C C

V I X�

� �22

0 R L C

E V V V� �

� �22

0 0 L C

E I R X X� �

Impedance � �22

L C

Z R X X� �

0

I is max, when L C

# � #

1

2

2

fL

fC

� ��

( )I0 max

I

#r

#

1

2

f

LC

��

(7)


23. Modes of propagation

(i) Ground wave propagation : Travel along curved earth’s surface from transmitter to the receiver.

(ii) Sky wave propagation : Radio waves travel skywards and if its frequency is below certain critical frequency

(typically 30 MHz). It is returned to the earth by ionosphere.

(iii) Space wave : In space wave propagation, radio waves travel in a straight line from transmitting antenna to the

receiving antenna.

Sky wave : The sky wave below critical frequency travels from the transmitting antenna to receiving antenna

via ionosphere. The ionosphere consists of layers of air molecules which have become positively charged by

removal of electrons by sun’s ultraviolet radiations. On striking the earth, the sky wave bounces back to the

ionosphere where it is again gradually refracted and returned earthwards as if by reflection. This continues

until it is completely attenuated.

24. In 56

26

Fe nucleus, there are 26 protons and 30 neutrons

Mass defect � �56

26

26 30 Fep n

m m m� �

= 0.528461 u

Binding energy = (mass defect)c2

$ 0.528461 931.5 MeV 492.26 MeV� �

�Binding Energy 492.26 MeV

8.76 MeV

Nucleon 56

� �

(8)


25. Consider the -decay of 238

92

U

238 234 4

92 90 2

U Th He��

Neutron-proton ratio before -decay =

238 92

1.587

92

�

Neutron-proton ratio after -decay =

234 90

1.6

90

�

Thus, neutron-proton ratio increases during -decay.

Nuclear Force : Force between nucleons is called nuclear forces. It is one of the four fundamental forces in

nature.

Properties :

(i) It is an attractive force.

(ii) It is independent of the interacting nucleons.

(iii) It is a short range force.

(iv) It is a non-central force.

OR

(a)� �

2

2

13.6

eVn

Z

E

n

�

For Li , 3Z

��

2

122.4

eVn

E

n

�

For transition from n = 1 to n = 3

3 1

hc

E E E� � �%

114.2 Å% �

(b) No. of spectral lines = 3

26. During solar flare, a large number of electrons and protons are ejected from the sun, some of them get trapped in

the earth’s magnetic field and move in helical paths along the field lines. The magnetic field lines come closer to

each other near the magnetic pole. Hence, density of charges increases near the poles. These particles collide

with atoms and molecules of the atmosphere. Excited oxygen atom emit green light and erected nitrogen atoms

emit pink light.

(9)


SECTION-D

27. (a) Consider point P on the screen, path difference

2 1

x S P S P� �

2

2 2

2

2

d

S P y D

⎛ ⎞� � �⎜ ⎟

⎝ ⎠

2

2 2

1

2

d

S P y D

⎛ ⎞� �⎜ ⎟

⎝ ⎠Q

P

S

D

y

d

S1

S2

2 2

2 1

2S P S P yd �

� �2 1 2 1

2

2

yd

S P S P S P S P D

D

� & &

yd

x

D

⇒ � �

For constructive Interference,

x n%� �

yd

n

D

⇒ % �

� �0,1,....

n D

y n

d

%⇒ � �

For destructive interference

� �2 1

2

x n

%� � �� 2 1

0,1,....

2

n D

y n

d

� %⇒ � �

Hence, fringe width

D

d

%�

(b) Fresnel distance,

2

F

a

Z �%

40 mF

Z �

OR

Huygens Principle : Every point on wavefront may be considered as a source that produces secondary wavelets.

These wavelets propagate in the forward direction with a speed equal to speed of wave motion. The surface which

touches these wavelets at any later instant is the position of new wavefront, called secondary wavefront.

1 2

AD CE

v v

�

sin

AD

i

CD

� from �CAD

sin

CE

r

CD

� from �CED

1

2

sin

sin

vi AD CD

r CD CE v

� � �

(10)


As 1 2

2 1

,

vc

v

v

��

2

1

sin

sin

i

r

��

sin i� = constant (Snell’s law)

Convex Lens

Refracted wavefrontIncident wavefrontRefracted

wavefront

Incident

wavefront

28. (a) V-I characteristic

VF(V)

VR (V)

IF (mA)

IR ( A)�

Minority

CarriersForward Bias

Reverse

Breakdown

Breakdown

Majority

Carriers

V (Knee Voltage)K

80

60

40

20

–80 –60 –40 –20

–1

–2

Forward Bias

P

n

V

Reverse Bias

P

n

V

(b)

1242 eV nm

0.207 eV

6000 nm

g

hc

E � � �%

Since g

E E' , the photodiode cannot detect wavelength of 6000 nm.

(11)


OR

(a) Transistor as a common - emitter amplifier.

Circuit Diagram.

ac inputac output

VBB

BE

BIB

CIC

VCE

RL

VCC

I

E

Operation :

(i) With no signal input

CE CC C L

V V I R� �

(ii) With signal applied to the emitter base circuit for positive half cycle, increases the forward bias resulting

in an increase in collector current. During negative half cycle, the input signal opposes the forward bias

of the input circuit, thereby reducing the emitter and consequently the collector current.

dc current gain C

B

I

I

� �

(b)3

9 V

30 A

300 10

CC BE CC

B

B B

V V V

I

R R

� & � � ��

(as BE CC

V V'' )

� �50 30 A 1.5 mAC B

I I� � � � �

� �� 3 3

9 V 1.5 10 A 2 10 6 VCE CC C C

V V I R

� � � � � �

29. (a) Magnetic field at a point on Axial line : End on position

� �0

2

4

S

m

B

r a

�⎛ ⎞� ⎜ ⎟�⎝ ⎠ �

�

� �0

2

4

N

m

B

r a

�⎛ ⎞� ⎜ ⎟�⎝ ⎠

�

2a

r

–m +m

S N

� �0

2

2 2

4

4

P S N

m ra

B B B

r a

��

� � �

Since r >> a

0

3

2

4

P

M

B

r

�⎛ ⎞⎛ ⎞� � ⎜ ⎟⎜ ⎟� ⎝ ⎠⎝ ⎠

�

Magnetic field at a point on equatorial line

� �0

2 2

4

N

m

B

r a

�⎛ ⎞� ⎜ ⎟�⎝ ⎠ �

�

� �0

2 2

4

S

m

B

r a

�⎛ ⎞� ⎜ ⎟�⎝ ⎠ �

�

S

B

P

B

N

B

S N

P

(12)


� �0

3/2

2 2

2

4

P N S

am

B B B

r a

�⎛ ⎞� � � ⎜ ⎟�⎝ ⎠ �

� � �

r >> a

� �0

3

4

P

M

B

r

�⎛ ⎞� ⎜ ⎟�⎝ ⎠

�

axial equatorial

2B B��

(b) As m

C

T

( �

'

'

m

m

T

T

(⇒ �

(

' 200 KT⇒ �

OR

(a) Biot Savart’s law,

dB

�

the magnetic field at point P

0

2

sin

4

Idl

dB

r

� �� r

dl

�

P

I

⇒net

2 sindB dB� ��

⇒

0

net 2

2 sin

4

Idl

dB

r

��

�

⇒

0

net 2

2

4

Idl a

dB

r r

��

dB

dB

x

r

��

a

x

I⇒

0

net 3

2

4

Ia

dB dl

r

��∫ ∫

⇒ � �

2

0

net 3/2

2 2

Ia

B

a x

��

(b)

420 A

30 div 6 10 A

div

g

I

�⎛ ⎞� � � �⎜ ⎟

⎝ ⎠

Shunt required 0.015

1

g

G

S

I

I

� � �

G

S

Ammeter

� � �

(13)

Chemistry (Code-A) Solutions of Sample Question Paper for Class XII

SECTION-A

1. 3-Bromo-5-fluoro-3, 5-dimethylheptane

2. (i) It is a zero order reaction.

(ii)

0

1/2

[R]

t

2K

�

3. Out of the five electron pairs around the central Br-atom, the two lone pairs are at equatorial positions to minimise

the repulsive interactions. Hence, BrF3

has T-shaped structure.

F Br

F

F

4. Dispersed phase – milk fats

Dispersion medium – water

5. Number of A-atoms per unit cell

1

8(corners) 1

8

� � �

Number of B-atoms per unit cell

1

6 3

2

� � �

Hence, the formula of the compound is AB3

.

6. ABaq.

��⇀�↽��

A–

(aq.) + B+

(aq.)

Initially 1 0 0

At equilibrium 0.5 0.5 0.5

van’t Hoff factor 1 1.5� � �

7. NaCN is used as a depressant for ZnS and prevents it from coming with froth.

8. 6XeF4

+ 12H2

O �� 4Xe + 2XeO3

+ 24HF + 3O2

SECTION-B

9. The structural formula of 4-methylpent-3-en-2-one is

H C — C — CH C — CH3 3

O

—

CH3

Due to the presence of H C — C — 3

O

group, it will respond positive to iodoform test.

SOLUTIONS

(14)

Solutions of Sample Question Paper for Class XII Chemistry (Code-A)

10.Osmotic pressure may be defined as the excess pressure which must be applied to the solution side to just

prevent the osmosis. Osmotic pressure is directly proportional to temperature as well as pressure.

11. 2Cu2

S + 3O2

�� 2Cu2

O + 2SO2

Cu2

S + 2Cu2

O �� 6Cu + SO2

12. The two components of starch are amylose and amylopectin. Amylose is a linear polymer of -D-(+)-Glucose

while amylopectin is heavily branched polymer of -D(+)-Glucose.

13. (a) H—C — (CHOH) — CH — OH CH CH CH CH CH CH4 2 3 2 2 2 2 3

��

O

Reduction

HI/P

n-Hexane

(b)(CHOH) + 5(CH CO) O (CHOCOCH ) + 5CH COOH

4 3 2 3 4 3��

——

CH OH2

Glucose

CHO

——

CH OCOCH2 3

Pentacetyl glucose

CHO

14.H SO H+ :OSO H

2 4 3��

H C — CH — O – H + H H C —CH — O — H3 2 3 2

+��

—

+

H

Slow

H C — CH + H O3 2 2

+

H — C — C — H + :OSO H H C = CH + H SO3 2 2 2 4

��

——

H

H

—

H

)Fast

15. (a) Hofmann’s Bromamide reaction:

H C — C — NH + Br + 4KOH H C – NH + 2KBr + K CO + 2H O3 2 2 3 2 2 3 2

��Heat

O

(b) Gabriel phthalimide reaction:

—

CO

CO

—

NH ��KOH(alc.)

–H O2

—

CO

CO

—

NK ��C H –I2 5

–KI

—

CO

CO

—

N — CH CH2 3

H /H O

+

2

—

COOH

COOH

+ H C — CH — NH3 2 2

(15)


16. (a) Distinction can be made by Lucas reagent. Treat both the solutions separately with Lucas reagent which is

a mixture of HCl(g) and anhy. ZnCl2

. The compound that shows turbidity instantly is 2-methylpropan-2-ol

while the solution that shows turbidity after 5 minutes of heating, is Butan-1-ol.

(b) Distinction can be made by litmus test. Add a few drops of blue litmus separately to the solutions of both

the compounds. The solutions which changes the colour of the blue litmus to red is that of phenol while

the other is benzyl alcohol.

17. (a) Aniline does not undergo Friedel Crafts reaction. It being a Lewis base, coordinates with anhy. AlCl3

.

—

H N AlCl2 3

��

The amino group now, is not in a position to activate the benzene ring towards electrophilic substitution,

that is alkylation or acylation. Also, AlCl3

does not remain free to generate the carbonation from alkyl or

acyl halide.

(b) The –NH

2

group over benzene ring makes it so electron-rich site that Br2

molecule itself gets polarized

under its influence. This helps in the generation of the electrophile Br+

, even in the absence of a halogen

carrier.

OR

(a) During acylation aniline gives CH – C – NH3

O

, which is a resonance stabilized compound.

(b) 3°-amine does not contain replaceable hydrogen on N-atom.

18. The solution of acetone and chloroform will show negative deviation from their ideal behaviour.

Cl–C–H..........O=C

Cl

Cl

�(+) �(–)CH

3

CH3

There is formation of hydrogen bond between acetone and chloroform molecules, so the escaping tendency of

both components is lowered.

SECTION-C

19. We know that, 3

A

z M

,

N a

��

here ‘a’ is edge length

23 8 3

6.23 6.023 10 (4.00 10 )

z 4.002 4

60

� � � ��

� The cubic unit cell is ‘face-centered cubic’.

(16)


Now, in fcc lattice,

a

r

2 2

�

8

4 10

2 2

��

= 1.414 ×10–8

cm

20. (a) Terylene is a condensation polymer and its monomers are Ethylene glycol (HOCH2

CH2

OH) and terephthalic

acid HOOC COOH

(b) Teflon is an addition polymer and its monomer is tetrafluoroethylene (CF2

= CF2

)

(c) Its monomer unit is isoprene i.e., CH –C=CH–CH3 3

CH3

21. (a) These two compounds are ionization isomers to each other and on treating the aq. solutions of the two with

AgNO3

(aq), [CO(NH3

)5

(SO4

)]Br gives yellow ppts of AgBr while the other compound does not.

(b) The electronic configuration of Fe(Z = 26) is [Ar], 3d6

, 4s2

while that of Fe(II) is [Ar], 3d6

since F–

is a

weak field ligand that does not cause pairing up of electrons, sp3

d2

hybridization takes place.

[FeF ] :6

4–

3d 4s 4p 4d

sp d3 2

hybridization

The structure of the complex is octahedral and it is a high-spin complex. On the other hand, CN–

is a strong

field ligand that causes greater crystal field splitting and hence, pairing up of electrons.

[Fe(CN) ] :6

4–

3d 4s 4p

d sp2 3

This way, d2

sp3

hybridization takes place and the octahedral structure is diamagnetic.

22. (a)

(i) The adsorbate molecules

are held to the surface of

adsorbent by weak van

der Waals forces.

(ii) It is not specific in nature.

(iii)It forms multimolecular

layers.

(i) Adsorbate molecules are

held to the surface of

adsorbent by strong

chemical forces (chemical

bonds).

(ii) It is highly specific.

(iii) It forms monomolecular

layers.

Physisorption Chemisorption

(17)


(b)

(i) When dispersion medium

likes dispersed phase,

the sol is called lyophilic.

(ii) They are quite stable and

are not precipitated easily.

(iii) They are reversible in

nature.

Lyophilic sol Lyophobic sol

(i) When dispersion medium

dislikes dispersed phase,

the sol is called lyophobic.

(ii) They are easily precipitated

by addition of a small

amount of a suitable

electrolyte.

(iii) They are reversible in

nature.

23. (a) ICl is polar due to electronegativity difference between I and Cl. But I2

is nonpolar covalent compound.

So stability of polar bond is less than covalent bond and ICl is more reactive than I2

.

(b)

N 1 odd electron (incomplete octet of N)

O

N — N

O

O

O

(Complete octet of N)

O

O

At lower temperature NO2

is converted into N2

O4

due to incomplete octet of N in NO2

.

(c) In H3

PO3

, oxidation number of P is +3 and in H3

PO4

, P present in higher oxidation number +5, so H3

PO4

is not further oxidised.

24. (a)

F

FF

F

Xe

O

sp d

3 2

, Square pyramidal

(b)

F

F

FP

F

F

Triangular bipyramidal

(c)

S

O

H

O

S

OO

O O

O O

H

sp

3

-Hybridised Sulphur

(18)


OR

(a) Ca P + 6H O 3Ca(OH) + 2PH3 2 2 2 3

��Cal. Phosphate Phosphine

(b) XeF + NaF Na [XeF ] [Sodium heptafluoroxenate (vi)]6 7

�� + –

(c) Cu + 2H2

SO4

(conc.) �� CuSO4

+ SO2

+ 2H2

O

25. KP

= Ae–EP

/RT

; P � Presence of catalyst

KA

= Ae–EA

/RT

; A � Absence of catalyst

P A(E E )/RT E/RTP

A

K

e e

K

��

P A A

10

A

K E E

Ine log 10

K RT 2.303RT

� �

P P A

A

K (E E ) (45 55)

Antillog Antilog

K 2.303RT 2.303R300K

⎡ ⎤ ⎡ ⎤� �⎢ ⎥⎢ ⎥

⎣ ⎦ ⎣ ⎦

P

A

K 10

Antilog 1.003

K 2.303 8.314 300

� ��

P AK K�

26. (a) KCN is ionic compound and generate C

–

N ion in solution, the attack takes place mainly through carbon atom

due to C — C bond strength is higher than C – N bond strength.

(b) Resonating structure of chlorobenzene

CI CI

)CI

)CI

)CI

—

— — — — — — —

Lone pair of Cl is conjugated with double bonds of ring. So the electron density of ortho and para

positions increases by resonance and it undergoes electrophilic substitution at ortho and para positions

while Cl is deactivating group.

(c)CH = CH – CH – Cl CH = CH – CH + Cl

2 2 2 2��

Allyl Chloride Allyl (more stable carbocation

by resonance)

Nu)

CH – CH – CH – Cl CH – CH – CH + Cl3 2 2 3 2 2

��n-Propylchloride 1° carbocation

Nu)

Intermediate allyl carbocation is more stable than 1° carbocation. So the allyl chloride is more reactive towards

nucleophilic substitution.

(19)


27. Deficiency diseases by vitamin – A � Xerophthalmia, Night blindness

C � Scurvy (Bleeding gums)

Vitamin-A � Retinol

Vitamin-C � Ascorbic acid

SECTION-D

28. (a) (i) CH — CH + CH — CH CH — CH — CH — CH CH — CH — CH 3 2 3 2 3 2

��

O

H

O

Aldol

(Condensations)

Ba(OH)2

OH O

LiAlH4

OH

CH — OH2

Butan-1-3-diol

(ii) CH — C — CH + CH — Mg — I CH CH — C — OH 3 3 3 3 3

��

O

C

CH3

CH3

OMgI

–Mg(OH)I

H O2

CH3

CH3

Acetone Methyl

Mag. Iodide

Adduct

tert-butyl alcohol

(iii)

COOH

��

( HNO +H SO )3 2 4

COOH

NO2

Benzoic Acid Meta ntiro benzoic acid

(Conc.)

(b) (i) Clemmensen reduction

CH — CHO + 4[H] CH — CH + H O3 3 3 2

��Zn–Hg

�Ethanal Ethane

(ii) Cannizzaro reaction

2HCHO + NaOH CH OH + HCOONa��3

Formaldehyde Methanol

�

OR

(a) (i)

CH — CH CH — CH CH — CH — Cl CH — CH — CH — CH3 3 3 3 2 3 2 2 3

��

Ethanal

Clemmenson

reductionEthane

Mono

Chlorination

Wurtz

Reaction

O

Zn–Hg (Conc. HCl) Cl /U.V.2

Na

Cl /U.V.2

CH — CH — CH — CH CH — CH — CH = CH CH — CH — CH — CH3 2 2 2 3 2 2 3 2 3

*�� *��

Cl

(CH ) COK3 3

(i) B H ,THF2 6

(ii) H O /OH2 2

–

Butan-1-ol

OH

(20)


(ii)

CHO

Benzaldehyde

��[O]

KMnO4

COOH

��NaOH

CaO

��Cl – CH – CH = CH2 2

AlCl3

CH – CH = CH2 2

��B H , THF2 6

H O2 2

CH – CH2 2

HO – CH2

3-Phenyl Propanol

(iii)

CH – CH2 3

Ethyl Benzene

��Baeyer’s reagent

KMnO4

COOH

Benzoid Acid

��Soda Lime

NaOH + CaO

Decarboxylation

Benzene

(b) (i) Aldol condensation

CH — CH + CH — CHO CH — CH — CH — CHO3 2 3 2

��

O

Ba(OH)2

OH

Ethanal Aldol

OH

(ii) HVZ reaction

CH — C — OH CH — C — OH CH — C — OH 3 2

��

O

�Br /P

2

Br

O

Br /P2

�

Br

Br

O

Br /P2

� CBr — C — OH2

O

29. (a) H+

+ e–

� H2

(g)

Ecell

= 2

pH0.0591

– log

1 H�

�⎡ ⎤

⎣ ⎦

or –0.118 =

0.0591 1

– log

1 H�

�⎡ ⎤

⎣ ⎦

� [H+

] = 10–2

mol/L

� pH = 2

(b)At anode H (g) 2H + 2e

At cathode 2H + 2e +1/2 O (I) H O(l)

Net reaction

2

2 2

��

��

+ –

+ –

H (g) +1/2 O (g) H O(l)2 2 2

��

(21)


(c) Anode Zn–Hg

Cathode Carbon Paste

��

Mercury cell

Overall reaction,

E = 1.35 V0

Zn(Hg) + HgO(s) ZnO(s) + Hg(l)��

In the final equation of reaction, no solvent or ion is involved therefore, the emf does not drop easily.

OR

(a) E°cell

= E°cathode

– E°anode

= 0.34 – (–0.25)

= 0.59 V

� �G° = –2 × 96500 × 0.59 J

= –113870 J

= –113.9 kJ

Now, �G° = –2.303 RT log KC

� logKC

=

–113900

–2.303 8.314 298� � = 19.96

� KC

= antilog (19.96)

� KC

� 1020

(b) (i) At cathode

Cu+2

(aq) + 2e–

�� Cu(s)

At anode

OH (aq) OH + e�� –

4OH �� 2H2

O + O2

(ii) 2H SO 2H + 2HSO2 4 4

+

At anode

2HSO H S O + 2e4 2 2 8�� –

H S O + H O 2H SO + ½ O2 2 8 2 2 4 2

��

At cathode

2H + 2e H

) – �� +2

Net reaction

H O H +1/2 O2 2 2

��Electric

Current

(22)


(c) W = ZQ

W W 9

Q 96500 96500

Z E 9

� � � � �

Q 96500 coulomb�

30. (i) Cl – C – NO2

Cl

Cl

.

Which is also called as tear gas. When it reaches in eye it irritates gland and brings tears.

(ii) Social responsibility and social justice.

(iii) 1, 1, 1 Trichloro-1-nitromethane

OR

(i) I2

< F2

< Br2

< Cl2

: Increasing bond dissociation energy

(ii) HF < HCl < HBr < HI : Increasing acid strength

(iii) SbH3

< AsH3

< PH3

< NH3

: Increasing base strength

(iv) H2

S < H2

Se < H2

Te < H2

O : Increasing boiling point

(v) HCl < HBr < HI < HF : Increasing boiling point

� � �

(23)

Mathematics (Code-A) Solutions of Sample Question Paper for Class XII

SECTION-A

1. � �1 1 11 3

sin ,cos ,tan 3

2 6 2 6 3

⎛ ⎞� � �⎛ ⎞ � � �⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

0

6 6 3

� � ��

2. 12 × 27 = HCF(12, 27) = 3

3. Cofactor of a31

= (–1)3 + 1

[(4 × 1) – (2 × 1)] = 2

4. For singular matrix

1 2

0

3 1

x

x

��

$ (x + 1) – 2(3 – x) = 0

$ x + 1 – 6 + 2x = 0

5

3

x⇒ �

5. � �� 1 1 ( 1)x x x dx � �∫

3

2

( 1)( 1) ( 1)

3

x

x x dx x dx x C� � � � �∫ ∫

6.ˆ ˆ ˆ

3 2 2a b i j k� � � �

�

| | 9 4 4 17a b� � � � ��

�

7. Power of

2

2

1

d y

dx

� � degree of given differential equation

8. x = 4

9.ˆ ˆ ˆ ˆ ˆ( ) (2 3 4 )r i j i j k� � � % � �

�

10.

2 2 2

2(1) 3(2) 4(3) 8 12 4

29 292 3 4

� � ��

SOLUTIONS

(24)

Solutions of Sample Question Paper for Class XII Mathematics (Code-A)

SECTION-B

11. � �2 2cos 2sin

dy d

d d

� � � � ��

� �2 2sin 2 2cos

dx d

d d

� � � � ��

2sin

2 2cos

dy

dyd

dxdx

d

��

�

Now,

( /2)

dy

dx ��

⎛ ⎞

⎜ ⎟

⎝ ⎠

2(1) 2

1

2 2(0) 2

� � �

12. For continuity left hand limit at x = 0 should be equal to right hand limit at x = 0.

0

lim (0 1)

x

a a

��

=

3

0

sin

sin

coslim

x

x

x

x

x

��

2

2

0 0

sin

sin (1 cos ) sin 2 1 12

lim lim . . .

4 cos 2cos

2

x x

x

x x x

xx x xx x

� ��

⎡ ⎤⎛ ⎞

⎜ ⎟⎢ ⎥ ⎝ ⎠⎢ ⎥� � �⎢ ⎥

⎢ ⎥⎣ ⎦

1

2

a⇒ �

13. Put sin–1

x = �

2

and sin

1

dx

d x

x

� � � �

sin d⇒ � � �∫

Apply integration by parts, we get

cos cos d� � � � �∫

1 2

sin cos sin 1C x x x C

⇒ � � � � � � �

OR

sin( )

sin

a x

dx

x

∫

sin cos cos sin

sin sin

a x a x

dx

x x

⎛ ⎞� ⎜ ⎟

⎝ ⎠∫

sin cot cosa xdx adx� ∫ ∫

= sina�log|sinx| – xcosa + C

(25)


14.

1 1 13 8 84

sin sin cos

5 17 85

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ �⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠

Now,1 1

3 3

sin tan

5 4

⎛ ⎞ ⎛ ⎞�⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

$ 1 18 8

sin tan

17 15

⎛ ⎞ ⎛ ⎞�⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

$ 1 184 13

cos tan

85 84

⎛ ⎞ ⎛ ⎞�⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

Now we have to prove,

1 1 13 8 13

tan tan tan

4 15 84

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ �⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠

L.H.S.

1

3 8

4 15tan

3 8

1 ·

4 15

⎛ ⎞⎜ ⎟

� ⎜ ⎟

⎜ ⎟�⎜ ⎟

⎝ ⎠

=

1

45 32

60tan

60 24

60

⎛ ⎞

⎜ ⎟

⎜ ⎟�⎜ ⎟⎜ ⎟

⎝ ⎠

=

113

tan

84

⎛ ⎞

⎜ ⎟

⎝ ⎠

R.H.S. proved.

15. Given fog = gof = IR

$ g(x) is inverse of f(x)

2 3

3 – 2

x

y

x

�⇒ �

$ 3y – 2xy = 2x + 3 $ 2x(1+y) = 3(y – 1)

$

3 1

2 1

y

x

y

⎛ ⎞�⎜ ⎟�⎝ ⎠

13 1

( ) ( )

2 1

x

g x f x

x

⎛ ⎞⇒ � �

⎜ ⎟�⎝ ⎠

Domain of g(x), x + 1 , 0

1x⇒ ,

Domain of � �( ) 1g x R�

(26)


OR

At x = x1

, f(x1

) = 2x

1

+ 7

and at x = x2

, f(x2

) = 2x

2

+ 7

Let f(x1

) = f(x2

)

$ 2x

1

+ 7 = 2x

2

+ 7

$ x

1

= x2

is the only solution

Hence f(x) is an one-one function, also range of f is (–-, -)

Hence f is onto

Finding inverse of f(x),

y = 2x + 7

7

2

y

x

⇒ �

17

( )

2

x

f x

⇒ �

16. sec sec tan

dy

y x x x

dx

� �

Comparing with ( ) ( )

dy

yP x Q x

dx

⎛ ⎞� �⎜ ⎟⎝ ⎠

sec ln(sec tan )

I.F (sec tan )

xdxx x

e e x x

�∫� � � �

2 2

(sec tan ) (sec tan )y x x x x dx⇒ � � ∫

(sec tan )y x x x C⇒ � � �

17.ˆ ˆ

8 6a b c i j� � � ��

� �

Squaring both sides

2 2ˆ ˆ( ) (8 6 )a b c i j� � � �

��

$ ˆ ˆ ˆ ˆˆ( )( ) (8 6 )(8 6 )a b c a b c i j i j� � � � � � ��

� � �

$ 2 2 2

| | | | | | 2( ) 64 36a b c a b b c c a� � � � � � � � � ��

� � � � � �

$ 25 64 9 2( ) 100a b b c c a� � � � � � � � ��

� � � �

$ 2( ) 2a b b c c a� � � � � ��

� � � �

$1a b b c c a� � � � � �

� ��

18.

1

11

x

dx

I

e

��∫ … (i)

Using property ( ) ( )

b b

a a

f x dx f a b x dx� � ∫ ∫

1 1

1 11 1

x

x x

dx e dx

I

e e

⇒ � �� ∫ ∫ …(ii)

(27)


Adding (i) and (ii), we get

1

1

1

2

1

x

x

e

I dx

e

⎛ ⎞�� ⎜ ⎟⎜ ⎟�⎝ ⎠

∫

1

1

2I dx

⇒ � ∫

1

1

2I x ⇒ �

1 ( 1) 2

1

2 2

I

⇒ � � �

OR

0

| cos | | sin |x x dx

�

�∫

/2

0 /2 0

cos cos sinxdx xdx xdx

� � �

�

� �∫ ∫ ∫

/2

0 /2 0

(sin ) ( sin ) ( cos )x x x

� � ��

$ 1 + (–(–1)) + (1 – (–1)

$ 1 + 1 + 2 = 4

19.2

3 6

dy

x

dx

� = Slope of line

3x

2

– 6 = – 3

$ 3x

2

= 3 $ x = ±1

for x = 1 y = –2

and x = –1 y = 8

Point (1, –2) satisfies the line but (–1, 8) doesn’t, so required point (1, –2)

OR

Let ( )f x x�

Let x = 25 and �x = 0.2 then

�y = (x + �x)1/2

– x1/2

�y = (25.2)1/2

– (25)1/2

$ 25.2 25 5y y� � � � � �

Now approximating �y

dy

y x

dx

⎛ ⎞� � �⎜ ⎟

⎝ ⎠

1

(0.2)

2 x

�(as )y x�

1

0.2 0.02

2 5

� � ��

25.2 5 0.02 5.02⇒ � � �

(28)


20. x + 2y + z = 1

x + y – z = 2

2x + 3y + z = 1

1 2 1 1

1 1 –1 2

2 3 1 1

x

y

z

⎡ ⎤ ⎡ ⎤ ⎡ ⎤

⎢ ⎥ ⎢ ⎥ ⎢ ⎥�⎢ ⎥ ⎢ ⎥ ⎢ ⎥

⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

AX = B, X = A

–1

B

Now,

| A |

= 1(1 + 3) – 2 (1 + 2) + 1(3 – 2) = 4 – 6 + 1 = –1

1

4 1 –3 –4 –1 3

adj.

1 –3 –1 2 = 3 1 –2

1 1 –1 –1 –1 1

A

A

A

⎡ ⎤ ⎡ ⎤

⎢ ⎥ ⎢ ⎥� � ⎢ ⎥ ⎢ ⎥

⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

1

–4 –1 3 1 3

= 3 1 –2 2 3

–1 –1 1 1 2

x

x y A B

z

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤

⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥� � �⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥

⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

� x = –3, y = 3, z = –2

21. Vector direction of lines perpendicular to both given lines is

ˆ ˆ ˆ

ˆ ˆ ˆ ˆ ˆ ˆ2 2 4 (2 12) (2 16) (6 8) 10 14 2

4 3 1

i j k

i j k i j k� � � �

Line passing through (2, 1, 2) and parallel to ˆ ˆ ˆ

( 10 14 2 )i j k � in vector form is

ˆ ˆ ˆ ˆ ˆ ˆ(2 2 ) ( 10 14 2 )r i j k i j k� � � � % �

�

22. Let the coin tossed n times

Probability of getting at least one tail = 1 – (Probability of getting no tail)

1

1

2

n

⎛ ⎞� ⎜ ⎟

⎝ ⎠

Given that

1

1 0.9

2

n

⇒ .

1

0.1

2

n

⇒ .

1

2

0.1

n

⇒ .

2 10

n

⇒ .

4n⇒ .

So coin must be tossed 4 times.

(29)


SECTION-C

23. Let ABCD be a rectangle inscribed in the circle of given radius r.

AB = x and BD = y

Area = x × y

Also, x2

+ y2

= 4r

2

For maximum area,

2

2

0, 0

dA d A

dx dx

� '

� �1

2 22

4 0

dA d

x r x

dx dx

⎡ ⎤

� �⎢ ⎥

⎢ ⎥⎣ ⎦

A B

C D

y

x

r

r

2 3

2 2 4

(8 4 )

0

2 4

r x x

r x x

⇒ �

2x r⇒ � , also

2

2

0

d A

dx

' for 2 x r� . And for 2 , 2x r y r� �

Hence for maximum area ABCD must be square

24. Let the plane be Ax + By + Cz + 1 = 0

Passing through (1, 2, 1), (2, 3, 1) and (2, 1, 0), we get

A + 2B + C + 1 = 0 …(i)

2A + 3B + C + 1 = 0 …(ii)

2A + B + 1 = 0 …(iii)

Given A = –1, B = 1 and C = –2

$ Required plane –x + y – 2z + 1 = 0

$�x – y + 2z = 1

Distance from point (1, 1, 1) is

2 2

1 1 2 1 1

61 1 4

d

� � ��

25. Required area = area of ABCD

1

2

0

2 ((– 2) )x x dx� � ∫

1

2 3

0

2 2

2 3

x x

x

⎛ ⎞/ � ⎜ ⎟⎜ ⎟

⎝ ⎠

1 1

2 2 0

2 3

⎛ ⎞/ � ⎜ ⎟

⎝ ⎠

(–2, 0)

(2, 0)

(1, 1)(–1, 1)

y

x

=

+ 2

y x = – + 2

A

B

C

0

5

2 2

6

⎛ ⎞/ ⎜ ⎟

⎝ ⎠

12 5 7

2 units

6 3

⎛ ⎞/ �⎜ ⎟

⎝ ⎠

(30)


OR

Area of shaded region required

1 2

2

0 1

( 1) ( 1)x dx x dx� � �∫ ∫

1 2

3 2

0 1

3 2

x x

x x

⎛ ⎞ ⎛ ⎞

� � � �⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

P(0, 1)

Q

(1, 2)

(0, 2)

x = 2

x = 1

1 1 23

1 2 2 1 units

3 2 6

� � � � �

26. Let number of nuts be ‘x’ and bolts be ‘y’

� L.P.P. is

Maximise z = 20x + 10y

such that 1.5x + 3y � 42

and 3x + y � 24

B(4, 12)A(0, 14)

C (8, 0)O

3 + = 24x y1.5 +3 = 42x y

x

y

x . 0, y . 0

Vertices of feasible region are A(0, 14), B(4, 12), C(8, 0)

Z(A) = Rs. 140, Z(B) = Rs. 200, Z(C) = Rs. 160

For maximum profit

Number of nuts = 4

Number of bolts = 12

27. Given matrix can be written as

1 3 2 1 0 0

3 0 1 0 1 0

2 1 0 0 0 1

A

⎡ ⎤ ⎡ ⎤

⎢ ⎥ ⎢ ⎥�⎢ ⎥ ⎢ ⎥

⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦

Applying C2

� C2

+ 3C

1

, C3

��C3

– 2C

1

1 0 0 1 3 2

3 9 5 0 1 0

2 7 4 0 0 1

A

⎡ ⎤ ⎡ ⎤

⎢ ⎥ ⎢ ⎥⇒ �

⎢ ⎥ ⎢ ⎥

⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦

2 2 3

1 0 0 1 1 2

2 3 1 5 0 1 0

2 1 4 0 2 1

C C C A

⎡ ⎤ ⎡ ⎤

⎢ ⎥ ⎢ ⎥� � ⇒ �⎢ ⎥ ⎢ ⎥

⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

3 3 2

1 0 0 1 1 3

5 3 1 0 0 1 5

2 1 1 0 2 9

C C C A

⎡ ⎤ ⎡ ⎤

⎢ ⎥ ⎢ ⎥� ⇒ � ⎢ ⎥ ⎢ ⎥

⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦

2 2 3 3 3

1 0 0 1 2 3

, – 3 1 0 0 4 5

2 0 1 0 7 9

C C C C C A

⎡ ⎤ ⎡ ⎤

⎢ ⎥ ⎢ ⎥� � � ⇒ � ⎢ ⎥ ⎢ ⎥

⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦

(31)


1 1 3 2 2

1 0 0 5 2 3

2 , 3 1 0 10 4 5

0 0 1 18 7 9

C C C C C A

⎡ ⎤ ⎡ ⎤

⎢ ⎥ ⎢ ⎥� � � ⇒ �⎢ ⎥ ⎢ ⎥

⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

1 1 2

1 0 0 1 2 3

3 0 1 0 2 4 5

0 0 1 3 7 9

C C C A

⎡ ⎤ ⎡ ⎤

⎢ ⎥ ⎢ ⎥� ⇒ � ⎢ ⎥ ⎢ ⎥

⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

1

1 2 3

2 4 5

3 7 9

A

⎡ ⎤

⎢ ⎥� � ⎢ ⎥

⎢ ⎥⎣ ⎦

28.

3 4x ydy

e

dx

�

4 3y x

e dy e dx⇒ �

Integrating both sides

4 3y x

e dy e dx�∫ ∫

4 3

4 3

y x

e e

C⇒ � �

4 3

4 3

y x

e e

C⇒ � � Putting x = 0, y = 0 we get

1

12

c �

4 3

1

0

4 3 12

y x

e e � �

4 3

3 – 4 1 0

y x

e e⇒ � �

29. Let Selected student of section A

Eventof student that student is passed

Student of section B

A

E

B

///

( )

( ) ( )

E

P P A

A A

P

E EE

P P A P P B

A B

⎛ ⎞ �⎜ ⎟

⎛ ⎞ ⎝ ⎠�⎜ ⎟

⎛ ⎞ ⎛ ⎞⎝ ⎠ � � �⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

45 30 3 2

0.75, 0.5, ( ) , ( )

60 40 5 5

E E

P P P A P B

A B

⎛ ⎞ ⎛ ⎞� � � � � �⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

3

0.75

2.25 95

3 2 3.25 13

0.75 0.5

5 5

A

P

E

�⎛ ⎞ � � �⎜ ⎟

⎛ ⎞ ⎛ ⎞⎝ ⎠ � � �⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

(32)


OR

Bag A Bag B Bag C

2R + 3W 2W + 3B 2B + 3R

P(A) = Probability of selecting bag A =

1

3

P(B) = Probability of selecting bag B =

1

3

P(C) = Probability of selecting bag C =

1

3

P(W) = Probability of selecting white ball.

( ) ( ). ( ). ( ).

W W W

P W P A P P B P P C P

A B C

⎛ ⎞ ⎛ ⎞ ⎛ ⎞� � �⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠

1 3 1 2 1 1 2 3 2 1

0

3 5 3 5 3 5 15 15 3

��

� � �

Physics (Code-A) Sample Question Paper for Class XII

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Transcript of Physics (Code-A) Sample Question Paper for Class XII