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

(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,

251 1 6

5

DM

f

When image is formed at infinity,

255

5

DM

f

6. KEmax

= eVSP

3 eV3 volt.

SPV

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

9040

90

C

C

72 FC⇒

Physics (Theory) - Class XII (Code-A)

SOLUTIONS

(2)

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

10. (a) Outside the shell,

enc

0

qE ds

∫�

�

�

2

0

4q

E r⇒

R r R>

2

04

qE

r⇒

(b) Inside the shell, Take Gaussian surface

enc

0

qE 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

Li

SI unit is henry.

Consider, air cored solenoid

I IBAN

2

0ni r nl

2 2

0L 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 enclosedB dl I ∫� �

�

Magnetic field due to a toroid

r

Current

in each

turn is I

0 enclosedB dl I ∫� �

�

02B r NI

0

2

NIB

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.

tanh

f

when h = 1, 1

tanf

Power of a lens 1

tanPf

f

h F

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

When lenses are in contact

eq 1 2P P P

14.

Magnification :

(a) For strained eye, 1

o e

L DM

f f

⎡ ⎤ ⎢ ⎥⎣ ⎦

(b) For relaxed eye,

o e

LDM

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 = 2fL

f = 500 Hz

(b)1

0.10 F2

CX C

fC ⇒

OR

15 FC

1212.3

2C

XfC

rms

rms0.52 A

C

VI

X

SECTION-C

17.0

3,

4

I dl rdB B

r

��

� �

directed outwards

0

24

I dldB

r

r

dl

I

0 0

24 2

Idl IB dB

r r

∫ ∫

For half coil 0

4

IB

r

(5)


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

Numerically, Q

CV

.

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

0

E

VE

d

–

–

–

–

–

–

–

–

–

+

+

+

+

+

+

+

+

+

E

P

Q –Q

0

dV

0AQ

CV d

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

Cd 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 eEV

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

fv x

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

v d x

Thus, d x x

fd x x

20x dx fd⇒

24

2

d d fdx

⇒

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

i.e. 1 4 0f

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

0sinE E t

L R C

0 0,

R L LV I R V I X and

0C CV I X

22

0 R L CE V V V

22

0 0 L CE I R X X

Impedance 22

L CZ R X X

0I is max, when

L C

12

2fL

fC

( )I0 max

I

r

1

2f

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

26Fe nucleus, there are 26 protons and 30 neutrons

Mass defect 56

2626 30 Fe

p nm 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 MeVNucleon 56

(8)


25. Consider the -decay of 238

92U

238 234 4

92 90 2U Th He

Neutron-proton ratio before -decay = 238 92

1.58792

Neutron-proton ratio after -decay = 234 90

1.690

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

13.6eV

n

ZE

n

For Li , 3Z

2

122.4eV

nE

n

For transition from n = 1 to n = 3

3 1

hcE 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 1x S P S P

2

2 2

2

2

dS P y D

⎛ ⎞ ⎜ ⎟⎝ ⎠

2

2 2

1

2

dS P y D

⎛ ⎞ ⎜ ⎟⎝ ⎠ Q

P

S

D

y

d

S1

S2

2 2

2 12S P S P yd

2 1 2 1

2

2

ydS P S P S P S P D

D

ydx

D⇒

For constructive Interference,

x n yd

nD

⇒

0,1,....n D

y nd

⇒

For destructive interference

2 12

x n

2 10,1,....

2

n Dy n

d

⇒

Hence, fringe width D

d

(b) Fresnel distance,

2

F

aZ

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

sinAD

iCD

from CAD

sinCE

rCD

from CED

1

2

sin

sin

vi AD CD

r CD CE v

(10)


As 1 2

2 1

,

vcv

v

2

1

sin

sin

i

r

sin i = constant (Snell’s law)

Convex Lens

Refracted wavefrontIncident wavefront Refracted

wavefront

Incident

wavefront

28. (a) V-I characteristic

VF(V)

VR (V)

IF (mA)

IR ( A)

MinorityCarriers Forward Bias

Reverse Breakdown

Breakdown

Majority Carriers

V (Knee Voltage)K

80

60

40

20

–80 –60 –40 –20

–1

–2

Forward Bias

Pn

V

Reverse Bias

Pn

V

(b)1242 eV nm

0.207 eV6000 nm

g

hcE

Since gE 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

C IC

VCE

RL

VCC

I

E

Operation :

(i) With no signal input

CE CC C LV 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 V30 A

300 10

CC BE CC

B

B B

V V VI

R R

(as BE CC

V V )

50 30 A 1.5 mAC BI I

3 39 V 1.5 10 A 2 10 6 V

CE CC C CV V I R

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

0

24

S

mB

r a

⎛ ⎞ ⎜ ⎟⎝ ⎠

�

0

24

N

mB

r a

⎛ ⎞ ⎜ ⎟⎝ ⎠

�

2a

r

–m +m

S N

0

22 2

4

4P S N

m raB B B

r a

� � �

Since r >> a

0

3

2

4P

MB

r

⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

�

Magnetic field at a point on equatorial line

0

2 24

N

mB

r a

⎛ ⎞ ⎜ ⎟⎝ ⎠

�

0

2 24

S

mB

r a

⎛ ⎞ ⎜ ⎟⎝ ⎠

�

SB

PB

NB

S N

P

(12)


0

3/22 2

2

4P N S

amB B B

r a

⎛ ⎞ ⎜ ⎟⎝ ⎠

� � �

r >> a

0

34

P

MB

r

⎛ ⎞ ⎜ ⎟⎝ ⎠

�

axial equatorial2B 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

IdldB

r

r

dl

P

I

⇒net

2 sindB dB �

⇒0

net 22 sin4

IdldB

r

�

⇒0

net 224

Idl adB

r r

dB

dB

x

r

a

x

I⇒

0

net 3

2

4

IadB dl

r

∫ ∫

⇒

2

0

net 3/22 2

IaB

a x

(b)420 A

30 div 6 10 Adiv

gI

⎛ ⎞ ⎜ ⎟⎝ ⎠

Shunt required 0.015

1

g

GS

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) 18

Number of B-atoms per unit cell 1

6 32

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 + 12H

2O 4Xe + 2XeO

3 + 24HF + 3O

2

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.

Chemistry (Theory) - Class XII (Code-A)

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. 2Cu2S + 3O

2 2Cu

2O + 2SO

2

Cu2S + 2Cu

2O 6Cu + SO

2

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 COOH4 3 2 3 4 3

——

CH OH2

Glucose

CHO

——

CH OCOCH2 3

Pentacetyl glucose

CHO

14. H SO H+ :OSO H2 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 Br

2 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 36.23 6.023 10 (4.00 10 )

z 4.002 460

�

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

(16)


Now, in fcc lattice, a

r2 2

84 10

2 2

= 1.414 ×10–8 cm

20. (a) Terylene is a condensation polymer and its monomers are Ethylene glycol (HOCH2CH

2OH) and terephthalic

acid HOOC COOH

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

2)

(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(NH

3)5(SO

4)]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, sp3d2 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, d2sp3 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

OO

O

(Complete octet of N)

O

O

At lower temperature NO2 is converted into N

2O

4 due to incomplete octet of N in NO

2.

(c) In H3PO

3, oxidation number of P is +3 and in H

3PO

4, P present in higher oxidation number +5, so H

3PO

4

is not further oxidised.

24. (a)

F

FF

F

Xe

O

sp d3 2

, Square pyramidal

(b)

F

F

FP

F

F

Triangular bipyramidal

(c)S

O

H

OS

OO

O O

O O

H

sp3

-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 + 2H2SO

4(conc.) CuSO

4 + SO

2 + 2H

2O

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

Ke e

K

P A A

10

A

K E EIne 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 10Antilog 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 + Cl2 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 + HCOONa3

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

Clemmensonreduction

EthaneMono

ChlorinationWurtz

ReactionO

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-olOH

(20)


(ii)

CHO

Benzaldehyde

[O]KMnO

4

COOH

NaOH

CaOCl – CH – CH = CH

2 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

– log1 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 2H2O + O

2

(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 9Q 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 < F

2 < Br

2 < Cl

2 : Increasing bond dissociation energy

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

(iii) SbH3 < AsH

3 < PH

3 < NH

3 : Increasing base strength

(iv) H2S < H

2Se < H

2Te < H

2O : 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 3sin ,cos ,tan 3

2 6 2 6 3

⎛ ⎞ ⎛ ⎞ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

06 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 20

3 1

x

x

(x + 1) – 2(3 – x) = 0

x + 1 – 6 + 2x = 0

5

3x⇒

5. 1 1 ( 1)x x x dx ∫

32( 1)( 1) ( 1)

3

xx x dx x dx x C ∫ ∫

6. ˆ ˆ ˆ3 2 2a b i j k �

�

| | 9 4 4 17a b �

�

7. Power of

2

21

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

MathematicsMathematicsMathematicsMathematicsMathematics - Class XII (Code-A)

SOLUTIONS

(24)

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

SECTION-B

11. 2 2cos 2sindy d

d d

2 2sin 2 2cosdx d

d d

2sin

2 2cos

dy

dy d

dxdx

d

Now,

( /2)

dy

dx

⎛ ⎞⎜ ⎟⎝ ⎠

2(1) 2

12 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

sinsin

coslimx

xx

x

x

2

20 0

sinsin (1 cos ) sin 2 1 12

lim lim . . .4 cos 2cos

2

x x

x

x x x

xx x xx x

⎡ ⎤⎛ ⎞⎜ ⎟⎢ ⎥ ⎝ ⎠⎢ ⎥

⎢ ⎥⎢ ⎥⎣ ⎦

1

2a⇒

13. Put sin–1x =

2

and sin

1

dxd x

x

sin d⇒ ∫Apply integration by parts, we get

cos cos d ∫

1 2sin cos sin 1C x x x C

⇒

OR

sin( )

sin

a xdx

x

∫

sin cos cos sin

sin sin

a x a xdx

x x

⎛ ⎞ ⎜ ⎟⎝ ⎠∫

sin cot cosa xdx adx ∫ ∫= sinalog|sinx| – xcosa + C

(25)


14.1 1 13 8 84

sin sin cos5 17 85

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

Now,1 13 3

sin tan5 4

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

1 18 8sin tan

17 15

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

1 184 13cos tan

85 84

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Now we have to prove,

1 1 13 8 13tan tan tan

4 15 84

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

L.H.S.1

3 8

4 15tan

3 81 ·

4 15

⎛ ⎞⎜ ⎟ ⎜ ⎟

⎜ ⎟⎜ ⎟⎝ ⎠

=

1

45 32

60tan

60 24

60

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

= 1 13

tan84

⎛ ⎞⎜ ⎟⎝ ⎠

R.H.S. proved.

15. Given fog = gof = IR g(x) is inverse of f(x)

2 3

3 – 2

xy

x

⇒

3y – 2xy = 2x + 3 2x(1+y) = 3(y – 1)

3 1

2 1

yx

y

⎛ ⎞ ⎜ ⎟⎝ ⎠

1 3 1( ) ( )

2 1

xg x f x

x

⎛ ⎞⇒ ⎜ ⎟⎝ ⎠

Domain of g(x), x + 1 0

1x⇒

Domain of ( ) 1g x R

(26)


OR

At x = x1, f(x

1) = 2x

1 + 7

and at x = x2, f(x

2) = 2x

2 + 7

Let f(x1) = f(x

2)

2x1 + 7 = 2x

2 + 7

x1 = x

2 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

yx

⇒

1 7( )

2

xf x

⇒

16. sec sec tandy

y x x xdx

Comparing with ( ) ( )dy

yP x Q xdx

⎛ ⎞ ⎜ ⎟⎝ ⎠

sec ln(sec tan )I.F (sec tan )

xdx x xe e x x

∫

2 2(sec tan ) (sec tan )y x x x x dx⇒ ∫

(sec tan )y x x x C⇒

17. ˆ ˆ8 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

dxI

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 dxI

e e

⇒ ∫ ∫ …(ii)

(27)


Adding (i) and (ii), we get

1

1

12

1

x

x

eI dx

e

⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠∫

1

1

2I dx

⇒ ∫

1

12I x ⇒

1 ( 1) 21

2 2I

⇒

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 6dy

xdx

= Slope of line

3x2 – 6 = – 3

3x2 = 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

dyy x

dx

⎛ ⎞ ⎜ ⎟⎝ ⎠

1(0.2)

2 x

(as )y x

10.2 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 3adj.

1 –3 –1 2 = 3 1 –2

1 1 –1 –1 –1 1

AA

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

12

n

⎛ ⎞ ⎜ ⎟⎝ ⎠

Given that 1

1 0.92n

⇒

10.1

2n

⇒

12

0.1

n⇒

2 10n

⇒

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 = 4r2

For maximum area,

2

20, 0

dA d A

dx dx

1

2 2 24 0dA d

x r xdx dx

⎡ ⎤ ⎢ ⎥

⎢ ⎥⎣ ⎦

A B

C D

y

x

rr

2 3

2 2 4

(8 4 )0

2 4

r x x

r x x

⇒

2x r⇒ , also

2

20

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 4d

25. Required area = area of ABCD

1

2

0

2 ((– 2) )x x dx ∫

12 3

0

2 22 3

x xx

⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠

1 12 2 0

2 3

⎛ ⎞ ⎜ ⎟⎝ ⎠

(–2, 0)

(2, 0)

(1, 1)(–1, 1)

yx

=

+ 2

y x = – + 2

A

B

C

0

52 2

6

⎛ ⎞ ⎜ ⎟⎝ ⎠

12 5 72 units

6 3

⎛ ⎞ ⎜ ⎟⎝ ⎠

(30)


OR

Area of shaded region required

1 2

2

0 1

( 1) ( 1)x dx x dx ∫ ∫

1 23 2

0 13 2

x xx x

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

P(0, 1)

Q(1

, 2)

(0, 2)

x = 2

x = 1

1 1 231 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 24B(4, 12)A(0, 14)

C (8, 0)O

3 + = 24x y 1.5 +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 C

2 + 3C

1, C

3 C

3 – 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 y

dye

dx

4 3y xe dy e dx⇒

Integrating both sides

4 3y xe dy e dx∫ ∫

4 3

4 3

y xe e

C⇒

4 3

4 3

y xe e

C⇒ Putting x = 0, y = 0 we get 1

12c

4 31

04 3 12

y xe e

4 33 – 4 1 0

y xe e⇒

29. Let Selected student of section A

Eventof student that student is passed

Student of section B

A

E

B

( )

( ) ( )

EP P A

A AP

E EEP P A P P B

A B

⎛ ⎞ ⎜ ⎟⎛ ⎞ ⎝ ⎠⎜ ⎟

⎛ ⎞ ⎛ ⎞⎝ ⎠ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

45 30 3 20.75, 0.5, ( ) , ( )

60 40 5 5

E EP P P A P B

A B

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

30.75

2.25 95

3 2 3.25 130.75 0.5

5 5

AP

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 PA B C

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

1 3 1 2 1 1 2 3 2 10

3 5 3 5 3 5 15 15 3

� � �

Documents

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