PHYSICS (PAPER-2)

SECTION – 1 : (Maximum Marks : 18)

This section contains SIX questions.

Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is correct.

For each question, darken the bubble corresponding to the correct option in the ORS.

For each questions, marks will be awarded in one of the following categories :

Full Marks : +3 If only the bubble corresponding to the correct option is darkened.

Zero Marks : 0 If none of the bubbles is darkened.

Negative Marks : –1 In all other cases.

1 : ( : 18)

bl [kaM esa iz'u gSaA

izR;sd iz'u esa pkj fodYi (A), (B), (C) rFkk (D) gSaA ftuesa dsoy ,d gh lgh gSA

izR;sd iz'u ds fy, vks-vkj-,l- ij lgh mÙkj fodYi ds vuq:i cqycqys dks dkyk djsaA

izR;sd iz'u ds fy, vad fuUufyf[kr ifjfLFkr;ksa esa ls

iw.kZ vad : +3 ;fn flQZ lgh fodYi ds vuq:i cqycqys dks dkyk fd;k gSA

'kwU; vad : 0 ;fn fdlh Hkh cqycqys dks dkyk ugha fd;k gSA

_.k vad : –1 vU; lHkh ifjfLFkr;ksa esaA

1. A gas is enclosed in a cylinder with a movable frictionless piston. Its initial thermodynamic state at

pressure Pi = 105 Pa and volume Vi = 10

–3 m

3 changes to a final state at Pf = (1/32) × 10

5 Pa and

Vf = 8 × 10–3

m3 in an adiabatic quasi-static process, such that P

3V

5 = constant. Consider another

thermodynamic process that brings the system from the same initial state to the same final state in two

steps : an isobaric expansion at Pi followed by an isochoric (isovolumetric) process at volume Vf. The

amount of heat supplied to the system in the two-step process is approximately.

,d [kks[kys csyu] ftlesa ,d ?k"kZ.k&jfgr pyk;eku fiLVu yxk gS] esa ,d xSl can gSA fudk; dh izkjfEHkd Å"ekxfrdh

voLFkk (thermodynamic state) esa xSl dk ncko Pi = 105 Pa ,oa vk;ru Vi = 10

–3 m

3 gSA ,d :}ks"e LFkSfrddYi

(adiabatic quasi-static) dh izfØ;k] ftlesa P3V

5 = fLFkjkad gS] ls fudk; vafre Å"ekxfrdh voLFkk Pf = (1/32) × 10

5 Pa ,oa

Vf = 8 × 10–3

m3 esa ifjofrZr gks tkrk gSA ,d nwljh Å"ekxfrdh izfØ;k esa ogh izkjfEHkd ,oa vafre voLFkk,a nks pj.kksa esa

iw.kZ dh tkrh gSa : igys pj.k esa Pi ij leku ncko of) (isobaric expansion) ds ckn nwljs pj.k esa ,d leku vk;ru

izfØ;k(isochoric /isovolumetric process) Vf vk;ru ij gksrh gSA nks pj.kksa okyh izfØ;k esa fudk; dks nh xbZ

Å"ek dh ek=kk yxHkx gSA

(A) 112 J (B) 294 J (C) 588 J (D) 813 J

Ans. (C)

Sol. ���✁✂✄☎✆✝☎✞✂✁✞✟✆

105

5110

32�

10–3 8 × 10

–3

P (pascal)

V(m3)

PV5/3

= C ✁ ✂ = 5/3

✄Q1 = nCP✄T

= R

n T1☎✆

☎ ✝ =

P V

1

☎ ✆

☎ ✝ = 5 3

5

310 (7 10 )

51

3

✞

✟ ✠✡ ☛☞ ✌ ✍

✟ ✠✎✡ ☛

☞ ✌

= 2 25 357 10 10 J

2 2✏ ✏ ✑ ✏

✄Q2 = nCV✄T = nR ( P)V

T1 1

✆✆ ✒

☎ ✝ ☎ ✝ =

5 311 10 8 10

32

51

3

✓✔ ✕✖ ✗ ✗✘ ✙

✚ ✛

✖

= 231 8 10

332 2

✝ ✜✜ ✜ = 293

10 J8

✢✣

✄Q = ✄Q1 + ✄Q2 = 35 93

2 8

✤ ✥✦✧ ★

✩ ✪ × 10

2

✄Q2 = 140 93

1008

✢✣ =

4700

8 = 587.5 Joule

2. An accident in a nuclear laboratory resulted in deposition of a certain amount of radioactive material of

half-life 18 days inside the laboratory. Tests revealed that the radiation was 64 times more than the

permissible level required for safe operation of the laboratory. What is the minimum number of days

after which the laboratory can be considered safe for use ?

,d ukfHkdh; iz;ksx'kkyk esa nq?kZVuk dh otg ls jsfM;ks,fDVo inkFkZ dh dqN ek=kk tek gks x;h] ftldh v/kkZ;q 18

fnuksa dh gSA ijh{k.k ls irk pyk fd iz;ksx'kkyk esa fofdj.k dk Lrj lqjf{kr Lrj ls 64 xq.kk T;knk FkkA U;wure

fdrus fnuksa ds ckn iz;ksx'kkyk dke djus ds fy, lqjf{kr gksxh \

(A) 64 (B) 90 (C) 108 (D) 120

Ans. (C)

Sol. H0

t / TA A 2

✫✬

H00

t / TAA 2

64

✭✑ 6 = H

t

Tt = 6TH = 108 days

✮✮✮✯✰✱✲✳✴✲✵✰✯✵✶✳

3. There are two Vernier calipers both of which have 1 cm divided into 10 equal divisions on the main

scale. The Vernier scale of one of the calipers (C1) has 10 equal divisions that correspond to 9 main

scale divisions. The Vernier scale of the other caliper (C2) has 10 equal divisions that correspond to 11

main scale divisions. The readings of the two calipers are shown in the figure. The measured values (in

cm) by calipers C1 and C2, respectively, are

nks ofuZ;j dSfyilZ bl rjg ls gSa fd muds eq[; iSekus dk 1 cm, 10 leHkkxksa esa foHkkftr gSA ,d dSfyij (C1) ds

ofuZ;j iSekus ij 10 cjkcj Hkkx gSa tks fd eq[; iSekus ds 9 Hkkxksa ds cjkcj gSA nwljs dSfyij (C2) ds ofuZ;j iSekus

ij Hkh 10 cjkcj Hkkx gSa tks fd eq[; iSekus ds 11 Hkkxksa ds cjkcj gSaA nksuksa dSfyilZ ds iBuksa dks fp=k esa n'kkZ;k x;k

gSA C1 rFkk C2 n~okjk ekis x, lgh eku (cm esa) Øe'k% gSa

C1

2 3 4

C2

2 3 4

0 5 10

0 5 10

(A) 2.87 and 2.87 (B) 2.87 and 2.86 (C) 2.87 and 2.83 (D) 2.85 and 2.82

(A) 2.87 ,oa 2.87 (B) 2.87 ,oa 2.86 (C) 2.87 ,oa 2.83 (D) 2.85 ,oa 2.82

Ans. (C)

Sol. For vernier C1

10 VSD = 9 MSD = 9 mm

1 VSD = 0.9 mm

� LC = 1MSD – 1VSD = 1mm – 0.9 mm = 0.1 mm

Reading of C1 = MSR + (VSR)(L.C.) = 28mm + (7)(0.1)

Reaing of C1 = 28.7 mm = 2.87 cm

For vernier C2 : the vernier C2 is abnormal,

So we have to find the reading from basics.

The point where both of the marks are matching :

distance measured from main scale = distance measured from vernier scale

28mm + (1mm)(8) = (28 mm + x) + (1.1 mm) (7)

solving x = 0.3 mm

So reading of C = 28 mm + 0 3 mm = 2 83 cm

✁✁✁✂✄☎✆✝✞✆✟✄✂✟✠✝

ofuZ;j C1 ds fy;s 10 VSD = 9 MSD = 9 mm

1 VSD = 0.9 mm

� LC = 1MSD – 1VSD = 1mm – 0.9 mm = 0.1 mm

C1 dk ikB~;kad = MSR + (VSR)(L.C.) = 28mm + (7)(0.1)

C1 dk ikB~;kad = 28.7 mm = 2.87 cm

ofuZ;j C2 ds fy;s : ofuZ;j C2 vlkekU; gS

blfy;s bldk ikB~;kad fuEu ds vk/kkj ij Kkr djrs gSa

og fcUnq tgk¡ nksuksa fpUg lqesfyr gksrs gSa:

eq[; iSekus ls ekih x;h nwjh = ofuZ;j iSekus ls ekih x;h nwjh

28mm + (1mm)(8) = (28 mm + x) + (1.1 mm) (7)

gy djus ij x = 0.3 mm

vr% C2 dk ikB~;kad = 28 mm + 0.3 mm = 2.83 cm

4. A smaller object is placed 50 cm to the left of a thin convex lens of focal length 30 cm. A convex

spherical mirror of radius of curvature 100 cm is placed to the right of the lens at a distance of 50 cm.

The mirror is tilted such that the axis of the mirror is at an angle ✁ = 30º to the axis of the lens, as

shown in the figure.

,d NksVh oLrq dks 30 cm Qksdl nwjh (focal length) okys ,d irys mÙky (convex) ysal dh ckbZa vksj 50 cm dh nwjh

ij j[kk x;k gSA 100 cm dh oØrk f=kT;k okys ,d mÙky xksykdkj niZ.k dks ysal dh nkbZa vksj 50 cm dh nwjh ij j[kk

x;k gSA niZ.k dks bl rjg ls >qdk;k x;k gS fd niZ.k dk v{k ysal ds v{k ls ✁ = 30º dk dks.k cukrk gS] tSlk fp=k esa

n'kkZ;k x;k gSA

(0, 0)

50 cm

R = 100 cm

(–50, 0)

f = 30 cm

x

(50 50 3, 50)✂ ✄

☎

If the origin of the coordinate system is taken to be at the centre of the lens, the coordinates (in cm) of

the point (x, y) at which the image is formed are

;fn funsZ'kkad i)fr dk ewy fcUnq ysal ds e/; esa gks rks tgk¡ izfrfcac cuk gS ml fcUnq dk funsZ'kkad (x, y), lsaVhehVj esa]

D;k gksaxak\

(A) (125/3, 25 / 3 ) (B) (25, 25 3) (C) (50 25 3, 25)✆ (D) (0, 0)

Ans. (B)

✝✝✝✞✟✠✡☛☞✡✌✟✞✌✍☛

Sol.

o x

50–x 25

I1

I2

30°

30°

y

For I1 � u = –50 f = +30 v = +75cm

final image is I2

vafre izfrfcEc I2 gSA

y

50 – x = tan 60 = 3 y = 50 3 – 3 x

y + 3 x = 50 3

A and B both option satisfy this, nksuksa fodYi A rFkk B bldks larq"V djrs gSa,

size of image > size of object

izfrfcEc dk vkdkj > oLrq dk vkdkj

✁ ✂22y 50 – x✄ sin30° > 25 sin30°

y2 + (50 – x)

2 > 625

for option (A) 625 625

3 9☎ < 625

fodYi (A) ds fy;s 625 625

3 9☎ < 625

for option (B) 625(3) + 625 > 625

so B is correct.

fodYi (B) ds fy;s 625(3) + 625 > 625 vr% B lgh gSA

5. The electrostatic energy of Z protons uniformly distributed throughout a spherical nucleus of radius R is

given by

R4

e)1–Z(Z

5

3E

0

2

✆✝✞

The measured masses of the neutron, H11 , N15

7 , and O158 are 1.008665 u, 1.007825 u, 15.000109 u

and 15.003065 u, respectively. Given that the radii of both the N157 and O15

8 nuclei are same,

1 u = 931.5 MeV/c2

(c is the speed of light) and ✟ ✠02 4/e ✡☛ =1.44 MeV fm. Assuming that the difference

between the binding energies of N157 and O15

8 is purely due to the electrostatic energy, the radius of

either of the nuclei is (1 fm = 10–15

m)

☞☞☞✌✍✎✏✑✒✏✓✍✌✓✔✑

f=kT;k R okys ,d xksykdkj ukfHkd (nucleus) esa Z izksVksu leku:i ls forfjr gSA ,sls ukfHkd dh fLFkj fon~;qr~

ÅtkZ uhps lehdj.k esa nh xbZ gS

R4

e)1–Z(Z

5

3E

0

2

�✁✂

U;wVªkWu] H11 , N15

7 ,oa O158 ukfHkdksa (nuclei) ds ekis x;s nzO;eku Øe'k% 1.008665 u, 1.007825 u, 15.000109 u

,oa 15.003065 u gSA N157 vkSj O15

8 ukfHkdksa dh f=kT;k;sa leku nh xbZ gSA 1 u = 931.5 MeV/c2

(tgka ij c

izdk'k dh xfr gS) vkSj ✄ ☎02 4/e ✆✝ = 1.44 MeV fmA ;fn N15

7 vkSj O158 dh ca/kd ÅtkZvksa dk varj flQZ fLFkj

fon~;qr~ ÅtkZ ds dkj.k gS] rks nksuksa esa ls fdlh Hkh ukfHkd dh f=kT;k D;k gksxh \(1 fm = 10–15

m)

(A) 2.85 fm (B) 3.03 fm (C) 3.42 fm (D) 3.80 fm

Ans. (C)

Sol. E = 2

0

3 Z(Z 1)e

5 4 R

✞✟✠

15 15 1

8 7 1n O N H✡ ☛☛☞ ✡

Q = ✌ ✍15 15 18 7 1

n O N HM M m M✎ ✏ ✏ C

2

= 0.003796 × 931.5

= 3.5359 MeV

✑E = 2

0

3 e 1(8 7 7 6)

5 4 R✒ ✒ ✒ ✞ ✒

✟✠

= 3 1

(1.44 MeV fm) 145 R✓ ✓ ✓ = 3.5359 MeV

R = 3.42 fm

6. The ends Q and R of two thin wires, PQ and RS, are soldered (joined) together. Initially each of the

wires has a length of 1m at 10 °C. Now the end P is maintained at 10 °C, while the end S is heated and

maintained at 400 °C. The system is thermally insulated from its surroundings. If the thermal

conductivity of wire PQ is twice that of the wire RS and the coefficient of linear thermal expansion of PQ

is 1.2 × 10–5

K–1

, the change in length of the wire PQ is.

,d irys rkj PQ ds Nksj Q dks vU; irys rkj RS ds Nksj R ij Vkadk yxkdj (soldered) tksM+k x;k gSA

10 °C ij nksuksa rkjksa dh yEckbZ 1m gSA vc bl fudk; ds Nksj P rFkk Nksj S dks Øe'k% 10 °C rFkk 400 °C ij

fLFkj j[kk tkrk gSA ;g fudk; pkjksa vksj ls Å"ekjks/kh gSA ;fn rkj PQ dh Å"e pkydrk rkj RS dh Å"e

pkydrk ls nqxuh gS rFkk rkj PQ dk js[kh; Åf"er o`f) xq.kkad (coefficient of linear thermal expansion)

1.2 × 10–5

K–1

gS] rc rkj PQ dh yEckbZ esa ifjorZu dk eku gS

(A) 0.78 mm (B) 0.90 mm (C) 1.56 mm (D) 2.34 mm

Ans. (A) ✔✔✔✕✖✗✘✙✚✘✛✖✕✛✜✙

Sol.

P Q,R S 1m 1m

10°C 140°C 400°C

2k, �1 k, �2

d✁ = dx✂1 (✄ – 10)

☎✁ = d✆ ✝ ✞ – 10 130

x 1

✟✠ ✡ ☛☞ ✌✍ ✎✏ ✄ = 10 + 130x

☎✁ = ✑ ✒1

1

0

130x dx✓✔ ☎✁ = 130 ✂1 2x

2

☎✁ = 130 × 1.2 × 10–5

× 1

2 = 78 × 10

–5 = 0.78mm Ans. (A)

SECTION – 2 : (Maximum Marks : 32) This section contains EIGHT questions.

Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four

option(s) is(are) correct.

For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS.

For each question, marks will be awarded in one of the following categories :

Full Marks : +4 If only the bubble(s) corresponding to all the correct option(s) is(are) darkened.

Partial Marks : +1 For darkening a bubble corresponding to each correct option, provided NO

incorrect option is darkened.

Zero Marks : 0 If none of the bubbles is darkened.

Negative Marks : –2 In all other cases.

For example, if (A), (C) and (D) are all the correct options for a question, darkening all these three will

result in +4 marks ; darkening only (A) and (D) will result in +2 marks and darkening (A) and (B) will

result in –2 marks, as a wrong option is also darkened.

2 : ( : 32)bl [kaM esa ç'u gSaAizR;sd iz'u esa mÙkj fodYi (A), (B), (C) rFkk (D) gSaA ftuesa ls fodYi lgh gSA

izR;sd iz'u ds fy, vks-vkj-,l- ij lkjs lgh mÙkj ¼mÙkjksa½ ds vuq:i cqycqys ¼cqycqyksa½ dks dkyk djsaA

izR;sd iz'u ds fy, vad fuUufyf[kr ifjfLFkr;ksa esa ls fdlh ,d ds vuqlkj fn;s tk;saxs %

iw.kZ vad : +4 ;fn flQZ lgh fodYi ¼fodYiksa½ ds vuq:i cqycqys ¼cqycqyksa½ dks dkyk fd;k gSA

vkaf'kd vad : +1 çR;sd lgh fodYi ds vuq:i cqycqys dks dkyk djus ij] ;fn dksbZ xyr fodYi dkyk

ugha fd;k gSA

'kwU; vad : 0 ;fn fdlh Hkh cqycqys dks dkyk ugha fd;k gSA

_.k vad : –2 vU; lHkh ifjfLFkfr;ksa esa

mnkgkj.k % ;fn ,d ç'u ds lkjs lgh mÙkj fodYi (A), (C) rFkk (D) gSa] rc bu rhuksa ds vuq:i cqycqyksa dks dkys

djus ij +4 vad feysaxs flQZ (A) vkSj (D) ds vuq:i cqycqyksa dks dkys djus ij +2 vad feysaxs rFkk (A) vkSj (B)

ds vuq:i cqycqyksa dks dkys djus ij –2 vad feysaxs D;ksafd ,d xyr fodYi ds vuq:i cqycqys dks Hkh dkyk fd;k

x;k gSA✕✕✕✖✗✘✙✚✛✙✜✗✖✜✢✚

7. Light of wavelength �ph falls on a cathode plate inside a vacuum tube as shown in the figure., The work

function of the cathode surface is ✁ and the anode is a wire mesh of conducting material kept at a

distance d from the cathode. A potential difference V is maintained between the electrodes. If the

minimum de Broglie wavelength of the electrons passing through the anode is �e, which of the following

statement(s) is (are) true ?

�ph rjaxnS/;Z dk izdk'k fuokZr uyhdk (vacuum tube ) ds vanj ,d dSFkksM ij fxjrk gS] tSlk fp=k esa n'kkZ;k x;k gSA dSFkksM dh lrg dk dk;ZQyu ✁ gS ,oa ,uksM] tks dh ,d pkydh; inkFkZ ds rkjksa dh tkyh gS] dSFkksM ls d nwjh ij fLFkr gSA ,ysDVªksM+ksa ds chp dk foHkokUrj V fLFkj gSA ;fn ,uksM dks ikj djus okys bysDVªkWuksa dh U;wure " n czksXyh" (de Broglie) rjaxnS/;Z �e gS , fuEufyf[kr esa ls dkSulk@ dkSuls dFku lR; gS@gSa \

Electrons

Light

V – +

(A) �e increases at the same rate as �ph for �ph < hc/✁.

(B) For large potential difference (V >> ✁/e), �e is approximately halved if V is made four times.

(C) �e is approximately halved , if d is doubled

(D) �e decreases with increase in ✁ and �ph.

(A) vxj �ph < hc/✁ gS rks �ph ds lkFk �e,d leku nj ls c<+sxkA

(B) mPp foHkokUrj (V >> ✁/e) ij vxj V dks pkj xquk c<+k;k tk, rks �e yxHkx vk/kk gks tk,xkA

(C) d dks nqxuk djus ij �e yxHkx vk/kk gks tk,xkA

(D) ✁ vkSj �ph dks c<+kus ij �e de gksxkA

Ans. (B)

Sol. �e = Min. de broglie wave length U;wure fM&czksXyh rjaxnS/;Z

ph

hc

✂ = ✁ +

e

hc

✄+ eV

hc pn2pn

–1d

☎ ✆✝ ✞✟✝ ✞✟✠ ✡

= hc e2e

1– d

☛ ☞✌✍ ✎✍ ✎✌✏ ✑

2ph ph

2e e

d

d

✒ ✒✓

✒ ✒(A wrong vlR; gS)

V >> e

✔

Energy of electron bysDVªkWu dh ÅtkZ = eV

2P

2m = E � =

h 1

p 2mE✕

P = 2mE Ans (B)

✖✖✖✗✘✙✚✛✜✚✢✘✗✢✣✛

8. Two thin circular discs of mass m and 4m, having radii of a and 2a, respectively , are rigidly fixed by a

massless, rigid rod of length 24a�l through their centers. This assembly is laid on a firm and flat

surface, and set rolling without slipping on the surface so that the angular speed about the axis of the

rod is ✁. The angular momentum of the entire assembly about the point 'O' is L✂

(see the figure). Which

of the following statement (s) is (are) true ?

m rFkk 4m nzO;eku okyh nks iryh o`rkdkj pf=kdk,¡ (discs), ftudh f=kT;k;sa Øe'k% a rFkk 2a gSa] ds dsUnzksa dks

24a�l yEckbZ dh nzO;eku&jfgr nz<+ (rigid) MaMh ls tksM+k x;k gSA bl lewg dks ,d etcwr lery lrg ij

fyVk;k x;k gS vkSj fQlyk;s fcuk bl rjg ls ?kqek;k x;k gS fd bl dks.kh; xfr MaMh ds v{k ds fxnZ ✁ gSA iwjs

lewg dk fcUnq 'O' ds fxnZ dks.kh; laosx L✂

gS (fp=k nsf[k;sa)A fuEufyf[kr esa ls dkSulk@dkSuls dFku lR; gS@gSaA

l

l

✄m

4m

2a

O

z

a

(A) The magnitude of the z-component of L✂

is 55 ma2 ✁.

(B) The magnitude of angular momentum of the assembly about its centre of mass is 17 ma2

2

☎.

(C)The magnitude of angular momentum of centre of mass of the assembly about the point O is 81ma2 ✁✆

(D) The centre of mass of the assembly rotates about the z-axis with an angular speed of 5

✝.

(A) L✂

ds z-?kVd dk ifjek.k 55 ma2 ✁ gS

(B) iwjs lewg dk mlds lagfr&dsUnz ds fxnZ dks.kh; laosx dk ifjek.k 17 ma2

2

☎ gS

(C) iwjs lewg dk lagfr&dsUnz dk fcUnq O ds fxnZ dks.kh; laosx dk ifjek.k 81ma2 ✁ gS

(D) iwjs lewg ds lagfr&dsUnz z-v{k ds fxnZ dks.kh; osx 5

✝ ls ?kwe jgk gS

Ans. (BD)

Sol. ✞

✟

✟

✟ O

a 4✠/5

CMm

P

4m

✠

✡

cos ✟ = 5

24

a22☛

☞✌

✌

z

✡✍✍✍✎✏✑✒✓✔✒✕✏✎✕✖✓

(D) Velocity of point P : a� = ✁✂ then fcUnw P dk osx : a� = ✁✂ rks

✂ = ✄

☎a = Angular velocity of C.M. w.r.t point O.

✂ = ✄

☎a= nzO;eku dsUnz dk fcUnq O ds lkis{k dks.kh; osx

Angular velocity of C.M. w.r.t z axis = ✂ cos ✆

nzO;eku dsUnz dks z v{k ds lkis{k dks.kh; osx = ✂ cos ✆

�C.M. – z = a 24 a 24

5 524a

✝ ✝✞

✟

�CM – z = 5

☎

(B) LD – CM = 2

ma17

2

)a2(m4

2

ma 222 ✠✡✠☛✠

(C) LCM – O = (5m) 2 29 9 81m 81m a

5 5 5 5

☞ ✝✌ ✍☞ ✞ ✞ ✎✏ ✑

✒ ✓

✟ ✟ ✟ ✟

✟

LCM – O = 5

ma2481

5

am81 2 ✔✕

✔✖

(A) LZ = LCM–0 cos✆ – LD–CM sin✆

= 2

281 24 24 17ma 1a m

5 5 2 24

✝✝✎ ✗ ✎ =

2 281 24ma 17ma

25 2 24

✘ ✔ ✔✙

9. In the circuit shown below, the key is pressed at time t = 0. Which of the following statement(s) is (are)

true?

uhps fn[kk;s x, ifjiFk esa le; t = 0 ij cVu (key) dks nck;k x;k gSA fuEufyf[kr esa ls dkSulk@dkSuls dFku

lR; gS@gSa\

40✚F

A

5V

V –

+

25k✛

20✚F 50k ✛

+ –

key (A) The voltmeter displays –5 V as soon as the key is pressed, and displays +5 V after a long time

(B) The voltmeter will display 0 V at time t = ✁n 2 seconds

(C) The current in the ammeter becomes 1/e of the initial value after 1 second

(D) The current in the ammeter becomes zero after a long time

(A) cVu dks nckrs gh oksYVehVj –5 V fn[kkrk gS tcfd yEcs le; ds ckn oks +5 V fn[kkrk gS

(B) le; t = ✁n 2 seconds ij oksYVehVj 'kwU; oksYV fn[kkrk gS

(C) 1 second ds ckn vehVj esa /kkjk izkjfEHkd /kkjk dk 1/e xq.kk gksrh gS

(D) yacs le; ds ckn vehVj esa /kkjk 'kU; gks tkrh gS

✜✜✜✢✣✤✥✦✧✥★✣✢★✩✦

Ans. (ABCD)

Sol.

Q1

A

5V

V–

+

25k�

50k �

+ –

key

✁ ✁✂

✁✄Q2

q1 = (200 × 10–3

)

t–

11– e☎ ✆✝ ✞✝ ✞✟ ✠

; q2 = (100 × 10–3

)

t–

11– e☎ ✆✝ ✞✝ ✞✟ ✠

1q

C= (50 × 10

3) 2dq

dt

✡ ☛ ✡ ☛–3 –t

–6

200 10 1– e

40 10

☞

☞ = (50 × 10

3)(100 × 10

–3)[e

–t]

(1 – e–t)

610

20 = 50 × 10

3(e

–t)

1

2 = e

–t

t = ✌n2

I = I1+ I2 = (200 × 10–3

)(e–t) + (100 × 10

–3)e

–t

= 100 × 10–3

[2e–t + e

–t]

= (300 × 10–3

) e–t

= –3300 10

e

✍ ✎✏✑ ✒✑ ✒✓ ✔

At t = ✕ ij ✖ I = 0

10. In an experiment to determine the acceleration due to gravity g, the formula used for the time period of

a period of a periodic motion is 7(R – r)

T 25g

✗ ✘ . The values of R and r are measured to be (60 1)✙

mm and ✚ ✛10 1✜ mm respectively. In five successive measurements, the time period is found to be

0.52 s, 0.56 s, 0.57 s, 0.54 s and 0.59 s. The least count of the watch used for the measurement of time

period is 0.01 s. Which of the following statement(s) is (are) true?

(A) The error in the measurement of r is 10%

(B) The error in the measurement of T is 3.57%

(C) The error in the measurement of T is 2%

(D) The error in the determined value of g is 11%

✢✢✢✣✤✥✦✧★✦✩✤✣✩✪✧

xq:Roh; Roj.k g ds fu/kkZj.k ds ,d iz;ksx esa iz;qDr vkorhZ&xfr dk le;dky dk lw=k 7(R – r)

T 25g

� ✁ gSA R

rFkk r dk ekik x;k eku Øe'k% (60 1)✂ mm, rFkk (10 1)✂ mm gSaA yxkrkj ik¡p ekiu esa ekik x;k le;dky

0.52 s, 0.56 s, 0.57 s, 0.54 s rFkk 0.59 s gSaA le;dky ds ekiu ds fy, iz;ksx esa yk;h x;h ?kM+h dk vYiRekad

0.01 s gSA fuEufyf[kr esa ls dkSulk@dkSuls dFku lR; gS@gSa\

(A) r ds ekiu esa =kqfV 10% gS

(B) T ds ekiu esa =kqfV 3.57% gS

(C) T ds ekiu esa =kqfV 2% gS

(D) g ds fudkys x;s eku esa =kqfV 11% gS

Ans. (ABD)

Sol.

S. No. T absolute error = |T – Tmean|

(1) 0.52 0.04

(2) 0.56 0.00

(3) 0.57 0.01

(4) 0.54 0.02

(5) 0.59 0.03

Tmean = 2.78

5

Tmean = 0.56

(✄T)mean = 0.02

% error in T (T esa izfr'kr =kqfV ) = 0.02

0.56×100 = 3.57%

According to the question iz'u ds vuqlkj

g ☎ 2T

R – r

dg dT dR dr2

g t R – r

✆✝ ✆

dg

g = 2(3.57%) +

1 1

60 – 10

✞×100%

dg

g = 11% Ans. (A,B,D)

11. A rigid wire loop of square shape having side of length L and resistance R is moving along the x-axis

with a constant velocity v0 in the plane of the paper. At t = 0, the right edge of the loop enters a region

of length 3L where there is a uniform magnetic field B0 into the plane of the paper; as shown in the

figure. For sufficiently large v0, the loop eventually crosses the region. Let x be the location of the right

edge of the loop. Let v(x), ✟(x) and F(x) represent the velocity of the loop, current in the loop, and force

on the loop, respectively, as a function of x. Counter-clockwise current is taken as positive.

✠✠✠✡☛☞✌✍✎✌✏☛✡✏✑✍

,d oxhZ; vkdfr okyk rkj dk nz<+ Qank] ftlds Hkqtk dh yackbZ L ,oa izfrjks/k R gS] x-v{k dh fn'kk esa ,d fLFkj

xfr v0 ls bl dkxt ds Iysu ij (plane of the paper) xfreku gSA le; t = 0 ij Qans dk nkfguk fdukjk 3L

yackbZ ds fLFkj pqacdh; {ks=k B0 esa izos'k djrk gSA pqacdh; js[kkvksa dh fn'kk dkxt ds Iysu ds yacor~ vanj dh vksj

gS ¼tSlk fp=k esa n'kkZ;k x;k gS½A v0 dk eku i;kZIr gksus ij varrksxRok Qank pqacdh; {ks=k dks ikj djrk gSA eku

fyft, dh Qans dh nkfguh Hkqtk LFkku x ij gSA Qans dh xfr] Qans esa /kkjk ,oa Qans ij cy dh x ij fuHkZjrk dks

Øe'k% v(x), �(x) ,oa F(x) ls fu:fir fd;k x;k gSA okeorZ /kkjk dks iksftfVo ysaAx x x x x x x x x x x x x x x x x x x x x

x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x

x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x

x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x

x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x

x x x x x x x x x x x x x x x x x x x x x

L

v0

0 L 2L 3L x

4L

R

Which of the following schematic plot(s) is(are correct? (Ignore gravity)

fuEufyf[kr esa ls dkSulk@dkSuls O;oLFkk fp=k lgh gS@ gSa \ ¼xq:Rokd"kZ.k ux.; ekus½

(A)

0 x

L 2L 3L 4L

F(x)

(B)

0 x

L 2L 3L 4L

✁(x)

(C)

0 x

L 2L

3L 4L

✂(x)

(D)

0 x

L 2L 3L 4L

v(x)

Ans. (CD)

Sol. while entering izos'k djrs le; i.e. x < L

i =r

vB✄vB☎

i

i

f

✆✆✆✝✞✟✠✡☛✠☞✞✝☞✌✡

F = i�B = R

vB 22✁

a2m

f =

mR

vB 22✁ = Kv =

vdv

dx✂

✄☎✆✄x

0

v

v

dxKdv0

✝ v = v0 – kx

f = x)kxV(R

B0

22

✞✟✠✡✟✁

i = R

B)kxv( 0

☛☞ = i0 – ✌x

for 3 L > x > L ds fy;s f = 0 i = 0 v = constant. fu;r

4L > × > 3 L

i =r

vB✍

vB✎ R

f

f = i�B = R

vB 22✁

a = vmR

B 22✁

= kv = dx

vdv☞

v = 0v ' kx✏

f = ' '.x✑ ✏✒

i = 0'i ' x✓ ✔

f

o L 3L 4L

I

L 3 L 4 L

v

3L 4L L

✕✕✕✖✗✘✙✚✛✙✜✗✖✜✢✚

12. A block with mass M is connected by a massless spring with stiffness constant k to a rigid wall

and moves without friction on a horizontal surface. The block oscillates with small amplitude A

about an equilibrium position x0. Consider two cases : (i) when the block is at x0 ; and (ii) when

the block is at x = x0 + A. In both the cases, a particle with mass m (< M) is softly placed on the

block after which they stick to each other. Which of the following statement(s) is(are) true about

the motion after the mass m is placed on the mass M ?

(A) The amplitude of oscillation in the first case changes by a factor ofM

m M�, whereas in the second

case it remains unchanged

(B) The final time period of oscillation in both the cases is same

(C) The total energy decreases in both the cases

(D) The instantaneous speed at x0 of the combined masses decreases in both the cases

,d nzO;eku&jfgr fLiazx] ftldk nz<+rk xq.kkad (stiffness constant) k gS] ds ,d Nksj ij M nzO;eku dk ,d

xqVdk tqM+k gS] rFkk nwljs Nksj dks nz<+ nhokj ls tksM+k x;k gSA ;g xqVdk ,d lery ?k"kZ.k&jfgr lrg ij ,d

larqfyr fLFkfr x0 ds fxnZ NksVs vk;ke A ls nksyu djrk gSA ;gk¡ nks ifjfLFkfr;ka ekfu, % (i) tc xqVdk x0 ij gS

vkSj (ii) tc xqVdk x = x0 + A ij gSA nksuksa ifjfLFkfr;ksa esa nzO;eku m (< M) ds ,d d.k dks xqVds ij /khjs ls bl

izdkj j[kk tkrk gS dh og rqjar xqVds ls fpid tkrk gSA d.k dks xqVds ds Åij j[kus ds ckn xfr ds ckjs esa

fuEufyf[kr esa ls dkSulk@dkSuls dFku lR; gS@gSa \

(A) igyh ifjfLFkfr esa nksyu dk vk;keM

m M� HkkT; (factor) ls ifjofrZr gksrk gS] tcfd nwljh ifjfLFkfr esa

;g vifjofrZr jgrk gSA

(B) nksuksa ifjfLFkfr;ksa esa nksyu dk vafre le;dky leku gSA

(C) nksuksa ifjfLFkfr;ksa esa lEiw.kZ ÅtkZ de gks tkrh gSA

(D) lfEefyr nzO;ekuksa dh x0 ij rkR{kf.kd xfr nksuksa ifjfLFkfr;ksa esa de gks tkrh gSA

Ans. (ABD)

Sol. Case-1

M V1

Just before m is placed

X0

M V2

Just after m is placed

X0

m

Case-2

M

Just before m is placed

X0+A M

Just after m is placed

m

Case-I ✁✁✁✂✄☎✆✝✞✆✟✄✂✟✠✝

M V1

m dks j[kus ds Bhd igys

X0

M V2

m dks j[kus ds Bhd ckn

X0

m

Case-II

M

m dks j[kus ds Bhd igys

X0+A M

m dks j[kus ds Bhd ckn

m

In case-1 esa

MV1 = (M + m)V2

V2 = M

M m

� ✁✂ ✄☎✆ ✝

V1

k

M m✞A2 =

M k

M m M

✟ ✠✡ ☛

☞✌ ✍A1

A2 = M

M m✞A1

In case-2 esa

A2 = A1

T = 2✎M m

K

✞in both case. nksuksa fLFkfr;ksa esa

Total energy decreases in first case where as remain same in 2nd

case.

Instantaneous speed at x0 decreases in both case.

izFke fLFkfr esa dqy ÅtkZ ?kVrh gS tcfd f}rh; fLFkfr esa fu;r jgrh gSA

x0 ij rkR{kf.kd pky nksuksa fLFkfr;ksa esa ?kVrh gSA

13. While conducting the Young's double slit experiment, a student replaced to two slits with a large

opaque plate in the x-y plane containing two small holes that act as two coherent point sources (S1, S2)

emitting light of wavelength 600 nm. The student mistakenly placed the screen parallel to the x-z plane

(for z > 0) at a distance D = 3 m from the mid-point of S1S2, as shown schematically in the figure. The

distance between the sources d = 0.6003 mm. The origin O is at the intersection of the screen and the

line joining S1S2. Which of the following is(are) true of the intensity pattern on the screen ?✏✏✏✑✒✓✔✕✖✔✗✒✑✗✘✕

,d fon~;kFkhZ us ;ax nks fLyV okys iz;ksx (Young's double slit experiment) djrs le; nks fLyVksa dh txg ,d

cM+h lery vikjn'khZ iêh dks x-y ry ij j[k fn;kA bl iêh esa nks NksVs fNnz gSa tks 600 nm rjaxnS/;Z izdk'k

mRiUu djus okys nks dyklac) fcUnq L=kksrksa (S1, S2) ds leku gSaA fon~;kFkhZ us xyrh ls insZ (screen) dks x-z ry

(z > 0) ds lekukUrj S1S2 ds e/; fcUnq ls D = 3 m dh nwjh ij j[k fn;k] tSlk dh O;oLFkk&fp=k esa fn[kk;k x;k

gSA L=kksrksa ds chp fd nwjh d = 0.6003 mm gSA S1S2 dks tksM+us okyh js[kk tgk¡ insZ ls feyrh gS ogk¡ ij ewyfcUnq

O gSA insZ ij rhozrk izfr:i (intensity pattern) ds fy, fuEufyf[kr esa ls dkSulk@dkSuls lR; gS@gSa \

Scre

en

z

Oy

x

D

d S1 S2

(A) Semi circular bright and dark bands centered at point O

(B) Hyperbolic bright and dark bands with foci symmetrically placed about O in the x-direction

(C) The region very close to the point O will be dark

(D) Straight bright and dark bands parallel to the x-axis

(A) fcUnq O ij dsfUnzr v/kZorh; nhIr rFkk vnhIr ifê;k¡

(B) x-fn'kk esa fcUnq O ds fxnZ lefer Qksdlksa ds lkFk vfrijoyf;d (Hyperbolic) nhIr rFkk vnhIr ifê;k¡

(C) fcUnq O dk fudVre {ks=k vnhIr gksxk

(D) x-v{k ds lekukUrj nhIr rFkk vnhIr lh/kh ifê;k¡

Ans. (AC)

Sol. from theory fringes will be semi circular

and 2

11000

d�✁

✂

at 0 ✄x = 2

1000✂

�✂

so at 0 it will be dark

F;ksjh ds vk/kkj ij fÝUtsa v)Zo`Ùkkdkj gksaxh

rFkk 2

11000

d�✁

✂

0 ij ✄x = 2

1000✂

�✂

vr% 0 ij ;g dkyh gksxh

☎☎☎✆✝✞✟✠✡✟☛✝✆☛☞✠

14. Consider two identical galvanometers and two identical resistors with resistance R. If the internal

resistance of the galvanometers RC < R/2, which of the following statement(s) about any one of the

galvanometers is(are) true ?

(A) The maximum voltage range is obtained when all the components are connected in series

(B) The maximum voltage range is obtained when the two resistors and one galvanometer are

connected in series, and the second galvanometer is connected in parallel to the first galvanometer

(C) The maximum current range is obtained when all the components are connected in parallel

(D) The maximum current range is obtained when the two galvanometers are connected in series, and

the combination is connected in parallel with both the resistors.

nks ,dleku xsYosuksehVj rFkk ,dleku izfrjks/k R okys nks izfrjks/kd fn;s x;s gSaA ;fn xsYosuksehVj dk vkarfjd

izfrjks/k RC < R/2 gS] rks fdlh Hkh ,d xsYosuksehVj ds ckjs esa fn;s x, fuEufyf[kr dFkuksa esa ls dkSulk@dkSuls lR;

gS@gSa \

(A) izkIr fd xbZ oksYVrk ifjlj (voltage range) vf/kdre gksxh tc lHkh ?kVd Js.kh esa tqM+s gq, gSa

(B) izkIr fd xbZ oksYVrk ifjlj vf/kdre gksxh tc nks izfrjks/kd rFkk ,d xsYosuksehVj Js.kh esa tqM+s gSa rFkk nwljk

xsYosuksehVj igys xsYosuksehVj ds lekukUrj esa tqM+k gSA

(C) izkIr fd xbZ /kkjk ifjlj (current range) vf/kdre gksxh tc lHkh ?kVd lekukUrj esa tqM+s gSa

(D) izkIr fd xbZ /kkjk ifjlj vf/kdre gksxh tc nks xsYosuksehVj Js.kh esa tqM+s gSa rFkk ;s la;kstu izfrjks/kdksa ds lkFk

lekukUrj esa tqM+k gSA

Ans. (BC)

Sol. Suppose current for full deflection (i.e., For maximum range) through galvanometer is �g.

(A)

G G ✁g

R R

Total potential difference V1 = 2Ig(R + RC)

(B)

2✂g

R R G

G✂g

✂g

Total potential difference V2 = 2�g ✄☎

✆✝✞

✟✠ R2

2

RC = �g (RC + 4R)

Now since 2RC < R

So V1 < 3�gR

while V2 > 4R�g

So V2 > V1 ✡✡✡☛☞✌✍✎✏✍✑☞☛✑✒✎

(C)

�1 G

�g

G

�

�

�g

✁gRC = ✁R

✁ = R

RCg✂

✁1 = ✄☎

✆✝✞

✟✠✡

R

R12 C

g

(D)

☛2G

☛gG

R

☛

☛

2✁gRC = ✁R

✁ = R

R2 C

g✂

✁2 = ✁g ✄☎

✆✝✞

✟✠

R

R41 C

We can be see ✁1 > ✁2

SECTION – 3 : (Maximum Marks : 12)

This section contains TWO paragraphs.

Based on each paragraph, there are TWO Questions.

Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is correct.

For each question, darken the bubble corresponding to the correct option in the ORS.

For each questions, marks will be awarded in one of the following categories :

Full Marks : +3 If only the bubble corresponding to the correct option is darkened.

Zero Marks : 0 In all other cases.

3 : ( : 12)

☞ bl [kaM esa gSaA

☞ izR;sd vuPNsn ij nks iz'u fn;s x;s gSaA

✌✌✌✍✎✏✑✒✓✑✔✎✍✔✕✒

izR;sd iz'u esa pkj fodYi (A), (B), (C) rFkk (D) gSaA ftuesa dsoy ,d gh lgh gSA

izR;sd iz'u ds fy, vks-vkj-,l- ij lgh mÙkj fodYi ds vuq:i cqycqys dks dkyk djsaA

izR;sd iz'u ds fy, vad fuUufyf[kr ifjfLFkr;ksa esa ls

iw.kZ vad : +3 ;fn flQZ lgh fodYi ds vuq:i cqycqys dks dkyk fd;k gSA

'kwU; vad : 0 vU; lHkh ifjfLFkr;ksa esaA

Paragraph for Question Nos. 15 to 16

15 16

A frame of reference that is accelerated with respect to an inertial frame of reference is called a non-

inertial frame of reference. A coordinate system fixed on a circular disc rotating about a fixed axis with a

constant angular velocity � is an example of a non-inertial frame of reference. The relationship between

the force rotF✁

experienced by a particle of mass m moving on the rotating disc and the force inF✁

experienced by the particle in an inertial frame of reference is,

✂ ✄ ☎ ✆rot in rotF F 2m m r✝ ✞ ✟ ✠ ✡ ✞ ✡✠✠✡☛ ☛ ☛☛ ☛ ☛☛

,

Where rot☞✌

is the velocity of the particle in the rotating frame of reference and r✍

is the position vector of

the particle with respect to the centre of the disc.

Now consider a smooth slot along a diameter of a disc of radius R rotating counter-clockwise with a

constant angular speed � about its vertical axis through its center. We assign a coordinate system with

the origin at the center of the disc, the x-axis along the slot, the y-axis perpendicular to the slot and the

z-axis along the rotation axis ✎ ✏k✑ ✒ ✑✓

. A small block of mass m is gently placed in the slot at

✔ ✕ ˆr R / 2 i✖✗

at t = 0 and is constrained to move only along the slot.

,d funsZ'k ra=k tks ,d tM+Roh; funsZ'k ra=k dh rqyuk esa Rofjr gks] vtM+Roh; funsZ'k ra=k dgykrk gSA fLFkj dks.kh;

osx � ls ?kwerh gqbZ fMLd ij c) (fixed) funsZ'k ra=k vtM+Roh; ra=k dk ,d mnkgj.k gSA m nzO;eku dk ,d d.k

?kwerh gqbZ fMLd ij xfreku gSA xfreku d.k fMLd ij c) funsZ'k ra=k ds lkis{k cy rotF✁

rFkk ,d tM+Roh; funsZ'k

ra=k ds lkis{k cy inF✁

dks eglwl djrk gSA rotF✁

vkSj inF✁

ds chp dk laca/k fuEufyf[kr lehdj.k esa fn;k x;k gS

✂ ✄ ☎ ✆rot in rotF F 2m m r✝ ✞ ✟ ✠ ✡ ✞ ✡✠ ✠ ✡☛ ☛ ☛☛ ☛ ☛ ☛

,

;gk¡ ij rot☞✌

?kwers gq, funsZ'k ra=k esa d.k dk osx gS rFkk r✍d.k dk fMLd ds e/; fcUnq ds lkis{k fLfFkfr lfn'k

(position vector) gSA

Ekfu, fd R f=kT;k dh ,d fMLd] ftlesa O;kl ds lekukUrj ,d ?k"kZ.kjfgr [kk¡pk gS] ,d fLFkj dks.kh; xfr � ls

vius v{k ij okekorZ fn'kk esa ?kwe jgh gSA ,d funsZ'k ra=k ekfu, ftldk ewyfcUnw fMLd ds e/; fcUnq ij gS ,oa

x-v{k [kk¡ps ds lekukUrj gS] y-v{k [kk¡ps ds vfHkyEc ij gS ,oa z-v{k ?kweus okyh v{k ds lekukUrj gS ✎ ✏k✑ ✒ ✑✓

A

m nzO;eku okys ,d NksVs xqVds dks le; t = 0 ij ✔ ✕ ˆr R / 2 i✖✗

fcUnq ij /khjs ls bl rjg ls j[kk tkrk gS fd oks

flQZ [kk¡ps esa gh py ldsA✘✘✘✙✚✛✜✢✣✜✤✚✙✤✥✢

�

R

R/2

m

15. The distance r of the block at time t is :

Lke; t ij xqVds dh nwjh r dk eku gS %

(A) R

cos2 t2

✁ (B) ✂ ✄2 t 2 tRe e

4

☎ ✆ ☎✝ (C) R

cos t2

✁ (D) ✞ ✟t tRe e

4

☎ ✆☎✝

Ans. (D)

Sol. mr✠2 = ma

a = r✠2

2rdr

vdv ✡☛ ; ☞ ☞✌✍v

0

r

2/R

2 rdrvdv

v = 4

Rr

22 ✎✏

✑✑ ✒✓✔

t

0

r

2/R2

2

dt

4

Rr

dr.....(1)

Assume ekuk : r = ✕sec2

R

dr = 2

Rsec ✖ tan ✖ d✖ ; ✗✗ ✘✙

✚

✚✚✚ t

022

dt

tan4

R

dtansec2

R

; ✠t = ✛✛✜

✢✣✣✤

✥ ✦✧R

Rr4

R

r2n

22

★

✩ ✪tt ee4

Rr ☎✆☎ ✝✫

16. The net reaction of the disc on the block is :

xqVds ij fMLd dh usV izfrfØ;k (net reaction) gS %

(A) 2 ˆ ˆm R sin t j – mgk✬ ✬ (B) 2 ˆ ˆm Rcos t j – mgk✭ ✬ ✬

(C) ✮ ✯2 t t1 ˆ ˆm R e e j mgk2

☎ ✆☎✁ ✰ ✝ (D) ✂ ✄2 2 t 2 t1 ˆ ˆm R e e j mgk2

☎ ✆ ☎✁ ✰ ✝✱✱✱✲✳✴✵✶✷✵✸✳✲✸✹✶

Ans. (C)

Sol.

r vr

i

)j(y

)k(�

k)irk(mk)iv(m2FFrotinrot ✁✂✂✁✄✁✂✄☎

✆✆

irm)j(mv2Fimr 2rotin

2 ✝✞✟✝✞✠✝✡

jmv2F rin ✝✠✡

......(1)

☛ ☞tt ee4

Rr ✌✍✌ ✎✏

✑ ✒ttr ee

4

Rv

dt

dr ✌✍✌ ✓✔✓✏✏ ; ✕ ✖ jee4

Rm2F tt

in ✓✔✓✏ ✌✍✌✗

; ✘ ✙jee2

mRF tt

2

in✚✛✚ ✜✢✣

✤

Also reaction is due to disc surface then

bl izdkj pdrh dh lrg }kjk Hkh izfrfØ;k vkjksfir gks rks

✥ ✦ kmgjee2

mRF tt

2

reaction ✧✜✢✣ ✚✛✚✤

Paragraph for Question Nos. 17 to 18

17 18

Consider an evacuated cylindrical chamber of height h having rigid conducting plates at the ends and

an insulating curved surface as shown in the figure. A number of spherical balls made of a light weight

and soft material and coated with a conducting material are placed on the bottom plate. The balls have

a radius r << h. Now a high voltage source (HV) is connected across the conducting plates such that

the bottom plate is at +V0 and the top plate at –V0. Due to their conducting surface the balls will get

charged, will become equipotential with the plate and are repelled by it. The balls will eventually collide

with the top plate, where the coefficient of restitution can be taken to be zero due to the soft nature of

the material of the balls. The electric field in the chamber can be considered to be that of a parallel plate

capacitor. Assume that there are no collisions between the balls and the interaction between them is

negligible. (Ignore gravity)

h ÅapkbZ okys fuokZfrr (evacuated) ,d csyukdkj d{k ds nksuksa Nksjksa ij nks nz< (rigid) pkyd iV~Vhdka, gSa vkSj

mldk oØiz"V vpkyd gS] tSlk dh fp=k esa n'kkZ;k x;k gSA de Hkkj okyh eqyk;e inkFkZ ls cuh gq;h dbZ xksykdkj

xksfy;k¡] ftudh lrg ij ,d pkyd inkFkZ dh ijr p<+h gS] uhps okyh ifV~Vdk ij j[kh gqbZ gSaA bu xksfy;ksa dh ★★★✩✪✫✬✭✮✬✯✪✩✯✰✭

f=kT;k r << h gSA vc ,d mPp oksYVrk dk lzksr (HV) bl rjg ls tksMk tkrk gS fd uhps okyh ifV~Vdk ij +V0 ,oa

Åij okyh ifV~Vdk ij –V0 dk foHko vk tkrk gSA pkyd ijr ds dkj.k xksfy;k¡ vkosf'kr gksdj ifV~Vdk ds lkFk

lefoHko gks tkrh gSa ftlds dkj.k os ifV~Vdk ls izfrdf"kZr gksrh gSaA varrksxRok xksfy;k¡ Åijh ifV~Vdk ls Vdjkrh

gSa] tgk¡ ij xksfy;ksa ds inkFkZ dh eqyk;e izdfr ds dkj.k izR;oLFkku xq.kkad (coefficient of restitution) dks 'kwU;

fy;k tk ldrk gSA d{k esa fon~;qr {ks=k dks lekukUrj ifV~Vdk okys la/kkfj=k ds leku ekuk tk ldrk gSA xksfy;ksa

dh ,d nwljs ls ikjLifjd fØ;k ,oa Vdjko dks ux.; ekuk tk ldrk gSA (xq:Rokd"kZ.k ux.; gSA)

A

HV

–

+

17. Which one of the following statements is correct?

(A) The balls will execute simple harmonic motion between the two plates

(B) The balls will bounce back to the bottom plate carrying the opposite charge they went up with

(C) The balls will bounce back to the bottom plate carrying the same charge they went up with

(D) The balls will stick to the top plate and remain there

fuEufyf[kr esa ls dkSulk dFku lR; gS\

(A) xksfy;k¡ nksuksa ifV~Vdkvksa ds chp ljy vkorZ xfr fu"ikn djsaxh

(B) xksfy;k¡ ftl vkos'k ds lkFk Åij tkrh gSa mlds foijhr vkos'k ds lkFk mNydj fupyh ifV~Vdk ij okil vk

tkrh gSA

(C) xksfy;k¡ ftl vkos'k ds lkFk Åij tkrh gSa mlh vkos'k ds lkFk mNydj fupyh ifV~Vdk ij okil vk tkrh gSa

(D) xksfy;k¡ Åijh ifV~Vdk ij fpiddj ogha jg tkrh gSa

Ans. (B)

18. The average current in the steady state registered by the ammeter in the circuit will be

(A) proportional to V02 (B) proportional to V0

1/2

(C) proportional to the potential V0 (D) zero

ifjiFk esa yxk, vehVj esa LFkk;h voLFkk esa vkSlr /kkjk

(A) V02 ds lekuqikrh gksxh (B) V0

1/2 ds lekuqikrh gksxh

(C) V0 ds lekuqikrh gksxh (D) dk eku 'kwU; gksxk

Ans. (A)

Sol. (17–18)

���✁✂✄☎✆✝☎✞✂✁✞✟✆

2q

02 VE

h�

–V0

+V0 +q

–q

2q

2qjust before collision

¼VDdj ds Bhd igys½just after collision

¼VDdj ds Bhd ckn½

Kq

r = Vo = q =

V ro

K

1

2

qE 2t

m

✁ ✂✄ ☎✆ ✝

= h

1

2

0V r

K

02 V

hmt2 = h

t2 =

2

2

0

h

V

mk

r

t = 0

h mk

V r

During every collision 2q charge will flow from circuit.

izR;sd VDdj ds nkSjku 2q vkos'k ifjiFk ls izokfgr gksrk gSA

Average current vkSlr /kkjk Iavg = 2q

t =

2

02V r r

h mk k

2

avg 0I V✞

17. The balls will bounce back to the bottom plate carrying the opposite charge they went up with.

xksfy;k¡ ftl vkos'k ds lkFk Åij tkrh gSa mlds foijhr vkos'k ds lkFk mNydj fupyh ifV~Vdk ij okil vk

tkrh gSA

18. 2

avg 0I V✞

✟✟✟✠✡☛☞✌✍☞✎✡✠✎✏✌

PART : II CHEMISTRY

SECTION – 1 : (Maximum Marks : 18)

This section contains SIX questions.

Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is correct.

For each question, darken the bubble corresponding to the correct option in the ORS.

For each questions, marks will be awarded in one of the following categories :Full Marks : +3 If only the bubble corresponding to the correct option is darkened. Zero Marks : 0 If none of the bubbles is darkened. Negative Marks : –1 In all other cases.

19. The geometries of the ammonia complexes of Ni2+

, Pt2+

and Zn2+

, respectively, are

(A) octahedral, square planar and tetrahedral

(B) square planar, octahedral and tetrahedral

(C) tetrahedral, square planar and octahedral

(D) octahedral, tetrahedral and square planar

Ni2+

, Pt2+

rFkk Zn2+ ds veksfu;k ladqyksa dh T;kfefr;k¡ Øe'k% gSa

(A) v"VQydh;, oxZ leryh rFkk prq"Qydh;

(B) oxZ leryh] v"VQydh;, rFkk prq"Qydh;

(C) prq"Qydh;] oxZ leryh rFkk v"VQydh;

(D) v"VQydh;] prq"Qydh; rFkk oxZ leryh

Ans. (A)

Sol. Ni2+

with NH3 shows CN=6 forming [Ni(NH3)6]2+

(Octahedral)

Pt2+

with NH3 shows CN = 4 forming [Pt(NH3)4]2+

(5d series CMA, square planner)

Zn2+

with NH3 shows CN = 4 forming [Zn(NH3)4]2+

(3d10

configuration, tetrahedral)

NH3 ds lkFk Ni2+

, CN=6 n'kkZrk gS rFkk [Ni(NH3)6]2+

cukrk gSSA (v"VQydh;)

NH3 ds lkFk Pt2+

, CN = 4 n'kkZrk gS rFkk [Pt(NH3)4]2+

cukrk gSSA (5d Js.kh CMA, oxZ leryh)

NH3 ds lkFk Zn2+

, CN = 4 n'kkZrk gS rFkk [Zn(NH3)4]2+

cukrk gSSA (3d10

foU;kl prq"Qydh;)

���✁✂✄☎✆✝☎✞✂✁✞✟✆

20. The major product of the following reaction sequence is

O

i) HCHO (excess)/NaOH, heat

ii) HCHO/H+(catalytic amount)

fuEufyf[kr vfHkfØ;k vfHkØe dk eq[; mRikn gS

O

i) HCHO (vf/kd ek=kk esa)/NaOH, Å"ek

ii) HCHO/H+(mRizsjd ek=kk)(catalytic amount)

(A)

O O

(B)

O O OH

(C)

O OHO

(D)

O OH

OH

Ans. (A)

Sol.

O

HCHO/OH�

Cross aldol

O

OH HCHO/OH✁

Cross Cannizzaro

OH OH

+ HCOOH

HCHO / H✂ Nucleophilic addition

O O

Sol.

O

HCHO/OH✄

ØkWl ,YMksy

O

OH HCHO/OH☎

ØkWl dSfutkjks

OH OH

+ HCOOH

HCHO / H✆ukfHkdLusgh ;ksx

O O ✝✝✝✞✟✠✡☛☞✡✌✟✞✌✍☛

21. The correct order of acidity for the following compounds is

fuEufyf[kr ;kSfxdksa dh vEyrk dk lgh Øe gS

CO2H

OH HO

I

CO2H

OH

II

CO2H

OH III

CO2H

OH

IV

(A) I > II > III > IV (B) III > I > II > IV

(C) III > IV > II > I (D) I > III > IV > II

Ans. (A)

Sol. Due to ortho effect, ortho substituted benzoic acid is more acidic than meta & para isomers.

vkWFkksZ izHkko ds dkj.k] vkWFkksZ izfrLFkkfi csUtksbd vEy dh vEyh;rk mlds esVk ,oa isjk leko;oh ls vf/kd gksrh gSA

or ;k

Due to strong hydrogen bond in conjugate base of ortho hydroxybenzoic acid, it is more acidic than

its meta & para isomers.

vkWFkksZ gkbMªkWDlh csUtksbd vEy ds l;qXeh {kkj esa izcy gkbMªkstu cU/k ds dkj.k vEyh;rk mlds esVk ,oa isjk

leko;oh ls vf/kd gksrh gSA

22. For the following electrochemical cell at 298 K,

Pt(s) | H2 (g, 1 bar) | H+ (aq, 1 M) || M

4+(aq) | M

2+(aq) | Pt(s)

Ecell= 0.092 V when

� ✁

� ✁

2+ aqM

4 aqM

✂ ✄☎ ✆✝ ✞✂ ✄✟☎ ✆✝ ✞

= 10x.

Given : 4 2

0

/M ME ✠ ✠ = 0.151 V; 2.303

RT

F = 0.059 V

The value of x is

(A) –2 (B) –1 (C) 1 (D) 2

298 K ij fuEufyf[kr oS|qr&jklk;fud lsy (electrochemical cell),

Pt(s) | H2 (g, 1 bar) | H+ (aq, 1 M) | M

4+(aq) M

2+(aq) | Pt(s)

ds fy, Ecell= 0.092 V tc

� ✁

� ✁

2+ aqM

4 aqM

✂ ✄☎ ✆✝ ✞✂ ✄✟☎ ✆✝ ✞

= 10x.

eku yhft, dh : 4 2

0

/M ME ✠ ✠ = 0.151 V; 2.303

RT

F= 0.059 V, rc

x dk eku D;k gksxk \

(A) –2 (B) –1 (C) 1 (D) 2

✡✡✡☛☞✌✍✎✏✍✑☞☛✑✒✎

Ans. (D)

Sol. Ecell = E°cell 2

059.0– log10

24

22

pH]M[

]H][M[�

��

0.092 = 0.151 = 2

059.0 log10 10

x

✁ x = 2

23. The qualitative sketches I, II and III given below show the variation of surface tension with molar

concentration of three different aqueous solution of KCl, CH3OH and CH3(CH2)11 OSO3

– Na

+ at room

temperature. The correct assignment of the sketches is

uhps fn;s xq.kkRed js[kkfp=k I, II rFkk III lkekU; rki ij KCl, CH3OH rFkk CH3(CH2)11 OSO3

– Na

+ ds rhu fHké

tyh; foy;uksa dh eksyj lkanzrk (concentration ) ds lkFk i"B ruko (surface tension) ds ifjorZu dks n'kkZrs gSaA

js[kkfp=kksa dk lgh fufnZ"Vhdj.k D;k gS \

Surf

ace tensio

n

Concentration

I

Surf

ace tensio

n

Concentration

II

Surf

ace tensio

n

Concentration

III

(A) I : KCl II : CH3OH III : CH3(CH2)11 OSO3

– Na

+

(B) I : CH3(CH2)11 OSO3

– Na

+ II : CH3OH III : KCl

(C) I : KCl II : CH3(CH2)11 OSO3

– Na

+III : CH3OH

(D) I : CH3OH II : KCl III : CH3(CH2)11 OSO3

– Na

+

Ans. (D)

Sol. Impurities affect surface tension appreciably. It is observed that impurities which tend to concentrate on

surface of liquids, compared to its bulk lower the surface tension.

Substance like detergents, soaps (CH3(CH2)11SO3–Na

+) decreases the surface tension sharply.

Those like alcohol (eg. –CH3OH, C2H5OH) lower the surface tension slightly. This can also be related to

the fact that CH3OH has smaller dielectric constant. Dielectric constant is directly proportional to

surface tension. So, on adding CH3OH in water, overall dielectric constant decreases and surface

tension decreases.

Inorganic impurities present in bulk of a liquid such as KCl tend to increase the surface tenstion of

water.

: v'kqf);k¡ i"V ruko ij miq;Dr izHkko Mkyrh gS ;g izsf{kr gksrk gS fd v'kqf);k¡ tks cYd dh vis{kk æoksa dh lrg ij

lkfUær gksrh gS rks i"V ruko dks de dj nsrh gSA

inkFkZ tSls viektZd] lksi (CH3(CH2)11SO3–Na

+) i`"V ruko dks rsth ls ?kVkrh gSA

,YdksgkWy (mnkgj.k –CH3OH, C2H5OH) i`"V ruko dks FkksM+k de djrh gSA bldks bl rjg Hkh dgk tk ldrk gS

CH3OH dk ijkoS|qr LFkjkad de gSA ijkoS|qr LFkjkad i`"V ruko ds lekuqikrh gksrk gSA blfy, ty esa CH3OH

feykus ij ijkoS|qr fLFkjkad ?kVrk gS rFkk i`"V ruko ?kVrk gSA

✂✂✂✄☎✆✝✞✟✝✠☎✄✠✡✞

24. In the following reaction sequence in aqueous solution, the species X, Y and Z, respectively, are

Ag+

X

Clear solution

Ag+

Y

white precipitate

with time

Y

black precipitate

S2O3 2–

(A) [Ag(S2O3)2]3–

, Ag2S2O3, Ag2S (B) [Ag(S2O3)3]5–

, Ag2SO3,Ag2S

(C) [Ag(SO3)2]3–

, Ag2S2O3, Ag (D) [Ag(SO3)3]3–

, Ag2SO4,Ags

tyh; foy;u esa fuEufyf[kr vfHkfØ;k vfHkØe es] Lih'kht+ (sequence) X, Y rFkk Z, Øe'k% gSa

Ag+

X

lkQ foy;u

Ag+

Y

lQsn vo{ksi

le; ds lkFk Y

dkyk vo{ksi

S2O3 2–

(A) [Ag(SO2O3)2]3–

, Ag2S2O3, Ag2S (B) [Ag(S2O3)3]5–

, Ag2SO3,Ag2S

(C) [Ag(SO3)2]3–

, Ag2S2O3, Ag (D) [Ag(SO3)3]3–

, Ag2SO4,Ags Ans. (A)

Sol. Ag+ + S2

�23O ✁✁✂ [Ag(S2O3)2]

3– ✄✄☎

+AgAg2S2O3 (✆ ) ✁✁✁✁✂

(with time) Ag2S (✆ )

(X) White ppt. (Y) Black (Z)

Ag+ + S2

�23O ✁✁✂ [Ag(S2O3)2]

3– ✄✄☎

+AgAg2S2O3 (✆ )

( )✄✄✄✄✄☎

le; d s lkFk Ag2S (✆ )

(X) 'osr vo{ksi (Y) dkyk (Z)

SECTION – 2 : (Maximum Marks : 32)

This section contains EIGHT questions.

Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these fouroption(s) is(are) correct.

For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS.

For each question, marks will be awarded in one of the following categories :Full Marks : +4 If only the bubble(s) corresponding to all the correct option(s) is(are) darkened. Partial Marks : +1 For darkening a bubble corresponding to each correct option, provided NO

incorrect option is darkened. Zero Marks : 0 If none of the bubbles is darkened. Negative Marks : –2 In all other cases.

For example, if (A), (C) and (D) are all the correct options for a question, darkening all these three willresult in +4 marks ; darkening only (A) and (D) will result in +2 marks and darkening (A) and (B) willresult in –2 marks, as a wrong option is also darkened.

25. Extraction of copper from copper pyrite (CuFeS2) involves

(A) crushing followed by concentration of the ore by froth-flotation

(B) removal of iron as slag

(C) self-reduction step to produce 'blistercopper' following evolution of SO2

(D) refining of 'blister copper' by carbon reduction✝✝✝✞✟✠✡☛☞✡✌✟✞✌✍☛

dkWij ikbjkbV (CuFeS2) ls dkWij ¼rk¡ck½ ds fu"d"kZ.k esa D;k lafyIr gS ¼gSa½ \

(A) nyu rFkk Qsu&Iyou froth-flotation }kjk v;Ld dk lkanz.k

(B) yksgs dk /kkrqey ds :i esa fu"dklu

(C) SO2 fudkl ds i'pkr QQksysnkj rk¡csa (blistercopper) ds mRikn ds fy; Lo%&vip;u dk ix

(D) dkcZu vip;u }kjk 'QQksysnkj rk¡csa ' dk 'kks/ku

Ans. (ABCD)

Sol. Extraction of Copper :

CuFeS2 (Ore)

Crushing of Ore Concentrated Ore

Roasting with O2

Cu2S + FeS Copper matte

O2 blast

+ SiO2

SO2 + Cu + FeSiO3 Blister Slag copper

Refining by poling Add coal & stirred with green pole.

Pitch copper

Further refining

By Electrolysis

Pure copper

Conc. By

froth flotation

CuFeS2 (Ore)

v;Ld dk nyu lkfUær v;Ld

O2 ds lkFk

Cu2S + FeS

dkWij esV

O2 blast

+ SiO2

SO2 + Cu + FeSiO3

QQksysnkj èkkrqey rk¡ck

ikWfyx }kjk kksèku dksy dks feykrs gS rFkk gjh NM ds lkFk fgykrs gSA

fiap dkWijoS|qrvi?kVu }kjk

iqu% kksèku

kq) dkWij

Qsu Iyou }kjk lkUæ.k

���✁✂✄☎✆✝☎✞✂✁✞✟✆

26. According to Molecular orbital Theory,(A) C2

2– is expected to be diamagnetic

(B) O22+

is expected to have a longer bond length than O2

(C) N2+ and N2

– have the same bond order

(D) He2+ has the same energy as two isolated He atoms

v.kq d{kd fl}kUr (Molecular orbital Theory) ds vuqlkj(A) C2

2– izR;kf'kr :i ls izfrpqEcdh; (diamagnetic) gS

(B) O22+

dh vkca/k yEckbZ (bond length) izR;kf'kr :i ls O2 dh vkca/k yEckbZ ls yEch gS(C) N2

+ rFkk N2

– dh vkca/k dksfV (bond length) leku gS(D) He2

+ dh ÅtkZ nks ,dy (isolated) He ijek.kqvksa dh ÅtkZ ds leku gS

Ans. (AC)

Sol. (A) �2

2C Total no. of electrons = 14 so it is diamagnetic

(B) ✁2

2O Bond order = 3; 2O Bond order = 2

✂ Bond length in ✁2

2O is less than bond length in O2.

(C ) Bond order of ✁

2N = 2.5

Bond order of ✁

2He = 1/2

✂ Some energy is released during the formation of ✁

2He from two isolated He atoms.

Sol. (A) �2

2C bysDVªkWuksa dh dqy la[;k = 14 blfy, ;g izfrpqEcdh; gSA

(B) ✁2

2O ca/k Øe = 3; 2O ca/k Øe = 2

✂ ✁2

2O esa ca/k yEckbZ O2 dh rqyuk esa de gksrh gSA

(C ) ✁

2N dk cU/k Øe = 2.5✁

2He dk cU/k Øe = 1/2

✂ nks He ijek.kqvksa ls ✁

2He ds fuekZ.k ds nkSjku dqN ÅtkZ eqDr gksrh gSA

27. Mixture(s) showing positive deviation from Raoult's law at 35ºCis(are)(A) carbon tetrachloride + methanol (B) carbon disulphide + acetone (C) benzene + toluene (D) phenol + aniline

feJ.k tks 35ºC ij jkmYV fu;e (Raoult's law) ls /kukRed fopyu iznf'kZr djrk gS ¼djrs gS½(A) dkcZu VsVªkDyksjkbM + esFksukWy (B) dkcZu MkblYQkbM + ,lhVksu(C) csUthu + VkWYohu (D) QhukWy + ,fuyhu

Ans. (AB)

Sol. (A) CCl4 + CH3OH ✄ Positive deviation from Raoult’s law

(B) CS2 +

CH3 CH3

C

O

✄ Positive deviation from Raoult’s law

(C) C6H6 + C7H8 ✄ Ideal solution

(D)

OH

+

NH2

✄ Negative deviation from Raoult’s law.

☎☎☎✆✝✞✟✠✡✟☛✝✆☛☞✠

Sol. (A) CCl4 + CH3OH � jkÅYV fu;e ls /kukRed fopyu

(B) CS2 +

CH3 CH3

C

O

� jkÅYV fu;e ls /kukRed fopyu

(C) C6H6 + C7H8 � vkn'kZ foy;u

(D)

OH

+

NH2

� jkÅYV fu;e ls _.kkRed fopyu

28. For 'invert sugar', the correct statement(s) is(are)

(Given : specific rotations of (+)-sucrose, (+)-maltose, L-(–)-glucose and L-(+)-fructose in aqueous

solution are +66º, +140º, –52º and +92º, respectively)

(A) 'invert sugar' is prepared by acid catalyzed hydrolysis of maltose

(B) 'invert sugar' is an equimolar mixture of D-(+)-glucose and D-(–)-fructose

(C) specific rotation of 'invert sugar' is –20º

(D) on reaction with Br2 water, 'invert sugar' forms saccharic acid as one of the products

'vior 'kdZjk' ('invert sugar') ds fy;s lgh dFku gS@gSa

(fn;k gS% (+)-lwØkst (sucrose), (+)-ekWYVkst (maltose), L-(–)-Xywdkst (glucose) rFkk L-(+)-ÝqDVkst (fructose) dk

tyh; foy;u esa fof'k"V /kzqo.k ?kw.kZu (specific rotation) Øe'k% +66º, +140º, –52º rFkk +92º gS½

(A) 'vio`r 'kdZjk' ekWYVkst ds vEy&mRizsfjr (acid catalyzed) ty&vi?kVu (hydrolysis) ls cuk;k tkrk gSS

(B) 'vio`r 'kdZjk' D-(+)-Xywdkst rFkk D-(–)-ÝqDVkst dk lev.kqd (equimolar) feJ.k gS

(C) 'vio`r 'kdZjk' dk fof'k"V /kzqo.k ?kw.kZu–20º gS

(D) Br2 ty ls vfHkfØ;k djus ij 'vio`r 'kdZjk' mRiknksa esa ls ,d mRikn ds :i esa] lSdsfjd vEy (saccharic

acid) cukrh gS

Ans. (BC)

Sol. Sucrose ✁✁✁✁ ✂✁Hydrolysis

D(+)-glucose + D(–) fructose

[✄] = +52° [✄] = –92°

[✄]mix = 0.5 × (+52°) + 0.5 × (–92°)

= –20°

lwØksl ✁✁✁✁ ✂✁Hydrolysis

D(+)-Xywdksl + D(–) ÝDVksl

[✄] = +52° [✄] = –92°

[✄]mix = 0.5 × (+52°) + 0.5 × (–92°)

= –20° ☎☎☎✆✝✞✟✠✡✟☛✝✆☛☞✠

29. The CORRECT statement(s) for cubic close packed (ccp) three dimensional structure is(are)

(A) The number of the neighbours of an atom present in the topmost layer is 12

(B) The efficiency of atom packing is 74%

(C) The number of octahedral and tetrahedral voids per atom are 1 and 2, respectively

(D) The unit cell edge length is 2 2 times the radius of the atom

?kuh; fufoM ladqfyr (cubic close packed) (ccp) f=kfoeh; lajpuk ds fy;s lgh dFku gS@gSa

(A) ,d ijek.kq tks lokZsPp ijr (topmost layer) esa mifLFkr gS mlds fudVre izfrosf'k;ksa ¼iM+ksfl;ksa½ dh la[;k 12 gSA

(B) ijek.kq dh ladqyu {kerk 74% gS

(C) v"VQydh; rFkk prq"Qydh; fjfä;ksa dh la[;k izfr ijek.kq Øe'k% 1 rFkk 2 gSa

(D) ,d dksf"Bdk ds dksj (unit cell edge) dh yEckbZ ijek.kq dh f=kT;k dk 2 2 xquk gSA

Ans. (BCD)

Sol. (A) For any atom in top most layer, coordination number is not 12 since there is no layer above top

most layer

(B) Fact

(C) Fact

(D) 2 a = 4R

So a = 2 2 R

(A) fdlh Hkh ijek.kq ds fy, lcls Åijh ijr esa] lgla;kstd la[;k 12 ugha gks ldrh D;ksfd Åijh dksbZ ijr ugha

gksrh gS

(B) rF;

(C) rF;

(D) 2 a = 4R

blfy, a = 2 2 R

30. Among the following, reaction(s) which gives(give) tert-butyl benzene as the major product is(are)

fuEufyf[kr esa VVZ&C;wfVy csUthu (tert-butyl benzene) eq[; mRikn ds :i esa nsaus okyh vfHkfØ;k ¼;sa½ gS ¼gSa½

(A)

Br

NaOC2H5 (B)

Cl

AlCl3

(C) H2SO4

(D) OH

BF3�OEt2

Ans. (B,C,D)

✁✁✁✂✄☎✆✝✞✆✟✄✂✟✠✝

Sol. Cl

H2SO4

OH

BF3�OEt2

31. Reagent(s) which can be used to bring about the following transformation is(are)

COOH

C O O

O

H O

COOH

C O O

OH O

(A) LiAlH4 in (C2H5)2O (B) BH3 in THF

(C) NaBH4 in C2H5OH (D) Raney Ni/H2 in THF

fuEufyf[kr :ikUrj.k ds fy;s fdu vfHkdkjd ¼vfHkdkjdksa½ (Reagent(s)) dk mi;ksx fd;k tk ldrk gS ¼gSa½\

COOH

C O O

O

H O

COOH

C O O

H O

(A) (C2H5)2O esa LiAlH4 (B) THF esa BH3

(C) C2H5OH esa NaBH4 (D) THF esa jkus (Raney) Ni/H2

Ans. (C,D)

Sol. NaBH4/C2H5OH & Raney Ni/H2 in T.H.F do not reduce acid (–COOH), ester (–COOR) & epoxide

O

.

gy NaBH4/C2H5OH ,oa jSus Ni/H2 ,T.H.F vEy (–COOH), ,LVj (–COOR) & bikWDlkbMO

dks vipf;r ugh

djrs gSA

✁✁✁✂✄☎✆✝✞✆✟✄✂✟✠✝

32. The nitrogen containing compound produced in the reaction of HNO3 with P4O10

(A) can also be prepared by reaction of P4 and HNO3

(B) is diamagnetic

(C) contains one N-N bond

(D) react with Na metal producing a brown gas

HNO3 dh P4O10 ds lkFk vfHkfØ;k esa mRikfnr ukbVªkstu vUrfoZ"V ;kSfxd

(A) P4 rFkk HNO3 dh vfHkfØ;k ls Hkh cuk;k tk ldrk gSA

(B) izfrpqEcdh; (diamagnetic) gS

(C) esa ,d N–N cU/k vUrfoZ"V gS

(D) Na /kkrq ls vfHkfØ;k dj ,d Hkwjh (brown) xSl mRikfnr djrk gSA

Ans. (B,D)

Sol. 4HNO3 + P4O10 �✁ 2N2O5 + 4HPO3

So, nitrogen containing compound is N2O5

(A) P4 + 20HNO3 �✁ 4H3PO4 + 20NO2 + 4H2O

N2O5 Can't be prepared by reaction of P4 & HNO3.

(B) N2O5 is diamagnetic.

(C) Structure of N2O5 is:

N

O O

O

O N

O

(It does not contain N–N bond)

(D) N2O5 + Na �✁ NaNO3 + NO2 ✂

. 4HNO3 + P4O10 �✁ 2N2O5 + 4HPO3

blfy, ukbVªkstu ;qDr ;kSfxd N2O5 gS

(A) P4 + 20HNO3 �✁ 4H3PO4 + 20NO2 + 4H2O

N2O5 dks P4 & HNO3 dh vfHkfØ;k }kjk ugh cuk;k tk ldrkA

(B) N2O5 izfrpqEcdh; gSA

(C) N2O5 dh lajpuk gS&

N

O O

O

O N

O

blesa N–N cU/k vuqifLFkr gSA

(D) N2O5 + Na �✁ NaNO3 + NO2 ✂ ✄✄✄☎✆✝✞✟✠✞✡✆☎✡☛✟

SECTION – 3 : (Maximum Marks : 12)

This section contains TWO paragraphs.

Based on each paragraph, there are TWO Questions.

Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is correct.

For each question, darken the bubble corresponding to the correct option in the ORS.

For each questions, marks will be awarded in one of the following categories :Full Marks : +3 If only the bubble corresponding to the correct option is darkened. Zero Marks : 0 In all other cases.

Paragraph 1

Thermal decomposition of gaseous X2 to gaseous X at 298 K takes place according to the following

equation :

X2(g) 2X(g)

The standard reaction Gibbs energy, �rGº, of this reaction is positive. At the start of the reaction, there

is one mole of X2 and no X. As the reaction proceeds, the number of moles of X formed is given by ✁.

Thus. ✂equilibrium is the number of moles of X formed at equilibrium. The reaction is carried out at a

constant total pressure of 2 bar. Consider the gases to behave ideally.

(Given : R = 0.083 L bar K–1

mol–1

)

1

298 K ij xSlh; (gaseous) X2 dk xSlh; X esa Å"ek&vi?kVu (thermal decomposition) fuEufyf[kr lehdj.k

X2(g) 2X(g)

ds vuqlkj gksrk gSA bl vfHkfØ;k dh ekud vfHkfØ;k fxCl ÅtkZ (standard reaction Gibbs energy), �rGº,

/kukRed gSA vfHkfØ;k ds izkjEHk esa X2 dk 1 eksy gS rFkk X ugha gSA tSls&tSls ;g vfHkfØ;k c<+rh gS] fufeZr X ds

eksyksa dh la[;k ✁ }kjk nh tkrh gSA bl izdkj] lkE;okLFkk ij fufeZr X ds eksyksa dh la[;k ✂equilibrium gSA vfHkfØ;k 2

bar ds fLFkj dqy nkc ij dh tkrh gSA eku ysa fd xSlsa vkn'kZ O;ogkj djrh gSA

(fn;k x;k gS : R = 0.083 L bar K–1

mol–1

)

33. The equilibrium constant Kp for this reaction at 298 K, in terms of ✂equilibrium , is

298 K ij bl vfHkfØ;k dk ✂equilibrium ds in esa lkE;oLFkk fLFkjkad (equilibrium constant) Kp D;k gksxk\

(A)

2

equilibrium

equilibrium

8

2

✄

☎✄(B)

2

equilibrium

2

equilibrium

8

4

✄

☎✄(C)

2

equilibrium

equilibrium

4

2

✄

☎✄(D)

2

equilibrium

2

equilibrium

4

4

✄

☎✄

Ans. (B)

✆✆✆✝✞✟✠✡☛✠☞✞✝☞✌✡

Sol. Paragraph-1

X2 (g) 2 X (g)

Initial mole 1 0

t = teq. (1–�) 2�

Given 2� = ✁equilibrium

So � = equilibrium

2

✂

Total mole at equilibrium = (1 + �) = (1+eq

2

✄☎ ✆✝ ✞✟ ✠

Px2

=

eq.

totaleq.

12 P

12

✡

✡

☛ ☞✌✍ ✎

✍ ✎✍ ✎✏✍ ✎✑ ✒

= eq

totaleq

2P

2

✓

✓

✔ ✕✖✗ ✘

✙✗ ✘✚ ✛ =

eqtotal

eq

2P

2

✓

✓

✔ ✕✖✗ ✘

✙✗ ✘✚ ✛

PX(g) = eq

totaleq

P

12

✡

✡

☛ ☞✍ ✎✍ ✎✍ ✎✏✍ ✎✑ ✒

= eq

eq

2

2

✓

✓

✔ ✕✗ ✘

✙✗ ✘✚ ✛PTotal

So KP = ✜ ✢

✜ ✢

2

2

Px

Px =

✣ ✤

2

eq.total

eq.

eq.total

eq.

2P

2

2P

2

✥

✥

✥

✥

✦ ✧★✩ ✪

✫✩ ✪✬ ✭✦ ✧✮✩ ✪★✩ ✪✫✬ ✭

KP =

2eq.

2eq.

4

4 –

✯

✯× PTotal =

2eq

2eq

8

4 –

✰

✰

✱ ✲✳ ✴✳ ✴✵ ✶

So Ans. is = B

vuqPNsn-1

X2 (g) 2 X (g)

izkjfEHkd eksy 1 0

t = teq. (1–�) 2�

fn;k gS : 2� = ✁lkE;

vr% � = ✷

2lkE; ✸✸✸✹✺✻✼✽✾✼✿✺✹✿❀✽

lkE; ij dqy eksy = (1 + �) = (1+eq

2

✁✂ ✄☎ ✆✝ ✞

Px2

=

eq.

totaleq.

12 P

12

✟

✟

✠ ✡☛☞ ✌

☞ ✌☞ ✌

✍☞ ✌✎ ✏

= eq

totaleq

2P

2

✑

✑

✒ ✓✔✕ ✖

✗✕ ✖✘ ✙ =

eqtotal

eq

2P

2

✑

✑

✒ ✓✔✕ ✖

✗✕ ✖✘ ✙

PX(g) = eq

totaleq

P

12

✟

✟

✠ ✡☞ ✌☞ ✌☞ ✌

✍☞ ✌✎ ✏

= eq

eq

2

2

✑

✑

✒ ✓✕ ✖

✗✕ ✖✘ ✙PTotal

vr% KP =✚ ✛

✚ ✛

2

2

Px

Px =

✜ ✢

✣

✣

✣

✣

✤ ✥✦✧ ★

✩✧ ★✪ ✫✤ ✥✬✧ ★✦✧ ★✩✪ ✫

2

eq.

eq.

eq.

eq.

2P

2

2P

2

dqy

dqy

KP =

2eq.

2eq.

4

4 –

✭

✭× PTotal =

2eq

2eq

8

4 –

✮

✮

✯ ✰✱ ✲✱ ✲✳ ✴

vr% Ans. is = B

34. The INCORRECT statement among the following, for this reaction, is

(A) Decrease in the total pressure will result in formation of more moles of gaseous X

(B) At the start of the reaction, dissociation of gaseous X2 takes place spontaneously

(C) ✵equilibrium = 0.7

(D) KC < 1

bl vfHkfØ;k ds fy;s fueu esa lsa vlR; dFku gS

(A) dqy nkc ds ?kVus ds ifj.kke Lo:i xSlh; X ds vf/kd eksy cusaxs

(B) vfHkfØ;k ds izkjEHk esa xSlh; X2 dk fo;kstu Lor% izofrZr (spontaneously) gksrk gS

(C) ✵equilibrium = 0.7

(D) KC < 1

Ans. (C)

Sol. (A) Correct statement.

As on decrease in pressure reaction move indirection where no. of gaseous molecules increase.

(B) Correct statement

At the start of reaction QP < KP so dissociation of X2 take place spontaneousely.

(C) Incorrect statement as KP =

2eq

2eq

8

4

✭

✭✶ =

✷ ✸

✷ ✸

2

2

8x 0.7

4 0.7✹

> 1 but

✺✺✺✻✼✽✾✿❀✾❁✼✻❁❂✿

(D) Correct statement.

As �G

0 > 0 & ✁G

0 = –RTlnKP

✁G0 > 1, So KP should be less than 1.

So K < 1

KP - KC(RT)✂ng.

(RT > 1)

KC = PK

RT

KC < KP So KC < 1

Sol. (A) lgh dFku % tSls&tSls nkc de gksrk tkrk gS vfHkfØ;k xSlh; v.kqvksa dh la[;k esa o`f) dh rjQ c<rh gS

(B) lgh dFku % vfHkfØ;k dh 'kq:vkr esa QP < KP vr% X2 dk fo;kstu Lor% izofrZr gksrk gSA

(C) xyr dFku D;ksafd KP =

2eq

2eq

8

4

✄

✄☎ =

✆ ✝

✆ ✝

2

2

8x 0.7

4 0.7✞> 1,

(D) lgh dFku %

bl rjg �G0 > 0 & ✁G

0 = –RTlnKP

✁G0 > 1, vr% KP dk eku 1 ls de gksuk pkfg,A

vr% K < 1

KP - KC(RT)✂ng.

(RT > 1)

KC = PK

RT

KC < KP So KC < 1

✟✟✟✠✡☛☞✌✍☞✎✡✠✎✏✌

Paragraph 2

Treatment of compound O with KMnO4/H+ gave P, which on heating with ammonia gave Q. The

compound Q on treatment with Br2/NaOH produced R. On strong heating, Q gave S, which on further

treatment with ethyl 2-bromopropanoate in the presence of KOH followed by acidification, gave a

compound T.

2

;kSfxd O dh KMnO4/H+ ls fØ;k us P fn;k] ftlus veksfu;k ds lkFk xeZ djus Q fn;kA ;kSfxd Q us Br2/NaOH

ds lkFk fØ;k djus ij R mRikfnr fd;kA izcy :i ls xeZ djus ij Q us S fn;k ftlus ,fFky 2-czkseksizksisuksvksV

(ethyl 2-bromopropanoate) ds lkFk dh KOH mifLFkfr esa vkxs fØ;k dh ftlds i'pkr vEyhdj.k us ;kSfxd T

fn;kA

(O)

35. The compound R is

;kSfxd R gS

(A)

NH2

NH2 (B)

O

Br

Br

O

(C)

O

NHBr

O

NHBr (D)

O

O

NBr

Ans. (A)

36. The compound T is

(A) glycine (B) alanine (C) valine (D) serine

;kSfxd T gS

(A) Xykblhu (glycine) (B) ,ykuhu (alanine)

(C) oSyhu (valine) (D) lsjhu (serine)

Ans. (B)

���✁✂✄☎✆✝☎✞✂✁✞✟✆

Sol.

COOH KMnO4/H

�

COOH

NH3/✁

O

C–NH2

C–NH2

O

O

H3O�

O

N–CH–C–OC2H5 CH3–CH–C–OC2H5

KOH

OBr O

O

NH

NH2

NH2

CH3–CH–COOH

NH2

Br2/NaOH

✁

CH3

O

Alanine

(O) (P) (Q) (R)

(S)

(T)

Sol.

COOH KMnO4/H

✂

COOH

NH3/✄

O

C–NH2

C–NH2

O

O

H3O✂

O

N–CH–C–OC2H5 CH3–CH–C–OC2H5

KOH

O Br O

O

NH

NH2

NH2

CH3–CH–COOH

NH2

Br2/NaOH

✄

CH3

O

,sysuhu

(O) (P) (Q) (R)

(S)

(T)

☎☎☎✆✝✞✟✠✡✟☛✝✆☛☞✠

MATHEMATICS

37. The value of dxe1

xcosx2

2–

x

2

�

✁

✁ ✂ is equal to

dxe1

xcosx2

2–

x

2

�

✁

✁ ✂dk eku gS&

(A) 2–4

2✄(B) 2

4

2

☎✄(C) ✆2

– e✝/2 (D) ✆2

+ e✝/2

Ans. (A)

Sol. ✞ =/ 2 2

x

– /2

x cos xdx

(1 e )

✟

✟ ✠✡ ☛ ✞ = / 2 2 2

x – x

0

x cos x x cos xdx

1 e 1 e

☞ ✌ ✍✎✏ ✑✎ ✎✒ ✓✔

✞ =

/ 2

2 2 / 2

0

0

x cos xdx (x sinx – 2x(– cosx) (2)(– sinx))✟ ✟✕ ✠✡ =

2 2

– 2 – (0) – 24 4

✌ ✍✖ ✖✗✏ ✑✒ ✓

38. Let bi > 1 for i = 1,2,….,101. Suppose logeb1,logeb2,…,logeb101 are in Arithmetic progression (A.P.) with

the common difference loge 2. Suppose a1, a2,…,a101 are in A.P. such that a1 = b1 and a51= b51. If

t = b1 + b2 + …. + b51 and s = a1 + a2 + … + a51, then

(A) s > t and a101 > b101 (B) s > t and a101 < b101

(C) s < t and a101 > b101 (D) s < t and a101 < b101

ekukfd i = 1,2,….,101 ds fy, bi > 1 gSA eku yhft, fd logeb1,logeb2,…,logeb101 lkoZvarj (common

difference) loge 2 okyh lekarj Js.kh (A.P.) esa gSaA eku yhft;s fd a1, a2,…,a101 lekarj Js.kh esa bl izdkj gSa fd

a1 = b1 rFkk a51= b51 ;fn t = b1 + b2 + …. + b51 rFkk s = a1 + a2 + … + a51 gSa, rc

(A) s > t vkSj a101 > b101 (B) s > t vkSj a101 < b101

(C) s < t vkSj a101 > b101 (D) s < t vkSj a101 < b101

Ans. (B)

Sol. loge b1, logeb2, logeb3, ...... logeb101 are in A.P. lekUrj Js.kh esa gS

b1, b2, b3, ..........., b101 are in G.P. xq.kksÙkj Js.kh esa gS

Given fn;k x;k gS : loge(b2) – loge(b1) = loge(2) ☛ 2

1

b

b= 2 = r (common ratio of G.P. xq.kksÙkj Js.kh dk lkoZ

vuqikr) ✘✘✘✙✚✛✜✢✣✜✤✚✙✤✥✢

a1, a2, a3, ......... a101 are in A.P. lekUrj Js.kh esa gS

a1 = b1 = a

b1 + b2 + b3 + ........ b51 = t ,

S = a1 +a2 + ...... + a51

t = sum of 51 terms of G.P.(t = xq.kksÙkj Js.kh ds 51 inks dk ;ksx) = b1 51(r – 1)

r –1=

51a(2 – 1)

2 – 1 = a(2

51 –1)

s = sum of 51 terms of A.P. (s = lekUrj Js.kh ds 51 inks dk ;ksx.) = 51

2[2a1 + (n–1)d] =

51

2 (2a + 50d)

Given fn;k x;k gS a51 = b51

a + 50d = a(2)50

50d = a(250

– 1)

Hence vr% s = a51

2 [2

50 + 1] � s = 49 51

a 51 .22

✁ ✂✄☎ ✆

✝ ✞

s = 49 49 512 4.2 47.2

2

✁ ✂✄ ✄☎ ✆✝ ✞

� s = 51 49 53a (2 – 1) 47.2

2

✁ ✂✄ ✄☎ ✆✝ ✞

s – t = 49 53a 47 . 2 +

2

✁ ✂☎ ✆✝ ✞

Clearly Li"Vr % s > t

a101 = a1 + 100d = a + 2a.250

– 2a = a (251

– 1)

b101 = b1 r 100

= a.2100

Hence vr% b101 > a101

39. Let P =

✟✟✟

✠

✡

☛☛☛

☞

✌

1416

014

001

and ✍ be the identity matrix of order 3. If Q = [qij] is a matrix such that P50

– Q= ✍,

then

21

3231

q

qq ✎ equals

ekuk fd P =

✟✟✟

✠

✡

☛☛☛

☞

✌

1416

014

001

vkSj ✍ rhu dksfV (order 3) dk rRled vkO;wg (identity matrix) gSA ;fn Q = [qij] ,d

vkO;wg bl izdkj gS fd P50 – Q = ✍ gS, rc

21

3231

q

qq ✎dk eku gS&

(A) 52 (B) 103 (C) 201 (D) 205

Ans. (B)

✏✏✏✑✒✓✔✕✖✔✗✒✑✗✘✕

Sol. P =

1 0 0

4 1 0

16 4 1

� ✁✂ ✄✂ ✄✂ ✄☎ ✆

✝ P2 =

1 0 0

8 1 0

16(1 2) 8 1

� ✁✂ ✄✂ ✄✂ ✄✞☎ ✆

, P3 =

1 0 0

12 1 0

16(1 2 3) 12 1

� ✁✂ ✄✂ ✄✂ ✄✞ ✞☎ ✆

P50

=

1 0 0

4 50 1 0

16(1 2 ... 50) 4 50 1

� ✁✂ ✄✟✂ ✄✂ ✄✞ ✞ ✞ ✟☎ ✆

2

✠P

50 =

1 0 0

200 1 0

20400 200 1

� ✁✂ ✄✂ ✄✂ ✄☎ ✆

✡ P50

– Q = ☛

✝

1 0 0

200 1 0

20400 200 1

� ✁✂ ✄✂ ✄✂ ✄☎ ✆

–

11 12 13

21 22 23

31 32 33

q q q

q q q

q q q

� ✁✂ ✄✂ ✄✂ ✄☎ ✆

=

1 0 0

0 1 0

0 0 1

� ✁✂ ✄✂ ✄✂ ✄☎ ✆

✝ 200 – q21 = 0 ✝ q21 = 200

20400 – q31 = 0 ✝ q31 = 20400 and vkSj 200 – q32 = 0 ✝ q32 = 200

☞ 31 32

21

q q

q

✌=

20400 200

200

✍ = 103

40. The value of ✎✏ ✑

✒✓✔

✕✖ ✗✘✗

✑✒✓✔

✕✖ ✗✙✘✗

13

1k

6

k

4sin

6

)1k(

4sin

1 is equal to

✎✏ ✑

✒✓✔

✕✖ ✗✘✗

✑✒✓✔

✕✖ ✗✙✘✗

13

1k

6

k

4sin

6

)1k(

4sin

1 dk eku gS&

(A) 33 ✚ (B) ✛ ✜332 ✢ (C) ✣ ✤132 ✢ (D) ✛ ✜322 ✌

Ans. (C)

Sol. 13

k 1

ksin – (k – 1)

4 6 4 6

ksin sin sin (k – 1)

6 4 6 4 6

✥

✦ ✧★ ★ ★ ★✩ ✪ ✩ ✪✫ ✫✬ ✭ ✬ ✭✮ ✯✰ ✱ ✰ ✱✲ ✳✩ ✪★ ★ ★ ★ ★✩ ✪ ✩ ✪✫ ✫✬ ✭✬ ✭ ✬ ✭✰ ✱ ✰ ✱✰ ✱

✴ = 13

k 1

k2 cot (k – 1) – cot

4 6 4 6✵✶ ✷✸ ✸ ✸ ✸✶ ✷ ✶ ✷✹ ✹✺ ✻✺ ✻ ✺ ✻✼ ✽ ✼ ✽✼ ✽

✾

= 13

2 cot – cot4 4 6

✶ ✷✸ ✸ ✸✶ ✷✹✺ ✻✺ ✻✼ ✽✼ ✽ =

292 1– cot

12

✶ ✷✸✶ ✷✺ ✻✺ ✻✼ ✽✼ ✽

=5

2 1– cot12

✶ ✷✸✶ ✷✺ ✻✺ ✻✼ ✽✼ ✽

= 2 (1–(2– 3 )) =2 (–1 + 3 )

= 2 ( 3 –1)

41. Let P be the image of the point (3, 1, 7) with respect to the plane x – y + z = 3. Then the equation of the

plane passing through P and containing the straight line 1

z

2

y

1

x ✿✿ is

ekuk fd fcUnq (3, 1, 7) dk] lery x – y + z = 3 ds lkis{k (with respect to), izfrfcEc (image) P gSA rc fcUnq P

ls xqtjus okys vkSj ljy js[kk 1

z

2

y

1

x ✿✿ dks /kkj.k djus okys lery dk lehdj.k gS&

(A) x + y – 3z = 0 (B) 3x + z = 0 (C) x – 4y + 7z = 0 (D) 2x – y = 0

Ans. (C)

❀❀❀❁❂❃❄❅❆❄❇❂❁❇❈❅

Sol.

•

P (–1,5,3)

x – y + z = 3

Q (3,1,7)

x – 3

1 =

y – 1

–1 =

z – 7

1 =

–2(6)–4

3�

x =– 1, y = 5, z = 3 P (–1, 5, 3)

a(x + 1) + b(y – 5) + c (z – 3) = 0

a + 2b + c = 0 ...................(i)

a – 5b – 3c = 0

a

–1 =

b

4 =

c

–7

–(x + 1) + 4 (y – 5) – 7 (z – 3) = 0

–x + 4y – 7z = 0

x – 4y + 7z = 0

42. Area of the region ✁ ✂159xy5,3xy:R)y,x( 2 ✄☎✄☎✆✝ is equal to

{ks=k (region) ✁ ✂159xy5,3xy:R)y,x( 2 ✄☎✄☎✆✝ dk {ks=kQy (area) gS&

(A) 6

1(B)

3

4(C)

2

3(D)

3

5

Ans. (C)

y2 = x + 3

y2 = –x – 3

D(1, 2)

C(1, 0)B(–3, 0) (–4, 0)A

E(–4, 1)

x = 6 5y = x + 9

Area ABE (under parabola) =

3

4

x 3

✞

✞✟ ✟✠ dx =

2

3 ✡✡✡☛☞✌✍✎✏✍✑☞☛✑✒✎

Area BCD (under parabola) =

1

3

x 3

�✁✂ dx =

16

3

Area of trapezium ACDE = 1

2(1 + 2)5 =

15

2

Required area = 15 16 2 3

2 3 3 2✄ ✄ ☎

Hindi

y2 = x + 3

y2 = –x – 3

D(1, 2)

C(1, 0)B(–3, 0) (–4, 0)A

E(–4, 1)

x = 6 5y = x + 9

ABE dk {ks=kQy (ijoy; ds vUnj) =

3

4

x 3

�

�✆ ✆✂ dx =

2

3

BCD dk {ks=kQy (ijoy; ds vUnj) =

1

3

x 3

�✁✂ dx =

16

3

leyEc prqHkZqt ACDE dk {ks=kQy = 1

2(1 + 2)5 =

15

2

vHkh"V {ks=kQy = 15 16 2 3

2 3 3 2✄ ✄ ☎

43. Let f : ✝✞✟

✠✡☛☞ 2,

2

1 ✌ R and g : ✝✞

✟✠✡☛☞ 2,

2

1 ✌ R be functions defined by f(x) = [x

2 – 3] and

g(x) = |x| f(x) + |4x – 7| f(x), where [y] denotes the greatest integer less than or equal to y for y ✍ R. Then

(A) f is discontinuous exactly at three points in ✝✞✟

✠✡☛☞ 2,

2

1

(B) f is discontinuous exactly at four point in ✝✞✟

✠✡☛☞ 2,

2

1

(C) g is NOT differentiable exactly at four points in ✎✏✑✒

✓✔ ☞ 2,

2

1

(D) g is NOT differentiable exactly at five points in ✎✏✑✒

✓✔ ☞ 2,

2

1.

✕✕✕✖✗✘✙✚✛✙✜✗✖✜✢✚

ekuk fd Qyu f : �✁

✂✄☎

✆✝ 2,

2

1✞ R vkSj g : �

✁

✂✄☎

✆✝ 2,

2

1✞ R, f(x) = [x

2 – 3] vkSj g(x) = |x| f(x) + |4x – 7| f(x) ls

ifjHkkf"kr gSa] tgk¡ y ✟ R ds fy, y ls de ;k y ds cjkcj ds egÙke iw.kkZad (greatest integer less than or equal to

y) dks [y] n~okjk n'kkZ;k x;k gSA rc

(A) �✁

✂✄☎

✆✝ 2,

2

1esa f Bhd rhu (exactly three) fcUnqvksa ij vlarr (discontinuous) gSA

(B) �✁

✂✄☎

✆✝ 2,

2

1esa f Bhd pkj (exactly four) fcUnqvksa ij vlarr gSA

(C) ✠✡

☛☞✌

✍✝ 2,

2

1esa g Bhd pkj (exactly four) fcUnqvksa ij vodyuh; (differentiable) ugha gSA

(D) ✠✡

☛☞✌

✍✝ 2,

2

1esa g Bhd ik¡p (exactly five) fcUnqvksa ij vodyuh; (differentiable) ugha gSA

Ans. (B,C)

Sol. f(x) = [x2 – 3] = [x

2] – 3 =

1–3 – x 1

2

–2 1 x 2

–1 2 x 3

0 3 x 2

1 x 2

✎✏ ✑✒

✒✒ ✏ ✑✒

✏ ✑✒✒

✏ ✑✒✒ ✓✔

g(x) = |x| f(x) + |4x–7| f(x)

= (|x| + |4x – 7|) [x2 – 3] =

1(–x – 4x – 7)(–3) – x 0

2

(x – (4x – 7))(–3) 0 x 1

(x – (4x – 7))(–2) 1 x 2

(x – (4x – 7))(–1) 2 x 3

((x – (4x – 7))(0) 3 x 7 / 4

(x (4x – 7))(0) 7 / 4 x 2

(x (4x – 7))(1) x 2

✕✖ ✗✘

✘✖ ✗✘

✘✖ ✗✘

✘ ✖ ✗✘✘ ✖ ✗✘

✙ ✖ ✗✘✘ ✙ ✚✘✛

=

115x 21 – x 0

2

9x – 21 0 x 1

6x – 14 1 x 2

3x – 7 2 x 3

0 3 x 2

5x – 7 x 2

✜✢ ✣ ✤✥

✥✣ ✤✥

✥✣ ✤✥

✥ ✣ ✤✥✥ ✣ ✤✥

✦✥✧

★★★✩✪✫✬✭✮✬✯✪✩✯✰✭

Now graph of given function is vc fn;s x;s Qyu dk xzkQ

f(x) –1

–1/2 21

–2

3 2

–3

1

g(x) –1/2

1 2 2 3

–21

–12

–8

6 2 14�

27/2

21

3 2 7✁

3

3 3 7✂

Clearly F is not discontinuous at exactly 4 point in [–1/2, 2] and g is not differentiable at 4 points in (–1/2, 2) Hence Ans. is BC

Li"Vr;% F [–1/2, 2] esa Bhd pkj fcUnqvksa ij vlrr~ rFkk g (–1/2, 2) esa Bhd pkj fcUnqvksa ij vodyuh; ugh gS

44. Let 1 2 3ˆ ˆ ˆu u i u j u k✄ ☎ ☎ be a unit vector in R

3 and ✆ ✝

1 ˆ ˆ ˆw i j 2k6

✞ ✟ ✟ . Given that there exists a vector ✠✡

in R3 such that u 1☛☞ ✄

✌ and ✍ ✎ˆ ˆw. u 1☛☞ ✄

✌. Which of the following statements(s) is (are) correct?

(A) There is exactly one choice for ✠✡

(B) There are infinitely many choices for such ✠✡

(C) If u lies in the xy-plane then 1 2u u✄

(D) If u lies in the xz-plane then 1 32 u u✄

Ekkuk fd R3 esa 1 2 3

ˆ ˆ ˆu u i u j u k✄ ☎ ☎ ,d ek=kd lfn'k (unit vector) gS vkSj ✆ ✝1 ˆ ˆ ˆw i j 2k6

✞ ✟ ✟ gSA fn;k gqvk gS

fd R3 esa lfn'k ✠

✡dk vfLrRo bl izdkj gS fd u 1☛☞ ✄

✌vkSj ✍ ✎ˆ ˆw. u 1☛☞ ✄

✌gSA fuEufyf[kr esa ls dkSu lk ¼ls½

dFku lgh gS ¼gSa½\

(A) bl izdkj ds ✠✡

ds fy, Bhd ,d (exactly one) p;u laHko gSA

(B) bl izdkj ds ✠✡

ds fy, vUkUr (infinitely) p;u laHko gSA

(C) ;fn u xy- lery ij gS rc 1 2u u✄ gSA

(D) ;fn u xz-lery ij gS rc 1 32 u u✄ gSA

Ans. (B,C)

✏✏✏✑✒✓✔✕✖✔✗✒✑✗✘✕

Sol. ˆ| u v | 1� ✁✂

|v| sin ✄ = 1 ✄ is angle between u & v☎

(✄, u & v☎ds e/; dks.k gSA)

Also rFkk, ˆ ˆw.(u v) 1� ✁✂

ˆ| w | ˆ|u | | v |✂

sin ✄ cos ✆ = 1 ✆ is angle between w & ˆ(u v)✝✞

(✆✟ w vkSj ˆ(u v)✝✞

ds e/; dks.k gSA)

1.1(1) cos ✆ = 1 ✠ 0✡ ☛ ☞ ✠ ˆ ˆu v w✌ ☛ ✍✎

where tgk¡ ✏ > 0

1 2 3

x y z

ˆ ˆ ˆi j k

u u u

v v v

= 6

✑✒ ✓ˆ ˆ ˆi j 2k✔ ✔ (u2vz – u3vy) i + (u3vx – u1vz) j + (u1vy – u2vx) k =

6

✑✒ ✓ˆ ˆ ˆi j 2k✔ ✔

(B) v✕

is a vector such that ˆ(u v)✝✞

is parallel to w .

(B) v✕bl izdkj gS fd ˆ(u v)✝

✞, w ds lekUrj gSA

(C) u3 = 0 ✠ u2vz = 6

✑ & –u1vz =

6

✑ ✠ 2 1|u | | u |✖

(D) |u1| = 2|u3| ( ✗ u2 = 0)

45. Let P be the point on the parabola y2 = 4x which is at the shortest distance from the center S of the

circle x2 + y

2 – 4x –16y +64 = 0. Let Q be the point on the circle dividing the line segment SP internally.

Then

(A) SP = 2 5

(B) SQ : QP =✘ ✙5 1 : 2✚

(C) the x-intercept of the normal to the parabola at P is 6

(D) the slope of the tangent to the circle at Q is 1

2

Ekkuk fd ijoy; (parabola) y2 = 4x ij P ,d ,slk fcUnq gS tks o`Ùk x2

+ y2 – 4x –16y +64 = 0 ds dsUnz fcUnq S

ls U;wure nwjh ij gSA ekuk fd o`Ùk ij fcUnq Q ,slk gS fd og js[kk[kaM SP dks vkarfjd foHkkftr djrk gSA rc

(A) SP = 2 5

(B) SQ : QP =✘ ✙5 1 : 2✚

(C) ijoy; ds fcaUnq P ij vfHkyEc (normal) dk x-var%[k.M 6 gSA

(D) o`Ùk ds fcUnq Q ij Li'kZjs[kk dh <ky (slope) 1

2gSA

Ans. (A,C,D) ✛✛✛✜✢✣✤✥✦✤✧✢✜✧★✥

Sol. S

(2, 8)

Q

P(t2, 2t)

y2 = 4x

let any point P(t2, 2t) on parabola.

As we know shortest distance between two curves lies along their common normal.

The common normal will pass through centre of circle.

Slope of normal to the parabola y2 = 4x at P = –t,

2

2t 8

t 2

�

�

= – t

✁ t3 = 8 ✁ t = 2 ✂ P(4, 4)

(i) equation of normal at P(4, 4) ✁ y = –2x + 12 ✁ x-intercept = 6

(ii) slope of tangent at Q = slope of tangent at P = 1

2

(iii) SQ 2 1 5 1

QP 42 5 2 5 1

✄☎ ☎ ☎

✆ ✆

Hindi. S

(2, 8)

Q

P(t2, 2t)

y2 = 4x

ekuk ijoy; ij fcUnq P(t2, 2t) gSA

ge tkurs gS fd nks oØks ds e/; U;wure nwjh buds mHk;fu"B vfHkyEc ds laxr gksrh gSA

mHk;fu"B vfHkyEc o`Ùk ds dsUnz ls xqtjsxk

ijoy; y2 = 4x ds vfHkyEc dh P = –t ij izo.krk

2

2t 8

t 2

�

�

= – t

✁ t3 = 8 ✁ t = 2 ✂ P(4, 4)

(i) P(4, 4) ij vfHkyEc dk lehdj.k ✁ y = –2x + 12 ✁ x-vUr%[k.M= 6

(ii) Q ij Li'kZ js[kk dh izo.krk = P ij Li'kZ js[kk dh izo.krk = 1

2

(iii) SQ 2 1 5 1

QP 42 5 2 5 1

✄☎ ☎ ☎

✆ ✆

✝✝✝✞✟✠✡☛☞✡✌✟✞✌✍☛

46. Let a, b � R and f : R ✁ R be defined by f(x) = a cos (|x3 –x|) + b|x| sin(|x

3 +x|). Then f is

(A) differentiable at x = 0 if a = 0 and b = 1

(B) differentiable at x = 1 if a = 1 and b = 0

(C) NOT differentiable at x = 0 if a = 1 and b = 0

(D) NOT differentiable at x = 1 if a = 1 and b = 1

ekuk fd a, b � R vkSj f : R ✁ R, f(x) = a cos (|x3 –x|) + b|x| sin(|x

3 +x|) ls ifjHkkf"kr gSA rc f

(A) x = 0 ij vodyuh; (differentiable) gS ;fn a = 0 vkSj b = 1

(B) x = 1 ij vodyuh; gS ;fn a = 1 vkSj b = 0

(C) x = 0 ij vodyuh; ugha gS ;fn a = 1 vkSj b = 0

(D) x = 1 ij vodyuh; ugha gS ;fn a = 1 vkSj b = 1

Ans. (A,B)

Sol. at x = 0, x = 0 is repeated root of g(x) = |x| sin|x3 + x|

hence f(x) is differentiable

& at x = 1 ✂ a cos|x3 – x| = acos(x

3 – x)

as cos(–✄) = cos(✄) ☎ f(x) is differentiable

Hindi. x = 0 ij x = 0, g(x) = |x| sin|x3 + x| dk iqujko`Ùk ewy gSA

vr% f(x) vodyuh; gSA

x = 1 ij

✂ a cos|x3 – x| = acos(x

3 – x)

pwafd cos(–✄) = cos(✄) ☎ f(x) vodyuh; gSA

47. Let f(x) =

n

x

2

22

2222

n

n

n

nx.....

4

nx)nx(!n

n

nx.....

2

nx)nx(n

lim

✆✆✆✆✆

✝

✞

✟✟✟✟✟

✠

✡

✆✆✝

✞✟✟✠

✡☛✆✆

✝

✞✟✟✠

✡☛☛

✆✝

✞✟✠

✡☛✆

✝

✞✟✠

✡☛☛

☞✌ , for all x > 0. Then

ekuk fd lHkh x > 0 ds fy, f(x) =

n

x

2

22

2222

n

n

n

nx.....

4

nx)nx(!n

n

nx.....

2

nx)nx(n

lim

✆✆✆✆✆

✝

✞

✟✟✟✟✟

✠

✡

✆✆✝

✞✟✟✠

✡☛✆✆

✝

✞✟✟✠

✡☛☛

✆✝

✞✟✠

✡☛✆

✝

✞✟✠

✡☛☛

☞✌gSA rc

(A) f )1(f2

1✍✎

✏

✑✒✓

✔ (B) ✎

✏

✑✒✓

✔✕✎

✏

✑✒✓

✔

3

2f

3

1f (C) f✖(2) ✗ 0 (D)

)2(f

)2(f

)3(f

)3(f ✘✙

✘

Ans. (B,C)

✚✚✚✛✜✢✣✤✥✣✦✜✛✦✧✤

Sol. f(x) =

x / n

2n

2 2 2n2n

2 2 2 2 2

x x 1 x 1n 1 .......

n n 2 n nlim

x x 1 x 1n! n 1 .......

n n 2 n n

�✁

✂ ✄☎ ✆☎ ✆ ☎ ✆✝ ✝✞ ✞ ✞✟ ✠✟ ✠ ✟ ✠✝ ✝✡ ☛✡ ☛ ✡ ☛☞ ✌☎ ✆☎ ✆ ☎ ✆✝ ✝✞ ✞ ✞✟ ✠✟ ✠ ✟ ✠✝ ✝✟ ✠✟ ✠ ✟ ✠✡ ☛✡ ☛ ✡ ☛✍ ✎

✏nf(x) =nlim✑✒

x

n

n n 2

2r 1 r 1

x 1 rx 1n n

n r rn✓ ✓

✔ ✕✖ ✗✖ ✗✘ ✘✙ ✚ ✙✛ ✜✢ ✣✛ ✜ ✛ ✜✤ ✥✘ ✘✤ ✥✦ ✧★ ★✩ ✩ =

nlim✑✒

n n 2 2

2r 1 r 1

x rx r xn 1 n 1

n n n✓ ✓

✔ ✕✖ ✗✖ ✗✘ ✘✙ ✚ ✙✛ ✜✢ ✣✛ ✜ ✛ ✜✤ ✥✘ ✘✤ ✥✦ ✧★ ★✩ ✩

✏n(f(x)) = x

1

0

n✪ ✩ (1 + xy)dy – x

1

0

n✪ ✩ (1 + x2y

2)dy

Let xy = t

✏n(f(x)) =

x

0

n✪✩ (1 + t)dt –

x

0

n✪✩ (1 + t2)dt

f '(x)

f(x) =

2

1 xn

1 x

✫✬ ✭✮ ✯✫✰ ✱

✲

f '(2)

f(2) =

3n

5

✬ ✭✮ ✯✰ ✱

✲ < 0 ✳ f'(2) < 0

f '(3)

f(3) =

4n

10

✬ ✭✮ ✯✰ ✱

✲ = 2

n5

✬ ✭✮ ✯✰ ✱

✲ ✳ f '(2)

f(2)

f '(3)

f(3)✴

Now f '(x)

f(x) > 0 in (0, 1) and

f '(x)

f(x) < 0 in (1 , ✵)

vc (0, 1) esa f '(x)

f(x)> 0 rFkk (1 , ✵) esa

f '(x)

f(x) < 0

0 1

f(x) is increasing in (0, 1) & decreasing in [1, ✵) (as f(x) is positive)

f(x), (0, 1) esa o)Zeku rFkk [1, ✵) esa ákleku (D;ksfd f(x) /kukRed gSA)

hence vr% f(1) ✶ 1

f2

✬ ✭✮ ✯✰ ✱

and rFkk 1 2

f f3 3

✬ ✭ ✬ ✭✷✮ ✯ ✮ ✯✰ ✱ ✰ ✱

✸✸✸✹✺✻✼✽✾✼✿✺✹✿❀✽

48. Let f : R � (0, ✁) and g : R � R be twice differentiable functions such that f " and g" are continuous

functions on R. Suppose f '(2) = g(2) = 0, f "(2) ✂ 0 and g'(2) ✂ 0, If 2x

lim✄ )x('g)x('f

)x(g)x(f= 1, then

(A) f has a local minimum at x = 2 (B) f has a local maximum at x = 2

(C) f "(2) > f(2) (D) f(x) – f "(x) = 0 for at least one x ☎ R

Ekkuk fd f : R � (0, ✁) vkSj g : R � R ,sls nks ckj vodyuh; (twice differentiable) Qyu gSa fd R ij f " vkSj

g" lrr~ (continuous) Qyu gSaA eku yhft;s fd f'(2) = g(2) = 0, f "(2) ✂ 0 vkSj g'(2) ✂ 0 gSaA ;fn

2xlim✄ )x('g)x('f

)x(g)x(f= 1 gS] rc

(A) x = 2 ij f LFkkuh; fuEure (local minimum) gSA

(B) x = 2 ij f dk LFkkuh; mPpre (local maximum) gSA

(C) f "(2) > f(2)

(D) de ls de ,d x ☎ R ds fy, f(x) – f "(x) = 0 gSA

Ans. (A,D)

Sol. x 2

f(x) g(x)lim 1

f '(x) g'(x)✆✝

✞x 2

f(x) g(x)lim

f '(x) g'(x)✆

0

0

✟ ✠✡ ☛☞ ✌

Indeterminant form as f'(2) = 0, g(2) = 0 ✞ Using L.H.

x 2

f '(x) g(x) g'(x)f(x)lim

f "(x) g'(x) g"(x)f '(x)✆

✍

✍ =

f '(2) g(2) g'(2)f(2)

f "(2) g'(2) g"(2)f '(2)

✍

✍ =

g'(2)f(2)1

f "(2)g'(2)✝ ✎ f"(2) = f(2)

and f'(2) = 0 & range of f(x) ☎ (0, ✁)

so f"(2) = f(2) = +ve so f(x) has point of minima at x = 2

and f(2) = f"(2) so f(x) = f"(x) have atleast one solution in x ☎ R

Hindi. x 2

f(x) g(x)lim 1

f '(x) g'(x)✆✝

✞x 2

f(x) g(x)lim

f '(x) g'(x)✆

0

0

✟ ✠✡ ☛☞ ✌

vfu/kkZj~; :i D;ksfda f'(2) = 0, g(2) = 0 ✞ L.H. ds mi;ksx ls

x 2

f '(x) g(x) g'(x)f(x)lim

f "(x) g'(x) g"(x)f '(x)✆

✍

✍ =

f '(2) g(2) g'(2)f(2)

f "(2) g'(2) g"(2)f '(2)

✍

✍ =

g'(2)f(2)1

f "(2)g'(2)✝

✎ f"(2) = f(2) rFkk f'(2) = 0 & f(x) dk ifjlj ☎ (0, ✁) vr% f"(2) = f(2) = +ve

vr% f(x) dk x = 2 ij fufEu"B fcUnq gSA

rFkk f(2) = f"(2) vr% f(x) = f"(x) dk x ☎ R esa de ls de ,d gy gSA ✏✏✏✑✒✓✔✕✖✔✗✒✑✗✘✕

49. Let a, �, ✁ ✂ R. Consider the system of linear equations

ax + 2y = �

3x – 2y = ✁

Which of the following statement(s) is(are) correct ?

(A) if a = – 3, then the system has infinitely many solutions for all values of � and ✁

(B) If a ✄ –3, then the system has a a unique solution for all values of � and ✁

(C) If � + ✁ = 0, the the system has infinitely many solutions for a = –3

(D) If � + ✁ ✄ 0, then the system has no solution for a = –3

Ekkuk fd a, �, ✁ ✂ R gSaA bu jSf[kd lehdj.kksa ds fudk; (system of linear equations) ij fopkj dhft,

ax + 2y = �

3x – 2y = ✁

fuEufyf[kr esa ls dkSu lk ¼ls½ dFku lgh gS ¼gSa½ ?

(A) ;fn a = – 3, rc � rFkk ✁ ds lHkh ekuksa ds fy, fudk; ds vuUr (infinitely many) gy gSaA

(B) ;fn a ✄ –3, rc � rFkk ✁ ds lHkh ekuksa ds fy, fudk; dk vf}rh; (unique) gy gSaA

(C) ;fn � + ✁ = 0, rc a = –3 ds fy, fudk; ds vuUr gy gSaA

(D) ;fn � + ✁ ✄ 0, rc a = –3 ds fy, fudk; dk dksbZ gy ugha gSaA

Ans. (B,C,D)

Sol. ax + 2y = �

3x – 2y = ✁

(A) a = – 3 gives fn;k x;k gS

� = –✁

or ;k � + ✁ = 0 not for all �, ✁ (lHkh �, ✁ ds fy, ugha gS☎

(B) a ✄ –3 ✆ ✝ ✄ 0 where tgk¡ ✝ =a 2

3 2✞ = –2a – 6

✟ (B) is correct lgh gS

(C) correct lgh

(D) if ;fn � ✠ ✁ ✄ 0

✆ –3x + 2y = � ........(1)

& rFkk 3x – 2y = ✁ ........(2)

inconsistent vlaxr ✆ (D) correct lgh ✡✡✡☛☞✌✍✎✏✍✑☞☛✑✒✎

50. Let a, b �Rand a2 + b

2 ✁ 0. Suppose S =

1z R : z ,t R,t 0

a ibt

✂ ✄☎ ✆ ☎ ✝✞ ✟

✠✡ ☛, where i = –1 .

If z = x + iy and z � S☞ then (x, y) lies on

(A) the circle with radius 1

2aand centre

1,0

2a

✌ ✍✎ ✏✑ ✒

for a > 0 , b ✁ 0

(B) the circle with radius –1

2a and centre

1– ,0

2a

✌ ✍✎ ✏✑ ✒

for a < 0, b ✁ 0

(C) the x-axis for a ✁ 0, b = 0

(D) the y-axis for a = 0, b ✁ 0

ekuk fd a, b � R vkSj a2 + b

2 ✁ 0 gSA eku yhft, fd S =

1z R : z ,t R,t 0

a ibt

✂ ✄☎ ✆ ☎ ✝✞ ✟

✠✡ ☛, tgk¡ i = –1 .

;fn z = x + iy vkSj z � S gS☞ rc (x, y)

(A) ml o`Ùk ij gS ftldh f=kT;k 1

2a vkSj dsanz fcUnq

1,0

2a

✌ ✍✎ ✏✑ ✒

gS tc a > 0 , b ✁ 0

(B) ml o`Ùk ij gS ftldh f=kT;k 1

2a✓ vkSj dsanz fcUnq

1,0

2a

✌ ✍✔✎ ✏✑ ✒

gS tc a < 0 , b ✁ 0

(C) x- v{k ij gS tc a ✁ 0, b = 0

(D) y- v{k ij gS tc a = 0, b ✁ 0

Ans. (A,C,D)

Sol. x + iy = 2 2 2

a – ibt

a b t✕

x = 2 2 2

a

a b t✕.........(1) ; y =

2 2 2

–bt

a b t✕.........(2)

If ;fn a = 0, b ✁ 0, x = 0 ✖ (D) ✖ If ;fn a ✁ 0, b = 0, y = 0 ✖ (C)

a2 + b

2t2 =

a

x& a

2 + b

2t2 =

bt

y

✗ ✘

a

x =

bt

y

✗ ✖ t =

ay

bx

✙........(3)

Putting (3) in (1) ((3) ls (1) esa j[kus ij)

2 22 2

2 2

a yx a b a

b x

✚ ✛✜ ✢ ✣✤ ✥✤ ✥

✦ ✧;

2 22

2

a yx a a

x

✚ ✛✜ ✣✤ ✥✤ ✥

✦ ✧ ✖ a

2(x

2 + y

2) = ax ✖ x

2 + y

2 –

1

ax = 0

circle with centre 1

, 02a

✌ ✍✎ ✏✑ ✒

dsUnz okyk o`Ùk ; radius f=kT;k =2

210 0

2a

★✩ ✪✫ ★✬ ✭

✮ ✯ =

1

2a✰✰✰✱✲✳✴✵✶✴✷✲✱✷✸✵

Paragraph # 1

Football teams T1 and T2 have to play two games against each other. It is assumed that the outcomes

of the two games are independent. The probabilities of T1 winning, drawing and losing a game against

T2 are 1

2,

1

6and

1

3, respectively. Each team gets 3 points for a win, 1 point for a draw and 0 point for a

loss in a game. Let X and Y denote the total points scored by teams T1 and T2, respectively, after two

games.

QqVckWy nyksa T1 rFkk T2 dks ,d nwljs ds fo:) nks [ksy (games) [ksyus gSA ;g eku fy;k x;k gS fd nksuksa [ksyksa

ds ifj.kke ,d nwljs ij fuHkZj ugha djrsA ny T1 ds ny T2 ds fo:) ,d [ksy esa thrus] cjkcj gksus vkSj gkjus dh

izkf;drk Øe'k% 1

2,

1

6 vkSj

1

3 gSaA izR;sd ny thrus ij 3 vad] cjkcjh ij 1 vad vkSj gkjus ij 0 vad vftZr

djrk gSA ekuk fd nks [ksyksa ds i'pkr ny T1 vkSj ny T2 ds }kjk vftZr dqy vad Øe'k% X vkSj Y gSaA

51. P (X > Y) is

P (X > Y) dk eku gS&

(A)1

4(B)

5

12(C)

1

2(D)

7

12

Ans. (B)

Sol. P(X > Y) = T1T1 + DT1 + T1D (Where T1 represents wins and D represents draw)

= 1 1 1 1 1 1

. . .2 2 6 2 2 6

� � = 5

12 ✁ (B) is correct

Hindi. P(X > Y) = T1T1 + DT1 + T1D (tgk¡ T1 thr rFkk D cjkcjh dks n'kkZrk gSA)

= 1 1 1 1 1 1

. . .2 2 6 2 2 6

� � = 5

12✁ (B) lgh gSA

52. P (X = Y) is

P (X = Y) dk eku gS&

(A)11

36(B)

1

3(C)

13

36(D)

1

2

Ans. (C)

Sol. P(X = Y) = DD + T1T2 + T2T1 = 1 1 1 1 1 1

. . .6 6 2 3 3 2

� � = 1 1

36 3� =

39

36 3✂ =

13

36✁ (C) is correct lgh gSA

✄✄✄☎✆✝✞✟✠✞✡✆☎✡☛✟

Paragraph-2

Let F1(x1, 0) and F2(x2, 0), for x1 < 0 and x2 > 0, be the foci of the ellipse 9

x2

+ 8

y2

= 1. Suppose a

parabola having vertex at the origin and focus at F2 intersects the ellipse at point M in the first quadrant

and at point N in the fourth quadrant.

ekuk fd F1(x1, 0) vkSj F2(x2, 0) (ftlesa x1 < 0, x2 > 0) nh?kZo`Ùk (ellipse) 9

x2

+ 8

y2

= 1 dh ukfHk;k¡ (Foci) gSA

ekuk fd ,d ijoy; (parabola) ftldk 'kh"kZ (vertex) ewyfcUnq (origin) ij vkSj ukfHk (focus) F2 ij gS] nh?kZo`Ùk

dks çFke prqFkkZa'k (first quadrant) esa M ij vkSj prqFkZ prqFkkZa'k (fourth quadrant) esa N ij çfrPNsfnr djrk gSA

53. The orthocentre of the triangle F1MN is

f=kHkqt F1MN dk yEcdsUæ (orthocentre) gS&

(A) �✁✂✄

☎✆

0,10

9– (B) �

✁✂✄

☎✆

0,3

2 (C) �

✁✂✄

☎✆

0,10

9 (D) �

✁✂✄

☎✆

6,3

2

Ans. (A)

Sol. Ellipse : 2 2x y

19 8

✝ ✞ ………..(1)

foci of ellipse are (±1, 0)

Equation of parabola having vertex (0, 0) and focus (1, 0) is y2 = 4x …………….(2)

y2

= 4x

M

F2(1,0)

F1(–1,0)

N

R

y

x

from equation (1) & (2) 2x 4x

19 8

✝ ✞ ✟ 2x2 + 9x – 18 = 0 ✟ x =

3

2, –6 (rejected)

✠ M 3

, 62

✆ ✂✄ �☎ ✁

and N 3

, 62

✆ ✂✡✄ �☎ ✁

Equation of altitude from vertex M 3

, 62

✆ ✂✄ �☎ ✁

is ☛ ☞y 6✌ = 5 3

x22 6

✆ ✂✡✄ �☎ ✁

put y = 0 we get ✟ 12 3x

5 2

✍ ✎ ✍ ✟ x = 9

10

✍ ✟ orthocenter of ✏ F1 MN is

9,0

10

✡✆ ✂✄ �☎ ✁

✑✑✑✒✓✔✕✖✗✕✘✓✒✘✙✖

Hindi. nh?kZo`Ùk: 2 2x y

19 8

� ✁ ………..(1)

nh?kZo`Ùk dh ukfHk;k¡ (±1, 0) gSaA

(0, 0) 'khZ"kZ rFkk (1, 0) ukfHk okys ijoy; dk lehdj.k y2 = 4x …………….(2)

y2

= 4x

M

F2(1,0)

F1(–1,0)

N

R

y

x

lehdj.k (1) rFkk (2) ls 2x 4x

19 8

� ✁ ✂ 2x2 + 9x – 18 = 0 ✂ x =

3

2, –6 (vLohdk;Z)

✄ M 3

, 62

☎ ✆✝ ✞✟ ✠

rFkk N 3

, 62

☎ ✆✡✝ ✞✟ ✠

M 3

, 62

☎ ✆✝ ✞✟ ✠

ls 'kh"kZyEc dk lehdj.k ☛ ☞y 6✌ =5 3

x22 6

☎ ✆✡✝ ✞✟ ✠

y = 0 j[kus ij

✂ 12 3x

5 2

✍ ✎ ✍ ✂ x = 9

10

✍

✏ F1 MN dk yEcdsUnz 9

,010

✡☎ ✆✝ ✞✟ ✠

gSA

Ans. (A)

54. If the tangents to the ellipse at M and N meet at R and the normal to the parabola at M meets the x-axis

at Q, then the ratio of area of the triangle MQR to area of the quadrilateral MF1NF2 is

;fn nh?kZo`Ùk ds fcUnqvksa M vkSj N ij Li'kZ js[kk,a (tangents) R ij feyrh gS vkSj ijoy; ds fcUnq M ij vfHkyEc

x-v{k dks Q ij feyrk gS] rc f=kHkqt MQR ds {ks=kQy vkSj prqHkZqt (quadrilateral) MF1NF2 ds {ks=kQy dk

vuqikr (ratio) gS&

(A) 3 : 4 (B) 4 : 5 (C) 5 : 8 (D) 2 : 3

Ans. (C)

✑✑✑✒✓✔✕✖✗✕✘✓✒✘✙✖

Sol. Equation of tangent at point M 3

, 62

� ✁✂ ✄☎ ✆

to the ellipse is x(3 / 2) y 6

19 8

✝ ✞

put y = 0 ✟ R is (6, 0)

Equation of normal to the parabola at point M 3

, 62

� ✁✂ ✄☎ ✆

is ✠ ✡y 6☛ = 6 3

x2 2

☞ ✌✍ ✍✎ ✏✑ ✒

put y = 0 ✟ Q is 7

,02

� ✁✂ ✄☎ ✆

Now 1 2

Area of MQR

Area of quadrilateral MFNF

✓=

1 5. . 6

52 21 8

.2.2 62

✔ Ans. (C)

Hindi. nh?kZo`Ùk ds fcUnq M 3

, 62

� ✁✂ ✄☎ ✆

ij Li'kZ js[kk dk lehdj.k x(3 / 2) y 6

19 8

✝ ✞

y = 0 j[kus ij ✟ R, (6, 0) gSA

M 3

, 62

� ✁✂ ✄☎ ✆

ij ijoy; ds vfHkyEc dk lehdj.k ✠ ✡y 6☛ =6 3

x2 2

☞ ✌✍ ✍✎ ✏✑ ✒

gSA

y = 0 j[kus ij ✟ Q7

,02

� ✁✂ ✄☎ ✆

gSA

vc 2

MQR

MF NF1

✕ dk {k=s kQyprqHkZqt dk {k=s kQy

=

1 5. . 6

52 21 8

.2.2 62

✔ Ans. (C)

✖✖✖✗✘✙✚✛✜✚✢✘✗✢✣✛