c JEE (Main) 2017 - Matrix JEE Academy · 2009-08-31 · Matrix JEE Academy : Opposite Reliance...

Date: 02/April/2017Time: 3 Hours. Max. Marks: 360

VERY IMPORTANT :A. The question paper consists of 3 parts (Mathematics, Physics & Chemistry). Please fill the OMR answer

Sheet accordingly and carefully.

B. Please ensure that the Question Paper you have received contains All the questions in each Section andPages. If you found some mistake like missing questions or pages then contact immediately to the Invigilator.

INSTRUCTIONS

1. All questions are single correct type questions. Each of these questions has four choices (A), (B), (C)and (D) out of which ONLY ONE is correct.For each question, you will be awarded 4 marks if you have darkened only the bubble corresponding tothe correct answer and zero mark if no bubble are darkened. In all other cases, (–¼) minus onefourthmark will be awarded.

2. Indicate the correct answer for each question by filling appropriate bubble in your answer sheet.3. Use of Calculator, Log Table, Slide Rule and Mobile is not allowed.

USEFUL DATA

Atomic weights: Al = 27, Mg = 24, Cu = 63.5, Mn = 55, Cl = 35.5, O = 16, H = 1, P = 31, Ag = 108, N = 14,Li = 7, I = 127, Cr = 52, K=39, S = 32, Na = 23, C = 12, Br = 80, Fe = 56, Ca = 40, Zn = 65.5, Ti = 48,Ba = 137, U = 238, Co= 59, B =11, F = 19, He = 4, Ne = 20, Ar = 40 , Mo = 96, Ni = 58.5, Sr = 87.5,Hg = 200.5 , Tl = 204, Pb = 207 [Take : ln 2 = 0.69, ln 3 = 1.09, e = 1.6 × 10–19, m

e= 9.1 × 10–31 kg ]

Take g = 10 m/s2 unless otherwise stated

cCode

Paper

JEE (Main) 2017

2Matrix JEE Academy : Opposite Reliance Petrol Pump, Piprali Road, Sikar Ph. 01572-241911, Mob. 97836-21999, 97836-31999

JEE (MAIN) 2017MatrixJEE Academy

PHYSICS

SECTION-ISINGLE CORRECT CHOICE TYPE

Q.61 to Q.90 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct.

61. An observer is moving with half the speed of light towards a stationary microwave source emitting waves at

frequency 10 GHz. What is the frequency of the microwave measured by the observer ?

(speed of light = 3 × 108 ms–1)

,d izs{kd izdk'k xfr dh vk/kh xfr ls 10 GHz vkofr ds ,d fLFkj lw{e rjax (microwave) L=kksr dh rjQ tk jgk

gSa izs{kd }kjk ekih x;h lw{e rjax dh vko`fr dk eku gksxk &

(izdk'k dh pky = 3 × 108 ms–1)

(1) 12.1 GHz (2) 17.3 GHz (3) 15.3 GHz (4) 10.1 GHz

Sol.

v1

cv ' vv

1c

11

2v ' v 3v1

12

v' = 10 × 1.73 = 17.3 GHz

62. The following observations were taken for determining surface tension T of water by capillary method :

diameter of capillary, D = 1.25 × 10–2 m rise of water, h = 1.45 × 10–2 m.

Using, g = 9.80 m/s2 and the simplified relation 3rhg

T 10 N / m2

, the possible error in surface tension is

closest to :

fuEu izs{k.kksa dks dsf'kdk fof/k ls ikuh dk i`"B ruko T ukius ds fy;s fd;k tkrk gSA

dsf'kdk dk O;kl] D = 1.25 × 10–2 m ikuh dk p<+ko] h = 1.45 × 10–2 m.

g = 9.80 m/s2 rFkk ljyhd`r lEcU/k 3rhg

T 10 N / m2

, dks mi;ksxh djrs gq, i`"B ruko esa lEHkkfor =kqfV dk

fudVre eku gksxk &

(1) 1.5% (2) 2.4% (3) 10% (4) 0.15%

Sol. % error in T = 100r h g

r h g

= 0.005 0.01 0.01

000.625 1.4 9.8

= 1.5%



PHYSICS

63. Some energy levels of a molecule are shown in the figure. The ratio of the wavelength r = 1/

2, is given by :

,d v.kq ds dqN ÅtkZ Lrjksa dks fp=k esa fn[kk;k x;k gSA rjaxnS/;Z ds vuqikr r = 1/

2 dk eku gksxk &

(1) 2

r3

(2) 3

r4

(3) 1

r3

(4) 4

r3

Sol. E1 = E So,

1 =

E2 =

3

ESo,

2 =

1

2

1

3

64. A body of mass m = 10–2 kg is moving in a medium and experiences a frictional force F = –k 2. Its initial

speed is 0 = 10 ms–1. If, after 10 s, its energy is 20

1m

8 , the value of k will be :

m = 10–2 kg nzO;eku dk ,d fi.M ,d ek/;e esa tk jgk gS vkSj ,d ?k"kZ.k cy F = –k 2 dk vuqHko djrk gSA

fi.M dk izkjfEHkd osx 0 = 10 ms–1 gSA ;fn 10 s ds ckn mldh ÅtkZ 20

1m

8 gS] rks k dk eku gksxk &

(1) 10–3 kg s–1 (2) 10–4 kg m–1 (3) 10–1 kg m–1 s–1 (4) 10–3 kg m–1

Sol. F = –KV2 m = 10–2 kg

V0 = 10 m/s K

i =

20

1

2mV

kf =

20

1

8mV =

4iK

then, 5m / s2

if

VV

a = –100 KV2

2100dv

KVdt

5

2

10 0

100t

dvKdt

V



PHYSICS

11000

10K K = 10–4 kg m–1

65. Cp and C

v are specific heats at constant pressure and constant volume respectively. It is observed that

Cp – C

v = a for hydrogen gas

Cp – C

v = b for nitrogen gas

The correct relation between a and b is :

fLFkj nkc rFkk fLFkj vk;ru ij fof'k"V Å"ek;sa Øe'k% Cp rFkk C

v gSaA ik;k tkrk gS fd

gkbMªkstu ds fy;s] Cp – C

v = a

ukbVªkstu ds fy;s] Cp – C

v = b

a vkSj b ds chp dk lgh lEcU/k gksxk &

(1) a = b (2) a = 14 b (3) a = 28 b (4) 1

a b14

Sol. Cp – C

v =

w

R

MM

w molecular weight

For H2

Cp – C

v =

2

Ra

For N2 =

28

Rb

14

6

a

b

a = 14b

66. The moment of inertia of a uniform cylinder of length l and radius R about its perpendicular bisector is I.

What is the ratio l/R such that the moment of inertia is minimum ?

,d f=kT;k R rFkk yEckbZ l ds ,d leku csyu dk mlds vfHkyEc f}Hkktd ds lkis{k tM+Ro vk?kw.kZ I gSA tM+Ro

vk?kw.kZ ds fuEure eku ds fy;s vuqikqr l/R D;k gksxk \

(1) 3

2(2) 1 (3)

3

2(4)

3

2

Sol.2 2

12 4

ml mRI Mass of object is constant.

R2l = constant

r2l = constant

2 20

12 4

dI l dl Rm

dR dR

2RldR + R2dl = 0



PHYSICS

2 20

6 4

l l R

R

2dl l

dR R

2

3 2

l R

R

2

2

3

2

l

R

3

2

l

R

67. A radioactive nucleus A with a half life T, decays into a nucleus B. At t = 0, there is no nucleus B.

At sometime t, the ratio of the number of B to that of A is 0.3. Then, t is given by :

,d jsfM;ks,fDVo ukfHkd A ftldh v)Z&vk;q T gS] dk {k; ,d ukfHkd& B esa gksrk gSA le; t = 0 ij dksbZ Hkh

ukfHkd&B ugha gSA ,d le; t ij ukfHkdksa B rFkk A dh la[;k dk vuqikr 0.3 gS] rks t dk eku gksxk &

(1) log1.3

t Tlog 2

(2) t T log(1.3) (3) T

tlog(1.3)

(4) T log 2

t2 log1.3

Sol. NA = N

0e–t N

B = N

0(1 – e–t)

10.3

tB

tA

N e

N e

1 0.3t te e

1 = 1.3 te

0 = ln(1.3) – t

ln(1.3)

ln 2

Tt

log1.3

log 2t T

68. Which of the following statements is false ?

(1) In a balanced wheatstone bridge if the cell and the galvanometer are exchanged, the null point is

disturbed

(2) A rheostat can be used as a potential divider

(3) Kirchhoff's second law represents energy conservation

(4) Wheatstone bridge is the most sensitive when all the four resistances are of the same order of magnitude

fuEufyf[kr esa ls dkSu lk dFku xyr gS &

(1) ,d larqfyr OghVLVksu lsrq esa] lsy ,d xSYosuksehVj dks vkil esa cnyus ij 'kwU; fo{ksi fcUnq izHkkfor gksrk gSA

(2) ,d /kkjk fu;=kad dks foHko foHkktd dh rjg mi;ksx dj ldrs gSA

(3) fdjpkWQ dk f}rh; fu;e ÅtkZ ds laj{k.k dks n'kkZrk gSA

(4) OghVLVksu lsrq dh lqxzghrk lcls vf/kd rqY; gksrh gS tc pkjksa izfrjks/kksa dk ifjek.k rqY; gksrk gSA



PHYSICS

Sol. Null point will not disturbed in a balanced wheatstone bridge if the cell and the galvanometer are exchanged.

69. A capacitance of 2 µF is required in an electrical circuit across a potential difference of 1.0 kV. A large

number of 1µF capacitors are available which can withstand a potential difference of not more than 300 V.

The minimum number of capacitors required to achieve this is :

,d fo|qr ifjiFk esa ,d 2 µF /kkfjrk ds la/kkfj=k dks 1.0 kV foHkokUrj ds fcUnqvksa ds chp yxkuk gSA 1µF /kkfjrk

ds cgqr lkjs la/kkfj=k tks fd 300 V foHkokUrj rd ogu dj ldrs gSa] miyC/k gSA

mijksDr ifjiFk dks izkIr djus ds fy;s U;wure fdrus la/kkfj=kksa dh vko';drk gksxh \

(1) 16 (2) 24 (3) 32 (4) 2

Sol. To required equivalent capacitance 2 µF.

4 capacitor in a series and this series is

70. In the given circuit diagram when the current reaches steady state in the circuit, the charge on the capacitor

of capacitance C will be :

fn;s x;s ifjiFk esa tc /kkjk fLFkjkoLFkk esa igq¡p tkrh gS rks /kkfjrk C ds la/kkfj=k ij vkos'k dk eku gksxk &

(1) 1

2

rCE

(r r) (2) 2

2

rCE

(r r ) (3) 1

1

rCE

(r r) (4) CE

Sol. After steady state

A B

2

Ei

r r

Voltage difference across the capacitor i.e. between point A and B.

2

2

ErV

r r

Charge on capacitor

2

2

CErq

r r



PHYSICS

71.

In the above circuit the current in each resistance is :

Åij fn;s x;s ifjiFk esa izR;sd izfrjks/k esa /kkjk dk eku gksxk &

(1) 0.25 A (2) 0.5 A (3) 0 A (4) 1 A

Sol. O (Assume)

O 6

6 4 2

24

So current in each resistor will be zero.

72. In amplitude modulation, sinusoidal carrier frequency used is denoted by c and the signal frequency is denotd

by m. The bandwidth (

m) of the signal such that

m<<

c. Which of the following frequencies is not

contained in the modulated wave ?

vk;ke ekWMqyu esa T;koØh; okgd vkofr dks c ls rFkk flXuy vkofr dks

m ls n'kkZrs gSA flXuy dh cS.M pkSM+kbZ

(m) dks bl izdkj pqurs gS fd

m<<

c fuEu esa ls dkSulh vko`fr ekWMqfyr rjax esa ugha gksxh \

(1) c

(2) m

+c

(3) c +

m(4)

m

Sol. Let c(t) = AC sin

ct represent carrier wave and m(t) = A

msin

mt represent the message or the modulating

signal where m = 2ƒ

m is the angular frequency of the message signal. The modulated signal cm (t) can be

written as cm (t) = (A

C + A

m sin

mt) sin

ct

mC c c

C

AA 1 sin t sin t

A

Note that the modulated signal now contains the message signal. From Eq. (i), we can write,c

m (t) = A

c sin

ct + A

c sin

mt sin

ct ..........(ii)

Here = Am/A

c is the modulation index; in practice, is kept 1 to avoid distortion.

Using the trignomatric relation sin A sin B = 1/2 (cos (A – B) – cos (A + B),we can write c

m (t) of Eq. (ii) as

c cm c c c m c m

A Ac (t) A sin t cos( )t cos( )t

2 2

......(iii)

In amplitude modulated wave, the frequencies contained are c –

m,

c,

c +

m.

The frequency of m is not contained in A.M. wave

73. In a common emitter amplifier circuit using an n-p-n transistor, the phase difference between the input and

the output voltages will be :



PHYSICS

n-p-n VªkWftLVj ls cuk;s gq, ,d mHk;fu"B mRltZd izo/kZd ifjiFk esa fuosf'kr rFkk fuxZr foHkoksa ds chp dykarj dk

eku gksxk &

(1) 90º (2) 135º (3) 180º (4) 45º

Sol. Since output voltage is out of phase with input voltage therefore phase difference will be 180.

74. A copper ball of mass 100 gm is at a temperature T. It is dropped in a copper calorimeter of mass 100 gm,

filled with 170 gm of water at room temperature. Subsequently, the temperature of the system is found to be

75ºC. T is given by : (Given : room temperature = 30ºC, specific heat of copper = 0.1 cal/gmºC)

100 gm nzO;eku okyk rkacs ds ,d xksys dk rkieku T gSA mls ,d 170 gm ikuh ls Hkjs gq, 100 gm ds rkacs ds

dSyksjhehVj] tksfd dejs ds rkieku ij gS] esa Mky fn;k tkrk gSA rRi'pkr~ fudk; dk rkieku 75ºC ik;k tkrk gSA

T dk eku gksxk : (fn;k gS : dejs dk rkieku = 30ºC, rkacs dh fof'k"V Å"ek = 0.1 cal/gmºC)

(1) 885ºC (2) 1250ºC (3) 825ºC (4) 800ºC

Sol. Heat lost = Heat gain

100(0.1) (T – 75) = 100 × 0.1 × 45 + 170 × 1 × 45

10T – 750 = 450 + 7650

T = 885°C

75. In a Young's double slit experiment, slits are separated by 0.5 mm and the screen is placed 150 cm away. A

beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes on

the screen. The least distance from the common central maximum to the point where the bright fringes due to

both the wavelengths coincide is :

;ax ds ,d f}f>jh iz;ksx esa] f>fj;ksa ds chp dh nwjh 0.5 mm ,oa insZ dh f>jh ls nwjh 150 cm gSA ,d izdk'k iaqt]

ftlesa 650 nm vkSj 520 nm dh nks rjaxnS/;Z gSa] dks insZ ij O;rhdj.k fÝUt cukus esa mi;ksx djrs gSA mHk;fu"B

dsUnzh; mfPp"B ls og fcUnq] tgk¡ nksuksa rjaxnS/;Z dh nhIr fÝUtsa lEikrh gksrh gS] dh U;wure nwjh gksxh &

(1) 7.8 mm (2) 9.75 mm (3) 15.6 mm (4) 1.56 mm

Sol. For 1 fringe width

11

D

d

For 2 fringe width

22

D

d

1

2

5

4

4

1 = 5

2

So, forth bright fringe of 1 wavelength conside with 5th bright fringe of

2.

So, distance from central maxima = 41 = 7.8 mm



76. An electric dipole has a fixed dipole moment p

, which makes angle with respect to x-axis. When

subjected to an electric field 1

Ê Ei

, it experiences a torque 1

ˆT k

. When subjected to another electric

field 2 1

Ê 3E j

it experiences a torque 2 1T T

. The angle is :

,d fo|qr f}/kzqo dk fLFkj f}/kzqo vk?kw.kZ p

gS tks fd x-v{k ls dks.k cukrk gSA fo|qr {ks=k 1

Ê Ei

esa j[kus ij

;g cy vk?kw.kZ 1

ˆT k

dk vuqHko djrk gSA fo|qr {ks=k 2 1

Ê 3E j

esa j[kus ij ;g cy vk?kw.kZ 2 1T T

dk

vuqHko djrk gSA dks.k dk eku gksxk &

(1) 45º (2) 60º (3) 90º (4) 30º

Sol.1 1T P E

= ˆsinPE k = k ...(i)

2 2T P E

= ˆ3 cosP E k = – k ...(ii)

Eq. (i) divided by (ii)

tan1

3

= 60°

77. A slender uniform rod of mass M and length l is pivoted at one end so that it can rotate in a vertical plane

(see figure). There is negligible friction at the pivot. The free end is held vertically above the pivot and then

released. The angular acceleration of the rod when it makes an angle with the vertical is :

,d nzO;eku M rFkk yEckbZ l dh iryh ,oa ,d leku NM+ dk ,d /kqjkxzLr gS ftlls fd og ,d Å/okZ/kj lery esa

?kwe ldrh gSA NM+ ds nwljs fljs dks /kqjh ds Åij Å/okZ/kj j[kdj NksM+ fn;k tkrk gSA tc NM+ Å/oZ ls dks.k

cukrh gS rks mldk dks.kh; Roj.k gksxk &

(1) 2g

sin3

l

(2) 3g

cos2

l

(3) 2g

cos3

l

(4) 3g

sin2

l

Sol.

sin�

mg

L2



2sin

2 3

MgL MLI

3 sin

2

g

L

78. An external pressure P is applied on a cube at 0ºC so that it is equally compressed from all sides. K is the

bulk modulus of the material of the cube and is its coefficient of linear expansion. Suppose we want to

bring the cube to its original size by heating. The temperature should be raised by :

0ºC ij j[ks gq, ,d ?ku ij ,d nckc P yxk;k tkrk gS ftlls og lHkh rjQ ls cjkcj laihfMr gksrk gSA ?ku ds

inkFkZ dk vk;ru izR;kLFkrk xq.kkad K ,oa js[kh; izlkj xq.kkad gSA ;fn ?ku dks xeZ djds ewy vkdkj esa ykuk gS

rks mlds rkieku dks fdruk c<+kuk iM+sxk \

(1) P

K(2)

3

PK

(3) 3PK (4)

P

3 K

Sol. K =

P

V

V

V P

V K

Due to increase in temp. (T) V = 3V T

33

PVV T

K

3

PT

Ka

79. A diverging lens with magnitude of focal length 25 cm is placed at a distance of 15 cm from a converging

lens of magnitude of focal length 20 cm. A beam of parallel light falls on the diverging lens. The final image

formed is :

(1) Virtual and at a distance of 40 cm from convergent lens

(2) Real and at a distance of 40 cm from the divergent lens

(3) Real and at a distance of 6 cm from the convergent lens

(4) Real and at a distance of 40 cm from convergent lens

,d 25 cm ifjek.k dh Qksdl nwjh ds vilkjh ysUl dks ,d 20 cm ifjek.k dh Qksdl nwjh ds vfHklkjh ysUl ls

15 cm dh nwjh ij j[kk tkrk gSA ,d lekarj izdk'k iaqt vilkjh ysal ij vkifrr gksrk gSA ifj.kkeh izfrfcEc

gskxk&

(1) vkHkklh vkSj vfHklkjh ysal ls 40 cm nwjh ij

(2) okLrfod vkSj vfHklkjh ysal ls 40 cm nwjh ij

(3) okLrfod vkSj vfHklkjh ysal ls 6 cm nwjh ij



(4) okLrfod vkSj vilkjh ysal ls 40 cm nwjh ij

Sol. For diversing lens image is formed at focus of lens. So distance of object for conversing lens is (25 + 15).

For conversing lens

1 1

40 20V

1 1

40V

V = 40 cm

V is +ve i.e. image is real.

80. An electron beam is accelerated by a potential difference V to hit a metallic target to produce X-rays. It

produces continuous as well as characteristic X-rays. If min

is the smallest possible wavelength of X-ray in

the spectrum, the variation of log min

with log V is correctly represented in :

X-fdj.ksa mRiUu djus ds fy;s ,d bysDVªkWu fdj.kiaqt dks foHkokUrj V ls Rofjr djds /kkrq dh IysV ij vkifrr

fd;k tkrk gSA blesa fofoDr ,oa vfojr X-fdj.ksa mRiUu gksrh gSA ;fn X-fdj.ksa LisDVªe esa U;wure laHko rjaxnS/;Z

min

gS rks min

dks log V ds lkFk cnyko fdl fp=k esa lgh fn[kk;k x;k gS \

(1) (2)

(3) (4)

Sol.min

hceV

min

hc

eV taking log on both sides.

minlog log loghc

Ve

81. The temperature of an open room of volume 30 m3 increases from 17ºC to 27ºC due to the sunshine. The

atmospheric pressure in the room remains 1 × 105 Pa. If ni and n

f are the number of molecules in the room

before and after heating, then nf – n

i will be :

lw;Z dh fdj.kksa ls ,d [kqys gq, 30 m3 vk;ru okys dejs dk rkieku 17ºC ls c<+dj 27ºC gks tkrk gSA dejs ds



vUnj ok;qeaMyh; nkc 1 × 105 Pa gh jgrk gSA ;fn dejs ds vUnj v.kqvksa dh la[;k xeZ gksus ls igys ,oa ckn esa

Øe'k% ni o n

f gS rks n

f – n

i dk eku gksxk &

(1) 1.38 × 1023 (2) 2.5 × 1025 (3) –2.5 × 1025 (4) –1.61 × 1023

Sol. i

A

niPV RT

N f

f

A

nPV RT

N

A

i

PVNni

RT A

f

f

PVNn

RT

Ti = 290 T

f = 300

P = 105Pa V = 30 m3

So, nf – n

i = –2.5 × 1025

82. In a coil of resistance 100 , a current is induced by changing the magnetic flux through it as shown in the

figure. The magnitude of change in flux through the coil is :

pqEcdh; ¶yDl ds cnyus ls 100 izfrjks/k dh dq.Myh esa izsfjr /kkjk dks fp=k esa n'kkZ;k x;k gSA dq.Myh ls xqtjus

okys ¶yDl esa cnyko dk ifjek.k gksxk &

(1) 225 Wb (2) 250 Wb (3) 275 Wb (4) 200 Wb

Sol. E = iR

E = d

dt

.d E dt

d iRdt R idt

idt = area under i-t curve.

= 250 wb

83. When a current of 5 mA is passed through a galvanometer having a coil of resistance 15 , it shows full

scale deflection. The value of the resistance to be put in series with the galvanometer to convert it into a

voltmeter of range 0 – 10 V is :

15 ds dq.Myh izfrjks/k ds xSYosuksehVj ls tc 5 mA dh /kkjk izokfgr dh tkrh gS rks og iw.kZ Ldsy fo{ksi n'kkZrk

gSA bls 0 – 10 V ijkl ds foHkoekih esa cnyus ds fy;s fdl eku ds izfrjks/k dks xSYosuksehVj ds lkFk Js.kh Øe esa



yxkuk gksxk &

(1) 2.045 × 103 (2) 2.535 × 103 (3) 4.005 × 103 (4) 1.985 × 103

Sol.V

ir R

3105 10

15 R

2000 = 15 + R

R = 1985 = 1.985 × 103

84. A time dependent force F = 6t acts on a particle of mass 1 kg. If the particle starts from resty, the work done

by the force during first 1 sec. will be :

1 kg nzO;eku dk ,d d.k] ,d le; ij fuHkZj cy F = 6t dk vuqHko djrk gSA ;fn d.k fojkekoLFkk ls pyrk gS rks

igys 1 s esa cy }kjk fd;k x;k dk;Z gksxk &

(1) 22 J (2) 9 J (3) 18 J (4) 4.5 J

Sol. F = 6t M = 1kg

a = 6t S = t3

23ds

tdt

ds = 3t2dt

w = .F ds

w = 1

2

06 .3t t dt

w = 418 t dt

w = 18

4.5J4

85. A magnetic needle of magnetic moment 6.7 × 10–2 Am2 and moment of inertia 7.5 × 10–6 kg m2 is performing

simple harmonic oscillations in a magnetic field of 0.01 T. Time taken for 10 complete oscillations is :

,d pqEcdh; vk?kw.kZ 6.7 × 10–2 Am2 ,oa tM+Ro vk?kw.kZ 7.5 × 10–6 kg m2 okyh pqEcdh; lqbZ ,d 0.01 T rhozrk ds

pqEcdh; {ks=k esa ljy vkorhZ nksyu dj jgh gSA 10 iwjs nksyu dk le; gksxk &

(1) 8.89 s (2) 6.98 s (3) 8.76 s (4) 6.65 s

Sol. = –MB

= –MB

I

MB

I

22 6.65sec

IT

MB

86. The variation of acceleration due to gravity g with distance d from centre of the earth is best represented by :



(R = Earth's radius)

i`Foh ds dsUnz ls nwjh d ds lkFk xq:Roh; Roj.k g dk cnyko fuEu esa ls fdl xzkQ esa lcls lgh n'kkZ;k x;k gS \

(R = iFoh f=kT;k)

(1) (2)

(3) (4)

Sol. For x < R

3

GMg x

R

So, graph will be 3.

87. A body is thrown vertically upwards. Which one of the follwing graphs correctly represent the velocity vs time?

,d fi.M dks Å/okZ/kj Åij dh rjQ Qsadk tkrk gSA fuEu esa ls dkSu lk xzkQ le; ds lkFk osax dks lgh n'kkZrk gS \

(1) (2) (3) (4)

Sol. During motion acceleration is always constant & negative.

88. A particle A of mass m and initial velocity collides with a particle B of mass 2

M which is a t rest. The collision

is heat on, and elastic. The ratio of the de-Broglie wavelengths A to

B after the collision is :

nzO;eku m ,oa vkjfEHkd osx ds ,d d.k A dh VDdj nzO;eku 2

M ds fLFkj d.k B ls gksrh gSA ;g VDdj lEeq[k ,oa

izR;kLFk gSA VDdj ds ckn fM&czksXyh rjaxnS/;Z A ,oa

B dk vuqikr gksxk &

(1) 2A

B

(2)

2

3A

B

(3)

1

2A

B

(4)

1

3A

B

Sol. From momentum conservation

1 fP P

mV = mv1 + 2

2

mv

v = v1 +

2

2

V...(i)



e = 1 = 2 1V V

V

V2 – V

1 = V ...(ii)

After solving equation (i) and (ii)

V1 =

3

VSo, P

A =

3

mV

V2 =

4

3

VP

B =

4 3

3 2 3

mV mV

A

A

h

P B

B

h

P

2A B

B A

P

P

89. A particle is executing simple harmonic motion with a time period T. At time t = 0, it is at its position of

equilibrium. The kinetic energy - time graph of the particle will look like :

,d d.k] vkorZdky T ls ljy vkorZ xfr dj jgk gSA le; t = 0 ij og lkE;koLFkk dh fLFkfr esa gSA fuEu esa ls dkSu

lk xzkQ le; ds lkFk xfrt ÅtkZ dks lgh n'kkZrk gS &

(1) (2)

(3) (4)

Sol. Tiem period of SHM = T

So, then time period of K.E. is 2

T,

then graph will be 3.

90. A man grows into a giant such that his linear dimensions increase by a factor of 9. Assuming that his density

remains same, the stress in the leg will change by a factor of :

,d euq";] ,d fo'kkydk; ekui esa bl izdkj ifjofrZr gksrk gS fd mldh js[kh; foek;sa 9 xquk c<+ tkrh gSA ekuk fd

mlds ?kuRo esa dksbZ ifjorZu ugha gksrk gS rks mlds Vk¡x esa izfrcy fdrus xquk gks tk;sxk \©1

(1) 1

9(2) 81 (3)

1

81(4) 9

Sol. Stress = .F Mg v L A

g gA A A A

Stress LSo, 9 times



CHEMISTRY



CODE : C

1. The freezing point of benzene decrease by 0.45°C when 0.2 g of acetic acid is added to 20 g of benzene. If

acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be:

(Kf for benzene = 5.12 K kg mol–1)

tc ,flfVd ,flM dk 0.2 g csathu ds 20 g esa feyk;k tkrk gS rks csathu dk fgekad 0.45°C ls de gks tkrk gSA ;fn

,flfVd ,flM csathu esa laxqf.kr gksdj Mkbej (f}r;) cukrk gS rks ,flfVd ,flM dk izfr'krrk laxq.ku gksxk &

(csathu ds fy, Kf = 5.12 K kg mol–1)

(1) 96.6% (2) 64.6% (3) 80.4% (4) 74.6%

Sol. Tf = ik

f m

0.45 = i × 5.12 × 0.2

100060 20

0.45 6 270 135i

5.12 572 256

1351 1

256 2

1 135

256 2

121 1210.946

2 256 128

Hence percentage association of acetic acid is 94.6%

2. On treatment of 100 mL of 0.01 M solution of CoCCl3.6H

2O with excess AgNO

3, 1.2 × 1022 ions are

pprecipitated. The complex is -

CoCCl3.6H

2O ds 0.1 M foy;u ds 100 mL dks AgNO

3 ds vkf/kD; esa vfHkd`r djus ij 1.2 × 1022 vk;u vo{ksfir

gksrs gSaA ladqy gS &

(1) [Co(H2O)

5Cl]Cl

2.H

2O (2) [Co(H

2O)

4Cl

2]Cl.2H

2O

(3) [Co(H2O)

3Cl

3].3H

2O (4) [Co(H

2O)

6]Cl

3

Sol. CoCl3 . 6H

2O + excess AgNO

3 AgCl

No of ions precipitated = 1.2 × 1022

Hence No of AgCl precipitated = 22.2 10

2

= 6 × 1021



CHEMISTRY

moles of AgCl precipitated = 21

23

6 10

6 10

= 10–2

= 10 milimoleSo 10 milimole of AgCl are precipitated from 10 milimole of CoCl

3 . 6H

2O

Hence for mole of complex is [Co(H2O)

4Cl

2] Cl . 2H

2O

3. Which of the following compounds will form significant amount of meta product during mono-nitration

reaction?

eksuksukbVªs'ku vfHkfØ;k esa fuEu esa ls dkSu lk ;kSfxd esVk mRikn dh egRoiw.kZ ek=kk mRiUu djsxk \

(1)

NHCOCH3

(2)

OH

(3)

OCOCH3

(4)

NH2

Sol.

NH2

H

+

Acidification converts aniline to anilinium ion which deactivates the benzene ring which makes its meta

directing.

4. The products obtained when chlorine gas reacts with cold and dilute aqueous NaOH are :

(1) Cl– and ClO2– (2) ClO– and ClO

3– (3) ClOl

2– and ClO

3– (4) Cl– and ClO–

tc Dyksjhu xSl BaMs ,oa ruq tyh; NaOH ds lkFk vfHkfØ;k djrh gS rks izkIr gksus okys mRikn gksaxs &

(1) Cl– rFkk ClO2– (2) ClO– rFkk ClO

3– (3) ClOl

2– rFkk ClO

3– (4) Cl– rFkk ClO–

Sol. Cl2 + 2NaOH NaCl + NaOCl + H

2O

cold and

dilute

5. Both lithium and magnesium display several similar properties due to the diagonal relationship; however, the

one which is incorrect, is:

(1) nitrates of both Li and Mg yield NO2 and O

2 on heating

(2) both form basic carbonates

(3) both form soluble bicarbonates

(4) both form nitrides

fod.kZ lEcU/k ds dkj.k] yhfFk;e rFkk eSXuhf'k;e nksuksa dbZ ,d tSls xq.k iznf'kZr djrs gSa fQj Hkh] og ,d tks xyr gS&

(1) yhfFk;e rFkk eSXuhf'k;e] nksuksa ds gh ukbVªsV xje djus ij NO2 rFkk O

2 nsrs gSA

(2) nksuksa {kkjh; dkcksZusV cukrs gSA

(3) nksuksa ?kqyu'khy ckbdkcksZusV cukrs gSA



CHEMISTRY

(4) nksuksa ukbVªkbM cukrs gSA

Sol. (i) 4LiNO3 2Li

2O + 4NO

2 + O

2

Mg(NO3)

2 MgO + NO

2 + O

2

6. A water sample has ppm level concentration of following anions :

F– = 10; SO4

2– = 100; NO3

– = 50

The anion/anions that make/makes the water sample unsuitable for drinking is/are :

(1) only SO42– (2) only NO

3– (3) both SO

42– and NO

3– (4) only F–

,d ty izfrn'kZ esa ih- ih- ,e- (ppm) Lrj dh fuEu _.kk;uksa dh lkUnzrk gSA

F– = 10; SO4

2– = 100; NO3

– = 50

og@os _.kk;u tks ty izfrn'kZ dks ihus ds fy, vuqi;qDr cukrk gS@ cukrs gSa@gSa &

(1) ek=k SO42– (2) ek=k NO

3– (3) SO

42– rFkk NO

3– nksuksa (4) ek=k F–

Sol. Limiting value

7. The formation of which of the following polymers involves hydrolysis reaction?

(1) Terylene (2) Nylon 6 (3) Bakelite (4) Nylon 6, 6

fuEu cgqydksa esa ls dkSu ls cgqyd esa ty vi?kVu vfHkfØ;k lfUufgr gS \

(1) Vsjhyhu (2) ukbykWu 6 (3) csdsykbV (4) ukbykWu 6, 6

Sol. Theory

8. The Tyndall effect is observed only when following conditions are satisfied : [surface chemistry]

(a) The diameter of the dispersed particles is much smaller than the wavelength of the light used.

(b) The diameter of the dispersed particle is not much smallar than the wavelength of the light used.

(c) The refractive indices of the dispersed phase and dispersion medium are almost similar in magnitude.

(d) The refractive indices of the dispersed phase and dispersion medium differ greatly in magnitude.

(1) (b) and (c) (2) (a) and (d) (3*) (b) and (d) (4) (a) and (c)

fVUMy izHkko rHkh fn[kk;h iM+sxk tc fuEu 'krZ larq"V gksrh gS &

(a) ifj{ksfir d.kksa dk O;kl] iz;qDr izdk'k ds rajxnS/;Z dh rqyuk esa cgqr NksVh gksA

(b) ifj{ksfir d.kksa dk O;kl] iz;qDr izdk'k ds rjaxnS/;Z dh rqyuk esa cgqr NksVk ugha gksA

(c) ifj{ksfir izkoLFkk rFkk ifj{ksi.k ek/;e ds viorZukad ifjek.k yxHkx ,d tSls gksA

(d) ifj{ksfir izkoLFkk rFkk ifj{ksi.k ek/;e ds viorZukad ifjek.k cgqr fHkUu gksA

(1) (b) rFkk (c) (2) (a) rFkk (d) (3) (b) rFkk (d) (4) (a) rFkk (c)

Sol. Tyndall effect is observed when (1) Diameter or size of dispersed phase particles is not much smaller than

the wavelength of light used

(2) the refractive indices of the dispersed phase and dispersion medium differ greatly in magnitude



CHEMISTRY

9. pKa of a weak acid (HA) and pK

b of a weak base (BOH) are 3.2 and 3.4, respectively. The pH of their salt

(AB) solution is :

,d nqcZy vEy (HA) dk pKa rFkk ,d nqcZy {kkjd (BOH) dk pK

b Øe'k% 3.2 rFkk 3.4 gSaA muds yo.k (AB) ds foy;u

dk pH gksxkA

(1) 1.0 (2) 7.2 (3) 6.9 (4) 7.0

Sol.H kw ka kb1

p p p p2

H 1p 14 3.2 3.4

2

= 6.9

10. The major product obtained in the following reaction is:

fuEu vfHkfØ;k esa izkIr eq[; mRikn gS &

(1) (2)

(3) (4)

Sol.

O

COOH

O

DIBAL - HOH

CHO

COOH



CHEMISTRY

11. Which of the following compounds will behave as a reducing sugar in an aqueous KOH solution?

,d tyh; KOH foy;u esa fuEu esa ls dkSu lk ;kSfxd ,d vipk;d 'kdZjk ds :i esa O;ogkj djsxk \

(1) (2)

(3) (4)

Sol.

CH OH2

HO

OOH

CH OH2

OCOCH3KOH

CH OH2

HO

OOH

CH OH2

OH

12. The correct sequence of reagent for the following conversion will be :

fuEu :ikUrj.k ds fy, vfHkdeZdksa dk lgh Øe gksxk &

(1) [Ag(NH3)

2]+OH–, CH

3MgBr, H+/CH

3OH (2) [Ag(NH

3)

2]+OH–, H+/CH

3OH, CH

3MgBr,

(3) CH3MgBr, H+/CH

3OH, [Ag(NH

3)

2]+OH– (4) CH

3MgBr, [Ag(NH

3)

2]+OH–, H+/CH

3OH

Sol.

O

CHO

CH3OH

CH3HO

CH3

O

COO

O

COOCH3

H /CH OH+3

CH MgBr(excess)

3

[Ag(NH ) ] OH2+

3(tollens test)



13. Which of the following species is not paramagnetic?

fuEu esa ls dkSu lh Lih'kht vuqpqEcdh; ugha gS \

(1) B2

(2) NO (3) CO (4) O2

Sol. B2 = 2

1s 2

1s 2

2s 2

2s 1

2px 1

2py = Paramagnetic

21s

*21s

22s

*22s

22pz

22pz

22py *1

2px *0

2py = Paramagnetic

Co = is like N2 hence diamagnetic O

2 is paramagnetic

14. Which of the following, upon treatment with tert-BuONa followed by addition of bromine water, fails to

decolourize the colour of bromine?

fuEu esa ls dkSu] tert-BuONa ds lkFk vfHkdr djus rFkk czksehu ty ds feykus ij] czksehu ds jax dks jaxghu djus esa vleFkZ

gksrk gS \

(1) (2) (3) (4)

Sol. No -hydrogen present in 2nd structure.

15. Which of the following reactions is an example of a redox reaction?

fuEu esa ls dkSu lh vfHkfØ;k vip;ksip; ¼fjMkWDl½ vfHkfØ;k dk mnkgj.k gS \

(1) XeF6 + 2H

2O XeO

2F

2 + 4HF (2) XeF

4 + O

2F

2 XeF

6 + O

2

(3) XeF2 + PF

5 [XeF]+ PF

6– (4) XeF

6 + H

2O XeOF

4 + 2HF

Sol.4 1 6 0

4 2 2 2Xe F O F Xe F6 O

In above reachion Xe is oxidised and O is reduced

16. U is equal to :

(1) Isothermal work (2) Isochoric work (3) Isobaric work (4) Adiabatic work

U ftlds cjkcj gS] og gS &

(1) lerkih dk;Z (2) le&vk;rfud dk;Z (3) lenkch dk;Z (4) :)ks"e dk;Z

Sol. In case of adiabatic process

U = q + w

U = w

17. Which of the following molecules is least resonance stabilized?

fuEu esa ls dkSu v.kq vuqukfnd :i ls U;wure fLFkj gS \

(1) (2) (3) (4)

CHEMISTRY



Sol.

18. The increasing order of the reactivity of the following halides for the SN1 reaction is :

SN1 vfHkfØ;k ds fy, fuEu gSykbMksa dh vfHkfØ;kRedrk dk c<+rk Øe gS &

(1) (II) < (III) < (I) (2) (III) < (II) < (I) (3) (II) < (I) < (III) (4) (I) < (III) < (II)

Sol. III > I > II (Stability of carbocation)

19. 1 gram of a carbonate (M2CO

3) on treatment with excess HCl produces 0.01186 mole of CO

2. The molar

mass of M2CO

3 in g mol–1 is :

,d dkcksZusV (M2CO

3) ds 1 xzke dks HCl ds vkf/kD; esa vfHkfØr fd;k tkrk gS vkSj mlls 0.01186 eksy CO

2 iSnk gksrh

gSA M2CO

3 dk eksyj nzO;eku g mol–1 esa gS &

(1) 11.86 (2) 1186 (3) 84.3 (4) 118.6

Sol. M2CO

3 + 2HCl 2MCl + H

2O + CO

2

Moles of M2CO

3 = moles of CO

2

Let molar mass of M2CO

3 is 'x' then

10.01186

x

1x 84.3

0.01186

20. Sodium salt of an organic acid 'X' produces effervescence with conc. H2SO

4. 'X' reacts with the acidified

aqueous CaCl2 solution to give a white precipitate which decolourises acidic solution of KMnO

4.

,d dkcksZfud vEy dk lksfM;e yo.k 'X' lkUnz H2SO

4 ds lkFk cqncqnkgV nsrk gSA 'X' vEyh; tyh; CaCl

2 ds lkFk

vfHkfØ;k djrk gS vkSj lQsn vo{ksi nsrk gS tks KMnO4 ds vEyh; foy;u dks jaxghu cuk nsrk gSA X gS &

(1) Na2C

2O

4(2) C

6H

5COONa (3) HCOONa (4) CH

3COONa

Sol. Na2C

2O

4 + H

2SO

4 CO + CO

2 + H

2O + Na

2SO

4

Na2C

2O

4 + CaCl

2 CaC

2O

4(white ppt.) + 2NaCl

CaC2O

4 + KMnO

4 CO

2 + Mn2+ (colourless)

21. The most abundant elements by mass in the body of a healthy human adult are:

Oxygen (61.4%); Carbon (22.9%); Hydrogen (10.0%); and Nitrogen (2.6%).

The weight which 75 kg person would gain if all 1H atoms are replaced by 2H atoms is :

,d LoLFk equ"; ds 'kjhj esa ek=kk dh n`f"V ls cgq;k;r ls feyus okys rRo gS% vkWDlhtu (61.4%$); dkcZu (22.9%);

gkbMªkstu (10.0%); rFkk ukbVªkstu (2.6%) A 75 kg otu okys ,d O;fDr ds 'kjhj ls lHkh 1H ijek.kqvksa dks 2H

ijek.kqvksa ls cny fn;k tk; rks mlds Hkkj esa tks o`f) gksxh] og gS &

CHEMISTRY



(1) 10 kg (2) 15 kg (3) 37.5 kg (4) 7.5 lg

Sol. Mass of 1H in 75 kg = 75 × 10

100

= 7.5 kg

moles of 1H atoms = 37.5 10

1

Mass of 2H atoms = 7.5 × 103 × 2

= 15 × 103

= 15 kg

So gain in weight = 15 – 7.5

= 7.5 kg

22. The major product obtained in the following reactions i:

fuEu vfHkfØ;k esa izkIr gksus okyk eq[; mRikn gS &

(1) (–)C6H

5CH(OtBu)CH

2C

6H

5(2) ( )C

6H

5CH(OtBu)CH

2C

6H

5

(3) C6H

5CH = CHC

6H

5(4) (+)C

6H

5CH(OtBu)CH

2C

6H

5

Sol. -elimination

Ph – CH = CH – Ph

23. Given :

C(graphite)

+ O2(g) CO

2(g);

rH° = –393.5 kJ mol–1

H2(g) +

1

2O

2(g) H

2O(l);

rH° = –285.8 kJ mol–1

CO2(g) + 2H

2O(l) CH

4(g) + 2O

2(g);

rH° = +890.3 kJ mol–1

Based on the above thermochemical equations, the value of rH° at 298 K for the reaction:

C(graphite)

+ 2H2(g) CH

4(g) will be:

fn;k x;k gS :

C(graphite)

+ O2(g) CO

2(g);

rH° = –393.5 kJ mol–1

H2(g) +

1

2O

2(g) H

2O(l);



rH° = –285.8 kJ mol–1

CO2(g) + 2H

2O(l) CH

4(g) + 2O

2(g);

rH° = +890.3 kJ mol–1

Åij fn;s x;s Å"ejklk;fud lehdj.kksa ds vk/kkj ij 298 K ij vfHkfØ;k

C(graphite)

+ 2H2(g) CH

4(g) ds

rHº dk eku gksxk &

(1) –144.0 kJ mol–1 (2) +74.8 kJ mol–1 (3) +144.0 kJ mol–1 (4) –74.8 kJ mol–1

Sol. (1) C(graphite)

+ O2(g) CO

2(g); H = –393.5 kJ mol–1

(2) H2(g) +

1

2O

2(g) H

2O(l); H = –285.8 kJ mol–1

(3) CO2(g) + 2H

2O(l) CH

4(g) + 2O

2(g); H = +890.3 kJ mol–1

(4) C(graphite)

+ 2H2(g) CH

4(g) H

IV = 1 + 2 × II + III

H = –393.5+2 (–285.8) + 890.3

= –74.8 kJ/mole

24. In the following reactions, ZnO is respectively acting as a/an :

(a) ZnO + Na2O Na

2ZnO

2

(b) ZnO + CO2 ZnCO

3

(1) acid and base (2) base and acid (3) base and base (4) acid and acid

fuEu vfHkfØ;kvksa esa ZnO Øe'k% dk;Z djsxk &

(a) ZnO + Na2O Na

2ZnO

2

(b) ZnO + CO2 ZnCO

3

(1) vEy rFkk {kkjd (2) {kkjd rFkk vEy (3) {kkjd rFkk {kkjd (4) vEy rFkk vEy

Sol. ZnO + Na2O Na

2ZnO

2

acid base

ZnO + CO2 ZnCO

3

base acid



25. The radius of the second Bohr orbit for hydrogen atoms is:

(Plank's Const. h = 6.6262 × 10–34 Js;

mass of electron = 9.1091 × 10–31 kg;

charge of electron e = 1.60210 × 10–19 C;

permitivity of vacuum

0 = 8.854185 × 10–12 kg–1 m–3 A2)

gkbMªkstu ijek.kq ds f}rh; cksj d{kk dk v)ZO;kl gksxk %

(IySad fLFkjkad h = 6.6262 × 10–34 Js;

bysDVªkWu dk nzO;eku = 9.1091 × 10–31 kg;

bysDVªkWU dk vkos'k e = 1.60210 × 10–19 C;

fuokZr dk ijoS|qrkad

0 = 8.854185 × 10–12 kg–1 m–3 A2)

(1) 2.12 Å (2) 1.65 Å (3) 4.76 Å (4) 0.529 Å

Sol. r = 0.529 × 2n

2

= 0.529 × 4

= 2.116 Å

26. Two reactions R1 and R

2 have identical pre-exponential factors. Activation energy of R

1 exceeds that of R

2 by

10 kJ mol–1. If k1 and k

2 are rate constant for reactions R

1 and R

2 respectively at 300 K, then ln(k

2/k

1) is equal

to : (R = 8.314 J mol–1 K–1)

nks vfHkfØ;kvksa R1 rFkk R

2 ds iwoZ pj?kkrkad xq.kd ,d tSls gSA R

1 dh lafØ;.k ÅtkZ R

2 ds lafØ;.k ÅtkZ ls 10 kJ mol–

1 T;knk gSA ;fn vfHkfØ;k R1 rFkk R

2 ds fy, 300 K ij nj fu;rakd Øe'k% k

1 rFkk k

2 gksa rks ln(k

2/k

1) fuEu esa ls fdlds

cjkcj gksxk &

(1) 4 (2) 8 (3) 12 (4) 6

Sol. k1 =

3a2

E 10 10

RTAe

k2 =

a2E

RTAe

310 10

1 RT

2

ke

k

310 10

1 RT

2

kn e

k

l

3 32

1

k 10 10 10 10n 4

k RT 8.314 300

l



27. A metal crystallises in a face centred cubic structure. If the edge length of its unit cell is 'a', the closest approach

between two atoms in metallic crystal will be :

,d /kkrq Qyd dsfUnzr ?ku lajpuk esa fØ:Vyh; gksrh gSA ;fn blds ,dd lsy dh dksj yEckbZ 'a' gS] rks /kkfRod fØLVy

esa nks ijek.kqvksa ds chp lfUudVre nwjh gksxh &

(1) 2

a(2) 2a (3) 2 2 a (4) 2 a

Sol. 2a 4rClosest distance of approach betweentwo atoms = 2r

2a a

2r2 2

28. The group having isoelectronic species is :

og xzqi ftlesa lebysDVªkWuh Lih'khy gS &

(1) O–, F–, Na+, Mg2+ (2) O2–, F–, Na+, Mg2+ (3) O–, F–, Na, Mg+ (4) O2–, F–, Na, Mg2+

Sol. Isoelectronic species are

O2–, F–, Na+, Mg2+ = each has 10 e–

29. Given:

32 / /

1.36 , 0.74o o

Cl Cl Cr CrE V E V

2 3 22 7 4

O / /Mn1.33 , 1.51o o

Cr Cr MnOE V E V

Among the following, the strongest reducing agent is :

fn;k x;k gS :

32 / /

1.36 , 0.74o o

Cl Cl Cr CrE V E V

2 3 22 7 4

O / /Mn1.33 , 1.51o o

Cr Cr MnOE V E V

fuEu esa ls izcyre vipk;d gS &

(1) Cl– (2) Cr (3) Mn2+ (4) Cr3+

Sol. 32 / /

1.36 , 0.74o o

Cl Cl Cr CrE V E V

2 3 22 7 4

O / /Mn1.33 , 1.51o o

Cr Cr MnOE V E V

Less the value of reduction potential better will be the reducing agent.

Hence Cr is strongest reducing agent.



30. 3-Methyl-pent-2-ene on reaction with HBr in presence of peroxide forms an additon product. The number of

possible stereoisomers for the procut is :

(1) Four (2) Six (3) Zero (4) Two

ijkDlkbM dh mifLFkfr esa] 3-esfFky-isUV-2-bZu HBr ds lkFk vfHkfØ;k djus ij ,d ladyu mRikn cukrk gSA mRikn

ds fy, lEHko f=kfoe leko;fo;ksa dh la[;k gksxh &

(1) pkj (2) N% (3) 'kwU; (4) nks

Sol. C – C = C

C

– C – C

HBr/H O22

C C C C C

C

HBr (anti-Markonikoff addition)

No. of stereoisomer = 22 = 4.



MATHS



31. The integral

3

4

4

dx

1 cos x

is equal to

lekdyu

3

4

4

dx

1 cos x

cjkcj gS &

(1) 4 (2) –1 (3) –2 (4) 2

Sol.dx

1 cos x =

3

4

2

4

dx

x2cos

2

3

4

4

xtan

2

= 2

32. Let nnI tan x dx, (n 1) . If I

4 + I

6 = a tan5x + bx5 + C, where C is a constant of integration, then the

ordered pair (a, b) is equal to

ekuk nnI tan x dx, (n 1) gSA ;fn I

4 + I

6 = a tan5x + bx5 + C gS] tgk¡ C ,d lekdyu vpj gSa] rks Øfer

;qXe (a, b) cjkcj gS &

(1) 1

, 15

(2)

1,0

5

(3) 1

,15

(4) 1

,05

Sol. 4 24 6I I tan (x) 1 tan x dx

=5(tan x)

C5

1a ,b 0

5

33. The area (in sq. units) of the region {(x, y) : x 0, x + y 3, x2 4y and y 1 + x } is

{ks=k {(x, y) : x 0, x + y 3, x2 4y rFkk y 1 + x }dk {ks=kQy (oxZ bdkb;ksa) esa gS &

(1) 7

3(2)

5

2(3)

59

12(4)

3

2



MATHS

Sol. 1 22 2

0 1

x xArea 1 x dx 3 x dx

4 4

= 5

2

(1, 2)

(2, 1)

34. A box contains 15 green and 10 yellow balls. If 10 balls are randomly drawn, one-by-one, with replace-

ment, then the variance of the number of green balls drawn is

,d cDls esa 15 gjh rFkk 10 ihyh xsansa gSA ;fn ,d&,d djds ;knPN;k] izfrLFkkiuk lfgr] 10 xsansa fudkyh tk,¡] rks

gjh xsanksa dh la[;k dk izlj.k gS &

(1) 4 (2) 6

25(3)

12

5(4) 6

Sol. Binomial prob distribution

n = 10, 15 3

p25 5

r2 = np(1 – p) = 12

5

35. If dy

2 sin x y 1 cos x 0dx

and y(0) = 1, then y2

is equal to

;fn dy

2 sin x y 1 cos x 0dx

rFkk y(0) = 1 gS] rks y2

cjkcj gS &

(1) 1

3 (2)

4

3(3)

1

3(4)

2

3

Sol. dy

2 sin x y 1 cos xdx

dy cos xdx

y 1 2 sin x

ln y 1 ln C ln(2 sin x)

Cy 1

2 sin x

y(0) 1 C = 4

y + 1 = 4

2 sin x

1y

2 3



MATHS

36. Let be a complex number such that 2 + 1 = z where z 3 . If 2 2

2 7

1 1 1

1 1 3k

1

, then k is

equal to

ekuk ,d lfEeJ la[;k ,slh gS fd 2 + 1 = z tgk¡ z 3 gSA ;fn 2 2

2 7

1 1 1

1 1 3k

1

gS] rks k cjkcj

gS &

(1) –1 (2) 1 (3) –z (4) z

Sol.1 i 3

2 2

, 21 0 , 3 1

k

R1 R

1 + R

2 + R

3

then expand

3k = 3(2 – )

= 3i 3

k i 3 z

37. Let ˆ ˆ â 2i j 2k

and ˆ ˆb i j

. Let c

be a vector such that | c a |

= 3, a b c 3

and the angle

between c

and a b

be 30°. Then a.c

is equal to

ekuk ˆ ˆ â 2i j 2k

rFkk ˆ ˆb i j

gSA ekuk c

,d ,slk lfn'k gS fd | c a |

= 3, a b c 3

rFkk c

vkSj

a b

ds chp dk dks.k 30° gS] rks a.c

cjkcj gS &

(1) 5 (2) 1

8(3)

25

8(4) 2

Sol. ˆ ˆ â b 2i 2 j k

a b 3

c a 3

a b c sin 30 3

2 2c a 2a.c 9

c 2

c.a 2



MATHS

38. The radius of a circle, having minimum area, which touches the curve y= 4 – x2 and the lines, y = |x| is

U;wure {ks=kQy okys ,sls o`r] tks oØ y= 4 – x2 rFkk js[kkvksa y = |x| dks Li'kZ djrk gS] dh f=kT;k gS &

(1) 4 2 1 (2) 4 2 1 (3) 2 2 1 (4) 2 2 1

Sol.

Center will be (0, 4 – r)

Distance of (0, 4 – r) from L : x – y = 0

radius = r

4 rr

2

4 – r = 2r

r = 4 2 1

39. If for 1

x 0,4

, the derivative of

1

3

6x xtan

1 9x

is x.g(x) , then g(x) equals :

;fn 1

x 0,4

ds fy,

1

3

6x xtan

1 9x

dk vodyu x.g(x) gS] rks g(x) cjkcj gS &

(1) 3

3x

1 9x(2) 3

3

1 9x(3) 3

9

1 9x(4) 3

3x x

1 9x

Sol. f(x) = 1 1

3

6tan 2 tan 3

1 9

x xx x

x

3

9'(x)

1 9

xf

x

= x.g x

g(x) = 3

9

1 9x

40. If two different numbers are taken from the set {0, 1, 2, 3, ...., 10}; then the probability that their sum as

well as absolute difference are both multiple of 4, is

;fn leqPp; {0, 1, 2, 3, ...., 10} esa ls nks fofHkUu la[;k,¡ fudkyh xbZ] rks muds ;ksxQy rFkk muds varj ds

fujis{k eku] nksuksa ds pkj ds xq.kd gksus dh izkf;drk gS &



(1) 14

45(2)

7

55(3)

6

55(4)

12

55

Sol. n(S) = 11C2 = 55

Both will be either from {2, 6, 10} or {0, 4, 8}

Prob. = 3 3

2 2 6

55 55

C C

41. 3x

2

cot x cos xlim

( 2x)

equals

3x

2

cot x cos xlim

( 2x)

cjkcj gS &

(1) 1

8(2)

1

4(3)

1

24(4)

1

16

Sol. x = /2 + h

3 30 0

tanh sinh tan sinh

8 8

h h

hLim Lim

h h =

1

16

42. The value of (21C1 – 10C

1) + (21C

2 – 10C

2) + (21C

3 – 10C

3) + (21C

4 – 10C

4) + ...... + (21C

10 – 10C

10) is

(21C1 – 10C

1) + (21C

2 – 10C

2) + (21C

3 – 10C

3) + (21C

4 – 10C

4) + ...... + (21C

10 – 10C

10) dk eku gS &

(1) 220 – 29 (2) 220 – 210 (3) 221 – 211 (4) 221 – 210

Sol. 21 21 21 210 1 2 10....C C C C – 10 10 10 21 10

0 1 10 0 0....C C C C C

2110 20 102

2 2 22

43. For three events A, B and C, P (Exactly one of A or B occurs) = P(Exactly one of B or C occurs) =

P(Exactly one of C or A occurs) = 1

4 and P(All the three events occur simultaneously) =

1

16.

Then the probability that at least one of the events occurs, is

rhu ?kVukvksa A, B rFkk C ds fy, P (A vFkok B esa ls dsoy ,d ?kfVr gksrh gS) = P(B vFkok C esa ls dsoy ,d

?kfVr gksrh gS) = P(C vFkok A esa ls dsoy ,d ?kfVr gksrh gS) = 1

4 rFkk P(lHkh rhu ?kVuk,¡ ,d lkFk ?kfVr gksrh

gS) = 1

16.

rks izkf;drk fd de ls de ,d ?kVuk ?kfVr gks] gS &

(1) 7

64(2)

3

16(3)

7

32(4)

7

16



Sol.a bd

gef

ch

A B

C

a + b + e + f = 1

4

b + d + c + f = 1

4

1

16g

c + e + a + d = 1

4

Add

2(a + b + c + d + e + f) = 3

4

Req. prob. = a + b + c + d + e + f + 0 = 3 1 7

8 16 16

44. Let a vertical tower AB have its end A on the level ground. Let C be the mid-point of AB and P be a point

on the ground such that AP = 2AB. If BPC , then tan is equal to

ekuk ,d Å/okZ/kj ehukj AB ,slh gS fd mldk fljk A Hkwfe ij gSA ekuk AB dk e/; fcUnq C gS rFkk Hkwfe ij fLFkr

fcUnq P ,slk gS fd AP = 2AB ;fn BPC gS] rks tan cjkcj gS &

(1) 2

9(2)

4

9(3)

6

7(4)

1

4

Sol.1

tan4

1tan( )

2

B

A P

C

h

h

4h

tan tan 1

1 tan tan 2

tan = 2/9

45. The eccentricity of an ellipse whose centre is at the origin is 1

2. If one of its directrices is x = –4, then the

equation of the normal to it at 3

1,2

is

,d nh?kZo`r] ftldk dsUnz ewy fcUnq ij gS] dh mRdsUnzrk 1

2 gSA ;fn mldh ,d fu;rk x = –4 gS] rks blds fcUnq

31,

2

ij mlds vfHkyac dk lehdj.k gS &

(1) 4x + 2y = 7 (2) x + 2y = 4 (3) 2y – x = 2 (4) 4x – 2y = 1

Sol.1

2e

4a

e a = 2



b2 = a2 (1 – e2) b2 = 3

2 2

14 3

x y

Equation of normal

2 2

1 1

1a x b y

x y

4 31

1 3 / 2

x y

4x – 2y = 1

46. If, for a positive integer n, the quadratic equation, x(x + 1) + (x + 1) (x + 2) + ....... + (x + n 1 ) (x + n) =

10 n has two consecutive integral solutions, then n is equal to

;fn fdlh /kuiw.kkZad n ds fy,] f}?kkrh lehdj.k x(x + 1) + (x + 1) (x + 2) + ....... + (x + n 1 ) (x + n) = 10 n

ds nks Øfer iw.kkZadh; gy gS] rks n cjkcj gS &

(1) 10 (2) 11 (3) 12 (4) 9

Sol.1

( 1)( ) 10

n

r

x r x r n

nx2 + n2x + r(r – 1) = 10n

nx2 + n2x + 2( 1)

103

n nn

x2 + nx + 2 31

03

n

| | 1 D

a n = 1

47. The following statement (p q) [(~p q) q] is

(1) equivalent to p ~ q (2) a fallacy

(3) a tautology (4) equivalent to ~p q

fuEu dFku (p q) [(~p q) q]

(1) p ~ q ds lerqY; gS (2) ,d gsRokHkkl gS

(3) ,d iqu:fDr gS (4) ~p q ds lerqY; gS

Sol. ( ) ( ) p q P q q

( ) (( ) ( ) ) p q p q q

tautology



MATHS

48. The normal to the curve y(x – 2) (x – 3) = x + 6 at the point where the curve intersects the y-axis passes

through the point

oØ y(x – 2) (x – 3) = x + 6 ds ml fcUnq ij] tgk¡ oØ y-v{k dks dkVrh gS] [khapk x;k vfHkyac fuEu esa ls fdl

fcUnq ls gksdj tkrk gS \

(1) 1 1

,2 3

(2)

1 1,

2 3

(3) 1 1

,2 2

(4)

1 1,

2 2

Sol. Point (0, 1)

y(x – 2) (x – 3) = x + 6

dy

dx

(x – 2) (x – 3) + (2x – 5)y = 1

Slope of tangent at (0, 1) = 1Equation of norm.y – 1 = –1(x – y)

x + y = 1

49. For any three positive real numbers a, b and c, 9 (25a2 + b2) + 25 (c2 – 3ac) = 15b (3a + c). Then

(1) a, b and c are in A.P. (2) a, b and c are in G.P.

(3) b, c and a are in G.P. (4) b, c and a are in A.P.

fdUgha rhu /kukRed okLrfod la[;kvksa a, b rFkk c ds fy, 9 (25a2 + b2) + 25 (c2 – 3ac) = 15b (3a + c) gS] rks &

(1) a, b rFkk c lekUrj Js<+h gS (2) a, b rFkk c xq.kksrj Js<+h gS

(3) b, c rFkk a xq.kksrj Js<+h gS (4) b, c rFkk a lekUrj Jss<+h gS

Sol. x2 + y2 + z2 = xy + yz + zx

x = y = z

15a = 3b = 5c

5 3

b ca

a = b = 5

c = 3

50. If the image of the point P(1, – 2, 3) in the plane, 2x + 3y – 4z + 22 = 0 measured parallel to the line ,

x y z

1 4 5 is Q, then PQ is equal to

;fn fcUnq P(1, – 2, 3) dk lery 2x + 3y – 4z + 22 = 0 esa og izfrfcac tks js[kk x y z

1 4 5 ds lekarj gS] Q gS]

rks PQ cjkcj gS &



MATHS

(1) 42 (2) 6 5 (3) 3 5 (4) 2 42

Sol. Equation of PQ

(1, 2,3) (1, 4,5)r

N(1 + , 4 – 2, 3 + 5)2(1 + ) + 3(4 – 2) – 4(3 + 5) + 22 = 0

6 = 6 N(2, 2, 8)Q(3, 6, 13)

PQ = 2 42

51. If 5 (tan2x – cos2x) = 2cos 2x + 9, then the value of cos 4x is

;fn 5 (tan2x – cos2x) = 2cos 2x + 9 gS] rks cos 4x dk eku gS &

(1) 2

9(2)

7

9 (3)

3

5 (4)

1

3

Sol.1 cos 2 1 cos 2

5 2cos 2 91 cos 2 2

x xx

x

cos2x = – 1

3

cos4x = 2cos22x – 1 = – 7

9

52. Let a, b, c R. If f(x) = ax2 + bx + c is such that a + b + c = 3 and f(x + y) = f(x) + f(y) + xy, x, y R,

then 10

n 1

f (n)

is equla to

ekuk a, b, c R A ;fn f(x) = ax2 + bx + c ,slk gS fd a + b + c = 3 gS rFkk lHkh x, y R ds fy, f(x + y) =

f(x) + f(y) + xy gS] rks 10

n 1

f (n)

cjkcj gS &

(1) 190 (2) 255 (3) 330 (4) 165



Sol. f(1) = 3

Put y = 1 f(x + 1) – f(x) = 3 + x

(a(x + 1)2 + b(x + 1) + c) – (ax2 + bx + c) = 3 + x

1

2a b =

5

2 c = 0

f(n) = 2 5

2

n n

210

1

5330

2n

n n

Alternate Solution :f(1) = 3f(2) = f(1) + f(1) + 1 3 + 3 + 1 = 7f(3) = f(2) + f(1) + 2 7 + 3 + 2 = 12f(4) = 18f(5) = 25f(6) = 33f(7) = 42f(8) = 52f(9) = 63f(10) = 75

add = 330

53. The distance of the point (1, 3, –7) from the plane passing through the point (1, – 1, – 1) having normal

perpendicular to both the lines x 1 y 2 z 4

1 2 3

and

x 2 y 1 z 7

2 1 1

is

,d lery tks fcUnq (1, –1, –1) ls gksdj tkrk gS rFkk ftldk vfHkyEc nksuksa js[kkvksa x 1 y 2 z 4

1 2 3

rFkk

x 2 y 1 z 7

2 1 1

ij yEc gS] dh fcUnq (1, 3, –7) ls nwjh gS &

(1) 5

83(2)

10

74(3)

20

74(4)

10

83



Sol. Normal vector of plane

ˆ ˆ ˆi j k

1 2 3

2 1 1

ˆ ˆ ˆ5i 7 j 3k

equation of plane 5x + 7y + 3z = –5

Distance = 5(1) 7(3) 3( 7) 5

83

= 10

83

54. If S is the set of distinct values of 'b' for which the following system of linear equations

x + y + z = 1

x + ay + z = 1

ax + by + z = 0

has no solution, then S is

(1) a finite set containing two or more elements (2) a singleton

(3) an empty set (4) an infinite set

;fn S, 'b' dh mu fofHkUu ekuksa dk leqPp; gS ftuds fy, fuEu jSf[kd lehdj.k fudk;

x + y + z = 1

x + ay + z = 1

ax + by + z = 0

dk dksbZ gy ugha gS] rks S :

(1) ,d ifjfer leqPp; gS ftlesa nks ;k vf/kd vo;o gS

(2) ,d gh vo;o okyk leqPp; gS

(3) ,d fjDr leqPp; gS

(4) ,d vifjfer leqPp; gS

Sol. D = 2

1 1 1

1 a 1 (a 1) 0

a b 1

a = 1If a = 1 two planes are parallesSo for no solutionThird plane must be parallel

b = 1



55. If 2 3

A4 1

, then adj (3A2 + 12A) is equal to

;fn 2 3

A4 1

gS] rks adj (3A2 + 12A) cjkcj gS &

(1) 51 84

63 72

(2) 72 63

84 51

(3) 72 84

63 51

(4) 51 63

84 72

Sol. (3A2 + 12A) = 3A (A + 4I)

72 63

84 51

adj (3A2 + 12A) = 51 63

84 72

56. A hyperbola passes through the point P 2, 3 and has foci at (±2, 0). Then the tangent to this hyper-

bola at P also passes through the point

,d vfrijoy; fcUnq P 2, 3 ls gksdj tkrk gS] rFkk mldh ukfHk;k¡ (±2, 0) ij gSa] rks vfrijoy; ds fcUnq P

ij [khaph xbZ Li'kZ js[kk ftl fcUnq ls gksdj tkrh gS] og gS &

(1) 3, 2 (2) 2, 3 (3) 3 2, 2 3 (4) 2 2,3 3

Sol. S1(2, 0) S

2(–2, 0)

P 2, 3

|PS1 – PS

2| = 2a = 2

a = 12ae = 4 e = 2Equation of hyperbola

2 2x y1

1 3

Equation of tangent at P

y2x 1

3

57. Let k be an integer such that the triangle with vertices (k, –3k), (5, k) and (–k, 2) has area 28 sq. units.

Then the orthocentre of this triangle is at the point.

ekuk k ,d ,slk iw.kkZad gS fd f=kHkqt] ftlds 'kh"kZ (k, –3k), (5, k) rFkk (–k, 2) gSa] dk {ks=kQy 28 oxZ bdkbZ gS] rks

f=kHkqt ds yEc&dsUnz ftl fcUnq ij gS] og gS &

(1) 3

1,4

(2)

12,

2

(3) 1

2,2

(4)

31,

4



Sol.

k 3k 1

5 k 1 56

k 2 1

k = 2

A(2, –6)

C(2, –2)B(5, 2)

orthocenter 1

2,2

58. Twenty meters of wire is available for fencing off a flower-bed inthe form of a circular sector. Then the

maximum area (in sq. m) of the fl,ower-bed, is

,d Qwyksa dh D;kjh] tks ,d o`r ds f=kT; [kaM ds :i esa gS] dh ?ksjkcanh djus ds fy, chl ehVj rkj miyC?k gSA rks

Qwyksa dh D;kjh dk vf/kdre {ks=kQy (oxZ eh- esa) gS &

(1) 25 (2) 30 (3) 12.5 (4) 10

Sol. r

r r

r + r + r = 20

20 2r

r

21A r

2

r(20 r)A

r

dA0 r 5

dr

= 2

Area = 21

r 252

59. The function 1 1

f : R ,2 2

defined as 2

xf (x)

1 x

, is

(1) surjective but not injective (2) neither injective nor surjective

(3) invertible (4) injective but not surjective



Qyu 1 1

f : R ,2 2

, tks 2

xf (x)

1 x

}kjk ifjHkkf"kr gS &

(1) vkPNknh gS ijUrq ,dSdh ugha gS (2) u rks vkPNknh vkSj u gh ,dSdh gS

(3) O;qRØe.kh; gS (4) ,dSdh gS ijUrq vkPNknh ugha gS

Sol. 2

xy

1 x

It is many one

Range of f(x) is 1 1

,2 2

Therefore it will be onto

60. A man X has 7 friends, 4 of them are ladies and 3 are men. His wife Y also has 7 friends, 3 of them are

ladies and 4 are men. Assume X and Y have no common friends. Then the total number of ways in which X

and Y together can throw a party inviting 3 ladies and 3 men, so that 3 friends of each of X and Y are in this

party, is

,d O;fDr X ds 7 fe=k gS] ftuesa 4 efgyk,¡ gS rFkk 3 iq:"k gS] mldh iRuh Y ds Hkh 7 fe=k gSa] ftuesa 3 efgyk,¡

rFkk 4 iq:"k gSA ;g ekuk x;k fd X rFkk Y dk dksbZ mHk;fu"B fe=k ugha gSA rks mu rjhdksa dh la[;k ftuesa X rFkk

Y ,d lkFk 3 efgykvksa rFkk 3 iq:"kksa dks ikVhZ ij cqyk,a fd X rFkk Y izR;sd ds rhu&rhu fe=k vk;sa gS &

(1) 469 (2) 484 (3) 485 (4) 468

Sol.

X Y

Man (3) Ladies(4) Man(4) Ladies(3) No of ways

0 3 3 0 4C3 4C

3

1 2 2 1 3C1 4C

2 4C

3 3C

1

2 1 1 2 3C2 4C

1 4C

1 3C

2

3 0 0 3 3C3 3C

3

485

c JEE (Main) 2017 - Matrix JEE Academy · 2009-08-31 · Matrix JEE Academy : Opposite Reliance...

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Transcript of c JEE (Main) 2017 - Matrix JEE Academy · 2009-08-31 · Matrix JEE Academy : Opposite Reliance...