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1 Matrix JEE Academy : Piprali Road, Sikar Ph. 01572-241911, Mob. 97836-21999, 97836-31999 Matrix JEE Academy JEE (MAIN ONLINE) 2019 PHYSICS 10 APRIL 2019 [Phase : II] JEE MAIN PAPER ONLINE KTG & Thermodynamics 1. One mole of an ideal gas passes through a process where pressure and volume obey the relation 2 0 0 V 1 P P 1 2 V . Here P 0 and V 0 are constants. Calculate the change in the temperature of the gas if its volume changes from V 0 to 2V 0 . ,d vkn'kZ xS l dk ,d eksy ,d ,s ls iz Øe ls xqtjrk gS ftlesa nkc rFkk vk;ru lw=k 2 0 0 V 1 P P 1 2 V ls lEcfU/kr gSaA ;gk¡ P 0 rFkk V 0 fu;rkad gSaA ;fn xSl dk vk;ru V 0 ls 2V 0 gks rk gS rks blds rkieku dk cnyko gksxk : (1) 0 0 PV 1 4 R (2) 0 0 PV 5 4 R (3) 0 0 PV 1 2 R (4) 0 0 PV 3 4 R A. 2 sol. If 0 1 0 1 0 P 1 V V P P 1 2 2 If 0 2 0 2 0 7P 1 1 V 2V P P 1 2 4 8 1 1 2 2 PV PV PV T T nR nR nR 0 0 0 0 1 1 2 2 PV 7P V 1 1 T PV PV nR nR 2 4 0 0 5P V 4nR Waves on a String 2. The correct figure that shows, schematically, the wave pattern produced by superposition of two waves of frequencies 9 Hz and 11 Hz, is : 11 Hz rFkk 9 Hz vko`fÙk dh nks rjaxks a ds v/;kjksi.k dks fuEu esa dkS u fp=k ;ks tukc) rjhds ls lgh n'kkZrk gS\ (1) (2)

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Page 1: Matrix JEE (MAIN ONLINE) 2019 JEE Academy · 2020. 9. 29. · Matrix JEE Academy : Piprali Road, Sikar Ph. 01572-241911, Mob. 97836-21999, 97836-31999 1 Matrix JEE Academy JEE (MAIN

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PHYSICS

10 APRIL 2019 [Phase : II]

JEE MAIN PAPER ONLINE

KTG & Thermodynamics

1. One mole of an ideal gas passes through a process where pressure and volume obey the relation

2

00

V1P P 1

2 V

. Here P

0 and V

0 are constants. Calculate the change in the temperature of the gas if its

volume changes from V0 to 2V

0.

,d vkn'kZ xSl dk ,d eksy ,d ,sls izØe ls xqtjrk gS ftlesa nkc rFkk vk;ru lw=k 2

00

V1P P 1

2 V

ls lEcfU/kr gSaA

;gk¡ P0 rFkk V

0 fu;rkad gSaA ;fn xSl dk vk;ru V

0 ls 2V

0 gksrk gS rks blds rkieku dk cnyko gksxk :

(1) 0 0P V1

4 R(2)

0 0P V5

4 R(3)

0 0P V1

2 R(4)

0 0P V3

4 R

A. 2

sol. If 01 0 1 0

P1V V P P 1

2 2

If 02 0 2 0

7P1 1V 2V P P 1

2 4 8

1 1 2 2P V P VPVT T

nR nR nR

0 0 0 01 1 2 2

P V 7P V1 1T P V P V

nR nR 2 4

0 05P V

4nR

Waves on a String

2. The correct figure that shows, schematically, the wave pattern produced by superposition of two waves of

frequencies 9 Hz and 11 Hz, is :

11 Hz rFkk 9 Hz vkofÙk dh nks rjaxksa ds v/;kjksi.k dks fuEu esa dkSu fp=k ;kstukc) rjhds ls lgh n'kkZrk gS\

(1) (2)

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(3) (4)

A. 3

sol. Beat frequency = |f1 – f

2| = 11 – 9 = 2 Hz

Gravitation

3. A spaceship orbits around a planet at a height of 20 km from its surface. Assuming that only gravitational field

of the planet acts on the spaceship, what will be the number of complete revolutions made by the spaceship in

24 hours around the planet?

[Given: Mass of planet = 8 × 1022 kg, Radius of planet = 2 × 106 m, Gravitational constant G = 6.67 × 10–11

Nm2/kg2]

,d xzg dh lrg ls 20 km Å¡pkbZ ij ,d varfj{k;ku xzg ds ifjr% d{kk esa ?kwe jgk gSA ;fn ;ku ij flQZ xzg dk xq:Roh; {ks=k

izHkkoh gS rks ;ku }kjk 24 hrs esa yxk;s x;s iwjs pDdjksa dh la[;k dk eku gksxk :

[fn;k gS, xzg dk nzO;eku = 8 × 1022 kg, xzg dh f=kT;k = 2 × 106 m, xq:Roh; fu;rkad G = 6.67 × 10–11 Nm2/kg2]

(1) 17 (2) 13 (3) 11 (4) 9

A. 3

sol.2 r GM

T , vv r

3r rT 2 r 2

GM GM

3 12

11 22

(202) 10T 2 sec

6.67 10 8 10

T = 7812.2 s

T 2.17hr 11 revolvtions

Kinematics

4. A plane is inclined at an angle = 30° with respect to the horizontal. A particle is projected with a speed u =

2 ms–1, from the base of the plane, making an angle = 15° with respect to the plane as shown in the figure. The

distance from the base, at which the particle hits the plane is close to : (Take g = 10 ms–2)

,d lery {kSfrt ls = 30° dk dks.k cukrk gSA ,d d.k dks bl lery ds vk/kkj ls xfr u = 2 ms–1 ls lery ls = 15°

ds dks.k ij fp=kkuqlkj iz{ksfir fd;k tkrk gSA ml fcUnq] tgk¡ d.k lery ij fxjrk gS] dh vk/kkj ls nwjh dk lfUudV eku gksxk:

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(g = 10 ms–2 yhft,A)

(1) 18 cm (2) 20 cm (3) 14 cm (4) 26 cm

A. 2

sol. Time of flight 2u sin

(T)g cos

(2)(2sin15) 4 sin15T

g cos30 10 cos30

Range (R) = (2cos15)T–1

2g sin 30(T)2

2

2

4 sin15 1 16 sin 15(2cos15) 10sin 30

10 cos30 3 100 cos 30

16 3 160.1952m 20 cm

60

NLM

5. A bullet of mass 20 g has an initial speed of 1 ms–1, just before it starts penetrating a mud wall of thickness 20

cm. If the wall offers a mean resistance of 2.5 × 10–2 N, the speed of the bullet after emerging from the other

side of the wall is close to :

20 cm eksVkbZ dh feV~Vh dh nhokj Hksnus ls Bhd igys 20 g nzO;eku dh ,d xksyh dh pky 1 ms–1 gSA ;fn nhokj 2.5 × 10–2 N

dk vkSlr vojks/k yxkrh gS] rks nhokj ds nwljs rjQ ls fuxZr xksyh dh pky dk lfUudV eku gksxk :

(1) 0.4 ms–1 (2) 0.7 ms–1 (3) 0.3 ms–1 (4) 0.1 ms–1

A. 2

sol. v2 = u2 – 2aS

22 2

3

2.5 10 20v (1) (2)

20 10 100

2 1v 1

2

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1v m/ s 0.7 m/ s

2

Rotation

6. The time dependence of the position of a particle of mass m = 2 is given by 2ˆ ˆr(t) 2 ti 3 t j

. Its angular

momentum, with respect to the origin, at time t = 2 is :

nzO;eku m = 2 ds ,d d.k dh fLFkfr] le; (t) ds vuqlkj] 2ˆ ˆr(t) 2 ti 3 t j

gSA bl d.k dk ewy fcUnq ds lkis{k t = 2 ij

dks.kh; laosx gksxk :

(1) ˆ ˆ34(k i) (2) ˆ ˆ48(i j) (3) ˆ36k (4) ˆ48k

A. 4

sol. 2ˆ ˆr 2ti – 3t j

dr ˆ ˆv 2i – 6tjdt

dv ˆa –6jdt

ˆF ma 12 j

ˆ ˆr (at t 2) 4i –12 j

ˆ ˆ ˆ ˆ ˆL m(r v) 2(4i –12 j) (2i –12 j) 48k

Semiconductors

7. The figure represents a voltage regulator circuit using a Zener diode. The breakdown voltage of the Zener

diode is 6 V and the load resistance is, RL = 4 k. The series resistance of the circuit is R

i = 1 k. If the battery

voltage VB varies from 8 V to 16 V, what are the minimum and maximum values of the current through Zener

diode?

fp=k esa T+ksuj Mk;ksM ls cuk;k gqvk oksYVst fu;a=k.k ifjiFk fn[kk;k x;k gSA T+ksuj Mk;ksM dh Hkatu oksYVrk 6 V rFkk yksM

izfrjks/k RL = 4 kgSA Js.kh izfrjks/k R

i = 1 kgSA ;fn lsy dk foHko V

B, 8 V ls 16 V ds chp cnyrk gS rks T+ksuj Mk;ksM

dh /kkjk ds U;wure rFkk vf/kdre eku D;k gksaxs\

(1) 0.5 mA; 6 mA (2) 0.5 mA; 8.5 mA (3) 1.5 mA; 8.5 mA (4) 1 mA; 8.5 mA

A. 2

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sol.

1

3 1I 8 6 0.5 mA

2 2

2

3I 16 6 8.5 mA

2

Rotation

8. A metal coin of mass 5 g and radius 1 cm is fixed to a thin stick AB of negligible mass as shown in the figure.

The system is initially at rest. The constant torque, that will make the system rotate about AB at 25 rotations per

second in 5 s, is closed to :

5 g nzO;eku rFkk 1 cm f=kT;k ds /kkrq ds ,d flDds dks ,d iryh ux.; nzO;eku dh NM+ AB ls fp=kkuqlkj tksM+k tkrk gSA

;g fudk; vkjEHk esa fLFkjkoLFkk esa gSA bls AB ds ifjr% 5 s rd 25 pDdj izfr lsd.M dh xfr ls ?kqekus ds fy, fu;r cy

vk?kw.kZ dk lfUudV eku gksxk :

(1) 4.0 × 10–6 Nm (2) 7.9 × 10–6 Nm (3) 2.0 × 10–5 Nm (4) 1.6 × 10–5 Nm

A. 3

sol. = 0 = t

25 × 2 = ()5 = I

25

mR4

3 45

(5 10 )(10 )104

= 2.0 × 10–5 Nm

Wave Optics

9. In a Young’s double slit experiment, the ratio of the slit’s width is 4 : 1. The ratio of the intensity of maxima to

minima, close to the central fringe on the screen, will be :

;ax ds ,d f}f>jh iz;ksx esa fLyV dh pkSM+kb;ksa dk vuqikr 4 : 1 gSA LØhu ij dsUnzh; fÝUt ds fudV ns[kh x;h mPpre rFkk

U;wure izdk'k rhozrk dk vuqikr gksxk :

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(1) 4

3 1 :16 (2) 4 : 1 (3) 25 : 9 (4) 9 : 1

A. 4

sol. I1 = 4I

0

I2 = I

0

2

1 2max

2

min1 2

I II 9

I 1I I

Nuclear Physics

10. Two radioactive substances A and B have decay constants 5 and respectively. At t = 0, a sample has the

same number of the two nuclei. The time taken for the ratio of the number of nuclei to become 2

1

e

will be :

nks jsfM;ks/kehZ inkFkksZa A rFkk B ds {k; fu;rkad] Øe'k%] 5 rFkk gSaA t = 0 ij ,d uewus esa bu nks ukfHkdksa dh cjkcj la[;k

gSA ukfHkdksa dh la[;k dk vuqikr 2

1

e

gksus esa yxs le; dk eku gksxk :

(1) 1

2(2)

1

4(3)

1

(4)

2

A. 1

sol. Nx(at t) = N

0e–5t

Ny(at t) = N

0e–t

4 tx2

y

N 1e

N e

4 t 2

2 1t

4 2

Fluid Mechanics

11. A cubical block of side 0.5 m floats on water with 30% of its volume under water. What is the maximum weight

that can be put on the block without fully submerging it under water? [Take, density of water = 103 kg/m3]

0.5 m Hkqtk yEckbZ dk ,d ?kukdkj xqVdk ikuh esa rSjrk gS ftlls mldk 30% vk;ru ikuh esa Mwck gSA bl xqVds ds Åij vf/

kdre fdruk Hkkj] xqVds dks fcuk iwjh rjg Mqck;s] j[kk tk ldrk gS\ [fn;k gS : ikuh dk ?kuRo = 103 kg/m3]

(1) 30.1 kg (2) 46.3 kg (3) 87.5 kg (4) 65.4 kg

A. 3

sol. Given (50)3 × 30

100 × (1) × g = M

cubeg ......(i)

Let m mass should be placedHence (50)3 × (1) × g = (M

cube + m)g ......(ii)

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equation (ii) – equation (i)

mg = (50)3 × g(1 – 0.3) = 125 × 0.7 × 103 g

m = 87.5 kg

Geometrical Optics

12. The graph shows how the magnification m produced by a thin lens varies with image distance v. What is the

focal length of the lens used?

fn;s x;s xzkQ esa ,d irys ysal ds vko/kZu m dks izfrfcEc dh nwjh v ds lkFk n'kkZ;k x;k gSA bl ysal dh Qksdl nwjh D;k gksxh\

(1) 2b

ac(2)

2b c

a(3)

a

c(4)

b

c

A. 4

sol. As the graph between magnification (m) and image distance (v) varies linearly, then

m = k1v + k

2

1 2

vk v k

u

2

1

k1k

u v

2

1

k 1k

v u

Clearly, k1 =

1

f and k

2 = 1 here

1 bf

slope of m v graph c

Atomic Structure

13. In Li++, electron in first Bohr orbit is excited to a level by a radiation of wavelength . When the ion gets

deexcited to the ground state in all possible ways (including intermediateemissions), a total of six spectral lines

are observed. What is the value of ? (Given: h = 6.63 × 10–34 Js; c = 3 × 108 ms–1)

Li + + vk;u esa bysDVªkWu dks mldh izFke cksgj d{kk ls rjaxnS/;Z ds fofdj.k ls ,d Å¡ph d{kk esa mÙksftr dj fn;k tkrk gSA

tc ;g vk;u viuh U;wure ÅtkZ voLFkk esa lHkh laHko rjhdksa (e/;orhZ mRltZuksa dks feykdj) ls vkrk gS rks dqy 6 LisDVªy

ykbusa ik;h tkrh gSaA dk eku D;k gksxk? (fn;k gS : h = 6.63 × 10–34 Js; c = 3 × 108 ms–1)

(1) 11.4 nm (2) 12.3 nm (3) 9.4 nm (4) 10.8 nm

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A. 4

sol.hc

E

2 1(13.4)(3) 1 eV

16

1242 16nm 10.8 nm

(13.4) (9)(15)

Friction

14. Two blocks A and B of masses mA = 1 kg and m

B = 3 kg are kept on the table as shown in figure. The

coefficient of friction between A and B is 0.2 and between B and the surface of the table is also 0.2. The

maximum force F that can be applied on B horizontally, so that the block A does not slide over the block B is:

[Take g = 10 m/s2]

nzO;eku mA = 1 kg rFkk m

B = 3 kg ds nks xqVdksa, A rFkk B, dks fp=kkuqlkj ,d est ij j[kk x;k gSA A rFkk B ds chp ?k"kZ.k

xq.kkad 0.2 ,oa B rFkk est ds chp Hkh ?k"kZ.k xq.kkad 0.2 gSA xqVds B ij yxk;s x;s {kSfrt cy F dk vf/kdre eku, ftlls xqVdk

A xqVdk B ds Åij ugha fQlys] gksxk : [fn;k gS, g = 10 m/s2]

(1) 40 N (2) 12 N (3) 16 N (4) 8 N

A. 3

sol. c

F fa

M m

F (0.2)4 10 F 8a

4 4

We have F 8

(0.2)104

F – 8 8

F 16

Electrostatics

15. In free space, a particle A of charge 1 C is held fixed at a point P. Another particle B of the same charge and

mass 4 g is kept at a distance of 1 mm from P. If B is released, then its velocity at a distance of 9 mm from P

is : 9 2 2

0

1Take 9 10 Nm C

4

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fuokZr esa ,d 1 C vkos'k ds ,d d.k A dks fcUnq P ij n`<+ j[kk gSA mlh vkos'k rFkk 4 g nzO;eku ds nwljs d.k B dks P ls

1 mm nwjh ij j[kk gSA B dks NksM+us ij P ls 9 mm nwjh ij mldh xfr dk eku gksxk :

9 2 2

0

19 10 Nm C

4

fn;k gS

(1) 2.0 × 103 m/s (2) 3.0 × 104 m/s (3) 1.5 × 102 m/s (4) 1.0 m/s

A. Bonus

sol.

Conservation of energy

21 2 1 2

1 2

kq q kq q 1mv

r r 2

2 1 2

1 2

2kq q 1 1v

m r r

9 129

9 3

2 9 10 10 11 4 10

4 10 10 9

v = 40 × 104 m/s = 6.32 × 104 m/s

Fluid Mechanics

16. Water from a tap emerges vertically downwards with an initial speed of 1.0 ms–1. The cross-sectional area of

the tap is 10–4 m2. Assume that the pressure is constant throughout the stream of water and that the flow is

streamlined. The cross-sectional area of stream, 0.15 m below the tap would be: (Take g = 10 ms–2)

,d uy ls ikuh Å/okZ/kj uhps dh vksj 1.0 ms–1 dh vkjfEHkd xfr ls fudyrk gSA uy ds vuqizLFk dkV dk {ks=kQy 10–4 m2

gSA ikuh dh /kkjk esa nkc dks fu;r rFkk cgko dks /kkjkjs[kh; ekfu;sA uy ls 0.15 m uhps /kkjk dk vuqizLFk dkV dk {ks=kQy

gksxk: (g = 10 ms–2 yhft,)

(1) 1 × 10–5 m2 (2) 5 × 10–5 m2 (3) 5 × 10–4 m2 (4) 2 × 10–5 m2

A. 2

sol. Using Bernoullie’s equation 22 1v v 2gh

Equation of continuityA

1V

1 = A

2V

2

(1 cm2)(1m/s) = (A2) 2 15

(1) 2 10100

2

2

1A (ln cm )

2

5 22A 5 10 m

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KTG & Thermodynamics

17. When heat Q is supplied to a diatomic gas of rigid molecules, at constant volume its temperature increases by

T. The heat required to produce the same change in temperature, at a constant pressure is:

n`<+ v.kqvksa okyh ,d f}ijek.kqd xSl dks tc Q Å"ek fu;r vk;ru ij nh tkrh gS rks mlds rkieku esa T dh o`f) gksrh gSA

blh rkieku o`f) dks fu;r nkc ij lqfuf'pr djus ds fy;s vko';d Å"ek gksxh :

(1) 3

Q2

(2) 2

Q3

(3) 7

Q5

(4) 5

Q3

A. 3

sol. Heat supplied at constant volume

Q = nCVT

and heat supplied at constant pressureQ

1 = nC

pT

p1

v

CQ

Q C

1

7Q (Q)

5

Electrostatics

18. A simple pendulum of length L is placed between the plates of a parallel plate capacitor having electric field E,

as shown in figure. Its bob has mass m and charge q. The time period of the pendulum is given by:

L yEckbZ ds ,d ljy nksyd dks fp=kkuqlkj ,d lekUrj IysV la/kkfj=k ds e/;] ftlesa fo|qr {ks=k E gS] esa j[kk gSA blds yksyd

dk nzO;eku m rFkk vkos'k q gSA bl nksyd dk vkorZdky gksxk :

(1) L

2qE

gm

(2) L

2qE

gm

(3) 2

2

L2

qEg

m

(4) 2 2

2

2

L2

q Eg

m

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A. 3

sol.eff

Lt 2

g

2

2eff

qEg g

m

2

2

Lt 2

qEg

m

Sound Waves

19. A source of sound S is moving with a velocity of 50 m/s towards a stationary observer. The observer measures

the frequency of the source as 1000 Hz. What will be the apparent frequency of the source when it is moving

away from the observer after crosssing him? (Take velocity of sound in air is 350 m/s)

,d /ofu L=kksr S 50 m/s dh xfr ls ,d fLFkj Jksrk dh rjQ c<+ jgk gSA Jksrk dks /ofu dh vko`fÙk 1000 Hz lqukbZ nsrh gSA

tc L=kksr mlh xfr ls Jksrk dks ikj djds mlls nwj tkrk gS rks Jksrk }kjk lquh x;h /ofu dh vko`fÙk dk eku gksxk :

(ekuk ok;q esa /ofu dh xfr = 350 m/s)

(1) 857 Hz (2) 1143 Hz (3) 807 Hz (4) 750 Hz

A. 4

sol.0

app act

s

V Vf f

V V

act

350 01000 f

350 ( 50)

and act

350f ' f

350 50

act

1000 300f

400

actf 750Hz

Dual Nature of Radiation & Matter

20. A 2 mW laser operates at a wavelength of 500 nm. The number of photons that will be emitted per second is:

[Given Planck’s constant h = 6.6 × 10–34 Js, speed of light c = 3.0 × 108 m/s]

,d 2 mW ysT+kj dh rjaxnS/;Z 500 nm gSA blls fudyus okys izfr lsd.M QksVkWuksa dh la[;k gksxh :

[fn;k gS, Iykad fu;rkad h = 6.6 × 10–34 Js, izdk'k dh pky c = 3.0 × 108 m/s]

(1) 2 × 1016 (2) 1.5 × 1016 (3) 1 × 1016 (4) 5 × 1015

A. 4

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sol.hc

E

Let no. of photons per sec. is N

hcN 2mW

3

19

2 2 5000 10N

hC 12420 1.6 10

N = 5 × 1015

Current Electricity

21. Space between two concentric conducting spheres of radii a and b (b > a) is filled with a medium of resistivity

. The resistance between the two spheres will be:

f=kT;kvksa a rFkk b (b > a) ds nks ledsUnzh; pkyd xksyksa ds chp ,d izfrjks/kdrk dk inkFkZ Hkj fn;k tkrk gSA bu xksyksa ds

chp izfrjks/k dk eku gksxk :

(1) 1 1

4 a b

(2)

1 1

2 a b

(3)

1 1

2 a b

(4)

1 1

4 a b

A. 1

sol.

2

( )(dx)dR

4 x

R dR b

2

a

dx

4 x

1 1.

4 a b

Elasticity

22. In an experiment, brass and steel wires of length 1 m each with areas of cross section 1 mm2 are used. The

wires are connected in series and one end of the combined wire is connected to a rigid support and other end

is subjected to elongation. The stress required to produce a net elongation of 0.2 mm is,

[Given, the Young’s Modulus for steel and brass are, respectively, 120 × 109 N/m2 and 60 × 109 N/m2]

,d iz;ksx esa, ihry rFkk LVhy ds nks rkjksa dk iz;ksx fd;k x;k gS ftuesa izR;sd dh yEckbZ 1 m rFkk vuqizLFk dkV dk {ks=kQy

1 mm2 gSaA bu rkjksa dks Js.khØe esa tksM+rs gSa rFkk la;qDr rkj ds ,d fljs dks n`<+ LrEHk ls tksM+rs gSa ,oa nwljs fljs dks [khapk tkrk

gSA 0.2 mm dh dqy o`f) ds fy;s izfrcy dk eku gksxk :

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[fn;k gS, LVhy rFkk ihry ds ;ax izR;kLFkrk xq.kkad, Øe'k% 120 × 109 N/m2 rFkk 60 × 109 N/m2 gSa]

(1) 1.8 × 106 N/m2 (2) 1.2 × 106 N/m2 (3) 4.0 × 106 N/m2 (4) 0.2 × 106 N/m2

A. Bonus

sol.

Corresponding to the stress ()

Total elongation 1 2

net

1 2

L L

Y Y

l

1 2

1 2

Y Y

Y Y

l

3 9120 600.2 10 10

180

6

2

N8 10

m (Answer is not matching)

EMI

23. A coil of self inductance 10 mH and resistance 0.1 is connected through a switch to a battery of internal

resistance 0.9 . After the switch is closed, the time taken for the current to attain 80% of the saturation value

is: [take ln5 = 1.6]

10 mH LoizsjdRo ,oa 0.1 izfrjks/k dh ,d dqaMyh dks ,d dqath ds lkFk ,d 0.9 vkarfjd izfrjks/k ds lsy ls tksM+rs gSaA

dqath dks can djus ds i'pkr~ bl ifjiFk esa /kkjk dk eku lar`Ir /kkjk ds 80% gksus esa yxk le; gksxk : [fn;k gS : ln5 = 1.6]

(1) 0.002 s (2) 0.324 s (3) 0.103 s (4) 0.016 s

A. 4

sol.

Rt

LsatI I 1 e

Here R = RL = r = 1

t

.01sat sat0.8I I 1 e

100t4

1 e5

100t 1

e5

100t = ln5

1

t ln 5100

= 0.016 sec

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Magnetic Field & Force

24. The magnitude of the magnetic field at the center of an equilateral triangular loop of side 1 m which is

carrying a current of 10 A is: [Take 0 = 4 × 10–7 NA–2]

1 m Hkqtk okys ,d leckgq f=kHkqtkdkj oy; esa 10 A /kkjk izokfgr gksrh gSA blds dsUnz ij pqEcdh; {ks=k ds ifjek.k dk eku

gksxk : [0 = 4 × 10–7 NA–2 yhft,]

(1) 18 T (2) 1 T (3) 3 T (4) 9 T

A. 1

sol.

1d (a sin 60)

3

a 3 ad

3 2 2 3

00

IB 3 (sin 60 sin 60)

4 d

0

0

7

3 I

a4

2 3

I9

2 a

9 2 10 10

1

18 T

Magnetic Field & Force

25. A square loop is carrying a steady current I and the magnitude of its magnetic dipole moment is m. If this square

loop is changed to a circular loop and it carries the same current, the magnitude of the magnetic dipole moment

of circular loop will be :

,d oxkZdkj oy; esa /kkjk I izokfgr djus ij blds pqEcdh; f}/kqzo vk?kw.kZ dk ifjek.k m gksrk gSA ;fn bl oxkZdkj oy; dks

eksM+dj ,d o`Ùkkdkj oy; esa ifjofrZr fd;k tk;s vkSj mlesa ogh /kkjk izokfgr dh tk, rks bl o`Ùkkdkj oy; ds pqEcdh;

f}/kqzo vk?kw.kZ dk ifjek.k gksxk :

(1) 2m

(2)

4m

(3)

m

(4)

3m

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A. 2

Sol.

2r = 4a 2a

r

m = (I) a2

m1 = (I) r2

m1=(I)()

2

2

4a

2

1

4Iam

1

4mm

Fluid Mechanics

26. A submarine experiences a pressure of 5.05 × 106 Pa at a depth of d1 in a sea. When it goes further to a depth

of d2, it experiences a pressure of 8.08 × 106 Pa. Then d

2 – d

1 is approximately (density of water = 103 kg/m3

and acceleration due to gravity = 10 ms–2) :

leqnz esa d1 xgjkbZ ij ,d iuMqCch 5.05 × 106 Pa dk nkc vuqHko djrh gSA tc ;g iuMqCch vkSj xgjkbZ d

2 ij tkrh gS, rks

8.08 × 106 Pa dk nkc vuqHko djrh gS] rc d2 – d

1 dk fudVre eku gksxk

(fn;k gS : ikuh dk ?kuRo = 103 kg/m3 rFkk xq:Roh; Roj.k = 10 ms–2) :

(1) 600 m (2) 500 m (3) 300 m (4) 400 m

A. 3

sol. P = P2 – P

1 = gH

3.03 × 106 = 103 × 10 × H

H 300 m

Elasticity

27. The elastic limit of brass is 379 MPa. What should be the minimum diameter of a brass rod if it is to support a

400 N load without exceeding its elastic limit?

ihry dh izR;kLFkrk lhek 379 MPa gSA 400 N cy dks fcuk izR;kLFkrk lhek ikj fd;s lg ldus okyh ihry dh NM+ dk

U;wure O;kl D;k gksxk\

(1) 0.90 mm (2) 1.16 mm (3) 1.00 mm (4) 1.36 mm

A. 2

sol. Stress 6 2

2

400379 10 N / m

r

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2

6

400r

379 10

2r 1.15 mm

Dual Nature of Radiation & Matter

28. Light is incident normally on a completely absorbing surface with an energy flux of 25 Wcm–2. If the surface has

an area of 25 cm2, the momentum transferred to the surface in 40 min time duration will be :

,d lEiw.kZ vo'kks"kd i`"B ij 25 Wcm–2 ÅtkZ izokg (flux) dk izdk'k yEcor~ vkifrr gksrk gSA ;fn i`"B dk {ks=kQy 25 cm2

gS rks 40 min le;kUrjky esa ml ij gqvk laosx varj.k (transfer) gksxk :

(1) 3.5 × 10–6 Ns (2) 6.3 × 10–4 Ns (3) 5.0 × 10–3 Ns (4) 1.4 × 10–6 Ns

A. 3

sol.E

PC

8

(25 25) 40 60

3 10

N-s

= 5 × 10–3 N-s

Rotation

29. A solid sphere of mass M and radius R is divided into two unequal parts. The first part has a mass of 7M

8 and

is converted into a uniform disc of radius 2R. The second part is converted into a uniform solid sphere. Let I1

be the moment of inertia of the disc about its axis and I2 be the moment of inertia of the new sphere about its

axis. The ratio I1/I

2 is given by :

nzO;eku M rFkk f=kT;k R ds ,d Bksl xksys dks nks vleku fgLlksa esa ck¡Vk tkrk gSA 7M

8 nzO;eku ds igys fgLls dks ,d 2R

f=kT;k dh ,dleku fMLd esa cnyk tkrk gSA cps gq;s fgLls ls ,d ,dleku Bksl xksyk cuk;k tkrk gSA ekukfd I1 fMLd dk

mldh v{k ds ifjr% tM+Ro vk?kw.kZ gS rFkk I2 u;s xksys dk mlds v{k ds ifjr% tM+Ro vk?kw.kZ gSA vuqikr I

1/I

2 gksxk :

(1) 140 (2) 185 (3) 285 (4) 65

A. 1

sol.

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2 21

7M 1 14I (2 R) MR

8 2 8

22

2 MI r

5 8

2 2

2

2 M R MRI

5 8 4 80

3

We have

4 Mr

3 8

Rr

2

1

2

I 14 80140

I (8)

Units & Dimensions

30. In the formula X = 5YZ2, X and Z have dimensions of capacitance and magnetic field, respectively. What are

the dimensions of Y in SI units ?

lw=k X = 5YZ2 esa X rFkk Z dh foek;sa] Øe'k%] /kkfjrk rFkk pqEcdh; {ks=k gSaA SI bdkbZ esa Y dh foek D;k gksxh\

(1) [M–2L–2T6A3] (2) [M–1L–2T4A2] (3) [M–2L0T–4A–2] (4) [M–3L–2T8A4]

A. 4

sol. X = 5YZ2

2

XY

Z

2 2 2

2 2

Q [A T ]X C

E [ML T ]

X = [M–1L–2T4A2]

FZ B

IL

Z = [MT–2A–1]

1 2 4 2

2 1 2

[M L T A ]Y

[MT A ]

Y = [M–3L–2T8A4]

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CHEMISTRY

10 APRIL 2019 [Phase : II]

JEE MAIN PAPER ONLINE

1. The correct statement is :

(1) Zincite is a carbonate ore.

(2) Zone refining process is used for the refining of titanium.

(3) Aniline is a froth stabilizer.

(4) Sodium cyanide cannot be used in the metallurgy of silver.

lgh dFku gS :

(1) ftalkbV ,d dkcksZusV v;Ld gSA

(2) tksu ifj"dj.k izØe VkbVsfu;e ds ifj"dj.k ds fy, iz;qDr gksrk gSA

(3) ,fuyhu ,d Qsu&LFkk;hdkjd gSA

(4) lksfM;e lk;ukbM dk mi;ksx flYoj (pk¡nh) ds /kkrqdeZ esa ugha dj ldrs gSaA

A. 3

sol. Ti is refined by Van Arkel method. Ag is leached by dilute solution of NaCN. Zincite is ZnO. Aniline is a froth

stabilizer.

2. The pH of a 0.02 M NH4Cl solution will be [given K

b(NH

4OH) = 10–5 and log2 = 0.301]

0.02 M NH4Cl foy;u dk pH gksxk : [fn;k x;k gS : K

b(NH

4OH) = 10–5 rFkk log2 = 0.301]

(1) 2.65 (2) 5.35 (3) 4.35 (4) 4.65

A. 2

sol. NH+4 + H

2O NH

4OH + H+

0.02 – x x x 14

9h 5

10K 10

10

0.02

2

h

xK

0.02

10–9 × 2 × 10–2 = x2

6x 20 10

6pH log 20 10

pH = 5.35

3. The INCORRECT statement is :

(1) The spin-only magnetic moments of [Fe(H2O)

6]2+ and [Cr(H

2O)

6]2+ are nearly similar.

(2) The gemstone, ruby, has Cr3+ ions occupying the octahedral sites of beryl.

(3) The spin-only magnetic moment of [Ni(NH3)

4(H

2O)

2]2+ is 2.83 BM.

(4) The color of [CoCl(NH3)

5]2+ is violet as it absorbs the yellow light.

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xyr dFku gS :

(1) [Fe(H2O)

6]2+ rFkk [Cr(H

2O)

6]2+ ds pØ.k &pqEcdh; vk?kw.kZ yxHkx ,d tSls gSaA

(2) tseLVku] :ch] esa Cr3+ vk;u gksrk gS tks csfjy ds v"VQydh; LFky esa mifLFkr jgrk gSA

(3) [Ni(NH3)

4(H

2O)

2]2+ dk pØ.k &pqEcdh; vk?kw.kZ 2.83 BM gSA

(4) tc [CoCl(NH3)

5]2+ ihyk izdk'k 'kksf"kr djrk gS rks bldk jax cSaxuh gks tkrk gSA

A. 2

sol. Ruby is aluminium oxide (Al2O

3) containing about 0.5 – 1% Cr3+ ions which are randomly distributed in the

position normally occupied by Al3+ ions.

4. Number of stereo centers present in linear and cyclic structures of glucose are respectively :

XyqdksT+k ds jSf[kd rFkk pØh; lajpukvksa esa mifLFkr f=kfoe dsUnzksa dh la[;k Øe'k% gksxh %

(1) 5 & 5 (2) 4 & 4 (3) 5 & 4 (4) 4 & 5

A. 4

sol.

CHO

H

HO

H

OH

H

OH

H OH

CH OH2

*

*

*

*

4 stereogenic centres

H H

H H

OH

OH

OH*

*

O

OH

H*

*

*

CH OH2

5 sterogenic centres

5. The difference between H and U (H – U), when the combustion of one mole of heptane( ) is carried

out at a temperature T, is equal to :

tc ,d eksy gsIVsu (l) dk ngu T rki ij fd;k tkrk gS rks H rFkk U dk vUrj (H – U), fuEu ds cjkcj gksxk :

(1) –3RT (2) 4RT (3) 3RT (4) –4RT

A. 4

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sol. 7 16 2 2 2(g) (g) ( )( )

C H 11O 7CO 8H O

H – U = ng RT

ng = –4

H – U = –4RT

6. Which one of the following graphs between molar conductivity m versus C is correct?

eksyj pkydrk m rFkk C ds chp cus xzkQksa esa dkSulk lgh gS\

(1) NaCl

KCl

C

m (2)

NaCl

KC

l

C

m (3)

NaClKCl

C

m(4)

NaClKCl

C

m

A. 3

sol. KCl is more conducting than NaCl

NaCl

KCl

C

m

7. The crystal field stabilization energy (CFSE) of [Fe(H2O)

6]Cl

2 and K

2[NiCl

4], respectively, are :

[Fe(H2O)

6]Cl

2 rFkk K

2[NiCl

4] dh fØLVy {ks=k LFkk;hdj.k ÅtkZ (CFSE) Øe'k% gSa :

(1) –2.40 and –1.2

t(2) –0.6

0 and –0.8

t

(3) –0.40 and –0.8

t(4) –0.4

0 and –1.2

t

A. 3

sol. [Fe(H2O)

6]2+ t

2g4 e

g2 CFSE = –0.4

0

[NiCl4]2– e4 t

24 CFSE = –0.8

t

8. For the reaction,

2SO2(g) + O

2(g) 2SO

3(g),

H = –57.2 kJ mol–1 and Kc = 1.7 × 1016.

Which of the following statement is INCORRECT?

(1) The equilibrium constant is large suggestive of reaction going to completion and so no catalyst is required.

(2) The addition of inert gas at constant volume will not affect the equilibrium constant.

(3) The equilibrium will shift in forward direction as the pressure increases.

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(4) The equilibrium constant decreases as the temperature increases.

vfHkfØ;k 2SO2(g) + O

2(g) 2SO

3(g) ds fy, H = –57.2 kJ mol–1 rFkk K

c = 1.7 × 1016.

fuEu esa ls dkSulk dFku xyr gS\

(1) lkE; fLFkjkad cM+k gksuk crkrk gS fd vfHkfØ;k iw.kZrk dks tk jgh gS vkSj mRizsjd dh vko';drk ugha gSA

(2) fLFkj vk;ru ij, fuf"Ø; xSl ds feykus ij lkE; fLFkjkad izHkkfor ugha gksxkA

(3) tc nkc c<+rk gS rks lkE; vxz fn'kk esa foLFkkfir gksrh gSA

(4) tc rki c<+rk gS rks lkE; fLFkjkad ?kVrk gSA

A. 1

sol. 2 2 32SO (g) O (g) 2SO (g)

Kc = 1.7 × 1016 i.e. reaction goes to completion.

Equilibrium constant has no relation with catalyst. Catalyst only affects the rate with which a reaction proceeds.For the given reaction, catalyst V

2O

5 is used to speed up the reaction (Contact process).

9. A hydrated solid X on heating initially gives a monohydrated compound Y. Y upon heating above 373 K

leads to an anhydrous white powder Z. X and Z, respectively are :

(1) Baking soda and dead burnt plaster.

(2) Baking soda and soda ash.

(3) Washing soda and soda ash.

(4) Washing soda and dead burnt plaster.

,d ty;ksftr Bksl X xeZ djus ij izkjEHk esa ,d ,dy&ty;ksftr ;kSfxd Y nsrk gSA 373 K ds Åij Y dks xeZ djus ij

,d futZy lQsn ikmMj Z feyrk gSA X rFkk Z Øe'k% gSa :

(1) csfdax lksMk rFkk iw.kZnX/k IykLVj

(2) csfdax lksMk rFkk lksMk ,s'k

(3) okf'kad lksMk rFkk lksMk ,s'k

(4) okf'kax lksMk rFkk iw.kZnX/k IykLVj

A. 3

sol. 2 3 2 2 3 2 2(X) (Y)

Na CO .10H O Na CO .H O 9H O

2 3 2 2 3 2(Y) (Z)

Na CO .H O Na CO H O

X = Washing soda

Z = Soda ash

10. Which of these factors does not govern the stability of a conformation in acyclic compounds?

(1) Angle strain (2) Steric interactions

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(3) Electrostatic forces of interaction (4) Torsional strain

vpØh; ;kSfxdksa esa buesa dkSulk dkjd la:i.kksa ds LFkkf;Ro ds fy;s ugha ykxw gksxk\

(1) dks.kh; fod`fr (2) f=kfoeh vU;ksU;fØ;k

(3) vU;ksU;fØ;k dk fLFkj oS|qr cy (4) ejksM+h fod`fr

A. 1

sol. Angle strain is not present in acyclic compounds.

11. The highest possible oxidation states of uranium and plutonium, respectively, are :

;wjsfu;e rFkk IyqVksfu;e dh mPpre lEHko vkWDlhdj.k voLFkk;sa Øe'k% gSa :

(1) 6 and 7 (2) 7 and 6 (3) 6 and 4 (4) 4 and 6

A. 1

sol. Maximum oxidation state shown by

Uranium = + 6

Plutonium = + 7

12. The correct option among the following is :

(1) Colloidal medicines are more effective because they have small surface area.

(2) Colloidal particles in lyophobic sols can be precipitated by electrophoresis.

(3) Brownian motion in colloidal solution is faster if the viscosity of the solution is very high.

(4) Addition of alum to water makes it unfit for drinking.

fuEu esa ls lgh fodYi gS :

(1) dksykbMh vkS"kf/k;k¡ T;knk izHkko'kkyh gSa D;ksafd mudk i`"Bh; {ks=kQy NksVk gksrk gSA

(2) nzofojkxh lkWy esa dksykbMh d.k oS|qr d.k lapyu }kjk vo{ksfir fd;s tk ldrs gSaA

(3) dksykbMh foy;u esa ;fn foy;u dh ';kurk cgqr T;knk gS rks czkmfu;u xfr rhozrj gksrh gSA

(4) ikuh esa fQVfdjh feykus ls og (ikuh) ihus ds v;ksX; gks tkrk gSA

A. 2

sol. Electrophoresis is used to coagulate lyophobic colloids.

13. The major product obtained in the given reaction is :

nh xbZ vfHkfØ;k esa izkIr eq[; mRikn gS %

CH3 CH2 CHCH2

CH3

Cl

OAlCl3 Product mRikn

(1)

H3C O

CH3

(2)

H3C CHCHCH2

CH3O

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(3)

H3C CH2 CHCH2 CH2

O

(4) CH3

CH3

O

A. 1

sol.

O

O

AlCl3

AlCl3Cl :

14. The number of pentagons in C60

and trigons (triangles) in white phosphorus, respectively, are :

C60

esa iapHkqtksa rFkk lQsn QkLQksjl esa f=kHkqtksa (f=kdks.kksa) dh la[;k Øe'k% gSa :

(1) 20 and 3 (2) 12 and 3 (3) 12 and 4 (4) 20 and 4

A. 3

sol. Pentagons in C60

= 12

Triangles in P4 = 4

15. For the reaction of H2 with I

2, the rate constant is 2.5 × 10–4 dm3 mol–1 s–1 at 327°C and 1.0 dm3 mol–1 s–1 at

527°C. The activation energy for the reaction, in kJ mol–1 is : (R = 8.314 J K–1 mol–1)

I2 ds lkFk H

2 dh vfHkfØ;k ds fy;s nj fu;rkad 327°C ij 2.5 × 10–4 dm3 mol–1 s–1 rFkk 527°C ij 1.0 dm3 mol–1 s–1

gSA vfHkfØ;k dh lfØ;.k ÅtkZ (kJ mol–1 esa) gksxh : (R = 8.314 J K–1 mol–1)

(1) 150 (2) 59 (3) 72 (4) 166

A. 4

sol.a2

1 1 2

EK 1 1log

K 2.303R T T

a4

E1log

2.5 10 8.314 2.303

aE 2003.6

8.314 2.303 600 800

Ea = 165.4 kJ/mol

166 kJ/mol

16. The major product ‘Y’ in the following reaction is :

fuEu vfHkfØ;k esa eq[; mRikn ‘Y’ gS &

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Cl EtONaHeat

HBrX Y

(1) HO

(2) Br

(3) Br

(4) Br

A. 3

sol.

Br

HBr

Cl

NaOEt

17. The major product ‘Y’ in the following reaction is :

fuEu vfHkfØ;k esa eq[; mRikn ‘Y’ gS &

Ph

O

CH3

X YNaOCl (i) SOCl 2

(ii) aniline

(1) Ph

ONH2

(2)

Ph

ON

(3)

Ph

O

HN

(4)

PhO

NH2

A. 3

sol.

O

C CH3

O

C OH

SOCl2

1. NaOCl

2. H

OC Cl

Ph – C – NH – Phaniline

O

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18. 1 g of a non-volatile non-electrolyte solute is dissolved in 100 g of two different solvents A and B whose

ebullioscopic constants are in the ratio of 1 : 5. The ratio of the elevation in their boiling points, b

b

T (A)

T (B)

, is :

tc ,d vok"i'khy oS|qr&vuqi?kV~; ds 1 g dks nks vyx&vyx foyk;dksa (A rFkk B), ftuds bC;wfy;ksLdksfid fLFkjkad

1 : 5 vuqikr esa gSa, ds 100 g esa ?kksyk tk;s rks muds DoFkukadksa ds mUu;u dk vuqikr b

b

T (A)

T (B)

, gksxk :

(1) 1 : 5 (2) 10 : 1 (3) 5 : 1 (4) 1 : 0.2

A. 1

sol. Tb = k

B × m

b A

b B

k 1

k 5

b bA A

b bB B

T k 1

T k 5

19. Compound A(C9H

10O) shows positive iodoform test. Oxidation of A with KMnO

4/KOH gives acid

B(C8H

6O

4). Anhydride of B is used for the preparation of phenolphthalein. Compound A is:

;kSfxd A(C9H

10O) ldkjkRed vk;MksQkeZ ijh{k.k iznf'kZr djrk gSA KMnO

4/KOH ds lkFk A dk vkWDlhdj.k ,d vEy

B(C8H

6O

4) nsrk gSA B ds ,ugkbMªkbM dks QsukYQFkSyhu dks cukus ds fy, iz;ksx djrs gSaA ;kSfxd A gS :

(1)

CH3

CH3

O

(2) CH3

CH3

O

(3)

CH3

O(4)

CH2 –C –H

CH3

O

A. 2

sol.

C –CH 3

CH3

O

C –ONa

CH3

O

+ CHl3

I + NaOH2

KMnO4

C –OH

C –OH

O

C

O

+ H2O

O O

C

O

anhydrideC H O8 6 4 pthalic

It is used for preparationof phenolphthaleinindicator

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20. The ratio of the shortest wavelength of two spectral series of hydrogen spectrum is found to be about 9. The

spectral series are :

(1) Paschen and Pfund (2) Brackett and Pfund

(3) Lyman and Paschen (4) Balmer and Brackett

gkbMªkstu LisDVªe ds nks LisDVªeh Jsf.k;ksa ds y?kqre rjaxnS/;Z dk vuqikr yxHkx 9 ik;k x;kA LisDVªeh Jsf.k;k¡ gSa :

(1) ik'psu rFkk QqUM (2) cSzdsV rFkk QqUM

(3) ykbeu rFkk ik'psu (4) ckej rFkk czSdsV

A. 3

sol. Shortest wavelength means n2 =

Lyman series 2L 2

L

1 11312 1

1

Paschen series 2P 2

P

1 11312 1

3

L P

P L

9

21. The correct match between Item-I and Item-II is:

Item I Item II

(a) High density polythene (I) Peroxide catalyst

(b) Polyacrylonitrile (II) Condensation at high temperature and pressure

(c) Novolac (III) Ziegler Natta Catalyst

(d) Nylon 6 (IV) Acid or base catalyst

en–I rFkk en–II ds chp lgh lqesy gS :

en – I en – II

(a) mPp ?kuRo ikWyhFkhu (I) ijkWDlkbM mRizsjd

(b) ikWyh,fØyksukbVªkby (II) mPp rki rFkk nkc ij la?kuu

(c) uksoksysd (III) ftxyj&ukVk mRizsjd

(d) uk;yku 6 (IV) vEy vFkok {kkjd mRizsjd

(1) (a) (III), (b) (I), (c) (IV), (d) (II)

(2) (a) (III), (b) (I), (c) (II), (d) (IV)

(3) (a) (IV), (b) (II), (c) (I), (d) (III)

(4) (a) (II), (b) (IV), (c) (I), (d) (III)

A. 1

sol. a. HDPE – Ziegler-Natta Catalyst

b. Polyacrylonitrile – Peroxide Catalystc. Novolac – Catalysed by acid or based. Nylon-6 – Condensation at High T and P

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22. Air pollution that occurs in sunlight is :

(1) Fog (2) Oxidising smog (3) Acid rain (4) Reducing smog

og ok;q iznw"k.k tks lw;Z ds izdk'k esa gksrk gS] gS :

(1) QkWx (2) vkWDlhdkjd /kwedqgk (3) vEyh; o"kkZ (4) vipk;h LekWx (/kwedqgk)

A. 2

sol. Air pollution caused by sunlight is photochemical smog and it is oxidising.

23. The increasing order of nucleophilicity of the following nucleophiles is :

fuEu ukfHkdjkfx;ksa ds ukfHkdjkfxrk dk c<+rk Øe gS %

(a) 3 2CH CO (b) H2O (c) 3 3CH SO (d) OH

(1) (d) < (a) < (c) < (b) (2) (b) < (c) < (d) < (a)

(3) (a) < (d) < (c) < (b) (4) (b) < (c) < (a) < (d)

A. 4

sol. Greater the negative charge Present on a nucleophilic centre greater would be its nucleophilicity.

–OH > CH3–C–O >CH –S–O > H O– –

3 2

O O

O

24. The correct order of the first ionization enthalpies is :

izFke vk;uu ,UFkSfYi;ksa dk lgh Øe gS %

(1) Mn < Ti < Zn < Ni (2) Ti < Mn < Zn < Ni

(3) Ti < Mn < Ni < Zn (4) Zn < Ni < Mn < Ti

A. 3

sol. Order for I.E. is

Ti < Mn < Ni < Zn

25. The noble gas that does NOT occur in the atmosphere is :

og mRd"V xSl tks ok;qeaMy esa mifLFkr ugha gS] gksxh %

(1) Ne (2) Kr (3) He (4) Ra

A. 4

sol. Radon is not present in atmosphere.

26. Which of the following is NOT a correct method of the preparation of benzylamine from cyanobenzene?

fuEu esa ls dkSu lk;ukscsathu ls csafty,ehu ds cukus dk lgh rjhdk ugha gS\

(1) (i) SnCl2 + HCI(gas) (ii) NaBH

4

(2) H2/Ni

(3) (i) LiAlH4

(ii) H3O+

(4) (i) HCl/H2O (ii) NaBH

4

A. 4

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sol.

HClH O2

or

C

O

NH2 COOH

(depending upon temperature)

CN

NaBH4

No reaction

27. The minimum amount of O2(g) consumed per gram of reactant is for the reaction:

(Given atomic mass : Fe = 56, O = 16, Mg = 24, P = 31, C = 12, H = 1)

vfHkdkjd ds izfrxzke ds fy, O2(g) dh yxus okyh vYire ek=kk fuEu esa ls fdl vfHkfØ;k ds fy, gksxh\

(fn;k x;k ijek.kq nzO;eku : Fe = 56, O = 16, Mg = 24, P = 31, C = 12, H = 1)

(1) 2Mg(s) + O2(g) 2MgO(s)

(2) 4Fe(s) + 3O2(g) 2Fe

2O

3(s)

(3) C3H

8(g) + 5O

2(g) 3CO

2(g) + 4H

2O(I)

(4) P4(s) + 5O

2(g) P

4O

10(s)

A. 2

sol. (1) 2 Mg + O2 2 MgO

1 g requires 32

48 g = 0.66 g of O

2

(2) 4Fe + 3O2 2Fe

2O

3

1 g Fe requires = 0.43 g of oxygen

(3) C3H

8 + 5O

2 3CO

2 + 4H

2O

1 g of C3H

8 requires = 3.6 g of O

2

(4) P4 + 5O

2 P

4O

10

1 g of P requires = 1.3 g of oxygen

28. Points I, II and III in the following plot respectively correspond to (Vmp

: most probable velocity)

Dis

trib

uti

on f

unct

ion,

f(v)

Speed, v I II III

(1) Vmp

of N2 (300 K); V

mp of O

2 (400 K);V

mp of H

2 (300 K)

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(2) Vmp

of H2 (300 K); V

mp of N

2 (300 K);V

mp of O

2 (400 K)

(3) Vmp

of N2 (300 K); V

mp of H

2 (300 K);V

mp of O

2 (400 K)

(4) Vmp

of O2 (400 K); V

mp of N

2 (300 K);V

mp of H

2 (300 K)

vkys[k esa fcUnq I, II rFkk III Øe'k% buls lEcfU/kr gSa, (Vmp

: izkf;dre osx)

Dis

trib

uti

on f

unct

ion,

f(v)

Speed, v I II III

(1) N2 dk V

mp (300 K); O

2 dk V

mp (400 K); H

2 dk V

mp (300 K)

(2) H2 dk V

mp (300 K); N

2 dk V

mp (300 K); O

2 dk V

mp (400 K)

(3) N2 dk V

mp (300 K); H

2 dk V

mp (300 K);O

2 dk V

mp (400 K)

(4) O2 dk V

mp (400 K); N

2 dk V

mp (300 K); H

2 dk V

mp (300 K)

A. 1

sol. mp

2RTV

M

as T

M increases, V

mp increases

From curve

(Vmp

)I < (V

mp)

II < (V

mp)

III

2 2 2

mp mp mpN O H

300 400 300V , V , V

28 32 2

2 2 2

mp mp mpN O HV V V

(Under given Condition)

29. The correct statements among (a) to (d) are :

(a) Saline hydrides produce H2 gas when reacted with H

2O.

(b) Reaction of LiAlH4 with BF

3 leads to B

2H

6.

(c) PH3 and CH

4 are electron - rich and electron - precise hydrides, respectively.

(d) HF and CH4 are called as molecular hydrides.

(1) (a), (c) and (d) only.

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(2) (c) and (d) only.

(3) (a), (b) and (c) only.

(4) (a), (b), (c) and (d).

(a) ls (d) ds chp] lgh dFku gSa :

(a) yo.k gkbMªkbM~l H2O ds lkFk vfHkfØ;k djus ij H

2 xSl nsrs gSaA

(b) BF3 ds lkFk LiAlH

4 dh vfHkfØ;k ls B

2H

6 curk gSA

(c) PH3 rFkk CH

4 Øe'k% bysDVªkWu&lEiUu rFkk bysDVªkWu&ifj'kq) gkbMªkbM~l gSaA

(d) HF rFkk CH4 vkf.od gkbMªkbM dgs tkrs gSaA

(1) dsoy (a), (c) rFkk (d)

(2) dsoy (c) rFkk (d)

(3) dsoy (a), (b) rFkk (c)

(4) (a), (b), (c) rFkk (d)

A. 4

sol. – With water saline hydrides produce H2 gas

– 3LiAlH4 + 4BF

3 2B

2H

6 + 3LiF + 3AlF

3

– PH3 is electron rich while CH

4 is electron precise hydride

– HF and CH4 are molecular hydrides

30. In chromatography, which of the following statements is INCORRECT for Rf ?

(1) The value of Rf cannot be more than one.

(2) Higher Rf value means higher adsorption.

(3) Rf value is dependent on the mobile phase.

(4) Rf value depends on the type of chromatography.

ØksesVksxzkQh esa, Rf ds fy;s fuEu dFkuksa esa ls dkSulk xyr gS?

(1) Rf dk eku 1 ls vf/kd ugha gks ldrk gSA

(2) mPprj Rf eku dk vFkZ gS mPprj vf/k'kks"k.kA

(3) Rf dk eku xfr'khy izkoLFkk ij fuHkZj djrk gSA

(4) Rf dk eku ØksesVksxzkQh ds izdkj ij fuHkZj djrk gSA

A. 2

sol. Rf represents retardation factor in chromatography.

f

Distance moved by the substance from base lineR

Distance moved by the solvent from baseline

– Higher Rf value means lower adsorpation

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MATHS

10 APRIL 2019 [Phase : II]

JEE MAIN PAPER ONLINE

MOD

1. Let f(x) = loge(sinx), (0 < x < ) and g(x) = sin–1 (e–x), (x 0). If is a positive real number such that a = (fog)'

() and b = (fog)(), then :

ekuk f(x) = loge(sinx), (0 < x < ) rFkk g(x) = sin–1 (e–x), (x 0) gSaA ;fn ,d /kukRed okLrfod la[;k ds fy,

a = (fog)' () rFkk b = (fog)() gS, rks :

(1) a2 – b – a = 1 (2) a2 + b + a = 0

(3) a2 – b – a = 0 (4) a2 + b – a = –22

A. 1

sol. f(x) = ln(sin x), g(x) = sin–1 (e–x)

f(g(x)) = ln(sin(sin–1 e–x))

= –x

– = b

f'(g()) = a

i.e.,a = –1

a2 – b + 1 = –2 + 2 + 1 = –a

Solution of Triangle

2. The angles A, B and C of a triangle ABC are in A.P. and a : b = 1 : 3 . If c = 4 cm, then the area (in sq. cm)

of this triangle is :

,d f=kHkqt ABC ds dks.k A, B rFkk C lekUrj Js.kh esa gSa rFkk a : b = 1 : 3 gSA ;fn c = 4 cm gS, rks bl f=kHkqt dk {ks=kQy

(oxZ lseh esa) gS :

(1) 2

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

4

3

A. 3

sol. A, B, C, are in A.P

2B = A + C

B3

Area = 1

2(4x) sin 60º

3x

Now cos 2 216 x 3x

60º8x

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4x = 16 – 2x2

x = 2 (as –4 is rejected)

Hence, area = 2 3 sq. cm

Determinant

3. The sum of the real roots of the equation

x 6 1

2 3x x 3

3 2x x 2

= 0, is equal to :

lehdj.k

x 6 1

2 3x x 3

3 2x x 2

= 0, ds okLrfod ewyksa dk ;ksxQy gS :

(1) 6 (2) 0 (3) –4 (4) 1

A. 2

sol.

x 6 1

2 3x x 3 0

3 2x x 2

x(–3x2 – 6x – 2x2 + 6x) –6(–3x + 9 – 2x – 4) –(4x – 9xA) = 0

x(–5x2) –6(–5x + 5) – 4x + 9x = 0

x3 – 7x + 6 = 0All the roots are real

Sum of real roots 0

01

Complex Number

4. If z and w are two complex numbers such that |zw| = 1 and arg(z) – arg(w) = 2

, then :

;fn z rFkk w nks ,slh lfEeJ la[;k,¡ gSa fd |zw| = 1 rFkk arg(z) – arg(w) = 2

, rks :

(1) 1 i

zw2

(2) zw i (3)

1 izw

2

(4) zw i

A. 4

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sol. |zw| = 1 ...(i)

zarg

w 2

...(ii)

z z

0w w zw zw

from (i) zz ww 1

2

zw 1 zw i

from (ii) –arg ( z ) – argw = 2

arg zw2

Hence, zw i

Circle

5. The locus of the centes of the circles, which touch the circle, x2 + y2 = 1 externally, also touch the y-axis and lie

in the first quadrant, is :

,sls o`Ùkksa, tks o`Ùk x2 + y2 = 1 dks cká Li'kZ djrs gSa, y-v{k dks Hkh Li'kZ djrs gSa rFkk izFke prqFkkZa'k esa fLFkr gSa] ds dsUnzksa dk

fcUnqiFk gS :

(1) y 1 2x, x 0 (2) x 1 4y, y 0 (3) x 1 2y, y 0 (4) y 1 4x, x 0

A. 1

sol. Let centre of required circle is (h, k).

OO' = r + r'

2 2h k 1 h h2 + k2 = 1 + h2 + 2hk2 = 1 + 2h

Locus is y 1 2x

Sequence & Progression

6. The sum 1 +3 3 3 3 31 2 1 2 3

1 2 1 2 3

+ ... +

3 3 3 31 2 3 ... 15 1

1 2 3 ... 15 2

(1 + 2 + 3 + ... + 15) is equal to :

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;ksxQy 1 +3 3 3 3 31 2 1 2 3

1 2 1 2 3

+ ... +

3 3 3 31 2 3 ... 15 1

1 2 3 ... 15 2

(1 + 2 + 3 + ... + 15) cjkcj gS :

(1) 1860 (2) 620 (3) 660 (4) 1240

A. 2

sol.3 3 3 3 31 2 1 2 3

S 1 ....15 terms1 2 1 2 3

3 3 3

n

n (n 1)1 2 ...n n(n 1)4T

n(n 1)1 2 ...n 22

15 152

n 1 n 1

1 1 15(16)(31) 15(16)S n n 680

2 2 6 2

1 15(16)680 680 60 620

2 2

Binomial Theorem

7. The smallest natural number n, such that the coefficient of x in the expansion of n

2

3

1x

x

is nC

23, is :

og U;wure izkd`r la[;k n, ftlds fy, n

2

3

1x

x

ds izlkj esa x dk xq.kkad nC

23 gS, gS :

(1) 58 (2) 35 (3) 38 (4) 23

A. 3

sol.

n

2

3

1x

x

General term Tr+1

= nCr(x2)n–r

r

3

1

x

nCr . x2n–5r

for coefficiant of x, 2n – 5r = 1Given nC

r = nC

23

r = 23 or n – r = 23

n = 58 or n = 38Minimum value is n = 38

Statistics

8. If both the mean and the standard deviation of 50 observations x1, x

2, .... x

50 are equal to 16, then the mean of

(x1 – 4)2, (x

2 – 4)2, ... (x

50 – 4)2

is :

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;fn 50 izs{k.kksa x1, x

2, .... x

50 dk ek/; rFkk ekud fopyu nksuksa 16 gSa, rks (x

1 – 4)2, (x

2 – 4)2, ... (x

50 – 4)2

dk ek/; gS:

(1) 380 (2) 480 (3) 400 (4) 525

A. 3

sol.1 2 50x x .....x

1650

2 2 22 21 2 50x x .....x

16 1650

2 2 2 21 2 502(16) 50 x x .....x

Required mean =

2 2 2

1 2 50x 4 x 4 .... x 4

50

2 216 100 4 (50) 8 16 50

50

= 162(2) + 16 – 8(16) = 400

Indefinite Integration

9. If 2 25 x xx e dx g(x)e c , where c is a constant of integration, then g(–1) is equal to :

;fn 2 25 x xx e dx g(x)e c gS, tgk¡ c ,d lekdyu vpj gS, rks g(–1) cjkcj gS :

(1) –1 (2) 1

2 (3) 1 (4)

5

2

A. 4

sol.25 xI x .e dx

Put –x2 = t –2xdx = dt

2 tt 2t .e dt 1

I e (t 2 t 2) c( 2) 2

4 21

g(x) (x 2x 2)2

5

g( 1)2

Sequence & Progression

10. Let a, b and c be in G.P. with common ratio r, where a 0 and 0 < r 1

2. If 3a, 7b and 15c are the first three

terms of an A.P., then the 4th term of this A.P. is :

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ekuk a, b rFkk c xq.kksÙkj Js<+h esa gSa ftldk lkoZvuqikr r gS, tgk¡ a 0 vkSj 0 < r 1

2gSA ;fn 3a, 7b rFkk 15c ,d lekUrj

Js<+h ds izFke rhu in gSa, rks bl lekUrj Js<+h dk pkSFkk in gS :

(1) 2

a3

(2) a (3) 7

a3

(4) 5a

A. 2

sol. Let b = ar, c = ar2

AP : 3a, 7ar, 15ar2

14ar = 3a + 15ar2

15r2 – 14r + 3 = 0

r = 1

3 or r =

3

5 rejected

Fourth term = 15ar2 + 7ar – 3a= a(15r2 + 7r – 3)

15 7a 3

9 3

= a

Area Under Curve

11. The area (in sq. units) of the region bounded by the curves y = 2x and y = |x + 1|, in the first quadrant is :

oØksa y = 2x rFkk y = |x + 1| }kjk izFke prqFkkZa'k esa ifjc) {ks=k dk {ks=kQy (oxZ bdkb;ksa esa) gS :

(1) e

3 1

2 log 2 (2)

1

2(3) log

e 2 +

3

2(4)

3

2

A. 1

sol.

Area = 1

x

0

(x 1) 2 dx

12 x

0

x 2x

2 ln 2

1 2 11

2 ln 2 ln 2

3 1

2 ln 2

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3 D

12. A perpendicular is drawn from a point on the line x 1 y 1 z

2 1 1

to the plane x + y + z = 3 such that the foot

of the perpendicular Q also lies on the plane x – y + z = 3. Then the co-ordinates of Q are :

js[kk x 1 y 1 z

2 1 1

ds ,d fcUnq ls lery x + y + z = 3 ij ,d yEc bl izdkj Mkyk x;k fd bldk yacikn Q, lery

x – y + z = 3 ij Hkh fLFkr gS] rks Q ds funsZ'kkad gSa :

(1) (1, 0, 2) (2) (2, 0, 1) (3) (4, 0, –1) (4) (–1, 0, 4)

A. 2

sol.

Let Q be () = 3 ...(i)– = 3 ...(ii) = 3 and = 0Equating DR’s of PQ :

2 1 1 3

1 1 1

3 2, 2 1

Substituting in equation (i) we get

5 3 3

0 Point is Q(2, 0, 1)

Tangent & Normal

13. If the tangent to the curve y = 2

x

x 3, x R, x 3 , at a point (, ) (0, 0) on it is parallel to the line

2x + 6y – 11 = 0, then :

;fn oØ y = 2

x

x 3, x R, x 3 ds ,d fcUnq (, ) (0, 0) ij [khaph xbZ Li'kZ js[kk] js[kk 2x + 6y – 11 = 0

ds lekUrj gS, rks :

(1) |6 + 2| = 19 (2) |2 + 6| = 19 (3) |6 + 2| = 9 (4) |2 + 6| = 11

A. 1

sol. 2

xy

x 3

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2 2

2 2 2 2

dy (x 3) x(2 x) x 3

dx (x 3) (x 3)

2

2 2( , )

dy 3 1

dx ( 3) 3

3(2 + 3) = (2 – 3)2 ...(i)i.e. 2 = 9

Also, 3 63

13,

2

Which satisfies |6 + 2| = 19

Differential Equation

14. Let y = y(x) be the solution of the differential equation, dy

dx+ y tanx = 2x + x2 tan x, x ,

2 2

, such that

y(0) = 1. Then :

ekuk y = y(x), vody lehdj.k dy

dx+ y tanx = 2x + x2 tan x, x ,

2 2

, tcfd y(0) = 1 gS, dk gy gS] rks:

(1)

2

y y 24 4 2

(2) y y 2

4 4

(3) y ' y ' 24 4

(4) y ' y ' 2

4 4

A. 4

sol.2dy

y tan x 2x x tan xdx

P = tanx, Q = 2x + x2tan x

tan xdx ln|secx|I.F. e e | secx |

2y(sec x) (2 x x tanx)secxdx

2x tan x sec xdx 2x sec xdx

2x sec x 2x sec x dx 2x sec xdx = x2secx + cAs y (0) = 1, c = 1 y = x2 + cosx

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At 2 1

x , y4 4 16 2

2 1y

4 16 2

y y 04 4

dy2x sin x

dx

1 1y ' , y '

4 2 4 22 2

y ' y ' 24 4

Sequence & Progression

15. Let a1, a

2, a

3, ..... be an A.P. with a

6 = 2. Then the common difference of this A.P., which maximises the product

a1 a

4 a

5, is :

ekuk a1, a

2, a

3, ..... ,d lekUrj Js<+h gS ftlesa a

6 = 2 gS, rks bl lekUrj Js<+h dk og lkoZvUrj tks xq.kuQy a

1 a

4 a

5 dks

vf/kdre djrk gS] gS :

(1) 2

3(2)

8

5(3)

3

2(4)

6

5

A. 1

sol. a + 5d = 2

Let A = a1 a

4 a

5 = a(a + 3d)(a + 4d)

= a(2 – 2d)(2 – d)A = (2 – 5d)(4 – 6d + 2d2)

dA0

dd

(2 – 5d)(–6 + 4d) + (4 – 6d + 2d2)(–5) = 0

15d2 – 34d + 16 = 0

8 2d ,

5 3

For 2

2

2 d Ad , 0

3 dd

Hence 2

d3

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3 D

16. If the plane 2x – y + 2z + 3 = 0 has the distances 1

3 and

2

3 units from the planes 4x – 2y + 4z + = 0 and

2x – y + 2z + = 0, respecitvely, then the maximum value of + is equal to :

;fn lery 2x – y + 2z + 3 = 0 dh leryksa 4x – 2y + 4z + = 0 rFkk 2x – y + 2z + = 0 ls nwfj;k¡ Øe'k% 1

3 rFkk

2

3 bdkb;k¡ gSa] rks + dk vf/kdre eku gS :

(1) 13 (2) 15 (3) 5 (4) 9

A. 1

sol. P1 : 2x – y + 2z + 3 = 0

P2 : 2x – y + 2z + 0

2

P3 : 2x – y + 2z + = 0

Given

max 8

31 2

3 13 29

Also, max

2 | 3 |5

3 9

max

13

Limit

17. If 2

x 1

x ax blim

x 1

= 5, then a + b is equal to :

;fn 2

x 1

x ax blim

x 1

= 5 gS, rks a + b cjkcj gS &

(1) 5 (2) –4 (3) 1 (4) –7

A. 4

sol.2

x 1

x ax blim 5

x 1

As limit is finite, 1 – a + b = 0

x 1

2x a 0lim 5 from

1 0

i.e., 2 – a = 5or a = – 3 b = –4 a + b = – 3 – 4 = – 7

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Quadratic Equation

18. The number of real roots of the equation 5 + |2x – 1| = 2x(2x – 2) is :

lehdj.k 5 + |2x – 1| = 2x(2x – 2) ds okLrfod ewyksa dh la[;k gS :

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

A. 3

sol. Let 2x – 1 = t

5 + | t | = (t + 1) (t – 1) | t | = t2 – 6For t > 0, t2 – t – 6 = 0 i.e., t = 3 or – 2 (rejected)For t < 0, t2 + t – 6 = 0i.e., t = – 3 or 2 (both rejected) 2x – 1 = 3

x = 2

Straight Line

19. Lines are drawn parallel to the line 4x – 3y + 2 = 0, at a distance 3

5from the origin. Then which one of the

following points lies on any of these lines ?

js[kk 4x – 3y + 2 = 0 ds lekUrj js[kk,¡ [khaph xbZ gSa tks ewyfcUnq ls 3

5dh nwjh ij gSa] rks fuEu esa ls dkSu&lk ,d fcUnq buesa

ls fdlh js[kk ij fLFkr gSa\

(1) 1 1

,4 3

(2)

1 1,

4 3

(3) 1 2

,4 3

(4) 1 2

,4 3

A. 3

sol. Let straight line be 4x – 3y + = 0

Given 3

5 5

3

Line is 4x – 3y + 3 = 0 or 4x – 3y – 3 = 0

Clearly 1 2

,4 3

satisfies 4x – 3y + 3 = 0

P & C

20. Suppose that 20 pillars of the same height have been erected along the boundary of a circular stadium. If the

top of each pillar has been connected by beams with the top of all its nonadjacent pillars, then the total number

of beams is :

ekuk ,d o`Ùkh; LVsfM;e dh lhek ij ,d gh Å¡pkbZ ds 20 [kEHks [kM+s fd, x, gSaA ;fn izR;sd [kEHks ds f'k[kj dks lHkh vlayXu

[kEHkksa ds f'k[kjksa ls dfM+;ksa (beams) }kjk tksM+k x;k gS, rks ,slh dfM+;ksa dh dqy la[;k gS :

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(1) 210 (2) 180 (3) 170 (4) 190

A. 3

sol. Required number of beams = 20C2 – 20

= 190 – 20 = 170

Tangent & Normal

21. A spherical iron ball of radius 10 cm is coated with a layer of ice of uniform thickness that melts at a rate of

50 cm3/min. When the thickness of the ice is 5 cm, then the rate at which the thickness (in cm/min) of the ice

decreases, is :

10 cm f=kT;k dh yksgs dh ,d xksykdkj xsan ds pkjksa vksj leku eksVkbZ dh cQZ dh rg p<+kbZ xbZ gS] tks 50 cm3/min dh nj ls

fi?ky jgh gSA tc cQZ dh eksVkbZ 5 cm gS, rc cQZ dh eksVkbZ ds ?kVus dh nj (cm/min) esa, gS :

(1) 5

6(2)

1

36(3)

1

9(4)

1

18

A. 4

sol.icedV

50dt

3 3

ice

4 4V 10 r 10

3 3

2dV 4 dr

3 10 rdt 3 dt

2 dr

4 10 rdt

At r = 5, 50 = 4(225)dr

dt

dr 50

dt 4 (225)

1cm / min

18

Probability

22. Minimum number of times a fair coin must be tossed so that the probability of getting at least one head is more

than 99% is :

,d U;k¸; flDds dks U;wure fdruh ckj mNkysa fd de ls de ,d fpÙk vkus dh izkf;drk 99% ls vf/kd gks?

(1) 8 (2) 6 (3) 5 (4) 7

A. 4

sol.

n1 99

12 100

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n1 1

2 100

n 7

Minimum value is 7.

Hyperbola

23. If 5x + 9 = 0 is the directrix of the hyperbola 16x2 – 9y2 = 144, then its corresponding focus is :

;fn vfrijoy; 16x2 – 9y2 = 144 dh fu;rk (directrix) 5x + 9 = 0 gS, rks bldk laxr ukfHkdsUnz gS :

(1) 5

,03

(2) 5

,03

(3) (–5, 0) (4) (5, 0)

A. 3

sol. 16x2 – 9y2 = 144

i.e. 2 2x y

19 16

Focus S'(–ae, 0)

9

x5

a = 3, b = 4

e2 = 1 + 16 25

9 9

5

S' 3 ,0 5,03

Determinant

24. Let be a real number for which the system of linear equations

x + y + z = 6

4x + y – z = – 2

3x + 2y – 4z = – 5

has infinitely many solutions. Then is a root of the quadratic equation :

ekuk ,d ,slh okLrfod la[;k gS ftlds fy, jSf[kd lehdj.k fudk;

x + y + z = 6

4x + y – z = – 2

3x + 2y – 4z = – 5

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ds vuUr gy gSa] rks ftl f}?kkr lehdj.k dk ,d ewy gS] og gS :

(1) 2 + 3 – 4 = 0 (2) 2 – – 6 = 0 (3) 2 + – 6 = 0 (4) 2 – 3 – 4 = 0

A. 2

sol. = 1 =

2 =

3 = 0

1 1 1

4 0

3 2 4

0 0 1

4 2 0

1 6 4

= 3

1

6 1 1

2 0

5 2 4

for = 3

2

1 6 1

4 2 0

3 5 4

for = 3

3

1 1 6

4 2 0

3 2 5

for = 3

For = 3, infinitely many solutions is obtained.

Definite Integration

25. The integral /3

2/3 4/3

/6

sec x cos ec x

dx is equal to :

lekdy /3

2/3 4/3

/6

sec x cos ec x

dx cjkcj gS :

(1) 37/6 – 35/6 (2) 35/3 – 31/3 (3) 35/6 – 32/3 (4) 34/3 – 31/3

A. 1

sol.

/32 4

3 3

/6

I sec x.cosec x dx

/3

2 43 3

/6

1.dx

cos x.sin x

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/3 /3 2

4 42 3 3

/6 /6

1 dx sec xdx

cos x. tan x tan x

Let tan x = t

313

3 1433

1

3

3 t

I t dt1

16

16

13 3

3

1 16 63 3 3

1 16 63 3 3

7 56 63 3

Parabola

26. If the line ax + y = c, touches both the curves x2 + y2 = 1 and y2 = 4 2x , then |c| is equal to :

;fn js[kk ax + y = c, nksuksa oØksa x2 + y2 = 1 rFkk y2 = 4 2x , dks Li'kZ djrh gS, rks |c| cjkcj gS :

(1) 1

2(2)

1

2(3) 2 (4) 2

A. 4

sol. Tangent on 2y 4 2x is 2yt x 2t

As it is tangent on circle also,

2

2

2t1

1 t

2t4 = 1 + t2 i.e. t2 = 1

Equation is y x 2

Hence | c | = 2

Ellipse

27. The tangent and normal to the ellipse 3x2 + 5y2 = 32 at the point P(2, 2) meet the xaxis at Q and R, respectively.

Then the area (in sq. units) of the triangle PQR is :

nh?kZo`Ùk 3x2 + 5y2 = 32 ds fcUnq P(2, 2) ij [khaph xbZ Li'kZ js[kk rFkk vfHkyEc, x–v{k dks Øe'k% Q rFkk R ij dkVrs gSa] rks

f=kHkqt PQR dk {ks=kQy (oxZ bdkb;ksa esa) gS:

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(1) 16

3(2)

14

3(3)

34

15(4)

68

15

A. 4

sol. For 2 23x 5y

132 32

Tangent at P is

3(2) x 5(2) y1

32 32

3x 5y1

16 16

16Q ,0

3

Normal at P is 32x 32y 32 32

3(2) 5(2) 3 5

4R ,0

5

area of PQR = 1

2 (PQ) (PR) =

1 136 136.

2 3 5

68

15

Vector

28. The distance of the point having position vector ˆ ˆ ˆi 2 j 6k from the straight line passing through the point

(2, 3, –4) and parallel to the vector, ˆ ˆ ˆ6i 3j 4k is :

,d fcUnq ftldk fLFkfr lfn'k ˆ ˆ ˆi 2 j 6k gS, dh ,d ljy js[kk, tks fcUnq (2, 3, –4) ls gksdj tkrh gS rFkk lfn'k

ˆ ˆ ˆ6i 3j 4k ds lekUrj gS, ls nwjh gS :

(1) 7 (2) 4 3 (3) 6 (4) 2 13

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A. 1

sol. Equation of I is x 2 y 3 z 4

6 3 4

(6 + 2, 3 + 3, –4 –4)Let M (6 + 2, 3 + 3, –4 – 4)DR’s of PM is < 6+ 3, 3+ 1, –4– 10 >

(6 + 3)(6) + (3 + 1)(3) + (–4– 10)(–4) = 0

= –1 i.e. M (–4, 0, 0)

PM = 9 4 36 7

ITF

29. If cos–1x – cos–1y

2 = , where – 1 x 1, –2 y 2, x

y

2, then for all x, y, 4x2 – 4xy cos + y2 is equal

to :

;fn cos–1x – cos–1y

2 = , tgk¡ – 1 x 1, –2 y 2, x

y

2 gS, rks lHkh x, y, , ds fy,, 4x2 – 4xy cos + y2 cjckj

gS :

(1) 2 sin2 (2) 4 sin2 – 2x2y2 (3) 4 sin2 + 2x2y2 (4) 4 sin2

A. 4

sol.1 1 y

cos x cos2

2

1 2xy ycos 1 x . 1

2 4

2 21 x 4 yxy

cos2 2

2 2xy 1 x 4 y 2cos

(xy – 2 cos)2 = (1 – x2) (4 – y2) x2y2 + 4cos2 – 4xycos = 4 – y2 – 4x2 + x2y2

4x2 – 4xycos + y2 = 4sin2

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Mathematical Reasoning

30. The negation of the Boolean expression ~ s (~ r s) is equivalent to :

cwys O;atd ~ s (~ r s) dk fu"ks/ku fuEu esa ls fdl ds lerqY; gS\

(1) s r (2) r (3) ~ s ~ r (4) s r

A. 1

sol. ~s (~r s)

(~s ~r) (~s s)

(~s ~r) ( (~s s) is tautology)

~(s r)Hence its negation is s r