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JEE (MAIN ONLINE) 2019

PHYSICS09 APRIL 2019 [Phase : II]

JEE MAIN PAPER ONLINEGravitation

1. A test particle is moving in a circular orbit in the gravitational field produced by a mass density 2

K(r)

r .

Identify the correct relation between the radius R of the particle’s orbit and its period T :(1) T/R is a constant (2) TR is a constant (3) T/R2 is a constant (4) T2/R3 is a constant

,d ijh{k.k d.k nzO;eku ?kuRo 2

K(r)

r ls mRiUu xq:Roh; {ks=k esa ,d o`Ùkkdkj d{kk esa ?kwe jgk gSA d.k ds d{k dh

f=kT;k R rFkk blds vkorZdky T ds chp lgh lEcU/k gksxk :(1) T/R fu;r gSA (2) TR fu;r gSA (3) T/R2 fu;r gSA (4) T2/R3 fu;r gSA

A. 1

sol.

R2

(in) 2

0

kM 4 r dr.

r

M(in)

= 4KR

2

2

4 KR VmG. m v 4 GK

R R

2 R 2 .RT

v 4 GK

T

R = constant

KTG & Thermodynamics

2. The specific heats, CP and C

V of a gas of diatomic molecules, A, are given (in units of J mol–1 K–1) by 29 and

22, respectively. Another gas of diatomic molecules, B, has the corresponding values 30 and 21. If they are

treated as ideal gases, then :

(1) A is rigid but B has a vibrational mode.

(2) A has a vibrational mode but B has none.

(3) Both A and B have a vibrational mode each.

(4) A has one vibrational mode and B has two.

,d f}ijek.kqd xSl A ds v.kqvksa dh fof'k"V Å"ek;sa (J mol–1 K–1 dh bdkbZ esa) CP rFkk C

V, Øe'k% 29 vkSj 22 gSaA nwljh

f}ijek.kqd xSl B ds v.kqvksa ds fy, laxr eku 30 vkSj 21 gSaA ;fn bUgsa vkn'kZ xSl ekuk tk;s rks :

(1) A n`<+ gS fdUrq B esa ,d dEiu fo/kk gSA

(2) A esa ,d dEiu fo/kk gS fdUrq B esa dksbZ dEiu fo/kk ugha gSA

(3) A rFkk B nksuksa esa ,d&,d dEiu fo/kk;sa gSaA

(4) A esa ,d dEiu fo/kk rFkk B esa nks dEiu fo/kk;sa gSA

A. 2


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sol. For (A) CP = 29, C

V = 22

For (B) CP = 30 C

V = 21

A

A

P

A

V

C 291.31

C 22

When A has vibrational degree of freedom, then A 9 / 7 1.29

B

B

P

B

V

C 301.42

C 21

B has no vibrational degree of freedom

Kinematics

3. The position vector of a particle changes with time according to the relation 2 2ˆ ˆr t 15t i 4 20t j

. What

is the magnitude of the acceleration at t = 1 ?

,d d.k dk fLFkfr&lfn'k le; ds lkFk fuEu lw=k ls cnyrk gS]

2 2ˆ ˆr t 15t i 4 20t j

t = 1 ij d.k ds Roj.k dk ifjek.k gksxk :

(1) 50 (2) 100 (3) 40 (4) 25

A. 1

sol. 2 2ˆ ˆ ˆr 15t i 4 j 20t j

dr ˆ ˆ30ti 40tjdt

2

2

d r ˆ ˆa 30i 40jdt

a =

22 2 2

2

d r30 40 50 m / s

dt

Dual Nature of Radiation & Matter

4. 50 W/m2 energy density of sunlight is normally incident on the surface of a solar panel. Some part of incident

energy (25%) is reflected from the surface and the rest is absorbed. The force exerted on 1 m2 surface area will

be close to (c = 3 × 108 m/s) :

,d lksyj iSuy dh lrg ij 50 W/m2 ÅtkZ ?kuRo dk lw;Z dk izdk'k vfHkyEcor~ vkifrr gksrk gSA vkifrr ÅtkZ dk dqN

Hkkx (25%) lrg ls ijkofrZr gks tkrk gS rFkk cpk gqvk Hkkx vo'kksf"kr gks tkrk gSA lrg ds 1 m2 {ks=kQy ij yxus okyk cy

gksxk : (c = 3 × 108 m/s) :

(1) 20 × 10–8 N (2) 35 × 10–8 N (3) 15 × 10–8 N (4) 10 × 10–8 N

A. 1


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sol.W

Pc

and pressure = I/c

h h 5hP P

4 4

for one photon

5 Nh.

4 tA

= pressure

But 2Nhc50W / m

. tA

5 50

pressure4 c

2

8n

5 50 1mF 20 10 N

4 c

Current Electricity

5. A metal wire of resistance 3 is elongated to make a uniform wire of double its previous length. This new wire

is now bent and the ends joined to make a circle. If two points on this circle make an angle 60° at the centre,

the equivalent resistance between these two points will be :

3 izfrjks/k okys ,d /kkrq ds rkj dks [khapdj mldh iqjkuh yEckbZ dk nksxquk ,d leku rkj cuk;k x;k gSA bl u;s rkj dks

eksM+dj rFkk nksuksa fljsa tksM+dj ,d oÙk cukrs gSaA ;fn bl oÙk ds nks fcUnq dsUnz ls 60° dk dks.k cukrs gSa rks bu nksuksa fcUnqvksa ds

chp rqY; izfrjks/k gksxk :

(1) 5

3 (2)

5

2 (3)

7

2 (4)

12

5

A. 1

sol.

0

LR 3

A

Now if Lf = 2L

Then f

AA

2

f

2L 2R 12

A

1

12R 2

6

R2 = 10

eq

1 1 1 6

R 2 10 10


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eq

1 5

R 3

Rotation

6. A thin smooth rod of length L and mass M is rotating freely with angular speed 0 about an axis perpendicular

to the rod and passing through its center. Two beads of mass m and negligible size are at the center of the rod

initially. The beads are free to slide along the rod. The angular speed of the system, when the beads reach the

opposite ends of the rod, will be :

nzO;eku M rFkk yEckbZ L dh ,d iryh NM+ dks.kh; pky 0 ls NM+ ds yEcor~ rFkk mlds dsUnz ls tkus okyh v{k ds ifjr%

Lora=k :i ls ?kwe jgh gSA nzO;eku m rFkk ux.; vkdkj dh nks ef.kdk;sa vkjEHk esa NM+ ds dsnz ij gSaA ;g ef.kdk;sa NM+ ij pyus

dks Lora=k gSaA ef.kdk;sa tc NM+ ds foijhr fljksa ij igq¡prh gS] rks bl foU;kl dh dks.kh; pky gksxh :

(1) 0M

M 3m

(2)

0M

M m

(3)

0M

M 6m

(4)

0M

M 2m

A. 3

sol. Initial angular momentum = Final Angular Momentum

2 2 2

0

ML ML mL2

12 12 4

0M

M 6m

Rotation

7. Moment of inertia of a body about a given axis is 1.5 kg m2. Initially the body is at rest. In order to produce a

rotational kinetic energy of 1200 J, the angular acceleration of 20 rad/s2 must be applied about the axis for a

duration of :

,d fi.M dk fn;s x;s v{k ds ifjr% tM+Ro vk?kw.kZ 1.5 kg m2 gSA vkjEHk esa fi.M fojkekoLFkk esa gSA 1200 J dh ?kw.kZu xfrt

ÅtkZ mRiUu djus ds fy;s] mlh v{k ds ifjr% 20 rad/s2 dk dks.kh; Roj.k fdrus le;kUrjky rd yxkuk gksxk\

(1) 2.5 s (2) 2 s (3) 5 s (4) 3 s

A. 2

sol. I = 1.5 ; rad/s2

t = 20t

21E I 1200

2

211.5 (20t) 1200 J

2

t = 2 s

8. The resistance of a galvanometer is 50 ohm and the maximum current which can be passed through it is 0.002


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A. What resistance must be connected to it in order to convert it into an ammeter of range 0 – 0.5 A?

,d /kkjkekih dk izfrjks/k 50 ohm gS rFkk blls vf/kdre 0.002 A /kkjk izokfgr gks ldrh gSA bldks 0 – 0.5 A ijkl ds

vehVj esa ifjofrZr djus ds fy;s blesa fdruk izfrjks/k tksM+uk pkfg;s?

(1) 0.2 ohm (2) 0.002 ohm (3) 0.5 ohm (4) 0.02 ohm

A. 1

sol.

We haveI

g R

g = (0.5 – I

g) S

Ig) S

0.002 50

S 0.20.5

Atomic Structure

9. A He+ ion is in its first excited state. Its ionization energy is :

,d He+ vk;u viuh izFke mÙksftr voLFkk esa gSA bldh vk;uu ÅtkZ gksxh :

(1) 13.60 eV (2) 6.04 eV (3) 48.36 eV (4) 54.40 eV

A. 1

sol.2

0 0n 02

E z E 4E : E

n 4

To ionise it E0 energy must be supplied.

E0 = 13.6 eV.

Sound Waves

10. Two cars A and B are moving away from each other in opposite directions. Both the cars are moving with a

speed of 20 ms–1 with respect to the ground. If an observer in car A detects a frequency 2000 Hz of the sound

coming from car B, what is the natural frequency of the sound source in car B?

(speed of sound in air = 340 ms–1)

nks dkj A rFkk B ,d&nwljs ls nwj foijhr fn'kk esa tk jgh gSaA nksuksa dkj i`Foh ds lkis{k 20 ms–1 dh pky ls py jgh gSaA ;fn

dkj A esa cSBk izs{kd] dkj B ls vkus okyh /ofu dh vko`fÙk 2000 Hz ikrk gS rks dkj B esa /ofu L=kksr dh okLrfod vkofÙk gS :

(/ofu dh ok;q esa pky = 340 ms–1)

(1) 2150 Hz (2) 2300 Hz (3) 2060 Hz (4) 2250 Hz

A. 4

sol.S O

20 m/s 20 m/s


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00 0

s

(v u ) (v 20)f .f .f

(v u ) (v 20)

0

3202000 .f

360

0

2000 9f 2250 Hz

8

Geometrical Optics

11. A convex lens of focal length 20 cm produces images of the same magnification 2 when an object is kept at two

distances x1 and x

2 (x

1 > x

2) from the lens. The ratio of x

1 and x

2 is :

20 cm Qksld nwjh ds ,d mÙky ysal ls fdlh oLrq ds izfrfcEc dk vko/kZu 2 gh gksrk tc oLrq dks ysal ls nks nwfj;ksa x1 rFkk x

2

(x1 > x

2) ij j[krs gSaA x

1 vkSj x

2 dk vuqikr gS :

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

A. 1

sol. 1 1 1

v u f

v = (2u)for v = 2u

1 1 1 3 1

2u u 20 2u 20

u1 = 30 cm

for v = –2u

and1 1 1

u 2u 20

1 1

2u 20 u

2 = 10

30

3.10

Kinematics

12. The position of a particle as a function of time t, is given by x(t) = at + bt2 – ct3 where a, b and c are constants.

When the particle attains zero acceleration, then its velocity will be :

,d d.k dh fLFkfr le; 't' ds Qyu esa fuEu gS :

x(t) = at + bt2 – ct3

tgk¡ a, b rFkk c fu;rkad gSA tc d.k dk Roj.k 'kwU; gS] rc mldk osx gksxk :

(1) 2b

a4c

(2) 2b

a3c

(3) 2b

a2c

(4) 2b

ac

A. 2

sol. x = at + bt2 – ct3


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v = 2x a 2bt 3ct

a = x 2b 6ct

For x 0 b

t3c

2b bv x a 2b 3c

3c 3c 3c

2 2 2b 2b b

v a a3c 3c 3c

Capacitance

13. The parallel combination of two air filled parallel plate capacitors of capacitance C and nC is connected to a

battery of voltage, V. When the capacitors are fully charged, the battery is removed and after that a dielectric

material of dielectric constant K is placed between the two plates of the first capacitor. The new potential

difference of the combined system is :

ok;q ls Hkjs nks lekUrj IysV la/kkfj=kksa] ftudh /kkfjrk,¡ C rFkk nC gSa, ds lekUrj la;kstu dks V oksYVrk dh cSVjh ls tksM+k x;k

gSA tc la/kkfj=k iw.kZr;k vkosf'kr gks tkrk gSa rks cSVjh dks gVk fn;k tkrk gS vkSj rRi'pkr~ igys la/kkfj=k dh nksuksa IysVksa ds chp

ijkoS|qrkad K dk ijkoS|qr inkFkZ j[k nsrs gSaA la;qDr la;kstu ds fy;s u;k foHkokUrj gS :

(1) n 1 V

K n

(2) V

K n(3) V (4)

nV

K n

A. 1

sol. Initially

Q = CV (1 + n)

Ceq

= (K + n)C

CV

nCVCV(1 n) V(1 n)

V'(K n)C (K n)

Communication Systems

14. The physical sizes of the transmitter and receiver antenna in a communication system are :

(1) Inversely proportional to modulation frequency

(2) Inversely proportional to carrier frequency

(3) Proportional to carrier frequency

(4) Independent of both carrier and modulation frequency

,d lapkj O;oLFkk ds fy;s izs"kd rFkk vfHkxzkgh ,saVhuk ds HkkSfrd vkdkj gksaxs :

(1) ekMqyu vko`fÙk ds O;qRØekuqikrh

(2) okgd vko`fÙk ds O;qRØekuqikrh


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(3) okgd vkofÙk ds lekuqikrh

(4) okgd rFkk ekMqyu vko`fÙk nksuksa ij fuHkZj ugha djrkA

A. 2

sol. Size of antenna depends on wavelength of carrier wave.

COM, Momentum & Collision

15. A wedge of mass M = 4m lies on a frictionless plane. A particle of mass m approaches the wedge with speed

v. There is no friction between the particle and the plane or between the particle and the wedge. The maximum

height climbed by the particle on the wedge is given by :

M = 4m nzO;eku dk ,d ost (wedge) vkdkj dk xqVdk ,d ?k"kZ.kghu lrg ij j[kk gSA m nzO;eku dk ,d d.k xqVds dh

vksj, v pky ls vkrk gSA d.k vkSj lrg ;k d.k vkSj xqVds ds chp dksbZ ?k"kZ.k ugha gSA d.k ds }kjk xqVds ds Åij p<+h x;h

vf/kdre Å¡pkbZ gksxh :

(1)

2v

g(2)

22v

5g(3)

22v

7g(4)

2v

2g

A. 2

sol. at maximum height they attain same speed

mv = (4m + m)v'

Common speed v

v '5

From energy conservation

221 v 1

mgh 5m. mv2 25 2

mgh = 2 21 1 1 4

mv 1 mv .2 5 2 5

2 22mv 2v

h5 mg 5g

Fluid Mechanics

16. A wooden block floating in a bucket of water has 4

5 of its volume submerged. When certain amount of an oil

is poured into the bucket, it is found that the block is just under the oil surface with half of its volume under

water and half in oil. The density of oil relative to that of water is

ckYVh esa rSjrs gq,, ,d ydM+h ds xqVds ds vk;ru dk 4

5 Hkkx ikuh esa Mwck gqvk gSA tc ckYVh esa dqN rsy Mkyrs gSa rks ik;k tkrk

gS fd xqVdk rsy dh lrg ls Bhd uhps rFkk bldk vk/kk fgLlk rsy ds vUnj rFkk vk/kk ikuh ds vUnj gSA ikuh ds lkis{k rsy dk

?kuRo gksxk :

(1) 0.5 (2) 0.8 (3) 0.7 (4) 0.6

A. 4


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sol.4

V g v g5

(Initial condition) ....(i)

0

v vV g g g

2 2 (Final condition)

oil 4

2 2 5

oil 4 1 3

2 5 2 10

oil

30.6

5

Heat Transfer

17. Two materials having coefficients of thermal conductivity '3K' and 'K' and thickness 'd' and '3d', respectively,

are joined to form a slab as shown in the figure. The temperatures of the outer surfaces are ‘ 2 ’ and ‘ 1 ’

respectively, ( 2 > 1 ). The temperature at the interface is:

fn[kk;s x;s fp=kkuqlkj '3K' rFkk 'K' Å"ek pkydrk xq.kkad ,oa, Øe'k% 'd' rFkk '3d' eksVkbZ okys nks inkFkksZa dks tksM+dj ,d

ifV~Vdk cuk;h x;h gSA muds ckgjh lrgksa ds rkieku Øe'k% ‘ 2 ’ vkSj ‘ 1 ’ gSa ( 2 > 1 )A varji"B dk rkieku gS :

(1) 1 25

6 6

(2)

1 29

10 10

(3)

1 22

3 3

(4)

1 2

2

A. 2

sol. 2 1

3KA KAH

d 3d

2 19

10 10

Errors & Significant Digits

18. The area of a square is 5.29 cm2. The area of 7 such squares taking into account the significant figures is :

,d oxZ dk {ks=kQy 5.29 cm2 gSA ,sls lkr oxksZa dk {ks=kQy mfpr lkFkZd vadksa esa gksxk :

(1) 37.03 cm2 (2) 37.0 cm2 (3) 37.030 cm2 (4) 37 cm2

A. 2

sol. 5.29 × 7 = 37.03 cm2

But answer should be in 3 significant digits. So area = 37.0

Geometrical Optics

19. Diameter of the objective lens of a telescope is 250 cm. For light of wavelength 600 nm. coming from a distant


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object, the limit of resolution of the telescope is close to :

,d nwjn'khZ ds vfHkn`';d ysUl dk O;kl 250 cm gSA ,d nwj fLFkr oLrq ls vkus okys rjaxnS/;Z 600 nm ds izdk'k ds fy;s

nwjn'khZ dh foHksnu lhek gksxh] yxHkx :

(1) 1.5 × 10–7 rad (2) 3.0 × 10–7 rad (3) 2.0 × 10–7 rad (4) 4.5 × 10–7 rad

A. 2

sol.1.22

D

9

71.22 600 10100 2.92 10

250

× 10–7 rad

Dual Nature of Radiation & Matter

20. A particle 'P' is formed due to a completely inelastic collision of particles 'x' and 'y' having de-Broglie wavelengths

‘ x ’ and ‘ y ’ respectively. If x and y were moving in opposite directions, then the de-Broglie wavelength of

'P' is

nks d.k 'x' rFkk 'y', ftudh Mh&czkfXy rjaxnS/;Z Øe'k% ‘ x ’ rFkk ‘ y ’ gSa, ds lEiw.kZ vizR;kLFk la?kV~V ls ,d d.k 'P' cuk gSA

;fn d.k 'x' rFkk 'y' foijhr fn'kkvksa esa xfr'khy Fks] rks 'P' dh Mh&czkfXy rjaxnS/;Z gS :

(1) x y

x y

(2) x y (3) x y (4) x y

x y

A. 1

sol. 1

x

hP

2

y

hP

1 2

x y

1 1P P P h

h

P

x y

h 1 1h

y x

x y

| |1

x y

x y

.

| |


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Magnetic Field & Force

21. A moving coil galvanometer has a coil with 175 turns and area 1 cm2. It uses a torsion band of torsion constant

10–6 N-m/rad. The coil is placed in a magnetic field B parallel to its plane. The coil deflects by 1° for a current

of 1 mA. The value of B (in Tesla) is approximately

,d py dq.Myh /kkjkekih esa 175 Qsjkas okyh rFkk 1 cm2 {ks=kQy dh ,d dq.Myh yxh gSA blesa ejksM+kad 10–6 N-m/rad okys

,d ejksM+ cS.M dk iz;ksx gksrk gSA bl dq.Myh dks ,d pqEcdh; {ks=k B esa j[krs gSa tks fd blds lery ds lekUrj gSA 1 mA

/kkjk ds fy;s dq.Myh esa fo{ksi 1° gSA B dk eku (VsLyk esa) yxHkx gS :

(1) 10–4 (2) 10–2 (3) 10–1 (4) 10–3

A. 4

sol. NIAB = K

175 × 1 × 10–3 × 1 × 10–4 × 610

B180

410

B 9.97 10 T180 175

B = 10–3 T

Current Electricity

22. In a conductor, if the number of conduction electrons per unit volume is 8.5 × 1028 m–3 and mean free time is

25 fs (femto second), it’s approximate resistivity is (me = 9.1 × 10–31 kg)

fdlh pkyd esa ;fn pkyd bysDVªkWuksa dh la[;k izfr ,dkadh vk;ru 8.5 × 1028 m–3 gS vkSj ek/; eqDr le; 25 fs

(QsEVks&lsds.M) gS rks mldh djhch izfrjks/kdrk gS : (me = 9.1 × 10–31 kg)

(1) 10–5 m (2) 10–6 m (3) 10–7 m (4) 10–8 m

A. 4

sol. J = evd d

eEv

m

e.eE.

J Em

2.eJ .E E

m

2

m

e

( = 1/)

31

28 19 2 15

9.1 10m

8.5 10 (1.6 10 ) 25 10

( 59 53) 89.110 1.67 10 m

8.5 2.56 25


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Geometrical Optics

23. A thin convex lens L (refractive index = 1.5) is placed on a plane mirror M. When a pin is placed at A, such that

OA = 18 cm, its real inverted image is formed at A itself, as shown in figure. When a liquid of refractive index

l is put between the lens and the mirror, the pin has to be moved to A', such that OA' = 27 cm, to get its

inverted real image at A' itself. The value of l will be :

1.5 viorZukad ds ,d irys mÙky ysUl L dks] fdlh lery niZ.k M dh lrg ij j[krs gSaA tc ,d fiu dks A ij j[krs gSa] rc

bl d k okLr fod fd Ur q mYVk i zfr fcEc] fn[ kk; s fp=kkuql kj A ij gh curk gSA fn;k gS OA = 18 cmA viorZukad l ds ,d nzo

dks ysUl rFkk niZ.k ds chp Mkyus ij] fiu ds okLrfod ,oa mYVs izfrfcEc dks A' ij gh ikus ds fy, fiu dks A' rd bl izdkj

mBkrs gSa fd OA' = 27 cmA l dk eku gksxk :

(1) 2 (2) 4

3(3) 3 (4)

3

2

A. 2

sol. fL = 18 cm

1 20.5

18 R R = 18 cm

1

2

1 1( 1)

f 18

1 1( 1) 1 11 1

27 18 18 18

2 = 3(2 ) = 6 – 3

1

4

3

Electromagnetic Induction

24. Two coils 'P' and 'Q' are separated by some distance. When a current of 3 A flows through coil 'P', a magnetic

flux of 10–3 Wb passes through Q. No current is passed through 'Q'. When no current passes through 'P' and

a current of 2A passes through 'Q', the flux through ‘P’ is

nks dq.Mfy;k¡ 'P' rFkk 'Q' dqN nwjh ij j[kh gSaA tc dq.Myh 'P' esa 3A dh /kkjk izokfgr gksrh gS rks dq.Myh 'Q' ls 10–3 Wb


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dk pqEcdh; ¶yDl xqtjrk gSA 'Q' esa dksbZ /kkjk ugha gSA tc 'P' esa dksbZ /kkjk ugha gS rFkk 'Q' ls 2A /kkjk izokfgr gksrh gS] rks ‘P’

ls xqtjus okyk ¶yDl gksxk :

(1) 6.67 × 10–3 Wb (2) 6.67 × 10–4 Wb (3) 3.67 × 10–3 Wb (4) 3.67 × 10–4 Wb

A. 2

sol. Q = Mi

10–3 = M(3)

P = M(2)

3

P10

3 2

4 4P

2010 6.67 10 Wb.

3

Calorimetry

25. A massless spring (k = 800 N/m), attached with a mass (500 g) is completely immersed in 1 kg of water. The

spring is stretched by 2 cm and released so that it starts vibrating. What would be the order of magnitude of the

change in the temperature of water when the vibrations stop completely? (Assume that the water container and

spring receive negligible heat and specific heat of mass = 400 J/kg K, specific heat of water = 4184 J/kg K)

500 g nzO;eku ls tqM+h ,d nzO;eku jfgr fLizax (k = 800 N/m) dks 1 kg ikuh esa iw.kZr;k Mqck;k x;k gSA fLizax dks 2 cm

yEckbZ ls [khapdj NksM+us ij nksyu vkjEHk gks tkrs gSaA tc nksyu iw.kZr;k :d tkrs gSa rc ikuh ds rkieku esa cnyko dh dksfV

gksxh : (ekuk fd ikuh ds ik=k vkSj fLizax dks feyh Å"ek ux.; gS rFkk nzO;eku dh fof'k"V Å"ek = 400 J/kg K, ikuh dh fof'k"V

Å"ek = 4184 J/kg K)

(1) 10–5 K (2) 10–1 K (3) 10–3 K (4) 10–4 K

A. 1

sol.21

k x E(dissipated)2

1 2 2 16

800 J2 100 100 100

16 1400 T 1 4184 T

100 2

16

(200 4184) T 4384 T100

516

T 3.6 10 K4384 100

COM, Momentum & Collision

26. A particle of mass ‘m’ is moving with speed ‘2v’ and collides with a mass ‘2m’ moving with speed ‘v’ in the

same direction. After collision, the first mass is stopped completely while the second one splits into two particles

each of mass ‘m’, which move at angle 45° with respect to the original direction. The speed of each of the


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moving particle will be

nzO;eku ‘m’ dk ,d d.k pky ‘2v’ ls tkrs gq;s ,d nzO;eku ‘2m’ ds d.k tks blh fn'kk esa pky ‘v’ ls tk jgk gS] ls la?kV~V

djrk gSA la?kV~V ds ckn igyk d.k fLFkj voLFkk esa vk tkrk gS rFkk nwljk d.k ,d gh nzO;eku ‘m’ ds nks d.kksa esa foHkkftr gks

tkrk gSA ;s nksuksa d.k vkjfEHkd fn'kk ls 45° ds dks.k ij tkrs gS :

izR;sd pyk;eku d.k dh xfr dk eku gksxk :

(1) v

2 2 (2) v

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

A. 4

sol. Initial momentum · Pi = 2mv + 2mv = 4 mv

Let v' be the speed of l particle

mv '

2 4mv2

v ' 2 2 v

Waves on a String

27. A string 2.0 m long and fixed at its ends is driven by a 240 Hz vibrator. The string vibrates in its third harmonic

mode. The speed of the wave and its fundamental frequency is

nksuksa fljksa ls c¡/kh gqbZ 2.0 m yEch ,d Mksjh 240 Hz ds ,d dfEi=k ls pkfyr gSA Mksjh vius rhljs xq.kko`Ùkh (harmonic) esa

daiu djrh gSA rjax dh pky vkSj bldh ewy vko`fÙk gSa :

(1) 320 m/s, 120 Hz (2) 320 m/s, 80 Hz (3) 180 m/s, 80 Hz (4) 180 m/s, 120 Hz

A. 2

sol.2

2 3

4

m3

4

v 240 320 m / s3

3rd harmonicf

n = nf

0


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0

240f 80 Hz

3

Semiconductors

28. The logic gate equivalent to the given logic circuit is

fn;s x;s ykWftd ifjiFk dk rqY; ykWftd xsV gS :

(1) OR (2) NAND (3) AND (4) NOR

A. 1

sol. y A.B A B

y = A + B

OR gate

Electromagnetic Induction

29. A very long solenoid of radius R is carrying current I(t) = kte–t (k > 0), as a function of time (t 0). Counter

clockwise current is taken to be positive. A circular conducting coil of radius 2R is placed in the equatorial

plane of the solenoid and concentric with the solenoid. The current induced in the outer coil is correctly depicted,

as a function of time, by :

R f=kT;k dh vR;kf/kd yEch ifjukfydk esa izokfgr /kkjk I(t) = kte–t (k > 0) le; ds Qyu (t 0) ds :i esa gSA okekorZ

fn'kk esa /kkjk dks /kukRed fy;k x;k gSA 2R f=kT;k okyh ,d oÙkkdkj dq.Myh dks ifjukfydk ds ledsUnzh; rFkk blds e/;orhZ

lery esa j[krs gSaA cká dq.Myh esa izsfjr /kkjk dks le; ds Qyu esa lgh :i ls n'kkZus okyk xzkQ gS :

(1) (2)

(3) (4)

A. 2


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sol. i = te–t.k

k1i = k

1te–t + k

1te–t

t t

1 1

dE k e k te

dt

= –k1e–t (1 – t)

Induced current t1

EI k e (1 t)

r

Electrostatics

30. Four point charges –q, +q, +q and –q are placed on y-axis at y = –2d, y = –d, y = +d and y =+2d, respectively.

The magnitude of the electric field E at a point on the x-axis at x = D, with D >> d, will behave as:

pkj fcUnq vkos'kksa –q, +q, +q vkSj –q dks y-v{k ij] Øe'k% y = –2d, y = –d, y = +d rFkk y =+2d ij j[kk x;k gSA x-v{k

ij mifLFkr ,d fcUnq x = D, tgk¡ D >> d gS, ij fo|qr {ks=k ds ifjek.k E dk O;ogkj gksxk :

(1) 3

1E

D (2)

1E

D (3) 4

1E

D (4) 2

1E

D

A. 3

sol. Consider as a dipole

3 3

2 2 2 22 20 0

2qD 2qDE ( ve x)

4 D 4y 4 (D y )

3 332 20 2 2

2qD 1 1| E |

4 D2y y

1 1D D

2 2 2

2 2 2 40 0

2q 3 4y 3 y 9dy| E | 1 . 1

4 D 2 D 2 D 4 D


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CHEMISTRY09 APRIL 2019 [Phase : II]

JEE MAIN PAPER ONLINE1. The major product of the following reaction is :

fuEu vfHkfØ;k dk eq[; mRikn gS &

CH OH2

OH

CO Et2

H SO (cat.)2 4

CHCl3

(1)

O

CO Et2

(2)

OH

O

O

(3)

O

COOH

(4)

OH

OEt

O

A. 2

sol.

O

OHOH

C–O–Et

H SO (cat.)2 4

CHCl3

O

O

OH

Acid catalysed intramolecular esterification

O

OH

C–O–Et

O

OHOH

H

+O–H

–H

O

OH

O

2. The peptide that gives positive ceric ammonium nitrate and carbylamine tests is

og isIVkbM tks ldkjkRed lsfjd veksfu;e ukbVªsV rFkk dkfcZy,sehu ijh{k.k nsrk gS] og gS %(1) Ser - Lys (2) Lys - Asp (3) Gln - Asp (4) Asp - Gln

A. 1

sol. Ceric ammonium nitrate test is given by alcohol. Only serine(ser) contain –OH group.

3. Which of the following compounds is a constituent of the polymer – HN – C – NH – CH – 2

O

n?

(1) Formaldehyde (2) Ammonia (3) Methylamine (4) N-Methyl urea

fuEu esa ls dkSulk ,d ;kSfxd] cgqyd – HN – C – NH – CH – 2

O

n dk la?kVd gS\

(1) QkesZYMhgkbM (2) veksfu;k (3) esfFky ,sehu (4) N-esfFky ;wfj;k

A. 1


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sol. [NH – C – NH – CH ]2

O

- urea-formaldehyde resin

Monomers : Urea and formaldehyde

4. During compression of a spring the work done is 10 kJ and 2 kJ escaped to the surroundings as heat. The

change in internal energy, U (in kJ) is

,d fLizax dks laihfMr djus esa fd;k x;k dk;Z 10 kJ gS rFkk 2 kJ Å"ek ds :i esa okrkoj.k dks pyk tkrk gSA vkarfjd ÅtkZ

esa ifjorZu U (kJ esa) gksxk :

(1) 12 (2) –12 (3) 8 (4) –8

A. 3

sol. w = 10 kJ

q = – 2 kJ

U = q + w = 10 – 2 = 8 kJ

5. What would be the molality of 20% (mass/mass) aqueous solution of KI (molar mass of KI = 166 g mol–1)

KI ds 20% (nzO;eku/nzO;eku) tyh; foy;u dh eksyyrk D;k gksxh \ (KI dk eksyj nzO;eku = 166 g mol–1)

(1) 1.48 (2) 1.51 (3) 1.08 (4) 1.35

A. 2

sol. 20% W/W KI solution

i.e. 100 g solution contains 20 g KI

Mass of solvent = 100 – 20 = 80 g

20 1000

Molality166 80

1.51 molar

6. A solution of Ni(NO3)2 is electrolysed between platinum electrodes using 0.1 Faraday electricity. How many

mole of Ni will be deposited at the cathode?

0.1 QSjkMs fo|qr dk iz;ksx djrs gq,, IysfVue bysDVªksMksa ds chp, Ni(NO3)2 ds foy;u dks fo|qr vi?kfVr fd;k x;kA dSFkksM

ij Ni dk fdruk eksy fu{ksfir gksxk\

(1) 0.20 (2) 0.15 (3) 0.10 (4) 0.05

A. 4

sol. 0.1 F of electricity is passed through Ni(NO3)2 solution

Amount of Ni deposited = 0.1 eq

Moles 0.1

0.052


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7. HF has highest boiling point among hydrogen halides, because it has

(1) Strongest hydrogen bonding (2) Lowest dissociation enthalpy

(3) Strongest van der Waals’ interactions (4) Lowest ionic character

HF dk DoFkukad gkbMªkstu gSykbMksa esa mPpre gksrk gS] bldk dkj.k gS :

(1) izcyre gkbMªkstu vkcU/ku (2) fuEure fo;kstu ,UFkSYih

(3) izcyre okUMj okYl vU;ksU; fØ;k (4) fuEure vk;fud y{k.k

A. 1

sol. HF has highest boiling point among the hydrogen halides due to strong H-bonding between HF molecules.

8. The one that is not a carbonate ore is

(1) Bauxite (2) Calamine (3) Siderite (4) Malachite

og ,d tks dkcksZusV v;Ld ugha gS] og gS :

(1) ckWDlkbV (2) dsykekbu (3) flMsjkbV (4) esykdkbV

A. 1

sol. Bauxite AlOx (OH)3–2x

Calamine ZnCO3

Siderite FeCO3

Siderite FeCO3

Malachite CuCO3 . Cu(OH)2

9. Noradrenaline is a / an

(1) Neurotransmitter (2) Antihistamine (3) Antacid (4) Antidepressant

ukj,sMªhusfyu gS ,d :

(1) ra=kdh; lapkjd (2) izfrfgLVkfeu (3) izfrvEy (4) izfr&volknd

A. 1

sol. Noradrenaline is neurotransmitter.

10. Among the following species, the diamagnetic molecule is

fuEu Lih'khT+k esa] izfrpqEcdh; v.kq gS :

(1) CO (2) NO (3) O2 (4) B2

A. 1

sol. Molecule No. of unpaired electrons

NO 1

CO Zero

O2 2

B2 2

Diamagnetic species is CO

11. The correct statements among I to III regarding group 13 element oxides are,


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(I) Boron trioxide is acidic.

(II) Oxides of aluminium and gallium are amphoteric.

(III) Oxides of indium and thallium are basic.

(1) (II) and (III) only (2) (I) and (II) only (3) (I), (II) and (III) (4) (I) and (III) only

xzqi–13 rRoksa ds vkWDlkbMksa ls lEcfU/kr I ls III esa ls lgh dFku gSa :

(I) cksjkWu VªkbvkWDlkbM vEyh; gSA

(II) ,Y;wehfu;e rFkk xSfy;e ds vkWDlkbM mHk;/kehZ gSaA

(III) bufM;e rFkk FkSfy;e ds vkWDlkbM {kkjh; gSaA

(1) dsoy (II) rFkk (III) (2) dsoy (I) rFkk (II) (3) (I), (II) rFkk (III) (4) dsoy (I) rFkk (III)

A. 3

sol. B2O3 is an acidic oxide

Al2O3 and Ga2O3 are amphoteric oxide

In2O3 and Tl2O are basic oxide

12. The major products A and B for the following reactions are, respectively

fuEufyf[kr vfHkfØ;kvksa ds eq[; mRikn A rFkk B Øe'k% gSa :

IO

KCNDMSO

[A]H /Pd2 [B]

(1) CN

OCH NH2 2

O

;

(2) I

CH –NH2 2

;

HO CNI

HO

(3) I

CH –NH2 2

;

HO CNH

HO

(4) CN

OCH NH2 2

OH

;

A. 4


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sol.I

O

KCNDMSO

CNO

CH NH2 2

OH

H ,Pd2

13. Increasing order of reactivity of the following compounds for SN1 substitution is

SN1 izfrLFkkiu ds fy, fuEu ;kSfxdksa dh vfHkfØ;k'khyrk dk c<+rk Øe gS :

(A)

CH3

CH3

CH Cl2 – (B) H C3 Cl

(C) Cl

H CO3

(D) Cl

(1) (B) < (C) < (A) < (D) (2) (B) < (C) < (D) < (A)

(3) (B) < (A) < (D) < (C) (4) (A) < (B) < (D) < (C

A. 3

sol. SN1 reaction proceeds via formation of carbocation.

CH2

OCH3

(C)

CH2

(D)

>

On comparing (A) and (B), in (A) there is formation of tertiary carbocation CCH3

CH3

CH3

after rearrangement

while (B) is primary.

So, (C) > (D) > (A) > (B).

14. Which of the following potential energy (PE) diagrams represents the SN1 reaction?

fLFkfrt ÅtkZ (PE) dk fuEu esa ls dkSulk vkjs[k SN1 vfHkfØ;k dks vfHkO;Dr djrk gS :


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(1)

PE

Progress of reaction

(2)

PE


(3)

PE


(4)

PE


A. 4

sol. In SN1 reaction, formation of carbocation (1st step) is rate determining step (RDS)

Correct graph is given in option-4.

15. Molal depression constant for a solvent is 4.0 K kg mol–1. The depression in the freezing point of the solvent

for 0.03 mol kg–1 solution of K2SO4 is (Assume complete dissociation of the electrolyte)

,d foyk;d ds fy, eksyy voueu fLFkjkad 4.0 K kg mol–1 gSA K2SO4 ds 0.03 mol kg–1 foy;u ds fy, foyk;d ds

fgekad esa fxjkoV gksxh, (eku yhft, fo|qr vi?kV~; dk fo;kstu iw.kZ :is.k gS)

(1) 0.36 K (2) 0.18 K (3) 0.12 K (4) 0.24 K

A. 1

sol. 22 4 4K SO 2K SO

i (Van't Hoff Factor) = 3

Tf = iKfm

= 3 × 4 × 0.03= 0.36 K

16. Consider the given plot of enthalpy of the following reaction between A and B.

A + B C + D

Identify the incorrect statement

20

15

10

5

Enth

alpy (

kJ

mol

)–

1

Reaction Coordinate

D

A + B C

(1) Activation enthalpy to form C is 5 kJ mol–1 less than that to form D


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(2) D is kinetically stable product

(3) Formation of A and B from C has highest enthalpy of activation

(4) C is the thermodynamically stable product

fuEufyf[kr A ,oa B ds chp vfHkfØ;k dh ,UFkSYih ds fn;s x;s IykV ij fopkj dhft,A

A + B C + D

rFkk xyr dFku dks crkb;sA

20

15

10

5

Enth

alpy (

kJ

mol

)–

1

Reaction Coordinate

D

A + B C

(1) C dks cukus esa lafØ;.k ,UFkSYih, D dks cukus esa yxus okyh lafØ;.k ,UFkSYih ls 5 kJ mol–1 de gSA

(2) D xfrdr% LFkk;h mRikn gSA

(3) C ls A rFkk B ds cuus esa lafØ;.k ,UFkSYih mPpre gSA

(4) C Å"ekxfrdh; :i ls fLFkj mRikn gSA

A. 1

sol. Activation enthalpy to form C is 5 kJ more than that to form D.

Ea = (A + B) C 15 KJ/mole

Ea = (A + B) D 10 KJ/mole

17. In an acid-base titration, 0.1 M HCl solution was added to the NaOH solution of unknown strength. Which of

the following correctly shows the change of pH of the titration mixture in this experiment?

,d vEy {kkjd vuqekiu esa, 0.1 M HCl foy;u dks ,d vKkr lkeF;Z okys NaOH ds foy;u esa feyk;k x;kA bl iz;ksx

esa, fuEu esa ls dkSu vuqekiu feJ.k ds pH–ifjorZu dks lgh&lgh iznf'kZr djrk gS\

pH

V (mL)(A)

pH

V (mL)(B)

pH

V (mL)(C)

pH

V (mL)(D)

(1) A (2) C (3) B (4) D

A. 1

sol. The pH of NaOH is more than 7 and during the titration it decreases so graph (1) is correct


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18. Hinsberg’s reagent is

fgalcxZ vfHkdeZd gS %

(1) C6H5SO2Cl (2) (COCl)2 (3) C6H5COCl (4) SOCl2

A. 1

sol. Hinsberg’s reagent is benzenesulphonyl chloride

SO Cl2

19. The layer of atmosphere between 10 km to 50 km above the sea level is called as

(1) Stratosphere (2) Mesosphere (3) Thermosphere (4) Troposphere

leqnz ry ls Åij 10 km ls 50 km ds chp dh ok;qeaMy irZ dks dgk tkrk gS :

(1) LVsªVksLQh;j (2) eslksLQh;j (3) FkeksZLQh;j (4) VªksiksLQh;j

A. 1

sol. Between 10-50 km above sea level lies stratosphere.

20. Assertion : For the extraction of iron, haematite ore is used.

Reason : Haematite is a carbonate ore of iron.

(1) Only the reason is correct

(2) Only the assertion is correct

(3) Both the assertion and reason are correct and the reason is the correct explanation for the assertion

(4) Both the assertion and reason are correct, but the reason is not the correct explanation for the assertion

dFku : vk;ju ds fu"d"kZ.k ds fy, gsekVkbV v;Ld iz;qDr gksrk gSA

dkj.k : gsekVkbV vk;ju dk dkcksZusV v;Ld gSA

(1) dsoy dkj.k lgh gSA

(2) dsoy dFku lgh gSA

(3) dFku rFkk dkj.k nksuksa lR; gSa vkSj dkj.k] dFku dh lgh O;k[;k djrk gSA

(4) dFku rFkk dkj.k nksuksa lR; gSa ijUrq dkj.k] dFku dh lgh O;k[;k ugha djrk gSA

A. 2

sol. For the extraction of iron, haematite ore in used.

Haematite = Fe2O3

21. At a given temperature T, gases Ne, Ar, Xe and Kr are found to deviate from ideal gas behaviour. Their

equation of state is given as RT

PV b

at T..

Here, b is the van der Waal’s constant. Which gas will exhibit steepest increase in the plot of Z (compression

factor) vs P?


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fn;s x;s rki T ij ;g ik;k x;k fd Ne, Ar, Xe rFkk Kr xSlsa vkn'kZ xSl O;ogkj ls fopfyr gksrh gSaA mudk voLFkk lehdj.k

bl izdkj fn;k gS RT

PV b

; fn;s x;s T ij

;gk¡ b okUMjokYl fLFkjkad gSA dkSulh xSl Z (laihMudkjd) rFkk P ds IykV esa lokZf/kd [kM+h o`f) iznf'kZr djsxh?

(1) Kr (2) Ar (3) Xe (4) Ne

A. 3

sol.m

RTP

V b

PVm – Pb = RT

MPV Pb

1RT RT

Pb

Z 1RT

Slope of Z vs P curve (straight line) b

RT

Higher the value of b, more steeper will be the curve and b size of gas molecules

22. The amorphous form of silica is

(1) Quartz (2) Tridymite (3) Kieselguhr (4) Cristobalite

flfydk dk vfØLVyh; :i gS :

(1) DokV~Zl (2) VªkbMkbekbV (3) fdtsyxwj (4) fØLVkscsykbV

A. 3

sol. Quartz, tridymite and cristobalite are crystalline forms of silica.

Kieselguhr is an amorphous form of silica.

23. The correct statements among I to III are

(I) Valence bond theory cannot explain the color exhibited by transition metal complexes.

(II) Valence bond theory can predict quantitatively the magnetic properties of transition metal complexes.

(III) Valence bond theory cannot distinguish ligands as weak and strong field ones.

(1) (II) and (III) only (2) (I), (II) and (III) (3) (I) and (II) only (4) (I) and (III) only

I ls III esa ls lgh dFku gSa :

(I) laØe.k /kkrq ladjksa }kjk iznf'kZr jax dks la;kstdrk vkcU/k fl)kUr le>k ugha ldrkA

(II) laØe.k /kkrq ladjksa ds pqEcdh; xq.kksa dh ek=kkRed izkxqfDr la;kstdrk vkcU/k fl)kUr dj ldrk gSA

(III) la;kstdrk vkcU/k fl)kUr nqcZy rFkk izcy {ks=k ds fyxsUMksa ds chp vUrj ugha crk ldrkA

(1) dsoy (II) rFkk (III) (2) (I), (II) rFkk (III) (3) dsoy (I) rFkk (II) (4) dsoy (I) rFkk (III)

A. 4

sol. Valence bond theory cannot predict quantitatively the magnetic properties of transition metal complex.


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24. In the following reaction carbonyl compound + MeOH HCl acetal

Rate of the reaction is the highest for

(1) Acetone as substrate and methanol in excess

(2) Propanal as substrate and methanol in stoichiometric amount

(3) Propanal as substrate and methanol in excess

(4) Acetone as substrate and methanol in stoichiometric amount

fuEu vfHkfØ;k esa dkcksZfuy ;kSfxd + MeOH HCl ,flVy

vfHkfØ;k dh nj fuEu esa ls fdlds fy, mPpre gS\

(1) ,lhVksu voLrj ds :i esa rFkk esFksukWy vkf/kD; esa

(2) izksisuy voLrj ds :i esa rFkk esFksukWy LVkWbfd;ksehVªh ek=kk esa

(3) izksisuy voLrj ds :i esa rFkk esFksukWy vkf/kD; esa

(4) ,lhVksu voLrj ds :i esa rFkk esFksukWy LVkWbfd;ksehVªh ek=kk esa

A. 3

sol.

O

CH C CH3 3– – CH CH CHO3 2– –

Generally, aldehydes are more reactive than ketones in nucleophilic addition reactions.

Rate of reaction with alcohol to form acetal and ketal is

O

CH C CH3 3– –CH CH CHO >3 2– –

25. The structures of beryllium chloride in the solid state and vapour phase, respectively, are

(1) Chain and dimeric (2) Dimeric and dimeric(3) Dimeric and chain (4) Chain and chain

csjhfy;e DyksjkbM dh lajpuk,a Bksl voLFkk rFkk ok"i izkoLFkk esa Øe'k% gSa :

(1) Üka[kyk rFkk f}yd (2) f}yd rFkk f}yd (3) f}yd rFkk Üka[kyk (4) Üka[kyk rFkk Üka[kyk

A. 1

sol. BeCl2 in vapour phase exist as dimer (below 1200 K temperature)

BeCl2 in solid state has chain structure.

26. p-Hydroxybenzophenone upon reaction with bromine in carbon tetrachloride gives

dkcZu VsVªkDyksjkbM esa czksehu ds lkFk vfHkfØ;k djus ij p-gkbMªkDlh csatksQsuksu nsrk gS :

(1)

Br

HO

O

(2)

HO

O

Br

(3)

HO

O

Br(4)

Br

HO

O


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A. 4

sol.

C OHBr /CCl2 4

O

C

O

Br

–M group +M group

OH

Product will formed as per –OH group (+M group)

27. The maximum possible denticities of a ligand given below towards a common transition and inner-transition

metal ion, respectively, are

VªkfUt'ku rFkk buj&VªkfUt'ku /kkrq ds izfr uhps fn;s x;s fyxS.M dh vf/kdre lEHko nfUrdrk;sa Øe'k% gSa :

OOC

OOCN N N

COO COO

COO

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

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

A. 1

sol. The maximum possible denticities of the given ligand towards transition metal ion is 6 and towards inner

transition metal ion is 8.

28. 10 mL of 1 mM surfactant solution forms a monolayer covering 0.24 cm2 on a polar substrate. If the polar head

is approximated as a cube, what is its edge length?

1 mM i`"B lafØ;d foy;u dk 10 mL ,d iksyj voLrj ij ,d eksuksys;j cukdj 0.24 cm2 ?ksjrk gSA ;fn iksyj gsM dks

,d ?kud :i esa ekuk tk;s rks blds dksj dh yEckbZ D;k gksxh\

(1) 2.0 pm (2) 2.0 nm (3) 0.1 nm (4) 1.0 pm

A. 1

sol. No. of surfactant molecule 6 × 1023 × 31010

1000

6 × 1018 molecule

Let edge length = a cm

Total surface area of surfactant = 6 × 1018 a2

0.24 = 6 × 1018 a2

a = 2 × 10–10 cm = 2 pm

29. Which one of the following about an electron occupying the 1s orbital in a hydrogen atom is incorrect?

(The Bohr radius is represented by a0)

(1) The probability density of finding the electron is maximum at the nucleus

(2) The electron can be found at a distance 2a0 from the nucleus


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(3) The magnitude of the potential energy is double that of its kinetic energy on an average

(4) The total energy of the electron is maximum when it is at a distance a0 from the nucleus

gkbMªkstu ijek.kq ds 1s d{kd esa mifLFkr bysDVªkWu ds ckjs esa fuEu esa ls dkSulk lgh ugha gS\

(cksj f=kT;k dks a0 }kjk iznf'kZr fd;k x;k gSA)

(1) bysDVªkWu ds ik;s tkus dh izkf;drk ?kuRo ukfHkd ij lokZf/kd gSA

(2) bysDVªkWu, ukfHkd ls 2a0 dh nwjh ij ik;k tk ldrk gSA

(3) vkSlru] fLFkfrt ÅtkZ dk eku blds xfrt ÅtkZ ds eku dk nqxquk gSA

(4) bysDVªkWu dh dqy ÅtkZ mPpre rc gksxh tc og ukfHkd ls a0 nwjh ij gSA

A. 4

sol. The total energy of the electron is minimum when it is at a distance a0 from the nucleus for 1 s orbital.

30. The maximum number of possible oxidation states of actinoides are shown by

(1) Berkelium (Bk) and californium (Cf) (2) Neptunium (Np) and plutonium (Pu)

(3) Actinium (Ac) and thorium (Th) (4) Nobelium (No) and lawrencium (Lr)

,DVhUok;Mksa dh lEHko vkWDlhdj.k voLFkkvksa dh mPpre la[;k fuEu esa ls fdlds }kjk iznf'kZr gksrh gS\

(1) cdsZfy;e (Bk) rFkk dsyhQksfuZ;e (Cf) (2) usIV;wfu;e (Np) rFkk IyqVksfu;e (Pu)

(3) ,DVhfu;e (Ac) rFkk Fkksfj;e (Th) (4) ukscsfy;e (No) rFkk ykjsfUl;e (Lr)

A. 2

sol. Actinoids Oxidation state shown

Th +4

Ac +3

Pu +3, +4, +5, +6, +7

Np +3, +4, +5, +6, +7

Bk +3, +4

Cm +3, +4

Lr +3

Maximum oxidation state is shown by (Np and Pu)


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MATHS

09 APRIL 2019 [Phase : II]

JEE MAIN PAPER ONLINE

Height & Distance

1. Two poles standing on a horizontal ground are of heights 5 m and 10 m respectively. The line joining their tops

makes an angle of 15° with the ground. Then the distance (in m) between the poles, is:

{kSfrt /kjkry ij [kM+s nks [kEcksa dh Å¡pkbZ Øe'k% 5 m rFkk 10 m gSA muds f'k[kjksa dks feykus okyh js[kk /kjkry ls 15° dk

dks.k cukrh gS] rks [kEcksa ds chp dh nwjh (m esa) gS :

(1) 5 2 3 (2) 10 3 1 (3) 5 3 1 (4) 52 3

2

A. 1

sol.

5 3 15 5tan15º d

d tan15º 3 1

5(4 2 3)

2

= 5( 2 3)

Determinant

2. If the system of equations 2x + 3y – z = 0, x + ky – 2z = 0 and 2x – y + z = 0 has a non-trivial solution (x, y,

z), then x y z

ky z x is equal to:

;fn lehdj.k fudk; 2x + 3y – z = 0, x + ky – 2z = 0 rFkk 2x – y + z = 0 dk ,d vrqPN (non-trivial) gy (x, y, z)

gS, rks x y z

ky z x cjkcj gS :

(1) 1

2(2) –4 (3)

3

4(4) –

1

4

A. 1


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sol.

2 3 19

0 1 k 2 0 k2

2 1 1

Equations are 2x + 3y – z = 0 ....(i)

2x – y + z = 0 ....(ii)2x + 9y – 4z = 0 ....(iii)

By (i) – (ii) 2y z z 4x and 2x y 0

x y z 1 1 9k 4

y z x 2 2 2

1

2

Quadratic Equation

3. If m is chosen in the quadratic equation (m2 + 1) x2 – 3x + (m2 + 1)2 = 0 such that the sum of its roots is

greatest, then the absolute difference of the cubes of its roots is:

;fn f}?kkrh; lehdj.k (m2 + 1) x2 – 3x + (m2 + 1)2 = 0 esa m bl izdkj fy;k x;k gS] fd blds ewyksa dk ;ksxQy vf/kdre

gS] rks blds ewyksa ds ?ku dk fujis{k vUrj gS :

(1) 8 3 (2) 10 5 (3) 4 3 (4) 8 5

A. 4

sol. Sum of roots = 2

3

m 1

For maximum m = 0Hence equation becomes x2 – 3x + 1 = 0

3, 1, | | 5

3 3 2 2| | ( )( ) | 5(9 1) 8 5

3 D

4. The vertices B and C of a ABC lie on the line, x 2 y 1 z

3 0 4

such that BC = 5 units. Then the area (in

sq. units) of this triangle, given that the point A(1, –1, 2), is:

ABC ds 'kh"kZ B rFkk C js[kk x 2 y 1 z

3 0 4

ij fLFkr gSa rFkk BC = 5 bdkbZ gSA ;fn fn;k gS fd fcUnq A(1, –1, 2)

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

(1) 5 17 (2) 34 (3) 6 (4) 2 34

A. 2


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sol.

Area of 1

ABC BC AD2

Given BC = 5 so we need perpendicular distance of A from line BC.

Let a point D on BC = (3)

ˆ ˆ ˆAD (3 3) i 2 j (4 2) k

Also AD & BC

should be perpendicular AD.BC 0

(3)3 + 2(0) + (4) 4 = 0

179 9 16 8 0

25

Hence, 1 68

D ,1,25 25

2 2

21 68| AD | 1 (2) 2

25 25

2 224 18

425 25

2 2 2

2

(24) 4(25) (18)

25

2

576 2500 324

25

2

3400

25

34.10 2 34

25 5

Area of triangle = 1

| BC | | AD |2

1 2 345 34

2 5

Differential Equation

5. If cosx dy

dx–y sinx = 6x, 0 x

2

and y 0

3

, then y

6

is equal to :


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;fn cosx dy

dx–y sinx = 6x, 0 x

2

rFkk y 0

3

gS, rks y

6

cjkcj gS :

(1) 2

2

(2)

2

4 3

(3)

2

2 3

(4)

2

2 3

A. 4

sol. cos xdy – (sin x)ydx = 6x dx

d(y cos x) 6x dx y cos x = 3x2 + C

As

2 21 3y 0 (0) C C

3 2 9 3

y cos x = 2

23x3

For y6

2 23 3y

2 36 3

2 23y 3y

2 12 2 3

Indefinite Integration

6. If esecx(secx tanxf (x) + sec × tanx + sec2x )) dx = esecxf(x) + C, then a possible choice of f(x) is :

;fn esecx(secx tanxf (x) + sec × tanx + sec2x )) dx = esecxf(x) + C, rks f(x) dk ,d laHko fodYi gS :

(1) secx – tanx –1

2(2) secx + tanx +

1

2(3) secx + xtanx –

1

2(4) x sec x + tanx +

1

2

A. 2

sol.secx 2e (sec x tan x f (x) (sec x tan x sec x))dx = secxe f(x) C

We know that

g(x) g(x)e ((g '(x)f (x)) f '(x))dx e f (x) C

2f (x) ((sex x tan x) sec x)dx

f (x) sec x tan x C

Binomial Theorem

7. If some three consecutive coefficients in the binomial expansion of (x + 1)n in powers of x are in the ratio

2:15:70, then the average of these three coefficients is:

;fn (x + 1)n ds x dh ?kkrksa esa f}in izlkj esa dksbZ rhu Øekxr xq.kkad 2:15:70 ds vuqikr esa gSa, rks bu rhu xq.kkadksa dk vkSlr


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gS :

(1) 625 (2) 964 (3) 232 (4) 227

A. 3

sol. Given nCr–1

: nCr : nC

r+1 = 2 : 15 : 70

n nr 1 r

n nr r 1

C C2 15&

C 15 C 70

r 2 r 1 3

&n r 1 15 n r 14

15r = 2n – 2r + 2 & 14r + 14 = 3n – 3r

17r = 2n + 2 & 17r = 3n – 14

i.e., 2n + 2 = 3n – 14 n = 16 & r = 2

16 16 161 2 3C C C 16 120 560

Mean3 3

696232

3

Straight Line

8. A rectangle is inscribed in a circle with a diameter lying along the line 3y = x + 7. If the two adjacent vertices

of the rectangle are (–8, 5) and (6, 5), then the area of the rectangle (in sq. units) is :

,d o`Ùk] ftldk ,d O;kl js[kk 3y = x + 7 ds vuqfn'k gS] ds vUrxZr ,d vk;r cuk;k x;k gSA ;fn vk;r ds nks layXu 'kh"kZ

(–8, 5) rFkk (6, 5) gSa, rks vk;r dk {ks=kQy (oxZ bdkb;ksa esa) gS :

(1) 56 (2) 84 (3) 72 (4) 98

A. 2

sol. Given situation

Perpendicular bisector of AB will pass from centre.

Equation of perpendicular bisector x = –1


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Hence centre (–1, 2)

Let 6 5

D ( , ) 1 & 22 2

& D

|AD| = 6 & |AB| = 14

Area = 6 × 14 = 84

Vector

9. If a unit vector a

makes angles 3

with i,

4

with j and (0, ) with k , then a value of is :

;fn ,d ek=kd lfn'k a

, i ls 3

, j ls

4

rFkk k ls (0, ) dks.k cukrk gS, rks dk ,d eku gS :

(1) 5

12

(2)

2

3

(3)

4

(4)

5

6

A. 2

sol. Let cos , cos , cos be direction cosines of a

Hence, by given data

cos cos ,cos cos & cos cos3 4

2 2 2cos cos cos 1

3 4

2 1 1 2cos cos , or

4 2 3 3

Continuity & Diff

10. If the function f(x) = a | x | 1, x 5

b | x | 3, x 5

is continuous at x = 5, then the value of a – b is:

;fn Qyu f(x) = a | x | 1, x 5

b | x | 3, x 5

x = 5 ij larr gS, rks a – b dk eku gS :

(1) 2

5(2)

2

5

(3)

2

5(4)

2

5

A. 4

sol. L.H.L x 5lim b | 5 | 3 (5 ) b 3

f(5) = R.H.L. x 5lim a | 5 | 1 a(5 ) 1

For continuity LHL = RHL

(5 ) b 3 (5 ) a 1

2 = (a – b)(5 – )


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2

a b5

Complex Number

11. Let z C be such that | z | < 1. If = 5 3z

5(1 z)

, then

ekuk z C bl izdkj gS fd | z | < 1. ;fn = 5 3z

5(1 z)

, rks :

(1) 5 Re() > 4 (2) 5 Re() > 1 (3) 4 Im() > 5 (4) 5 Im() < 1

A. 2

sol.5 3z

5w 5wz 5 3z5 5z

5– 5 = z(3 + 5)

5( 1)

z3 5

Given |z| < 1

5|| < |3 + 5|

25( 1) 9 25 15 15

(usin |z|2 = z z )

16 < 40

2

5

2

2Re( )5

1

Re( )5

3 D

12. Let P be the plane, which contains the line of intersection of the planes, x + y + z – 6 = 0 and 2x + 3y + z + 5

= 0 and it is perpendicular to the xy-plane. Then the distance of the point (0, 0, 256) from P is equal to :

ekuk P ,d lery gS ftlesa leryksa x + y + z – 6 = 0 rFkk 2x + 3y + z + 5 = 0 dh izfrPNsnu js[kk varfoZ"V gS rFkk ;g

xy-ry ds yacor gS] rks fcUnq (0, 0, 256) dh P ls nwjh cjkcj gS :

(1) 63 5 (2) 17

5(3) 205 5 (4)

11

5

A. 4

sol. Let the plane be

P (2x + 3y + z + 5) + (x + y + z – 6) = 0

As the above plane is perpendicular to xy plane

ˆ ˆ ˆ ˆ((2 )i (3 ) j (1 ) k).k 0

P x + 2y + 11 = 0


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Distance from (0, 0, 256)

0 0 11 11

5 5

Continuity & Diff

13. If f(x) = [x] – x

4

, x R, where [x] denotes the greatest integer function, then :

(1) x 4lim

f(x) exists but x 4lim

f(x) does not exist

(2) f is continuous at x = 4

(3) x 4lim

f(x) exists but x 4lim

f(x) does not exist

(4) Both x 4lim

f(x) and x 4lim

f(x) exist but are not equal

;fn f(x) = [x] – x

4

, x R gS, tgk¡ [x] egÙke iw.kkZad Qyu gS] rks :

(1) x 4lim

f(x) dk vfLrRo gS ijUrq x 4lim

f(x) dk vfLrRo ugha gSA

(2) x = 4 ij f larr gSA

(3) x 4lim

f(x) dk vfLrRo gS ijUrq x 4lim

f(x) dk vfLrRo ugha gSA

(4) x 4lim

f(x) rFkk x 4lim

f(x) nksuksa dk vfLrRo gS ijUrq og cjkcj ugha gSaA

A. 2

sol. L.H.Lx 4

xlim[x] 3 0 3

4

x x

x 4 [x] 3& 1 04 4

R.H.L x 4

x x xlim[x] 4 1 3 x 4 [x] 4 & 1 1

4 4 4

4f (4) [4] 4 1 3

4

LHL = f(4) = RHLHence f(x) is continuous at x = 4

Circle

14. The common tangent to the circles x2 + y2 = 4 and x2 + y2 + 6x + 8y – 24 = 0 also passes through the point :

oÙkksa x2 + y2 = 4 rFkk x2 + y2 + 6x + 8y – 24 = 0 dh mHk;fu"B Li'kZ js[kk fuEu esa ls fdl fcUnq ls gksdj tkrh gS\

(1) (–6, 4) (2) (–4, 6) (3) (4, –2) (4) (6, –2)

A. 4

sol. In given situation 1 2c c 1 2d | r r |


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x + y + 6x + 8y – 24 = 02 2

x + y = 42 2

Common tangentS

1 – S

2 = 0

6x + 8y – 20 = 0 3x + 4y – 10 = 0

Hence (6, –2) lies on it

Function

15. The domain of the definition of the function f(x) =2

1

4 x + log

10(x3 – x) is :

f(x) =2

1

4 x + log

10(x3 – x) }kjk ifjHkkf"kr Qyu dk izkar gS :

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

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

A. 2

sol. For domain denominator 0

4 – x2 0 x 2 ....(1)

and x3 – x > 0

x(x – 1)(x + 1) > 0

–1 0 1

x ( 1,0) (1, ) ....(2)

Hence domain is intersection of (1) & (2) i.e.

x ( 1,0) (1, 2) (2, )

Sequence & Progression

16. The sum of the series 1 + 2 × 3 + 3 × 5 + 4 × 7 + ... upto 11th term is :

Js.kh 1 + 2 × 3 + 3 × 5 + 4 × 7 + ... ds 11osa in rd ;ksxQy gS :

(1) 916 (2) 946 (3) 945 (4) 915

A. 2


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sol. 1 + 2.3 + 3.5 + 4.7 + ……

Lets break the sequence as shown

S

1 (2.3 3.5 4.7 ......)

We find 10

10n 1

S (n 1)(2n 1)

102

n 1

(2n 3n 1)

2n(n 1)(2 n 1) 3n(n 1)n(n 10)

6 2

2.10.11.21 3.10.1110

6 2

= 770 + 165 + 10 = 945

Hence required sum = 1 + 945 = 946

Parabola

17. The area (in sq. units) of the smaller of the two circles that touch the parabola, y2 = 4x at the point (1, 2) and

the x-axis is :

ijoy; y2 = 4x dks fcUnq (1, 2) ij Li'kZ djus okys rFkk x-v{k dks Li'kZ djus okys nks o`Ùkksa esa ls NksVs o`Ùk dk {ks=kQy (oxZ

bdkb;ksa esa) gS :

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

A. 3

sol.

The circle and parabola will have common tangent at P(1, 2)

Equation of tangent to parabola

(x 1)y (2) 4 2y 2x 2

2

Let equation of circle be (using family of circles)

(x – x1)2 + (y – y

1)2 + T = 0

c (x – 1)2 + (y – 2)2 + (x – y + 1) = 0

Also circle touches x-axis y-coordinate of centre = radius

c x2 + y2 + ()x + (–)y + () = 0


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2 24 2 4

( 5)2 2 2

2

24 45 4 4 4 20

4

8 64 64

2

4 4 2

4 4 2 (Other value forms bigger circle)

Hence centre of circle (2 2 2,4 2 2)

Radius 4 2 2

Area 2(4 2 2) 8 (3 2 2)

Matrices

18. The total number of matrices

A =

0 2y 1

2x y 1

2x y 1

, (x, y R, x y) for

Which ATA = 3I3 is :

vkO;wgksa A =

0 2y 1

2x y 1

2x y 1

, (x, y R, x y)

ftuds fy, ATA = 3I3 gS, dh dqy la[;k gS :

(1) 6 (2) 3 (3) 4 (4) 2

A. 3

sol.T

0 2x 2x 0 2y 1

A A 2y y y 2x y 1 3I

1 1 1 2x y 1

2

2

8x 0 0 3 0 0

0 6y 0 0 3 0

0 0 3 0 0 3

8x2 = 3, 6y2 = 3


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3 1x , y

8 2

Total combinations of (x, y) = 2 × 2 = 4

Mathematical Reasoning

19. If p (q r) is false, then truth values of p, q, r are respectively :

;fn p (q r) lR; ugha gS] rks p, q, r ds lR; eku Øe'k% gSa :

(1) T, F, F (2) F, F, F (3) T, T, F (4) F, T, T

A. 1

sol. For p (q r) to be F

(q r) must be F & p must be T

for (q r) to be F q and r must b F

p, q, r must be T, F, F

Statistics

20. The mean and the median of the following ten numbers in increasing order

10, 22, 26, 29, 34, x, 42, 67, 70, y are 42 and 35 respectively, then y

x is equal to :

o/kZeku Øe esa fuEu nl la[;kvksa 10, 22, 26, 29, 34, x, 42, 67, 70, y ds ek/; rFkk ekf/;dk Øe'k% 42 rFkk 35 gSa] rks y

x

cjkcj gS :

(1) 7

3(2)

8

3(3)

7

2(4)

9

4

A. 1

sol.ix x y 300

Mean 42 x y 120n 10

5 6T T 34 xMedian 35 x 36 & y 84

2 2

y 84 7Hence

x 36 3


21. Some identical balls are arranged in rows to form an equilateral triangle. The first row consists of one ball, the

second row consists of two balls and so on. If 99 more identical balls are added to the total number of balls

used in forming the equilateral triangle, then all these balls can be arranged in a square whose each side contains

exactly 2 balls less than the number of balls each side of the triangle contains. Then the number of balls used to

form the equilateral triangle is :


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dqN ,d tSlh xsansa iafDr;ksa esa bl izdkj j[kh xbZ gSa fd og ,d leckgq f=kHkqt cukrh gSA igyh iafDr esa ,d xsan gS] nwljh iafDr

esa nks xsansa gSa rFkk blh izdkj vU; iafDr;ksa esa xsansa gSaA leckgq f=kHkqt cukus esa yxh dqy xsanksa esa ;fn ,d tSlh 99 xsansa vkSj tksM+ nh

tk;sa rks bu lkjh xsanksa dks ,d ,sls oxZ ds vkdkj esa j[kk tk ldrk gS ftldh izR;sd Hkqtk esa f=kHkqt dh izR;sd Hkqtk ls Bhd nks

xsansas de gSa] rks leckgq f=kHkqt cukus esa yxh xsanksa dh la[;k gS :

(1) 157 (2) 225 (3) 262 (4) 190

A. 4

sol. Balls used in equilateral triangle n(n 1)

2

Here side of equiolateral triangle has n-balls No. of balls in each side of square is = (n – 2)

Given 2n(n 1)99 (n 2)

2

n2 + n + 198 = 2n2 – 8n + 8

n2 – 9n – 190 = 0

n2 – 19n + 10n – 190 = 0

(n – 19) (n + 10) = 0

n = 19

Balls used to form triangle

n(n 1) 19 20190

2 2

Sets and Relations

22. Two newspapers A and B are published in a city. It is known that 25% of the city population reads A and 20%

reads B while 8% reads both A and B. Further, 30% of those who read A but not B look into advertisements

and 40% of those who read B but not A also look into advertisements, while 50% of those who read both A

and B look into advertisements. Then the percentage of the population who look into advertisements is :

,d 'kgj esa nks lekpkj i=k A rFkk B izdkf'kr gksrs gSaA ;g Kkr gS fd 'kgj dh 25% tula[;k A i<+rh gS rFkk 20% B i<+rh

gS tcfd 8% A rFkk B nksuksa dks i<+rh gSA blds vfrfjDr, A i<+us rFkk B u i<+us okyksa esa 30% foKkiu ns[krs gSa vkSj B i<+us

rFkk A u i<+us okyksa esa Hkh 40% foKkiu ns[krs gSa, tcfd A rFkk B nksuksa dks i<+us okyksa esa ls 50% foKkiu ns[krs gSa] rks tula[;k

esa foKkiu ns[kus okyksa dk izfr'kr gS :

(1) 13.9 (2) 13 (3) 12.8 (4) 13.5

A. 1

sol. A

82025

B


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n(A only) = 25 – 8 = 17%

n(B only) = 20 – 8 = 12%

% of people from A only who read advertisement = 17 × 0.3 = 5.1%

% of people from B only who read advertisement = 12 × 0.4 = 4.8%

% of people from A & B both who read advertisement = 8 × 0.5 = 4%

Total % of people who read advertisement = 5.1 + 4.8 + 4 = 13.9%

Tangent & Normal

23. A water tank has the shape of an inverted right circular cone, whose semi-vertical angle is tan–11

2

. Water is

poured into it at a constant rate of 5 cubic metre per minute. Then the rate (in m/min.), at which the level of

water is rising at the instant when the depth of water in the tank is 10 m; is :

,d ikuh dh Vadh mYVs yac o`Ùkh; 'kadq ds vkdkj dh gS, ftldk v/kZ&'kh"kZ dks.k tan–11

2

gSA blesa ikuh 5 ?ku ehVj izfr

feuV dh leku nj ls Mkyk tkrk gS] rks Vadh esa ikuh dh xgjkbZ 10 m gksus ij og nj (eh-@fe- esa), ftl ij ikuh dh lrg c<+

jgh gS] gS :

(1) 2

(2)

1

15(3)

1

10(4)

1

5

A. 4

sol.

Given 3dv5m / min

dt

21V r h

3 ....(i) (where r is radius and h is height at any time)

Also r 1 dh 2dr

tan h 2rh 2 dt dt

....(ii)

Differentiate eqn. (i), we get

2dV 1 dr dh 1 1 dh2r h r 100 25

dt 3 dt dt 3 2 dt

at h = 10, r = 5


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75 dh5

3 dt

dh 1

m / mindt 5

Definite Integration

24. The value of the integral

1

0

x cot–1(1 – x2 + x4) dx is

lekdy 1

0

x cot–1(1 – x2 + x4) dx dk eku gS :

(1) 4

– log

e2 (2)

2

– log

e2 (3)

1

2 2

log

e2 (4)

1

4 2

log

e2

A. 4

sol. 1 1

1 2 4 1

4 2

0 0

1x cot 1 x x dx x tan

1 x x

1 2 21

2 2

0

x (x 1)x tan dx

1 x (x 1)

1 11 2 1 2

0 0

x tan x dx x tan (x 1)dx

Put x2 = t 2xdx = dt, (For 1st integration)

put x2 – 1 = k 2xdx = dk (For 2nd integration)

1 01 1

0 1

1 1x tan tdt 1tan kdk

2 2

1 1 11

2

0 0 0

1 t 1t tan t dt

2 1 t 2

0 01

2

1 1

kk tan k dx

1 k

1

2

0

1 1 1In 1 t

2 4 2 2

0

2

1

10 In 1 k

4 2

1 1

In2 10 In28 4 8 4

1

In24 2

Area Under Curve

25. The area (in sq. units) of the region 2y

A (x, y) : x y 42

is


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{ks=k 2y

A (x, y) : x y 42

dk {ks=kQy (oxZ bdkb;ksa eas) gS :

(1) 18 (2) 16 (3) 53

3(4) 30

A. 1

sol.

Hence area 4

2

xdy

4 2

2

yy 4 dy

2

42 3

2

y y 64 84y 8 16 2 8

2 6 6 6

32 4 40 14 5424 6 18

3 3 3 3 3

Straight Line

26. If the two lines x + (a – 1) y = 1 and 2x + a2y = 1 (a R – {0, 1}) are perpendicular, then the distance of their

point of intersection from the origin is

;fn nks js[kk;sa x + (a – 1) y = 1 rFkk 2x + a2y = 1 (a R – {0, 1}) yacor~ gSa] rks muds izfrPNsnu fcUnq dh ewy fcUnq ls

nwjh gS :

(1) 2

5(2)

2

5(3)

2

5(4)

2

5

A. 1

sol. For perpendicular m1 m

2 = – 1

2

1 21

a 1 a

2 = a2(1 – a)

a3 – a2 + 2 = 0


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(a + 1) (a2 + 2a + 2) = 0

a 1

Hence lines are x – 2y = 1 and 2x + y = 1

Intersection point 3 1

,5 5

Distance from origin = 9 1 10 2

25 25 25 5

Trig Ratio

27. The value of sin10° sin30° sin50° sin70° is

sin10° sin30° sin50° sin70° dk eku gS &

(1) 1

18(2)

1

32(3)

1

16(4)

1

36

A. 3

sol. sin(60° + A). sin(60° – A) sinA = 1

4sin 3A

Hence, sin10° sin50° sin70°

= sin10° sin(60° – 10°) sin(60° + 10°) 1

4 sin30°

Hence, sin10° sin30° sin50° sin70° 1

4 sin230°

1

16

Ellipse

28. If the tangent to the parabola y2 = x at a point (, ), ( > 0) is also a tangent to the ellipse, x2 + 2y2 = 1, then

is equal to

;fn ijoy; y2 = x ds ,d fcUnq (, ), ( > 0) ij, Li'kZ js[kk, nh?kZo`Ùk x2 + 2y2 = 1 dh Hkh Li'kZ js[kk gS, rks cjkcj gS:

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

A. 2

sol. Let tangent in terms of m to parabola and ellipse

i.e. y = mx 1

4m for parabola at point

2

1 1,

4m 2m

and 2 1y mx m

2 for ellipse on comparing

2 2

2

1 1 1 1m m

4m 2 16m 2

16m4 + 8m2 – 1 = 0

2 8 64 64 8 8 2 2 1m

2(16) 2(16) 4


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2

1 12 1

4m 2 14

4

Definite Integration

29. If f : R R is a differentiable function and f(2) = 6, then f (x)

x 26

2t dtlim

(x 2) is :

;fn f : R R ,d vodyuh; Qyu gS rFkk f(2) = 6 gS, rks f (x)

x 26

2t dtlim

(x 2) gS :

(1) 0 (2) 2f'(2) (3) 12f'(2) (4) 24f'(2)

A. 3

sol. Using L' Hospital rule and Leibnitz theorem

f (x)

6

x 2

2tdt

lim(x 2)

x 2

2f (x)f '(x) 0lim

1

2f(2)f '(2) = 12f '(2)


30. If the sum and product of the first three terms in an A.P. are 33 and 1155, respectively, then a value of its 11th

term is

;fn ,d lekUrj Js<+h ds izFke rhu inksa dk ;ksxQy rFkk xq.kuQy Øe'k% 33 rFkk 1155 gS, rks blds 11osa in dk ,d eku gS:

(1) –36 (2) 25 (3) –25 (4) –35

A. 3

sol. Let terms be a – d, a, a + d

3a = 33 11

Product of terms

(a – d) a (a + d) = 11 (121 – d2) = 1155

121 – d2 = 105 d = 4

if d = 4

1

2

3

T 7

T 11

T 15

T11

= T1 + 10d = 7 + 10(4) = 47

if d = – 4


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1

2

3

T 15

T 11

T 7

T11

= T1 + 10d = 15 + 10(–4) = –25

Matrix JEE (MAIN ONLINE) 2019 JEE Academy PHYSICS · Matrix JEE Academy : Piprali Road, Sikar Ph....

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