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Physics Paper With Solution

JEE(MAIN)-2020

(Memory Based)Online Test Paper

PART : PHYSICS Single Choice Type (,dy fodYih; izdkj)

This section contains 20 Single choice questions. Each question has 4 choices (1), (2), (3) and (4) for its answer, out of which Only One is correct. bl [k.M esa 20 ,dy fodYih iz'u gSaA izR;sd iz'u ds 4 fodYi (1), (2), (3) rFkk (4) gSa] ftuesa ls flQZ ,d lgh gSA

1. A block of mass m is suspended from a pulley in form of a circular disc of mass m & radius R. The systemis released from rest, find the angular velocity of disc when block has dropped by height h. (there is noslipping between string & pulley),d m nzO;eku dk CykWd f?kjuh ls jLlh ds }kjk tqM+k gqvk gSA f?kjuh pdrh ds :i esa gSA ftldk nzO;eku m rFkk

f=kT;k R gSA ;fn fudk; dks fojke ls NksM+k tkrk gSA rc CykWd }kjk h Å¡pkbZ uhps vkus ij pdrh dk dks.kh; osx

gksxk (f?kjuh rFkk pdrh esa dksbZ fQlyu ugha gS)

m

m

(1) 3gh4

R1 (2)

3gh2

R1 (3)

3gh2R (4)

3gh4R

Ans. (1)

Sol. mgh = 22

21mv

21

v = R (no slipping dksbZ fQlyu ugha)

mgh = 22

22

2mR

21Rm

21

mgh = 22Rm43

3gh4

R1

R3gh4

2

2. Three point masses 1kg, 1.5 kg, 2.5 kg are placed at the vertices of a triangle with sides 3cm, 4cm and5cm as shown in the figure. The location of centre of mass with respect to 1kg mass is :

2.5 kg

5cm4 cm

1 kg 3 cm

1.5 kg

(1) 0.6 cm to the right of 1 kg and 2 cm above 1 kg mass (2) 0.9 cm to the right of 1kg and 2 cm above 1 kg mass (3) 0.9 cm to the left of 1kg and 2 cm above 1kg mass (4) 0.9 cm to the right of 1 kg and 1.5 cm above 1kg mass

1

fp=k esa fn[kk, vuqlkj rhu fcUnq nzO;eku ftuds nzO;eku Øe'k% 1kg, 1.5kg rFkk 2.5kg gS] ,d f=kHkqt ds dksuksa ij

j[ks gSA f=kHkqt dh Hkqtk,sa Øe'k% 3 cm, 4cm rFkk 5cm gSA rc nzO;eku dsUnz dh fLFkfr 1kg nzO;eku ds lkis{k gksxh :

2.5 kg

5cm4 cm

1 kg 3 cm

1.5 kg

(1) 1kg ds nka;h vksj 0.6 cm nwj rFkk 1kg nzO;eku ds 2 cm Åij

(2) 1kg ds nka;h vksj 0.9 cm nwj rFkk 1kg nzO;eku ds 2 cm Åij

(3) 1kg ds cka;h vksj 0.9 cm nwj rFkk 1kg nzO;eku ds 2 cm Åij

(4) 1 kg ds nka;h vksj 0.9 cm nwj rFkk 1kg nzO;eku ds 1.5 cm Åij

Ans. (2) Sol. Take 1kg mass at origin

1 kg nzO;eku dks ewy fcUnq ysus ij

2.5 kg

5cm4 cm

1 kg 3 cm

1.5 kg

Y

X

cm1 0 1.5 3 2.5 0X 0.9cm

5

cm1 0 1.5 0 2.5 4Y 2cm

5

3. In a single slit diffraction set up, second minima is observed at an angle of 60°. The expected position offirst minima is,dy fNnz foorZu izfØ;k esa] 2nd fufEu"B (minima) 60° ds dks.k ij fn[kkbZ nsrk gS] rks 1st fufEu"B (minima) dh

visf{kr fLFkfr gksxh &

(1) 25° (2) 20° (3) 30° (4) 45° Ans. (1) Sol. For 2nd minima f}rh; fufEu"B ds fy,

d sin = 2

sin = 23 (given)

43

d

… (i)

So for 1st minima is blfy, izFke fufEu"B ds fy, d sin =

sin = 43

d

(from equation (i) (lehdj.k (i) ls

= 25.65° (from sin table) (sin rkfydk ls) 25

2

4. There are two infinite plane sheets each having uniform surface charge density + C/m2. They areinclined to each other at an angle 30° as shown in the figure. Electric field at any arbitrary point P is:fp=k esa nks vuUr yEckbZ dh lery ifêdk gS ftudk le:i i"Bh; vkos'k ?kuRo + C/m2 gSA nksuksa ifêdk ,d nwljs

ls 30° dks.k ij >qdh gSA rc fp=kkuqlkj fdlh Hkh ;knfPNd fcUnq P ij fo|qr {ks=k gksxk:

30°+

+

P

Y

X

(1) 0

3 1ˆ ˆ1– y – x2 2 2

(2) 0

3 1ˆ ˆ1 y – x2 2 2

(3) 0

3 1ˆ ˆ1– y x2 2 2

(4) 0

3 1ˆ ˆ1 y x2 2 2

Ans. (1) Sol.

30° +

+

P60°

0 0 0

ˆ ˆE cos60 –x – sin60 y2 2 2

E

= 0

3 1ˆ ˆ1– y – x2 2 2

5. A parallel plate capacitor with plate area A & plate separation d is filled with a dielectric material of

dielectric constant given by k = k0(1 + x). Calculate capacitance of system: (given d << 1)

,d lekUrj iV~V la/kkfj=k ftldh lrg dk {ks=kQy A gS rFkk ifV~Vdkvksa ds e/; dh nwjh d gSA buds e/; ,d

ijkoS|qrkad inkFkZ Hkjk gSA inkFkZ dk ijkoS|qrkad fu;rkad k = k0(1 + x) gSA rc fudk; dh /kkfjrk gksxh:

(fn;k gS d << 1)

x

(1) 2200 d1d

Ak

(2)

2d1

dAk 00 (3) d1

d2Ak 00

(4)

2d1

d2Ak 00

Ans. (2)

3

Sol. Capacitance of element ?kVd dh /kkfjrk = dx

Ak 0

dxx

Capacitance of element ?kVd dh /kkfjrk, C' =dx

A)x1(k 00

d

0 00 )x1(Akdx

'C1

)d1(nAk

1C1

00

Given fn;k gS& d << 1

2dd

Ak1

C1 22

00

2d1

Akd

C1

00

2d1

dAk

C 00

6. A long solenoid of radius R carries a time dependent current I = I0 t(1 – t). A ring of radius 2R is placedcoaxially near its centre. During the time interval 0 t 1, the induced current R and the induced emf VR

in the ring vary as:(1) current will change its direction and its emf will be zero at t = 0.25sec.(2) current will not change its direction & emf will be maximum at t = 0.5sec(3) current will not change direction and emf will be zero at 0.25sec.(4) current will change its direction and its emf will be zero at t = 0.5sec.,d R f=kT;k dh ijhuyhdk ftlesa ,d le; fuHkZj /kkjk I = I0 t(1 – t) izokfgr gks jgh gSA ,d 2R f=kT;k dh oy;

ifjuyhdk ds lek{kh; :i ls mlds dsUnz ds ikl fLFkr gSA le; vUrjky 0 t 1, ds nkSjku oy; esa izsjhr /kkjk RvkSj izsfjr fo+|qr okgd cy VR ds laxr lgh fodYi gksxk

(1) /kkjk viuh fn'kk cnysxh vkSj fo|qr okgd cy le; t = 0.25sec ij 'kwU; gksxkA(2) /kkjk viuh fn'kk ugha cnysxh vkSj fo|qr okgd cy le; t = 0.5sec ij vf/kdre gksxkA

(3) /kkjk viuh fn'kk ugha cnysxh vkSj fo|qr okgd cy le; t = 0.25sec ij 'kwU; gksxkA(4) /kkjk viuh fn'kk cnysxh vkSj fo|qr okgd cy t = 0.5sec ij 'kwU; gksxkA

Ans. (4) Sol. I = I0t – I0t2

= BA = 0nIA

VR = dtd– = –0nAI0 (1 – 2t)

VR = 0 at t = 21

4

and rFkk R = loopofcetansisRe

VR

sèk dk izfrjkyqi

RV

e

1/2 t

7. If 10% of intensity is passed from analyser, then, the angle by which analyser should be rotated suchthat transmitted intensity becomes zero. (Assume no absorption by analyser and polarizer).;fn ,d èkqzod ls 10% rhozrk xqtjrh gks rks èkqzod dks fdrus dks.k ls ?kqek;s tk;s fd fuxZr rhozrk 'kwU; gks tk;sA

(eku fyft,s dh /kzqod ls vo'kks"k.k 'kwU; gS)(1) 60° (2) 18.4° (3) 45° (4) 71.6°

Ans. (B) Sol. I = I0 cos2

20

0 cos10

II

cos = 101 = 0.31 <

21 which is 0.707

So > 45° and rFkk 90 – < 45º so only one option is correct blfy, dsoy ,d fodYi lgh gksxkA

i.e. vFkkZr 18.4ºangle rotated should be ?kqek;k gqvk dks.k = 90° – 71.6° = 18.4°

8. Three moles of ideal gas A with34

CC

V

P is mixed with two moles of another ideal gas B with 35

CC

V

P .

The v

P

CC of mixture is (Assuming temperature is constant)

vkn'kZ xSl A ds 3 eksy ftlds fy, fd34

CC

V

P gS] dks nwljh vkn'kZ xSl B ds 2 eksy ftlds fy,35

CC

V

P gS] ds

lkFk feyk;k tkrk gS rks feJ.k dh v

P

CC

dk eku crkb, (rkieku dks ,d&leku eku ekfu;sA)

(1) 1.5 (2) 1.42 (3) 1.7 (4) 1.3 Ans. (2)

Sol.

1–Rn

1–Rn

1–Rn

1–Rn

CnCnCnCn

2

2

1

1

2

22

1

11

V2V1

P2P1mixture

21

21

on rearranging we get iqu O;ofLFkr djus ij,

1–n

1–n

1–nn

2

2

1

1

mix

21

; 3/2

23/1

31–

5mix

1–5

mix = 9 + 3 = 12

1251

1217

mixure ; mix = 1.42

5

9. Given magnetic field equation is B = 3 × 10–8 sin(t + kx + ) j , then appropriate equation for electric field(E) will be :fn;s x;s pqEcdh; {kS=k dh lehdj.k B = 3 × 10–8 sin(t + kx + ) j gS] rc fo|qr {kS=k (E) dh mfpr lehdj.k gksxhA

(1) 20 × 10–9 sin (t + kx + ) k (2) 9 sin (t + kx + ) k(3) 16 × 10–9 sin (t + kx + ) k (4) 3 × 10–9 sin (t + kx + ) k

Ans. (2)

Sol. 0

0

EB

= C (speed of light in vacuum fuokZr esa izdk'k dh pky)

E0 = B0C = 3 × 10–8 × 3 × 108

= 9 N/C So blfy, E = 9 sin (t + kx + )

10. There is a LCR circuit, If it is compared with a damped oscillation of mass m oscillating with force constantk and damping coefficient 'b'. Compare the terms of damped oscillation with the devices in LCR circuit.,d LCR ifjiFk fn;k tkrk gSA bldh ,d voefUnr nksyu ds lkFk rqyuk dh tkrh gS ftldk æO;eku m, nksyu dk

cy fu;rkad 'k' gS vkSj voefUnr fu;rkad b gS] rc voefUnr nksyu dh bdkbZ;ksa dks LCR ifjiFk ls rqyuk djksA

(1) L m , C 1k

, R b (2) L m , C k, R b

(3) L k , C b, R m (4) L 1m

, C 1k

, R 1b

Ans. (1) Sol. In damped oscillation voefUnr nksyu esa

ma + bv + kx = 0

2

2

d x dxm b kxdtdt

= 0 …(i)

In the circuit ifjiFk esa

– iR – diLdt

– qc

= 0

2

2

d q dq 1L R .qdt cdt

= 0 …(ii)

Comparing equation (i) and (ii) rqyuk djus ij

m = L, b = R, k = 1c

11. A lift can hold 2000kg, friction is 4000N and power provided is 60HP. (1 HP = 746W) Find the maximumspeed with which lift can move up.,d fy¶V 2000kg nzO;eku mBk ldrh gS] ;fn ?k"kZ.k 4000N gS rFkk nh xbZ 'kfDr 60HP gSA (1 HP = 746W) rcfy¶V dh vf/kdre pky D;k gksxh] ftlls og ÆÅij tk ldsA

6

(1) 1.9 m/s (2) 1.7 m/s (3) 2 m/s (4) 1.5 m/s Ans. (1) Sol. 4000 × V + mg × V = P

20000400074660

= V

V = 1.86 m/s. 9.1 m/s.

12. A H–atom in ground state has time period T = 1.6 × 10–16 sec. find the frequency of electron in first excitedstate,d H–ijek.kq ds bysDVªkWu dk fuEure voLFkk esa vkorZdky T = 1.6 × 10–16 sec. gSA rc bysDVªkWu dh izFke mÙksftr

voLFkk esa vkorhZ gksxh &

(1) 7.8 × 1014 (2) 7.8 × 1016 (3) 3.7 × 1014 (4) 3.7 × 1016 Ans. (1)

Sol. T 2

32

zn

zn

zn

vr

81

nn

TT

32

31

2

1

T2 = 8T1 = 8 × 1.6 × 10–16 = 12.8 × 10–16

f2 = 16–108.121

7.8 × 1014

13. Magnification of compound microscope is 375. Length of tube is 150mm. Given that focal length ofobjective lens is 5mm, then value of focal length of eyepiece is:,d la;qDr lq{en'khZ dh vko/kZu {kerk 375 gSA uyh dh yEckbZ 150 mm gSA fn;k x;k gS vfHkn`';d ySUl dh Qksdl

nwjh fo = 5mm rc vfHkus=k ysal dh Qksdl nwjh fe dk eku gksxk ?(1) 2mm (2) 22mm (3) 12mm (4) 33mm

Ans. (2) Sol. Case-I

If final image is at least distance of clear vision ;fn vfUre izfrfcEc fudVre Li"B nf"V ij gks rks

M.P. =

e0 fD1

fL ; 375 =

ef251

5150

ef251

30375

ef25

30345

fe = 345750 = 2.17 cm; fe 22 mm

Case-II If final image is at infinity ;fn vfUre izfrfcEc vuUr ij gks

M.P. =

e0 fD

fL

= 375

fe 22 mm

7

14. 1 litre of a gas at STP is expanded adiabatically to 3 litre. Find work done by the gas. Given = 1.40 and31.4= 4.65,d xSl ds 1 yhVj dks STP ij 3 yhVj rd izlkfjr fd;k tkrk gSA xSl }kjk fd;k x;k dk;Z gksxkA fn;k gS =1.40 rFkk 31.4= 4.65(1) 100.8J (2) 90.5J (3) 45J (4) 18J

Ans. (2) Sol. P1 = 1atm, T1 = 273K

1 1 2 2P V P V

P2 = P1 1

2

VV

= 1atm 1.41

3

now work done vc fd;k x;k dk;Z = 1 1 2 2P V – P V– 1

= 88.7 J

Closest ans is 90.5 J

15. A string of length 60 cm, mass 6gm and area of cross section 1mm2 and velocity of wave 90m/s. Givenyoung's modulus is Y = 1.6 × 1011 N/m2. Find extension in string,d 60 cm yEch jLlh] ftldk nzO;eku 6gm vkSj vuqizLFk dkV dk {ks=kQy 1mm2 vkSj rjax dk osx 90m/s gSA(fn;k gS ;ax izR;kLFkrk xq.kkad Y = 1.6 × 1011 N/m2) jLlh esa izlkj crkb,

(1) 0.3 mm (2) 0.2 mm (3) 0.1 mm (4) 0.4 mm Ans. (1)

Sol. Tv

T = v2

2v YA

= 2v

AY

after substituting value of v,,A and Y we get v,,A rFkk Y dk eku j[kus ij

= 0.3 mm

16. Which of the following gate is reversiblebuesa ls dkSulk }kj mRØe.kh; gSA

(1) (2)

(3) (4) Ans. (1) Sol. A logic gate is reversible if we can recover input data from the output

eg. NOT gate ,d rdZ }kj mRØe.kh; gksxk ;fn fuxZr ladsr ds vk/kkj ij fuos'kh ladsr izkIr fd;k tk ldsA eg. NOT }kj

8

17. A thin uniform rod is of mass M and length L. Find the radius of gyration for rotation about an axis passing

through a point at a distance of L4

from centre and perpendicular to rod.

,d iryh ,dleku NM+ ftldk nzO;eku M vkSj yEckbZ L gSA rks NM+ ds dsUnz ls L4

nwjh vkSj NM+ ds yEcor v{k ds

lkis{k ?kq.kZu f=kT;k Kkr djsA

(1) 748

L (2) 548

L (3) 724

L (4) 1924

L

Ans. (1)

Sol.

L/4

22ML LM12 4

= MK2

2 2L L12 16

= K2

7K48

L

18. A satellite of mass 'M' is projected radially from surface of earth with speed 'u'. When it reaches a height

equal to radius of earth, it ejects a rocket of mass M10

and it itself starts orbiting the earth in circular path

of radius 2R, find the kinetic energy of rocket.,d nzO;eku 'M' ds mixzg dks i`Foh dh lrg ls 'u' pky ds lkFk f=kT; fn'kk eas iz{ksfir fd;k tkrk gSA tc og i`Foh

dh f=kT;k ds cjkcj Å¡pkbZ rd igq¡p tkrk gS] rc jkWdsV M10

nzO;eku dks fu"dkflr dj nsrk gS vkSj ;g Loa; 2R

f=kT;k ds oÙkh; iFk esa ifjØek djuk 'kq: dj nsrk gS] rks jkWdsV dh xfrt ÅtkZ gksxh

(1) 2 e119GM5M u

200R

(2) 2 e113GM5M u

200R

(3) 2 e119GMM u20 100R

(4) 2 e113GMM u20 200R

Ans. (1)

Sol. 2 2e eGM M GM M1 1Mu MvR 2 2R 2

M

u

R

R M

Rv

Me

Me

v = 2 eGMu

R

9

M/10Me

Vr

9M/10

VT

eGMV

2R

VT Transverse velocity of rocket VR Radial velocity of rocket

2R

eT

GMM 9MV10 10 2R

2 er

GMM V M u10 R

Kinetic energy xfrt ÅtkZ = 2 2 2e eT r

GM GM1 M MV V 81 100u 1002 10 20 2R R

= 2 e119GMM 100u20 2R

= 2 e119GM5M u

200R

19. The current 'i' in the given circuit isfn;s x;s ifjiFk esa /kkjk 'i' gS &

1V

2

1

1 i 2

(1) 0.2A (2) 0.3A (3) 0.5A (4) 0.25A Ans. (1)

10

Sol.

1V

2

1

1 i 2

I

I = 5.2

1 = 0.4A

i = 2I = 0.2A

20. A current carrying circular loop is placed in an infinite plane if i is the magnetic flux through the innerregion and o is magnitude of magnetic flux through the outer region, then,d /kkjkokgh oÙkkdkj ywi vuUr ry esa fLFkr gS] ;fn vkUrfjd {kS=k ls lEcaf/kr pqEcdh; ¶yDl i rFkk ckg~; {kS=k ls

lEcaf/kr pqEcdh; ¶yDl o gks rks(1) i > o (2) i < o

(3) i = – o (4) i = o Ans. 3 Sol. As magnetic field lines always form a closed loop, hence every magnetic field line creating magnetic flux

in the inner region must be passing through the outer region. Since flux in two regions are in opposite direction, i = – 0

Hindi. pawfd pqEcdh; {kS=k js[kk,saa cUn ywi dk fuekZ.k djrh gSA vr% izR;sd pqEcdh; {kS=k js[kk }kjk vkUrfjd {kS=k esa mRiUu

fd;k x;k pqEcdh; ¶yDl ckg~; {kS=k ls xqtjrk gSA vr% bu nksuksa {kS=kks esa mifLFkr ¶yDl foijhr fn'kk esa gksrk gSA

i = – 0

Numerical Value Type (la[;kRed izdkj) This section contains 5 Numerical value type questions.

bl [k.M esa 5 l[;kRed izdkj ds iz'u gSaA

21. Consider a loop ABCDEFA with coordinates A (0, 0, 0), B(5, 0, 0), C(5, 5, 0), D(0, 5, 0) E(0, 5, 5) andF(0, 0, 5). Find magnetic flux through loop due to magnetic field ˆ ˆB 3i 4k

ywi ABCDEFA dh dYiuk dhft, ftlds funsZ'kkad Øe'k% A (0, 0, 0), B(5, 0, 0), C(5, 5, 0), D(0, 5, 0)

E(0, 5, 5) rFkk F(0, 0, 5) gSA pqEcdh; {kS=k ˆ ˆB 3i 4k

ds dkj.k ywi ls lEcaf/kr pqEcdh; ¶yDl Kkr djksA

Ans. 175

11

Sol. = B.A

= ˆ ˆ(3i 4k) . ˆ ˆ(25i 25k)

= (3 × 25) + (4 × 25) = 175 weber

22. A Carnot's engine operates between two reservoirs of temperature 900K and 300K. The engine performs1200 J of work per cycle. The heat energy delivered by the engine to the low temperature reservoir in acycle is:,d dkuksZ±V batu nks rkih; fudk; ds e/; dk;Z djrk gS ftuds rkieku Øe'k% 900K rFkk 300K gSA batu izR;sd

pØ esa 1200J dk;Z djrk gSA ,d pØ esa batu fuEu rki okys fudk; dks fdruh Å"ek LFkkukUrjhr djrk gSA

Ans. 600

Sol. = hQ

W = 9003001 =

32

Qh = 23 W = 1800 J

QL = Qh – W = 600 J

23. A non-isotropic solid metal cube has coefficients of linear expansion as 5 × 10–5 /°C along the x-axis and5 × 10–6 /°C along y-axis and z-axis. If coefficient of volume expansion of the solid is C × 10–6 /°C thenthe value of C isvlekax Bksl /kkfRod ?ku dk rkih; js[kh; izlkj xq.kkad x–v{k ds vuqfn'k 5 × 10–5 /°C rFkk y ,oa z–v{k ds vuqfn'k

5 × 10–6 /°C gSA ;fn Bksl dk vk;ru izlkj xq.kkad C × 10–6 /°C gks] rks C dk eku Kkr djksA

Ans. 60 Sol. V = 22 + 1

= 10 × 10–6 + 5 × 10–5 = 60 × 10–6 /°C

24. A particle is released at point A. Find its kinetic energy at point P. (Given m = 1 kg and all surfaces arefrictionless),d d.k dks fcUnq A ls NksM+k tkrk gSA fcUnq P ij xfrt ÅtkZ Kkr djksA fn;k gS m = 1 kg rFkk lHkh lrg ?k"kZ.kjfgr

gSA

A

2m

B

C 1m

P

Ans. 10 Sol. KE = PE1 – PE2 = mgh1 – mgh2

= 1 × 10 × 2 – 1 × 10 × 1 = 10J

12

25. On a photosensitive metal of area 1 cm2 and work function 2eV, light of intensity 6.4 × 10–5 W/cm2 andwavelength 310 nm is incident normally. If 1 out of every 103 photons are successful, then number ofphotoelectrons emitted in one second is 10x. Find x,d izdk'k laosnh /kkrq dk {kS=kQy 1 cm2 rFkk dk;ZQyu 2eV gSA bl ij 6.4 × 10–5 W/cm2 rhozrk ;qDr 310 nmrajxnS/;Z dk izdk'k yEcor~ vkifrr gSA ;fn 103 QksVksu esa ls dsoy ,d QksVksu] ,d izdk'k bysDVªkWu mRltZu dj

ldrk gks rks ,d lsd.M esa mRlftZr QksVks bysDVªkWu dh la[;k 10x gS] rc x Kkr djksA

Ans. 11

Sol. Energy of photon. E = 1240310

= 4eV > 2eV (so photoelectric effect will take place)

= 4 × 1.6 × 10–19 = 6.4 × 10–19 Joule No. of photons falling per second

= 5

1419

6.4 10 1 106.4 10

No. of photoelectron emitted per second

= 14

113

10 1010

Hindi. QksVksu dh ÅtkZ E = 1240310

= 4eV > 2eV (vr% izdk'k fo|qr izHkko mRiUu gksxk)

= 4 × 1.6 × 10–19 = 6.4 × 10–19 Joule izfrlsd.M vkifrr QksVksu

= 5

1419

6.4 10 1 106.4 10

izfrlsd.M mRlftZr QksVksbysDVªkWu

= 14

113

10 1010

13