Download Solved Question Paper (Question + Solution) of Physics 7th January Shift-1 – JEE Mains 2020 by Concepts Made Easy (By Er. Ajay Kumar)

PAPER-1 (B.E./B. TECH.)

JEE (Main) 2020 COMPUTER BASED TEST (CBT)

Memory Based Questions & Solutions Date: 07 January, 2020 (SHIFT-1) | TIME : (9.30 a.m. to 12.30 p.m)

Duration: 3 Hours | Max. Marks: 300 SUBJECT : PHYSICS

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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 system is released from rest, find the angular velocity of disc when block has dropped by height h. (there is no slipping between string & pulley) m

(1)

1 R

4gh 3

Ans.

(1)

Sol.

mgh =

(2)

1 R

2gh 3

(3) R

2gh 3

(4) R

4gh 3

1 1 mv2  2 2 2 v = R (no slipping) 1 1 mR2 2 m2R 2   2 2 2 3 mgh = m2R 2 4 mgh =



4gh 3R



1 4gh R 3

Three point masses 1kg, 1.5 kg, 2.5 kg are placed at the vertices of a triangle with sides 3cm,4cm and 5cm as shown in the figure. The location of centre of mass with respect to 1kg mass is : 2.5 kg

5cm

4 cm

1 kg

Ans.

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 (2)

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PAGE # 1

Sol.

Take 1kg mass at origin 2.5 kg

Y 5cm

4 cm

1 kg

3 cm

1.5 kg

1 0  1.5  3  2.5  0  0.9cm 5 1 0  1.5  0  2.5  4   2cm 5

Xcm  Ycm 3.

Ans. Sol.

In a single slit diffraction set up, second minima is observed at an angle of 60°. The expected position of first minima is (1) 25° (2) 20° (3) 30° (4) 45° (1) For 2nd minima d sin = 2 sin =

3 (given) 2

 3  d 4 So for 1st minima is d sin = 



… (i)

 3  (from equation (i)) d 4  = 25.65° (from sin table)   25

sin =

There are two infinite plane sheets each having uniform surface charge density + C/m2. They are inclined to each other at an angle 30° as shown in the figure. Electric field at any arbitrary point P is:

+ P

30° X

+

Ans.

(1)

 2 0

 3 1  yˆ – xˆ   1–  2  2  

(2)

 2 0

 3 1  yˆ – xˆ   1   2  2  

(3)

 2 0

 3 1  yˆ  xˆ   1–  2  2  

(4)

 2 0

 3 1  yˆ  xˆ   1   2  2  

(1)

PAGE # 2

Sol.

+ 60°

P 30° +

E

     cos60  –xˆ    – sin60   yˆ  20  20 20 

 2 0

 3 1  yˆ – xˆ   1–  2  2  

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).



k 00 A 1   2 d2 d

(1)



(2)

Ans.

(2)

Sol.

Capacitance of element =

k 00 A  d  1   d  2 

(3)

k 00 A 1  d 2d

(4)

k 00 A  d  1   2d  2 

k 0 A dx

x dx Capacitance of element, C' =



1  C'

k 0

k 0 (1  x) 0 A dx

dx 0  0 A(1  x )

1 1  n(1  d) C k 0  0 A

Given d << 1 2 2  1 1  d   d  C k 0  0 A  2

   

1 d  d   1   C k 00 A  2 

C

k 00 A  d  1   d  2 

PAGE # 3

A long solenoid of radius R carries a time dependent current I = I0 t(1 – t). A ring of radius 2R is placed

Ans. Sol.

coaxially 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. (4) I = I0t – I0t2  = BA  = 0nIA d VR = – = –0nAI0 (1 – 2t) dt VR = 0 at t = and R =

1 2

VR Re sistan ce of loop

e 1/2

If 10% of intensity is passed from analyser, then, the angle by which analyser should be rotated such that transmitted intensity becomes zero. (Assume no absorption by analyser and polarizer). (1) 60° (2) 18.4° (3) 45° (4) 71.6°

Ans. Sol.

(B) I = I0 cos2 I0  I 0 cos2  10 1 cos = = 0.31 10  = 71.6° angle rotated should be = 90° – 71.6° = 18.4°

Three moles of ideal gas A with The

Ans.

cP 4 c 5  is mixed with two moles of another ideal gas B with P  . cV 3 cV 3

cP of mixture is (Assuming temperature is constant) cv

(1) 1.5 (2)

(2) 1.42

(3) 1.7

(4) 1.3

PAGE # 4

 mixture 

Sol.

n1c P1  n2c P2 n1c V1  n2c V2

n1 

1R  R  n2 2 1 – 1 2 – 1 n1R n R  2 1 – 1  2 – 1

on rearranging we get, n1  n 2 n1 n2    mix – 1 1 – 1  2 – 1 5  mix – 1



3 2  1/ 3 2 / 3

5 = 9 + 3 = 12  mix – 1

  mixure 

17 5  1 12 12

mix = 1.42 Given magnetic field equation is B = 3 × 10–8 sin(t + kx + ) ĵ

then appropriate equation for electric field (E) will be : (1) 20 × 10–9 sin (t + kx + ) k̂

(2) 9 sin (t + kx + ) k̂

(3) 16 × sin (t + kx + ) k̂ (2) E0 = C (speed of light in vacuum) B0

(4) 3 × 10–9 sin (t + kx + ) k̂

10–9

Ans. Sol.

E0 = B0C = 3 × 10–8 × 3 × 108 = 9 N/C So E = 9 sin (t + kx + ) 10.

Ans. Sol.

There is a LCR circuit , If it is compared with a damped oscillation of mass m oscillating with force constant k and damping coefficient 'b'. Compare the terms of damped oscillation with the devices in LCR circuit. 1 (1) L  m , C  , R  b (2) L  m , C  k, R  b k 1 1 1 (3) L  k , C  b, R  m (4) L  ,C ,R b m k (1) In damped oscillation ma + bv + kx = 0

d2 x dx b  kx = 0 dt dt 2

…(i)

PAGE # 5

In the circuit di q – iR – L – =0 dt c

d2 q dq 1 R  .q = 0 …(ii) 2 dt c dt Comparing equation (i) and (ii) 1 m = L, b = R, k = c L

11.

Ans. Sol.

A lift can hold 2000kg, friction is 4000N and power provided is 60HP. (1 HP = 746W) Find the maximum speed with which lift can move up. (1) 1.9 m/s (2) 1.7 m/s (3) 2 m/s (4) 1.5 m/s (1) 4000 × V + mg × V = P 60  746 =V 4000  20000 V = 1.86 m/s.  1.9 m/s.

Ans.

A H–atom in ground state has time period T = 1.6 × 10–16 sec. find the frequency of electron in first excited state (1) 7.8 × 1014 (2) 7.8 × 1016 (3) 3.7 × 1014 (4) 3.7 × 1016 (1)

Sol.

T

12.

r n 2 n n3    v z z z2

T1 n13 1   T2 n32 8 T2 = 8T1 = 8 × 1.6 × 10–16 = 12.8 × 10–16 1 f2 =  7.8 × 1014 12.8  10 –16 13.

Ans. Sol.

Magnification of compound microscope is 375. Length of tube is 150mm. Given that focal length of objective lens is 5mm, then value of focal length of eyepiece is: (1) 2mm (2) 22mm (3) 12mm (4) 33mm (2) Case-I If final image is at least distance of clear vision M.P. =

375 =

L f0

 D 1    f  e  

150  25  1   5  fe 

375 25  1 30 fe 345 25  30 fe

PAGE # 6

750 = 2.17 cm 345 fe  22 mm Case-II If final image is at infinity fe =

D   = 375 f   e fe  20 mm M.P. =

14.

Ans. Sol.

L f0

1 litre of a gas at STP is expanded adiabatically to 3 litre. Find work done by the gas. Given  = 1.40 and 31.4= 4.65 (1) 100.8J (2) 90.5J (3) 45J (4) 18J (2) P1 = 1atm, T1 = 273K

P1V1  P2 V2 V  P2 = P1  1   V2 



 1 = 1atm    3

1.4

P1V1 – P2 V2 = 88.7 J  –1 Most appropriate ans is 90.5 J now work done =

A string of length 60 cm, mass 6gm and area of cross section 1mm 2 and velocity of wave 90m/s. Given young's modulus is Y = 1.6 × 1011 N/m2. Find extension in string ,d 60 cm dh yEch jLlh] ftldk nzO;eku 6gm vkSj {ks=kQy 1mm2 vkSj rjax dk osx 90m/s gSA

15.

Ans.

(Y = 1.6 × 1011 N/m2) jLlh esa izlkj crkb, (1) 0.3 mm (2) 0.2 mm (1)

Sol.

v

(3) 0.1 mm

(4) 0.4 mm

T 

T = v2

v 2  Y A =

v 2 AY

after substituting valu of v,,A and Y we get = 0.3 mm

PAGE # 7

16.

Which of the following gate is reversible (1)

Ans. Sol.

17.

(2)

(3) (4) (1) A logic gate is reversible if we can recover input data from the output eg. NOT gate A thin uniform rod of mass M and length L. Find the radius of gyration for rotation about an axis passing L through a point at a distance of from centre and perpendicular to rod. 4 (1)

Ans.

7 L 48

(2)

5 L 48

(3)

7 L 24

(4)

19 L 24

(1)

Sol.

L/4 2

ML2  L  M   = MK2  4 12 L2 L2  = K2 12 16 K

18.

A satellite of mass 'M' is projected radially from surface of earth with speed 'u'. When is reaches a M height equal to radius of earth, it ejects a rocket of mass and it itself starts orbiting the earth in 10 circular path of radius 2R, find the kinetic energy of rocket. 119GMe  113GMe    (1) 5M  u2  (2) 5M  u2    200R  200R    (3)

Ans. Sol.

7 L 48

M  2 119GMe  u   20  100R 

(4)

M  2 113GMe  u   20  200R 

(1) GMeM 1 2 GMeM 1 2  Mu   Mv R 2 2R 2

PAGE # 8

Me u



u2 



GMe R VT

M/10 Vr 2R

9M/10

VT  Transverse velocity of rocket VR Radial velocity of rocket

GMe 2R

V

M 9M GMe VT  10 10 2R GMe M Vr  M u2  10 R

Kinetic energy = =





GMe  1M 2 M  GMe VT  Vr2   100u2  100  81  2 10 20  2R R 

119GMe  M 2  100u   20  2R 

119GMe   = 5M  u2   200R  

19.

The current 'i' in the given circuit is 1 i

2

 1

2 1V

Ans.

(1) 0.2A (1)

(2) 0.3A

(3) 0.5A

(4) 0.25A

PAGE # 9

Sol. 1 i

2

I  

1

2 1V 1 I= = 0.4A 2.5 I i = = 0.2A 2

20.

Ans. Sol.

A current carrying circular loop is placed in an infinite plane if i is the magnetic flux through the inner region and o is magnitude of magnetic flux through the outer region, then (1) i > o (2) i < o (3) i = – o (4) i = o (3) 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 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) and F(0, 0, 5). Find magnetic flux through loop due to magnetic field B  3iˆ  4kˆ

Ans.

175

Sol.

ˆ . (25iˆ  25k) ˆ  = B.A = (3iˆ  4k)

 = (3 × 25) + (4 × 25) = 175 weber

PAGE # 10

22.

Ans. Sol.

A carnot engine operates between two reservoirs of temperature 900K and 300K. The engine performs 1200 J of work per cycle. The heat energy delivered by the engine to the low temperature reservoir in a cycle is: 600 W 300 2 = = 1 = Qh 900 3

3 W = 1800 J 2 QL = Qh – W = 600 J Qh =

23.

Ans. Sol.

24.

A non-isotropic solid metal cube has coefficients of linear expansion as 5 × 10 –5 /°C along the x-axis and 5 × 10–6 /°C along the y and the z-axis. If coefficient of volume expansion of the solid is C × 10–6 /°C then the value of C is 60 V = 22 + 1 = 10 × 10–6 + 5 × 10–5 = 60 × 10–6 /°C A particle is released at point A. Find Kinetic energy at point P Given m = 1 kg and all surfaces are frictionless P

A 2m

Ans. Sol.

10 KE = PE1 – PE2 = mgh1 – mgh2 = 1 × 10 × 2 – 1 × 10 × 1 = 10J

25.

On a photosensitive metal of area 1 cm 2 and work function 2eV, light of intensity 6.4 × 10–5 W/cm2 in wavelength 310 nm fall normally. If 1 out of every 10 3 photons are successful, then number of photoelectrons emitted in one second is 10 x. Find x 11 1240 Energy of photon. E = = 4eV > 2eV (so photoelectric effect will take place) 310 = 4 × 1.6 × 10–19 = 6.4 × 10–19 Joule No. of photons falling per second

Ans. Sol.

6.4  10 5  1 19

 1014

6.4  10 No. of photoelectron per second

1014 103

 1011

PAGE # 11

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