JEE ADVANCED 2021 MOCK TEST @ UNACADEMY [ NITIN …

JEE ADVANCED 2021 MOCK TEST @ UNACADEMY [ NITIN SIR ]

POWERED BY [ INDIAN SCHOOL OF PHYSICS ]

Max marks : 75 Time : 75 mins

Numerical response type [ +4 , -1 ]

1. A series combination of 𝑛1 capacitors, each of value 1C , is charged by a

source of potential difference 4V. When another parallel combination of 𝑛2

capacitors, each of value 2C , is charged by a source of potential difference

V, it has the same (total energy) stored in it as the first combination has. The

value of 2C in terms of 1C , is 𝑥𝐶1𝑛1𝑛2

. Find the value of x.

1. 16

( ) ( )2 212 2

1

1 1. 42 2

C V n C Vn

=

12

1 2

16CCn n

=

INDIAN SCHOOL OF PHYSICS

2. A particle of mass 1 kg and charge 13

c is projected from a long distance

towards a non-conducting fixed spherical shell having same charge uniformly distributed on its surface. Find critical velocity(in m/s) of projection required by the particle if it just grazes the shell

Key 1.63

2. Apply conservation of energy and conservation of angular momentum between points at infinite separation and on the surface of the sphere


3. A non-uniform magnetic field B varies with x. It exist for 𝑥 ≥ 0, into the xy

plane. A particle having charge q and of mass m, enters the magnetic field at

the origin with speed 𝑣�̂�. It is seen that it travels along 𝑦 = 2𝑥2 curve where

y and x are in meters. Find the value of B(in tesla)at x = 0 (𝐺𝑖𝑣𝑒𝑛:𝑚 =

1𝑔𝑚, 𝑣 = 1𝑚/𝑠, 𝑞 = 1𝑚𝑐)

3. 4

Radius of curvature of the curve

2

2

1 dydxr

d ydx

+ =

2

222 , 4 , 4dy d yy x x

dx dx= = =

21 16 10,

4 4xr at x r+

= = =

4mv mVr BqB qr

= = =

y = 2x2

vy vxi

X

B

Y

Y


4. An infinite long conductor carrying current 𝐼1 is placed near to an infinite long plane sheet carrying current 𝐼2 as shown in figure. The current is distributed uniformly over the width of sheet b. If b=4a then the magnetic force per unit length between them is𝜇𝑜

4𝜋𝑝𝐼1𝐼2

𝑏 then p = ___________

[𝑙𝑛 1 0 = 2.303, 𝑙𝑜𝑔 2 = 0.301]

4

3.22

4. 0 1 21

2I IF4 b

=

ln(1 b / a)+


5. Three identical capacitors 1 2,C C and 3C have a capacitance of 1.0 F each

and they are uncharged initially. They are connected in a circuit as shown in

the figure and 1C is then filled completely with a dielectric material of

relative permittivity r . The cell electromotive force (emf) 0 8V V= . First the

switch 1S is closed while the switch 2S is kept open. When the capacitor 3C is

fully charged, 1S is opened and 2S is closed simultaneously. When all the

capacitors reach equilibrium, the charge on 3C is found to be 5 C . The

value of r =

5. 1.5 q cV=

1C KC=


6. In the xy − plane, the region 0y has a uniform magnetic field 1ˆB k and the

region 0y has another uniform magnetic field 2ˆB k . A positively charged

particle is projected from the origin along the positively charged particle is projected from the origin along the positive y – axis with speed 1

0v ms −= at 0t = , as shown in the figure. Neglect gravity in this problem. Let t T= be the

time when the particle crosses the x− axis from below for the first time. If 2 14B B= , the average speed of the particle, in 1ms− , along the x− axis in the

time interval T is y

x1

0v ms −=

1B

2B

6. 2

1 2

1 2av

D Dvt t

+=

+

( )1 2

1 2

2av

R Rv

t t+

=+

mvRqB

=


7. A particle having specific charge s is projected in xy-plane with a speed ‘v’.

There exists a uniform magnetic field in z-direction having a fixed

magnitude B0. The field is made to reverse its direction after every interval

of 0

2B

. Calculate the maximum separation (in m) between two positions of

the particle during its course of motion. (given 0

2vB

= metre) (neglect any

other force including gravity throughout the motion)

7. 8

0

4 8mvAB metreqB

= =


8. Half portion of a spherical capacitor is filled with a dielectric of dielectric

constant k = 2 and conductivity ‘ ’. The charge given to spherical capacitor

is Q0. Due to the conductivity of dielectric charge leaks the time constant for

the discharge circuit is 0n . Find the value of ‘n’.

8. 3

( )

( )02 1ab k

Cb a

+=

−

( )1 b aR

ab −

=

( )( )

( )02 12

ab k b aRC

b a ab

+ −

=−


9. A very long straight conductor has a circular cross-section of radius R and carries a current density J. Inside the conductor there is a cylindrical hole of radius 𝑎 whose axis is parallel to the axis of the conductor and a distance b from it. Let the z-axis be the axis of the conductor. And let the axis of the hole be at 𝑥 = 𝑏. Find the x-component of magnetic field on the y-axis at

𝑦 = 2𝑅. If your answer is 𝐵𝑥 = 𝜇0𝐽𝑅 (1𝐴− 𝑎2

𝐵𝑅2+𝑏2) fill value of |𝐴| + |𝐵|.

9. key : 8

1B due to total wire at 2y R= is = ( )

0

2 2i iR

1B due to total wire at 2y R= is

( )0

2 2i iR

( )

( )

20 0ˆ ˆ2 2 4

J R JRi iR

= =

` Horizontal component of magnetic field due to smaller wire 2B x


= ( )2 2

, ˆ.cos 02 4

i iR b

−

+ `

( )2

02 22 2

2 ˆ.42 4

J a R iR bR b

= −

++

( )

20

2 2ˆ

4Ja R i

R b

=+

10. Two identical rods (each of mass m) are welded on semicircular massless

ring of radius R. The composite system was placed on rough horizontal surface as shown in figure. The system is released from the position ring rolls on surface without slipping. When line joining AB becomes horizontal

angular velocity of rod AC is ,2xgR

then x is

10. key : 3 21 322 2 2cR gmg I

R = =


Paragraph for Question Nos. 11 to 12: [SCQ ] [+3,-1]

Consider a conducting, uniform, thin hemispherical shell of radius ‘R’, which is connected with two wires along its diameter, carrying current ‘I’ as shown. Neglect field of connecting wires.

11. What is magnetic field due to hemi-sphere at it’s center?

A) 𝜇0𝐼2𝑅

B) 𝜇0𝐼2𝜋𝑅

C) √2𝜇0𝐼4𝑅

D) Zero

12. If uniform magnetic field B, is applied in direction perpendicular to base of

hemisphere ,what will be torque on it about the centre of the base?

A) 𝐼𝑅2𝐵 B) 𝐼𝑅2𝐵𝜋

C) 𝜋𝐼𝑅2𝐵2

D) Zero

Key : 11 B 12 D


11. Magnetic field due to a half ring at it’s center 0

4i

R . Break hemisphere into

many half rings connected at junction with external wires. Current in each half ring Ii d

=

So 0 sin4net

iBR

=

0

0

sin4

I dR

=

=

=

0

2IBR

= 12. Force passes through centre.


Paragraph for Question Nos. 13 & 14 : [MCQ +4 , -1 ] A large parallel plate capacitor with uniform surface charge on the upper

plate & – on the lower is moving with a constant speed V as shown in

figure.

V

V

+ –

13. Magnetic field between the plates is –

A) Parallel to the plates and perpendicular to velocity

B)Parallel to velocity and perpendicular to plates

C) 𝜇0𝜎𝑉2

D) 𝜇0𝜎𝑉

14. Magnetic force per unit area on the upper plate, including its direction –

A) 𝜇0𝜎2𝑣2 B) 𝜇0𝜎2𝑣2

2

C) Upwards D) Downward

13 AD 14 BC


19: AD 20 : BC


Single choice correct [ + 3 , -1 ] 15. A small electric dipole P is placed on the X – axis at the point (1, 0). The

dipole vector forms an angle of 30with the X – axis. Consider a non – uniform electric field to have been applied in the region given by the vector �⃗� = 𝑥2�̂� + 𝑦2𝑗̂. What is the force acting on the dipole?

A) 2|�⃗� | 𝑐𝑜𝑠 3 0°(�̂� + 2𝑗̂) B) 2|�⃗� | 𝑐𝑜𝑠 3 0°(𝑖̂)

C) 2|�⃗� | 𝑐𝑜𝑠 3 0°(2𝑗̂) D) Data insufficient

1. B

. =

xEF Px


16. A neutral particle at rest in a uniform magnetic field decays into two charged

particles of different masses at point P, as shown in the figure. The energy

released goes to their kinetic energy and particles move in the plane of the

paper. Magnetic field is into the plane of paper. Select the diagram which

describes path followed by the particles most approximately.

A)

B)

C)

D)

2. B Conceptual


Multichoice multi correct [ + 5, -2 ]

17. A metal sphere of radius a is surrounded by a concentric metal sphere of

inner radius of b, where b > a. The space between the spheres is filled with a

material whose electrical conductivity varies with the electric field

strength E as 𝜎 = 𝑘𝐸 where k is a constant.

A potential difference V is maintained between spheres.

A) Current is 4𝜋𝑟2𝑘𝐸2, where (𝑎 < 𝑟 < 𝑏)

B) Current is 2𝜋𝑟2𝑘𝐸2,𝑤ℎ𝑒𝑟𝑒(𝑎 < 𝑟 < 𝑏)

C) Potential difference between spheres is 𝑉 = √ 𝐼4𝜋𝑘

ℓ𝑛 (𝑏𝑎) where I is total

current

D) Potential difference between spheres is 𝑉 = √ 𝐼2𝜋𝑘

ℓ𝑘 (𝑏𝑎) where I is total

current

12. AC

𝐼 =𝑑𝑉𝑑𝑅

=𝐸𝑑𝑟𝑑𝑟

𝜎4𝜋𝑟2

= 𝑘𝐸24𝜋𝑟2. . . . . . . . . . . . . . . (𝑖)

𝑑𝑉 = 𝐸. 𝑑𝑟 From (1)𝐸2 = 1

4𝜋𝑘𝑟2

⇒ 𝑑𝑉 = ∫ √ 𝐼4𝜋𝑘

1𝑟𝑑𝑟𝑏

𝑎


18. A conducting medium is shaped in a form of a quarter of an annulus of radii

‘b’ and 𝑎(𝑏 > 𝑎) and thickness ‘t’. Specific resistance of the medium is ‘ ’.

The CORRECT option(s) is/are

A) Resistance between faces LMNO and PQRS is 2𝜌𝜋𝑡

𝑙𝑛 𝑏𝑎

B) Resistance between faces LMRS and ONQP is 𝜋𝜌

2𝑡 𝑙𝑛(𝑏𝑎)

C) Resistance between faces MRQN and LSPO is 4𝑡𝜌𝑏2−𝑎2

D) Between faces LMNO & PQRS and LMRS & ONQP, resistance is same.

Key : AB

Between faces LMNO and PQRS 𝑑𝑅 = 𝜌𝑑𝑟

𝑡𝜋2𝑟

𝑅 = 2𝜌𝑡𝜋

𝑙𝑛 𝑏𝑎

Between faces LMRS and ONQP for an element at distance r from O.

𝑑𝑅 =𝜌𝜋

2𝑟

𝑡𝑑𝑟

So, 1𝑅

= ∫ 1𝑑𝑅

= ∫ 𝑡𝑑𝑟𝜌𝜋

2𝑟𝑏𝑎 = 2𝑡

𝜌𝜋𝑙𝑛 (𝑏

𝑎)

2 ln

Rbta

=


19. A small disc of mass ‘m’ is released on a parabolic curve in a vertical plane

such that gravity acts along negative y-axis. The equation of parabolic curve

is 𝑥2 = 2𝑎√3

𝑦, where ‘a’ is a positive constant. Frictional force between disc

and curve are sufficient for pure rolling. When disc is reached at 𝑥 = 𝑎 then

choose the CORRECT option(s).

A) Acceleration of disc along the trajectory is √3𝑔

B) Acceleration of disc along the trajectory is 𝑔√3

C) Frictional force between disc and curve is 𝑚𝑔2√3

D) Frictional force between disc and curve is 𝑚𝑔√3

Key : BC

𝑥2 = 2𝑎√3

𝑦

=> 2𝑥 = 2√3

𝑎 𝑑𝑦𝑑𝑥

𝑡𝑎𝑛 𝜃 = √3 𝑥𝑎 060 =

𝑎 =𝑔 𝑠𝑖𝑛 𝜃

1 + 𝐼𝑐𝑚𝑚𝑅2

=𝑔√3

𝑓 =𝑚𝑔 𝑠𝑖𝑛 𝜃

1 + 𝑚𝑅2

𝐼𝑐𝑚

=𝑚𝑔2√3


JEE ADVANCED 2021 MOCK TEST @ UNACADEMY [ NITIN …

Documents

Transcript of JEE ADVANCED 2021 MOCK TEST @ UNACADEMY [ NITIN …