Q51 / 51 · JEE Advanced 2023 (Paper 1) · Physics

A thin conducting rod MN of mass 20 gm, length 25 cm and resistance 10Ω is held on frictionless, long, perfectly conducting vertical rails as shown in the figure. There is a uniform magnetic field B0=4 T directed perpendicular to the plane of the rod-rail arrangement. The rod is released from rest at time t=0 and it moves down along the rails. Assume air drag is negligible. Match each quantity in List-I with an appropriate value from List-II, and choose the correct option.

[Given: The acceleration due to gravity g=10 ms−2 and e−1=0.4]

Figure

 

  List-I   List-II
P At t=0.2 s, the magnitude of the induced emf in Volt 1 0.07
Q At t=0.2 s, the magnitude of the magnetic force in Newton 2 0.14
R At t=0.2 s, the power dissipated as heat in Watt 3 1.20
S

The magnitude of terminal velocity of the rod in m s−1

4 0.12
    5 2.00

  • A
    P→5, Q→2, R→3, S→1
  • B
    P→3, Q→1, R→4, S→5
  • C
    P→4, Q→3, R→1, S→2
  • Correct answer
    D
    P→3, Q→4, R→2, S→5

Solution

Figure

Induced emf ε=Blv

⇒ Induced current i=εR=BlvR

Direction of induced current would be from N to M.

Now magnetic force acting on the rod due to this current would be in the upward direction. Therefore, we can write

⇒ mg-ilB=ma [Applying 2nd law]

⇒ mg-B2l2vR=mdvdt

⇒∫0v dvmg-B2l2vR=∫0tdtm ⇒ lnmg-B2l2vR0v-B2l2R=tm

⇒ mg-B2l2vRmg=e-B2l2mRt

Now, mg=20×10-3×10=0.2, B2l2R=42×0.25210=0.1 and B2l2mR=42×0.25220×10-3×10=5

Therefore,

⇒ 0.2-0.1v0.2=e-5t

⇒ v=2[1-e–5t]

⇒ At t=0.2 s, v=21-1e=2×0.6=1.2 m s-1

⇒ ε=Blv=4×0.25×1.2=1.2 V volts

Then, current at t=0.2 s will be i=4×0.25×1.210=0.12 A

and magnetic force =ilB=0.12×0.25×4=0.12 N

and power dissipated =Fv=0.12×1.2=0.144 W

also for maximum velocity(terminal velocity) t→∞, e-5t→0 and hence v=21-0=2 m s-1

⇒ Correct match is (D)

Hence, the correct answer is option: 'D'.

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