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Physics · Electromagnetic Induction

JEE Advanced 2023 — Paper 1 — Question 33

A thin conducting rod MN of mass 20 gm , length 25 cm and resistance 10Ω10 \Omega is held on frictionless, long, perfectly conducting vertical rails as shown in the figure. There is a uniform magnetic field B0=4 T\mathrm{B}_{0}=4 \mathrm{~T} directed perpendicular to the plane of the rod-rail arrangement. The rod is released from rest at time t=0t=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.[0pt] [Given: The acceleration due to gravity g=10 m s−2\mathrm{g}=10 \mathrm{~m} \mathrm{~s}^{-2} and e−1=0.4\mathrm{e}^{-1}=0.4 ]

LIST-ILIST-II
P) At t=0.2 st=0.2 \mathrm{~s}, the magnitude of the induced emf in Volt1) 0.07
Q) At t=0.2 st=0.2 \mathrm{~s}, the magnitude of the magnetic force in Newton2) 0.14
R) At t=0.2t=0.2 s, the power dissipated as heat in Watt3) 1.20
S) The magnitude of terminal velocity of the rod in ms−1\mathrm{m} \mathrm{s}^{-1}4) 0.12
5) 2.00
Question figure
  1. Option A:

    P→5,Q→2,R→3, S→1\mathrm{P} \rightarrow 5, \mathrm{Q} \rightarrow 2, \mathrm{R} \rightarrow 3, \mathrm{~S} \rightarrow 1

  2. Option B:

    P→3,Q→1,R→4, S→5\mathrm{P} \rightarrow 3, \mathrm{Q} \rightarrow 1, \mathrm{R} \rightarrow 4, \mathrm{~S} \rightarrow 5

  3. Option C:

    P→4,Q→3,R→1, S→2\mathrm{P} \rightarrow 4, \mathrm{Q} \rightarrow 3, \mathrm{R} \rightarrow 1, \mathrm{~S} \rightarrow 2

  4. Option D:

    P→3,Q→4,R→2, S→5\mathrm{P} \rightarrow 3, \mathrm{Q} \rightarrow 4, \mathrm{R} \rightarrow 2, \mathrm{~S} \rightarrow 5

    Correct

Answer: D

Step-by-step solution

mg−iℓB=mam g-i \ell B=m a i=BℓvRi=\frac{B \ell v}{R} mg−B2ℓ2Rv=mdvdt\mathrm{mg}-\frac{\mathrm{B}^{2} \ell^{2}}{\mathrm{R}} \mathrm{v}=\frac{\mathrm{mdv}}{\mathrm{dt}} dvdt=g−B2ℓ2mRv=g−cv\frac{d v}{d t}=g-\frac{B^{2} \ell^{2}}{m R} v=g-c v where c=B2ℓ2mR=5\mathrm{c}=\frac{\mathrm{B}^{2} \ell^{2}}{\mathrm{mR}}=5 V=2(1−e−5t)\mathrm{V}=2\left(1-\mathrm{e}^{-5 \mathrm{t}}\right) at t=0.2⇒v=1.20\mathrm{t}=0.2 \Rightarrow \mathrm{v}=1.20 at t=0.2⇒ Fm=0.12\mathrm{t}=0.2 \Rightarrow \mathrm{~F}_{\mathrm{m}}=0.12 P=i2R=0.14\mathrm{P}=\mathrm{i}^{2} \mathrm{R}=0.14 VT=2\mathrm{V}_{\mathrm{T}}=2 (p) →3\rightarrow 3, (q) →4\rightarrow 4, (r) →2\rightarrow 2, (s) →5\rightarrow 5

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2023
Paper
Paper 1
Subject
Physics
Chapter
Electromagnetic Induction
Topic
Motional EMF
A thin conducting rod MN of mass 20 gm , length 25 cm and resistance… | JEE Advanced 2023 PYQ with Solution · DhiX AI