JEE Advance - Physics (2023 - Paper 1 Online - No. 17)
A thin conducting rod $M N$ of mass $20 ~\mathrm{gm}$, length $25 \mathrm{~cm}$ and resistance $10 ~\Omega$ is held on frictionless, long, perfectly conducting vertical rails as shown in the figure. There is a uniform magnetic field $B_0=4 \mathrm{~T}$ 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 $g=10 \mathrm{~m} \mathrm{~s}^{-2}$ and $e^{-1}=0.4$ ]
[Given: The acceleration due to gravity $g=10 \mathrm{~m} \mathrm{~s}^{-2}$ and $e^{-1}=0.4$ ]
| List - I | List - II |
|---|---|
| (P) At $t=0.2 \mathrm{~s}$, the magnitude of the induced emf in Volt | (1) 0.07 |
| (Q) At $t=0.2 \mathrm{~s}$, the magnitude of the magnetic force in Newton | (2) 0.14 |
| (R) At $t=0.2 \mathrm{~s}$, the power dissipated as heat in Watt | (3) 1.20 |
| (S) The magnitude of terminal velocity of the rod in $\mathrm{m} \mathrm{s}^{-1}$ | (4) 0.12 |
| (5) 2.00 |
$P \rightarrow 5, Q \rightarrow 2, R \rightarrow 3, S \rightarrow 1$
$P \rightarrow 3, Q \rightarrow 1, R \rightarrow 4, S \rightarrow 5$
$P \rightarrow 4, Q \rightarrow 3, R \rightarrow 1, S \rightarrow 2$
$P \rightarrow 3, Q \rightarrow 4, R \rightarrow 2, S \rightarrow 5$
Explanation
From force equation
$$ \begin{aligned} & \mathrm{mg}-\mathrm{Bi} \ell=\frac{\mathrm{mdv}}{\mathrm{dt}} \\\\ & \mathrm{mg}-\frac{\mathrm{BBi} \ell}{\mathrm{R}} \times \ell=\frac{\mathrm{mdv}}{\mathrm{dt}} \\\\ & \frac{\mathrm{mgR}}{\mathrm{B}^2 \ell^2}-\mathrm{v}=\frac{\mathrm{mR}}{\mathrm{B}^2 \ell^2} \frac{\mathrm{dv}}{\mathrm{dt}} \\\\ & \frac{\mathrm{B}^2 \ell^2}{\mathrm{mR}} \int\limits_{\mathrm{t}=0}^{\mathrm{t}} \mathrm{dt}=\int\limits_0^{\mathrm{v}} \frac{\mathrm{dv}}{\left(\frac{\mathrm{mgR}}{\mathrm{B}^2 \ell^2}-\mathrm{v}\right)} \end{aligned} $$
Now $\frac{\mathrm{mgR}}{\mathrm{B}^2 \ell^2}=\frac{20 \times 10^{-3} \times 10 \times 10}{16 \times \frac{1}{16}}=2$
And $\frac{\mathrm{B}^2 \ell^2}{\mathrm{mR}}=\frac{16 \times \frac{1}{16}}{20 \times 10^{-3} \times 10}=\frac{1}{0.2}=5$
$\begin{aligned} & \therefore 5 \mathrm{t}=[-\ln (2-\mathrm{v})]_0^{\mathrm{v}} \\\\ & -5 \mathrm{t}=\ell \mathrm{n}\left[\frac{2-\mathrm{v}}{\mathrm{v}}\right] \\\\ & \therefore \mathrm{v}=2\left(1-\mathrm{e}^{-5 \mathrm{t}}\right)\end{aligned}$
At $\mathrm{t}=0.2 \mathrm{sec}$
$\begin{aligned} & \mathrm{v}=2\left(1-\mathrm{e}^{-5 \times 0.2}\right) \\\\ & \mathrm{v}=2(1-0.4) \\\\ & \mathrm{v}=1.2 \mathrm{~m} / \mathrm{s}\end{aligned}$
(P) Now at $\mathrm{t}=0.2 \mathrm{sec}$
The magnitude of the induced emf $=\mathrm{E}=\mathrm{Bv} \ell$
$$ =4 \times 1.2 \times \frac{1}{4}=1.2 \mathrm{Volt} $$
(Q) At $\mathrm{t}=0.2 \mathrm{sec}$, the magnitude of magnetic force $=\mathrm{BI} \ell \sin \theta$
$$ \begin{aligned} & =\mathrm{B} \times \frac{\mathrm{B} \ell \mathrm{v}}{\mathrm{R}} \times \ell \times \sin 90^{\circ} \\\\ & =\frac{4 \times 4 \times \frac{1}{4} \times 1.3 \times \frac{1}{4}}{10}=0.12 \text { Newton } \end{aligned} $$
(R) At t $=0.2 \mathrm{sec}$, the power dissipated as heat
$$ \begin{aligned} & P=i^2 R=\frac{v^2}{R}=\frac{1.2 \times 1.2}{10} \\\\ & P=0.144 \text { watt } \end{aligned} $$
(S) Magnitude of terminal velocity
At terminal velocity, the net force become zero
$$ \begin{aligned} & \therefore \mathrm{mg}=\mathrm{Bi} \ell \\\\ & \mathrm{mg}=\mathrm{B} \times \frac{\mathrm{B} \ell \mathrm{v}_{\mathrm{t}}}{\mathrm{R}} \times \ell \\\\ & \therefore \mathrm{v}_{\mathrm{T}}=\frac{\mathrm{mgR}}{\mathrm{B}^2 \ell^2}=\frac{20 \times 10^{-3} \times 10 \times 10}{16 \times \frac{1}{16}} \\\\ & \mathrm{v}_{\mathrm{T}}=2 \mathrm{~m} / \mathrm{s} \end{aligned} $$
Hence, Answer is (D)
$$ \begin{aligned} & \mathrm{mg}-\mathrm{Bi} \ell=\frac{\mathrm{mdv}}{\mathrm{dt}} \\\\ & \mathrm{mg}-\frac{\mathrm{BBi} \ell}{\mathrm{R}} \times \ell=\frac{\mathrm{mdv}}{\mathrm{dt}} \\\\ & \frac{\mathrm{mgR}}{\mathrm{B}^2 \ell^2}-\mathrm{v}=\frac{\mathrm{mR}}{\mathrm{B}^2 \ell^2} \frac{\mathrm{dv}}{\mathrm{dt}} \\\\ & \frac{\mathrm{B}^2 \ell^2}{\mathrm{mR}} \int\limits_{\mathrm{t}=0}^{\mathrm{t}} \mathrm{dt}=\int\limits_0^{\mathrm{v}} \frac{\mathrm{dv}}{\left(\frac{\mathrm{mgR}}{\mathrm{B}^2 \ell^2}-\mathrm{v}\right)} \end{aligned} $$
Now $\frac{\mathrm{mgR}}{\mathrm{B}^2 \ell^2}=\frac{20 \times 10^{-3} \times 10 \times 10}{16 \times \frac{1}{16}}=2$
And $\frac{\mathrm{B}^2 \ell^2}{\mathrm{mR}}=\frac{16 \times \frac{1}{16}}{20 \times 10^{-3} \times 10}=\frac{1}{0.2}=5$
$\begin{aligned} & \therefore 5 \mathrm{t}=[-\ln (2-\mathrm{v})]_0^{\mathrm{v}} \\\\ & -5 \mathrm{t}=\ell \mathrm{n}\left[\frac{2-\mathrm{v}}{\mathrm{v}}\right] \\\\ & \therefore \mathrm{v}=2\left(1-\mathrm{e}^{-5 \mathrm{t}}\right)\end{aligned}$
At $\mathrm{t}=0.2 \mathrm{sec}$
$\begin{aligned} & \mathrm{v}=2\left(1-\mathrm{e}^{-5 \times 0.2}\right) \\\\ & \mathrm{v}=2(1-0.4) \\\\ & \mathrm{v}=1.2 \mathrm{~m} / \mathrm{s}\end{aligned}$
(P) Now at $\mathrm{t}=0.2 \mathrm{sec}$
The magnitude of the induced emf $=\mathrm{E}=\mathrm{Bv} \ell$
$$ =4 \times 1.2 \times \frac{1}{4}=1.2 \mathrm{Volt} $$
(Q) At $\mathrm{t}=0.2 \mathrm{sec}$, the magnitude of magnetic force $=\mathrm{BI} \ell \sin \theta$
$$ \begin{aligned} & =\mathrm{B} \times \frac{\mathrm{B} \ell \mathrm{v}}{\mathrm{R}} \times \ell \times \sin 90^{\circ} \\\\ & =\frac{4 \times 4 \times \frac{1}{4} \times 1.3 \times \frac{1}{4}}{10}=0.12 \text { Newton } \end{aligned} $$
(R) At t $=0.2 \mathrm{sec}$, the power dissipated as heat
$$ \begin{aligned} & P=i^2 R=\frac{v^2}{R}=\frac{1.2 \times 1.2}{10} \\\\ & P=0.144 \text { watt } \end{aligned} $$
(S) Magnitude of terminal velocity
At terminal velocity, the net force become zero
$$ \begin{aligned} & \therefore \mathrm{mg}=\mathrm{Bi} \ell \\\\ & \mathrm{mg}=\mathrm{B} \times \frac{\mathrm{B} \ell \mathrm{v}_{\mathrm{t}}}{\mathrm{R}} \times \ell \\\\ & \therefore \mathrm{v}_{\mathrm{T}}=\frac{\mathrm{mgR}}{\mathrm{B}^2 \ell^2}=\frac{20 \times 10^{-3} \times 10 \times 10}{16 \times \frac{1}{16}} \\\\ & \mathrm{v}_{\mathrm{T}}=2 \mathrm{~m} / \mathrm{s} \end{aligned} $$
Hence, Answer is (D)
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