JEE Mains · Physics · STD 12 -Electromagnetic induction
- JEE Mains 2019
- Hard
- Numerical
A thin strip \(10\, cm\) long is on a \(U\) shaped wire of negligible resistance and it is connected to a spring of spring constant \(0.5\,Nm^{-1}\) (see figure). The assembly is kept in a uniform magnetic field of \(0.1\, T\). If the strip is pulled from its equilibrium position and released, the number of oscillations it performs before its amplitude decreases by a factor of \(e\) is \(N\). If the mass of the strip is \(50\, grams\), its resistance \(10\,\Omega \) and air drag negligible, \(N\) will be close to

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Answer & Solution
Correct Answer
(D) \(5000\)
Step-by-step Solution
Detailed explanation
\(\mathrm{T}_{0}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}=\frac{2 \pi}{\sqrt{10}}\)
\(\mathrm{A}=\mathrm{A}_{0} \mathrm{e}^{-1 / \gamma}\)
\(\therefore\) for \(\mathrm{A}=\frac{\mathrm{A}_{0}}{\mathrm{e}}, \mathrm{t}=\gamma\)
\(\mathrm{t}=\gamma=\frac{2 \mathrm{m}}{\mathrm{b}}=\frac{\frac{2 \mathrm{m}}{\mathrm{B}^{2} \ell^{2}}}{\frac{\mathrm{B}^{2} \ell^{2}}{\mathrm{R}}}=10^{4} \,\mathrm{s}\)
\(\therefore \) No of oscillation \(\frac{\mathrm{t}}{\mathrm{T}_{0}}=\frac{10^{4}}{2 \pi / \sqrt{10}} \approx 5000\)
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