A magnetic needle of magnetic moment $6 \times 10^{-2} \mathrm{Am}^2$ and moment of inertia $9.6 \times…

A magnetic needle of magnetic moment $6 \times 10^{-2} \mathrm{Am}^2$ and moment of inertia $9.6 \times 10^{-5} \mathrm{~kg} \mathrm{~m}^2$ performs simple harmonic motion in a magnetic field of 0.01 T . Time taken to complete 10 oscillations is [Take $\pi=3 \cdot 14$ ]
  1. 0.2512 s
  2. 2.512 s
  3. 25.12 s
  4. $251 \cdot 2 \mathrm{~s}$

Solution

Period of oscillation $T=2 \pi \sqrt{\frac{\mathrm{I}}{\mathrm{MB}}}$ $\begin{aligned} & =2 \pi \sqrt{\frac{9.6 \times 10^{-5}}{6 \times 10^{-2} \times 0.01}} \\ & =2.512 \mathrm{~s} \end{aligned}$ $\therefore \quad$ Time taken for 10 oscillations $=2.512 \times 10$ $=25.12 \mathrm{~s}$

Asked in: MHT CET 2024 (03 May Shift 1)

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