Simple Harmonic Motion Ques 4
- A magnetic needle of magnetic moment $6.7 \times 10^{-2} \mathbf{A m}^2$ and moment of inertia $7.5 \times 10^{-6} \mathrm{~kg} \mathbf{m}^2$ is performing simple harmonic oscillations in a magnetic field of $0.01 \mathrm{~T}$. Time taken for 10 complete oscillations is
(2017 Main)
(a) $8.89 \mathrm{~s}$
(b) $6.98 \mathrm{~s}$
(c) $8.76 \mathrm{~s}$
(d) $6.65 \mathrm{~s}$
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Answer:
Correct Answer: 4.( d )
Solution:
- Time period of oscillation is
$ \begin{aligned} T & =2 \pi \sqrt{\frac{I}{M B}} \\ \Rightarrow \quad T & =2 \pi \sqrt{\frac{7.5 \times 10^{-6}}{6.7 \times 10^{-2} \times 0.01}}=0.665 \mathrm{~s} \end{aligned} $
Hence, time for 10 oscillations is $t=6.65 \mathrm{~s}$.
BETA
