Q 12-05-079JEE MainJEE Main 2017 (2 Apr)Easy
A magnetic needle of magnetic moment $6.7\times10^{-2}\ \text{A m}^2$ and moment of inertia $7.5\times10^{-6}\ \text{kg m}^2$ is performing simple harmonic oscillations in a magnetic field of $0.01$ T. Time taken for $10$ complete oscillations is:
Answer: (B) $6.65$ s
$$T = 2\pi\sqrt{\frac{I}{mB}} = 2\pi\sqrt{\frac{7.5\times10^{-6}}{6.7\times10^{-2}\times0.01}} = 2\pi\sqrt{1.119\times10^{-2}}$$
$$T = 2\pi(0.1058) \approx 0.665\ \text{s}$$
Ten oscillations take $10T \approx 6.65$ s.
Solution by Sreeraj P, M.Sc Physics