Q149 / 160 · TS EAMCET 2023 (13 May Shift 2) · Physics
The period of oscillation of a bar magnet at a place is 2 s. At the same place, the period of oscillation of another identical bar magnet whose magnetic moment is 4 times to that of first magnet is
- A$4 \mathrm{~s}$
- Correct answerB$1 \mathrm{~s}$
- C$2 \mathrm{~s}$
- D$0.5 \mathrm{~s}$
Solution
Period of oscillation of a bar, $\mathrm{T}_1=2 \mathrm{~s}$
Let magnetic moment, $M_1=M$ then $M_2=4 M$
$$
\begin{aligned}
& \mathrm{T}=2 \pi \sqrt{\frac{\mathrm{I}}{\mathrm{MH}}} \\
& \frac{\mathrm{T}_1}{\mathrm{~T}_2}=\sqrt{\frac{\mathrm{I}}{\mathrm{M}_1 \mathrm{H}} \times \frac{\mathrm{M}_2 \mathrm{H}}{\mathrm{I}}} \\
& =\sqrt{\frac{4 \mathrm{M}}{\mathrm{M}}} \\
& \Rightarrow \frac{2}{\mathrm{~T}_2}=2 \\
& \Rightarrow \mathrm{T}_2=1 \mathrm{~s} .
\end{aligned}
$$
Hence, the correct answer is option: 'B'.