Q110 / 150 · MHT CET 2008 · Physics
The length of a magnet is large compared to its width and breadth. The time period of its oscillation in a vibration magnetometer is $2 \mathrm{~s}$. The magnet is cut along its length into three equal parts and three parts are then placed on each other with their like poles together. The time period of this combination will be
- A$2 \mathrm{~s}$
- Correct answerB$2 / 3 \mathrm{~s}$
- C$2 \sqrt{3} \mathrm{~s}$
- D$2 / \sqrt{3} \mathrm{~s}$
Solution
The time period of oscillations of magnet
$$
T=2 \pi \sqrt{\left(\frac{I}{M H}\right)}
$$
where $I=$ moment of inertia of magnet $=\frac{m L^{2}}{12}(m$, being the mass of magnet $)$ $M=$ pole strength $\times L$ and $H=$ horizontal component of earth's magnetic field. When the three equal parts of magnet are placed on one another with their like poles together, then
$$
\begin{aligned}
I^{\prime} &=\frac{1}{12}\left(\frac{m}{3}\right) \times\left(\frac{L}{3}\right)^{2} \times 3 \\
&=\frac{1}{12} \frac{m L^{2}}{9}=\frac{I}{9}
\end{aligned}
$$
and $\quad M^{\prime}=$ pole strength $\times \frac{L}{3} \times 3=M$
Hence, $T^{\prime}=2 \pi \sqrt{\left(\frac{I / 9}{M H}\right)}$
$$
\begin{aligned}
\Rightarrow \quad T^{\prime} &=\frac{1}{3} \times T \\
T^{\prime} &=\frac{2}{3} \mathrm{~s}
\end{aligned}
$$
Hence, the correct answer is option: 'B'.