Question

The time period of a freely suspended magnet is \(4\,s\). If it is broken in length into two equal parts and one part is suspended in the same way, then the time period will be:

• A. \(4\,s\)
• B. \(2\,s\)
• C. \(0.5\,s\)
• D. \(0.25\,s\)

Hint:

Cut along length: \(I\) decreases faster than \(M\) → time period decreases.

Answer 1

The Correct Option is B

To solve this problem, we need to understand the concept of the time period of a simple pendulum and how it changes when we alter the length of the magnet. The time period \( T \) of a simple magnet pendulum is given by the formula:

\(T = 2\pi \sqrt{\frac{I}{mB}}\)

Where:

• \(I\) is the moment of inertia of the pendulum.
• \(m\) is the mass of the pendulum.
• \(B\) is the magnetic field strength.

For a bar magnet, the moment of inertia \(I = \frac{1}{12} m L^2\), where \(L\) is the length of the magnet.

If the original magnet (length \(L\)) is broken into two equal parts, each part has a length of \(\frac{L}{2}\). The new moment of inertia for one part becomes:

\(I' = \frac{1}{12} \left(\frac{m}{2}\right) \left(\frac{L}{2}\right)^2 = \frac{1}{12} \left(\frac{m L^2}{8}\right) = \frac{1}{96} m L^2\)

So the new time period \(T'\) is:

\(T' = 2\pi \sqrt{\frac{I'}{(m/2)B}} = 2\pi \sqrt{\frac{\frac{1}{96} m L^2}{(m/2)B}}\)

Simplifying further:

\(T' = 2\pi \sqrt{\frac{1}{48} \frac{m L^2}{m B}} = 2\pi \sqrt{\frac{1}{48} \times \frac{L^2}{B}} = \frac{1}{\sqrt{2}} \times \left(2\pi \sqrt{\frac{I}{mB}}\right)\)

Since \(T = 4\,s\), the new period becomes:

\(T' = \frac{1}{\sqrt{2}} \times 4\,s = 2\,s\)

The time period of one part of the magnet, when suspended freely in the same way, will be \(2\,s\). Thus, the correct answer is \(2\,s\).