Question

The periodic time of a magnet in vibration magnetometer is \(T_0\). If a magnet is replaced by another magnet whose moment of inertia is three times and magnetic moment is one-third of the original magnet, then time period of new magnet will be

• A. \(\frac{T_0}{3}\)
• B. \(\frac{T_0}{\sqrt{3}}\)
• C. \(3\,T_0\)
• D. \(\sqrt{3}\,T_0\)

Hint:

T = 2 pi sqrt(I / (M B)).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
A magnet oscillating in a vibration magnetometer has time period \(T = 2\pi\sqrt{\frac{I}{MB_H}}\), where \(I\) is the moment of inertia, \(M\) the magnetic moment and \(B_H\) the horizontal field.

Step 2: Apply the changes:
New \(I' = 3I\) and \(M' = \frac M3\):
\[ T' = 2\pi\sqrt{\frac{3I}{(M/3)B_H}} = 2\pi\sqrt{\frac{9I}{MB_H}} = 3T_0 \]

Step 3: Why the other options are wrong.
\(\frac{T_0}{3}\) and \(\frac{T_0}{\sqrt3}\) invert the effect. \(\sqrt3\,T_0\) comes from changing only \(I\) or only \(M\), but both changes together multiply \(\frac IM\) by 9.

Final Answer:
The new time period is \(3T_0\), option (C). \[ \boxed{3T_0} \]