Period of oscillation of a bar magnet in earth's magnetic field is T . If its mass is reduced to one - fourth of the original mass , maintaining all other parameters same , the new time period will be

Options

A. T
B. 2 T
C. (T/2)
D. (T/4)

Explanation

To find the new time period of oscillation of a bar magnet when its mass is reduced to one-fourth of its original mass, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formula for Time Period (T)**:
The time period of a bar magnet in a magnetic field is given by the formula:
\[ T = 2\pi \sqrt{\frac{I}{mB}} \]
where \( I \) is the moment of inertia, \( m \) is the mass of the magnet, and \( B \) is the magnetic field.

2. **Identify the Change in Mass**:
The problem states that the mass of the magnet is reduced to one-fourth of its original mass. If the original mass is \( m \), the new mass \( m' \) will be:
\[ m' = \frac{m}{4} \]

3. **Substitute the New Mass into the Time Period Formula**:
The new time period \( T' \) can be expressed as:
\[ T' = 2\pi \sqrt{\frac{I}{m'B}} = 2\pi \sqrt{\frac{I}{\left(\frac{m}{4}\right)B}} = 2\pi \sqrt{\frac{4I}{mB}} \]

4. **Relate the New Time Period to the Original Time Period**:
Since the original time period \( T \) is:
\[ T = 2\pi \sqrt{\frac{I}{mB}} \]
We can express \( T' \) in terms of \( T \):
\[ T' = 2\pi \sqrt{4} \sqrt{\frac{I}{mB}} = 2 \cdot 2\pi \sqrt{\frac{I}{mB}} = 2T \]

5. **Conclusion**:
Therefore, the new time period \( T' \) when the mass is reduced to one-fourth is:
\[ T' = 2T \]

### Final Answer:
The new time period will be \( 2T \).