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 solve the problem, we need to analyze the relationship between the time period of oscillation of a bar magnet in a magnetic field and its mass. The formula for the time period \( T \) of a bar magnet in a magnetic field is given by:

\[ T = 2\pi \sqrt{\frac{I}{mB}} \]

Where:
- \( T \) = time period of oscillation
- \( I \) = moment of inertia of the magnet
- \( m \) = mass of the magnet
- \( B \) = magnetic field strength

### Step 1: Understand the effect of mass on the time period

Given that the mass of the magnet is reduced to one-fourth of its original mass, we denote the original mass as \( m \) and the new mass as:

\[ m' = \frac{m}{4} \]

### Step 2: Substitute the new mass into the time period formula

Since all other parameters (moment of inertia \( I \) and magnetic field \( B \)) remain constant, we can substitute \( m' \) into the time period formula:

\[ T' = 2\pi \sqrt{\frac{I}{m'B}} = 2\pi \sqrt{\frac{I}{\left(\frac{m}{4}\right)B}} = 2\pi \sqrt{\frac{4I}{mB}} \]

### Step 3: Relate the new time period to the original time period

Now we can express the new time period \( T' \) in terms of the original time period \( T \):

\[ T' = 2\pi \sqrt{\frac{4I}{mB}} = 2\pi \cdot 2 \sqrt{\frac{I}{mB}} = 2T \]

### Step 4: Conclusion about the new time period

Since we have derived that:

\[ T' = 2T \]

This means that the new time period \( T' \) is twice the original time period \( T \).

### Final Answer

The new time period will be:

\[ T' = 2T \]