The time period of oscillation of a bar magnet suspended horizontaliy along the magnetic meridian is `T_(0)`. If this magnet is replaced by another magnet of the same size and pole strength but with double the mass, the new time period will be

Options

A. `(T_(0))/(2)`
B. `(T_(0))/(sqrt(2))`
C. `sqrt(2) T_(0)`
D. `2 T_(0)`

Answer

To solve the problem, we need to determine how the time period of oscillation of a bar magnet changes when its mass is doubled while keeping its size and pole strength constant.

### Step-by-Step Solution:

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

2. **Determine Moment of Inertia**:
For a bar magnet, the moment of inertia \( I \) about its center is given by:
\[ I = m r^2 \]
where \( m \) is the mass of the magnet and \( r \) is the distance from the pivot to the center of mass.

3. **Initial Time Period**:
Let the initial mass of the magnet be \( m \). The initial time period \( T_0 \) can be expressed as:
\[ T_0 = 2\pi \sqrt{\frac{m r^2}{m B}} = 2\pi \sqrt{\frac{r^2}{B}} \]

4. **New Time Period with Double Mass**:
When the mass is doubled (new mass \( M = 2m \)), the new moment of inertia becomes:
\[ I' = 2m r^2 \]
The new time period \( T' \) can be expressed as:
\[ T' = 2\pi \sqrt{\frac{I'}{M' B}} = 2\pi \sqrt{\frac{2m r^2}{2m B}} = 2\pi \sqrt{\frac{r^2}{B}} \]

5. **Comparing the Two Time Periods**:
Notice that the expression for \( T' \) simplifies to:
\[ T' = 2\pi \sqrt{\frac{r^2}{B}} \cdot \sqrt{2} = \sqrt{2} \cdot T_0 \]
Thus, the new time period \( T' \) is:
\[ T' = \sqrt{2} T_0 \]

### Final Answer:
The new time period will be:
\[ T' = \sqrt{2} T_0 \]