The length of a magnet is large compared to its width and breadth. The time period of its oscillation in a vibration magnetometer is ` 2 s`. The magnet is cut along its length into three equal parts and these parts are then placed on each other with their like poles together . The time period of this combination will be

Options

A. 2s
B. `(2)/(3)s`
C. `2sqrt3s`
D. `(2)/(sqrt3)s`

Explanation

To solve the problem, we need to analyze how the time period of oscillation of a magnet changes when it is cut and reassembled.

### Step-by-Step Solution:

1. **Understand the Initial Conditions:**
- The initial time period \( T \) of the magnet is given as \( 2 \, \text{s} \).
- The magnet is cut into three equal parts along its length.

2. **Determine the Moment of Inertia:**
- The moment of inertia \( I \) of a uniform rod about its end is given by the formula:
\[ I = \frac{1}{3} m L^2 \]
- For the original magnet of mass \( M \) and length \( L \), the moment of inertia is:
\[ I = \frac{1}{3} M L^2 \]

3. **Calculate the Moment of Inertia After Cutting:**
- When the magnet is cut into three equal parts, each part has a mass of \( \frac{M}{3} \) and a length of \( \frac{L}{3} \).
- The moment of inertia for each part about its end is:
\[ I_{\text{part}} = \frac{1}{3} \left(\frac{M}{3}\right) \left(\frac{L}{3}\right)^2 = \frac{1}{3} \cdot \frac{M}{3} \cdot \frac{L^2}{9} = \frac{M L^2}{81} \]
- Since there are three parts, the total moment of inertia \( I' \) of the stacked parts is:
\[ I' = 3 \cdot I_{\text{part}} = 3 \cdot \frac{M L^2}{81} = \frac{M L^2}{27} \]

4. **Relate the Time Period to the Moment of Inertia:**
- The time period \( T \) of a magnet in a vibration magnetometer is given by:
\[ T = 2\pi \sqrt{\frac{I}{B}} \]
- Since the magnetic moment \( M \) remains unchanged, we can express the new time period \( T' \) as:
\[ T' = 2\pi \sqrt{\frac{I'}{B}} \]

5. **Calculate the New Time Period:**
- From the relationship between the original and new moments of inertia:
\[ \frac{T'}{T} = \sqrt{\frac{I'}{I}} \implies T' = T \sqrt{\frac{I'}{I}} \]
- Substitute \( I' = \frac{M L^2}{27} \) and \( I = \frac{M L^2}{3} \):
\[ \frac{I'}{I} = \frac{\frac{M L^2}{27}}{\frac{M L^2}{3}} = \frac{1}{9} \]
- Therefore:
\[ T' = T \sqrt{\frac{1}{9}} = T \cdot \frac{1}{3} \]
- Since \( T = 2 \, \text{s} \):
\[ T' = 2 \cdot \frac{1}{3} = \frac{2}{3} \, \text{s} \]

6. **Final Answer:**
- The new time period of the combination is \( \frac{2}{3} \, \text{s} \).