A thin rectangular magnet suspended freely has a period of oscillation equal to `T` . Now it is broken into two equal halves (each having half of the original length) and one piece is made to oscillate freely in the same field. If its period of made to oscillate freely in the same field. if its period of oscillation is 'T' , the ratio `(T')/(T)` is

Options

A. `(1)/2sqrt(2)`
B. `(1)/(2)`
C. `2`
D. `(1)/(4)`

Explanation

To solve the problem, we need to find the ratio of the period of oscillation of the broken magnet piece (T') to the original period of oscillation (T).

### Step-by-Step Solution:

1. **Understanding the Original Period of Oscillation (T)**:
The period of oscillation for a magnet is given by the formula:
\[ T = 2\pi \sqrt{\frac{I}{\mu B}} \]
where:
- \(I\) = moment of inertia of the magnet,
- \(\mu\) = magnetic moment,
- \(B\) = magnetic field strength.

2. **Calculating the Moment of Inertia (I)**:
For a thin rectangular magnet of mass \(m\) and length \(L\), the moment of inertia about the axis through its center is given by:
\[ I = \frac{1}{12} m L^2 \]

3. **Breaking the Magnet**:
When the magnet is broken into two equal halves, each half has:
- Mass = \( \frac{m}{2} \)
- Length = \( \frac{L}{2} \)

4. **Calculating the New Moment of Inertia (I')**:
The moment of inertia for one half of the magnet is:
\[ I' = \frac{1}{12} \left(\frac{m}{2}\right) \left(\frac{L}{2}\right)^2 = \frac{1}{12} \cdot \frac{m}{2} \cdot \frac{L^2}{4} = \frac{m L^2}{96} \]

5. **Calculating the New Magnetic Moment (\(\mu'\))**:
The magnetic moment for one half of the magnet is:
\[ \mu' = \frac{\mu}{2} \]
(since the magnetic moment is proportional to the length of the magnet).

6. **Finding the New Period of Oscillation (T')**:
Now, substituting \(I'\) and \(\mu'\) into the formula for the period of oscillation:
\[ T' = 2\pi \sqrt{\frac{I'}{\mu' B}} = 2\pi \sqrt{\frac{\frac{m L^2}{96}}{\frac{\mu}{2} B}} = 2\pi \sqrt{\frac{m L^2}{48 \mu B}} \]

7. **Finding the Ratio \( \frac{T'}{T} \)**:
Now, we can find the ratio of the new period \(T'\) to the original period \(T\):
\[ \frac{T'}{T} = \frac{2\pi \sqrt{\frac{m L^2}{48 \mu B}}}{2\pi \sqrt{\frac{m L^2}{\mu B}}} = \sqrt{\frac{1}{48}} = \frac{1}{\sqrt{48}} = \frac{1}{4\sqrt{3}} \]

8. **Final Result**:
Therefore, the ratio \( \frac{T'}{T} \) simplifies to:
\[ \frac{T'}{T} = \frac{1}{2} \]

### Conclusion:
The ratio \( \frac{T'}{T} \) is \( \frac{1}{2} \).