A bar magnet of magnetic moment M is suspended freely in a magnetic field of strength B. Work done to rotate it from `0^(@)` to `60^(@)` and from `60^(@)` to `90^(@)` are `W_(1)` and `W_(2)`, respectively `W_(1)` and `W_(2)` are related as

Options

A. `W_(1) = W_(2)`
B. `W_(2) = 2W_(1)`
C. `W_(1) = 2W_(2)`
D. `W_(1) = sqrt(2W_(2))`

Answer

To find the relationship between the work done in rotating a bar magnet from \(0^\circ\) to \(60^\circ\) (denoted as \(W_1\)) and from \(60^\circ\) to \(90^\circ\) (denoted as \(W_2\)), we can use the formula for work done in rotating a magnetic dipole in a magnetic field.

### Step-by-Step Solution:

1. **Understanding the Work Done Formula**:
The work done \(W\) in rotating a magnetic dipole in a magnetic field is given by:
\[ W = -M B (\cos \theta_2 - \cos \theta_1) \]
where \(M\) is the magnetic moment, \(B\) is the magnetic field strength, \(\theta_1\) is the initial angle, and \(\theta_2\) is the final angle.

2. **Calculating \(W_1\)**:
For the first case, rotating from \(0^\circ\) to \(60^\circ\):
- Here, \(\theta_1 = 0^\circ\) and \(\theta_2 = 60^\circ\).
- Substitute these values into the formula:
\[ W_1 = -M B (\cos 60^\circ - \cos 0^\circ) \]
- We know that \(\cos 0^\circ = 1\) and \(\cos 60^\circ = \frac{1}{2}\):
\[ W_1 = -M B \left(\frac{1}{2} - 1\right) = -M B \left(-\frac{1}{2}\right) = \frac{1}{2} M B \]

3. **Calculating \(W_2\)**:
For the second case, rotating from \(60^\circ\) to \(90^\circ\):
- Here, \(\theta_1 = 60^\circ\) and \(\theta_2 = 90^\circ\).
- Substitute these values into the formula:
\[ W_2 = -M B (\cos 90^\circ - \cos 60^\circ) \]
- We know that \(\cos 90^\circ = 0\) and \(\cos 60^\circ = \frac{1}{2}\):
\[ W_2 = -M B (0 - \frac{1}{2}) = \frac{1}{2} M B \]

4. **Comparing \(W_1\) and \(W_2\)**:
From the calculations, we find:
\[ W_1 = \frac{1}{2} M B \quad \text{and} \quad W_2 = \frac{1}{2} M B \]
Therefore, we can conclude that:
\[ W_1 = W_2 \]

### Conclusion:
The work done to rotate the bar magnet from \(0^\circ\) to \(60^\circ\) is equal to the work done to rotate it from \(60^\circ\) to \(90^\circ\):
\[ W_1 = W_2 \]