The work done in turning a magnet of magnetic moment 'M' by an angle of `90^(@)` from the meridian is 'n' times the corresponding work done to turn it through an angle of `60^(@)`, where 'n' is given by

Options

A. `1//2`
B. `2`
C. `1//4`
D. `1`

Explanation

To solve the problem, we need to find the relationship between the work done in turning a magnet of magnetic moment 'M' by 90 degrees and the work done in turning it by 60 degrees.

### Step-by-Step Solution:

1. **Understanding Work Done on a Magnetic Moment**:
The work done (W) in turning a magnetic moment (M) from an angle θ1 to θ2 in a magnetic field (B) is given by:
\[ W = MB \left( \cos(\theta_1) - \cos(\theta_2) \right) \]

2. **Calculating Work Done for 90 Degrees (W1)**:
- For turning the magnet from 0 degrees to 90 degrees:
- θ1 = 0 degrees, θ2 = 90 degrees
- Thus, we have:
\[ W_1 = MB \left( \cos(0) - \cos(90) \right) = MB \left( 1 - 0 \right) = MB \]

3. **Calculating Work Done for 60 Degrees (W2)**:
- For turning the magnet from 0 degrees to 60 degrees:
- θ1 = 0 degrees, θ2 = 60 degrees
- Thus, we have:
\[ W_2 = MB \left( \cos(0) - \cos(60) \right) = MB \left( 1 - \frac{1}{2} \right) = MB \left( \frac{1}{2} \right) = \frac{MB}{2} \]

4. **Setting Up the Relationship**:
According to the problem, the work done in turning the magnet by 90 degrees (W1) is 'n' times the work done in turning it by 60 degrees (W2):
\[ W_1 = n \cdot W_2 \]

5. **Substituting the Values of W1 and W2**:
Substitute the expressions for W1 and W2 into the equation:
\[ MB = n \cdot \frac{MB}{2} \]

6. **Solving for n**:
- Cancel MB from both sides (assuming M and B are not zero):
\[ 1 = \frac{n}{2} \]
- Multiply both sides by 2:
\[ n = 2 \]

### Conclusion:
The value of 'n' is 2. Thus, the work done in turning the magnet by 90 degrees is 2 times the work done in turning it by 60 degrees.