The moment of inertia of a disc of radius 0.5m about its geometric axis is 2kg-`m^(2)` . If a string is tied to its circumference and a force of 10 Newton is applied, the value of torque with respect to this axis will be:

Options

A. 2.5 N-m
B. 5 N-m
C. 10 N-m
D. 20 N-m

Answer

To find the torque applied to the disc with respect to its geometric axis, we can use the formula for torque (τ):

\[ \tau = F \times R \times \sin(\theta) \]

Where:
- \( F \) is the force applied,
- \( R \) is the radius at which the force is applied,
- \( \theta \) is the angle between the force vector and the radius vector.

### Step 1: Identify the given values
- The radius \( R = 0.5 \) m
- The force \( F = 10 \) N
- The angle \( \theta = 90^\circ \) (since the force is applied tangentially to the circumference of the disc)

### Step 2: Calculate the sine of the angle
Since the angle \( \theta \) is \( 90^\circ \):
\[ \sin(90^\circ) = 1 \]

### Step 3: Substitute the values into the torque formula
Now, substituting the values into the torque formula:
\[ \tau = F \times R \times \sin(\theta) = 10 \, \text{N} \times 0.5 \, \text{m} \times 1 \]

### Step 4: Perform the multiplication
Calculating the above expression:
\[ \tau = 10 \times 0.5 = 5 \, \text{N m} \]

### Conclusion
The value of the torque with respect to the geometric axis of the disc is:
\[ \tau = 5 \, \text{N m} \]