In the above question, the acceleration of the rim will be?

Options

A. 2T/M
B. 2T/MR
C. T/MR
D. T/M

Answer

To find the acceleration of the rim of a uniform disc when a tension \( T \) is applied to a string wound around it, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the System**: We have a uniform disc of mass \( m \) and radius \( r \) that is mounted on a frictionless horizontal axis. A light string is wound around the rim of the disc, and a tension \( T \) is applied to the string.

2. **Identify the Motion**: Since the disc is attached to an axis, it cannot translate; it can only rotate about that axis. The applied tension will create a torque that causes the disc to rotate.

3. **Calculate the Torque**: The torque \( \tau \) caused by the tension \( T \) can be calculated using the formula:
\[ \tau = r \times T \]
where \( r \) is the radius of the disc.

4. **Relate Torque to Angular Acceleration**: According to Newton's second law for rotation, the torque is also related to the moment of inertia \( I \) and angular acceleration \( \alpha \):
\[ \tau = I \alpha \]
For a uniform disc, the moment of inertia \( I \) is given by:
\[ I = \frac{1}{2} m r^2 \]

5. **Set the Two Expressions for Torque Equal**: We can set the two expressions for torque equal to each other:
\[ rT = I \alpha \]
Substituting the moment of inertia:
\[ rT = \left(\frac{1}{2} m r^2\right) \alpha \]

6. **Solve for Angular Acceleration**: Rearranging the equation to solve for \( \alpha \):
\[ \alpha = \frac{2T}{mr} \]

7. **Relate Angular Acceleration to Linear Acceleration**: The linear acceleration \( a \) of a point on the rim of the disc is related to the angular acceleration \( \alpha \) by the formula:
\[ a = r \alpha \]
Substituting the expression for \( \alpha \):
\[ a = r \left(\frac{2T}{mr}\right) \]

8. **Simplify to Find the Acceleration of the Rim**: The \( r \) in the numerator and denominator cancels out:
\[ a = \frac{2T}{m} \]

### Final Result:
The acceleration of the rim of the disc is:
\[ \boxed{\frac{2T}{m}} \]