A uniform solid cylindrical roller of mass '`m`' is being pulled on a horizontal surface with force `F` parallel to the surface and applied at its centre. If the acceleration of the cylinder is '`a`' and it is rolling without slipping, then the value of '`F`' is `:-`

Options

A. `(3)/(2)ma`
B. 2ma
C. `(5)/(3)ma`
D. ma

Answer

To solve the problem, we need to analyze the forces and torques acting on the uniform solid cylindrical roller.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Cylinder:**
- The applied force \( F \) is acting horizontally at the center of the cylinder.
- The frictional force \( f \) acts in the opposite direction to the applied force, preventing slipping.

2. **Apply Newton's Second Law:**
- The net force acting on the cylinder can be expressed as:
\[ F - f = ma \]
- Here, \( m \) is the mass of the cylinder and \( a \) is its linear acceleration.

3. **Apply the Torque Equation:**
- The torque \( \tau \) about the center of the cylinder due to the applied force \( F \) is given by:
\[ \tau = F \cdot r \]
- The frictional force \( f \) also creates a torque, but in the opposite direction:
\[ \tau_f = f \cdot r \]
- The net torque about the center is:
\[ \tau = F \cdot r - f \cdot r \]

4. **Relate Torque to Angular Acceleration:**
- The moment of inertia \( I \) for a solid cylinder about its center is:
\[ I = \frac{1}{2} m r^2 \]
- The angular acceleration \( \alpha \) is related to the linear acceleration \( a \) by:
\[ \alpha = \frac{a}{r} \]
- Thus, we can write the torque equation as:
\[ F \cdot r - f \cdot r = I \cdot \alpha \]
- Substituting the values gives:
\[ F \cdot r - f \cdot r = \frac{1}{2} m r^2 \cdot \frac{a}{r} \]
- Simplifying this, we get:
\[ F - f = \frac{1}{2} m a \]

5. **Combine the Equations:**
- We now have two equations:
1. \( F - f = ma \)
2. \( F - f = \frac{1}{2} m a \)
- From the first equation, we can express \( f \):
\[ f = F - ma \]
- Substituting this into the second equation:
\[ F - (F - ma) = \frac{1}{2} m a \]
- This simplifies to:
\[ ma = \frac{1}{2} m a \]
- Rearranging gives:
\[ F = \frac{3}{2} ma \]

### Final Answer:
Thus, the value of the force \( F \) is:
\[ F = \frac{3}{2} ma \]