A constant horizontal force F is applied on the top of a solid sphere and a hallow sphere of same mass and radiuis both kept on a sufficiently rough surface. Let `a_(S)` and `a_(H)` be their linear acceleration respectively, then

Options

A. `a_(S) = a_(H)`
B. `a_(S) gt a_(H)`
C. `a_(S) lt a_(H)`
D. data sufficient

Explanation

To solve the problem, we need to analyze the motion of both the solid sphere and the hollow sphere under the influence of a constant horizontal force \( F \). We will derive the linear accelerations \( a_S \) for the solid sphere and \( a_H \) for the hollow sphere step by step.

### Step 1: Understand the Forces Acting on the Spheres
Both spheres are subjected to a constant horizontal force \( F \) applied at their tops. Additionally, there will be a frictional force \( f \) acting at the point of contact with the surface, opposing the motion.

### Step 2: Write the Equation of Motion for the Solid Sphere
For the solid sphere, the net force acting on it can be expressed as:
\[ F - f = m a_{CM} \]
where \( a_{CM} \) is the linear acceleration of the center of mass of the solid sphere.

### Step 3: Write the Torque Equation for the Solid Sphere
The torque \( \tau \) about the center of the solid sphere due to the applied force \( F \) and the friction force \( f \) is given by:
\[ \tau = F \cdot R - f \cdot R \]
This torque is also related to the angular acceleration \( \alpha \) of the sphere:
\[ \tau = I \alpha \]
For a solid sphere, the moment of inertia \( I \) is:
\[ I = \frac{2}{5} m R^2 \]
Thus, we can write:
\[ F \cdot R - f \cdot R = \frac{2}{5} m R^2 \alpha \]

### Step 4: Relate Linear and Angular Acceleration
For pure rolling motion, the linear acceleration \( a_{CM} \) is related to the angular acceleration \( \alpha \) by:
\[ a_{CM} = R \alpha \]
Substituting this into the torque equation gives:
\[ F \cdot R - f \cdot R = \frac{2}{5} m R^2 \left(\frac{a_{CM}}{R}\right) \]
This simplifies to:
\[ F - f = \frac{2}{5} m a_{CM} \]

### Step 5: Substitute and Solve for Friction
Now we have two equations:
1. \( F - f = m a_{CM} \)
2. \( F - f = \frac{2}{5} m a_{CM} \)

From the first equation, we can express \( f \):
\[ f = F - m a_{CM} \]
Substituting this into the second equation:
\[ F - (F - m a_{CM}) = \frac{2}{5} m a_{CM} \]
This simplifies to:
\[ m a_{CM} = \frac{2}{5} m a_{CM} + F \]
Rearranging gives:
\[ \frac{3}{5} m a_{CM} = F \]
Thus, the acceleration of the solid sphere is:
\[ a_S = \frac{5F}{3m} \]

### Step 6: Repeat for the Hollow Sphere
For the hollow sphere, the moment of inertia is:
\[ I = \frac{2}{3} m R^2 \]
Following similar steps as above, we can derive:
\[ F - f = m a_{CM} \]
And the torque equation becomes:
\[ F \cdot R - f \cdot R = \frac{2}{3} m R^2 \alpha \]
Using the relation \( a_{CM} = R \alpha \), we can derive:
\[ F - f = \frac{2}{3} m a_{CM} \]
Solving this leads to:
\[ a_H = \frac{3F}{m} \]

### Step 7: Compare the Accelerations
Now we have:
- \( a_S = \frac{5F}{3m} \)
- \( a_H = \frac{3F}{m} \)

To compare \( a_S \) and \( a_H \):
\[ a_S = \frac{5F}{3m} < a_H = \frac{3F}{m} \]

### Conclusion
Thus, the linear acceleration of the solid sphere \( a_S \) is less than that of the hollow sphere \( a_H \).