1. Given data

• Mass of hollow cylinder: $m = 5\,\text{kg}$
• Radius: $R = 0.5\,\text{m}$
• Pulling force on string: $F = 40\,\text{N}$
• The cylinder is hollow, so its moment of inertia about its center is $I = mR^2$

Thus, $I = 5(0.5)^2 = 5 \times 0.25 = 1.25\,\text{kg m}^2$

2. Choose sign convention

Take rightward translation as positive and clockwise rotation as positive.

Since the string is pulled horizontally at the top, the force $F$ produces a clockwise torque.

Let friction at the ground be $f$. Its direction is to be determined by the equations.

3. Equation of translation

Horizontal forces on the cylinder:

• Pulling force $F$ to the right
• Friction $f$ at contact point

So, $F + f = ma \qquad (1)$

4. Equation of rotation about center

Taking clockwise as positive:

• Torque due to pulling force at the top: $+FR$
• Torque due to friction at bottom: friction to the right gives anticlockwise torque, so contribution is $-fR$

Hence, $FR - fR = I\alpha$ $R(F-f)=I\alpha \qquad (2)$

5. Rolling without slipping condition

For pure rolling, $a = \alpha R \qquad (3)$

6. Substitute using $I=mR^2$

From (2): $R(F-f)=mR^2\alpha$ $F-f=mR\alpha$

Using $a=\alpha R$, we get $F-f=ma \qquad (4)$

From (1): $F+f=ma \qquad (1)$

Comparing (1) and (4): $F+f = F-f$ $2f=0 \Rightarrow f=0$

So friction is actually zero.

Then from (1): $F=ma$ $40 = 5a$ $a = 8\,\text{m/s}^2$

Now, $\alpha = \frac{a}{R} = \frac{8}{0.5} = 16\,\text{rad/s}^2$

7. Match with options

$\boxed{\alpha = 16\,\text{rad/s}^2}$

So the correct option is A.

8. Comparison with stored answer

Stored correct answer: A

Our derived answer: A

They match.