Step 1: Analyzing the forces and torques.
The torque \( \tau \) applied to the solid cylinder is given by:
\[
\tau = P \times r
\]
where \( P = 90 \, \text{N} \) and \( r = 0.5 \, \text{m} \). Therefore:
\[
\tau = 90 \times 0.5 = 45 \, \text{N.m}
\]
Step 2: Moment of Inertia of a Solid Cylinder.
The moment of inertia \( I \) of a solid cylinder is:
\[
I = \frac{1}{2} M r^2
\]
where \( M \) is the mass of the cylinder and \( r \) is the radius. From the weight \( W = 100 \, \text{N} \), we have:
\[
M = \frac{W}{g} = \frac{100}{10} = 10 \, \text{kg}
\]
Thus, the moment of inertia is:
\[
I = \frac{1}{2} \times 10 \times (0.5)^2 = 1.25 \, \text{kg.m}^2
\]
Step 3: Angular acceleration.
Using the rotational form of Newton's second law:
\[
\tau = I \alpha
\]
where \( \alpha \) is the angular acceleration. Substituting the values:
\[
45 = 1.25 \times \alpha
\]
\[
\alpha = \frac{45}{1.25} = 36 \, \text{rad/s}^2
\]
Thus, the angular acceleration is \( 24 \, \text{rad/s}^2 \).
Correction: If the cylinder is also rolling without slipping, the friction force would affect the net torque. However, the problem does not provide enough information to account for friction.
Thus, the most straightforward solution (ignoring friction) yields \(36 \, \text{rad/s}^2\) , which is not among the options.