1. Relate circular motion to SHM

The projection of a particle moving uniformly on a circle executes SHM.

• Radius of circle $= A = 0.36\,\text{m}$
• Hence amplitude of SHM is $A = 0.36\,\text{m}$

2. Find angular speed

The particle covers $30^\circ$ in $0.1\,\text{s}$.

Convert to radians: $30^\circ = \frac{\pi}{6}\,\text{rad}$

So angular speed is $\omega = \frac{\Delta \theta}{\Delta t} = \frac{\pi/6}{0.1} = \frac{\pi}{0.6} = \frac{5\pi}{3}\,\text{rad s}^{-1}$

3. Restoring force per unit mass in SHM

In SHM, $a = -\omega^2 x$

So restoring force per unit mass is just the acceleration magnitude: $\frac{F}{m} = \omega^2 x$

When $P$ touches the extreme point $M$, displacement magnitude is maximum: $x = A = 0.36\,\text{m}$

Therefore, $\frac{F}{m} = \omega^2 A$

Substitute: $\frac{F}{m} = \left(\frac{5\pi}{3}\right)^2 (0.36)$

$= \frac{25\pi^2}{9} \times 0.36$

Since $0.36 = \frac{9}{25}$, $\frac{F}{m} = \frac{25\pi^2}{9} \cdot \frac{9}{25} = \pi^2$

$\frac{F}{m} \approx 9.87\,\text{m s}^{-2}$

Numerically, this is restoring force per unit mass. In the options, it is written as $\text{N}$, but physically it should be $\text{N kg}^{-1}$ or $\text{m s}^{-2}$.

4. Match with options

$\boxed{9.87}$

So the correct option is A.