(b)

Exactly at $10\,O'clock$ the hour hand has travelled $300^{\circ}$ from $120^{\prime}$ 'clock.

One hour $=60$ minute.

One minute hand moves $1^{\circ}$ and hour clock hand move $\left(\frac{30}{360}\right)^{\circ}=\left(\frac{1}{12}\right)^{\circ}$

Assuming we have made it to 10 O'clock and now the hour and the minute hand start moving spontaneously.

If the hands of the watch are symmetric with vertical line.

Supposing this happens when $x$ minutes have passed $x$ minutes $=(6 x)^{\circ}$ have been covered our hour hand would cover.

On subtracting this from $360^{\circ}$ to find the angle from 12 O'clock anti-clockwise, we get

$360^{\circ}-\left(300+\frac{x}{2}\right)^{\circ}=\left(60-\frac{x}{2}\right)^{\circ}$

So, they are symmetric.

$\therefore \quad 60-\frac{x}{2}=6 x$

$\Rightarrow \quad x=\left(\frac{120}{13}\right)^{\circ}=9min\,13.8s$

$\therefore \text { Time }=10 h\,9m\,14s$