Question

At what time between $10O^{\prime }$ clock and $11O$' clock are the two hands of a clock symmetric with respect to the vertical line (give the answer to the nearest second)?

• A. $10h9m13s$
• B. $10h9m14s$
• C. $10h9m22s$
• D. $10h9m50s$

Answer 1

Answer: (B)

Exactly at $10O$' clock the hour hand has travelled $300^{\circ }$ from $12O$'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 $10O$' 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 $=(6x{)}^{\circ }$ have been covered our hour hand would cover.
${\left[(6x)\left(\frac{1}{12}\right)\right]}^{\circ }={\left(\frac{1}{2}x\right)}^{\circ }$
$\therefore$ Our hand has covered ${\left(300+\frac{x}{2}\right)}^{\circ }$
On subtracting this from $360^{\circ }$ to find the angle from $12O$ '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.
$\therefore 60-\frac{x}{2}=6x$
$\Rightarrow x={\left(\frac{120}{13}\right)}^{\circ }=9\min 13.8s$
$\therefore$ Time $-10h9m14s$