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Q.
At what time between $10\, O'$ clock and $11\, O$' clock are the two hands of a clock symmetric with respect to the vertical line (give the answer to the nearest second)?
KVPYKVPY 2009
Solution:
Exactly at $10\, O$' clock the hour hand has travelled $300^{\circ}$ from $12\, O$'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.
$\left[(6 x)\left(\frac{1}{12}\right)\right]^{\circ}=\left(\frac{1}{2} x\right)^{\circ}$
$\therefore$ Our hand has covered $\left(300+\frac{x}{2}\right)^{\circ}$
On subtracting this from $360^{\circ}$ to find the angle from $12\, O$ '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.
$\therefore 60-\frac{x}{2}=6 x$
$\Rightarrow x=\left(\frac{120}{13}\right)^{\circ}=9\, \min 13.8\, s$
$\therefore$ Time $-10\, h\, 9\, m\, 14\, s$
