Question

Figure shows a torch producing a straight light beam falling on a plane mirror at an angle ${60^0}$. The reflected beam makes a spot P on the screen along the y-axis. If at $t = 0$ the mirror starts rotating about the hinge A with an angular velocity $\omega = {1^0}/sec$ clockwise, find the speed of the spot P on screen after $t = 15\sec $.

A. $\dfrac{\pi }{{15}}m{\sec ^{ - 1}}$
B. $\dfrac{\pi }{{30}}m{\sec ^{ - 1}}$
C. $\dfrac{{2\pi }}{{15}}m{\sec ^{ - 1}}$
D. $\dfrac{\pi }{{60}}m{\sec ^{ - 1}}$

Answer 1

Hint: In order to find the speed of spot P, we will find the relation between the distance of spot on y-axis at time $t = 15\sec $ and when mirror rotates with angular velocity of $\omega = {1^0}/sec$ then, the reflected ray from mirror will have the twice angular velocity and in time $t = 15\sec $ mirror will rotate by ${15^0}$ then the reflected ray will rotate by ${30^0}$ with the normal.

Complete step by step answer:
Let us first draw the diagram when after $t = 15\sec $ the reflected and incident ray will make an angle of ${30^0}$ so, from the geometry of figure we have, $ < PAB = \theta = {60^0}$ , let $BP$ be the distance on y axis of spot denoted by $y$.

In the right angle triangle $\Delta ABP$ we have,
$\tan \theta = \dfrac{{BP}}{{AP}}$
$\Rightarrow y = 3\tan \theta $
Differentiate above equation with respect to time, we get
$\dfrac{{dy}}{{dt}} = 3{\sec ^2}\theta .\dfrac{{d\theta }}{{dt}}$ $\{ \dfrac{d}{{d\theta }}(\tan \theta ) = {\sec ^2}\theta \} $
So we get, $\dfrac{{dy}}{{dt}} = 3{\sec ^2}\theta .\dfrac{{d\theta }}{{dt}}$

Now as we know, $\dfrac{{dy}}{{dt}}$ is nothing but the velocity of spot $P$ lets denoted as ${v_p}$. And since the angular velocity of mirror is $\omega = {1^0}per\sec ond$ the angular velocity of reflected ray will be $2\omega = {2^0}/sec$ and $\dfrac{{d\theta }}{{dt}}$ is nothing but this angular velocity which is $2\omega = {2^0}/sec$
Converting $2\omega = {2^0}/sec$ into radian angle we have,
${2^0}/sec = \dfrac{{2 \times \pi }}{{180}}$
And, from the diagram we have, this $\theta = {60^0}$
Put all the parameters value in equation $\dfrac{{dy}}{{dt}} = 3{\sec ^2}\theta \dfrac{{d\theta }}{{dt}}$ we get,
${v_p} = 3{\sec ^2}{60^0}(\dfrac{{2 \times \pi }}{{180}})$
$\Rightarrow {v_p} = 3 \times 4 \times (\dfrac{\pi }{{90}})$............$\{ \sec {60^0} = 2\} $
$\therefore {v_p} = \dfrac{{2\pi }}{{15}}m{\sec ^{ - 1}}$

Hence, the correct option is C.

Note:It must be remembered that, the angular speed at which mirror rotates, the reflected ray will rotate by twice the angle with which mirror rotate and hence twice the angular velocity of rotation of reflected ray, and angular velocity of the body is the differentiation of angular displacement with respect to time $\omega = \dfrac{{d\theta }}{{dt}}$.