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Question: Figure shows a torch producing a straight light beam falling on a plane mirror at an angle \({60^0}\...

Figure shows a torch producing a straight light beam falling on a plane mirror at an angle 600{60^0}. The reflected beam makes a spot P on the screen along the y-axis. If at t=0t = 0 the mirror starts rotating about the hinge A with an angular velocity ω=10/sec\omega = {1^0}/sec clockwise, find the speed of the spot P on screen after t=15sect = 15\sec .

A. π15msec1\dfrac{\pi }{{15}}m{\sec ^{ - 1}}
B. π30msec1\dfrac{\pi }{{30}}m{\sec ^{ - 1}}
C. 2π15msec1\dfrac{{2\pi }}{{15}}m{\sec ^{ - 1}}
D. π60msec1\dfrac{\pi }{{60}}m{\sec ^{ - 1}}

Explanation

Solution

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=15sect = 15\sec and when mirror rotates with angular velocity of ω=10/sec\omega = {1^0}/sec then, the reflected ray from mirror will have the twice angular velocity and in time t=15sect = 15\sec mirror will rotate by 150{15^0} then the reflected ray will rotate by 300{30^0} with the normal.

Complete step by step answer:
Let us first draw the diagram when after t=15sect = 15\sec the reflected and incident ray will make an angle of 300{30^0} so, from the geometry of figure we have, <PAB=θ=600 < PAB = \theta = {60^0} , let BPBP be the distance on y axis of spot denoted by yy.

In the right angle triangle ΔABP\Delta ABP we have,
tanθ=BPAP\tan \theta = \dfrac{{BP}}{{AP}}
y=3tanθ\Rightarrow y = 3\tan \theta
Differentiate above equation with respect to time, we get
dydt=3sec2θ.dθdt\dfrac{{dy}}{{dt}} = 3{\sec ^2}\theta .\dfrac{{d\theta }}{{dt}} ddθ(tanθ)=sec2θ\\{ \dfrac{d}{{d\theta }}(\tan \theta ) = {\sec ^2}\theta \\}
So we get, dydt=3sec2θ.dθdt\dfrac{{dy}}{{dt}} = 3{\sec ^2}\theta .\dfrac{{d\theta }}{{dt}}

Now as we know, dydt\dfrac{{dy}}{{dt}} is nothing but the velocity of spot PP lets denoted as vp{v_p}. And since the angular velocity of mirror is ω=10persecond\omega = {1^0}per\sec ond the angular velocity of reflected ray will be 2ω=20/sec2\omega = {2^0}/sec and dθdt\dfrac{{d\theta }}{{dt}} is nothing but this angular velocity which is 2ω=20/sec2\omega = {2^0}/sec
Converting 2ω=20/sec2\omega = {2^0}/sec into radian angle we have,
20/sec=2×π180{2^0}/sec = \dfrac{{2 \times \pi }}{{180}}
And, from the diagram we have, this θ=600\theta = {60^0}
Put all the parameters value in equation dydt=3sec2θdθdt\dfrac{{dy}}{{dt}} = 3{\sec ^2}\theta \dfrac{{d\theta }}{{dt}} we get,
vp=3sec2600(2×π180){v_p} = 3{\sec ^2}{60^0}(\dfrac{{2 \times \pi }}{{180}})
vp=3×4×(π90)\Rightarrow {v_p} = 3 \times 4 \times (\dfrac{\pi }{{90}})............sec600=2\\{ \sec {60^0} = 2\\}
vp=2π15msec1\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 ω=dθdt\omega = \dfrac{{d\theta }}{{dt}}.