Question

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

• A. $\frac{4\pi }{15}\text{m/s}$
• B. $\frac{7\pi }{15}\text{m/s}$
• C. $\frac{11\pi }{15}\text{m/s}$
• D. $\frac{2\pi }{15}\text{m/s}$

Answer 1

The correct option is D $\frac{2\pi }{15}\text{m/s}$

Given:
Angular velocity of plane mirror,
$\omega =1^{\circ }\text{/s}$ clockwise.



In $15\text{s}$, mirror will rotate by ${15}^{\circ }$. Thus, reflected ray will rotate by ${30}^{\circ }$ as its angular velocity will be $2^{\circ }\text{/s}$ clockwise.

At any time $t$,


From the diagram
$x=3\tan \theta$
$\Rightarrow \frac{dx}{dt}=3{\sec }^2\theta \frac{d\theta }{dt}$

At $t=15\text{s},\theta ={60}^{\circ }$

$\Rightarrow$ Speed of the spot at $t=15\text{s}$,
$v_p=3\times {\sec }^2\left({60}^{\circ }\right)\times 2^{\circ }$

[since $\frac{d\theta }{dt}=$ angular velocity of rotation of reflected ray]

$\Rightarrow v_p=3\times 4\times 2\times \frac{\pi }{180}=\frac{2\pi }{15}\text{m/s}$

Hence, option $\left(d\right)$ is the correct answer.