Question

A spotlight $S$ rotates in a horizontal plane with a constant angular velocity of $0.1rad{s}^{-1}$ The spot of light $P$ moves along the wall at a distance $3m$. What is the velocity of spot $P$ when $\theta ={45}^{\circ }$ ?

• A. $0.1m{s}^{-1}$
• B. $0.2m{s}^{-1}$
• C. $0.4m{s}^{-1}$
• D. $0.6m{s}^{-1}$

Answer 1

The correct option is D $0.6m{s}^{-1}$


From triangle $\tan \alpha =\frac{x}{3}$
$\Rightarrow \frac{dx}{dt}=3{\sec }^2\alpha \left[\frac{d\alpha }{dt}\right]\Rightarrow v=\left[3{\sec }^2\alpha \right]\omega Here,v=\frac{dx}{dt}=velocityofP$
and $\omega =\frac{d\alpha }{dt}=$ angular velocity of S
When $\theta ={45}^{\circ },$ we get $\alpha ={45}^{\circ }$
Also, $\omega =0.1rad{s}^{-1}$

$\Rightarrow v=3{\sec }^2\left({45}^{\circ }\right)\times 0.1\Rightarrow v=3\times 2\times 0.1=0.6m{s}^{-1}$