A cubical room is made by placing $6$ identical plane mirrors with the reflecting surface on the inside of the room. An insect crawls along the diagonal of the floor with constant speed of $40\text{cm/sec}$. What is the magnitude of velocity of image of insect w.r.t insect in one of the two adjacent walls?

The correct option is B $40\sqrt{2}\text{cm/s}$

Here $v=40\text{cm/s}$ = velocity of insect.
The angle $\theta ={45}^{\circ }$ from any of the two adjacent walls.


Thus, the magnitude of velocity of image w.r.t insect will be,
$v\sin {45}^{\circ }-(-v\sin {45}^{\circ })$
$\Rightarrow 2v\sin {45}^{\circ }=2\times 40\times \frac{1}{\sqrt{2}}=40\sqrt{2}\text{cm/s}$
Hence, in any of the two adjacent wall, the speed of image w.r.t insect is $40\sqrt{2}\text{cm/s}$.