Question

A particle revolves in clockwise direction (as seen from point A) in a circle C of radius $1\text{cm}$ and completes one revolution in $2s$. The axis of the circle and the principal axis of the mirror $M$ coincide. The radius of curvature of the mirror is $20\text{cm}$. Then, the direction of revolution (as seen from A) of the image of the particle and its speed is

• A. Clockwise, $1.57\text{cm/s}$
• B. Clockwise, $3.14\text{cm/s}$
• C. Anticlockwise, $1.57\text{cm/s}$
• D. Anticlockwise, $3.14\text{cm/s}$

Answer 1

The correct option is A Clockwise, $1.57\text{cm/s}$

If the object rotates in clockwise direction, the image also has to rotate in clockwise direction.

$m=\frac{f}{f-u}=\frac{10}{10+10}=\frac{1}{2}$

Therefore,
New radius $r^{\prime }=\frac{1}{2}\times 1=0.5cm$

Since the image completes one revolution in $2s$

$\frac{2\pi \times \frac{1}{2}}{v}=2$
$\Rightarrow$ Speed, $v=1.57cm/s$ in clockwise.