Q. A rocket is receding away from earth with velocity = 0.2c.The rocket emit signal or frequency $4 \times 10^{7}$ Hz. The apparent frequency of the signal produced by the rocket observed by the observer on earth will be
A) $3 \times 10^{6}Hz$
B) $4 \times 10^{6}Hz$
C) $2.4 \times 10^{7}Hz$
D) $5 \times 10^{7}Hz$
Explanation
(c) As $v = \frac{1}{2}\frac{\mathrm{\Delta}v}{v}c\ $
$\therefore\ 0.2c = \frac{1}{2}\frac{\mathrm{\Delta}v}{(4 \times 10^{7})}c$
$\mathrm{\Delta}v = 1.6 \times 10^{7}\ Hz$
As the rocket is receding away
$\therefore\ v^{'} = v - \mathrm{\Delta}v = 4 \times 10^{7} - 1.6 \times 10^{7}$
$= 2.4 \times 10^{7}\ Hz$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.