Q. A rocket standing vertically on a launch pad has to start moving up with practically zero velocity. If the initial mass of the rocket is $5 \times 10^{5}$ kg, then the rate of burning of the fuel should be [Take $g = 10\ ms^{- 2}$ and velocity of exhaust gases = 10 $kms^{- 1}$]
A) 10 $kgs^{- 1}$
B) 50$\ kgs^{- 1}$
C) 500$\ kgs^{- 1}$
D) 5000$\ kgs^{- 1}$
Explanation
(c) $v_{r}\frac{dm}{dt} = m_{0}g$ or $\frac{dm}{dt} = \frac{m_{0}g}{v_{r}}$
or $\frac{dm}{dt} = \frac{5 \times 10^{5}}{10 \times 10^{3}}$ kg $s^{- 1} = 500$ kg $s^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.