Q. A rocket of mass 100 kg burns 0.1 kg of fuel per second. If velocity of exhaust gas is ${1\ kms}^{- 1}$, then it lifts with an acceleration of
A) ${1000\ ms}^{- 2}$
B) ${100\ ms}^{- 2}$
C) ${10\ ms}^{- 2}$
D) ${1\ ms}^{- 2}$
Explanation
(d) $Given\ that,\ \frac{dm}{dt} = 0.1\ {kgs}^{- 1};$
mass of the rocket=100 kg
$and\ v = 1\ {kms}^{- 1} = 1000\ {ms}^{- 1}$
$Thrust\ on\ the\ rocket,\ F = v\frac{dm}{dt} = 1000 \times 0.1$
$Now,\ \ F = Ma$
$\therefore\ \ a = \frac{1000 \times 0.1}{100} = 1\ {ms}^{- 2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.