Question

A rocket of mass $1000\, kg$ is to be projected vertically upwards. The gases are exhausted vertically downwards with velocity $100 \,ms^{−1}$ with respect to the rocket. What is the minimum rate of burning of fuel, so as to just lift the rocket upwards against, the gravitational attraction? (Take $g = 10ms^{−2}$)

• A. $50\, kgs^{−1}$
• B. $100\, kgs^{−1}$
• C. $200\, kgs^{−1}$
• D. $400\, kgs^{−1}$

Answer 1

Answer: (B)

Given, the velocity of exhaust gases with respect to rocket $=100\, ms ^{-1}$
The minimum force on the rocket to lift it
$F_{\min }=m g=1000 \times 10=10000 N$
Hence, minimum rate of burning of fuel is given by'
$\frac{d m}{d t} =\frac{F_{\min }}{v}=\frac{10000}{100}$
$=100\, kgs ^{-1}$