Question

A rocket of mass $1000\, kg$ is to be projected vertically upwards. The gases are exhausted vertically downwards with velocity $100 \,m / s$ 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 $g=10\, m / s ^{2}$ ).

• A. 50 kg/s
• B. 100 kg/s
• C. 200 kg/s
• D. 400 kg/s

Answer 1

Answer: (B)

The force $(F)$ lifting the rocket is given by
$F=v_{r}\left(\frac{\Delta m}{\Delta t}\right)$
where $v_{r}$ is velocity of the gas relative to rocket,
$\frac{\Delta m}{\Delta t}$ is rate of change of mass.
Given, $F=M g=1000 \times 10=10,000\, N$
$v_{r}=100\, m / s$
$\therefore \frac{\Delta m}{\Delta t}=\frac{F}{v_{r}}$
$=\frac{10,000}{100}=100 \,kg / s$