Question

A rocket of mass 1000 kg is to be projected vertically upwards. The gases are exhausted vertically downwards with velocity $100m{s}^{-1}$ 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 $g=10m{s}^{-2}$) :

• A. $50kg{s}^{-1}$
• B. $100kg{s}^{-1}$
• C. $200kg{s}^{-1}$
• D. $400kg{s}^{-1}$

Answer 1

Answer: (B)

$100kg{s}^{-1}$

Given, the velocity of exhaust gases with respect to rocket $=100m{s}^{-1}$
The minimum force on the rocket to lift it
$F_{min}=mg=1000\times 10=10000N$

Hence, minimum rate of burning of fuel is given by
$\frac{dm}{dt}=\frac{F_{min}}{v}=\frac{10000}{100}$
$=100kg{s}^{-1}$