JEE Advanced 2020 Full Test 4 Paper 1

We have relation M‌

dv

dt

=Fext+vrel‌

dM

dt

In the question, we have the force by the gravity as the external force. As the rocket is to lift from the ground the minimum acceleration of the rocket to do so is, o, beyond which, with a slight increase it will zoom into the sky.
‌ Therefore, ‌‌

dv

dt

=0
Putting these into the equation, we obtain
0=Mg+v‌rel ‌(−‌

dM

dt

)
(‌

dM

dt

is rate of decrease of mass hence negative. Mg is in the direction of the relative velocity of the fuel with respect to the rocket)
Hence, ‌

dM

dt

=‌

M

V

g=‌

400×10

2×1(103)

=2kg∕sec