Question

If the thrust force on a rocket which is ejecting gases with a relative velocity of 300m/s is 210N. Then the rate of combustion of the fuel will be:
$A.10.7Kg/s$
$B.0.07Kg/s$
$C.1.4Kg/s$
$D.0.7Kg/s$

Answer 1

Hint: We will apply the basic concept of thrust force and its formula to solve the given problem. We know that the thrust is a reactive force. We will use the concept of mass rate to find the rate of combustion of the fuel.

Formula used:
We are using the following formula to get the correct answer:-
$T=v\dfrac{dm}{dt}$.

Complete step by step solution:
We know that the thrust is a reactive force which acts perpendicular to the surface.
From the problem given above we have the following parameters with us:-
Relative velocity, $v=300m/s$
Thrust force,$T=210N$
Mass of the combustion is denoted with $m$.
Time taken in the process is represented as $t$.
We have to find the rate of combustion which is represented with $\dfrac{dm}{dt}$.
We have the following formula of thrust force:-
$T=v\dfrac{dm}{dt}$………………. $(i)$
Using the given parameters and putting their values in equation $(i)$we get
$210=300\times \dfrac{dm}{dt}$
Solving further we get
$\dfrac{dm}{dt}=\dfrac{210}{300}$
$\dfrac{dm}{dt}=0.7Kg/s$

Hence, option $(D)$ is the correct one.

Additional Information:
SI unit of thrust is Newton just like any other force. Some of the applications of the thrust force are as follows:-
$(i)$ Motorboat is propelled due to the action of thrust force.
$(ii)$ Rocket is propelled by the action of thrust force.
$(iii)$ Aircraft is also propelled with the help of thrust force.
$(iv)$ Thrust force is widely used in the engineering field to design different structures and designs.

Note:
The main thing which should be kept in mind during analysis of thrust force is that it is a reactive agent and not an active agent. We cannot call every force as thrust because thrust always acts perpendicular to the force. Consideration of units is also very important.