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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
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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=vdmdt.

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 dmdt.
We have the following formula of thrust force:-
T=vdmdt………………. (i)
Using the given parameters and putting their values in equation (i)we get
210=300×dmdt
Solving further we get
dmdt=210300
dmdt=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.
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