If the force on a rocket moving with a velocity of $$300\,m/s$$ is $$345\,N,$$ then the rate of combustion of the fuel is
A.
$$0.55\,kg/s$$
B.
$$0.75\,kg/s$$
C.
$$1.15\,kg/s$$
D.
$$2.25\,kg/s$$
Answer :
$$1.15\,kg/s$$
Solution :
Thrust on the rocket is the force with which the rocket moves upwards. Thrust on rocket at time $$t$$ is given by $$F = - u\frac{{dm}}{{dt}}$$
The negative sign indicates that thrust on the rocket is in a direction opposite to the direction of escaping gases.
Here, velocity of the rocket $$u = 300\,m/s$$
and force $$F = 345\,N$$
∴ Rate of combustion of fuel $$ - \left( {\frac{{dm}}{{dt}}} \right) = \frac{F}{u} = \frac{{345}}{{300}} = 1.15\,kg/s$$
Releted MCQ Question on Basic Physics >> Impulse
Releted Question 1
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