If the force on a rocket moving with a velocity of 300 m/s is 210 N, then the rate of combustion of the fuel is

The correct option is 0.7 kg/s.

To find the rate of combustion of the fuel, we can use the relation between thrust force, relative exhaust velocity, and the rate of mass consumption (rate of combustion of fuel).

The thrust force $F$ acting on a rocket is given by the formula:
$F=v\cdot \frac{dm}{dt}$
where:
- $F$ is the force on the rocket (thrust), measured in Newtons (N),
- $v$ is the velocity of the exhaust gases relative to the rocket, measured in meters per second (m/s), and
- $\frac{dm}{dt}$ is the rate of combustion of the fuel (rate of mass consumption), measured in kilograms per second (kg/s).

We are given the following values in the problem:
- Force, $F=210\text{N}$
- Velocity, $v=300\text{m/s}$

By rearranging the formula to solve for the rate of combustion of fuel $\frac{dm}{dt}$:
$\frac{dm}{dt}=\frac{F}{v}$

Substitute the given values into the equation:
$\frac{dm}{dt}=\frac{210}{300}$

Calculate the value:
$\frac{dm}{dt}=0.7\text{kg/s}$

Thus, the rate of combustion of the fuel is 0.7 kg/s.