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

# If force on a rocket having exhaust velocity of $300\text{\hspace{0.17em}m/s}$ is 210 N, then rate of combustion of the fuel is

1. A

$0.7\text{\hspace{0.17em}kg/s}$

2. B

$1.4\text{\hspace{0.17em}kg/s}$

3. C

$0.07\text{\hspace{0.17em}kg/s}$

4. D

$10.7\text{\hspace{0.17em}kg/s}$

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### Solution:

Force $=\frac{\text{d}\overline{p}}{\text{dt}}$

$=\frac{\text{dm}\overline{v}}{\text{dt}}=\stackrel{\to }{v}\frac{\text{dm}}{\text{dt}}\to \left(1\right)$

using (1)

$210=300\frac{\text{dm}}{\text{dt}}$

$\frac{\text{dm}}{\text{dt}}=0.7\text{\hspace{0.17em}kg/s}$

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