A Machine gun fires 360 bullets per minute each bullet moves with a velocity of 600 m/s. If the power of gun is \[5.4kW\], mass of each bullet is :
A. 5 g
B. \[0.5\]g
C. \[0.5\times {{10}^{3}}g\]
D. \[0.0025g\]

Hint:As we know the power is known as work done per unit time. Where work done is calculated in the form of kinetic energy because here Bullets are fired with a velocity.

Complete answer:
As given Rate of Bullet Firing = \[\dfrac{360\text{ Bullets}}{\text{1 Minute}}\]
Velocity of bullet from gun (v) = 600 m/s
Power of machine gun = 5.4 kW
Let the mass of the bullet = m kg.
Power of the gun is calculated as:
\[\text{Power}=\text{Kinetic energy of each bullet}\times \text{No}\text{. of bullets fired per second}\]
Here work done is obtain in the form of kinetic energy
\[KE=\dfrac{1}{2}\text{mass of bullet}\times {{\left( \text{velocity of bullet} \right)}^{2}}\]
\[KE=\dfrac{1}{2}\times m\times {{\left( 600 \right)}^{2}}\text{ }\left( 1 \right)\]
No of bullets fired per second is calculated as:
\[\begin{gathered}
  & \dfrac{360\text{ Bullet}}{\min } \\
 & \Rightarrow \dfrac{360}{60\sec }\text{ }\left( 2 \right) \\
\end{gathered}\]
By substituting values from equation 1 and 2, we get,
\[\text{Power}=\text{Kinetic energy of each bullet}\times \text{No}\text{. of bullets fired per second}\]
\[\Rightarrow 5.4\times {{10}^{3}}=\dfrac{1}{2}\times m\times 3600\times \dfrac{360}{60}\]
\[\Rightarrow 5.4\times {{10}^{3}}=1800m\times 6\]
\[\Rightarrow m=\dfrac{5.4\times {{10}^{3}}}{1800\times 6}\]
\[\Rightarrow m=\dfrac{5400}{1800\times 6}\]
\[\Rightarrow m=\dfrac{3}{6}=0.5kg=0.5\times {{10}^{3}}g\]

Hence, the correct option is choice C.

Note:
Power is converted from kW to W by multiplying with 1000, that is, \[5.4\text{ kW}=5.4\times {{10}^{3}}\text{ W}\]. Also, time is converted from minutes to seconds where, 1 min = 60 sec, to calculate number of bullets fired per second. Since, all the given options are in grams, mass of the bullet is also converted to g by multiplying with 1000.