Initial fuel in a rocket is $1000kg.$ If it was given an acceleration of $20\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.$ the velocity of fuel with respect to the rocket is $500\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$. Find the rate of consumption of fuel.

The correct option is B

$60\raisebox{1ex}{$kg$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

Solution:

Step 1: Given information

Initial fuel$=1000kg.$

Acceleration$=20\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.$

Fuel's velocity$=500\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

Step 2: Calculate the rate of consumption of fuel

Know that, given a system with variable mass,

$TotalForce(Byrocket)-GravitationalForce=ActingForce$,

$f_T-mg=ma$…..(i)

Where,

$f_T=$Total force

$m=$Mass of fuel

$g=$Acceleration due to gravity

$a=$Acceleration produced by rocket

Therefore, from equation (i),

$m\frac{v_r}{t}-mg=ma$…..(ii)

Where, $(f_T=m\times a)$

Step 3: Solving by putting the given values in the equation (ii)

Therefore, from equation (ii),

$\left(\frac{dm}{dt}\right)v_r=mg+ma$

$\left(\frac{dm}{dt}\right)500=1000(10+20)$

$\left(\frac{dm}{dt}\right)=60kg/s$

Hence, the correct option is (B).