Q. Diwali rockets are ejecting 50 g of gases per second at a velocity of${\ 400\ ms}^{- 1}$. The accelerating force on the rocket will be
A) 22 dyne
B) 20 N
C) 20 dyne
D) 100 N
Explanation
(b) The accelerating force of the rocket
$= upward\ thrust = \frac{\mathrm{\Delta}m}{\mathrm{\Delta}t} \bullet v$
$Given,\ \frac{\mathrm{\Delta}m}{\mathrm{\Delta}t} = 50 \times 10^{- 3}\ {kgs}^{- 1},\ \ v = 400\ {ms}^{- 1}$
So, accelerating force $= 50 \times 10^{3} \times 400 = 20\ N$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.