Q. A machine which is 75% efficient uses 12 J of energy in lifting up a 1 kg mass through a certain distance. The mass is then allowed to fall through, that distance. The velocity of the ball at the end of its fall is
A) $\sqrt{24}\ ms^{- 1}$
B) $\sqrt{32}\ ms^{- 1}$
C) $\sqrt{18}\ ms^{- 1}$
D) $3\ ms^{- 1}$
Explanation
(c) Useful work $= \frac{75}{100} \times 12\ J = 9J$
Now, $\frac{1}{2} \times 1 \times v^{2} = 9$ or $v = \sqrt{18}\ ms^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.