Q. A 5 kg stone of relative density 3 is resting at the bed of a lake. It is lifted through a height of 5 m in the lake. If $g = 10ms^{2}$, then the work done is
A) $\frac{500}{3}J$
B) $\frac{350}{3}J$
C) $\frac{750}{3}J$
D) Zero
Explanation
(a) Relative density $= \ \frac{Weight\ in\ air}{Loss\ weight\ in\ water}$
$\therefore$ Loss of weight in water $= \frac{5 \times 10}{3}\ N$
Weight in water $= \left( 50 - \frac{50}{3} \right)\ N = \frac{100}{3}\ N$
Work done$= \frac{100}{3}N \times 5m = \frac{500}{3}\ J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.