Question

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 $g=10 \,ms ^{-2}$, then the work done is

• A. $\frac{500}{3} \,J$
• B. $\frac{350}{3} \,J$
• C. $\frac{750}{3} \,J$
• D. zero

Answer 1

Answer: (A)

Relative density $=\frac{\text { Weight in air }}{\text { Loss of weight in water }}$
$\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 $=\frac{100}{3} N \times 5 m =\frac{500}{3} \,J$