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

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) (500/3) J (B) (350/3) J (C) (750/3) J (D) zero

Tardigrade Admin · Accepted Answer

Relative density =( text Weight in air / text Loss of weight in water ) ∴ Loss of weight in water =(5 × 10/3) N Weight in water =(50-(50/3)) N =(100/3) N Work done =(100/3) N × 5 m =(500/3) J

Q. A $5 k g$ 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 $g = 10 m s^{- 2}$ , then the work done is

$\frac{500}{3} J$

$\frac{350}{3} J$

$\frac{750}{3} J$

zero

Relative density $= \frac{Weight in air}{Loss of weight in water}$
$∴$ Loss of weight in water $= \frac{5 \times 10}{3} N$
Weight in water $= (50 - \frac{50}{3}) N = \frac{100}{3} N$
Work done $= \frac{100}{3} N \times 5 m = \frac{500}{3} J$

Q. A 5kg stone of relative density 3 is resting at the bed of a lake. It is lifted through a height of 5m in the lake. If g=10ms−2, then the work done is

3500​J

3350​J

3750​J