JEE Main Physics 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
Answer: (A)
Solution
Apparent weight (accounting for buoyancy) = mg(1 − 1/RD) = 50(1 − 1/3) = 100/3 N. Work = (100/3)×5 = 500/3 J.