A stone of mass 10kg is lying at the bed of a lake 5m deep. If the relative density of stone is 2,the amount of work done to bring it to the top of the lake will be.

work done = mgh 

mg is weight so we need to find the apparent weight of the stone inside the water.
 
Apparent weight = original weight - weight of water replaced.

now,

volume of water replaced = volume of the stone (Archemedes' principle)

Volume of water replaced = $\frac{mass}{density}$

$$\begin{gathered} volume of water displaced = \frac{10}{2000} = 0.005m^3 \\ now, weight of water displaced = Volume of water displaced \times {\rho }_{water}\times g \\ {w}_{displaed}= 0.005\times 1000\times 10 = 50N \\ Apparent weight = w_A = 10\times 10-50 \\ {w}_A = 50N \\ now, \\ work done = w_A \times h \\ W = 50\times 5 \\ W = 250 J \end{gathered}$$