SSC JE Mechanical Engineering 25 Sep 2019 Shift 1 Paper

Weight of the cube $\left(\mathit{W}\right)=$ Density $\times$ Volume $\times \mathit{g}=500\times 1^3=\left(500\times \mathit{g}\right)\mathit{N}$ .......(1)
Buoyant Force $\left(\mathit{F}\right)=\left(\mathit{A}\times \mathit{h}\times {\mathrm{\rho }}_{\mathit{w}}\times \mathit{g}\right)\mathit{N}=\left(1^2\times \mathit{h}\times 1000\times \mathit{g}\right)$ $=\left(1000\times \mathit{g}\times \mathit{h}\right)\mathit{N}$ .......(2)
Buoyant force (F) = Weight of cube (W)
From equation 1 and 2,
500 × g = 1000 × g × h
h = 0.5 m