Question

a buoyant force of 200 gwt acts on a body dipped completely in water. If the apperent weight of the body is 100gwt, then find its actual weight

Answer 1

The buoyant force acting on the body,
$$\begin{gathered} {\rho }_{w}gV=200\mathrm{g}\mathrm{wt}.=200\times {10}^{-3}\times 10=2\mathrm{N} \\ \mathrm{Apparent}\mathrm{weight}=100gwt.=100\times {10}^{-3}\times 10=1\mathrm{N} \\ \mathrm{The}\mathrm{actual}\mathrm{weight}\mathrm{of}\mathrm{the}\mathrm{body}={\rho }_{b}gV \\ \mathrm{The}\mathrm{apparent}\mathrm{weight}=\mathrm{actual}\mathrm{weight}-\mathrm{buoyant}\mathrm{force} \\ \Rightarrow 1={\rho }_{b}gV-2 \\ \Rightarrow {\rho }_{b}gV=3 \\ \Rightarrow {\rho }_{b}gV=300\times {10}^{-3}\times 10=300\mathrm{g}\mathrm{wt}. \end{gathered}$$