AP EAPCET 24-Aug-2021 Shift 2 Paper

Given that, initially body is submerged into the water by $\frac{4}{5}$ th volume of body. i.e. Volume inside the water, $V_i = \frac{4}{5} V$ where, V be total volume of body. According to law of floatation, Weight of body = Buoyant force $Mg = V_i \rho g$ $\Rightarrow V \sigma = V_i \rho$ $\Rightarrow \sigma = \frac{V_i}{V} \rho = \frac{4}{5} \rho$ ...(i) where, $\sigma$ be the density of body. When oil is poured into water, $\left(\frac{V}{2}\right)$ volume of body is under water and oil. Let $\rho_o$ be density oil, then again by using condition of floatation, $Mg = F_{\text{walt}} + F_{\text{oil}}$ $V \sigma g = V_w \rho g + V_{\text{oil}} \rho_o \; g$ $V \sigma = \frac{V}{2} \rho + \frac{V}{2} \rho_o$ $V \times \frac{4}{5} \rho = \frac{V \rho}{2} + \frac{V \rho_o}{2}$ [From Eq. (i)] $\rho_0 = \left(\frac{4}{5} - \frac{1}{2}\right) \rho \times 2 = \frac{3}{5} \rho$ Hence, density of oil w.r.t. water is $\frac{3}{5}$.