A solid float such that its $(1 / 3)$ rd part is above water surface. Then, the density of solid is

A solid float such that its $(1 / 3)$ rd part is above water surface. Then, the density of solid is
  1. $744 \mathrm{~kg} \mathrm{~m}^{-3}$
  2. $\frac{1000}{3} \mathrm{~kg} \mathrm{~m}^{-3}$
  3. $\frac{2000}{3} \mathrm{~kg} \mathrm{~m}^{-3}$
  4. $910 \mathrm{~kg} \mathrm{~m}^{-3}$

Solution

Given that, $\left(\frac{1}{3}\right) \operatorname{rd}$ part of body is above the water surface.
$\therefore$ Volume of body outside the water $=\frac{1}{3}$(volume of body) $V_o=\frac{1}{3} V$ $\therefore$ Volume of body inside, $V_i=V-V_0=V-\frac{V}{3}=\frac{2 V}{3}$ Let mass of body $=M$ Density of body $=\sigma$ Density of water, $\rho=10^3 \mathrm{~kg} \mathrm{~m}^{-3}$ According to the law of floatation, body will float into the liquid (water) when weight of body is balanced by Buoyant force. $\therefore \quad W=F_B$ $\Rightarrow \quad M g=V_i \rho g$ $\begin{gathered}V \sigma g=V_i \rho g \\ V \sigma=V_i \rho \\ \sigma=\frac{V_i \rho}{V}\end{gathered}$ Substituting the values, we get $\begin{aligned} \sigma & =\frac{2 V}{3 V} \rho \\ & =\frac{2}{3} \times 10^3=\frac{2000}{3} \mathrm{~kg} \mathrm{~m}^{-3}\end{aligned}$ Hence, density of liquid, $\sigma=\frac{2000}{3} \mathrm{~kg} \mathrm{~m}^{-3}$

Asked in: AP EAMCET 2021 (24 Aug Shift 2)

Practice more Mechanical Properties of Fluids questions on Aicharya