A wooden block floating in a bucket of water has $(4 / 5)$ th of its volume submerged. When certain amount…
A wooden block floating in a bucket of water has $(4 / 5)$ th of its volume submerged. When certain amount of an oil is poured into the bucket, it is found that the block is just under the oil surface, with half of its volume under water and the other half in oil. The density of oil used relative to that of water is
$\frac{1}{4}$
$\frac{3}{5}$
$\frac{2}{5}$
$\frac{5}{3}$
Solution
Given that, initially body is submerged into the water by $\left(\frac{4}{5}\right)$ th volume of body.
i.e. Volume inside the water, $V_i=\frac{4}{5} \mathrm{~V}$
where, $V$ be total volume of body.
According to law of floatation,
Weight of body = Buoyant force
$M g=V_i \rho g$
$V \sigma g=V_i \rho g$
$\Rightarrow \quad V \sigma=V_i \rho$
$\Rightarrow \quad \sigma=\frac{V_i}{V} \rho=\frac{4}{5} \rho$ ...(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,
$\begin{aligned} M g & =F_{\text {water }}+F_{\text {oil }} \\ V \sigma g & =V_w \rho g+V_{\text {oil }} \rho_o g \\ V \sigma & =\frac{V}{2} \rho+\frac{V}{2} \rho_o\end{aligned}$
$V \times \frac{4}{5} \rho=\frac{V \rho}{2}+\frac{V \rho_o}{2}$[From Eq. (i)]
$\Rightarrow \quad \rho_o=\left(\frac{4}{5}-\frac{1}{2}\right) \rho \times 2=\frac{3}{5} \rho$
Hence, density of oil w.r.t. water is $\frac{3}{5}$.