Question

A body $P$ floats in water with half its volume immersed. Another body $Q$ floats in a liquid of density $\frac{3}{4}th$ of the density of water with two-third of the volume immersed. The ratio of density of $P$ to that of $Q$ is

• A. $1:2$
• B. $1:1$
• C. $2:3$
• D. $3:4$
• E. $5:1$

Answer 1

Answer: (B)

Let, for body $P$ , volume $=V_{P}$
Given, Immersed volume $=\frac{V_{P}}{2}$
For body $Q$ volume $=V_{Q}$
Immersed volume $=\frac{2}{3}V_{Q}$
According to Archimedes principle,
$\because $ Weight of body $= \, $ weight of fluid displaced
For body $P,$
$ V_{p} \rho_{p} g=\left(\frac{1}{2} V_{p}\right) \rho_{\omega g} $
So, $\frac{\rho_{P}}{\rho_{w}}=\frac{1}{2} \ldots( i )$
For body $Q$,
$ V_{Q} \rho_{Q} g=\left(\frac{2}{3} V_{Q}\right)\left(\frac{3}{4}\right) \rho_{w} g $
$\left(\because\right.$ density of liquid $\left.=\frac{3}{4} \rho_{w}\right)$ So, $\frac{\rho_{Q}}{\rho_{w}}=\frac{1}{2} \ldots$ (ii)
From Equations (i) and (ii),
$ \frac{\rho_{P}}{\rho_{Q}}=1 \text { or } \rho_{P}: \rho_{Q}=1: 1 $