Question

A block of wood floats in water with ${\left(\frac{4}{5}\right)}^{th}$of its volume submerged. If the same block just floats in a liquid, the density of the liquid (in Kg m${}^{-3}$) is

• A. 1250
• B. 600
• C. 400
• D. 800

Answer 1

Answer: (D)

According to Archimedes’ principle Weight of the body = Weight of the liquid displaced
Let V be volume of the block.
In water
$V_{{\rho }_{block}}g=\left(\frac{4}{5}V\right){\rho }_{water}g$
${\rho }_{block}=\frac{4}{5}{\rho }_{water}$ ...(i)
In liquid, $V_{{\rho }_{block}}g=V_{{\rho }_{liquid}}g$
${\rho }_{block}={\rho }_{liquid}$ ...(ii)
From (i) and (ii), we get
${\rho }_{liquid}=\frac{4}{5}{\rho }_{water}=\frac{4}{5}\times 10^{3}Kg{m}^{-3}$
= 800 Kg m${}^{-3}$