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)

800

Answer 2

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$^{-3}$