Question

A body floats with $\frac{1}{3}$ of its volume outside water and $\frac{3}{4}$ of its volume outside another liquid. The density of the other liquid is

• A. $\frac{9}{4}\times {10}^{3}kg/m^3$
• B. $\frac{4}{9}\times {10}^{3}kg/m^3$
• C. $\frac{8}{3}\times {10}^{3}kg/m^3$
• D. $\frac{3}{4}\times {10}^{3}kg/m^3$

Answer 1

The correct option is C $\frac{8}{3}\times {10}^{3}kg/m^3$

Let D = density of the body, d = density of water
$VDg{V}^{\prime }dg$
$\downarrow \downarrow$
Weight of upward
the object thrust

For water
$VDg=\frac{2}{3}V\times {10}^3\times g$
$D=\frac{2}{3}\times {10}^{3}kg/m^3$
When object is dipped in liquid
$V\times \frac{2}{3}\times {10}^3\times g=\frac{V}{4}\times d\times g$
$d=\frac{8}{3}\times {10}^{3}kg/m^3$