Question

A cube floats in water with 1/3rd parts it outside the surface of water and it floats in liquid with 3/sth part is outside the liquid then the density of liquid is

• A. 8/3
• B. 2/3
• C. 4/3
• D. 5/3

Answer 1

Answer: (D)

5/3

Archimedes principle states that the upward buoyant force that is exerted on a body immersed in a fluid, whether fully or partially submerged is equal to the weight of the fluid that the body displaces and acts in the upward direction at the centre of mass of the displaced fluid.

So, for an object to float

Weight of object$=F_B$ (Buoyant force)

${\rho }_0-$ density of an object

$V_0-$ volume of object

${\rho }_l-$ density of liquid

$V_{dis}-$ Volume of disp. liquid

$V_{above}-$ Volume above the liquid

${\rho }_{\omega }-$ density of water

$\Rightarrow {\rho }_0{V}_{0}g={\rho }_l{v}_{dis}g\Rightarrow {\rho }_l=\frac{{\rho }_0{V}_0}{V_{dis}}{\rho }_l=\frac{{\rho }_0{V}_0}{V_0-V_{above}}\to \left(i\right)$

A/Q cube in water

$V_{above}=1/3V_{)}$

From $\left(i\right)$

${\rho }_{\omega }={\rho }_0\frac{V_0}{V_0-\frac{1}{3}{V}_0}\Rightarrow {\rho }_{\omega }=\frac{3}{2}{\rho }_0\to \left(ii\right)$

Cube in liquid

$V_{above}=3/5V_0$

From $\left(i\right)$

$\Rightarrow {\rho }_l={\rho }_0\frac{V_0}{V_0-3/5V_0}\Rightarrow {\rho }_l=\frac{5}{2}{\rho }_0\to \left(iii\right)$

From $\left(ii\right)\&\left(iii\right)$

${\rho }_l=\frac{5}{2}\times \frac{2}{3}{\rho }_{\omega }\Rightarrow {\rho }_l=\frac{5}{3}{\rho }_{\omega }$

As

${\rho }_{\omega }=1$ So,

${\rho }_l=5/3$