Question

Calculate the buoyant force (in $N$) acting on a $250\text{g}$ rubik's cube of side length $8\text{cm}$, if only half of it is submerged in kerosene. The rubik's cube is hanging by a thread. Density of kerosene is $0.8\text{g}{\text{cm}}^{-3}$. Take $\text{g}=10\text{m}{s}^{-2}$.

• A. 2.048

Answer 1

The correct option is A 2.048
Given:
Density of kerosene, ${\rho }_k=0.8g{cm}^{-3}=800kg{m}^{-3}$
Cube's side length, $l=8cm=0.08m$
Cube's mass, $m=250g=0.25kg$

Volume of cube is:
$V=l^3$

Half of the cube is submerged. Volume submerged is equal to the volume of kerosene displaced. So, volume of kerosene displaced is:
$V_k=\frac{V}{2}$
$V_k=\frac{l^3}{2}$

Using definition of density, mass of kerosene displaced is:
$m_k={\rho }_k{V}_k$
$m_k={\rho }_k\frac{l^3}{2}$

Using definition of weight, weight of kerosene displaced is:
$W_k=m_{k}g$
$W_k={\rho }_k\frac{l^3}{2}g$

According to Archimedes principle, the weight of the kerosene displaced is equal to the buoyant force.
$F_b=W_k$
$F_b={\rho }_k\frac{l^3}{2}g$
$F_b=800\times \frac{{0.08}^3}{2}\times 10$
$F_b=2.048N$