Question

A cylinder is made up of a material of relative density $2$ with a height of $5{\text{cm}}$ and area of cross-section $5{\text{c}}{{\text{m}}^2}$. On immersing it in a liquid, it loses half of its weight. Find the density of the liquid.

Answer 1

Hint: When the cylinder is immersed in the liquid, it displaces the liquid and loses weight. This loss in weight is due to the buoyant force acting on it. The buoyant force will be equal to the weight of the liquid that was displaced during the immersion of the cylinder.

Formulas used:
The mass of the cylinder is given by, $m = \rho \times V$ where $\rho $ is the density of the cylinder and $V$ is the volume of the cylinder.
The volume of a cylinder is given by, $V = A \times h$ where $A$ is the area of the cylinder and $h$ is the height of the cylinder.
The force of buoyancy is given by, ${F_b} = {\rho _l} \times V \times g$ where ${\rho _l}$ is the density of the liquid, $V$ is the volume of the liquid that got displaced and $g$ is the acceleration due to gravity.
The weight of the cylinder is given by, $W = mg$ where $m$ is the mass of the cylinder and $g$ is the acceleration due to gravity.

Complete step by step answer:
The area of cross-section of the cylinder is $A = 5{\text{c}}{{\text{m}}^2}$ and its height is $h = 5{\text{cm}}$.
Then the volume of the cylinder will be
$\Rightarrow V = A \times h = 5 \times 5 = 25{\text{c}}{{\text{m}}^3}$.
The relative density of the cylinder is given to be $R = 2$ and the density of water is known to be ${\rho _w} = 1{\text{gc}}{{\text{m}}^{ - 3}}$.
Then the density of the cylinder will be
$\Rightarrow {\rho _c} = 2{\text{gc}}{{\text{m}}^{ - 3}}$.

The weight of the cylinder is given by, $W = mg$ -------- (1)
where $m$ is the mass of the cylinder and $g$ is the acceleration due to gravity.
Now the mass of the cylinder will be $m = {\rho _c} \times V$ and on substituting for $V = 25{\text{c}}{{\text{m}}^3}$ and ${\rho _c} = 2{\text{gc}}{{\text{m}}^{ - 3}}$ we get the mass as
$\Rightarrow m = 2 \times 25 = 50{\text{g}}$
Substituting the value for $m = 50{\text{g}}$ in equation (1) we get, $W = 50 \times g$
It is mentioned that on immersion in the liquid, half of the weight of the cylinder gets lost. This loss in weight equals the force of buoyancy ${F_b}$.
$ \Rightarrow {F_b} = \dfrac{W}{2} = \dfrac{{50 \times g}}{2} = 25 \times g$
Thus the buoyant force is obtained to be ${F_b} = 25 \times g$.

The buoyant force acting on the cylinder is equal to the weight of the liquid displaced.
i.e.,
$\Rightarrow {F_b} = {\rho _l} \times V \times g$ -------- (2)
where ${\rho _l}$ is the density of the liquid, $V$ is the volume of the liquid that got displaced and $g$ is the acceleration due to gravity.
Substituting for $V = 25{\text{c}}{{\text{m}}^3}$ and ${F_b} = 25 \times g$ in equation (2) we get,
$\Rightarrow 25 \times g = {\rho _l} \times 25 \times g$
$ \Rightarrow {\rho _l} = \dfrac{{25 \times g}}{{25 \times g}} = 1{\text{gc}}{{\text{m}}^{ - 3}}$

Thus the density of the liquid is ${\rho _l} = 1{\text{gc}}{{\text{m}}^{ - 3}}$.

Note:
The relative density of the cylinder $R = 2$ refers to the density of the cylinder with respect to that of reference material and this reference material is usually water i.e.,
$R = \dfrac{{{\rho _c}}}{{{\rho _w}}}$.
$ \Rightarrow {\rho _c} = R \times {\rho _w} = 2 \times 1 = 2{\text{gc}}{{\text{m}}^{ - 3}}$
When the cylinder is immersed in the liquid, the volume of the liquid displaced will be equal to the volume of the cylinder.