Question

A leak proof cylinder of length $1\,{\text{m}}$ , made of a metal which has low coefficient of expansion is floating vertically in water at $0^\circ {\text{C}}$ such that its height above the water surface is $20\,{\text{cm}}$ . When the temperature of water is increased to $4^\circ {\text{C}}$ , the height of the cylinder above the water surface becomes $21\,{\text{cm}}$ . The density of water at $T = 4^\circ {\text{C}}$ , relative to the density at $T = 0^\circ {\text{C}}$ is close to:
(A) $1.26$
(B) $1.03$
(C) $1.04$
(D) $1.01$

Answer 1

First of all, we will calculate the length of the immersed part for both the cases. Then will calculate the buoyant force for the two cases separately and will manipulate accordingly to obtain the result.

Complete step by step answer:
In the given question, we are supplied with the following details:
The length of the cylinder is $1\,{\text{m}}$ .
Initial temperature of the water is $0^\circ {\text{C}}$ .
Final temperature of the water is $4^\circ {\text{C}}$ .
The cylinder is floating vertically in the water initially where the length of $20\,{\text{cm}}$ above the water.
But when the temperature of the water increased to $4^\circ {\text{C}}$ then the length of the tank which is above the water is $21\,{\text{cm}}$ .
We are asked to find the relative density of water at $T = 4^\circ {\text{C}}$ , relative to the density at $T = 0^\circ {\text{C}}$ .
To begin with, we will first find the length of the cylinder which is below the water at the two different temperatures.
Since, the total length of the cylinder is $100\,{\text{cm}}$ .
So, the length of the cylinder below the water at $0^\circ {\text{C}}$ is:
$\Rightarrow\left( {100 - 20} \right)\,{\text{cm}} \\\ \Rightarrow 80\,{\text{cm}} \\\ \Rightarrow 0.8\,{\text{m}} \\\$
The length of the cylinder below the water at $4^\circ {\text{C}}$ is:
$\Rightarrow\left( {100 - 21} \right)\,{\text{cm}} \\\ \Rightarrow 79\,{\text{cm}} \\\ \Rightarrow 0.79\,{\text{m}} \\\$
We know that the weight (downward force) of the immersed part of the cylinder is balanced by the buoyant force (upward thrust) of the water.
So, we will calculate the magnitude of buoyant force at the two temperatures separately:
$\Rightarrow w = {\rho _0}A \times {l_0} \times g$ …… (1)
Where,
$w$ indicates the magnitude of the buoyant force.
${\rho _0}$ indicates density of water at $0^\circ {\text{C}}$ .
$A$ indicates the cross-sectional area.
${l_0}$ indicates the length of the immersed part of the cylinder at $0^\circ {\text{C}}$ .
$g$ indicates the acceleration due to gravity.
Substituting the required values in the equation (1), we get:
$\Rightarrow w = {\rho _0}A \times {l_0} \times g$
$\Rightarrow w = {\rho _0}A \times 0.80 \times g$ …… (2)
Now, we calculate the buoyant force at temperature $4^\circ {\text{C}}$ :
$\Rightarrow w = {\rho _4}A \times {l_4} \times g$
$\Rightarrow w = {\rho _4}A \times 0.79 \times g$ …… (3)
Now, we divide equation (3) by equation (2):
$\Rightarrow\dfrac{w}{w} = \dfrac{{{\rho _4}A \times 0.79 \times g}}{{{\rho _0}A \times 0.80 \times g}} \\\ \Rightarrow\dfrac{{{\rho _4}}}{{{\rho _0}}} = \dfrac{{0.80}}{{0.79}} \\\ \Rightarrow\dfrac{{{\rho _4}}}{{{\rho _0}}} = 1.012 \\\ \Rightarrow\dfrac{{{\rho _4}}}{{{\rho _0}}} \sim 1.01 \\\$
Hence, the relative density of water at $T = 4^\circ {\text{C}}$ , relative to the density at $T = 0^\circ {\text{C}}$ is $1.01$ .
The correct option is D.

Note: It is important to note that the buoyant force is always calculated by taking the length of the immersed part of the body, taken into account. The part above the liquid has no significance in calculating the buoyant force. The upward thrust will remain the same for the two temperatures as the body is the same for both the cases.