Question

A leak proof cylinder of length $1m$, made of metal which has very low coefficient of expansion is floating in water at $0^{\circ }C$ such that its height above the water surface is $20cm$. When the temperature of water is increases to $4^{\circ }C$, the height of the cylinder above the water surface becomes $21cm$. The density of water at $T=4^{\circ }C$ relative to the density at $T=0^{\circ }C$ is close to

• A. $1.01$
• B. $1.03$
• C. $1.26$
• D. $1.04$

Answer 1

The correct option is A

$1.01$

Step 1: Given data and drawing the diagram of situation

Length of cylinder, $L=1m$

$=100cm\left[As,1m=100cm\right]$

Height of cylinder above the water surface at $0^{\circ }C$, $l_1=20cm$

Height of cylinder above the water surface at $4^{\circ }C$, $l_2=21cm$

Let $V_1$ be the volume of the cylinder which is immersed into the water initially.

Let $V_2$ be the volume of the cylinder which is immersed into the water finally.

Step 2: Find the length of the cylinder under the water

Length of cylinder under the water surface at $0^{\circ }C$, ${\left(l\text{'}\right)}_1=\left(100-20\right)cm$

$=80cm$

Length of cylinder under the water surface at $0^{\circ }C$, ${\left(l\text{'}\right)}_2=\left(100-21\right)cm$

$=79cm$

Step 3: Find the ratio of the density of water at ${\mathbf{4}}^{\mathbf{\circ }}\mathbf{}\mathit{C}\mathbf{}\mathbf{\left(}{\mathit{\rho }}_{{\mathbf{0}}^{\mathbf{\circ }}\mathbf{C}}\mathbf{\right)}$ to the ratio of the density of water at ${\mathbf{0}}^{\mathbf{\circ }}\mathbf{}\mathit{C}\mathbf{}\mathbf{\left(}{\mathit{\rho }}_{{\mathbf{4}}^{\mathbf{\circ }}\mathbf{C}}\mathbf{\right)}$

To find this write equation for the equilibrium first

Weight of the cylinder is balanced by the Buoyant force applied by the water

Equation of the equilibrium for initial condition

$$\begin{gathered} mg=\left({\rho }_{0^{\circ }\mathrm{C}}\right)g{V}_1\left[As\rho gVisbuoyantforce\right] \\ mg=\left({\rho }_{0^{\circ }\mathrm{C}}\right)gA{\left(l^{\text{'}}\right)}_1 \\ mg=\left({\rho }_{0^{\circ }\mathrm{C}}\right)gA\times 80......................\left(1\right) \end{gathered}$$

Again, for the final condition

$$\begin{gathered} mg=\left({\rho }_{4^{\circ }\mathrm{C}}\right)g{V}_2 \\ mg=\left({\rho }_{4^{\circ }\mathrm{C}}\right)gA{\left(l^{\text{'}}\right)}_2 \\ mg=\left({\rho }_{4^{\circ }\mathrm{C}}\right)gA\times 79......................\left(2\right) \end{gathered}$$

On dividing equation $\left(1\right)$ and $\left(2\right)$, we will get

$$\begin{gathered} \frac{{\rho }_{0^{\circ }\mathrm{C}}}{{\rho }_{4^{\circ }\mathrm{C}}}=\frac{80}{79} \\ \frac{{\rho }_{0^{\circ }\mathrm{C}}}{{\rho }_{4^{\circ }\mathrm{C}}}=1.01 \end{gathered}$$

Hence, option (A) is correct.