Question

A cylinder of height $1\text{m}$ is floating in water at $0^{\circ }\text{C}$ with $20\text{cm}$ height in air. When the temperature of water is raised to $4^{\circ }\text{C}$, height of the cylinder in air becomes $21\text{cm}$. By assuming the expansion of the cylinder to be negligible and that the cylinder stays vertical, find the ratio of density of water at $4^{\circ }\text{C}$ to density of water at $0^{\circ }\text{C}$.

• A. 1.01
• B. 1.03
• C. 2.01
• D. 1.04

Answer 1

The correct option is A 1.01

Given, height of cylinder $\left(h\right)=1\text{m}=100\text{cm}$
From the data given in the question we can say that,
Length of cylinder submerged in water at $0^{\circ }\text{C}$ is $l_1=80\text{cm}$
Length of cylinder submerged in water at $4^{\circ }\text{C}$ is $l_2=79\text{cm}$
Pictorial representation of situation given in the question is shown below.


From Archimedes principle, we can say that
Buoyant force = weight of the fluid displaced.
i.e $F_b=V_f{\rho }_{f}g$
and for vertical equilibrium of cylinder,
weight $mg=V_s{\rho }_{s}g=F_b=V_f{\rho }_{f}g$
At $0^{\circ }\text{C}$:
$V_s{\rho }_{s}g=V_f{\rho }_{f}g$
$\Rightarrow \frac{{\rho }_s}{{\rho }_f}=\frac{V_f}{V_s}=\frac{\text{Height of cylinder immersed}}{\text{Total height of cylinder}}$
$\Rightarrow \frac{{\rho }_s}{{\rho }_f}=\frac{l_1}{h}=\frac{80}{100}....\left(1\right)$

Similarly, at $4^{\circ }\text{C}$:
$V_s{\rho }_{s}g=V_f^{\prime }{\rho }_f^{\prime }g$
where ${\rho }_f^{\prime }$ is the new density of fluid and $V_f^{\prime }$ is the new volume displaced
$\Rightarrow \frac{{\rho }_s}{{\rho }_f^{\prime }}=\frac{\text{Height of cylinder immersed}}{\text{Total height of cylinder}}$
$\Rightarrow \frac{{\rho }_s}{{\rho }_f^{\prime }}=\frac{l_2}{h}=\frac{79}{100}...........\left(2\right)$
So, from $\left(1\right)$ and $\left(2\right)$
$\frac{{\rho }_f^{\prime }}{{\rho }_f}=\frac{80}{79}=1.01$
Thus, option (a) is the correct answer.