Question

A leakproof cylinder of length $1\text{m}$, made of a metal which has very 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 $T=4^{\circ }C$, relative to the density at $T=0^{\circ }C$ is close to:

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

Answer 1

The correct option is C $1.01$


Here,
$h_1=20\text{cm}:h_2=100\text{cm}$

When cylinder is floating in water at $0^{\circ }C$:

The buoyant force is given by,

$F_b=A(h_2-h_1){\rho }_{0}g$

$\Rightarrow F_b=A(100-20){\rho }_{0}g$

$\Rightarrow F_b=A\left(80\right){\rho }_{0}g$


When cylinder is floating in water at $4^{\circ }C$:

The buoyant force is given by,

$F_b=A(h_2-h_1){\rho }_{4}g$

$\Rightarrow F_b=A(100-21){\rho }_{4}g$

$\Rightarrow F_b=A\left(79\right){\rho }_{4}g$

As the buoyant force has to counter the weight $\left(mg\right)$ in both the cases, the buoyant force is equal in both the cases.

$\Rightarrow A\left(79\right){\rho }_{4}g=A\left(80\right){\rho }_{0}g$

$\therefore \frac{{\rho }_4}{{\rho }_0}=\frac{80}{79}=1.01$

Hence, option $\left(C\right)$ is correct.