Question

A metallic bob of weight $100\text{gm}$ is suspended in air. If it is immersed in a liquid at temperature of ${30}^{\circ }\text{C}$, it weighs $90\text{gm}$. When the temperature of the liquid is raised to ${100}^{\circ }\text{C}$, it weighs $90.2\text{gm}$. Calculate the coefficient of cubical expansion of the liquid. Given that coefficient of cubical expansion of the metal is $15\times {10}^{-6}/{}^{\circ }\text{C}$

• A. $3\times {10}^{-5}{/}^{\circ }\text{C}$
• B. $4\times {10}^{-4}{/}^{\circ }\text{C}$
• C. $3\times {10}^{-4}{/}^{\circ }\text{C}$
• D. $4\times {10}^{-5}{/}^{\circ }\text{C}$

Answer 1

The correct option is C $3\times {10}^{-4}{/}^{\circ }\text{C}$

Let $V_0$ be the volume of the metallic bob at ${30}^{\circ }\text{C}$.
So, loss in weight of bob in liquid at ${30}^{\circ }\text{C}$
$\left(\Delta w\right)=(100\text{gm}-90\text{gm})\times {10}^{-3}\times g$
$=10\times {10}^{-3}g$
$={10}^{-2}g\text{N}[\because g$ = acceleration due to gravity]
Weight of displaced liquid $=V_{30}{\rho }_{30}g={10}^{-2}g\text{N}---\left(I\right)$
where ${\rho }_{30}$ = density of liquid at ${30}^{\circ }\text{C}$

Now, Loss in weight in liquid at ${100}^{\circ }\text{C}$
$\left(\Delta w^1\right)=(100-90.2)\times {10}^{-3}\times g$
$=9.8\times {10}^{-3}g\text{N}$
Similarly, weight of displaced liquid $=V_{100}{\rho }_{100}g$
So, $V_{100}{\rho }_{100}g=9.8\times {10}^{-3}g\text{N}---\left(II\right)$
From eq. ($I$) & ($II$),
$\frac{V_{30}{\rho }_{30}g}{V_{100}{\rho }_{100}g}=\frac{{10}^{-2}g}{9.8\times {10}^{-3}g}=\frac{10}{9.8}$
$\Rightarrow \frac{V_{30}{\rho }_{30}}{V_{100}{\rho }_{100}}=\frac{10}{9.8}---\left(III\right)$

Now as we know, for metallic bob,
$V_{100}=V_{30}(1+\gamma \Delta T)$
$=V_{30}[1+15\times {10}^{-6}\times (100-30\left)\right]$
$=V_{30}[1+15\times {10}^{-6}\times 70]$
$\Rightarrow V_{100}=V_{30}\times 1.00105$
$\Rightarrow \frac{V_{100}}{V_{30}}=1.00105$
Similarly for liquid,
$\frac{V_{100}}{V_{30}}=(1+{\gamma }_l\times 70)$
$[{\gamma }_l$ = coefficient of cubical expansion of liquid ]
$\Rightarrow \frac{{\rho }_{30}}{{\rho }_{100}}=(1+{\gamma }_l\times 70)$ [$\because \rho \propto \frac{1}{V}$]
Putting the value of $\frac{V_{100}}{V_{30}}\&\frac{{\rho }_{30}}{{\rho }_{100}}$ in eq. $\left(III\right)$,
$\Rightarrow \frac{1}{1.00105}\times (1+{\gamma }_l\times 70)=\frac{10}{9.8}$
$\Rightarrow {\gamma }_l=3.068\times {10}^{-4}{/}^{\circ }\text{C}$
or ${\gamma }_l\approx 3\times {10}^{-4}{/}^{\circ }\text{C}$