Question

The coefficient of volumetric expansion of mercury is $8\times {10}^{-5}{}^{\circ }C$. A thermometer bulb has a volume ${10}^{-6}{m}^3$ and cross section of stem is $0.004c{m}^2$. Assuming that bulb is filled with mercury at $0^{\circ }C$, then the length of the mercury column at ${100}^{\circ }C$ is:

• A. 18.8 mm
• B. 9.2 mm
• C. 7.4 cm
• D. 4.5 cm

Answer 1

The correct option is D

4.5 cm

The volumetric expansion can be written as,
$V_2=V_1(1+\gamma \Delta \theta )$
where, $V_2$ is the final volume, $V_1$ is the initial volume, $\gamma$ is the coefficient of volumetric expansion and $\Delta \theta$ is the change in temperature.
or, $\text{Change in volume},V_2-V_1=V_1\gamma \Delta \theta$
As we know, $\text{Change in volume}=\text{Area}\times \text{Change in length}$
or, $V_2-V_1=\text{Area}\times \text{Change in length}$
or, $\text{Change in length}=\frac{V_2-V_1}{\text{Area}}$
or, $\text{Change in length}=\frac{V_1\gamma \Delta \theta }{A}$
Given, $V_1={10}^{-6}{m}^3$, $A=0.004c{m}^2=0.004\times {10}^{-4}{m}^2$, $\gamma =8\times {10}^{-5}{}^{\circ }C$,
$\Delta \theta =100-0={100}^{\circ }C$
Therefore,
$\text{Change in length}=\frac{{10}^{-6}\times 8\times {10}^{-5}\times 100}{0.004\times {10}^{-4}}$
or, $\text{Change in length}=2\times {10}^{-2}m=2cm$