Question

The coefficient of volumetric expansion of mercury is $18 \times 10^{-5} /^{\circ}C .$ A thermometer bulb has value of $10^{-6} m ^{3}$ and crossectional of stem is $0.002\, cm ^{2}$ assuming the bulb is tilled mercury at ${ }^{\circ} C$ the length of mercury at $100{ }^{\circ} C$ is:

• A. 18 cm
• B. 4.5 cm
• C. 2.25 cm
• D. 9 cm

Answer 1

Answer: (D)

The coefficient of volumetric expansion
$\gamma=18 \times 10^{-5} /{ }^{\circ} C$
Initial volume $V=10^{-6} m ^{3}$
Area of cross section $A=0.002\, cm ^{2}$
$=2 \times 10^{-7} m ^{2}$
Initial temperature $T_{1}=0^{\circ} C$
Final temperature $T_{2}=100^{\circ} C$
The final volume is $V=V\left[1+\left(T_{2}-T_{2}\right)\right]$
$=10^{-6}\left[1+18 \times 10^{-5}(100-0)\right]$
and $V=1.018 \times 10^{-6}$
Change in volume is $\Delta V=A \times \Delta l=V-V$
or $=2 \times 10^{-7} \times \Delta l$
$=1.018 \times 10^{-6}-10^{6}$
$2 \times 10^{-7} \times \Delta l=0.018 \times 10^{-6}$
Hence, $\Delta l=\frac{0.018 \times 10^{-6}}{2 \times 10^{-7}}$
$=0.09\, m =9\, cm$