Question

A mercury-in-glass thermometer has a stem of internal diameter 0.06 cm and contains 43 g of mercury. The mercury thread expands by 10 cm when the temperature changes from $0^{\circ }C$ to ${50}^{\circ }C$. Find the coefficient of cubical expansion of mercury. Relative density of mercury = 13.6 and ${\alpha }_{glass}=9\times {10}^{-6}{K}^{-1}$

• A. ${10}^{-5}{K}^{-1}$
• B. $2\times {10}^{-5}{K}^{-1}$
• C. $10.7\times {10}^{-5}{K}^{-1}$
• D. $20.7\times {10}^{-5}{K}^{-1}$

Answer 1

The correct option is D $20.7\times {10}^{-5}{K}^{-1}$

Apparent expansion of mercury
$=\frac{\pi }{4}(6\times {10}^{-4}{)}^2\times 10\times {10}^{-2}=9\pi \times {10}^{-9}{m}^3$
$Volumeofmercury=\frac{43\times {10}^{-3}}{13.6\times 1000}$
$=3.16\times {10}^{-6}{m}^3$
$\therefore 9\pi \times {10}^{-9}=3.16\times {10}^{-6}\times {\gamma }_a\times 50$
$(\because \Delta V=V_{\gamma }\times \Delta T)or{\gamma }_a=18\times {10}^{-5}{\gamma }_r={\gamma }_a+{\gamma }_g={\gamma }_a+3{\alpha }_g(\because {\gamma }_g=3{\alpha }_g)\therefore {\gamma }_r=18\times {10}^{-5}+3\times 9\times {10}^{-6}=20.7\times {10}^{-5}{K}^{-1}$