Question

A glass cylinder contains $m_0=100g$ of mercury at a temperature of $t_0=0^{o}C$. When temperature becomes $t_1={20}^{o}C$ the cylinder contains $m_1=99.7g$ of mercury. The coefficent of volume expension of mercury ${\lambda }_{He}=18times{10}^{-5}{/}^{o}C$. Assume that the temperature of the mercury is equal to that of the cylinder. The coefficent of linear expansion of glass $\alpha$ is

• A. ${10}^{-5}{/}^{o}C$
• B. $2\times {10}^{-5}{/}^{o}C$
• C. $3\times {10}^{-5}{/}^{o}C$
• D. $6\times {10}^{-5}{/}^{o}C$

Answer 1

The correct option is A ${10}^{-5}{/}^{o}C$

When the cylinder is heated, its volume increases as

$V_t=V_0(1+{\gamma }_g^{\prime }t)$ ...............(1)

If the densities of mercury at the temperatures $t_0$ and $t_1$ are denoted by ${\rho }_0$ and ${\rho }_1$

$m_0=V_0{\rho }_0$ and $m_1=V_1{\rho }_1$ ...........(ii)

$m_t=m_0$

${\rho }_t=\frac{{\rho }_0}{(1+\gamma hgt)}$ ............(iii)

${\gamma }_{glass}=\frac{m_t=(1+\gamma Hg{t}_t)-m_0}{m_0{t}_0}$

$=\frac{99.9(1+18\times {10}^{-5}\times 20)-100}{100\times 20}=2.946\times {10}^{-5}{/}^{0}C$

The coefficient of linear expansion of glass

${\alpha }_{glass}=\frac{{\gamma }_{glass}}{3}={10}^{-5}{/}^{0}C$