Question

A vertical glass tube, closed at the bottom, contains a mercury column of length $L_0$ at $0^{\circ }\text{C}$. If $\gamma$ is the coefficient of cubical expansion of mercury $\alpha$ the coefficient of linear expansion of glass, the length of the mercury column when the temperature rises to $t^{\circ }\text{C}$ is (assuming that $t$ not more than ${100}^{\circ }\text{C}$)

• A. $L_t=L_0[1+(\gamma -3\alpha \left)t\right]$
• B. $L_t=L_0[1+(\gamma +3\alpha \left)t\right]$
• C. $L_t=L_0[1+(\gamma +2\alpha \left)t\right]$
• D. $L_t=L_0[1+(\gamma -2\alpha \left)t\right]$

Answer 1

The correct option is D $L_t=L_0[1+(\gamma -2\alpha \left)t\right]$

If $V_0$ is the volume of mercury at $0^{\circ }\text{C}$ and $A_0$ the cross-sectional area of the tube at $0^{\circ }\text{C}$, then the length of the mercury column at $0^{\circ }\text{C}$ is
$L_0=\frac{V_0}{A_0}\dots \dots \left(1\right)$
The cross-sectional are at $t^{\circ }\text{C}$ is given by
$A_t=A_0(1+\beta t)$
where $\beta$ is the coefficient of superficial expansion of glass. Since $\beta =2\alpha$, we have
$A_t=A_0(1+2\alpha t)$
If $V_t$ is the volume of mercury at $t^{\circ }\text{C}$, the length of the mercury column at $t^{\circ }\text{C}$ is
$L_t=\frac{V_t}{A_t}=\frac{V_0(1+\gamma t)}{A_0(1+2\alpha t)}$
Using (1) we have
$L_t=L_0(1+\gamma t)(1+2\alpha t{)}^{-1}$
Since $t$ is not too high $(\sim {100}^{\circ }\text{C})$ and $\gamma$ are of the order of ${10}^{-5}.\gamma t$ and $\alpha t$ will be very small compared to unity.
Hence we can expand $(1+2\alpha t{)}^{-1}$ binomially and retain terms upto order $\alpha t$. Thus
$(1+2\alpha t{)}^{-1}=(1-2\alpha t)$
$\therefore L_t=L_0(1+\gamma t)(1-2\alpha t)$
or $L_t=L_0[1+(\gamma -2\alpha \left)t\right]$