A long mercury glass tube with a uniform capillary bore has in it a thread of mercury which is 1 m long at $0^{°}C$. What will be its length (in m) at ${100}^{°}C$ if the real coefficient of expansion of mercury is $0.000182{/}^{°}C$ and coefficient of cubical expansion of glass equal to $0.000025{/}^{°}C$. (round off to nearest integer)

Suppose $V_0$ and $V_t$ are the volumes of mercury thread at $0^{°}C$ and $t^{°}C$ respectively, then

 $V_t=V_0\left(1+{\gamma }_{r}t\right)$

Let $a_0$ and $a_t$ are the areas of cross-section of the tube at $0^{°}C$ and $t^{°}C$ respectively, then

 $a_t=a_0\left(1+\beta t\right)$

 $=a_0\left(1+2\alpha t\right)$                                               $\left\{\beta =2\alpha \right\}$

The length of the thread at $0^{°}C$,

 ${\mathcal{l}}_0=\frac{V_0}{a_0}$

The length of thread at $t^{°}C$,

 ${\mathcal{l}}_t=\frac{V_t}{a_t}$

 $=\frac{V_0\left(1+{\gamma }_{r}t\right)}{a_0\left(1+2\alpha t\right)}$

 $={\mathcal{l}}_0\left(1+{\gamma }_{r}t\right)\left(1-2\alpha t\right)$

Substituting the given values, we get

 ${\mathcal{l}}_t=1\left(1+0.000182\times 100\right)\left(1-2\times \frac{0.000025}{3}\times 100\right)$

 $\begin{array}{l}=1.0182\times 0.9983\\ =1.016m\end{array}$