The coefficient of apparent expansion of mercury in a glass vessel is $153 \times 10^{-6} /{ }^{\circ} C$ and in a steel vessel is $144 \times 10^{-6} /{ }^{\circ} C$. If $\alpha$ for steel is $12 \times 10^{-6} /{ }^{\circ} C$, then the coefficient of linear expansion $\alpha$ for glass is:

The coefficient of area expansion of a solid is $2 \times 10^{-5} \, ^{\circ}C^{-1}$. Its coefficient of linear expansion is:

Two metal rods $A$ and $B$ each of length $50 \ cm$ and diameter $4.0 \ mm$ are joined together at temperature $30^{\circ} C$. What is the change in length of the combined rod at $230^{\circ} C$ (in $mm$)? [Given linear expansion coefficients of rods $A$ and $B$ are respectively, $2.0 \times 10^{-5} /^{\circ} C$ and $1.0 \times 10^{-5} /^{\circ} C$]

Two rods,one of aluminium and the other of steel,having initial lengths $L_1$ and $L_2$ are connected together to form a single rod of length $(L_1+L_2)$. The coefficients of linear expansion of aluminium and steel are $\alpha_1$ and $\alpha_2$ respectively. If the length of each rod increases by the same amount,when their temperatures are raised by $t^{\circ}C$,then the ratio $\frac{L_1}{L_1+L_2}$ will be

We would like to prepare a scale whose length does not change with temperature. It is proposed to prepare a unit scale of this type whose length remains,say $10 \ cm$. We can use a bimetallic strip made of brass and iron each of different length whose length (both components) would change in such a way that the difference between their lengths remains constant. If $\alpha_{\text{iron}} = 1.2 \times 10^{-5} \ K^{-1}$ and $\alpha_{\text{brass}} = 1.8 \times 10^{-5} \ K^{-1}$,what should be the length of each strip?

Railway tracks are made of steel segments separated by small gaps to allow for linear expansion. $A$ segment of the track is $10 \ m$ long when laid at a temperature of $17^{\circ} C$. The maximum temperature that can be reached is $45^{\circ} C$. The increase in the length of the segment of the railway track is $x \times 10^{-5} \ m$. The value of $x$ is (given $\alpha_{\text{steel}} = 1.2 \times 10^{-5} /{ }^{\circ} C$).