Two different wires having lengths $L _{1}$ and $L _{2}$ and respective temperature coefficient of linear expansion $\alpha_{1}$ and $\alpha_{2},$ are joined end-to-end. Then the effective temperature coefficient of linear expansion is
$4 \frac{\alpha_{1} \alpha_{2}}{\alpha_{1}+\alpha_{2}} \frac{ L _{2} L _{1}}{\left( L _{2}+ L _{1}\right)^{2}}$
$2 \sqrt{\alpha_{1} \alpha_{2}}$
$\frac{\alpha_{1}+\alpha_{2}}{2}$
$\frac{\alpha_{1} L_{1}+\alpha_{2} L_{2}}{L_{1}+L_{2}}$
A steel rod with ${y}=2.0 \times 10^{11} \,{Nm}^{-2}$ and $\alpha=10^{-5}{ }^{\circ} {C}^{-1}$ of length $4\, {m}$ and area of cross-section $10\, {cm}^{2}$ is heated from $0^{\circ} {C}$ to $400^{\circ} {C}$ without being allowed to extend. The tension produced in the rod is ${x} \times 10^{5} \, {N}$ where the value of ${x}$ is ....... .
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 difference between their lengths remain constant. If ${\alpha _{iron}}$ $= 1.2 \times 10^{-5}\,K^{-1}$ and ${\alpha _{brass}}$ $= 1.8 \times 10^{-5}\,K^{-1}$ what should we take as length of each strip ?
Two rods of different materials having coefficient of linear expansion $\alpha_1$and $\alpha_2$ and Young's modulii $Y_1$ and $Y_2$ respectively are fixed between two rigid massive walls. The rods are heated such that they undergo the same increase in temperature. There is no bending of rods. If $\alpha_1:\alpha_2= 2 : 3$, the thermal stress developed in two rods are equal provided $Y_1 : Y_2$ is equal to
If the volume of a block of metal changes by $0.12 \%$ when it is heated thrugh $20^oC$, the coefficient of linear expansion (in $^oC^{-1}$) of the metal is
The coefficient of apparent expansion of mercury in a glass vessel is $153 \times 10^{-6} /{ }^{\circ} C$ and in steel vessel is $144 \times 10^{-6} /{ }^{\circ} C$. If $\alpha$ for steel is $12 \times 10^{-6} /{ }^{\circ} C ,$ then that of glass is