91Ó°ÊÓ

coefficient of thermal expansion is α = 1.1 x 10^-5 K^-1and

length of the steel bridge is L=1 km =1 x 10³ m.

Coefficient of thermal expansion of solid is given as

$\mathrm{Î\pm }=\frac{\left(\frac{\mathrm{Δ}L}{L}\right)}{\mathrm{Δ}T}$

So we can say change in length is given as

$\mathrm{Δ}L=\mathrm{Î\pm }×L×\mathrm{Δ}T......................\left(1\right)$

Lets assume the difference between cold winter temperature and hot day temperature is 40K

Substitute the values in the equation (1) we get

$$\begin{gathered} \mathrm{Δ}L=\mathrm{Î\pm }×L×\mathrm{Δ}T \\ \mathrm{Δ}L=(1.1×{10}^{-5}K{-1}^{})×(1×{10}^{3}m)×40K \\ \mathrm{Δ}L=0.44\mathrm{m} \end{gathered}$$

So the change in length is 0.44 m .

A dial thermometer with two metal strips with different value of α

A typical dial thermometer consists of two metal strip coils together with different values of α.

Metal with different value of α will expand differently with change of temperature.

So the coil will make a radial change by changing the temperature.

It is easier to notice the change and hence easier to measure the temperature.

The relationship is given

$\mathrm{β}={\mathrm{Î\pm }}_x+{\mathrm{Î\pm }}_y+{\mathrm{Î\pm }}_z$

Prove the relationship

For a non-isotropic solid, they will have different α values, i.e.,α_x , α_y , α_z

Which can be defined as, which are Coefficients of Linear expansion in all three directions are

$$\begin{gathered} {\mathrm{Î\pm }}_x=\frac{\mathrm{Δ}x}{x\mathrm{Δ}T}, \\ {\mathrm{Î\pm }}_y=\frac{\mathrm{Δ}y}{y\mathrm{Δ}T}, \\ {\mathrm{Î\pm }}_z=\frac{\mathrm{Δ}z}{z\mathrm{Δ}T} \end{gathered}$$

Where x,y and z is dimension of solid cube and Δx, Δy and Δz are changes in x, y and z respectively.

Coefficients of volume expansion is given by

$\mathrm{β}=\frac{\mathrm{Δ}V}{V\mathrm{Δ}T}$

Volume of rectangular solid is

V = xyz .......................................(1)

Differentiate this equation

$$\begin{gathered} \mathrm{Δ}V=yz\mathrm{Δ}x+x\mathrm{Δ}\left(yz\right) \\ \mathrm{Δ}V=yz\mathrm{Δ}x+xz\mathrm{Δ}y+xz\mathrm{Δ}y...........................\left(2\right) \end{gathered}$$

Divide equation (2) by V=x y z on both the side, we get

$\frac{\mathrm{Δ}V}{V}=\frac{\mathrm{Δ}x}{x}+\frac{\mathrm{Δ}y}{y}+\frac{\mathrm{Δ}z}{z}..............................\left(3\right)$

We know ${\mathrm{Î\pm }}_x\mathrm{Δ}T=\frac{\mathrm{Δ}x}{x},{\mathrm{Î\pm }}_y\mathrm{Δ}T=\frac{\mathrm{Δ}y}{y},{\mathrm{Î\pm }}_z\mathrm{Δ}T=\frac{\mathrm{Δ}z}{z}$

Substitute values in equation (3) ${\mathrm{Î\pm }}_x\mathrm{Δ}T=\frac{\mathrm{Δ}x}{x},{\mathrm{Î\pm }}_y\mathrm{Δ}T=\frac{\mathrm{Δ}y}{y}and{\mathrm{Î\pm }}_z\mathrm{Δ}T=\frac{\mathrm{Δ}z}{z}$

$$\begin{gathered} \frac{\mathrm{Δ}V}{V}={\mathrm{Î\pm }}_x\mathrm{Δ}T+{\mathrm{Î\pm }}_y\mathrm{Δ}T+{\mathrm{Î\pm }}_z\mathrm{Δ}T \\ \frac{\mathrm{Δ}V}{V}=\left({\mathrm{Î\pm }}_x+{\mathrm{Î\pm }}_y+{\mathrm{Î\pm }}_z\right)\mathrm{Δ}T \\ \frac{\mathrm{Δ}V}{V\mathrm{Δ}T}=\left({\mathrm{Î\pm }}_x+{\mathrm{Î\pm }}_y+{\mathrm{Î\pm }}_z\right)...............................\left(4\right) \end{gathered}$$

We know

From equation (4) and (5) we can conclude that

$\mathrm{β}=\left({\mathrm{Î\pm }}_x+{\mathrm{Î\pm }}_y+{\mathrm{Î\pm }}_z\right)$