A rail track made of steel having length 10 m is clamped on a railway line at its two ends

on a summer day due to rise in temperature by `20^@C,` it is deformed as shown in fig. Find x (displacement of the centre) if `alpha_(steel) = 1.2 xx 10^(-5)//^@C.`

Explanation

Consider the diagram,

Applying Pythagorus theorem in right angled triangle in figure
`((L+Delta L)/(2))^(2) = ((L)/(2))^(2) + x^(2)`
`rArr " " x = sqrt(((L+DeltaL)/(2))^(2)-((L)/(2))^(2))`
`= (1)/(2) sqrt((L+Delta L)^(2)-L^(2))`
`= (1)/(2) sqrt((L^(2)-Delta L^(2)+2LDelta L)-L^(2))`
`= (1)/(2) sqrt((Delta L^(2)+2LDelta L))`
As increase in length `Delta L` is very small, therefore, neglecting `(Delta L)^(2)`, we get
`x = (1)/(2) xx sqrt(2L Delta L)` ...(i)
But `Delta L = L alpha Delta t` ...(ii)
Substituting value of `Delta L` in Eq. (i) from Eq. (ii)
`x = (1)/(2)sqrt(2L xx L alpha Delta t) = (1)/(2)L sqrt(2alpha Delta t)`
`= (10)/(2) xx sqrt(2xx 1.2 xx 10^(-5) xx 20)`
`= 5 xx sqrt(4 xx 1.2 xx 10^(-4))`
`= 5 xx 2 xx 1.1 xx 10^(-2) =0.11 m = 11 cm`
Note Here we have assumed `Delta L` to be very small so that it can neglected compared to L.