A rail track made of steel having length `10 m` is clamped on a railway line at its two end (figure). On a summer day due to rise in temperatue by `20^(@)C`. It is deformed as shown in figure. Find `x` (displacement of the centre) if `alpha_("steel") = 1.2 xx 10^(-5)//^(@)C`

Options

A. `15 sqrt(30) xx 10^(-3) cm`
B. `10 sqrt(30) xx 10^(-3) cm`
C. `25 sqrt(30) xx 10^(-3) cm`
D. `20 sqrt(30) xx 10^(-3) cm`

Answer

D

Explanation

Consider the diagram

Applying Pythagoras therom in right angled triangle in figure
`((L+Delta L)/(2))^(2) = (L/2)^(2) +x^(2)`
`rArr x = sqrt(((L+Delta L)/(2))^(2) - (L/2)^(2)) = 1/2 sqrt((L+DeltaL)^(2)-L^(2))`
`rArr x = 1/2 sqrt((L^(2)+Delta L^(2)+2LDeltaL)-L^(2))`
`=1/2sqrt((DeltaL^(2)+2LDeltaL))`
As increase in length `Delta L` is very small, therefore, neglecting `(DeltaL)^(2)`, we get
`x=1/2xxsqrt(2 L Delta L)` ...(i)
But `Delta L = L alpha Delta t` ...(ii)
Substituing value of `DeltaL` in eq. (i) form eq. (ii)
`x = 1/2 sqrt(2LxxLalphaDelta t) = 1/2 L sqrt(2alpha Delta t)`
`=10/2 xx sqrt(2 xx 1.2xx10^(-5)xx20)`
`=5 xx sqrt(4xx1.2xx10^(-4)) = 20sqrt(30)xx10^(-3)cm`.