A rail track made of steel having length 10 m is clamped on a railway line at its two ends as shown in figure. On a summer day due to rise in temperature by `20^(@)C`, it is deformed as shown in figure. Find x (displacement of the centre) if `alpha_("steel")=1.2xx10^(-5)" "^(@)C^(-1)`

Explanation

By using Pythagoras theorem, for given right angle triangle,
`((L+DeltaL)/(2))^(2)=(L/2)^(2)+x^(2)`
`x=sqrt(((L+DeltaL)/(2))^(2)-(L/2)^(2))`
`=1/2(sqrt((L+DeltaL)^(2)-L^(2)))`

`=1/2[sqrt(L^(2)+2LDeltaL+DeltaL^(2)-L^(2))]`
`=1/2[sqrt(2LDeltaL+DeltaL^(2))]`
As `DeltaL` is very small, `DeltaL^(2)` is neglected
`x=1/2sqrt(2LDeltaL)`
But `DeltaL=LalphaDeltaT`
`:.x=1/2sqrt(2LxxLalphaDeltaT)`
`=1/2Lsqrt(2alphaDeltaT)`
`:.x=10/2xxsqrt(2xx1.2xx10^(-5)xx20)`
`=5xxsqrt(4.8xx10^(-4))`
`=5xx2xx1.1xx10^(-2)`
`=11xx10^(-2)`
`=0.11m=11cm`
Here, the increase in length of mercury is neglected as it is very small.