A rail track made of steel having length 10 m is clamped on a railway line at its two ends (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") = 1.2 xx 10^(-5)//^(@)C`.

Explanation

From figure, `X^(2)=((L+DeltaL)/(2))^(2)-((L)/(2))^(2)`
`rArrX=sqrt(((L+DeltaL)/(2))^(2)-((L)/(2))^(2)),sqrt(((L)/(2))^(2)+((DeltaL)/(2))^(2)-((L)/(2))^(2))`
Since `DeltaL` is aa small quantity, the with `(DeltaL)^(2)` beingvery small can be neglected.
`thereforeX=sqrt((LDeltaL)/(2))=sqrt((L(LalphaDeltaT))/(2))=Lsqrt((alphaDeltaT)/(2))`
Given , `L=10m, alpha=1.2xx10^(-5).^(@)C^(-1),DeltaT=20^(@)C`
Hence , `X=10sqrt((1.2xx10^(-5)xx20)/(2))=10xx1.1xx10^(-2)m=0.11m=11am`