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Question
A rail track made of steel having length 10 m is clamped on a raillway 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 αsteel = 1.2 × 10–5/°C.
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Solution
Consider the diagram
Applying Pythagoras theorem in the right-angled triangle in figure
`((L + ΔL)/2)^2 = (L/2)^2 + x^2`
⇒ `x = sqrt(((L + ΔL)/2)^2 = (L/2)^2)`
= `1/2 sqrt((L + ΔL)^2 - L^2)`
= `1/2 sqrt((L^2 + ΔL^2 + 2LΔL) - L^2)`
= `1/2 sqrt((ΔL^2 + 2LΔL)`
As increase in length ΔL is very small, therefore, neglecting (ΔL)2, we get
`x = 1/2 xx sqrt(2LΔL)` ......(i)
But ΔL = LαΔt ......(ii)
Substituting the value of ΔL in equation (i) from equation (ii)
`x = 1/2 sqrt(2L xx LαΔt)`
= `1/2 L sqrt(2αΔt)`
= `10/2 xx sqrt(2 xx 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
= 11 cm
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