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. `5 cm`
B. `20 cm`
C. `15 cm`
D. `11 cm`

Answer

D

Explanation

Diagram shows the deformation of a railway track due to rise in temperature

Applying Pythagoras theorem in right angles
tringle , `x^(2) = ((L + DeltaL)/(2))^(2) - ((L)/(2))^(2)`
rArr `x = sqrt(((L + DeltaL)/(2))^(2) - ((L)/(2))^(2))`
`x = sqrt(((L)/(2)^(2))+(2DeltaL)/(4)+((DeltaL)/(2))^(2)-((L)/(2))^(2))`
`(1)/(2) sqrt((L^(2) + DeltaL^(2) + 2LDeltaL) - L^(2)) = (1)/(2) = (1)/(2)sqrt((DeltaL^(2) + 2LDeltaL))`
As increase in length `DeltaL` is very small, therefore neglecting `(DeltaL)^(2)`, we get
`x = (sqrt(2LDeltaL))/(2)` ....(1)
But `Delta = LalphaDeltat` ........(ii)
According to the problelm, `L = 10 m`,
`alpha = 1.2 xx 10^(-5) ^(0)C^(-1), DeltaT = 20^(0)C`
Substituting value of `DeltaL` in Eq. `(i)` form Eq. `(ii)`
`x = (1)/(2)sqrt((2L xx LalphaDeltat)) = (1)/(2)Lsqrt(2alphaDeltat)`
`= 11cm`