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A wire of cross sectional area `4xx10^(-4) m^2` modulus of rigidity `2xx10^(11) N//m^2` and length 1 m is stretched between two vertical rigid poles. A mass of 1kg is suspended at its middle. Calculate the angle it makes with horizontal. Given `g = 10 ms^(-2)`

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Correct Answer - 0.005 rad
Refer to
for equilibrium position of mass M, `Mg = 2T sin theta…… (i)`
if `theta` is small, sin `theta ~~ theta = x//L….. (ii)`
`As, T = (YA)/(L) Delta L = (YA)/(L)[(L^2 + x^2)^(1//2)-L]`
From (i) `Mg = (2YA x^2)/(2L^2) xx sin theta = YA theta^2 xx theta = YA theta^3`
or `theta = ((Mg)/(YA))^(1//3) = [(1xx10)/(2xx 10^(11)xx4 xx10^(-4))]^(1//3)`
`~~ 0.005 rad`

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