A wire of cross -sectional area `4 xx 10^(-4) m^(2)` , modulus of elasticity `2 xx 10^(11) Nm^(-2)` and length 1 m is stretched between two rigid poles . A mass of 1 kg is suspended at its middle . Calculate the angle it makes with horizontal . Take g = `10 ms^(-2)`

Answer

0.005 rad

Explanation

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`