Figure shows a wire of length 2L and crosssectional area A, stretched horizontally between two clamps. When an object of mass M is suspended from the mid point of the wire, the downward displacement of the young's modulus of the material of the wire is x. Shown that `M = (YA x ^(3))/(gL ^(3)) ` (where `theta` is small)

Explanation

Here tensile stress `sigma _(n) = (F _(a))/( A.)`

From figure `A =A. cos theta and F _(n) = F cos theta `
`therefore ` tensile stress `sigma _(n ) = (F cos ^(2) theta )/( A) [ because A. = (A)/(cos theta )]" "...(1)`
Increase in length of wire `Delta L = x sin theta`
`therefore` Longitudinal strain `= (Delta L )/(L) = (x sin theta)/(L) " "...(2)`
Young modulus `Y = ("tensile stress")/("longitudinal strain ")`
`=( sigma _(n))/( ( Delta L )/( L )) =(F cos ^(2) theta. L )/( A. x sin theta) `
`therefore Y = (FL )/(Ax ) . cot theta . cos theta`
Since `theta` is small.
`cot theta = theta = cos theta =(L)/(x)`
`therefore Y = (FL ^(3))/( Ax ^(3))= ( Mg L ^(3))/( Ax ^(3)) `
`therefore M = (YA x ^(3))/(gL^(3))`