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)

Figure shows a wire of length 2L and crosssectional area A, stretched

1.	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)
Answer» SOLUTION :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))`