A steel wire of diameter `0.8mm` and length 1m is clamped firmly at two points A and B which are 1m apart and in the same plane. A body is hung from the middle point of the wire such that the middle point sags 1cm lower from the original position. Calculate the mass of the body. Given that Young's modulus of the material of wire is `2xx10^(11)N//m^(2)`.

Explanation

As shown in figure, for equilibrium of mass M,
`Mg=2Tsintheta`…….(i)

But from figure, `sintheta~=tantheta=((x)/(L))……(ii)`
any by definition of Young.s modulus,
`T=(YA)/(L)DeltaL=(YA)/(L)[(L^(2)+x^(2))^(1//2)-L]~-(YAx^(2))/(2L^(2))`.......(iii)
So substituting the values of `sintheta` and `T` from Eqns. (ii) and (iii) in Eqn, (i), we get
`Mg=2xx(YAx^(2))/(2L^(2))xx(x)/(L)`, i.e., `M=(YAx^(3))/(gL^(3))`.............(iv)
Now as here `2L=1m`, `x=1cm=10^(-2)m`
and `A=pir^(2)=pi[(0.8//2)xx10^(-3)]^(2)=pixx(4xx10^(-4))^(2)m^(2)`
So, `M=(2xx10^(11)xxpi(4xx10^(-4))^(2)xx(10^(-2))^(3))/(9.8xx(1//2)^(3))kg=82g`