A steel wire of diameter 0.8 mm and length 1 m is champed firmly at two points A and B which are 1n apart and in the same plane. A body is hung from the middle point of the wire such that the middle point sags 1 cm lower from the origional position. Calculate the mass of the body. Given the Young's modulus of the material of wing is `2 xx 10^(11) N//m^(2)`.

Explanation

As shown in figure, for equilibrium of maass M,
`Mg = 2T sin theta ……………..(i)`

Buf from figure, `sin theta cong tan theta = ((X)/(L)) ………. (ii)`
and by definition of Young modulus,
`T=(YA)/(L)DeltaL=(YA)/(L)[(L^(2)+x^(2))^(1//2)-L]=(YAx^(2))/(2L^(2))..........(iii)`
So substituting the values of `sin theta 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 = 1 m, x = 1 cm = 10^(-2)m`
and `A = pi r^(2) = pi [(0.8//2)xx10^(-3)]^(2) = pi xx (4 xx 10^(-4))^(2)m^(2)`
So, `M=(2xx10^(11)xxpi(4xx10^(-4))^(2)xx(10^(-2))6(3))/(9.8xx(1//2)^(3))kg = 82g`