A steel wire of diameter 0.8 mm and length 1m is clamped firmly at two points A and B
which are 1 m apart in the same horizontal line. A body is hung from the middle point
of the wire such that the middle point sags 1 cm from the original position. Calculate the
mass of the body.
[`Y=2 times 10^12 dyn.cm^-2`]

Explanation

In fig.1.136 C is the mid point of the wire AB. When a mass m is hung from C, it sags 1 cm and comes down to the point O.

`AB=1m=100cm,AC=CB=50cm,OC=1cm,OA=OB=sqrt(50^2+1^2)=50.01 cm,costheta=1/50.01,r=0.04cm`
Here, tension in the part OA or OB is T. The two horizontal components balance the weight mg.
`therefore2Tcostheta=mgor,T=(mg)/(2costheta)`
the length AC of the wire changes into AO.
`therefore` Elongation `=l=AO-AC=50.01-50=0.01cm`
Young.s modulus,
`Y=(TL)/(Al)=(mgL)/(2costheta.Al)`
or,`m=(2YAlcostheta)/(gL)`
`=(2times(2times10^12)times3.14times(0.04)^2times0.01)/(980times50)times1/50.01`
`=82.01g`