A wire of length 2 m with the area of cross section `10^(-6) m^(2)` is used to suspend a load of 980 N. Calculate (i) The stress developed in the wire (ii) the strain and (iii) the energy stored . Given `Y = 12 xx 10^(10) Nm^(-2)`.

Explanation

(i) Stress - `F/A = (980)/(10^(-6)) = 98 xx 10^(7) Nm^(-2)`
(ii) Strain = `("Stress")/(Y) = (98 xx 10^(7))/(12 xx 10^(10)) = 8.17 xx 10^(-3)` (no unit)
(iii) Since, volume = `2 xx 10^(-6) m^(3)`
Energy = `1/x ("stress" xx "strain") xx "volume" implies 1/2(98 xx 10^(7)) xx (8.17 xx 10^(-3)) xx 2 xx 10^(-6) = 8 "joule"`