Saved Bookmarks
| 1. |
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). |
|
Answer» SOLUTION :(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"` |
|