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    				| 1. | A steel wire of length 2.0 m/s is stretched through 2.0 mm. The cross sectional area of the wire is 4.0 mm^2. Calculate the elastic potential energy stored in the wire in the stretched condition. Young modulus of steel `=2.0x10^11Nm^-2` | 
| Answer» The strain in the wire `(/_l)/l=(2.0mm)/(2.0m)=10^-3` The stress in the wire `=YxxStrain` `=2.0xx10^1Nm^-2xx10^-3=2.0xx10^8Nm^-2` The volume of the wire `=(4xx10^-6m^2)xx(2.0m)` `=8.0xx106-6m^3` the elastic potential energy stored `=1/2xxstressxxstrainxxvolume` `=1/2xx2.0xx10^8Nm^-2xx8.0xx10^-6m^3` `=0.8J` | |