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If Young.s modulusof the material of a wireis 1.2xx10^(11) Nm^(-2) , calculatethe work done in stretching the wire of length 3 m and cross-sectional area 4mm^(2) whenit is suspendedvertically and a loadof 8 kg is attached to its lower end. |
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Answer» SOLUTION :`Y = 1.2 xx 10^11 Nm^(-2) , L = 3m, A = 4mm^2 = 4 xx 10^(-6) m^2 , M = 8 KG ` `Y =1.2 xx 10^11 Nm^(-2) , L = 3m ` , `A = 4mm^2 = 4 xx 10^(-6) m^2 , M = 8 kg ` work done = 1/2 x stretching force x increase in length. But increase in length, `e = (FL)/(AY)` The work done ` = 1/2 xx (F^2L)/(AY) = 1/2 xx ((mg)^2L)/(AY)` ` = 1/2 xx ((8 xx 9.8)^2 xx 3)/(4 xx 10^(-6) xx 1.2 xx 10^11)` ` = 0.0192 J` |
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