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A steel wire of length 2 m and 1.2 xx 10^(-7)m^(2) in cross sectional area is stretched by a force of 36 N. Calculate |
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Answer» Solution :stress, strain, increase in length, work done in stretching the wire `(Y=1.8xx10^(11)N//m^(2))` Stress `=("force")/("area")=(36)/(1.2xx10^(-7))=3xx10^(8)N//m^(2)` Stress Y= Strain `:. "Strain" =("stress")/(Y)=(3)/(1.8)xx10^(-3)=1.67xx10^(-3)` INCREASES of length `=Deltal=(l)/("strain")=1.67xx10^(-3)xx2=3.34xx10^(-3)m` Workdone `=(1)/(2)` (stress) (strain) volume `=(1)/(2)xx3xx10^(8)xx(5)/(3)xx10^(-3)(2xx1.2xx10^(-7))=6xx10^(-2)` joule. |
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