a) contraction in rod is given by
Δl=mglπr2Y\Delta l=\frac{mgl}{\pi r^2Y}Δl=πr2Ymgl
Putting value of all variable and constants according to data of question
Δl=5×9.8×50×10−23.14×1×10−4×3.5×1010\Delta l= \frac{5\times9.8\times50\times10^{-2}}{3.14\times1\times10^{-4}\times3.5\times10^{10}}Δl=3.14×1×10−4×3.5×10105×9.8×50×10−2
Δl=22.29×10−7m\Delta l=22.29\times10^{-7}mΔl=22.29×10−7m
b) stored energy is given
U=mgΔl2U=\frac{mg\Delta l}{2}U=2mgΔl
U=5×9.8×22.29×10−72U=\frac{5\times9.8\times22.29\times10^{-7} }{2}U=25×9.8×22.29×10−7
U=5.46×10−5JU=5.46\times10^{-5} JU=5.46×10−5J
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