Question #257253

A vertical brass rod of circular section is loaded by placing a 5kg weight on top of it. If it's length is 50cm, it's radius of cross section 1cm and the young modulus of the material is 3.5power 19


1
Expert's answer
2021-10-27T08:48:01-0400

a) contraction in rod is given by

Δl=mglπr2Y\Delta l=\frac{mgl}{\pi r^2Y}

Putting value of all variable and constants according to data of question

Δl=5×9.8×50×1023.14×1×104×3.5×1010\Delta l= \frac{5\times9.8\times50\times10^{-2}}{3.14\times1\times10^{-4}\times3.5\times10^{10}}

Δl=22.29×107m\Delta l=22.29\times10^{-7}m


b) stored energy is given

U=mgΔl2U=\frac{mg\Delta l}{2}

U=5×9.8×22.29×1072U=\frac{5\times9.8\times22.29\times10^{-7} }{2}

U=5.46×105JU=5.46\times10^{-5} J

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