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Question

An 8m long copper wire and a 4m long steel wire, each of cross-section 0.5 cm2 are fastened end to end and stretched by a 500 N force. The elastic potential energy of the system is (YCopper=1×1011Nm2;Ysteel=2×1011Nm2)


A

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B

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C

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D

4J

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Solution

The correct option is A


The potential energy of an elongated rod is given by U=F2L2AY. In this problem the total

potential energy is the sum of the individual potential energies. Therefore,

U=F22A(L1Y1+L2Y2)

On substitution,

U=0.25J=14J


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