Question

# A wire of resistance $$1\ ohm$$ is stretched to double its length. What is the new resistance?

Solution

## Given: a wire of resistance $$1\Omega$$is stretched  to double its length.To find the new resistanceSolution:Here, $$R=1\Omega, \text{Length}=L, \text{Cross sectional area, CSA}=A$$The formula used is,$$R=\rho\dfrac LA\\\implies \rho\dfrac LA=1..........(i)$$The new length, $$L'=2L$$And new CSA = $$A'$$So we have, Volume of the wire= $$AL$$Now after the act of stretching the volume remains constant.Therefore, $$AL=A'L'$$$$\implies AL=2LA'\\\implies A'=\dfrac A2$$Now the new resistance of the wire becomes$$R_{new}=\rho\dfrac {L'}{A'}\\\implies R_{new}=\rho\dfrac {2L}{\dfrac A2}\\\implies R_{new}=4\rho\dfrac LA$$$$R_{new}=4\times 1$$ by using equation (i)Therefore the new resistance is $$4\Omega$$Physics

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