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Question

A given wire of residence $$1\Omega $$ is stretched to double its length. What will be its new resistance.   


Solution

Given resistance$$=1\Omega$$ and initial length =L and final length =2 L.
Let Initial area$$A$$ and final area$${A}_{f}=\dfrac{A}{2}$$ 
Though the length and cross sectional area of the wire changes it volume will remain constant.
$${V}_{i} ={V}_{f}$$
Thus initial resistance$${R}_{i}=\rho \dfrac{L}{A}$$--------(1)  
Final resistance $${R}_{f}=\rho\dfrac{2L}{\dfrac{A}{2}}$$--------(2)
Computing 1 and 2 we get 
$${R}_{f}=4{R}_{i}$$
$${R}_{f}=4 \Omega$$

Physics

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