Question 672: Current Electricity

Question

A wire of resistance 4 $$\Omega $$ is stretched to twice its original length. The resistance of stretched wire would be

Accepted Answer

Resistance of a wire,
$$R = ho {l \over A} = 4\Omega $$   ...(i)
When wire is stretched twice, its new length be $$l'$$. Then
$$l' = 2l$$
On stretching volume of the wire remains constant.
$$ 	herefore lA = l'A'$$ where A' is the new cross-sectional area
$$ \Rightarrow A' = {l \over {l'}}A = {l \over {2l}}A = {A \over 2}$$
$$ 	herefore $$ Resistance of the stretched wire is
$$R' = ho {{l'} \over {A'}} = ho {{2l} \over {\left( {A/2} ight)}} = 4ho {l \over A}$$
$$ = 4\left( {4\Omega } ight) = 16\Omega $$

Answer

8 $$\Omega $$

Answer

16 $$\Omega $$

Answer

2 $$\Omega $$

Answer

4 $$\Omega $$

Current Electricity

A wire of resistance 4 $\Omega $ is stretched to twice its original length. The resistance of stretched wire would be

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