A conducting wire has length $^{'} L_{1}^{'}$ and diameter $^{'} d_{1}^{'}$ . After stretching the same wire length becomes $^{'} L_{2}^{'}$ and diameter $^{'} d_{2}^{'}$ . The ratio of resistances before and after stretching is

d_{2}^{4} : d_{1}^{4}

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d_{1}^{4} : d_{2}^{4}

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d_{2}^{2} : d_{1}^{2}

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d_{1}^{2} : d_{2}^{2}

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Solution

The correct option is A $d_{2}^{4} : d_{1}^{4}$
Resistance of body is directly proportional to length and resistivity of material & inversely propotional. to its cross sectional area.
$R = P \frac{L}{A}$
for same wire, $P$ is constant
$R_{1} = P \frac{L_{1}}{A_{1}}, R_{2} = P \frac{L_{2}}{A_{2}}$
$\frac{R_{1}}{R_{2}} = \frac{L_{1}}{A_{1}} \times \frac{A_{2}}{L_{2}} ⟶ (i)$
As same wire is stretched, so volume remains constant i.e,
$L_{1} A_{1} = L_{2} A_{2} ⟶ (i i)$
From $(i)$ & $(i i)$ ; $\frac{R_{1}}{R_{2}} = {(\frac{A_{2}}{A_{1}})}^{2}$
$\frac{R_{1}}{R_{2}} = {(\frac{d_{2}^{2}}{d_{1}^{2}})}^{2}$
$\Rightarrow \frac{R_{1}}{R_{2}} = \frac{d_{2}^{4}}{d_{1}^{4}}$

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