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If the diameter of a wire is doubled, the resistance changes by

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Decrease to half

Decrease to one-fourth

Increase to four times

Increase to double

answer is B.

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Detailed Solution

If the diameter of a wire is doubled, the resistance will decrease to one-fourth.
We know that,

R = \frac{ρl}{A}

⇒

R = \frac{ρl}{π \times {[\frac{d}{2}]}^{2}}

= \frac{ρl}{π} \times \frac{4}{d^{2}}

Let us assume that after doubling the diameter of the wire, the resistance of the wire is R’.
⇒

R^{'} = \frac{ρl}{A}

= \frac{ρl}{1} \times \frac{1}{{π (\frac{2 d}{2})}^{2}}

(d becomes 2 d)

⇒

R^{'} = \frac{ρl}{π} \times \frac{4}{4 d^{2}}

⇨

\frac{1}{4} \times \frac{ρl}{π} \times \frac{4}{d^{2}}

⇨

\frac{1}{4} R

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