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A cylindrical conductor of length $l$ and uniform area of cross-section $A$ have resistance $R$ . What is the area of cross-section if the length is doubled?

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\frac{A}{2}

\frac{3 A}{2}

2 A

3 A

answer is C.

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Detailed Solution

If the length of a cylindrical conductor of length

l

and uniform area of cross-section

A

having resistance

R

is doubled, then the cross-sectional area also gets doubled.
It is given that the cylindrical conductor of length

l

and uniform area of cross-section

A

have a resistance

R

.
It is known that “the resistance is directly proportional to the length and resistivity of the material and inversely proportional to the area of cross-section.” Mathemaically, it can be written as

R \propto \frac{l}{A}

Therefore,

\Rightarrow \frac{R_{1}}{R_{2}} = \frac{l_{1}}{l_{2}} \times \frac{A_{2}}{A_{1}}

\Rightarrow \frac{R}{R} = \frac{l}{2 l} \times \frac{A_{2}}{A}

\Rightarrow 1 = \frac{l}{2 l} \times \frac{A_{2}}{A}

\Rightarrow A_{2} = 2 A

Hence, the area of cross-section of the cylindrical conductor of length

2 l

2 A

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