The electric resistance of a certain wire of iron is R. If its length and radius are both doubled, then

Moderate

A

the resistance will be doubled and the specific resistance will be halved
B

the resistance will be halved and the specific resistance will remain unchanged
C

the resistance will be halved and the specific resistance will be doubled
D

the resistance and the specific resistance, will both remain unchanged

Solution

The formula for resistance of wire is
$R = \frac{ρ l}{A}$
where $ρ$ = specific resistance of the wire
$\Rightarrow R \propto \frac{l}{A}$
$\Rightarrow R \propto \frac{1}{r^{2}} (∵ A = π r^{2})$
$\begin{array}{l} ∴ \frac{R_{1}}{R_{2}} = \frac{l_{1}}{l_{2}} \times \frac{r_{2}^{2}}{r_{1}^{2}} . ... (i) \\ Given, l_{1} = l, l_{2} = 2 l, r_{1} = r, r_{2} = 2 r, R_{1} = R \end{array}$
Substituting these values in Eq. (i), we have
$\frac{R_{1}}{R_{2}} = \frac{l}{2 l} \times \frac{{(2 r)}^{2}}{r_{}^{2}}$
$\frac{R_{1}}{R_{2}} = 2 and R_{2} = \frac{R}{2}$
Therefore, resistance will be halved.
Now the specific resistance of the wire does not depend on the geometry of the wire, hence, it remains unchanged.
Aliter: $R \propto \frac{l}{r^{2}} \Rightarrow \frac{R_{1}}{R_{2}} = \frac{l_{2}^{}}{l_{1}} \times \frac{r_{1}^{2}}{r_{2}^{2}} = (\frac{2}{1}) \times {(\frac{2}{1})}^{2}$
$= \frac{1}{2}$
$\Rightarrow R_{2} = \frac{R_{1}}{2},$ specific resistance doesn't depend upon length and radius.

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