A wire of resistance 5 ohm is drawn out so that its new length is three times its original length. What is the resistance of the new wire ?

\frac{5}{3} Ω

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5 Ω

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15 Ω

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45 Ω

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Solution

The correct option is D $45 Ω$
We know that $V o l u m e = \frac{M a s s}{d e n s i t y}$

Let m be the mass of the given wire, d its density and $ρ$ its resistivity.

These quantities remain unchanged when the wire is drawn out.
Let $l_{1}$ be the length of the smaller wire and $r_{1}$ its corresponding radius of cross section. If $l_{2}$ and $r_{2}$ represent the length and radius of cross section of the elongated wire,

$⟹ m = π r_{1}^{2} l_{2} d$ for smaller wire and for elongated wire, $m = π r_{2}^{2} l_{2} d$ .
Since mass and density are constant
$⟹ r_{1}^{2} l_{1} = r_{2}^{2} l_{2}$

If $R_{1}$ is the resistance of the smaller wire and $R_{2}$ is the resistance of the elongated wire, we have,
$R_{1} = \frac{ρ l_{1}}{π r_{1}^{2}}$
$R_{2} = \frac{ρ l_{2}}{π r_{2}^{2}}$

Now, $\frac{R_{2}}{R_{1}} = \frac{l_{2}}{l_{1}} \times \frac{r_{1}^{2}}{r_{2}^{2}}$
It is given that, $l_{2} = 3 l_{1} .$
Therefore $r_{2}^{2} = \frac{r_{1}^{2}}{3}$
Substituting the values,
$\frac{R_{2}}{R_{1}} = 9$
Since, $R_{1} = 5 Ω$
$R_{2} = 45 Ω$ .

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