A cylindrical rod is reformed to half of its original length keeping volume constant. If its resistance before this change were R, then the resistance after reformation of rod will be.

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R / 4

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3 R / 4

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R / 2

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Solution

The correct option is B $R / 4$
The resistance of rod before reformation
$R_{1} = R = \frac{ρ l_{1}}{π r_{1}^{2}}$ $[∵ R = \frac{ρ l}{A} = \frac{ρ l}{π r^{2}}]$
Now the rod is reformed such that
$l_{2} = \frac{l_{1}}{2}$

$∴ π r_{1}^{2} l_{1} = π r_{2}^{2} l_{2}$ ( $∵$ Volume remains constant)

or $\frac{r_{1}^{2}}{r_{2}^{2}} = \frac{l_{2}}{l_{1}}$ ..(i)
Now the resistance of the rod after reformation

$R_{2} = \frac{ρ l_{2}}{π r_{2}^{2}}$ $∴ \frac{R_{1}}{R_{2}} = \frac{ρ l_{1}}{π r_{1}^{2}} / \frac{ρ l_{2}}{π r_{2}^{2}} = \frac{l_{1}}{l_{2}} \times \frac{r_{2}^{2}}{r_{1}^{2}}$

or $\frac{R_{1}}{R_{2}} = \frac{l_{1}}{l_{2}} \times \frac{l_{1}}{l_{2}} = {(\frac{l_{1}}{l_{2}})}^{2} = (2)^{2}$ (using (i))

$∴ R_{2} = \frac{R}{4}$

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