A cell of emf $E$ volt with no internal resistance is connected to a wire whose cross section changes. The wire has three sections of equal length. The middle section has a radius $a$ whereas the radius of the outer two sections is $2 a$ . The ratio of the potential difference across the section $A B$ to the potential difference across the section $B C$ is

5

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4

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\frac{1}{2}

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\frac{1}{4}

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Solution

The correct option is D $\frac{1}{2}$
Let the resistivity of the material be ' $ρ$ '.

Then the resistance of $A B = \frac{ρ l}{2 a}$

resistance of $B C = \frac{ρ l}{a}$

resistance of $C D = \frac{ρ l}{2 a}$

Total resistance of the material( $r$ ) $= \frac{2 ρ l}{a}$

Current( $i$ ) $= \frac{E}{r}$

$= \frac{E a}{2 ρ l}$

the potential difference across $A B = i \times \frac{ρ l}{2 a}$

$= \frac{E}{4}$

The potential difference across $B C = i \times \frac{ρ l}{a}$

$= \frac{E}{2}$

ratio of $A B$ to $B C = \frac{1}{2}$

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