A uniform metal wire of length has 10 resistance. Now this wire is stretched to a length and then bent to form a perfect circle. The equivalent resistance across any arbitrary diameter of that circle is [2024]

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Solution

Q.
A uniform metal wire of length $l$ has 10 $Ω$ resistance. Now this wire is stretched to a length $2 l$ and then bent to form a perfect circle. The equivalent resistance across any arbitrary diameter of that circle is [2024]

1 10Ω

2 5Ω

3 40Ω

4 20Ω

Ans.
(1)

The volume of wire remains same even after stretching the wire, i.e., $V' = V; A' l' = A l$

   $A' = \frac{A l}{l'} = \frac{A l}{2 l} = \frac{A}{2} (∵ l' = 2 l)$

and Resistance of wire is $R = \frac{ρ L}{A} = 10 Ω$

After stretching, resistance of circle becomes

$R' = ρ \frac{(2 l) (2)}{A} = 4 R = 40 Ω$

As the wire is bent at its mid-point to form a circle, thus, the effective resistance of semi-circle is $= \frac{R'}{2} = \frac{40}{2} = 20 Ω$

$∴$    Equivalent resistance across diameter of circle,

   $= 20 Ω ∥ 20 Ω$

   $\frac{1}{R_{eq}} = \frac{1}{20} + \frac{1}{20}$

$∴$ $R_{eq} = 10 Ω$

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