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

Question

In a meter bridge experiment, a resistance of $ 10 \, \Omega $ is balanced by a resistance $ X $ with a balance point at $ 40 \, \mathrm{cm} $. What is the value of $ X $?

Answer 1

Initial resistance, R₀, is 10 Ω. Upon stretching to length 2l, the resistance R₁ becomes $\frac{2l}{l}$R₀ = 20 Ω. Bending into a circle does not alter the total resistance, which remains 20 Ω.

The resistance of a semi-circle along any diameter is calculated as 20/2 = 10 Ω.

A uniform metal wire of length \(l\) has 10 Ω resistance. Now this wire is stretched to a length \(2l \) and then bent to form a perfect circle. The equivalent resistance across any arbitrary diameter of that circle is :

The Correct Option is A

Solution and Explanation

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