Q 12-03-110JEE MainJEE Main 2024 (31 Jan, Shift 2)Medium
The resistance per centimetre of a metre bridge wire is $r$, with $X\ \Omega$ resistance in the left gap. The balancing length from the left end is at $40\ \text{cm}$ with $25\ \Omega$ resistance in the right gap. Now the wire is replaced by another wire of $2r$ resistance per centimetre. The new balancing length for the same settings will be at
Answer: (D) $40\ \text{cm}$
At balance $\dfrac X{25} = \dfrac{r\,l}{r(100 - l)} = \dfrac{l}{100 - l}$. The resistance per cm cancels, so changing the wire does not move the balance point: it stays at $40\ \text{cm}$.
Solution by Sreeraj P, M.Sc Physics