The resistance per centimeter of a meter bridge wire is $\mathrm{r}$, with $\mathrm{X}\ \Omega$ resistance in left gap. Balancing length from left end is at $40 \mathrm{~cm}$ with $25\ \Omega$ resistance in right gap. Now the wire is replaced by another wire of $2 \mathrm{r}$ resistance per centimeter. The new balancing length for same settings will be at
A$20 \mathrm{~cm}$
B $10 \mathrm{~cm}$
C$80 \mathrm{~cm}$
D $40 \mathrm{~cm}$
JEE MAIN 2024, Diffcult
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D $40 \mathrm{~cm}$
d $\frac{25}{\mathrm{r} \ell_1}=\frac{\mathrm{x}}{\mathrm{r} \ell_2}$ $.....(i)$
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