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JEE MAIN 2024
310124 S2
Question
The resistance per centimeter of a meter bridge wire is r , with $\mathrm{X} \Omega$ resistance in left gap. Balancing length from left end is at 40 cm with $25 \Omega$ resistance in right gap. Now the wire is replaced by another wire of $2 r$ resistance per centimeter. The new balancing length for same settings will be at
Select the correct option:
A
20 cm
B
10 cm
C
80 cm
D
40 cm
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
$$ \begin{aligned} & \frac{25}{\mathrm{r} \ell_1}=\frac{\mathrm{X}}{\mathrm{r} \ell_2} \\ & \frac{25}{2 \mathrm{r} \ell^{\prime}}=\frac{\mathrm{X}}{2 \mathrm{r} \ell^{\prime}} \end{aligned} $$ From (i) and (ii) $$ \ell_2^{\prime}=\ell_2=40 \mathrm{~cm} $$
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