JEE MAIN - Physics (2024 - 31st January Evening Shift - No. 15)
The resistance per centimeter of a meter bridge wire is $$r$$, with $$X \Omega$$ 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 $$2 r$$ resistance per centimeter. The new balancing length for same settings will be at
10 cm
80 cm
40 cm
20 cm
Explanation
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$$\begin{aligned} & \frac{25}{\mathrm{r} \ell_1}=\frac{\mathrm{X}}{\mathrm{r} \ell_2} \quad \text{.... (i)}\\ & \frac{25}{2 \mathrm{r} \ell_1^{\prime}}=\frac{\mathrm{X}}{2 \mathrm{r} \ell^{\prime}{ }_2} \quad \text{.... (ii)} \end{aligned}$$
From (i) and (ii)
$$\ell_2^{\prime}=\ell_2=40 \mathrm{~cm}$$
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