Question 158: Current Electricity

Question

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

Accepted Answer

$$\begin{aligned}
& \frac{25}{\mathrm{r} \ell_1}=\frac{\mathrm{X}}{\mathrm{r} \ell_2} \quad 	ext{.... (i)}\
& \frac{25}{2 \mathrm{r} \ell_1^{\prime}}=\frac{\mathrm{X}}{2 \mathrm{r} \ell^{\prime}{ }_2} \quad 	ext{.... (ii)}
\end{aligned}$$
From (i) and (ii)
$$\ell_2^{\prime}=\ell_2=40 \mathrm{~cm}$$

Answer

10 cm

Answer

80 cm

Answer

40 cm

Answer

20 cm

Current Electricity

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