Current Electricity Question 23
Question 23 - 2024 (31 Jan Shift 2)
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 $2 \mathrm{r}$ resistance per centimeter. The new balancing length for same settings will be at
(1) $20 \mathrm{~cm}$
(2) $10 \mathrm{~cm}$
(3) $80 \mathrm{~cm}$
(4) $40 \mathrm{~cm}$
Show Answer
Answer: (4)
Solution:
$\frac{25}{\mathrm{r} \ell_{1}}=\frac{\mathrm{X}}{\mathrm{r} \ell_{2}}$
$\frac{25}{2 \mathrm{r} \ell_{1}^{\prime}}=\frac{\mathrm{X}}{2 \mathrm{r} \ell_{2}^{\prime}}$
From (i) and (ii)
$\ell_{2}^{\prime}=\ell_{2}=40 \mathrm{~cm}$
