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(D) Let the resistance in the left gap be $R_L$ and the resistance in the right gap be $R_R$. In the first case,the right gap has two equal resistors

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(D) Let the resistance in the left gap be $R_L$ and the resistance in the right gap be $R_R$. In the first case,the right gap has two equal resistors $r$ in series,so $R_R = r + r = 2r$. The balancing point is at $l_1 = 50 \ cm$. Using the meter bridge formula: $\frac{R_L}{R_R} = \frac{l_1}{100 - l_1} \implies \frac{R_L}{2r} = \frac{50}{50} = 1 \implies R_L = 2r$. In the second case,one resistor $r$ is removed from the right gap,so the new right resistance is $R_R' = r$. This removed resistor $r$ is connected in parallel to $R_L$. The new left resistance is $R_L' = \frac{R_L \cdot r}{R_L + r} = \frac{2r \cdot r}{2r + r} = \frac{2r^2}{3r} = \frac{2}{3}r$. Let the new balancing point be $l_2$. Then $\frac{R_L'}{R_R'} = \frac{l_2}{100 - l_2} \implies \frac{(2/3)r}{r} = \frac{l_2}{100 - l_2} \implies \frac{2}{3} = \frac{l_2}{100 - l_2}$. $2(100 - l_2) = 3l_2 \implies 200 - 2l_2 = 3l_2 \implies 5l_2 = 200 \implies l_2 = 40 \ cm$.

When the right gap of a meter bridge consists of two equal resistors in series,the balancing point is at $50 \ cm$. When one of the resistors in the right gap is removed and is connected in parallel to the resistor in the left gap,the balancing point is at: (in $cm$)

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