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(B) Let the resistance in the left gap be $R_1 = 2 \Omega$ and the right gap be $R_2 = 3 \Omega$. Let the balance length be $l_1$. According to the me

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(B) Let the resistance in the left gap be $R_1 = 2 \Omega$ and the right gap be $R_2 = 3 \Omega$. Let the balance length be $l_1$. According to the metre-bridge principle: $\frac{R_1}{R_2} = \frac{l_1}{100 - l_1} \implies \frac{2}{3} = \frac{l_1}{100 - l_1}$.Solving for $l_1$: $200 - 2l_1 = 3l_1 \implies 5l_1 = 200 \implies l_1 = 40 \ cm$.When a shunt $x$ is added in parallel to $3 \Omega$,the new resistance $R_2'$ is $\frac{3x}{3+x}$.The new balance length $l_2 = l_1 + 22.5 = 40 + 22.5 = 62.5 \ cm$.Applying the principle again: $\frac{R_1}{R_2'} = \frac{l_2}{100 - l_2} \implies \frac{2}{\frac{3x}{3+x}} = \frac{62.5}{100 - 62.5} = \frac{62.5}{37.5} = \frac{5}{3}$.Simplifying: $\frac{2(3+x)}{3x} = \frac{5}{3} \implies \frac{6+2x}{3x} = \frac{5}{3}$.Cross-multiplying: $3(6+2x) = 15x \implies 18 + 6x = 15x \implies 9x = 18 \implies x = 2 \Omega$.

In a metre-bridge experiment,the balance point is obtained when the gaps are closed by $2 \Omega$ and $3 \Omega$ resistors. $A$ shunt of $x \Omega$ is added in parallel to the $3 \Omega$ resistor to shift the balance point by $22.5 \ cm$. The value of $x$ is: (in $\Omega$)

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