Question

In a metre bridge experiment, the balance point is obtained if the gaps are closed by \(2 \Omega\) and \(3 \Omega\). A shunt of \(X \Omega\) is added to \(3 \Omega\) resistor to shift the balancing point by 22.5 cm. The value of X is _______ .

Hint:

For metre bridge problems:

• Use the balance condition: \[ \frac{l_1}{l_2} = \frac{R_1}{R_2}. \]
• Adjust resistance values with shunts using parallel resistance formulas.

Answer 1

Correct Answer: 2

1. Initial Balance Point: The ratio of resistances gives the balance length: \[ \frac{l_1}{l_2} = \frac{R_1}{R_2}, \] where \(l_1 + l_2 = 100 \, \text{cm}\).
2. Initial Condition: For \(R_1 = 2 \, \Omega\) and \(R_2 = 3 \, \Omega\): \[ \frac{l_1}{l_2} = \frac{2}{3}, \quad \text{so} \, l_1 = \frac{2}{5} \cdot 100 = 40 \, \text{cm}. \]
3. After Adding Shunt: The effective resistance of \(R_2\) with a shunt \(X\): \[ R_2' = \frac{R_2 X}{R_2 + X} = \frac{3X}{3 + X}. \] The new balance point shifts by \(22.5 \, \text{cm}\): \[ l_1' = 40 + 22.5 = 62.5 \, \text{cm}. \]
4. New Condition: The new ratio is: \[ \frac{l_1'}{l_2'} = \frac{R_1}{R_2'}. \] Substituting \(l_1' = 62.5\), \(l_2' = 37.5\), \(R_1 = 2 \, \Omega\), and \(R_2' = \frac{3X}{3 + X}\): \[ \frac{62.5}{37.5} = \frac{2}{\frac{3X}{3 + X}}. \]
5. Simplify: \[ \frac{5}{3} = \frac{2(3 + X)}{3X}. \] Cross-multiply and solve for \(X\): \[ 15X = 18 + 6X \implies 9X = 18 \implies X = 2 \, \Omega. \]