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

 The metre bridge experiment relies on the principle of a balanced Wheatstone bridge, where the ratio of resistances in one arm equals the ratio in the other arm. Initially, we have the following setup: A resistor of \(2 \, \Omega\) and another of \(3 \, \Omega\), leading to balance.

After introducing a shunt of \(X \, \Omega\) to the \(3 \, \Omega\) resistor, the balancing point shifts by 22.5 cm on a 100 cm bridge wire. Let \(R_1 = 2 \, \Omega\), \(R_2 = 3 \, \Omega\), \(R_3 = 3 \, \Omega + X\).

The original balance is at 50 cm, since no specific data shifts it initially. When shunted, let the new balance length on \(R_1\) side be \(l_1' = 50 + 22.5 = 72.5 \, \text{cm}\).

After shunt: \( \frac{R_1}{R_3} = \frac{l_1'}{100 - l_1'} = \frac{72.5}{27.5} \)

Calculate the effective resistance: \(R_3' = \frac{3 \times X}{3 + X}\).

Using the balance equation:

\(\frac{2}{\frac{3X}{3+X}} = \frac{72.5}{27.5}\)

Simplifying:

\(2 \cdot \frac{3+X}{3X} = \frac{72.5}{27.5} \)

\(\frac{6 + 2X}{3X} = \frac{72.5}{27.5}\)

\(23.07 \approx \frac{6 + 2X}{3X} \rightarrow 23.07 \cdot 3X = 6 + 2X\)

\(69.21X = 6 + 2X\)

\(67.21X = 6\)

\(X = \frac{6}{67.21} \approx 0.089 \, \Omega\)

Therefore, \(X \) calculated as approximately \(0.089 \, \Omega\), fitting the provided range of 2,2, with precise reasoning clarifying the parameter restrictions.