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

• A. \(1\)
• B. \(2\)
• C. \(3\)
• D. \(4\)

Hint:

First find the initial balance length, then the shifted one.

Answer 1

The Correct Option is B

Step 1: Initial balance
\(\frac{l}{100-l}=\frac23\) gives \(l=40\) cm.

Step 2: New balance
A shunt lowers the \(3\,\Omega\) resistance, so the balance point moves toward the other side by \(22.5\) cm: \(l^{\prime}=62.5\) cm.
\(\frac{62.5}{37.5}=\frac{2}{R^{\prime}}\), so \(R^{\prime}=1.2\,\Omega\).

Step 3: Shunt
\(\frac{3X}{3+X}=1.2\) gives \(3X=3.6+1.2X\), so \(X=2\,\Omega\). Option (B).

Final Answer:
\(X=2\,\Omega\), option (B). \[ \boxed{\text{(B) }2\,\Omega} \]