Step 1: Understanding the Question:
A metre bridge is initially balanced with \(2 \Omega\) and \(3 \Omega\) in the left and right gaps.
Then a shunt \(x\) is connected in parallel to the \(3 \Omega\) resistor, changing the balance point. We need to find \(x\).
Step 2: Key Formula or Approach:
Metre bridge balance condition: \(\frac{R_{1}}{R_{2}} = \frac{l}{100 - l}\).
Step 3: Detailed Explanation:
Initial case: \(R_{1} = 2 \Omega\), \(R_{2} = 3 \Omega\).
\[ \frac{2}{3} = \frac{l_{1}}{100 - l_{1}} \implies 200 - 2l_{1} = 3l_{1} \implies 5l_{1} = 200 \implies l_{1} = 40 \text{ cm} \]
New case: A shunt \(x\) is added to \(3 \Omega\). The new right gap resistance is \(R_{2}' = \frac{3x}{3 + x}\).
Since a parallel resistance reduces the total resistance, \(R_{2}'<3 \Omega\).
Thus, \(\frac{R_{1}}{R_{2}'}>\frac{R_{1}}{R_{2}}\), which means the balance point \(l_{2}\) must shift towards the right (increase).
New balance point \(l_{2} = l_{1} + 22.5 = 40 + 22.5 = 62.5 \text{ cm}\).
Applying the balance condition again:
\[ \frac{2}{R_{2}'} = \frac{62.5}{100 - 62.5} = \frac{62.5}{37.5} \]
\[ \frac{2}{R_{2}'} = \frac{5}{3} \implies R_{2}' = \frac{6}{5} = 1.2 \Omega \]
Now, equate this to the parallel combination formula:
\[ 1.2 = \frac{3x}{3 + x} \]
\[ 1.2(3 + x) = 3x \implies 3.6 + 1.2x = 3x \]
\[ 3.6 = 1.8x \implies x = \frac{3.6}{1.8} = 2 \Omega \]
Step 4: Final Answer:
The value of \(x\) is \(2\).