Question

In a balanced metre bridge, \(5\,\Omega\) is connected in the left gap and \(R\,\Omega\) in the right gap. When \(R\,\Omega\) is shunted with an equal resistance, the new balance point is at \(1.6\,\ell_1\), where \(\ell_1\) is the earlier balancing length. The value of \(\ell_1\) is

• A. \(25\,\text{cm}\)
• B. \(40\,\text{cm}\)
• C. \(35\,\text{cm}\)
• D. \(30\,\text{cm}\)

Hint:

In metre bridge problems, shunting a resistance always reduces its effective value and shifts the balance point.

Answer 1

The Correct Option is A

Step 1: Write balance condition for metre bridge.
\[ \frac{5}{R} = \frac{\ell_1}{100 - \ell_1} \]
Step 2: New resistance after shunting.
When \(R\) is shunted with an equal resistance, effective resistance becomes:
\[ R' = \frac{R}{2} \]
Step 3: New balance condition.
\[ \frac{5}{R/2} = \frac{1.6\ell_1}{100 - 1.6\ell_1} \Rightarrow \frac{10}{R} = \frac{1.6\ell_1}{100 - 1.6\ell_1} \]
Step 4: Divide the two balance equations.
\[ \frac{10/R}{5/R} = \frac{\dfrac{1.6\ell_1}{100 - 1.6\ell_1}}{\dfrac{\ell_1}{100 - \ell_1}} \] \[ 2 = \frac{1.6(100 - \ell_1)}{100 - 1.6\ell_1} \]
Step 5: Solve for \(\ell_1\).
\[ 2(100 - 1.6\ell_1) = 1.6(100 - \ell_1) \] \[ 200 - 3.2\ell_1 = 160 - 1.6\ell_1 \] \[ 40 = 1.6\ell_1 \Rightarrow \ell_1 = 25\,\text{cm} \]
Step 6: Conclusion.
The initial balancing length is \(25\,\text{cm}\).