Question

In a meter bridge experiment, the balance point is obtained at length \(\ell_1\) cm from left end when resistances in the left gap and right gap are \(5\Omega\) and \(R\Omega\) respectively. When the resistance \(R\) is shunted with equal resistance, the new balance point is at \(1.6\ell_1\). The resistance \(R\) in ohm is

• A. 25
• B. 15
• C. 10
• D. 20

Hint:

In meter bridge problems, balance condition is \(\frac{P}{Q} = \frac{\ell}{100-\ell}\). Shunting reduces resistance. Setting up ratios avoids solving for \(\ell\) explicitly if not needed, but here solving is straightforward.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
Meter bridge works on Wheatstone bridge principle. Balance condition: \(\frac{P}{Q} = \frac{\ell}{100-\ell}\). Here left gap \(P = 5\Omega\), right gap \(Q = R\). First balance at \(\ell_1\). Then \(R\) is shunted with equal \(R\) → parallel combination gives \(R/2\). New balance at \(1.6\ell_1\). Find \(R\).

Step 2: Key Formula or Approach:
First condition: \(\frac{5}{R} = \frac{\ell_1}{100-\ell_1}\).
Second condition: \(\frac{5}{R/2} = \frac{1.6\ell_1}{100-1.6\ell_1}\).
Simplify second: \(\frac{10}{R} = \frac{1.6\ell_1}{100-1.6\ell_1}\).
Divide second equation by first to eliminate \(R\) and \(\ell_1\).

Step 3: Detailed Explanation:
Let \(x = \ell_1\). From first: \(\frac{5}{R} = \frac{x}{100-x}\) ⇒ \(R = \frac{5(100-x)}{x}\).
From second: \(\frac{10}{R} = \frac{1.6x}{100-1.6x}\). Substitute \(R\) from first into second? Easier: divide second eqn by first eqn: \[ \frac{10/R}{5/R} = \frac{ \frac{1.6x}{100-1.6x} }{ \frac{x}{100-x} } \Rightarrow 2 = \frac{1.6x}{100-1.6x} \cdot \frac{100-x}{x}. \] Cancel \(x\): \[ 2 = 1.6 \cdot \frac{100-x}{100-1.6x}. \] Multiply both sides by \((100-1.6x)\): \[ 2(100-1.6x) = 1.6(100-x) \Rightarrow 200 - 3.2x = 160 - 1.6x. \] Bring terms: \(200 - 160 = 3.2x - 1.6x \Rightarrow 40 = 1.6x \Rightarrow x = 25\ \text{cm}\). Then \(R = \frac{5(100-25)}{25} = \frac{5 \times 75}{25} = 5 \times 3 = 15\ \Omega\).

Step 4: Final Answer:
The resistance \(R\) is \(15\ \Omega\), option (B).