Question

In meter bridge experiment, two resistances X and Y in the two gaps give a null point dividing the wire in the ratio \(2:3\). When each resistance is increased by \(30\,\Omega\), the null point divides the wire in the ratio \(5:6\). The resistance X and Y are respectively

• A. \(20\,\Omega\) , \(30\,\Omega\)
• B. \(22\,\Omega\) , \(33\,\Omega\)
• C. \(32\,\Omega\) , \(48\,\Omega\)
• D. \(40\,\Omega\) , \(60\,\Omega\)

Hint:

Balance condition: X/Y = l/(100 - l), equivalent to the ratio of the wire parts.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
In a meter bridge the null point divides the wire in the ratio of the resistances in the two gaps. Here the null point divides the wire in the ratio \(2:3\) in the first case, and \(5:6\) when each resistance is increased by 30 \(\Omega\).

Step 2: First balance:
\(\frac XY = \frac23\), so let \(X = 2k\) and \(Y = 3k\).

Step 3: Second balance:
\(\frac{X + 30}{Y + 30} = \frac56\):
\[ 6(2k + 30) = 5(3k + 30) \Rightarrow 12k + 180 = 15k + 150 \Rightarrow k = 10 \]

Step 4: Resistances:
\(X = 20\ \Omega\) and \(Y = 30\ \Omega\).

Step 5: Why the other options are wrong.
The pairs 22 and 33, 32 and 48, and 40 and 60 do keep the ratio 2:3, but adding 30 gives ratios \(\frac{52}{63}\), \(\frac{62}{78}\) and \(\frac{70}{90}\), none equal to \(\frac56\).

Final Answer:
X = 20 ohm and Y = 30 ohm, option (A). \[ \boxed{X=20\,\Omega,\ Y=30\,\Omega} \]