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: Test the option:
All options keep $X:Y = 2:3$, so check the second condition on each.

Step 2: Option (A):
$\frac{20 + 30}{30 + 30} = \frac{50}{60} = \frac56$. It satisfies the condition.

Step 3: Others fail:
$\frac{52}{63}$, $\frac{62}{78}$, $\frac{70}{90}$ are not $\frac56$. So option (A) is the pair.

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