In a meter bridge, as shown in the figure, it is given that resistance Y = 12.5\,\,\Omega and that the balance is obtained at a distance 39.5cm from end A (by Jockey J). After interchanging the resistances X and Y , a new balance point is found at a distance l_2 from end A. What are the value of X and l_2 ?

Current Electricity PYQ — JEE Main 2018 · 15 Apr · Shift 1 · Q49 with Solution

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Correct answer: A

For a meter bridge at balance point,

$\frac{X}{Y} = \frac{l}{100-l}$

where $l$ is the balancing length from end $A$ .

Given:

So,

$\frac{X}{12.5} = \frac{39.5}{100-39.5} = \frac{39.5}{60.5}$

Hence,

$X = 12.5\times \frac{39.5}{60.5}$

$X = 12.5\times 0.65289 \approx 8.16\,\Omega$

So,

$X = 8.16\,\Omega$

Now the balance condition becomes

$\frac{Y}{X} = \frac{l_2}{100-l_2}$

But interchanging the resistances in a meter bridge makes the new balancing length complementary:

$l_2 = 100 - l_1 = 100 - 39.5 = 60.5\,\text{cm}$

So,

$l_2 = 60.5\,\text{cm}$

We found:

This matches Option A.

Stored correct answer: A

Our derived answer is also A, so they agree.