QJEEM24Ph4e50: JEE Main 2024 Physics - Class 12 JEEM24Ph4e50 Solution: In a metre-bridge when a resistance in the left gap is and…

To solve this problem, let's first understand that a meter bridge setup is based on the principle of a Wheatstone bridge, in which two unknown resistances are in such a configuration that if the bridge is balanced, the ratio of the resistances on one side is equal to the ratio of resistances on the other side. In a balanced condition, no current flows through the galvanometer that is connected diagonally across the bridge.

We can express the condition of the balance as follows:

$\frac{R_1}{R_2} = \frac{L_1}{L_2}$

where $R_1$ is the known resistance $2 Ω$ , $R_2$ is the unknown resistance, $L_1$ is the balance length $40 cm$ and $L_2$ is the length of the remaining wire on the meter bridge (100 cm - 40 cm = 60 cm).

Let's calculate the initial unknown resistance ( $R_2$ ) using the balance condition:

$\frac{2}{R_2} = \frac{40}{60}$

$\Rightarrow R_2 = \frac{2 \times 60}{40} = 3 \Omega$

Now, when the unknown resistance $R_2$ is shunted with a 2 Ω resistor, the new combined resistance ( $R'_2$ ) can be calculated using the parallel resistance formula:

$\frac{1}{R'_2} = \frac{1}{R_2} + \frac{1}{2}$

$\Rightarrow \frac{1}{R'_2} = \frac{1}{3} + \frac{1}{2} = \frac{2 + 3}{6} = \frac{5}{6}$

Therefore, the new combined resistance ( $R'_2$ ) is:

$R'_2 = \frac{6}{5} \Omega$

If $L'_1$ is the new balance length and $L'_2$ is the remaining length, we now have:

$\frac{2}{\frac{6}{5}} = \frac{L'_1}{100 - L'_1}$

$\Rightarrow \frac{L'_1}{100 - L'_1} = \frac{5}{3}$

Now, solve for (L'_1):

$3L'_1 = 5(100 - L'_1)$

$\Rightarrow 3L'_1 = 500 - 5L'_1$

$\Rightarrow 8L'_1 = 500$

$\Rightarrow L'_1 = 62.5 \mathrm{~cm}$

The balance length has changed from 40 cm to 62.5 cm, so the change by $L'_1 - L_1$ is:

$62.5 - 40 = 22.5 \mathrm{~cm}$

Therefore, the balance length changes by 22.5 cm, which corresponds to Option B.