JEE Main 2024 — Current Electricity Question with Solution

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 $2\Omega$, $R_2$ is the unknown resistance, $L_1$ is the balance length $40cm$ 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 ($R_2^{\prime }$) can be calculated using the parallel resistance formula:

$$\frac{1}{R_2^{\prime }}=\frac{1}{R_2}+\frac{1}{2}$$

$$\Rightarrow \frac{1}{R_2^{\prime }}=\frac{1}{3}+\frac{1}{2}=\frac{2+3}{6}=\frac{5}{6}$$

Therefore, the new combined resistance ($R_2^{\prime }$) is:

$$R_2^{\prime }=\frac{6}{5}\Omega$$

If $L_1^{\prime }$ is the new balance length and $L_2^{\prime }$ is the remaining length, we now have:

$$\frac{2}{\frac{6}{5}}=\frac{L_1^{\prime }}{100-L_1^{\prime }}$$

$$\Rightarrow \frac{L_1^{\prime }}{100-L_1^{\prime }}=\frac{5}{3}$$

Now, solve for (L'_1):

$$3L_1^{\prime }=5(100-L_1^{\prime })$$

$$\Rightarrow 3L_1^{\prime }=500-5L_1^{\prime }$$

$$\Rightarrow 8L_1^{\prime }=500$$

$$\Rightarrow L_1^{\prime }=62.5\mathrm{cm}$$

The balance length has changed from 40 cm to 62.5 cm, so the change by $L_1^{\prime }-L_1$ is:

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

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