Q74 / 75 · JEE Main 2026 (24 January Shift 2) · Physics
In a meter bridge experiment to determine the value of unknown resistance, first the resistances $2 \Omega$ and $3 \Omega$ are connected in the left and right gaps of the bridge and the null point is obtained at a distance $l \mathrm{~cm}$ from the left. Now when an unknown resistance $x \Omega$ is connected in parallel to $3 \Omega$ resistance, the null point is shifted by 10 cm to the right of wire. The value of unknown resistance $x$ is $\_\_\_\_$ $\Omega$.
Correct answer: 6
Solution
Meter bridge null point condition: $\frac{R_1}{R_2} = \frac{l}{100-l}$
Initial: $R_1 = 2Ω$, $R_2 = 3Ω$, null at $l$ cm
$\frac{2}{3} = \frac{l}{100-l}$ gives $2(100-l) = 3l$, so $l = 40$ cm
When $xΩ$ connected in parallel with $3Ω$: $R_2' = \frac{3x}{3+x}Ω$
New null point at $(40+10) = 50$ cm from left:
$\frac{2}{R_2'} = \frac{50}{50} = 1$
$\frac{2(3+x)}{3x} = 1$ gives $6 + 2x = 3x$
$x = 6Ω$
Initial: $R_1 = 2Ω$, $R_2 = 3Ω$, null at $l$ cm
$\frac{2}{3} = \frac{l}{100-l}$ gives $2(100-l) = 3l$, so $l = 40$ cm
When $xΩ$ connected in parallel with $3Ω$: $R_2' = \frac{3x}{3+x}Ω$
New null point at $(40+10) = 50$ cm from left:
$\frac{2}{R_2'} = \frac{50}{50} = 1$
$\frac{2(3+x)}{3x} = 1$ gives $6 + 2x = 3x$
$x = 6Ω$
Hence, the correct answer is 6.