Question

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 \(l\) cm from the left end. Now, when an unknown resistance \(x\,\Omega\) is connected in parallel to \(3\,\Omega\), the null point is shifted by \(10\,\text{cm}\) to the right. The value of \(x\) is ________ \(\Omega\).

Hint:

Start by writing the balance condition for the first setup, $\dfrac{2}{3} = \dfrac{l}{100-l}$, and solve it to fix the value of l first. Then work out the new balance length after the shift, and use the new balance condition to write an equation connecting x with the combined resistance of 3 ohm and x in parallel. Remember that connecting x in parallel with the 3 ohm resistor always makes the combined resistance smaller than 3 ohms, which tells you which direction the null point should move.

Answer 1

Correct Answer: 6

Step 1: Write balance condition for meter bridge.
\[ \frac{2}{3} = \frac{l}{100 - l} \Rightarrow l = 40\,\text{cm} \]
Step 2: New null point position.
Shift is \(10\,\text{cm}\) to the right, so \[ l' = 50\,\text{cm} \]
Step 3: New resistance in right gap.
\[ R = \frac{3x}{3 + x} \]
Step 4: Apply balance condition again.
\[ \frac{2}{R} = \frac{50}{50} = 1 \Rightarrow R = 2 \]
Step 5: Solve for \(x\).
\[ \frac{3x}{3 + x} = 2 \Rightarrow 3x = 6 + 2x \Rightarrow x = 6 \]

Answer 2

Concept:

• When the null point of a meter bridge lands exactly at the 50 cm mark, the two gaps must have equal resistance — you don't need to write the general balance equation a second time.

Step 1: Find the first null point $l$.
$\dfrac{2}{3} = \dfrac{l}{100-l} \Rightarrow 200 - 2l = 3l \Rightarrow l = 40\ \text{cm}$

Step 2: Find the new null point after the shift.
The null point shifts 10 cm to the right: $l' = 40 + 10 = 50\ \text{cm}$.

Step 3: Use the midpoint shortcut.
Since $l' = 50\ \text{cm}$ is exactly the midpoint, the left and right gap resistances must now be equal — no need to write $\dfrac{2}{R} = \dfrac{l'}{100-l'}$ in general form.
So the new right-gap resistance $R = 2\ \Omega$ directly.

Step 4: Set up the parallel combination equation.
$3\ \Omega$ in parallel with $x$ gives $R$: $\dfrac{3x}{3+x} = 2$

Step 5: Solve for $x$.
$3x = 2(3+x) \Rightarrow 3x = 6 + 2x \Rightarrow x = 6$

Final Answer: $x = 6\ \Omega$