Question

When an unknown resistance $X$ is connected in the left gap of a meter bridge and a known resistance $R$ in the right gap, null point is obtained at $40\ \text{cm}$ from the left end. If a $2\ \Omega$ resistance is connected in series with $X$, the null point shifts towards right by $10\ \text{cm}$, with same resistance in right gap. The value of $X$ must be

• A. $1\ \Omega$
• B. $4\ \Omega$
• C. $2\ \Omega$
• D. $3\ \Omega$

Hint:

In meter bridge problems, write both balance conditions and solve them simultaneously to find the unknown resistance.

Answer 1

The Correct Option is B

Step 1: Write meter bridge balance condition.
For a meter bridge at balance condition: \[ \dfrac{X}{R} = \dfrac{l_1}{100 - l_1} \] Step 2: Use given initial balance length.
Given $l_1 = 40\ \text{cm}$, \[ \dfrac{X}{R} = \dfrac{40}{60} = \dfrac{2}{3} \] Step 3: New balance condition after adding resistance.
After adding $2\ \Omega$ in series, new resistance is $(X + 2)$ and new balance point is $50\ \text{cm}$. \[ \dfrac{X + 2}{R} = \dfrac{50}{50} = 1 \] Step 4: Solve the equations.
From Step 2, \[ R = \dfrac{3X}{2} \] Substitute in Step 3: \[ X + 2 = \dfrac{3X}{2} \] Step 5: Final calculation.
\[ 2X + 4 = 3X \Rightarrow X = 4\ \Omega \]