Question

Two resistors $2\,\Omega$ and $3\,\Omega$ are connected in the gaps of a bridge as shown in the figure. The null point is obtained with the contact of jockey at some point on wire $XY$. When an unknown resistor is connected in parallel with $3\,\Omega$ resistor, the null point is shifted by $22.5\,\text{cm}$ towards $Y$. The resistance of unknown resistor is ___ $\Omega$.

• A. $2$
• B. $3$
• C. $4$
• D. $1$

Hint:

In meter bridge problems, always calculate initial balance length before introducing any change.

Answer 1

The Correct Option is A

Step 1: Condition of balance in meter bridge.
At balance condition:
\[ \dfrac{R_1}{R_2} = \dfrac{l_1}{l_2} \] Step 2: Initial balance condition.
\[ \dfrac{2}{3} = \dfrac{l}{100 - l} \] Solving,
\[ l = 40\,\text{cm} \] Step 3: New balance length after shifting.
Null point shifts $22.5\,\text{cm}$ towards $Y$,
\[ l' = 40 + 22.5 = 62.5\,\text{cm} \] Step 4: New resistance in right gap.
\[ \dfrac{2}{R'} = \dfrac{62.5}{37.5} \Rightarrow R' = 1.2\,\Omega \] Step 5: Using parallel combination formula.
\[ \dfrac{1}{R'} = \dfrac{1}{3} + \dfrac{1}{X} \] \[ \dfrac{1}{1.2} = \dfrac{1}{3} + \dfrac{1}{X} \] \[ \dfrac{1}{X} = \dfrac{5}{6} - \dfrac{1}{3} = \dfrac{1}{2} \Rightarrow X = 2\,\Omega \] Step 6: Final conclusion.
The resistance of the unknown resistor is $2\,\Omega$, corresponding to option (1).