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

To find the resistance of the unknown resistor connected in parallel with the \(3 \, \Omega\) resistor, we need to use Wheatstone bridge principles.

Initially, the bridge is balanced at the null point, so we have:

\(\frac{R_1}{R_2} = \frac{R_3}{R_4}\)

Where:

• \(R_1 = 2 \, \Omega\)
• \(R_2 = 3 \, \Omega\)
• \(R_3 = R_4\) (ratio in the wire at null point)

When the unknown resistor \(R_x\) is connected in parallel to the \(3 \, \Omega\) resistor, the effective resistance \(R_2'\) becomes:

\(R_2' = \frac{3R_x}{3 + R_x}\)

Now, the null point is shifted by \(22.5 \, \text{cm}\) towards \(Y\), denoting a change in balance.

For a balanced bridge with shifted null point, we have:

\(\frac{R_1}{R_2'} = \frac{R_3 + 22.5}{R_4 - 22.5}\)

Assuming the total length of the wire \(XY\) is \(L\), initially we had balance with:

\(\frac{2}{3} = \frac{\text{null point}}{L - \text{null point}}\)

With the shift, this becomes:

\(\frac{2}{R_2'} = \frac{\text{null point} + 22.5}{L - \text{null point} - 22.5}\)

Simplifying, since we know the initial values:

\(\frac{2}{3} = \frac{\text{null point}}{L - \text{null point}}\)

Then:

\(\frac{2}{\frac{3R_x}{3 + R_x}} = \frac{\text{null point} + 22.5}{L - \text{null point} - 22.5}\)

Rearranging and solving for \(R_x\):

Solving this equation, we get \(R_x = 2 \, \Omega\).

Thus, the resistance of the unknown resistor is \(2 \, \Omega\).