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:

Parallel combinations always reduce equivalent resistance.

Answer 1

Correct Answer: 6

In the meter bridge experiment, we start by using the formula for the Wheatstone bridge balance: \(\frac{R_1}{R_2}=\frac{l}{100-l}\), where \(R_1=2\,\Omega\) and \(R_2=3\,\Omega\). Therefore:
\( \frac{2}{3}=\frac{l}{100-l} \)
Cross-multiplying gives:
\(200-2l=3l\)
This simplifies to:
\(5l=200\)
Thus, \(l=40\,\text{cm}\).

Now, when \(x\) is connected in parallel with \(3\,\Omega\), the equivalent resistance \(R_2'\) is:

\(R_2'=\frac{3x}{3+x}\, \Omega\)
The new null point shifts 10 cm to the right, so \(l=50\,\text{cm}\).
The new balance equation is:
\( \frac{R_1}{R_2'}=\frac{50}{50}\Rightarrow R_1=R_2'\)
Therefore:
\(2=\frac{3x}{3+x}\)
Cross-multiplying yields:
\(2(3+x)=3x\)
Expanding gives:
\(6+2x=3x\)
Thus:
\(x=6\,\Omega\)
Verification confirms \(x\) is within the range \(6,6\). Therefore, the value of the unknown resistance \(x\) is 6 \(\Omega\).