JEE Mains · Physics · STD 12 - Current electricity
- JEE Mains 2026
- Medium
- Numerical
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. Now when an unknown resistance \(x \Omega\) is connected in parallel to \(3 \Omega\) resistance, the null point is shifted by 10 cm to the right of wire. The value of unknown resistance \(x\) is _________ \(\Omega\).
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) 6
Step-by-step Solution
Detailed explanation
In case I
\(\frac{2}{3}=\frac{\ell}{(100-\ell)} \ldots \ldots(1)\)
\(\ell=40 cm\)
In case II
\(\frac{2}{ R }=\frac{\ell+10}{100-(\ell+10)}\)
Put \(\ell=40 cm \&\) solve
\(R =2 \Omega\)
\(\therefore \frac{3 x}{3+x}=2\)
\(x=6 \Omega\)
\(\frac{2}{3}=\frac{\ell}{(100-\ell)} \ldots \ldots(1)\)
\(\ell=40 cm\)
In case II
\(\frac{2}{ R }=\frac{\ell+10}{100-(\ell+10)}\)
Put \(\ell=40 cm \&\) solve
\(R =2 \Omega\)
\(\therefore \frac{3 x}{3+x}=2\)
\(x=6 \Omega\)
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