JEE Mains · Physics · STD 12 - Current electricity
- JEE Mains 2026
- Medium
- MCQ
Two resistors \(2 \Omega\) and \(3 \Omega\) are connected in the gaps of bridge as shown in 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 cm toward Y . The resistance of unknown resistor is _________ \(\Omega\).

Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) 2
Step-by-step Solution
Detailed explanation
Initially, \(\frac{2}{3}=\frac{x}{100-x}\)
\(\Rightarrow x =40 cm\)
Now when ' R ' connected in parallel
\(\frac{2}{\frac{3 R}{3+R}}=\frac{40+22.5}{60-22.5}=\frac{62.5}{37.5}\)
\(\therefore R =2 \Omega\)
\(\Rightarrow x =40 cm\)
Now when ' R ' connected in parallel
\(\frac{2}{\frac{3 R}{3+R}}=\frac{40+22.5}{60-22.5}=\frac{62.5}{37.5}\)
\(\therefore R =2 \Omega\)
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