JEE Advanced · Physics · 19. Current Electricity
A resistance of \(2 \Omega\) is connected across one gap of a meter-bridge (the length of the wire is \(100 \mathrm{~cm}\) ) and an unknown resistance, greater than \(2 \Omega\), is connected across the other gap. When these resistances are interchanged, the balance point shifts by \(20 \mathrm{~cm}\). Neglecting any corrections, the unknown resistance is
- A \(3 \Omega\)
- B \(4 \Omega\)
- C \(5 \Omega\)
- D \(6 \Omega\)
Answer & Solution
Correct Answer
(A) \(3 \Omega\)
Step-by-step Solution
Detailed explanation
\(R >2 \Omega\)
\(\therefore 100-x >x\)
\(\text {Applying }\frac{P}{Q}=\frac{R}{S}\)


We have
\(
\begin{aligned}
& \frac{2}{R}=\frac{x}{100-x} \\
& \frac{R}{2}=\frac{x+20}{80-x}
\end{aligned}
\)
Solving Eqs. (i) and (ii), we get
\(
R=3 \Omega
\)
\(\therefore\) Answer is (a).
\(\therefore 100-x >x\)
\(\text {Applying }\frac{P}{Q}=\frac{R}{S}\)


We have
\(
\begin{aligned}
& \frac{2}{R}=\frac{x}{100-x} \\
& \frac{R}{2}=\frac{x+20}{80-x}
\end{aligned}
\)
Solving Eqs. (i) and (ii), we get
\(
R=3 \Omega
\)
\(\therefore\) Answer is (a).
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