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JEE Advanced 2007Paper 1
Exam
JEE Advanced
Year
2007
Paper
Paper 1
Subject
Physics
Chapter
Current Electricity

JEE Advanced 2007 · Paper 1 · Physics · Current Electricity

A resistance of 2 is connected across one gap of a meter-bridge...

A resistance of 2Ω2 \Omega is connected across one gap of a meter-bridge (the length of the wire is 100 cm100 \mathrm{~cm} ) and an unknown resistance, greater than 2Ω2 \Omega, is connected across the other gap. When these resistances are interchanged, the balance point shifts by 20 cm20 \mathrm{~cm}. Neglecting any corrections, the unknown resistance isA resistance of 2 is connected across one gap of a meter-bridge (the length of the wire is 100 ~cm ) and an unknownA resistance of 2 is connected across one gap of a meter-bridge (the length of the wire is 100 ~cm ) and an unknownA resistance of 2 is connected across one gap of a meter-bridge (the length of the wire is 100 ~cm ) and an unknownA resistance of 2 is connected across one gap of a meter-bridge (the length of the wire is 100 ~cm ) and an unknown

Options

  1. A3Ω3 \Omega
  2. B4Ω4 \Omega
  3. C5Ω5 \Omega
  4. D6Ω6 \Omega

Correct answer

A. 3Ω3 \Omega

Step-by-step solution

R>2Ω∴100−x>x Applying PQ=RS\begin{aligned}& R & >2 \Omega \\ \therefore & 100-x & >x \\ \text{ Applying }& \frac{P}{Q}& =\frac{R}{S} \end{aligned}A resistance of 2 is connected across one gap of a meter-bridge (the length of the wire is 100 ~cm ) and an unknown\\A resistance of 2 is connected across one gap of a meter-bridge (the length of the wire is 100 ~cm ) and an unknownWehave\\ We have2R=x100−xR2=x+2080−x\begin{aligned}& \frac{2}{R}=\frac{x}{100-x}\\ & \frac{R}{2}=\frac{x+20}{80-x} \end{aligned}SolvingEqs.(i)Solving Eqs. (i) and (ii), we get R=3 \Omega∴\therefore Answer is (a).
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