MHT CET · Physics · Current Electricity
A resistance of \(20 \Omega\) is connected in the left gap of a metre bridge and an unknown resistance greater than \(20 \Omega\) is connected in the right gap. When these resistances are interchanged, the balance point shift by \(20 \mathrm{~cm}\). The unknown resistance is
- A \(25 \Omega\)
- B \(40 \Omega\)
- C \(35 \Omega\)
- D \(30 \Omega\)
Answer & Solution
Correct Answer
(D) \(30 \Omega\)
Step-by-step Solution
Detailed explanation
Let the unknown resistance be \(R\) and the balance point be at \(l \mathrm{~cm}\) initially.
using, \(\frac{R_2}{R_1}=\frac{l}{100-l}\)
Case 1: \(R_1=20 \Omega\) and \(R_2=R\)
Using equation (1) and (2) we get,
\(\frac{l}{100-l}=\frac{80-l}{l+20}\)
\(\Rightarrow l=40 \mathrm{~cm}\)
From equation (1),
\(\frac{20}{R}=\frac{40}{100-40}\)
\(\Rightarrow R=30 \Omega\)
using, \(\frac{R_2}{R_1}=\frac{l}{100-l}\)
Case 1: \(R_1=20 \Omega\) and \(R_2=R\)
Using equation (1) and (2) we get,
\(\frac{l}{100-l}=\frac{80-l}{l+20}\)
\(\Rightarrow l=40 \mathrm{~cm}\)
From equation (1),
\(\frac{20}{R}=\frac{40}{100-40}\)
\(\Rightarrow R=30 \Omega\)
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