MHT CET · Physics · Current Electricity
The resistances in the left and right gap of a balanced metre bridge are \(12 \Omega\) and \(36 \Omega\) respectively If the resistances are interchanged the balance point shifts by
- A \(25 \mathrm{~cm}\) towards right,
- B \(50 \mathrm{~cm}\) towards right,
- C \(25 \mathrm{~cm}\) towards left,
- D \(50 \mathrm{~cm}\) towards left,
Answer & Solution
Correct Answer
(B) \(50 \mathrm{~cm}\) towards right,
Step-by-step Solution
Detailed explanation
In the first case: \(R=12 \Omega\) and \(S=36 \Omega\)
In balanced condition, \(\frac{R}{S}=\frac{1}{100-l}\) or \(\frac{12}{36}=\frac{l}{100-l}\) or \(3 l=100-l\) or \(4 l=100 \Rightarrow l=25 \mathrm{~cm}\) Therefore, first balance point is \(l_1=25 \mathrm{~cm}\)
In the second case: \(R=36 \Omega\) and \(S=12 \Omega\)
In balanced condition, \(\frac{R}{S}=\frac{l}{100-l}\) or \(\frac{36}{12}=\frac{1}{100-1}\) or \(l=300-3 l\) or \(4 l=300 \Rightarrow l=l_2=75 \mathrm{~cm}\)
Therefore, second balance point is \(l_2=75 \mathrm{~cm}\)
The shift \(=l_2-l_1=(75-25) \mathrm{cm}=50 \mathrm{~cm}\)
In balanced condition, \(\frac{R}{S}=\frac{1}{100-l}\) or \(\frac{12}{36}=\frac{l}{100-l}\) or \(3 l=100-l\) or \(4 l=100 \Rightarrow l=25 \mathrm{~cm}\) Therefore, first balance point is \(l_1=25 \mathrm{~cm}\)
In the second case: \(R=36 \Omega\) and \(S=12 \Omega\)
In balanced condition, \(\frac{R}{S}=\frac{l}{100-l}\) or \(\frac{36}{12}=\frac{1}{100-1}\) or \(l=300-3 l\) or \(4 l=300 \Rightarrow l=l_2=75 \mathrm{~cm}\)
Therefore, second balance point is \(l_2=75 \mathrm{~cm}\)
The shift \(=l_2-l_1=(75-25) \mathrm{cm}=50 \mathrm{~cm}\)
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