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(B) In the first case: $R = 12 \ \Omega$ and $S = 36 \ \Omega$. In the balanced condition,$\frac{R}{S} = \frac{l_1}{100 - l_1}$. Substituting the valu

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$50 \ cm$ towards right

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(B) In the first case: $R = 12 \ \Omega$ and $S = 36 \ \Omega$. In the balanced condition,$\frac{R}{S} = \frac{l_1}{100 - l_1}$. Substituting the values: $\frac{12}{36} = \frac{l_1}{100 - l_1} \Rightarrow \frac{1}{3} = \frac{l_1}{100 - l_1}$. $100 - l_1 = 3l_1 \Rightarrow 4l_1 = 100 \Rightarrow l_1 = 25 \ cm$. In the second case,the resistances are interchanged: $R = 36 \ \Omega$ and $S = 12 \ \Omega$. In the balanced condition: $\frac{36}{12} = \frac{l_2}{100 - l_2} \Rightarrow 3 = \frac{l_2}{100 - l_2}$. $300 - 3l_2 = l_2 \Rightarrow 4l_2 = 300 \Rightarrow l_2 = 75 \ cm$. The shift in the balance point is $\Delta l = l_2 - l_1 = 75 \ cm - 25 \ cm = 50 \ cm$. Since $l_2 > l_1$,the balance point shifts $50 \ cm$ towards the right.

The resistances in the left and right gaps of a balanced metre bridge are $12 \Omega$ and $36 \Omega$ respectively. If the resistances are interchanged,the balance point shifts by:

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