In a meter bridge experiment,a resistance of | vedclass

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(C) In a meter bridge experiment,let the balancing length be $l$ cm from the left end.According to the principle of the meter bridge,for the first cas

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$11$

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(C) In a meter bridge experiment,let the balancing length be $l$ cm from the left end.According to the principle of the meter bridge,for the first case: $\frac{9}{x} = \frac{l}{100 - l}$ ...$(i)$When the resistances are interchanged,the new balancing length becomes $l + 10$ cm (since $x > 9$,the balancing point shifts towards the higher resistance side).For the second case: $\frac{x}{9} = \frac{l + 10}{100 - (l + 10)} = \frac{l + 10}{90 - l}$ ...(ii)Multiplying equation $(i)$ and (ii),we get: $\frac{9}{x} 	imes \frac{x}{9} = \frac{l}{100 - l} 	imes \frac{l + 10}{90 - l} = 1$$l(l + 10) = (100 - l)(90 - l)$$l^2 + 10l = 9000 - 100l - 90l + l^2$$10l = 9000 - 190l$$200l = 9000 \Rightarrow l = 45 	ext{ cm}$.Substituting $l = 45$ in equation $(i)$:$\frac{9}{x} = \frac{45}{100 - 45} = \frac{45}{55} = \frac{9}{11}$Therefore,$x = 11 \Omega$.

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