In a meter bridge experiment, the balance poi | vedclass

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(C) In a meter bridge, the balance condition is given by $\frac{P}{Q} = \frac{l}{100-l}$, where $P$ is the left resistance and $Q$ is the right resist

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$45$

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(C) In a meter bridge, the balance condition is given by $\frac{P}{Q} = \frac{l}{100-l}$, where $P$ is the left resistance and $Q$ is the right resistance.Initially, $\frac{15}{R} = \frac{l_1}{100-l_1}$  --- $(1)$When $R$ is shunted with an equal resistance $R$, the equivalent resistance in the right gap becomes $R' = \frac{R 	imes R}{R+R} = \frac{R}{2}$.The new balance point is at $1.6 l_1$. Substituting this into the balance condition:$\frac{15}{R/2} = \frac{1.6 l_1}{100 - 1.6 l_1} \implies \frac{30}{R} = \frac{1.6 l_1}{100 - 1.6 l_1}$ --- $(2)$From $(1)$, $\frac{1}{R} = \frac{l_1}{15(100-l_1)}$. Substituting this into $(2)$:$30 	imes \frac{l_1}{15(100-l_1)} = \frac{1.6 l_1}{100 - 1.6 l_1}$$2 	imes \frac{l_1}{100-l_1} = \frac{1.6 l_1}{100 - 1.6 l_1}$Dividing by $l_1$ (assuming $l_1 
eq 0$):$\frac{2}{100-l_1} = \frac{1.6}{100-1.6 l_1}$$2(100 - 1.6 l_1) = 1.6(100 - l_1)$$200 - 3.2 l_1 = 160 - 1.6 l_1$$40 = 1.6 l_1 \implies l_1 = \frac{40}{1.6} = 25 	ext{ cm}$.Substituting $l_1 = 25$ into $(1)$:$\frac{15}{R} = \frac{25}{100-25} = \frac{25}{75} = \frac{1}{3}$$R = 15 	imes 3 = 45 	ext{ } \Omega$.

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