In a meter bridge,the null point is located a | vedclass

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(C) In a meter bridge,the balance condition is given by $\frac{R}{S} = \frac{l}{100-l}$.Given $l = 20 \ cm$,so $\frac{R}{S} = \frac{20}{80} = \frac{1}

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$5 \ \Omega, 20 \ \Omega$

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(C) In a meter bridge,the balance condition is given by $\frac{R}{S} = \frac{l}{100-l}$.Given $l = 20 \ cm$,so $\frac{R}{S} = \frac{20}{80} = \frac{1}{4}$,which implies $S = 4R$.When $S$ is shunted with $60 \ \Omega$,the new resistance $S'$ is $\frac{S 	imes 60}{S + 60}$.The null point shifts by $5 \ cm$. Since $S$ decreases,the null point shifts towards the left,so $l' = 20 - 5 = 15 \ cm$.The new balance condition is $\frac{R}{S'} = \frac{15}{85} = \frac{3}{17}$.Substituting $S' = \frac{60S}{S+60}$,we get $\frac{R(S+60)}{60S} = \frac{3}{17}$.Since $S = 4R$,we substitute $R = \frac{S}{4}$ into the equation: $\frac{(S/4)(S+60)}{60S} = \frac{3}{17} \Rightarrow \frac{S+60}{240} = \frac{3}{17}$.$17S + 1020 = 720$,which gives $17S = -300$ (This implies the shift is towards the right,$l' = 25 \ cm$).Let's re-evaluate with $l' = 25 \ cm$: $\frac{R}{S'} = \frac{25}{75} = \frac{1}{3} \Rightarrow 3R = S' = \frac{60S}{S+60}$.Substituting $S = 4R$: $3R = \frac{60(4R)}{4R+60} \Rightarrow 3 = \frac{240}{4R+60} \Rightarrow 12R + 180 = 240 \Rightarrow 12R = 60 \Rightarrow R = 5 \ \Omega$.Then $S = 4 	imes 5 = 20 \ \Omega$.

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