In a meter bridge,the resistances R and S are | vedclass

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(A) Case $I$: For a balanced meter bridge,the ratio of resistances is equal to the ratio of the lengths of the wire segments.$\frac{R}{S} = \frac{40}{

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$83.33 \Omega$ and $125 \Omega$

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(A) Case $I$: For a balanced meter bridge,the ratio of resistances is equal to the ratio of the lengths of the wire segments.$\frac{R}{S} = \frac{40}{100-40} = \frac{40}{60} = \frac{2}{3}$$S = \frac{3R}{2} \quad \dots (i)$Case $II$: When a resistance of $10 \Omega$ is connected in parallel with $S$,the new resistance $S'$ is given by:$S' = \frac{10S}{10+S}$The new null point is at $90 	ext{ cm}$.$\frac{R}{S'} = \frac{90}{100-90} = \frac{90}{10} = 9$$\frac{R(10+S)}{10S} = 9 \Rightarrow R(10+S) = 90S \quad \dots (ii)$Substitute $S = \frac{3R}{2}$ from Eq. $(i)$ into Eq. $(ii)$:$R(10 + \frac{3R}{2}) = 90(\frac{3R}{2})$$10 + \frac{3R}{2} = 135$$\frac{3R}{2} = 125$$R = \frac{250}{3} \approx 83.33 \Omega$Now,find $S$ using Eq. $(i)$:$S = \frac{3}{2} 	imes \frac{250}{3} = 125 \Omega$Thus,$R = 83.33 \Omega$ and $S = 125 \Omega$.

$S.No.$	$R$	$l$	$100-l$	$S = \left( \frac{100-l}{l} \right)R$
$1$	$20\,\Omega$	$43$	$57$	$26.51\,\Omega$
$2$	$30\,\Omega$	$51$	$49$	$28.82\,\Omega$
$3$	$40\,\Omega$	$59$	$41$	$27.80\,\Omega$
$4$	$60\,\Omega$	$70$	$30$	$25.71\,\Omega$

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