In a meter bridge,the null point is found at | vedclass

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(A) In a meter bridge,the balance condition is given by $\frac{R}{S} = \frac{\ell}{100-\ell}$,where $\ell$ is the distance of the null point from end

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(A) In a meter bridge,the balance condition is given by $\frac{R}{S} = \frac{\ell}{100-\ell}$,where $\ell$ is the distance of the null point from end $A$.Case $1$: Null point is at $\ell = 25\, cm$.$\frac{R}{S} = \frac{25}{100-25} = \frac{25}{75} = \frac{1}{3}$$S = 3R$  .......$(1)$Case $2$: $A$ resistance of $10\,\Omega$ is connected in parallel with $S$. The new resistance $S'$ is given by $\frac{1}{S'} = \frac{1}{S} + \frac{1}{10}$,so $S' = \frac{10S}{S+10}$.The null point is now at the midpoint,i.e.,$\ell = 50\, cm$.$\frac{R}{S'} = \frac{50}{100-50} = 1$$R = S' = \frac{10S}{S+10}$ .......$(2)$Substitute $S = 3R$ from $(1)$ into $(2)$:$R = \frac{10(3R)}{3R+10}$$R(3R+10) = 30R$Since $R 
eq 0$,divide by $R$:$3R + 10 = 30$$3R = 20$$R = \frac{20}{3} \approx 6.67\,\Omega$.

In a meter bridge,the null point is found at a distance of $25\, cm$ from $A$. If now a resistance of $10\,\Omega$ is connected in parallel with $S$,the null point occurs at the midpoint of $AB$. The value of $R$ is .............. $\Omega$. (in $.67$)

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