In a meter bridge (Figure), the null point is found at a distance of $33.7 \; cm$ from $A$. If now a resistance of $12 \; \Omega$ is connected in parallel with $S$, the null point occurs at $51.9 \; cm$. Determine the values of $R$ and $S$.

(N/A) From the first balance point, we get:
$\frac{R}{S} = \frac{33.7}{100 - 33.7} = \frac{33.7}{66.3} \dots (i)$
After $S$ is connected in parallel with a resistance of $12 \; \Omega$, the equivalent resistance $S_{eq}$ is:
$S_{eq} = \frac{12S}{S + 12}$
The new balance condition gives:
$\frac{R}{S_{eq}} = \frac{51.9}{100 - 51.9} = \frac{51.9}{48.1}$
$\frac{R(S + 12)}{12S} = \frac{51.9}{48.1} \dots (ii)$
Substituting $\frac{R}{S} = \frac{33.7}{66.3}$ from $(i)$ into $(ii)$:
$\frac{33.7}{66.3} \cdot \frac{S + 12}{12} = \frac{51.9}{48.1}$
$\frac{S + 12}{S} = \frac{51.9}{48.1} \cdot \frac{66.3}{33.7} \cdot 12 \approx 2.15$
Solving for $S$, we get $S \approx 13.5 \; \Omega$.
Substituting $S$ back into $(i)$:
$R = S \cdot \frac{33.7}{66.3} = 13.5 \cdot 0.508 \approx 6.86 \; \Omega$.