In a meter bridge experiment, the balance point is found at $40$ cm from the left end. If a resistance of $10$ $\Omega$ is connected in series with the left resistor, the balance point shifts to $60$ cm. What is the value of the original left resistance?

Solution

1. Let $R$ be the initial resistance on the left and $S$ be the resistance on the right; the initial balance condition is $$\begin{array}{rl}\frac{R}{S}& =\frac{40}{60}\\ & =\frac{2}{3}\end{array}$$, which implies $S=1.5R$.
2. After adding $10\Omega$ in series with $R$, the new balance condition is $$\begin{array}{rl}\frac{R+10}{S}& =\frac{60}{40}\\ & =\frac{3}{2}\end{array}$$.
3. Substituting $S=1.5R$ into the second equation gives $\frac{R+10}{1.5R}=1.5$, which simplifies to $R+10=2.25R$.
4. Solving for $R$ yields $1.25R = 10$, resulting in $R=8\Omega$.
5. Hence the answer is (A).