Question

In the figure shown for gives of $R_1$ and $R_2$ the balance point for jockey is at $40\text{cm}$ from $A$ when $R_2$ is shunted by a resistance of $10\Omega$, balance point shifts to $50\text{cm from}A,$. Then $R_1,R_2$ are

• A. $5\Omega ;\frac{15}{2}\Omega$
• B. $20\Omega ;30\Omega$
• C. $10\Omega ;15\Omega$
• D. $\frac{10}{3}\Omega ;5\Omega$

Answer 1

The correct option is D $\frac{10}{3}\Omega ;5\Omega$

$\frac{R_1}{R_2}=\frac{l}{100-l}\Rightarrow \frac{R_1}{R_2}=\frac{40}{60}=\frac{2}{3};\Rightarrow R_1=\frac{2}{3}{R}_2$

When $R_2$ is shunted by $10\Omega$, the balance point shifts to $50\text{cm}$. i.e

$\frac{R_1}{\left(\frac{R_2\left(10\right)}{R_2+10}\right)}=\frac{50}{100-50}=1$

$\Rightarrow R_1=\frac{10R_2}{R_2+10}\Rightarrow \frac{2}{3}{R}_2=\frac{10R_2}{R_2+10}$

$\Rightarrow R_2+10=15;\Rightarrow R_2=5\Omega$

$\therefore R_1=\frac{2}{3}{R}_2=\frac{2}{3}\times 5=\frac{10}{3}\Omega$