Question

In the figure shown, for the given values of $R_1\text{and}{R}_2$, the balance point for jockey is at $40\text{cm}$ from A. When $R_2$ is shunted by a resistance of $10\Omega$, the balance point shifts to $50\text{cm}$. If the length of AB is $1\text{m}$, the values of $R_1\text{and}{R}_2$ are

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

Answer 1

The correct option is C $\frac{10}{3}\Omega ,5\Omega$

For the situation given in question, the new balance is at $50\text{cm}$ from A, after shunting $R_2$ with $10\Omega$.
The effective esistance in the right gap, after shunting, is,
$R_{eq}=\frac{10R_2}{10+R_2}$

Applying the expression for balanced condition,

$\frac{R_1}{R_{eq}}=\frac{50}{50}$

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

$R_1=\frac{10R_2}{10+R_2}....\left(i\right)$

From initial condition without shunting,

$\frac{R_1}{R_2}=\frac{l_1}{l_2}$

$\Rightarrow \frac{R_1}{R_2}=\frac{40}{100-40}=\frac{4}{6}$

$R_1=\frac{2R_2}{3}$
Putting this value in $\left(i\right)$,

$\frac{2R_2}{3}=\frac{10R_2}{10+R_2}$

$\frac{2}{3}=\frac{10}{10+R_2}$

$20+2R_2=30$

$\therefore R_2=5\Omega \text{and}{R}_1=\frac{2R_2}{3}=\frac{10}{3}\Omega$

Hence, $\left(A\right)$ is the correct answer.