Question

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

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

Answer 1

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

Applying condition for balance whetstone bridge when resistances are $R_1$ and $R_2$

$\frac{R_1}{l}=\frac{R_2}{100-l}$

Here, $l=40\text{cm}$

$\frac{R_1}{40}=\frac{R_2}{60}$

$3R_1=2R_2...........\left(1\right)$

Since, $R_2$ is shunted by $10\Omega$ means both are in parallel combination, so their equivalent resistance will be

$R_2^{\prime }=\frac{10R_2}{10+R_2}$

Now applying condition for balanced meter bridge when resistances are $R_1$ and $R_2^{\prime }$.

$\frac{R_1}{50}=\frac{R_2^{\prime }}{50}$

$R_1=\frac{10R_2}{10+R_2}$

$10R_1+R_1{R}_2=10R_2$

From equation $\left(1\right)$,

$10R_1+(R_1\times \frac{3}{2}{R}_1)=15R_1$

$\therefore R_1=\frac{10}{3}\Omega$ and $R_2=5\Omega$

Hence, option $\left(a\right)$ is correct.