Question

An unknown resistance $R_1$ is connected in series with resistance $10\Omega$. When the combination is connected in left gap of a metre bridge and $R_2$ is connected in right gap, the balance point from left end is at $50\text{cm}$. When the $10\Omega$ is removed, the balance point from left end becomes $40\text{cm}$. The value of $R_1$ is

• A. $10\Omega$
• B. $20\Omega$
• C. $50\Omega$
• D. $40\Omega$

Answer 1

The correct option is B $20\Omega$

Initially,

$\frac{R_1+10}{R_2}=\frac{50}{50}=1$

Finally,

$\frac{R_1}{R_2}=\frac{40}{60}=\frac{2}{3}$

So, $\frac{R_1+10}{R_1}=\frac{3}{2}$

$2R_1+20=3R_1$

$R_1=20\Omega$