Question

An unknown resistance $R_1$ is connected in series with a resistance of $10\Omega$ . This combination is connected to one gap of the meter bridge while a resistance $R_2$ is connected in the other gap. The balance point is at $50cm$ , Now, when the $10\Omega$ resistance is removed the balance point shifts to $40cm$ . The value of $R_1$ (in ohm) is

• A. $20\Omega$
• B. $10\Omega$
• C. $60\Omega$
• D. $40\Omega$

Answer 1

Answer: (A)

The balance condition of a meter bridge experiment
$\frac{R}{S}=\frac{l_1}{\left(100-l_1\right)}$
Here, $R=R_1,S=R_2$
$\therefore \frac{R_1}{R_2}=\frac{l_1}{\left(100-l_1\right)}$
case-I
$\frac{R_1+10}{R_2}=\frac{50}{50}$ ...(i)
$\Rightarrow R_1+10=R_2$
case-II
$\frac{R_1}{R_2}=\frac{40}{60}$
$\Rightarrow R_2=\frac{60}{40}{R}_1$ ...(ii)
So, Eqs. (i) and (ii) give
$R_1+10=\frac{60}{40}{R}_1$
$\Rightarrow R_1=20\Omega$