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 $50 \, cm$ , Now, when the $10 \, \Omega$ resistance is removed the balance point shifts to $40 \, cm$ . 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)
$⇒ \, \, \, R_{1}+10=R_{2} \, $
case-II
$\frac{R_{1}}{R_{2}}=\frac{40}{60}$
$⇒ \, \, R_{2}=\frac{60}{40}R_{1}$ ...(ii)
So, Eqs. (i) and (ii) give
$R_{1}+10=\frac{60}{40}R_{1}$
$\Rightarrow R_{1} = 20 \, \, \Omega $