Q 12-03-290JEE MainJEE Main 2017 (9 Apr)Easy
In a meter bridge experiment resistances are connected as shown in the figure. Initially resistance $P = 4\ \Omega$ and the neutral point $N$ is at $60$ cm from $A$. Now an unknown resistance $R$ is connected in series to $P$ and the new position of the neutral point is at $80$ cm from $A$. The value of unknown resistance $R$ is:
Answer: (C) $\dfrac{20}{3}\ \Omega$
Balance condition: $\dfrac{P}{Q} = \dfrac{l}{100 - l}$.
Initially: $\dfrac{4}{Q} = \dfrac{60}{40} \Rightarrow Q = \dfrac83\ \Omega$.
With $R$ in series with $P$:
$$\frac{4 + R}{8/3} = \frac{80}{20} = 4 \;\Rightarrow\; 4 + R = \frac{32}{3} \;\Rightarrow\; R = \frac{20}{3}\ \Omega$$
Solution by Sreeraj P, M.Sc Physics