Q 12-03-113JEE MainJEE Main 2023 (29 Jan, Shift 1)Medium
In a metre bridge experiment the balance point is obtained if the gaps are closed by $2\ \Omega$ and $3\ \Omega$. A shunt of $X\ \Omega$ is added to the $3\ \Omega$ resistor to shift the balancing point by $22.5\ \text{cm}$. The value of $X$ is ______.
Numerical value type. Enter your answer.
Answer: 2
Initially $\dfrac{2}{3}=\dfrac{l}{100-l}\Rightarrow l=40\ \text{cm}$.
The shunt lowers the right-hand resistance, so the balance point moves to $62.5\ \text{cm}$:
$$\frac{2}{R'}=\frac{62.5}{37.5}\ \Rightarrow\ R'=1.2\ \Omega$$
$$\frac{3X}{3+X}=1.2\ \Rightarrow\ 3X=3.6+1.2X\ \Rightarrow\ X=2\ \Omega$$
Solution by Sreeraj P, M.Sc Physics