Q 12-03-270JEE MainJEE Main 2019 (11 Jan, Shift 1)Medium
The resistance of the meter bridge AB in given figure is $4\ \Omega$. With a cell of emf $\varepsilon = 0.5$ V and rheostat resistance $R_h = 2\ \Omega$ the null point is obtained at some point J. When the cell is replaced by another one of emf $\varepsilon = \varepsilon_2$ the same null point J is found for $R_h = 6\ \Omega$. The emf $\varepsilon_2$ is:
Answer: (B) $0.3$ V
Current in the wire $= \dfrac{6}{4 + R_h}$ (the 6 V cell taken as ideal). Let the resistance of AJ be $r$.
With $R_h = 2\ \Omega$: $I = 1$ A, so $\varepsilon = 1\times r = 0.5$ V gives $r = 0.5\ \Omega$.
With $R_h = 6\ \Omega$: $I = 0.6$ A, so
$$\varepsilon_2 = 0.6\times0.5 = 0.3\ \text{V}$$
Solution by Sreeraj P, M.Sc Physics