JEE PYQ: Current Electricity Question 66
Question 66 - 2019 (11 Jan 2019 Shift 1)
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 $R_b = 2,\Omega$ 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 $R_b = 6,\Omega$. The emf $\varepsilon_2$ is:
(1) 0.4 V
(2) 0.3 V
(3) 0.6 V
(4) 0.5 V
Show Answer
Answer: (2)
Solution
At null point: $\varepsilon_1 = \frac{E}{4+R_{b1}} \times \frac{4}{L} \times l$ and $\varepsilon_2 = \frac{E}{4+R_{b2}} \times \frac{4}{L} \times l$. Dividing: $\frac{0.5}{\varepsilon_2} = \frac{4+6}{4+2} = \frac{10}{6}$. $\varepsilon_2 = 0.3$ V.
BETA
