The cells of e.m.f. $E_1$ and $E_2$ are connected as shown in figure and $E_1> E_2$. When a potentiometer is connected between point X and Y, the balancing length of potentiometer wire is 150 cm. On connecting the same potentiometer between X and Z, the balancing length is 50 cm. The ratio of emf $E_1$ and $E_2$ is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 3 : 4
emf ∝ balancing length
and,
Balancing length = 50 cm
$E_1∝150$ and $E_1-E_2∝50$
$∴\frac{E_1}{E_2-E_1}=\frac{150}{50}$
$3E_2=4E_1$
$\frac{E_1}{E_2}=\frac{3}{4}$