Two cells of e.m.f. $E_1$ and $E_2$ ($E_1 > E_2$) are connected as shown in figure.

When a potentiometer is connected between points A and B the balancing length of potentiometer wire is $412$ cm. When same potentiometer wire is connected between points A and C the balancing length is $103$ cm. The ratio $E_1 : E_2$ is

From the circuit diagram, the potential difference between points A and B is $V_{AB} = E_1$.

Given the balancing length for points A and B is $412 \text{ cm}$, we have:
$E_1 = 412k$
where $k$ is the potential gradient of the potentiometer wire.

The cells $E_1$ and $E_2$ are connected in opposition because their negative terminals are joined at point B. The potential difference between points A and C is:
$V_{AC} = E_1 - E_2$

Given the balancing length for points A and C is $103 \text{ cm}$, we have:
$E_1 - E_2 = 103k$

Dividing the first equation by the second equation:
$\dfrac{E_1}{E_1 - E_2} = \dfrac{412}{103}$

$\dfrac{E_1}{E_1 - E_2} = 4$

$E_1 = 4(E_1 - E_2)$

$E_1 = 4E_1 - 4E_2$

$3E_1 = 4E_2$

$\dfrac{E_1}{E_2} = \dfrac{4}{3}$

Answer: $4 : 3$