Two cells of emf E_1 and E_2 (E_1 > E_2 ) are connected individua… - Vidyadip

MCQ

Two cells of emf $E_1$ and $E_2\left(E_1 > E_2\right)$ are connected individually to a potentiometer and their corresponding balancing length are $625 \,cm$ and $500 \,cm$, then the ratio $\frac{E_1}{E_2}$ is ...........

✓
$5: 4$
B
$3: 1$
C
$5: 1$
D
$4: 5$

✓

Answer

Correct option: A.

$5: 4$

a
(a)

$E_1=k l_1$ $\quad k$ is potential gradient

$E_2=k l_2$

$\frac{E_1}{E_2}=\frac{l_1}{l_2}=\frac{625}{500}=\frac{25}{20}=\frac{5}{4}$

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