In the given potentiometer circuit arrangement, the balancing length $AC$ is measured to be $250 cm$ . When the galvanometer connection is shifted from point $(1)$ to point $(2)$ in the given diagram, the balancing length becomes $400 cm$ . The ratio of the emf of two cells, $\frac{E_{1}}{E_{2}}$ is:

\frac{5}{3}

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\frac{8}{5}

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\frac{3}{2}

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\frac{4}{3}

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Solution

The correct option is A $\frac{5}{3}$
According to the working of potentiometer,

At position $(1)$ ,

$E_{1} = k l_{1} . . . . (i)$

At position $(2)$ ,

$E_{1} + E_{2} = k l_{2} . . . . (i i)$

$\frac{(i)}{(i i)}$

$\Rightarrow \frac{E_{1}}{E_{1} + E_{2}} = \frac{l_{1}}{l_{2}} = \frac{250}{400} = \frac{5}{8}$

$\Rightarrow 8 E_{1} = 5 E_{1} + 5 E_{2}$

$\Rightarrow 3 E_{1} = 5 E_{2}$

$\Rightarrow \frac{E_{1}}{E_{2}} = \frac{5}{3}$

Hence, option $(A)$ is correct.

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In the given potentiometer circuit arrangement, the balancing length AC is measured to be 250 cm. When the galvanometer connection is shifted from point (1) to point (2) in the given diagram, the balancing length becomes 400 cm. The ratio of the emf of two cells, E1E2 is:

The correct option is A 53According to the working of potentiometer, At position (1), E1=kl1....(i) At position (2), E1+E2=kl2....(ii) (i)(ii) ⇒E1E1+E2=l1l2=250400=58 ⇒8E1=5E1+5E2 ⇒3E1=5E2 ⇒E1E2=53 Hence, option (A) is correct.