The length of a wire of a potentiometer is 100 cm, and the e.m.f. of its standard cell is E volt. It is employed to measure the e.m.f. of a battery whose internal resistance is 0.5 Ω. If the balance point is obtained at l = 30 cm from the positive end, the e.m.f. of the battery is where i is the current in the potentiometer wire. (a) (30E)/(100.5) (b) (30E)/((100-0.5)) (c) (30(E-0.5i))/(100) (d) (30E)/(100) - 0.5i, where i is the current in the potentiometer wire

Answer: (d). Taking E as the emf across the whole 100 cm wire: k = E/100 per cm. At balance no current flows through the battery, so its internal resistance does not matter. ε = k× 30 = (30E)/(100) So the answer is (30E)/(100), which is option (d) in the original paper.

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Potentiometer

The length of a wire of a potentiometer is $100\text{ cm}$ , and the $\text{e.m.f.}$ of its standard cell is $E$ volt. It is employed to measure the $\text{e.m.f.}$ of a battery whose internal resistance is $0.5\ \Omega.$ If the balance point is obtained at $l = 30\text{ cm}$ from the positive end, the $\text{e.m.f.}$ of the battery is

where $i$ is the current in the potentiometer wire.

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Answer: (d)

Taking $E$ as the emf across the whole 100 cm wire: $k = \dfrac{E}{100}$ per cm.
At balance no current flows through the battery, so its internal resistance does not matter.
$\varepsilon = k\times 30 = \dfrac{30E}{100}$
So the answer is $\dfrac{30E}{100}$ , which is option (d) in the original paper.

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