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

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A

$\frac{30 E}{100.5}$

B

$\frac{30 E}{100 - 0.5}$

C

$\frac{30 \left(E - 0.5 i\right)}{100}, \, Where \, i \, is \, the \, current \, in \, the \, $ $potentiometer \, wire.$

D

$\frac{30 E}{100}$

Solution:

$V∝l$
$⇒ \, \, \, \frac{V}{E}=\frac{l}{L}$
Where, $l$ = balance point
L= length of potentiometer wire.
Or $V=\frac{l}{L}E$
$or \, \, V=\frac{30 \times E}{100}=\frac{30}{100}E$

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