Q. The length of a wire of a potentiometer is $100 \, cm$ and the emf of its standard cell is $E$ volt. It is employed to measured 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

$0.2 \, E$

B

$0.3 \, E$

C

$0.4 \, E$

D

$0.5 \, E$

Solution:

Emf of a battery is given by, $\epsilon =xl$ , where $x\&l$ represents potential gradient of the potentiometer and balancing length of potentiometer.
$\epsilon \, =\frac{\epsilon _{0} 𝓁}{L} \, \therefore \epsilon \, =\frac{E}{100}\times 30$
$\therefore \, \epsilon \, = \, 0.3 \, E$
l = balance length
L = length of potentiometer wire

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