Q. A $2.0 \, V$ potentiometer is used to determine the internal resistance of a $1.5\, V$ cell. The balance point of the cell in the open circuit is obtained at $75\, cm$. When a resistor of $10 \, \Omega$ is connected across the cell, the balance point shifts to $60 \, cm$. The internal resistance of the cell is :

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A

$1.5\, \Omega$

B

$2.5\, \Omega$

C

$3.5\, \Omega$

D

$4.5\, \Omega$

Solution:

In a potentiometer internal resistance of cell is given as
$r=R\left(\frac{l_{1}}{l_{2}}-1\right)=10\left(\frac{75}{60}-1\right)=2.5 \,\Omega$

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