In a potentiometer arrangement, a cell of emf 1.5 V gives a balance point at 45.0 cm length of the wire. If the cell is replaced by another cell of emf 2.25 V. Where will the balance point shift to?
Answer & explanation
Correct answer: option 2
$ E_1 = \frac{V_0}{L} l_1$
$E_2 = \frac{V_0}{L} l_2$
$\frac{E_1}{E_2} = \frac{l_1}{l_2} = \frac{1.5}{2.25} = \frac{2}{3}$
$l_2 = \frac{3l_1}{2} = 67.5cm$