A potentiometer experiment is performed to compare the electromotive forces (EMFs) of two primary cells, Cell X and Cell Y. The potentiometer wire has a total length of 120 cm. When Cell X is connected to the potentiometer, the null point is found at a length of 72 cm. When Cell Y is connected instead of Cell X, the null point is found at a length of 90 cm. Given that the EMF of Cell X is 1.5V, calculate the EMF of Cell Y.
1.875V
17.60 V
2.50 V
20.35 V
The ratio of the EMFs of two cells is equal to the ratio of the lengths of wire on either side of the null point:
Substitute the given values:
Solving for the EMF of Cell Y:
Hence, the electromotive force of Cell Y is approximately 1.875 V.
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