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  • 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 conne

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.

Option: 1

1.875V


Option: 2

17.60 V


Option: 3

2.50 V


Option: 4

20.35 V


Answers (1)

best_answer

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:

\frac{EMF of cell 'X'}{EMF of cell 'Y'} = \frac{Length of cell 'X'}{Length of cell 'Y'}

Substitute the given values:

\frac{1.5}{EMF of cell 'Y'} = \frac{72}{90}

Solving for the EMF of Cell Y:

{EMF of cell 'Y'} = \frac{1.5 \times 90}{72} = 1.875 V

Hence, the electromotive force of Cell Y is approximately 1.875 V.

 

Posted by

Pankaj Sanodiya

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