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  • A potentiometer experiment is conducted to compare the emf of two primary cells, Cell X and Cell Y. The length of the potentiometer wire is 800 cm. Cell X is connected to the potentiometer and its

A potentiometer experiment is conducted to compare the emf of two primary cells, Cell X and Cell Y. The length of the potentiometer wire is 800 cm. Cell X is connected to the potentiometer and its balancing length is found to be 600 cm. When Cell Y is connected to the potentiometer, the balancing length becomes 480 cm. Determine the emf of Cell Y if the emf of Cell X is 1.8 V.

Option: 1

160V


Option: 2

1.2 V


Option: 3

1.44 V


Option: 4

20 V


Answers (1)

best_answer

Let EX be the emf of Cell X, EY be the emf of Cell Y, LX be the balancing length for Cell X, and LY be the balancing length for Cell Y.

According to the principle of a potentiometer:

\frac{E_{\mathrm{X}}}{L_{\mathrm{X}}}=\frac{E_{\mathrm{Y}}}{L_{\mathrm{Y}}}

Given EX = 1.8 V, LX = 600 cm, and LY = 480 cm, we can solve for EY:

Therefore, the emf of Cell Y is 1.44 V.

Therefore, the correct option is 3.

Posted by

Ajit Kumar Dubey

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