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  • Two primary cells, Cell A and Cell B, are to be compared in terms of their electromotive force (emf). A potentiometer experiment is set up for this purpose. The emf of Cell A is found to be 1.5 V.

Two primary cells, Cell A and Cell B, are to be compared in terms of their electromotive force (emf). A potentiometer experiment is set up for this purpose. The emf of Cell A is found to be 1.5 V. Calculate the emf of Cell B if the balancing length of the potentiometer wire is 400 cm for Cell A and 320 cm for Cell B.

Option: 1

160V


Option: 2

1.2 V


Option: 3

200 V


Option: 4

20 V


Answers (1)

best_answer

Let ECell A be the emf of Cell A, ECell B be the emf of Cell B, LA be the balancing length for Cell A, and LB be the balancing length for Cell B.

According to the principle of a potentiometer:

\frac{E_{\text {Cell A }}}{L_{\mathrm{A}}}=\frac{E_{\text {Cell B }}}{L_{\mathrm{B}}}

Given ECell A = 1.5 V, LA = 400 cm, and LB = 320 cm, we can solve for ECell B:

E_{\text {Cell B }}=\frac{E_{\text {Cell A }} \cdot L_{\mathrm{B}}}{L_{\mathrm{A}}}=\frac{1.5 \cdot 320}{400}=1.2 \mathrm{~V}

Therefore, the emf of Cell B is 1.2 V.

Therefore, the correct option is 2.

 

 

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