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Tamil Nadu Board of Secondary EducationHSC Science Class 12

In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63 cm,

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Question

In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63 cm, what is the emf of the second cell?

Numerical
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Solution

Emf of the cell1, ξ1 = 1.25 V

Balancing length of the cell, l1 = 35 cm = 35 x 10-2 m

Balancing length after interchanged, l2 = 63 cm = 63 x 10-2 m

Emf of the cell1, ξ2 = ?

The ration of emf’s, `ξ_1/ξ_2 = l_1/l_2`

The ration of emf’s, ξ2 = ξ1 = `(l_2/l_1)`

= 1. 25 `xx ((63 xx 10^-2)/(35 xx 10^-2)) = 12.5 xx 1.8` 

ξ2 = 2.25 V.

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Chapter 2: Current Electricity - Evaluation [Page 123]

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Samacheer Kalvi Physics - Volume 1 and 2 [English] Class 12 TN Board
Chapter 2 Current Electricity
Evaluation | Q IV. 14. | Page 123

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