In a potentiometer experiment, when three cells A, B and C are connected in series, the balancing length is found to be 420 cm. If cells A and B are connected in series the balancing length is 220 cm and for cells B and C connected in series the balancing length is 320 cm. The emf of cells A, B and C are respectively in the ratio of (a) 2: 3: 5 (b) 5: 4: 3 (c) 1: 1.2: 2 (d) 1.2: 1: 2

Answer: (c). Emf ∝ balancing length, so take A+B+C = 420, A+B = 220, B+C = 320 (in units of k). C = 420-220 = 200, A = 420-320 = 100, B = 220-100 = 120. A:B:C = 100:120:200 = 1:1.2:2 So the answer is (c) 1:1.2:2.

In a potentiometer experiment, when three cells A, B and C are… | MHT CET 2021 (24 Sep, Shift 1)

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Potentiometer

In a potentiometer experiment, when three cells $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are connected in series, the balancing length is found to be $420\text{ cm}$ . If cells $\mathrm{A}$ and $\mathrm{B}$ are connected in series the balancing length is $220\text{ cm}$ and for cells $\mathrm{B}$ and $\mathrm{C}$ connected in series the balancing length is $320\text{ cm}$ . The emf of cells $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are respectively in the ratio of

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Answer: (c)

Emf $\propto$ balancing length, so take $A+B+C = 420$ , $A+B = 220$ , $B+C = 320$ (in units of $k$ ).
$C = 420-220 = 200$ , $A = 420-320 = 100$ , $B = 220-100 = 120$ .
$A:B:C = 100:120:200 = 1:1.2:2$
So the answer is (c) $1:1.2:2$ .

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Potentiometer question from MHT CET 2021 (24 Sep, Shift 1)

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