To compare EMF of two cells using potentiometer the balancing lengths obtained are 200 cm and 150 cm. The least count of scale is 1 cm. The percent

Question

To compare EMF of two cells using potentiometer the balancing lengths obtained are 200 cm and 150 cm. The least count of scale is 1 cm. The percentage error in the ratio of EMFs is ______. [2026]

Smart Achievers · Accepted Answer

(4)

$ε = λ ℓ$

$Potential gradient$

$ε_{1} = λ ℓ_{1}$

$ε_{2} = λ ℓ_{2}$

$y = \frac{ε_{1}}{ε_{2}} = \frac{ℓ_{1}}{ℓ_{2}}$

$\frac{Δ y}{y} = \frac{Δ ℓ_{1}}{ℓ_{1}} + \frac{Δ ℓ_{2}}{ℓ_{2}}$

$\frac{Δ y}{y} = \frac{1}{200} + \frac{1}{150}$

$\frac{Δ y}{y} \times 100 = (\frac{1}{200} + \frac{1}{150}) \times 100$

$= (\frac{3 + 4}{600}) \times 100 = \frac{7}{6} = 1.16 %$

Solution

Q.
To compare EMF of two cells using potentiometer the balancing lengths obtained are 200 cm and 150 cm. The least count of scale is 1 cm. The percentage error in the ratio of EMFs is ______. [2026]

1 1.55

2 1.45

3 1.65

4 None of these

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