Current Electricity

JEE Main level, single correct, medium

315 of 380

Question 315 of 380

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 ____\_\_\_\_.

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Answer: (D) 1.16

ϵ=λℓ\epsilon = \lambda \ell (Potential gradient)

ϵ1=λℓ1\epsilon_1 = \lambda \ell_1
ϵ2=λℓ2\epsilon_2 = \lambda \ell_2

y=ϵ1ϵ2=ℓ1ℓ2y = \frac{\epsilon_1}{\epsilon_2} = \frac{\ell_1}{\ell_2}

Δyy=Δℓ1ℓ1+Δℓ2ℓ2\frac{\Delta y}{y} = \frac{\Delta \ell_1}{\ell_1} + \frac{\Delta \ell_2}{\ell_2}

Δyy=1200+1150\frac{\Delta y}{y} = \frac{1}{200} + \frac{1}{150}

Δyy×100=(1200+1150)×100=(3+4600)×100=76=1.16%\frac{\Delta y}{y} \times 100 = \left(\frac{1}{200} + \frac{1}{150}\right) \times 100 = \left(\frac{3+4}{600}\right) \times 100 = \frac{7}{6} = 1.16\%

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