To compare the EMF of two cells using a poten | vedclass

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(A) The $EMF$ of a cell is given by $\epsilon = \lambda \ell$, where $\lambda$ is the potential gradient and $\ell$ is the balancing length.For two ce

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$1.16$

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(A) The $EMF$ of a cell is given by $\epsilon = \lambda \ell$, where $\lambda$ is the potential gradient and $\ell$ is the balancing length.For two cells, $\epsilon_1 = \lambda \ell_1$ and $\epsilon_2 = \lambda \ell_2$.The ratio of the EMFs is $y = \frac{\epsilon_1}{\epsilon_2} = \frac{\ell_1}{\ell_2}$.The relative error in the ratio is given by $\frac{\Delta y}{y} = \frac{\Delta \ell_1}{\ell_1} + \frac{\Delta \ell_2}{\ell_2}$.Given $\ell_1 = 200 \ cm$, $\ell_2 = 150 \ cm$, and $\Delta \ell_1 = \Delta \ell_2 = 1 \ cm$.Substituting the values: $\frac{\Delta y}{y} = \frac{1}{200} + \frac{1}{150}$.The percentage error is $\left( \frac{\Delta y}{y} ight) 	imes 100 = \left( \frac{1}{200} + \frac{1}{150} ight) 	imes 100$.$= \left( \frac{3 + 4}{600} ight) 	imes 100 = \frac{7}{6} \approx 1.16 \%$.

To compare the $EMF$ of two cells using a potentiometer, the balancing lengths obtained are $200 \ cm$ and $150 \ cm$. The least count of the scale is $1 \ cm$. The percentage error in the ratio of the EMFs is . . . . . . .

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