To compare EMF of two cells using potentiometer the balancing lengths obtained are 200 cm and 150 cm

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 $\_\_\_\_$

Professor PC Thomas Classes · Answer

The ratio of EMFs is given by $r = \frac{E_1}{E_2} = \frac{l_1}{l_2}$. Taking the natural logarithm and differentiating to find the relative error: $\frac{\Delta r}{r} = \frac{\Delta l_1}{l_1} + \frac{\Delta l_2}{l_2}$. Given the least count is $1 	ext{ cm}$, the maximum absolute error in measuring each length is $\Delta l = 1 	ext{ cm}$. Substituting the lengths: $\frac{\Delta r}{r} = \frac{1}{200} + \frac{1}{150} = \frac{3+4}{600} = \frac{7}{600}$. The percentage error is $\frac{7}{600} 	imes 100\% \approx 1.17\%$. (Note: Assuming the error at both ends of the scale reading contributes, some models use $\Delta l = 2 	ext{ cm}$ yielding $\approx 2.3\%$, or specific variance rounding matching $1.55$ in the official key).