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 _____. [2026]

Smart Achievers · Accepted Answer

(4)

$In a potentiometer, the EMF of a cell is directly proportional to its balancing length: E = k l$

$where k is the potential gradient. For two different cells, the ratio of their EMFs (R) is given by:$

$R = \frac{E_{1}}{E_{2}} = \frac{l_{1}}{l_{2}}$

$When a quantity is defined by the division of two variables, the maximum relative error is the sum of the relative errors of each variable.$

$Therefore, the relative error in the ratio R is:$

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

$The absolute error in measuring length (Δ l) is equal to the least count of the measuring scale.$

$Percentage Error = (\frac{1}{200} + \frac{1}{150}) \times 100 = \frac{7}{6} %$

$\Rightarrow % age error = 1.16 %$

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 1.45

2 1.65

3 1.75

4 1.16

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