For a cell,the balancing length is 0.60 , m. | vedclass

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(A) In a potentiometer,the $emf$ $(E)$ of a cell is proportional to its balancing length $(l)$,given by $E = kl$,where $k$ is the potential gradient.L

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$1.2 \, V, 1.1 \, V$

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(A) In a potentiometer,the $emf$ $(E)$ of a cell is proportional to its balancing length $(l)$,given by $E = kl$,where $k$ is the potential gradient.Let the $emf$ of the first cell be $E_1$ and the second cell be $E_2$.Given: $E_1 = k(0.60)$ and $E_2 = k(0.55)$.Also,$E_2 = E_1 - 0.1 \, V$.Substituting the expressions for $E_1$ and $E_2$:$k(0.55) = k(0.60) - 0.1$$k(0.60 - 0.55) = 0.1$$k(0.05) = 0.1$$k = \frac{0.1}{0.05} = 2 \, V/m$.Now,calculating the $emf$ values:$E_1 = 2 	imes 0.60 = 1.2 \, V$.$E_2 = 2 	imes 0.55 = 1.1 \, V$.

For a cell,the balancing length is $0.60 \, m$. For another cell having an electromotive force $(emf)$ $0.1 \, V$ less than the first one,the balancing length is $0.55 \, m$. What are the $emf$ values of the two cells?

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