(B) In a potentiometer,the emf $E$ of a cell is directly proportional to the balancing length $l$,i.e.,$E \propto l$.Therefore,for two cells $E_1$ and

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(B) In a potentiometer,the emf $E$ of a cell is directly proportional to the balancing length $l$,i.e.,$E \propto l$.Therefore,for two cells $E_1$ and $E_2$ with balancing lengths $l_1$ and $l_2$,we have the relation: $\frac{E_1}{E_2} = \frac{l_1}{l_2}$.Given: $E_A = 1.08 \ V$,$l_A = 400 \ cm$,and $l_B = 440 \ cm$.Substituting the values into the formula:$\frac{1.08}{E_B} = \frac{400}{440}$$E_B = \frac{1.08 \times 440}{400}$$E_B = 1.08 \times 1.1 = 1.188 \ V$.Thus,the emf of cell $B$ is $1.188 \ V$.

Class 11 Chemistry Data Entry - Chemistry Level 0 of all question bank

Two cells $A$ and $B$ are connected in the secondary circuit of a potentiometer one at a time,and the balancing lengths are $400 \ cm$ and $440 \ cm$ respectively. The emf of cell $A$ is $1.08 \ V$. The emf of the second cell $B$ in volts is:

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A
$1.08$
B
$1.188$
C
$11.88$
D
$12.8$

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Two cells $A$ and $B$ are connected in the secondary circuit of a potentiometer one at a time,and the balancing lengths are $400 \ cm$ and $440 \ cm$ respectively. The emf of cell $A$ is $1.08 \ V$. The emf of the second cell $B$ in volts is:

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