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  4. In an experiment to find emf of a cell using potentiometer, the length of null point for a cell of emf 1.5 V is found to be 60 cm. If this cell is replaced by another cell of emf E, the length-of null point increases by 40 cm. The value of E is (x/10) V. The value of x is

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Q. In an experiment to find emf of a cell using potentiometer, the length of null point for a cell of emf $1.5 \,V$ is found to be $60 \,cm$. If this cell is replaced by another cell of emf $E$, the length-of null point increases by $40 \,cm$. The value of $E$ is $\frac{x}{10} V$. The value of $x$ is _______

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Solution:

$ \frac{E_1}{E_2}=\frac{l_1}{l_2} $
$ \frac{1.5}{E_2}=\frac{60}{60+40}=\frac{6}{10}=\frac{3}{5} $
$ E_2=\frac{5}{2}=\frac{x}{10} $
$ x=25$