Experimental Physics Question 4
Question 4 - 01 February - Shift 1
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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Answer: (25)
Solution:
Formula: Potentiometer
$\frac{E_1}{E_2}=\frac{l_1}{l_2}$
$\frac{1.5}{E_2}=\frac{60}{60+40}=\frac{6}{10}=\frac{3}{5}$
$E_2=\frac{5}{2}=\frac{x}{10}$
$x=25$
