Question 418: Current Electricity

Question

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

Accepted Answer

E
1
= 1.5 V, l
1
= 60 cm, l
2
= 40 cm + 60 cm = 100 cm
$E \propto l$
$$
\begin{aligned}
& \frac{E_{1}}{E_{2}}=\frac{l_{1}}{l_{2}} \\
& \frac{1.5}{E}=\frac{60}{100} \\
& E=\frac{150}{60}=\frac{5}{2}=\frac{25}{10} \\
& 	ext { so } x=25
\end{aligned}
$$

Current Electricity

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