Question

40.A cell of emf e1, in the secondary circuit gives null deflection for 1.5m length of potentiometer of wire length 10m.If another cell of emf e2 is connected in series with e1 then null deflection was obtained for 2.5m length.Then e1:e2 is

Answer 1

Dear student,

The emf of two cells can be computed using potentiometer as,
$$\begin{gathered} \frac{E_1}{E_2}=\frac{l_1}{l_2} \\ \mathrm{According}\mathrm{to}\mathrm{the}\mathrm{question},E_1=e_1\mathrm{and}{E}_2=e_1+e_2 \\ \Rightarrow \frac{e_1}{e_1+e_2}=\frac{1.5}{2.5}=\frac{3}{5} \\ \Rightarrow \frac{1}{1+{\displaystyle \frac{e_2}{e_1}}}=\frac{3}{5} \\ \Rightarrow \frac{e_2}{e_1}=\frac{2}{3} \\ \Rightarrow \frac{e_1}{e_2}=\frac{3}{2} \end{gathered}$$
Regards.