Question

A cell of emf e${}_1$ in the secondary circuit gives null deflection for 1.5m length of potentiometer of wire length 10m. If another cell of emf e${}_2$ is connected in series with e${}_1$then null deflection was obtained for 2.5 m length. Then e${}_1$: e${}_2$ is:

• A. 3 : 5
• B. 5 : 3
• C. 3 : 2
• D. 2 : 3

Answer 1

The correct option is C 3 : 2

Formula $\frac{e_1}{e_2}=\frac{l_1}{l_2}$

$\Rightarrow \frac{e_1}{e_1+e_2}=\frac{1.5}{2.5}$ ( here $e_2=e_1+e_2$ )

$\frac{e_1}{e_1+e_2}=\frac{3}{5}$

$5e_1=3e_1+3e_2$

$\Rightarrow 2e_1=3e_2$

$\Rightarrow \frac{e_1}{e_2}=\frac{3}{2}$