question_answer

Consider the two cells having emf\[{{E}_{1}}\]and\[{{E}_{2}}\]\[({{E}_{1}}>{{E}_{2}})\]connected as shown in the figure. A potentiometer is used to measure potential difference between P and Q and the balancing length of the potentiometer wire is 0.8 m. Same potentiometer is then used to measure potential difference between P and R and the balancing length is 0.2 m. Then, the ratio\[{{E}_{1}}/{{E}_{2}}\]is

A)  4/3               

B)  5/4

C)  5/3 

D)                 4/1

Correct Answer: A

Solution :

 Consider, the diagram shown below \[({{V}_{p}}-{{V}_{\theta }})={{E}_{1}}\propto {{L}_{1}}\]                  ...(i) \[{{V}_{p}}-{{V}_{R}}=({{V}_{p}}-{{V}_{Q}})+({{V}_{Q}}-{{V}_{R}})\] \[=({{E}_{1}}+(-{{E}_{2}}))={{E}_{1}}-{{E}_{2}}\propto {{L}_{2}}\]... (ii) where,\[{{L}_{1}}\]and\[{{L}_{2}}\]are the lengths of potentiometer From Eqs. (i) and (ii), we get \[\frac{{{E}_{1}}}{{{E}_{1}}-{{E}_{2}}}=\frac{{{L}_{1}}}{{{L}_{2}}}=\frac{0.8}{0.2}\Rightarrow \frac{{{E}_{1}}}{{{E}_{1}}-{{E}_{2}}}=\frac{8}{2}=4\] \[\Rightarrow \] \[{{E}_{1}}=4{{E}_{1}}-4{{E}_{2}}\] \[\Rightarrow \] \[4{{E}_{2}}=3{{E}_{1}}\Rightarrow \frac{{{E}_{1}}}{{{E}_{2}}}=\frac{4}{3}\]