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\[0.8m\]. Same potentiometer is then used to measure potential difference between P and R and the balancing length is \[0.2\text{ }m\]. 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 \[\frac{{{E}_{1}}}{{{E}_{1}}-{{E}_{2}}}=\frac{{{L}_{1}}}{{{L}_{2}}}=\frac{0.8}{0.2}\Rightarrow \frac{{{E}_{1}}}{{{E}_{1}}-{{E}_{2}}}=\frac{8}{2}=4\] \[\Rightarrow \] \[{{E}_{1}}=4{{E}_{1}}-4{{E}_{2}}\] \[\Rightarrow \] \[4{{E}_{2}}=3{{E}_{1}}\Rightarrow \frac{{{E}_{1}}}{{{E}_{2}}}=\frac{4}{3}\]