Two cells having unknown e.m.f.s $E_{1}$ and $E_{2}$ $(E_{1} > E_{2})$ are connected in a potentiometer circuit so as to assist each other. The null point is obtained at $490 \ cm$ from the higher potential end. When cell $E_{2}$ is connected so as to oppose cell $E_{1}$,the null point is obtained at $90 \ cm$ from the same end. The ratio of the e.m.f.s of the two cells $(\frac{E_{1}}{E_{2}})$ is:
- A$0.689$
- B$0.182$
- C$5.33$
- D$1.45$
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MediumNEET 2016
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