MHT CET · Physics · Current Electricity
In potentiometer experiment, cells of e.m.f. E and E 2 are connected in series (E \(_{1}>\mathrm{E}_{2}\) ), the balancing length is \(64 \mathrm{~cm}\) of the wire. If the polarity of \(\mathrm{E}_{2}\) is reversed, the balancing length becomes \(32 \mathrm{~cm}\). The ratio \(\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}\) is
- A \(1: 2\)
- B \(2: 1\)
- C \(1: 3\)
- D \(3: 1\)
Answer & Solution
Correct Answer
(D) \(3: 1\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} \frac{\mathrm{E}_{1}}{\mathrm{E}_{2}} &=\frac{\ell_{1}+\ell_{2}}{\ell_{1}-\ell_{2}} \\ &=\frac{490+90}{490-90}=\frac{580}{400}=1.45 \end{aligned}\)
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