MHT CET · Physics · Current Electricity
A potentiometer is used to measure the potential difference between \(\mathrm{A}\) and \(\mathrm{B}\), the null point is obtained at \(0.9 \mathrm{~m}\). Now potential difference between \(\mathrm{A}\) and \(\mathrm{C}\) is measured, the null point is obtained at \(0 \cdot 3 \mathrm{~m}\). The ratio \(\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}\) is \(\left(\mathrm{E}_{1}>\mathrm{E}_{2}\right)\)
- A \(3: 1\)
- B \(3: 2\)
- C \(2: 3\)
- D \(1: 3\)
Answer & Solution
Correct Answer
(C) \(2: 3\)
Step-by-step Solution
Detailed explanation
\(\frac{E_{a b}}{E_{a c}}=\frac{L_{a b}}{L_{a c}}=\frac{E_{1}}{E_{1}-E_{2}}=\frac{0.9}{0.3}=3\)
\(E_{1}=3 E_{1}-3 E_{2}\)
\(2 E_{1}=3 E_{2}\)
\(\frac{E_{1}}{E_{2}}=\frac{3}{2}\)
\(\therefore \frac{E_{2}}{E_{1}}=\frac{2}{3}\)
\(E_{1}=3 E_{1}-3 E_{2}\)
\(2 E_{1}=3 E_{2}\)
\(\frac{E_{1}}{E_{2}}=\frac{3}{2}\)
\(\therefore \frac{E_{2}}{E_{1}}=\frac{2}{3}\)
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