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 \(0.3 \mathrm{~m}\). The ratio \(\frac{E_2}{E_1}\) is \(\left(E_1>E_2\right)\)
- A \(3: 1\)
- B \(2: 3\)
- C 1:3
- D 3:1
Answer & Solution
Correct Answer
(B) \(2: 3\)
Step-by-step Solution
Detailed explanation
Position of null point is directly proportional to the potential drop, as the resistance of the wire is directly proportional to the length of the wire.
\(\begin{aligned} & \Rightarrow \frac{V_{A B}}{V_{A C}}=\frac{0.9 \mathrm{~m}}{0.3 \mathrm{~m}}=\frac{E_1}{E_1-E_2} \\ & \Rightarrow \frac{E_1}{E_1-E_2}=3---(1)\end{aligned}\)
Subtract one from each side of equation (1)
\(\begin{aligned} & \Rightarrow \frac{E_1}{E_1-E_2}-1=3-1 \\ & \Rightarrow \frac{E_2}{E_1-E_2}=2---(2)\end{aligned}\)
Take the ratio of equation (2) and (1),
\(\Rightarrow \frac{E_2}{E_1}=\frac{2}{3}\)
\(\begin{aligned} & \Rightarrow \frac{V_{A B}}{V_{A C}}=\frac{0.9 \mathrm{~m}}{0.3 \mathrm{~m}}=\frac{E_1}{E_1-E_2} \\ & \Rightarrow \frac{E_1}{E_1-E_2}=3---(1)\end{aligned}\)
Subtract one from each side of equation (1)
\(\begin{aligned} & \Rightarrow \frac{E_1}{E_1-E_2}-1=3-1 \\ & \Rightarrow \frac{E_2}{E_1-E_2}=2---(2)\end{aligned}\)
Take the ratio of equation (2) and (1),
\(\Rightarrow \frac{E_2}{E_1}=\frac{2}{3}\)
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