Search any question & find its solution
Question:
Answered & Verified by Expert
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 $\left(E_1>E_2\right)$
Options:
Solution:
2701 Upvotes
Verified Answer
The correct answer is:
$2: 3$
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),
$\Rightarrow \frac{E_2}{E_1}=\frac{2}{3}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.