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An unknown cell balances $300 \mathrm{~cm}$ of wire in a potentiometer circuit, but the balance length reduces to $280 \mathrm{~cm}$ when a resistance of $28 \Omega$ is added in parallel with the cell. A standard cell of emf $1.1 \mathrm{~V}$ balances $220 \mathrm{~cm}$ on the same potentiometer. The balance length $(in \, \, \mathrm{cm} )$ is $\mathrm{x} \times 144$ when $48 \Omega$ resistance is connected across the unknown cell then the value of $x$ is:
Option: 1
1

Option: 2
2

Option: 3
3

Option: 4
4

Answers (1)

The emf of the cell or the measured potential difference is proportional to the balance length on the potentiometer. The emf of the unknown cell is found to be 1.5 V and its internal resistance can be found as follows :

$\frac{1.5}{\mathrm{r}+28} \times 28=\text { drop across } 280 \mathrm{~cm} \text { of wire }$

$=1.4 \text { (in volt })$

$\therefore \quad r=2 \Omega$

The current through the $48 \Omega$ resistance, when it is connected across the unknown cell, is

$\mathrm{i}=\frac{1.5}{48+2} \mathrm{~A}=0.03 \mathrm{~A} \text {, and the balance length }=200 \times 48$

$\times 0.03$

$=288 \mathrm{~cm}$

Posted by

Ramraj Saini

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