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Two identical potentiometer wires $w_{1}$ and $w_{2}$ of equal length $l$ , connected to a battery of emf $\varepsilon _{p}$ and internal resistance $1\; \Omega$ through two switches $S_{1}$ and $S_{2}$ . A battery of emf $\varepsilon$ is balanced on these potentiometer wires one by one. If potentiometer wire $w_{1}$ is of resistance $2\; \Omega$ and balancing length is $\frac{l}{2}$ on it, when only $S_{1}$ is closed and $S_{2}$ is open. On closing $S_{2}$ and opening $S_{1}$ the balancing length On $w_{2}$ is found to be $\frac{2l}{3}$ ,then the resistance (in $\Omega$ ) of potentiometer wire $w_2$ .

Option: 1
$1\; \Omega$

Option: 2
$3\: \Omega$

Option: 3
$4\: \Omega$

Option: 4
$5\: \Omega$

Answers (1)

best_answer

$For \: \: w_1, \varepsilon=\frac{l}{2}\left[\left(\frac{\varepsilon_p}{1+2}\right) \frac{2}{1}\right]$ _______________(i)

$For \: \: w_2, \varepsilon=\frac{2l}{3}\left[\left(\frac{\varepsilon_p}{1+R}\right) \frac{R}{1}\right]$ ___________(ii)

Dividing eq. (i) by (ii) and on solving, we get, Resistance of wire $w_2=1 \Omega$

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SANGALDEEP SINGH

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