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A potentiometer wire of length $\mathrm{ 300 \mathrm{~cm}}$ is connected in series with a resistance $\mathrm{ 780 \Omega}$ and a standard cell of emf $\mathrm{4\mathrm{~V}}$ . A constant current flows through potentiometer wire. The length of the null point for cell of emf $\mathrm{20\; \mathrm{mV}}$ is found to be $\mathrm{ 60 \mathrm{~cm}.}$ The resistance of the potentiometer wire is _________ $\mathrm{ \Omega .}$
Option: 1
20

Option: 2
-

Option: 3
-

Option: 4
-

Answers (1)

best_answer

$\mathrm{\ell_{AB}=300\: cm}$

$\mathrm{E_1 =\phi \ell_1 }$

$\mathrm{E_1=\left(\frac{V_{A B}}{\ell_{A B}}\right) \ell_1 }$

$\mathrm{E_1=\left(\frac{IR_{A B}}{\ell_{A B}}\right) \ell_1 }$

$\mathrm{20 \times 10^{-3}=\left(\frac{4 R A B}{R_{A B}+780}\right) \times \frac{300}{30} }$

$\mathrm{0.1 R_{A B}+78=4 R_{A B} }$

$\mathrm{78=3.9 R_{A B} }$

$\mathrm{R_{A B}=20\Omega }$

The resistance of the potentiometer wire is $\mathrm{20\Omega }$

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