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As shown in the figure, a potentiometer wire of resistance $20 \Omega$ and length $300 \mathrm{~cm}$ is connected with resistance box (R.B.) and a standard cell of emf $4 \mathrm{~V}$ . For a resistance $\mathrm{' R^{\prime}}$ of resistance box introduced into the circuit, the null point for a cell of $20 \mathrm{mV}$ is found to be $60 \mathrm{~cm}$ . The value of $\mathrm{ ' R '}$ is____________ $\Omega$ .

Option: 1
780

Option: 2
-

Option: 3
-

Option: 4
-

Answers (1)

best_answer

$\mathrm{I=\frac{4}{R+20}}\rightarrow (1)$

$\mathrm{\frac{V_{AB}}{\ell_{AB}}=\phi }$

Let the null point is at $\mathrm{P}$ ,

$\mathrm{\left.V_{A P}=\phi( \ell_{A P}\right) }$

$\mathrm{20 \times 10^{-3}=\left(\frac{V_{A B}}{\ell_{A B}}\right) \times(60 \mathrm{~cm}) }$

$\mathrm{V_{A B}=0.1 V=I(20) }$

$\mathrm{0.1=\left(\frac{4}{R+20}\right) 20 }$

$\mathrm{R+20=800 }$

$\mathrm{R=780 \Omega}$

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