Voltmeter reads potential difference across the terminals of an old battery as $1.2\,volt$ , while a potentiometer reads $1.4\,volt$ . The internal resistance of battery is $40\,\Omega $ , then voltmeter resistance is .............. $\Omega$
Medium
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$\mathrm{E}=1.4 \mathrm{\,V} ; \mathrm{V}=1.2 \mathrm{\,V}$
$\Rightarrow \mathrm{ir}=1.2 \quad \Rightarrow \frac{\mathrm{Er}}{\mathrm{r}+\mathrm{R}}=1.2$
$\Rightarrow \frac{1.4 \mathrm{r}}{\mathrm{r}+40}=1.2 \Rightarrow \mathrm{r}=\frac{40 \times 1.2}{1.4-1.2}=240\, \Omega$
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