A voltmeter of resistance $20000\,\Omega$ reads $5$ volt. To make it read $20$ volt, the extra resistance required is
AIIMS 2016, Medium
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(c)
For a voltmeter, in which current $I_g$ passes,
when range is lower, $5= G \times I_g$
when range is higher, $20=(G+R) I_g$
where $G=$ resistance of voltmeter $=20000\,\Omega$
$\therefore \quad \operatorname{From}(i)$ and $(i i)$
$\frac{5}{20}=\frac{G}{G+R}$ or $\frac{1}{4}=\frac{G}{G+R}$
or $R=3 G=3 \times 20000=60000$
or extra resistance $=60000\,\Omega$ in series.
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