c
(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.
