A voltmeter has a range $0-V$ with a series resistance $R$. With a series resistance $2R$, the range is $0-V'$. The correct relation between $V$ and $V'$ is
Diffcult
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(d) For conversion of galvanometer (of resistances) into voltmeter, a resistance $R$ is connected in series.
${i_g} = \frac{{{V_1}}}{{R + G}}$ and ${i_g} = \frac{{{V_2}}}{{2R + G}}$
$==>$ $\frac{{{V_1}}}{{R + G}} = \frac{{{V_2}}}{{2R + G}}$ $==>$ $\frac{{{V_2}}}{{{V_1}}} = \frac{{2R + G}}{{R + G}} = \frac{{2(R + G) - G}}{{(R + G)}}$
$ = 2 - \frac{G}{{(R + G)}}$ $==>$ ${V_2} = 2{V_1} - \frac{{{V_1}G}}{{(R + G)}}$ $==>$ ${V_2} < 2{V_1}$
${i_g} = \frac{{{V_1}}}{{R + G}}$ and ${i_g} = \frac{{{V_2}}}{{2R + G}}$
$==>$ $\frac{{{V_1}}}{{R + G}} = \frac{{{V_2}}}{{2R + G}}$ $==>$ $\frac{{{V_2}}}{{{V_1}}} = \frac{{2R + G}}{{R + G}} = \frac{{2(R + G) - G}}{{(R + G)}}$
$ = 2 - \frac{G}{{(R + G)}}$ $==>$ ${V_2} = 2{V_1} - \frac{{{V_1}G}}{{(R + G)}}$ $==>$ ${V_2} < 2{V_1}$
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