A galvanometer, whose resistance is $50\,ohm,$ has $25$ divisions in it. When a current of $4\times 10^{-4}\,A$ passes through it, its needle (pointer) deflects by one division. To use this galvanometer as a voltmeter of range $2.5\,V,$ it should be connected to a resistance of.......$ohm$
JEE MAIN 2019, Medium
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${{\text{V}}_{\text{o}}} = {{\text{i}}_{{{\text{g}}_0}}}\left( {{{\text{R}}_{\text{G}}} + {\text{R}}} \right)$
${{\text{i}}_{{g_0}}} = 4 \times {10^{ - 4}} \times 25 = {10^{ - 2}}{\mkern 1mu} {\text{A}}$
$\mathrm{V}_{0}=2.5\, \mathrm{V}$
$R_{0}+R=\frac{V_{0}}{i_{0}}=\frac{2.5}{10^{-2}}=250$
$\Rightarrow \quad \mathrm{R}=200\, \Omega$
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