Physics - Magnetism Question with Solution | TestHub

Voltmeter rom galvanometer arrangement.
Given:$R_g=10\mathrm{\Omega }$and$I_g=2\mu A$
Required voltage range,$V=100\times {10}^{-3}\mathrm{v}\mathrm{o}\mathrm{l}\mathrm{t}\mathrm{s}$
$\Rightarrow V=I_g\left(R_g+R_v\right)={10}^{-1}$
$\Rightarrow \frac{{10}^{-1}}{2\times {10}^{-6}}=R_g+R_v\left(as,I_g=2\mu A\right)$
$\Rightarrow 5\times {10}^4\mathrm{\Omega }\approx R_v(R_v<{10}^5\mathrm{\Omega })\Rightarrow$Option$\left(\mathrm{A}\right)$is incorrect.

Galvanometer to Ammeter arrangement
$\Delta V_G=\Delta V_S$
$I_g\cdot R_g=\left(I-I_g\right)\cdot S$
$\Rightarrow S=\frac{I_g\cdot R_g}{I-I_g}=\frac{2\times {10}^{-6}\times 10}{{10}^{-3}-2\times {10}^{-6}}$
$\Rightarrow S=2\times {10}^{-5}\times {10}^3=2\times {10}^{-2}$
$\Rightarrow S=20m\mathrm{\Omega }\Rightarrow \left(\mathrm{C}\right)$is correct.
Circuit diagram for ohm's law

Resistance ammeter,
$R_A=\frac{20\times {10}^{-3}\times 10}{10}=20\times {10}^{-3}\mathrm{\Omega }$
$i=\frac{\epsilon }{\left(\frac{1000\times 50\times {10}^3}{51\times {10}^3}\right)}=\frac{51\epsilon }{5\times {10}^4}\left(\because R_A\to 0\right)$
$i^{\prime }=i\left(\frac{R_v}{51\times {10}^3}\right)=\frac{\epsilon }{1000}$
Measured resistance$\therefore R_m=\frac{i^{\prime }\times 1000}{i}=\frac{\epsilon }{51\epsilon }\times 5\times {10}^4=\frac{5\times {10}^4}{51}=980.4\mathrm{\Omega }$
If the voltmeter shows full scale deflection then
$\frac{\epsilon }{980}\times \left(\frac{1000}{51\times {10}^3}\right)\times 5\times {10}^4={10}^{-1}$
$\therefore \epsilon =999.6mv$
Since$i_A=1mA$so max reading of R can be$\frac{999.6mV}{1mA}=999.6\mathrm{\Omega }$