JEE Advanced 2019 — Paper 1 — Question 11

$V=I_{g}\left(G+R_{V}\right)$

$\mathrm{G}+\mathrm{R}_{\mathrm{V}} \simeq \mathrm{R}_{\mathrm{V}}=5 \times 10^{4} \Omega$

$I_{g} G=\left(I-I_{g}\right) S$

$\mathrm{S}=20 \mathrm{~m} \Omega$

$\mathrm{R}_{\text {ammeter }}=\frac{\mathrm{GS}}{\mathrm{G}+\mathrm{S}}=20 \times 10^{-3} \Omega$

$\mathrm{i}=\frac{\varepsilon}{\left(\frac{1000 \mathrm{R}_{\mathrm{V}}}{1000+\mathrm{R}_{\mathrm{V}}}\right)}=\frac{51 \varepsilon}{5 \times 10^{4}}\left(\mathrm{R}_{\mathrm{A}}\right.$ is small $)$

$\mathrm{i}_{1}=\mathrm{i}\left(\frac{1000+\mathrm{R}_{\mathrm{V}}}{1000}\right)=\frac{\varepsilon}{1000}$

$\mathrm{R}_{\text {measured }}=\frac{\mathrm{i}_{1}(1000)}{\mathrm{i}}=980.4 \Omega$