⇒ Current Electricity

Correct Option is (A)

Given: Resistance of the galvanometer = G

Range of the galvanometer $={\mathrm{V}}_{\mathrm{g}}$

The series resistance value to be used for converting the galvanometer into a voltmeter of range 0 to ${\mathrm{V}}_{{\mathrm{g}}^{\prime }}$ is,

$\mathrm{R}=\frac{{\mathrm{V}}_8}{{\mathrm{I}}_8}-\mathrm{G}$

Also,

$I_g=\frac{V_8}{G}$

To increase the measuring range to $\mathrm{V}$, the new resistance value

$\begin{array}{rl}{\mathrm{R}}^{\prime }& =\frac{\mathrm{V}}{\left(\frac{{\mathrm{V}}_{\mathrm{g}}}{\mathrm{G}}\right)}-\mathrm{G}\\ & =\frac{\mathrm{VG}}{{\mathrm{V}}_{\mathrm{g}}}-\mathrm{G}=\mathrm{G}(\frac{\mathrm{V}}{{\mathrm{V}}_{\mathrm{g}}}-1)\end{array}$

Correct Option is (D)

Given: $I_g=\frac{1}{100}I$

$\begin{array}{rl}& \therefore \frac{\mathrm{I}}{{\mathrm{I}}_{\mathrm{g}}}=100\text{..... (i)}\\ & \mathrm{S}=\frac{{\mathrm{GI}}_{\mathrm{g}}}{\mathrm{I}-{\mathrm{I}}_{\mathrm{g}}}=\frac{\mathrm{G}}{\frac{\mathrm{I}}{{\mathrm{I}}_{\mathrm{g}}}-1}\text{..... (ii)}\end{array}$

Substituting, (i) into (ii),

$\mathrm{S}=\frac{\mathrm{G}}{100-1}=\frac{\mathrm{G}}{99}$

Correct Option is (B)

$\begin{array}{rl}& \mathrm{R}=\frac{\mathrm{V}}{{\mathrm{I}}_{\mathrm{g}}}-\mathrm{G}\\ & \text{ Here, }{\mathrm{I}}_{\mathrm{g}}=\frac{50}{5}=10\mathrm{mA}\\ \therefore \mathrm{R}& =\frac{25}{10\times {10}^{-3}}-20\\ & =2480\mathrm{\Omega }\text{ in series }\end{array}$

Correct Option is (A)

$\begin{array}{rl}\frac{I_g}{I}& =\frac{S}{S+G}=\frac{0.1G}{0.1G+G}=\frac{1}{11}\\ \therefore I_g& =\frac{1}{11}I\end{array}$

Correct Option is (A)

Shunt resistance is given by

$S=\frac{I_{g}G}{I-I_g}$

In the first case $:3r=\frac{I_{g}G}{0.04-I_g}$ ..... (1)

In the second case : $r=\frac{I_{g}G}{0.08-I_g}$ ..... (2)

Dividing $\mathrm{Eq}.(1)$ by Eq.(2) we get

$3=\frac{0.08-{\mathrm{I}}_{\mathrm{g}}}{0.04-{\mathrm{I}}_{\mathrm{g}}}$

Solving, we get ${\mathrm{I}}_g=0.02\mathrm{A}$