A galvanometer has coil of resistance 50Ω and shows full deflection at 100μA. The resistance to be added for the galvanometer to work as an ammeter of range 10mA is nearly

Given,

$G=50\Omega$

$i_g=100\times {10}^{-6}A={10}^{-4}A$

$i=10\times {10}^{-3}A={10}^{-2}A$

The galvanometer can be converted into an ammeter of range 10 mA by adding a shunt resistor in parallel with it

$S=\frac{G}{\left({\displaystyle \frac{i}{i_g}}-1\right)}$

$S=\frac{50}{\left({\displaystyle \frac{{10}^{-2}}{{10}^{-4}}}-1\right)}$

$S=\frac{50}{100-1}$

$S\approx 0.5\Omega$

Hence the correct answer is 0.5Ω in parallel.