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(C) Given,resistance of galvanometer $R_g = 100 \Omega$. Full scale deflection current $I_g = 50 \mu A = 5 	imes 10^{-5} 	ext{ A}$. Required range o

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$0.5 \Omega$

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(C) Given,resistance of galvanometer $R_g = 100 \Omega$. Full scale deflection current $I_g = 50 \mu A = 5 	imes 10^{-5} 	ext{ A}$. Required range of ammeter $I = 10 	ext{ mA} = 10^{-2} 	ext{ A}$. To convert a galvanometer into an ammeter,a shunt resistance $R_s$ is connected in parallel. The formula for shunt resistance is $R_s = \frac{I_g}{I - I_g} 	imes R_g$. Substituting the values: $R_s = \frac{5 	imes 10^{-5}}{10^{-2} - 5 	imes 10^{-5}} 	imes 100$ $R_s = \frac{5 	imes 10^{-5}}{100 	imes 10^{-4} - 0.5 	imes 10^{-4}} 	imes 100$ $R_s = \frac{5 	imes 10^{-5}}{995 	imes 10^{-5}} 	imes 100$ $R_s = \frac{500}{995} \approx 0.5025 \Omega$. Thus,the resistance to be added is approximately $0.5 \Omega$.

$A$ galvanometer has a coil of resistance $100 \Omega$ showing a full scale deflection at $50 \mu A$. The resistance that should be added to use it as an ammeter of range $10 \text{ mA}$ is

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