A galvanometer has a 50 division scale. The b | vedclass

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(D) Let $I$ be the full scale deflection current and $V = 2\,V$ be the battery voltage.In case $1$, when $R_1 = 2400\,\Omega$, deflection $	heta_1 =

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Current sensitivity of galvanometer is $20\,\mu A/	ext{division}$.

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(D) Let $I$ be the full scale deflection current and $V = 2\,V$ be the battery voltage.In case $1$, when $R_1 = 2400\,\Omega$, deflection $	heta_1 = 40$ divisions:$\frac{40}{50} I = \frac{V}{G + R_1} \Rightarrow \frac{4}{5} I = \frac{2}{G + 2400} \dots (1)$In case $2$, when $R_2 = 4900\,\Omega$, deflection $	heta_2 = 20$ divisions:$\frac{20}{50} I = \frac{V}{G + R_2} \Rightarrow \frac{2}{5} I = \frac{2}{G + 4900} \dots (2)$Dividing $(1)$ by $(2)$:$\frac{4/5 I}{2/5 I} = \frac{G + 4900}{G + 2400} \Rightarrow 2 = \frac{G + 4900}{G + 2400}$$2G + 4800 = G + 4900 \Rightarrow G = 100\,\Omega$.Substituting $G = 100\,\Omega$ in $(1)$:$\frac{4}{5} I = \frac{2}{100 + 2400} = \frac{2}{2500} = \frac{1}{1250}$$I = \frac{5}{4} 	imes \frac{1}{1250} = \frac{1}{1000}\,A = 1\,mA$.Current sensitivity $= \frac{I}{50} = \frac{1\,mA}{50} = 0.02\,mA/	ext{division} = 20\,\mu A/	ext{division}$.For a deflection of $10$ divisions:$\frac{10}{50} I = \frac{V}{G + R} \Rightarrow \frac{1}{5} 	imes 10^{-3} = \frac{2}{100 + R}$$100 + R = 10000 \Rightarrow R = 9900\,\Omega$.Thus, option $D$ is correct.

$A$ galvanometer has a $50$ division scale. The battery has no internal resistance. It is found that there is a deflection of $40$ divisions when $R = 2400\,\Omega$. The deflection becomes $20$ divisions when the resistance taken from the resistance box is $4900\,\Omega$. Then we can conclude:

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