Two identical moving coil galvanometer have 10   Ω resistance and full scale deflection at 2…

1. The resistance of the Voltmeter will be $100\mathrm{ }\mathrm{k}\mathrm{\Omega }.$
2. The measured value of $\mathrm{R}$ will be $978\mathrm{ }\mathrm{\Omega }<\mathrm{R}<982\mathrm{ }\mathrm{\Omega }.$
3. The resistance of the Ammeter will be $0.02\mathrm{ }\mathrm{\Omega }$ (round off to $2^{\mathrm{n}\mathrm{d}}$ decimal place)
4. If the ideal cell is replaced by a cell having internal resistance of $5\mathrm{ }\mathrm{\Omega }$ then the measured value of $\mathrm{R}$ will be more than $1000\mathrm{ }\mathrm{\Omega }.$

Solution

Voltmeter rom galvanometer arrangement.
Given: $R_g=10\mathrm{\Omega }$ and $I_g=2\mu A$
Required voltage range, $V=100\times {10}^{-3}\mathrm{v}\mathrm{o}\mathrm{l}\mathrm{t}\mathrm{s}$
$\Rightarrow V=I_g\left(R_g+R_v\right)={10}^{-1}$
$\Rightarrow \frac{{10}^{-1}}{2\times {10}^{-6}}=R_g+R_v\left(as,I_g=2\mu A\right)$
$\Rightarrow 5\times {10}^4\mathrm{\Omega }\approx R_v(R_v<{10}^5\mathrm{\Omega })\Rightarrow$ Option $\left(\mathrm{A}\right)$ is incorrect.

Galvanometer to Ammeter arrangement
$\Delta V_G=\Delta V_S$
$I_g\cdot R_g=\left(I-I_g\right)\cdot S$
$\Rightarrow S=\frac{I_g\cdot R_g}{I-I_g}=\frac{2\times {10}^{-6}\times 10}{{10}^{-3}-2\times {10}^{-6}}$
$\Rightarrow S=2\times {10}^{-5}\times {10}^3=2\times {10}^{-2}$
$\Rightarrow S=20m\mathrm{\Omega }\Rightarrow \left(\mathrm{C}\right)$ is correct.
Circuit diagram for ohm's law

Resistance ammeter,
$R_A=\frac{20\times {10}^{-3}\times 10}{10}=20\times {10}^{-3}\mathrm{\Omega }$
$i=\frac{\epsilon }{\left(\frac{1000\times 50\times {10}^3}{51\times {10}^3}\right)}=\frac{51\epsilon }{5\times {10}^4}\left(\because R_A\to 0\right)$
$i^{\prime }=i\left(\frac{R_v}{51\times {10}^3}\right)=\frac{\epsilon }{1000}$
Measured resistance $\therefore R_m=\frac{i^{\prime }\times 1000}{i}=\frac{\epsilon }{51\epsilon }\times 5\times {10}^4=\frac{5\times {10}^4}{51}=980.4\mathrm{\Omega }$
If the voltmeter shows full scale deflection then
$\frac{\epsilon }{980}\times \left(\frac{1000}{51\times {10}^3}\right)\times 5\times {10}^4={10}^{-1}$
$\therefore \epsilon =999.6mv$
Since $i_A=1mA$ so max reading of R can be $\frac{999.6mV}{1mA}=999.6\mathrm{\Omega }$

Asked in: JEE Advanced 2019 (Paper 1)

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