A galvanometer of resistance $50\, \Omega$ is connected to a battery of $3\, V$ along with a resistance of $2950\, \Omega$ in series. A full scale deflection of $30$ divisions is obtained in the galvanometer. In order to reduce this deflection to $20$ divisions, the resistance in series should be.......$Ω$
AIPMT 2008, Diffcult
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Total initial resistance $=R_{G}+R_{1}=(50+2950)\, \Omega=3000\, \Omega$
$\varepsilon=3\, \mathrm{V}$
$\therefore$ Current $=\frac{3\, \mathrm{V}}{3000 \,\Omega}=1 \times 10^{-3} \,\mathrm{mA}$
If the deflection has to be reduced to $20$ divisions, current $i=1\, \mathrm{mA} \times \frac{2}{3}$ as the full deflection scale for
$1\,\mathrm{mA}=30$ divisions.
$3\, \mathrm{V}=3000\, \Omega \times 1 \,\mathrm{mA}=x\, \Omega \times \frac{2}{3} \,\mathrm{mA}$
$\Rightarrow x=3000 \times 1 \times \frac{3}{2}=4500\, \Omega$
But the galvanometer resistance $=50\, \Omega$
Therefore the resistance to be added
$=(4500-50) \,\Omega=4450\, \Omega$
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