Question

A galvanometer of resistance $50\mathrm{\Omega }$ is connected to a battery of $3\mathrm{V}$ along with a resistance of $2950\mathrm{\Omega }$ 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

A

$6050\mathrm{\Omega }$

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B

$4450\mathrm{\Omega }$

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C

$505\mathrm{\Omega }$

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D

$5550\mathrm{\Omega }$

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Solution

The correct option is B $4450\mathrm{\Omega }$Step 1. Given data A galvanometer of resistance $50\mathrm{\Omega }$ is connected to a battery of $3\mathrm{V}$ along with a resistance of $2950\mathrm{\Omega }$ in series which is shown in above fiugre. A full-scale deflection of $30$ divisions is obtained in the galvanometer. In order to reduce this deflection to $20$ divisions.We have to find the resistance in series that is to be connected to the above set up such that the deflection for $3\mathrm{V}$ connection across it corresponds to $20$ divisions.Step 2. Formula to be used.According to Kirchhoff's law, If ${\mathrm{V}}_{1},{\mathrm{V}}_{2},.....{\mathrm{V}}_{\mathrm{n}}$ shows the individual potential difference across them, then by Kirchhoff’s law of voltage we get, $\mathrm{V}={\mathrm{V}}_{1}+{\mathrm{V}}_{2}+.....+{\mathrm{V}}_{\mathrm{n}}$For the same is the effective resistance in series is $\mathrm{R}$, then current $\mathrm{I}$ in the circuit is given by ohms law i.e. $\mathrm{V}=\mathrm{IR}$Step 3. Calculate current and total initial resistance. Total initial resistance is,${\mathrm{R}}_{\mathrm{G}}+{\mathrm{R}}_{1}=50+2950$ $=3000\mathrm{\Omega }$ $\mathrm{\epsilon }=3\mathrm{V}$Therefore,Current $=\frac{3\mathrm{V}}{30000\mathrm{\Omega }}$ $=1×{10}^{-3}\mathrm{A}$Step 4. The value of current the resistance R to be connected in series.If the deflection has to be reduced to $20$ divisions, current $\mathrm{i}=1\mathrm{mA}×\frac{2}{3}$ as full deflection scale for $1\mathrm{mA}$ which is equal to $30$ divisions. $3\mathrm{V}=3000\mathrm{\Omega }×1\mathrm{mA}$ $=x\mathrm{\Omega }×\frac{2}{3}\mathrm{mA}$$⇒$ $\mathrm{x}=3000×1×\frac{3}{2}$ $=4500\mathrm{\Omega }$But the galvanometer resistance is $50\mathrm{\Omega }$.So, The resistance to be added $=\left(4500-50\right)\mathrm{\Omega }$ $=4450\mathrm{\Omega }$Hence, option $\mathrm{B}$ is correct answer.

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