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A galvanometer of resistance $50 \Omega$ is connected to a battery of $3 \mathrm{~V}$ 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
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Verified Answer
The correct answer is:
$4450 \Omega$
Current through the galvanometer
$$
\begin{aligned}
I=\frac{3}{(50+2950)} &=10^{-3} \mathrm{~A} \\
\text { Current for } 30 \text { divisions } &=10^{-3} \mathrm{~A} \\
\text { Current for } 20 \text { divisions } &=\frac{10^{-3}}{30} \times 20 \\
&=\frac{2}{3} \times 10^{-3} \mathrm{~A}
\end{aligned}
$$
For the same deflection to obtain for 20 divisions, let resistance added be $R$
$$
\begin{array}{lc}
\therefore \quad \frac{2}{3} \times 10^{-3}=\frac{3}{(50+1 R)} \\
\text { or } \quad R=4450 \Omega
\end{array}
$$
$$
\begin{aligned}
I=\frac{3}{(50+2950)} &=10^{-3} \mathrm{~A} \\
\text { Current for } 30 \text { divisions } &=10^{-3} \mathrm{~A} \\
\text { Current for } 20 \text { divisions } &=\frac{10^{-3}}{30} \times 20 \\
&=\frac{2}{3} \times 10^{-3} \mathrm{~A}
\end{aligned}
$$
For the same deflection to obtain for 20 divisions, let resistance added be $R$
$$
\begin{array}{lc}
\therefore \quad \frac{2}{3} \times 10^{-3}=\frac{3}{(50+1 R)} \\
\text { or } \quad R=4450 \Omega
\end{array}
$$
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