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

• A

  $5050 \Omega$
• B

  $5550 \Omega$
• C

  $6050 \Omega$
• Correct answer
  D

  $4450 \Omega$

Solution

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 $R$ $$ \begin{array}{lc} \therefore \quad \frac{2}{3} \times 10^{-3}=\frac{3}{(50+1 R)} \\ \text { or } \quad R=4450 \Omega \end{array} $$

Hence, the correct answer is option: 'D'.