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 \(I=\frac{3}{(50+2950)}=10^{-3} \mathrm{~A}\) Current for 30 divisions \(=10^{-3} \mathrm{~A}\) Current for 20 divisions \(=\frac{10^{-3}}{30} \times 20\) \(=\frac{2}{3} \times 10^{-3} \mathrm{~A}\) For the same deflection to obtain for 20 divisions, let resistance added be R \(\begin{aligned} & \therefore \frac{2}{3} \times 10^{-3}=\frac{3}{(50+1 \mathrm{R})} \\ & \text {or } \mathrm{R}=4450 \Omega \end{aligned}\)

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