A galvanometer of $50 \Omega$ resistance has 25 divisions. A current of $4 \times 10^{-4}$ A gives a deflection of one division. To convert this galvanometer into a voltmeter having a range of 25 V , it should be connected with a resistance of

• A

  $2500 \Omega$ as a shunt
• B

  $245 \Omega$ as a shunt
• C

  $2550 \Omega$ in series
• Correct answer
  D

  $2450 \Omega$ in series

Solution

Given, resistance of galvanometer $\left(\mathrm{R}_{\mathrm{g}}\right)=50 \Omega$ Number of divisions $=25 \mathrm{I}_{\mathrm{g}}=4 \times 10^{-4} \times 25=10^{-2} \mathrm{~A}$ R is the resistance connected in series of galvanometer to convert into voltmeter $\begin{aligned} \therefore V & =I_g\left(R_g+R\right) \\ 25 & =10^{-2}(50+R) \\ \Rightarrow R & =\frac{25}{10^{-2}}-50=2500-50 \Rightarrow R=2450 \Omega \text { in } \end{aligned}$ series.

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