question_answer

A galvanometer of resistance \[50\,\,\Omega \] is connected to a battery of 3 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) \[6050\,\,\Omega \]                      

D) \[4450\,\,\Omega \]

Correct Answer: D

Solution :

Current through the galvanometer \[I=\frac{3}{(50+2950)}={{10}^{-3}}A\] Current for 30 divisions \[={{10}^{-3}}A\] Current for 20 divisions\[=\frac{{{10}^{-3}}}{30}\times 20\] \[=\frac{2}{3}\times {{10}^{-3}}A\] For the same deflection to obtain for 20 divisions, let resistance added be R \[\therefore \]\[\frac{2}{3}\times {{10}^{-3}}=\frac{3}{(50+1R)}\]or\[R=4450\Omega \]