question_answer

A galvanometer of resistance \[50\,\,\Omega \] is connected to a battery of 3V 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 resitance in series should be:

A)  \[5050\,\,\Omega \]                               

B)  \[5550\,\,\Omega \]

C)  \[6050\,\,\Omega \]                   

D)  \[4450\,\,\Omega \]

Correct Answer: D

Solution :

Given \[{{R}_{1}}=50\Omega ,\,\,{{R}_{2}}=2950\,\,\Omega \] Current through the galvanometer is given by \[I=\frac{V}{{{R}_{1}}+{{R}_{2}}}\,=\frac{3}{(50+2950)}={{10}^{-3}}A\] Current for 30 divisions \[={{10}^{-3}}A\] Current for 20 divisions             \[=\frac{1{{0}^{-3}}}{30}\times 20\,=\frac{2}{3}\,\times {{10}^{-3}}A\] For the same deflection to obtain for 20 divisions, Let resistance added be R. \[\therefore \,\,\frac{2}{3}\,\times {{10}^{-3}}\,=\frac{3}{(50+1R)}\]