A galvanometer of resistance `50Omega` is connecter to a battery of 8 V along with a resistance of `3950Omega` in series. A full scale deflection on 30 div is obtained in the galvanometer. In order to reduce this deflection of 15 division, the resistance in series should be ______ `Omega`.

Options

A. 7900
B. 2000
C. 1950
D. 7950

Answer

D

Explanation

`I=V/(R+G)`
`I=8/(3950+50)=8/4000=1/500A`
Now, `I/2=V/(R^(1)+G)`
`1/(2xx500)=8/(R^(1)+50)`
`thereforeR^(1)+50=8xx1000`
`thereforeR^(1)=8000-50`
`thereforeR^(1)=7950Omega`