A galvanometer of resistance `50 Omega ` is connected to a battery of `3 V` along with 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 division the above series resistance should be

Options

A. `6050 Omega`
B. `4450 Omega`
C. `5050Omega`
D. `5550Omega`

Answer

B

Explanation

Current through galvanometer
`I=(3V)/(50Omega+2950Omega)=10^(-3)A`
Current for 30 division `=10^(-3)A`
Current for 20 division `=(20)/(30)xx10^(-3)`
`=(2)/(3)xx10^(-3)A=(3)/(50+R)`
`rArr" "R=4450Omega`