A galvanometer of resistance 50 `Omega` is connected to a battery of 3 V along with a resistance of 2950 `Omega` in series shows full-scale deflection of 30 divisions. The additional series resistance required to reduce the deflection to 20 divisions is

Options

A. `1500 Omega`
B. `4440 Omega`
C. `7400 Omega`
D. `2950 Omega`

Answer

A

Explanation

Current flowing in galvanometer,
`I=(3)/((50+2950))`
`I=10^(-3)A`
Current for 30 division `= 10^(-3)A`
Current for 20 division `= (10^(-3))/(30)xx20-(2)/(3)xx10^(-3)A`
Let the series resistance = R
`therefore (2)/(3)xx 10^(-3)=(3)/((50+R))`
`R = 4450 Omega`
Option (c ) is nearest option in the given options.