A galvanometer of resistance `50Omega` is connected to a b attery of 3V alongwith a resistance of `2950Omega` in series. A full scale deflection of 30 division is obtained in the galvanometer in order to reduce this deflection to 20 division. The resistance in sereis. should be:-

Options

A. `5050Omega`
B. `550Omega`
C. `6050Omega`
D. `4450Omega`

Answer

D

Explanation

Current through the galvanomete,
`l=(3)/((50+2950))=10^(-3)A`
Current for 30 divisions `= 10^(-3)A`
Current for 20 divisions `= (10^(-3))/(30)xx20 =(2)/(3)xx10^(-3)A`
For the same deflection to obtain for 20 divisions, let resistance added be R
`therefore " " (2)/(3)xx10^(-3)=(3)/((50+1R))`
`rArr " " R = 4450 Omega`