A galvanometer of resistance 50 Ω is connected to a . . .

Question : A galvanometer of resistance 50 Ω is connected to a battery of 3 V along with a resistance of 2950 Ω in series. A full scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20 divisions, the resistance in series should be 

a) 6050 Ω

b) 4450 Ω

c) 5050 Ω

d) 5550 Ω

Doubt by Eshani

Solution : 

G=50 Ω

R=2950 Ω

ε = 3 V

Total Resistance (R_eq)  = 50+2950 = 3000 Ω

Current drawn from the battery, I = ε/R_eq
I = 3/3000
I = 1/1000 = 10^-3 A = 1 mA

30 divisions are given by 1 mA  of current
1 division is given by 1/30 mA of current

20 divisions are given by (1/30)×20 mA of current

= 20/30 mA

= 2/3 mA

= 2/3000 A

= 1/1500 A

Let new resistance added in the series will be R'

So, 
ε=I'(G+R')
3=(1/1500)(50+R')
3=(50+R)/1500

3×1500 = 50+R

4500 = 50+R
4500-50 = R
4450 = R
R = 4450 Ω
So, the required resistance in the series will be of 4450 Ω. 

Hence, b) would be the correct option.