Correct answer is (c) 0.036 m^3/s
Easiest explanation: Given,
Interface among three wells,
B=100m
Radius of tube well=10cm=0.1m
coefficient of permeability is 4.15 * 10^-4 m/s
drawdown (H-h)=4m
aquifer=30m
radius of influence=245m
the discharge is given by,
\(q_1=q_2=q_3= \frac{2πkb(H-h)}{log_e \frac{R^3}{rB^2}} \)
\(q_1=q_2=q_3= \frac{2π*4.15 * 10^{-4}*30*4}{log_e \frac{245^3}{0.1*100^2}} \)
∴ q1=q2=q3=0.033 m^3/s.