Right choice is (c) 0.036 m^3/s
Explanation: Given,
Interface among two 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,
\(q_1=q_2=\frac{2πkb(H-h)}{log_e \frac{R^2}{rB}}\)
\(q_1=q_2=\frac{2π*4.15 * 10^(-4)*30*4}{log_e \frac{245^2}{0.1*100}} \)
∴ q1=q2=0.036 m^3/s.