Answer
$8.05\times 10^{-3}\; m^3/s$
Work Step by Step
The volume flow rate in the pipe is give by
$R_V=A_1V$
or, $R_V=\frac{\pi d^2V}{4}$
substituting the given values
or, $R_V=\frac{\pi\times (0.05)^2\times4.1}{4}\; m^3/s$
or, $\boxed{R_V=8.05\times 10^{-3}\; m^3/s}$
