$\Rightarrow r_{p}=10^{-2} m$

$d_{q}=4 \times 10^{-2} m$

$\Rightarrow r_{q}=2 \times 10^{-2} m$

The velocity of water flowing through the pipe is inversely proportional to the cross $-$ sectional area of tube, hence we can write

$\frac{v_{p}}{v_{q}}=\frac{a_{q}}{a_{p}}$

where, $v_{p}, v_{q}$ are velocities of water through both pipes and $a_{p}, a_{q}$ are the cross $-$  sectional area of tubes.

$\mathrm{Also}, a_{p}=\pi\left(r_{p}\right)^{2}$ and $a_{q}=\pi\left(r_{q}\right)^{2}$

$\frac{v_{p}}{v_{q}}=\frac{\pi\left(r_{q}\right)^{2}}{\pi\left(r_{p}\right)^{2}}$

$\frac{v_{p}}{v_{q}}=\frac{\left(2 \times 10^{-2}\right)^{2}}{\left(10^{-2}\right)^{2}}$

$\frac{v_{p}}{v_{q}}=4$

Hence, $v_{p}=4 \times v_{q}$