Rate of flow through both pipes will be same
$\text { i.e., } Q_1=Q_2$
$\frac{V_1}{t}=\frac{V_2}{t}$
$\frac{\pi r_1^2 l_1}{t}=\frac{\pi r_2^2 l_2}{t}$
$\left(\text { Where } \frac{l_1}{t}=V_P \text { and } \frac{l_2}{t}=V_Q\right)$
$\Rightarrow \frac{\pi d_1^2}{4} V_P=\frac{\pi d_2^2}{4} \times V_Q$
$\Rightarrow V_P=\left(\frac{d_2}{d_1}\right)^2 V_Q$
$\Rightarrow V_P=\left(\frac{4 \times 10^{-2}}{2 \times 10^{-2}}\right)^2 V_Q$
$V_P=4 V_Q$


(The bulk modulus of rubber $=9.8 \times 10^{8}\, {Nm}^{-2}$ Density of sea water $=10^{3} {kgm}^{-3}$
$\left.{g}=9.8\, {m} / {s}^{2}\right)$