Velocity of outflow (efflux) from hole  $=\sqrt{2gh}$.

Range of water jet, 

$x=\sqrt{2gh}$ $\times$ Time of fall 

$=\sqrt{2gh}\times \sqrt{\frac{2H}{g}}=2\sqrt{hH}$

Velocity $v$ of bucket is 

$v=\frac{dx}{dt}=\frac{d}{dt}2\sqrt{hH}$

$=2\sqrt{H}.\frac{d\sqrt{h}}{dt}$

$\Rightarrow v=2\sqrt{H}.\frac{1}{2\sqrt{h}}.\frac{dh}{dt}$

$=\sqrt{\frac{H}{h}}.\frac{dh}{dt}$  $...\left(\mathrm{i}\right)$

Now, using equation of continuity at area $A_0$ and area $A$, we have 

$A{v}_1=A_{0}V$

where, $v_1=\sqrt{2gh}$

and $V=\frac{dh}{dt}$.

$\therefore A\sqrt{2gh}=A_0\left(\frac{dh}{dt}\right)$

Substituting for $\frac{dh}{dt}$ from Eq $\left(\mathrm{i}\right)$, we get

$v=\sqrt{\frac{H}{h}}.\frac{A}{A_0}.\sqrt{2gh}$

$=\sqrt{\frac{5}{5}}.\left(\frac{5}{500}\right).\sqrt{2\times 10\times 5}$

$=\frac{1}{10} {\mathrm{ms}}^{-1}$

$=0.1 {\mathrm{ms}}^{-1}$