In the figure shown , R is equal to

 

$\mathrm{v}=\sqrt{2}\times \mathrm{g}\times$ distance of orifice from top surface of liquid 

$=\sqrt{2{\mathrm{gh}}_1}$

${\mathrm{t}}_{\mathrm{OP}}$ = free fall time 

(as vis horizontal)

$=\sqrt{\frac{2{\mathrm{h}}_2}{\mathrm{g}}}$

(Where, ${\mathrm{h}}_2$ = free fall height)

$\mathrm{R} ={\mathrm{vt}}_{\mathrm{OP}} =\left(\sqrt{2{\mathrm{gh}}_1}\right)\left(\sqrt{\frac{2{\mathrm{h}}_2}{\mathrm{g}}}\right)=2\sqrt{{\mathrm{h}}_1{\mathrm{h}}_2}$