Correct Answer - Option 2 : 2.756
\({h_f} = \frac{{fL{V^2}}}{{d \times 2g}}\)
So, \(V\alpha \surd d\)
Now, \(\frac{{{Q_1}}}{{{Q_2}}} = \frac{{{A_1}{V_1}}}{{{A_2}{V_2}}} = {\left( {\frac{{{d_1}}}{{{d_2}}}} \right)^{\frac{5}{2}}}\)
\(\Rightarrow \frac{{{Q_1}}}{{{Q_2}}} = {\left( {\frac{{15}}{{10}}} \right)^{\frac{5}{2}}} = 2.756\)