Correct Answer - Option 1 : 21%

__Concept:__

Head loss due to friction in a pipe:

\({h_f} = \frac{{f'L{V^2}}}{{2gd}}\)

h_{f} = head loss due to friction, f’ = friction factor, L = length of the pipe, V = average velocity of the fluid in the pipe, g = acceleration due to gravity, d = diameter of the pipe.

**Hence keeping all other parameters to be same, h**_{f} ∝ V^{2}

__Calculations:__

**Given:**

Final velocity (V_{2}) = 1.10 × Initial velocity (V_{1})

\(\frac{V_2}{V_1}=1.1\)

∵ h_{f} ∝ V^{2}

∴ \(\frac{h_{f2}}{h_{f1}}=\left(\frac{V_2}{V_1}\right)^2\)

\(\frac{h_{f2}}{h_{f1}}=(1.1)^2\)

\(\frac{h_{f2}}{h_{f1}}=1.21\)

Final head loss (h_{f2}) = 1.21 × Initial head loss (h_{f1})

**Hence head loss will increase by 21 %.**