Correct Answer - Option 2 : 0.62

**Concept:**

Head loss in the pipe due to friction is given as:

\(h_f = \frac {4f lV^2}{2gD}=\frac {8}{\pi ^2g}\frac {4flQ^2}{D^5} \)

where f = friction coefficient, 4f = friction factor, V = average velocity of flow, D = diameter of pipe, l = length of pipe

∴ head loss ∝ V2, head loss ∝ Q2, head loss ∝ l, \(head~loss \propto \frac 1{D^5}\)

**Calculation:**

**Given:**

d_{2} = 1.1d_{1}

From diameter proportional relation,

\(h_1 \propto \frac{1}{d_1^5}\)

\(h_2 \propto \frac{1}{d_2^5}\)

\(\Rightarrow h_2 \propto \frac{1}{(1.1d_1)^5}\)

\(\Rightarrow \frac{h_2}{h_1} = \frac{d_1^5}{(1.1d_1)^5}\)

\(\Rightarrow \frac{h_2}{h_1} = 0.621\)