Correct Answer - Option 3 : 20.0

Given that,

Pressure drop (ΔP_{1}) = 5 kg/m^{2}

Flow rate (Q) is doubled.

= Q_{2} = 2Q_{1}

The relation between pressure drop and flow rate is,

\(Q \propto \sqrt {{\rm{\Delta }}P}\)

\(\Rightarrow \frac{{{Q_1}}}{{{Q_2}}} = \;\sqrt {\frac{{{\rm{\Delta }}{{\rm{P}}_1}}}{{{\rm{\Delta }}{{\rm{P}}_2}}}}\)

\(\Rightarrow \frac{{{Q_1}}}{{2{Q_1}}} = \;\sqrt {\frac{5}{{{\rm{\Delta }}{{\rm{P}}_2}}}}\)

⇒ ΔP

_{2} = 5 × 4 = 20 kg/m

^{2}