Explanation

(c) From Bernoulli’s theorem,
$\ \ \ \ \ \ \rho gh = \frac{1}{2}\rho\left( v_{2}^{2} - v_{1}^{2} \right)$
$\Longrightarrow \ gh = \frac{1}{2}v_{1}^{2}\left( \left( \frac{v_{2}}{v_{1}} \right)^{2} - 1 \right)$
$\Longrightarrow \ gh = \frac{1}{2}v_{1}^{2}\left( \left( \frac{A_{1}}{A_{2}} \right)^{2} - 1 \right)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \therefore{(A}_{1}v_{1} = A_{2}v_{2})$
$\Longrightarrow \ \ \ \left( \frac{A_{1}}{A_{2}} \right)^{2} = 1 + \frac{2hg}{v_{1}^{2}}$
$\Longrightarrow \ \ \ \left( \frac{D_{1}}{D_{2}} \right)^{4} = 1 + \frac{2hg}{v_{1}^{2}}$
$\Longrightarrow \ \ \ D_{2} = \frac{D_{1}}{\left( 1 + \frac{2gh}{v_{1}^{2}} \right)^{1/4}}$
$= \frac{8 \times 10^{- 3}}{\left( 1 + \frac{2 \times 10 \times 0.2}{{(0.4)}^{2}} \right)^{1/4}} = 3.6 \times 10^{3}m$