The velocity of flow of a liquid through a pipe at a point of radius r is v. At a point where the radius is 2r, the mass of the liquid flowing out per second is (d = density of the liquid)

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$\frac{{πr}^{2} vd}{4}$

${πr}^{2} vd$

$2 {πr}^{2} vd$

${4πr}^{2} vd$

answer is C.

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Detailed Solution

$\begin{array}{l} r_{1} = r; V_{1} = v; V_{2} = ?; r_{2} = 2 r \\ \frac{V_{1}}{V_{2}} = \frac{r_{2}^{2}}{r_{1}^{2}}; \frac{v}{V_{2}} = \frac{4 r^{2}}{r^{2}}; V_{2} = \frac{v}{4} \\ \frac{m}{t} = {πr}_{2}^{2} v_{2} ρ = π 4 r^{2} \frac{v}{4} d = {πr}^{2} vd \end{array}$

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