  Question

A laminar stream is flowing vertically down from a tap of cross-section area $$1 { cm }^{ 2 }$$. At a distance $$10 cm$$ below the tap, the cross-section area of the stream has reduced to $$1/2 { cm }^{ 2 }$$. The volumetric flow rate of water from the tap in litre/min is

Solution

Given :     $$h = 10 cm = 0.1 m$$             $$A_1 = 1cm^2 = 10^{-4} m^2$$                    $$A_2 = 0.5cm^2 =0.5 \times 10^{-4} m^2$$$$\therefore$$ Velocity of the stream at a distance h below the tap       $$v = \sqrt{\dfrac{2gh}{1 - \frac{A_2^2}{A_1^2}}}$$$$\therefore$$        $$v = \sqrt{\dfrac{2(9.8) (0.1)}{1 - (0.5)^2}} = 1.62 m/s$$Volumetric flow of water at the tap is equal to that of at a distance h below the tap.$$\therefore$$   $$V_{tap} = v A_2 = 1.62 \times 0.5 \times 10^{-4} = 8.1 \times 10^{-5} m^3/s$$$$\implies V_{tap} = 8.1 \times 10^{-5} \times \dfrac{1000}{1/60} \approx 5$$ Litre/minPhysics

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