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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/min

Physics

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