A gardener waters the plants by a pipe of diameter 1 mm. The water comes out from the pipe atthe rate of 10 cm^{3}/sec. The reactionary force exerted on the hand of the gardener is?

Diameter of pipe = 1 mm

Radius will be, r = 0.5 mm = 0.0005 m

Area of cross-section of the pipe is, A = πr^{2}

Density of water is, d = 1000 kg/m^{3}

Volume of water coming out per sec is = 10 cm^{3}/s = 10^{-5} m^{3}/s

Mass of water coming out per sec is,

dm/dt = density × volume/sec

= (1000)(10^{-5}) kg/s = 10^{-2} kg/s

dm/dt = 10^{-2} kg/s

Now,

Volume/sec = Area × velocity

A = πr^{2}

^{ = }3.14 x (0.0005^{2})

=> 10^{-5} = (3.14) × (0.0005^{2}) × v

=> v = 12.74 m/s

Now, force = change in momentum/sec

=> F = d(mv)/dt

=> F = v(dm/dt) [velocity, ‘v’ is constant]

=> F = (12.74)(10^{-2})

=> F = 0.1274 N

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