^{th}December 2007

In the figure, a tall cylindrical empty tank with a small side hole of square shape near its bottom is shown. The area of the hole is A. Water flowing with a velocity ‘v’ through a pipe of circular cross section having the same area A (as that of the hole) starts falling vertically down in to the tank. Which one of the following statements is correct?

(a) No water can remain in the tank

(b) The height of water column in the tank will go on increasing

(c) The height of water column in the tank will increase initially and will remain steady at v/2g

(d) The height of water column in the tank will increase initially and will remain steady at √(v/2g)

(e)** **The height of water column in the tank will increase initially and will remain steady at v^{2}/2g.

The velocity with which water will flow out through the side hole in the tank will be small initially and hence the height of water column will go on increasing until the volume of water flowing into the tank becomes exactly equal to the volume flowing out through the side hole. The height of water column remains fixed at the value ‘h’ for attaining this condition. Since the area of the hole is equal to the area of cross section of the inlet pipe, the steady height condition is attained when the velocity of efflux (of water through the hole), as given by Torricelli’s theorem, is equal to the velocity of water through the inlet pipe.

Therefore we have, v = √(2gh) from which h = v^{2}/2g [ Option (e)].

Water is flowing steadily out through the end of a vertical pipe (fig) with a velocity of 4 ms^{–1}. At what distance from the end of the pipe will the area of cross section of the stream of water be (2/3)A where A is the area of cross section of the pipe?

(a) 0.5 m (b) 1 m (c) 1.5 m (d) 2.2 m (e)** **2.5 m ** **

From the equation of continuity, we have A_{1}v_{1} = A_{2}v_{2}.

Here A_{1} = A, A_{2} = 2A/3 and v_{1} = 4 ms^{–1}** **so that** **v_{2} = 6 ms^{–1}. This is the velocity of the stream at distance ‘x’ (let us say) below the end of the pipe, where the area of cross section of the stream reduces to (2/3)A** **

Since the stream is open to the atmosphere, the pressures are the same at all regions of the stream. The kinetic energy of the water particles will be increased because of an equal decrease in the gravitational potential energy.

Considering a mass m of water, we can write ½ mv_{2}^{2 }= ½ mv_{1}^{2} + mgx. [Note that we have taken the reference (zero) level for the gravitational potential energy at distance x below the end of the pipe].

This gives

v_{2} = √(v_{1}^{2 }+ 2gx). Or, 6 = √(4^{2 }+ 2gx).

Taking the acceleration due to gravity to be 10 ms^{–2}, x = 1m.

[If you blindly write Bernoulli’s equation,** **you will have P + ρgx + (½)ρv_{1} ^{2} =** **P + (ρg×0) + (½)ρv_{2}^{2} from which you will get v_{2} = √(v_{1}^{2 }+ 2gx). You will then proceed to find ‘x’].

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