## Tuesday, September 19, 2006

### Properties of Fluids

In any Medical and Engineering entrance test paper you will find a few questions based on the properties of fluids. Let us discuss some typical questions. The following simple question is taken from the A.F.M.C.2004 test paper:
Application of Bernoulli’s theorem can be seen in
(a) dynamic lift of aeroplane (b) hydraulic press (c) helicopter (d) none of these

The correct option is (a). The shape of the aeroplane wings is such that when it moves forward, the air molecules at the top of the wings have a greater velocity (relative to the wings) compared to the air molecules at the bottom. Therefore, in accordance with Bernoulli’s principle, the pressure at the top of the wings is less than that at the bottom. This results in a dynamic lift of the wings which balances the weight of the plane.
Let us now consider another question involving Bernoulli’s principle:

Water is flowing steadily through two horizontal pipes of radii 3cm and 6cm connected in series. The speed of water in the first pipe is 2m/s and the pressure of water in it is 2×104 pascal. The pressure of water in the second pipe will be nearly
(a) 2×104 pascal (b) 2.2×104 pascal (c) 2.4×104 pascal (d) 2.6×104 pascal (e) 3×104 pascal

The speed v2 of water in the second pipe is given (from the equation of continuity, a1v1 = a2v2 ) by v2 = 2×a1/a2 = 2×1/4 = 0.5 m/s. (The ratio of cross section areas is ¼ since the radius of the second pipe is twice that of the first).
As per Bernoulli’s principle, we have P1 + (ρv12 )/2 + ρgh = P2 + (ρv22)/2 + ρgh with usual notations. Therefore, P2 = P1 + ρ(v12- v22)/2 =2×104 + 1000(22 – 0.52)/2 = 2.1875×104 pascal. The correct option therefore is (b).
The following question involving surface tension may appear to be difficult for some of you. See how simple it is:
Two soap bubbles of radii ‘R’ and ‘r’ (R > r) are touching each other. The radius of curvature of the common surface at the region of contact is
(a) (R+r)/2 (b) R – r (c) (R – r)/2 (d) Rr/(R+r) (e) Rr/(R-r)
At the contact region, the radius of curvature (R') of the soap film is governed by the net excess of pressure 4T/r – 4T/R where T is the surface tension. Therefore, we have 4T/r – 4T/R = 4T/R' from which R' = Rr/R-r [option (e)].
Consider now the following M.C.Q.:

Two rain drops are falling through air with the same terminal velocity of 8cm/s. If they coalesce, the terminal velocity of the combined single drop will be (in cm/s)
(a) 8/√2 (b) 16 (c) 8×41/3 (d) 8×21/3 (e) 8√
2
On equating the apparent weight of a rain drop to the viscous force (air resistance) opposing the downward motion of the drop, we have, 4/3πr3(ρ-σ)g = 6πrηv, with usual notations. Therefore, the velocity, v α r2…………(1)
Since the drops have the same terminal velocity, it follows that they have the same radius. The radius (r1 ) of the single large drop (when the two drops coalesce) is given by
2× 4/3πr3 = 4/3 πr13. Therefore, r1=21/3 r.
The velocity of the combined single drop, v1 α [21/3 r]2 -------------- (2)
From equations (1)&(2), v1= 41/3v = 41/3 × 8, since v=8cm/s. The correct option therefore is (c).

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