**Numerical**

*The surface Tension of water in contact with air at 20°C is 0.0725 N/m. The pressure inside a droplet of water is to be 0.02 N/cm² greater than the outside pressure. Calculate the diameter of the droplet of water. (7 marks)*

**Solution**

Surface tension, σ = 0.0725 N/m

Pressure intensity, P = 0.02 N/m

Pressure intensity, P = 0.02 N/m

^{2}P = 4σ/d

Hence, the Diameter of the drop

d = 4 x 0.0725/200 = 1.45 mm

**Numerical**

*Find the surface tension in a soap bubble of 40 mm diameter when the inside pressure is 2.5 N/m² above atmospheric pressure. (7 marks)*

Answer: 0.0125 N/m

**Numerical**

*The pressure outside the droplet of water of diameter 0.04 mm is 10.32 N/cm² (atmospheric pressure). Calculate the pressure within the droplet if surface tension is given as 0.0725 N/m of water. (7 marks)*

Answer: 0.725 N/cm

^{2}**Numerical**

*An open lank contains water up to a depth of 2 m and above it an oil of specific gravity 0.9 for a depth of 1 m. Find the pressure intensity (i) at the interface of the two liquids, and (ii) at the bottom of the tank. (8 marks)*

**Solution**

See figure.

Pressure intensity at the interface of both liquids

= r

Pressure intensity at the bottom

= r

= (0.9 x 1000) x 9.81 x 1 + (1 x 1000) x 9.81 x 2

= 8829 + 19620 = 28449 N/m

= r

_{oil}gh_{1}= (0.9 x 1000) x 9.81 x 1 = 8829 N/m^{2}Pressure intensity at the bottom

= r

_{oil}gh_{1}+ r_{water}gh_{2}= (0.9 x 1000) x 9.81 x 1 + (1 x 1000) x 9.81 x 2

= 8829 + 19620 = 28449 N/m

^{2}**Numerical**

*A rectangular pontoon is 5 m long, 3 m wide and 1.20 m high. The depth of immersion of the pontoon is 0.80 m in seawater. If the centre of gravity is 0.6 m above the bottom of the pontoon, determine the meta-centric height. The density for seawater = 1025 kg/m*

^{3}.

**Solution**

Dimension of pontoon = 5 m x 3 m x 1.20 m

Depth of immersion = 0.8 m

Depth of immersion = 0.8 m

AG = 0.6 m

AB = (1/2) x Depth of immersion = (1/2) x .8 = 0.4 m

Density for seawater = 1025 kg/m

AB = (1/2) x Depth of immersion = (1/2) x .8 = 0.4 m

Density for seawater = 1025 kg/m

^{3}Meta-centre height GM. given by Equation is

GM = (I/V

_{sub}) - BGwhere I = M.O. Inertia of the plan of the pontoon about the y-y axis

= (1/12) x 5 x 3

^{3}= 45/4 m^{4}^{V}

_{sub}= 3 x 0.8 x 5.0 = 12.0 m

^{3}

Hence, GM = (45/4)(1/12) - 0.2 = 0.7375 m

**Numerical**

In a 100 mm diameter horizontal pipe, a venturimeter of 0.5 contraction ratio has been fixed. The head of water on the metre when there is no flow is 3 m (gauge). Find the rate of flow for which the throat pressure will be 2 metres of water absolute. The coefficientt of discharge is 0.97. Take atmospheric pressure head = 10.3 m of water. (10 marks)

**Solution**

Diameter of the pipe, d

_{1}= 100 mm = 10 cmArea, a

_{1}= (π/4)d_{1}^{2}= (π/4)(10)^{2}= 78.54 cm^{2}Diameter of the throat, d

_{2}= 0.5 x d_{1}= 0.5 x 10 = 5 cmArea, d

_{2}=(π/4)d_{2}^{2}= 19.635 cm^{2}Head of water for no flow = p

_{1}/ρg = 3 m (gauge) = 3 + 10.3 = 13.3 (absolute)Throat pressure head = p

_{2}/ρg = 2 m of water absolute=

= 29306.8 cm

^{3}/s = 29.306 litres/s.**Numerical**

An orifice meter with orifice diameter 15 cm is inserted in a pipe of 30 cm diameter. The pressure difference measured by a mercury oil differential manometer on the two sides of the orifice meter gives a reading of 50 cm of mercury. Find the rate of flow of oil of sp.gr. 0.9 when the coefficient of discharge of the orifice meter is 0.6. (10 marks)

**Solution**

A

_{0}= (π/4)d_{0}^{2}= (π/4)(0.15)^{2}= 0.0176 m^{2}A

_{0}= (π/4)d_{0}^{2}= (π/4)(0.15)^{2}= 0.0176 m^{2}Similarly, A

_{1}= 0.0706 m^{2}= 705.5 cm of oil

Discharge

=

= 0.137 m

^{3}/s---

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