Question

i need 2 and 3 solved

(1) A typical mud is 70 wt% sand, balance water. What is its density near RT if Puand = 2300 kg/mº? (2) An open-ended can 0.2 m tall is originally full of air, at 20°C. It is now immersed in water upside down, such that its top (now bottom) is 10 m below the water surface. The can is 0.2 m in height. Assuming the air stays at 20°C, how high will the water rise in the can? (3) A diver, 2 m tall and weighing 90 kg, is working in a vertical position with his head at 8 m depth. If we approximate his body as a cylinder, neglecting top and bottom, what is the total net force of water against him? If his body will dissolve 2.5 x 10 gair (g bl. dy-atm), how much extra air (L at STP) will his body eventually dissolve? Assume the body is essentially water. (4) An inclined manometer (shown) is measuring the P of a gas at point A (0.25 kPa gauge). It is desired that the reading length L above the reservoir level be 14 cm at this P. The manometer fluid is an oil of specific gravity 0.897. At what angle should the manometer be inclined? L-14 cm Oil level

Answer 1

Ans 2) According to Boyle's law,

P1 V1 = P2 V2

where, P1 = initial pressure of air inside cylinder

V1 =Initial Volume of air inside cylinder

P2 = Final pressure of air inside cylinder

V2 = Final volume of air inside cylinder

P1 = 101.325 kPa at 20^o C

V1 = Area x height = 0.2 A

   P2 = 101.325 + ($\dpi{100} \rho$ g H) = 101.325 + (1000 x 9.81 x 10) = 199.425 kPa

V2 = Area x height = A h

=> 101.325 x 0.2 A = 199.425 x A h

=> h = 0.102 m

Height of water = Total height - Height of air

= 0.20 - 0.102 = 0.098 m or 9.8 cm

Ans 3) Given,

Height = 2 m

Weight = 90 kg

Head = 8 m

Pressure experienced by man under water =  $\dpi{100} \rho$ g h

= 1000 x 9.81 x 8

= 78480 N/m²

Area of cylindrical body = 2$\dpi{100} \pi$r(r+h)

Assume radius as 0.1 m , Area = 0.2$\dpi{100} \pi$(2.1) =1.318 m²

Force experienced by body = Pressure x area

= 78480 x 1.318

= 103500 N or 103.5 kN

Now, air dissolve  rate = 2.5 x 10^-5gm / weight atm

   Pressure = 78480 N/m² or 0.775 atm

Weight of body = 90 kg or 90,000 gm

Extra air dissolved = 2.5 x 10^-5 x 90.000 x 0.775

= 1.74 gm

Density of air = 1.125 kg/m³ or 0.001125 g/cm³

=> Volume of 3.99 gm air = 1.74 / 0.001125

= 1546.67 cm³

or 1.546 L

Hence, extra air absorbed by body =1.546 L