TUTORIAL

HYDROSTATIC


1. Find the pressure due to the fluid at a depth of 76cm in still
Water (pw=1.00g/cm³)

P=pgh
=(0.001kg/m³)(9.8m/s²)(0.76m)
=7.448kpa = 7.5 kPa
(b)mecury.(p=13.6 g/cm³)
P=pgh
=(0.0136kg/m³)(9.8m/s²)(0.76m)
=1.01×10⁵N/m²=101.3×10³ N/m²

2. A weight piston confines afluid density p in a closed container. The combined weight of piston and weight is 200N, and the cross-sentional area of piston is A=8.0cm².Find the total pressure at point B if the mercury and h=25cm.What would an ordinary pressure gouge read at B?

P = F/A + Pgh (2000/8 x 10⁻⁴)+13600(9.8m/s²)(25×10⁻²)
=2500000+33354
= 2.8×10⁵Pa
P=2.8×10⁵Pa+Pm
=2.8×10⁵Pa+1.01×10³Nm
=3.8×10⁵Pa
Pm=(13.6g/cm³)(9.81)90.76)
=0. 136kg/m³)(9.8m/s²)(0.76m)
=1.01×10⁵Nm

3. A vertical test tube has 2.0 cm of oil (p=0.8g/cm³)flooting 8.0cm of water. What is the pressure at botton of the tube due to the fluid in it ?

P₁=Pgh
=800kg/m (9.81m/s²)(2×10⁻²)
=156.96
P₂=Pgh
=1000kg/m (9.81m/s²)(8×10⁻²)
=784.8
P₁+P₂ = 156.96+784.8
=0.94 kPa

4. The u-tube devuce connected to the tank. Wahat is the pressure in the tank if atmosphera pressure is 76cm of mercury? The density of mercury is 13.6g/cm³.

P= 76cm – 13.6g/cm³(9.81ms²)(5×10⁻²)
= 76cm – 13600kg/cm³(9.81ms²)(5×10⁻²)
= 101.325 kPa -66708
=94.65×10³
=95 kPa

5. The mass of a block of aluminiumis 25.0g. (a) what is its volume?(b0what will be the tension in a string that suspends the block when the block is totally submerged in water?the density of aluminium is 2700kg/m²?

V=m/p ft=vg(e-pal)
=0.025/2700 =(9.26×10⁻⁶×9.81)(1000-2700)
= 9.26 =0.154N

6. A solid aluminium cylinder with p= 2700kg/m³ has a measured of 67g in air and 45g when immersed in turpentine. Determine the density of turpentine.

FB=V₁P₁а
PT = FB / V1a
FB+Fr=mg
FB=mg-Fr
=(0.067×9.81)-(0.045×9.81)
=0.65727-0.4415
=0.21582
V1 = m/p = 0.067 / 2700
=2.481×10⁻⁵
P₁=0.21582
=(2.481×10⁻⁵)(9.81m/s²)
=8.9×10²kg/m³