Question

WITH COMPLETE SOLUTION

PROBLEM 1:

An infinite slope contains soil whose saturated unit weight is 18.639 kN/m^3 . There is a seepage through the soil and the groundwater table coincides with the ground surface. The cohesion and angle of friction are 30 kPa and 25°, respectively. Which of the following gives the factor of safety against sliding on the rock surface?

PROBLEM 2:

Calculate the excess pressure (above atmospheric) inside a soap bubble of radius 5 mm. The surface tension of the soap solution is known to be 0.042 N/m, in Pascals.

PROBLEM 3:

A tank contains oil (SG=0.80), gasoline (SG=0.90) and water(SG=1.05). What is the pressure at a depth of 1.20 m (in kg/m 2 ), if the depths of the liquids are 0.50 m, 0.60 m and 0.80 m respectively?

Answer 1

Ans) We know,when groundwater table coincides with ground surface, Factor of safety (FOS) can be given as :

FOS = C / ( H tan ) + ( / ) (tan / tan )

where, C = Cohesion = 30 kPa

= saturated unit weight = 18.639 kN/m3

= submerged unit weight = - = 18.639 - 9.81 = 8.829 kN/m3

H = Thickness of slope = 5 m (assume)

= Slope angle =  20 degree (assume)

= Friction angle = 25 degree

Putting values,

=> FOS = 30 / (18.639 x 5 x 20 tan 20) + (8.829/18.639) (tan 25 / tan 20)

=> FOS = 1 + 0.60

=> FOS = 1.60

.

Ans) We know, excess pressure inside soap bubble (P) ,

P = 4 S / R

where, S = Surface tension = 0.042 N/m

R = Radius of bubble = 5 mm or 0.005 m

Putting values,

=> P = 4  x 0.042 / 0.005

=> P = 33.6 Pa

.

Ans) Pressure at depth of 1.2 m can be given as,

=> P = g (0.5) + g (0.60) + g (0.1)

=> P = [(0.80 x 1000) x 9.81 x 0.5] + [0.90 x 1000) x 9.81 x 0.60] + [(1.05 x 1000) x 9.81 x 0.1]

=> P = 3924 + 5297.4 + 1030.05

=> P = 10251.45 N/ or 1045 kg/