The correct answer is: D. 1.0 kg/cm2 and 0.5 kg/cm2
The total stress at a depth of 5 m below the top level of water in a swimming pool is equal to the weight of the water above it. The weight of water is equal to its density times its depth. The density of water is 1000 kg/m3, and the depth is 5 m, so the total stress is 5000 kg/m2. This is equal to 5000/100 = 50 kg/cm2.
The effective stress is the total stress minus the pore water pressure. The pore water pressure is the pressure of the water in the pores of the soil. The pore water pressure at a depth of 5 m is equal to the weight of the water above it divided by the area of the pores. The area of the pores is very small, so the pore water pressure is very small. The effective stress is therefore equal to the total stress minus the pore water pressure, which is equal to 50 kg/cm2 – 0 = 50 kg/cm2.