A small sphere of radius 2cm is taken to a depth of 500m under the sea.  If the sphere shrinks to a radius of 1.98 cm, then Find the hydraulic stress of the sphere.   ×105×10 5 Pascals  Find the hydraulic strain of the sphere.

To solve this problem, we first need to understand the definitions of hydraulic stress and strain.

Hydraulic stress is the force per unit area exerted by a fluid against a surface. In this case, it's the pressure exerted by the sea water on the sphere.

Hydraulic strain is the change in volume of an object due to the applied stress, divided by the original volume of the object.

Step 1: Find the hydraulic stress

The pressure at a certain depth under the sea can be calculated using the formula P = ρgh, where ρ is the density of the fluid (sea water), g is the acceleration due to gravity, and h is the depth.

The density of sea water is approximately 1025 kg/m³, g is approximately 9.8 m/s², and the depth is 500 m.

So, P = 1025 kg/m³ * 9.8 m/s² * 500 m = 5.0225 * 10^6 Pa or 5.0225 * 10^5 kPa

Step 2: Find the hydraulic strain

The strain is the change in volume divided by the original volume.

The volume of a sphere can be calculated using the formula V = 4/3πr³.

The original volume of the sphere is V1 = 4/3π(2 cm)³ = 33.51 cm³.

The final volume of the sphere is V2 = 4/3π(1.98 cm)³ = 32.67 cm³.

So, the change in volume is ΔV = V2 - V1 = 32.67 cm³ - 33.51 cm³ = -0.84 cm³.

The strain is then ΔV/V1 = -0.84 cm³ / 33.51 cm³ = -0.025 or -2.5%.

So, the hydraulic stress is 5.0225 * 10^5 kPa and the hydraulic strain is -2.5%.