Question

A hollow sphere of mass 110 kg is chained to the bottom of a lake such that the sphere floats 30 meters above the bottom of the lake with the chain tight. The total depth of the lake is 122 meters at the location where the sphere is chained. The density of water is 1000kg/m3. The radius of the hollow sphere is 1 meter. 1. The density of rock is approximately 2300kg/m3. If a spherical rock of radius 25 cm is sitting on the bottom of the lake, find the apparent weight of the rock in the water.

Answer 1

Here, the apparent weight is equal to the rock’s weight minus the buoyant force.

Further, weight = Volume * Density rock * g
BF = Density of water * g * Volume of displaced water
Since the rock is sitting on the bottom of the lake, the volume of displaced water is equal to the volume of the rock.

Now,

V = 4/3 * π * 0.25^2 = π/12

So, weight = π/12 * 2300 * 9.8 = π * 1878⅓
BF = 1000 * 9.8 * π/12 = π * 816⅔
The apparent weight is the difference of these two forces.

Therefore,
W = π * 1878⅓ – π * 816⅔ = π * 1061⅔ = 3333.63 N
So, apparant weight of rock in the water = 3333.63 N