A 1.50 kg hollow steel ball is submerged in water. Its weight in water is 10.75 N. The volume of the cavity inside the ball is (density of steel is 7,900 kg/m³)

Question

Quizplus · Accepted Answer

First, we need to find the volume of the ball:
Weight in air = Weight in water + Buoyant force
Weight in air = 1.50 kg * 9.81 m/s^2 = 14.715 N
Buoyant force = 14.715 N - 10.75 N = 3.965 N
Buoyant force = weight of water displaced by the ball
Density of water = 1000 kg/m^3
Volume of water displaced = Buoyant force / Density of water = 0.003965 m^3
Volume of ball = Volume of water displaced = 0.003965 m^3

Next, we need to subtract the volume of the steel shell to find the volume of the cavity:
Density of steel = 7,900 kg/m^3
Mass of steel shell = 1.50 kg
Volume of steel shell = Mass of steel shell / Density of steel = 0.000190 m^3
Volume of cavity = Volume of ball - Volume of steel shell = 0.003965 m^3 - 0.000190 m^3 = 0.000213 m^3

Therefore, the answer is B.

A 150 Kg Hollow Steel Ball Is Submerged in Water