A $1.50 \mathrm{~kg}$ hollow steel ball is submerged in water. Its weight in water is $10.75 \mathrm{~N}$. The volume of the cavity inside the ball is (density of steel is $7,900 \mathrm{~kg} / \mathrm{m}^{3}$ )

Question

Quizplus · Accepted Answer

The correct answer is A.The weight of the ball in water is equal to the weight of the water displaced by the ball. The volume of the cavity inside the ball is equal to the volume of the water displaced by the cavity. The density of steel is $7,900 \mathrm{~kg} / \mathrm{m}^{3}$ and the density of water is $1,000 \mathrm{~kg} / \mathrm{m}^{3}$. Therefore, the volume of the cavity inside the ball is:$$V = \frac{W_{water}}{(\rho_{steel} - \rho_{water})g} = \frac{10.75 \mathrm{~N}}{(7,900 \mathrm{~kg} / \mathrm{m}^{3} - 1,000 \mathrm{~kg} / \mathrm{m}^{3})(9.81 \mathrm{~m} / \mathrm{s}^{2})} = 0.000213 \mathrm{~m}^{3}$$

A 1.50 kg1.50 \mathrm{~kg}1.50 kg Hollow Steel Ball Is Submerged in Water 10.75 N10.75 \mathrm{~N}10.75 N

A $1.50 \mathrm{~kg}$ Hollow Steel Ball Is Submerged in Water $10.75 \mathrm{~N}$