Answer to Question #88061 in Mechanics | Relativity for shanelle

Question

A large container of water is placed on a set of scales which initially read 0.485 kg. A ball is hung in the air from another set of scales which initially read 0.112 kg.

The ball is lowered into the large container until it is fully submerged.

The readings on both scales are now 0.496 kg and 0.101 kg respectively.

What is the density of the ball? (to 3 s.f and in kg m−3)

(ρwater = 1000 kg m−3)

Answer 1

$$W_1=\rho g V$$

$$W_2=\rho g V-\rho_w g V=(\rho-\rho_w)gV$$

Thus,

$$\frac{W_2}{W_1}=\frac{m_2}{m_1}=\frac{(\rho-\rho_w)}{\rho}=1-\frac{\rho_w}{\rho}$$

$$\frac{0.101}{0.112}=1-\frac{1000}{\rho}$$

$$\rho=10200 \frac{kg}{m^3}$$