The golden ring in the air weighs 0.04 N and in water 0.02 N. Identify the golden ring solid or hollow?

Given:

P1 = 0.04 Newton – the weight of the golden ring in the air;

P2 = 0.02 Newton – the weight of the gold ring in water;

g = 10 Newton / kilogram – acceleration of gravity;

ro1 = 1000 kg / m3 – water density;

ro2 = 19300 kg / m3 is the density of gold.

It is required to determine whether there are cavities in the ring.

Let’s find the Archimedean force acting on the ring:

F = P1 – P2 = 0.04 – 0.02 = 0.02 Newton.

Then the volume of the ring will be equal to:

V = F / (g * ro1) = 0.02 / (10 * 1000) = 0.02 * 10-4 m3.

A solid ring of gold of this volume should have a mass equal to:

m = V * ro2 = 0.02 * 10-4 * 19300 = 0.02 * 1.93 = 0.0386 kilograms.

The actual mass of the ring is:

m1 = P1 / g = 0.04 / 10 = 0.004 kilograms.

Since m> m1, it means there are cavities in the ring.

Answer: There are cavities in the ring.