In a glass of water, balanced on a beam balance, a steel ball weighing 250 g suspended

In a glass of water, balanced on a beam balance, a steel ball weighing 250 g suspended on a thread was lowered so that it did not touch the bottom. Determine the mass of the weight that can be used to balance the balance again.

m = 250 grams = 0.25 kilograms – the mass of the steel ball, which was lowered into a glass of water;

ro = 7800 kilograms / cubic meter – density of steel;

ro1 = 1000 kilograms / cubic meter – density of water;

g = 10 Newton / kilogram – acceleration of gravity.

It is required to determine m1 (kilogram) – the mass of the weight, with which you can balance the scales.

Find the volume of the steel ball:

V = m / ro = 0.25 / 7800 = 0.00003 m3.

Then the Archimedean force acting on the ball will be equal to:

A = ro1 * V * g = 1000 * 0.00003 * 10 = 0.3 Newtons.

Then the weight of the ball in water will be equal to:

P = m * g – A = 0.25 * 10 – 0.3 = 2.5 – 0.3 = 2.2 Newtons.

Accordingly, the mass of the load required to balance the scales will be equal to:

m1 = P / g = 2.2 / 10 = 0.22 kilograms.

Answer: in order to balance the scales, a weight equal to 0.22 kilograms is required.