A child rolls a ball on a level floor 3.5 m to another child. If the ball makes 12.0 revolutions, what is its diameter?

When an object having a circular cross-section revolves, the distance traveled by a point on its periphery in one revolution is similar to its circumference.

Thus, the distance that the ball covers in each revolution will be equal to its circumference.

Given data:

The distance covered by the ball is\(l = 3.5\;{\rm{m}}\).

The number of revolutions is \(N = 12.0\;{\rm{rev}}\).

The expression for the circumference of the ball is as follows:

\(c = \pi D\)

Here,\(D\)is the diameter of the ball.

Now, the diameter of the ball can be calculated by using the following expression.

\(\begin{aligned}{c}c = \frac{l}{N}\\\pi D = \frac{l}{N}\\D = \frac{l}{{\pi N}}\end{aligned}\)

Substitute the values in the above expression.

\(\begin{aligned}{l}D = \frac{{\left( {3.5\;{\rm{m}}} \right)}}{{\pi \left( {12.0\;{\rm{rev}}} \right)}}\\D = 9.28 \times {10^{ - 2}}\;{\rm{m}}\end{aligned}\)

Thus, the diameter of the ball is \(9.28 \times {10^{ - 2}}\;{\rm{m}}\).