4 balls each with a radius 1 inch fit snuggly into a 2x2x8 inch box. Is the total volume left over inside the box greater than or less than the volume of a fifth ball? Justify answer.
I got as far as figuring the volume of box is 32 cubic inches
I'm not sure about the balls?
Volume of 1 ball, of radius r
V=4πr^3/3
=4π1^3/3
=4π/3
Interstitial volume (empty space)
= 32 in.³ - 4V
You will find that the volume left over is just slightly less than half the volume of the box.
To determine whether the total volume left over inside the box is greater than or less than the volume of a fifth ball, we first need to find the volume occupied by the four balls.
Each ball has a radius of 1 inch. The formula for the volume of a sphere is given by V = (4/3)πr^3, where "V" represents volume and "r" represents radius.
So, for each ball, the volume is V_ball = (4/3)π(1^3) = (4/3)π cubic inches.
Since there are four balls, the total volume occupied by the balls is 4V_ball = 4(4/3)π = (16/3)π cubic inches.
Now, to determine if there is more or less volume left inside the box, we subtract the total volume occupied by the balls from the volume of the box.
Volume left over inside the box = Volume of the box - Total volume occupied by the balls
= 32 cubic inches - (16/3)π cubic inches.
To compare this volume with the volume of a fifth ball, we need to calculate the volume of a single ball.
The volume of the fifth ball is V_fifth_ball = (4/3)π(1^3) = (4/3)π cubic inches.
Now, we can compare the two volumes:
If the volume left over inside the box is greater than the volume of the fifth ball:
32 - (16/3)π > (4/3)π
If the volume left over inside the box is less than the volume of the fifth ball:
32 - (16/3)π < (4/3)π
To find the answer, we need to calculate the values and compare them.