Answer:
Answer: 5/12 or approx. 0.42
Step-by-step explanation:
The ball will displace as much water as its volume.
Meaning, the volume of the ball and the volume of the increase in water in the barrel will be equal.
Volume of a sphere:
[tex]V_s=\frac{4}{3} \pi r^3[/tex]
The radius is half the diameter. The diameter of the sphere is 10, thus:
[tex]r = d/2 = 10/2 = 5[/tex]
5^3 = 5*5*5 = 25*5 = 125
[tex]V_s=\frac{4}{3} \pi * 5^3=\frac{4}{3} \pi * 125[/tex]
The volume of a cylinder, which will be the shape of the increase in water, is the following:
[tex]V_c=\pi r^2 h[/tex]
We know the radius r, as we know the diameter.
[tex]r=d/2=40/2=20[/tex]
[tex]V_c=\pi r^2 h=\pi h * 20^2 = \pi h * 400[/tex]
We are looking for the height of the cylinder, h, as that will be the height of the water rise.
Since we know these 2 volumes are the same, we'll set them equal to each other and solve the equation:
[tex]V_s = V_c[/tex]
[tex]\frac{4}{3} * 125 * \pi = 400 * h * \pi[/tex]
We can get rid of pi by dividing both sides by pi:
[tex]\frac{4}{3} * 125 = 400 * h[/tex]
And now divide both sides by 400:
[tex]\frac{4}{3} * \frac{125}{400}= h[/tex]
Just reversing it and putting h to the left:
[tex]h = \frac{4}{3} * \frac{125}{400} = \frac{4 * 125}{3 * 400}[/tex]
I see that I can divide both the numerator and the denominator by 4:
[tex]h =\frac{125}{3 * 100}= \frac{125}{300}[/tex]
I'll now divide both the numerator and the denominator by 25 to simplify the expression:
[tex]h=\frac{125}{300} =\frac{125 / 25}{300 / 25} = \frac{5}{12}[/tex]
Answer: The water rises by 5/12 (five twelfths).
If I want an approximate decimal answer, I can simply run this expression in a calculator:
[tex]\frac{5}{12}[/tex] ≈ [tex]0.416666667[/tex] ≈ [tex]0.42[/tex]
