Answer :
To determine how much the water level rises in the cylinder when a marble is dropped into it, we need to follow a series of calculations involving volumes and areas. Let's break this down step by step.
### Step 1: Calculate the volume of the marble
The marble is spherical, and the formula for the volume [tex]\( V \)[/tex] of a sphere with radius [tex]\( r \)[/tex] is:
[tex]\[ V = \frac{4}{3} \pi r^3 \][/tex]
Given the radius of the marble [tex]\( r = 5 \)[/tex] mm:
[tex]\[ V = \frac{4}{3} \pi (5)^3 \][/tex]
[tex]\[ V = \frac{4}{3} \pi \cdot 125 \][/tex]
[tex]\[ V = \frac{500}{3} \pi \][/tex]
### Step 2: Calculate the cross-sectional area of the cylinder
The cross-sectional area [tex]\( A \)[/tex] of the cylinder (its base) with radius [tex]\( r = 9 \)[/tex] mm is:
[tex]\[ A = \pi r^2 \][/tex]
Given the radius of the cylinder [tex]\( r = 9 \)[/tex] mm:
[tex]\[ A = \pi (9)^2 \][/tex]
[tex]\[ A = \pi \cdot 81 \][/tex]
[tex]\[ A = 81 \pi \][/tex]
### Step 3: Calculate the rise in the water level
The rise in the water level [tex]\( h \)[/tex] when the marble is dropped into the cylinder can be found by dividing the volume of the marble by the cross-sectional area of the cylinder:
[tex]\[ h = \frac{V}{A} \][/tex]
Substituting the values we have calculated:
[tex]\[ h = \frac{\frac{500}{3} \pi}{81 \pi} \][/tex]
Simplify the expression:
[tex]\[ h = \frac{500}{3} \times \frac{1}{81} \][/tex]
[tex]\[ h = \frac{500}{243} \][/tex]
### Step 4: Convert the exact fraction to a decimal and round to the nearest tenth
[tex]\[ h \approx \frac{500}{243} \approx 2.0576 \][/tex]
Rounding to the nearest tenth, the rise in the water level is approximately:
[tex]\[ h \approx 2.1 \text{ mm} \][/tex]
Thus, the water level in the cylinder rises by approximately 2.1 mm.
### Step 1: Calculate the volume of the marble
The marble is spherical, and the formula for the volume [tex]\( V \)[/tex] of a sphere with radius [tex]\( r \)[/tex] is:
[tex]\[ V = \frac{4}{3} \pi r^3 \][/tex]
Given the radius of the marble [tex]\( r = 5 \)[/tex] mm:
[tex]\[ V = \frac{4}{3} \pi (5)^3 \][/tex]
[tex]\[ V = \frac{4}{3} \pi \cdot 125 \][/tex]
[tex]\[ V = \frac{500}{3} \pi \][/tex]
### Step 2: Calculate the cross-sectional area of the cylinder
The cross-sectional area [tex]\( A \)[/tex] of the cylinder (its base) with radius [tex]\( r = 9 \)[/tex] mm is:
[tex]\[ A = \pi r^2 \][/tex]
Given the radius of the cylinder [tex]\( r = 9 \)[/tex] mm:
[tex]\[ A = \pi (9)^2 \][/tex]
[tex]\[ A = \pi \cdot 81 \][/tex]
[tex]\[ A = 81 \pi \][/tex]
### Step 3: Calculate the rise in the water level
The rise in the water level [tex]\( h \)[/tex] when the marble is dropped into the cylinder can be found by dividing the volume of the marble by the cross-sectional area of the cylinder:
[tex]\[ h = \frac{V}{A} \][/tex]
Substituting the values we have calculated:
[tex]\[ h = \frac{\frac{500}{3} \pi}{81 \pi} \][/tex]
Simplify the expression:
[tex]\[ h = \frac{500}{3} \times \frac{1}{81} \][/tex]
[tex]\[ h = \frac{500}{243} \][/tex]
### Step 4: Convert the exact fraction to a decimal and round to the nearest tenth
[tex]\[ h \approx \frac{500}{243} \approx 2.0576 \][/tex]
Rounding to the nearest tenth, the rise in the water level is approximately:
[tex]\[ h \approx 2.1 \text{ mm} \][/tex]
Thus, the water level in the cylinder rises by approximately 2.1 mm.
