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 \(\( V \)\) of a sphere with radius \(\( r \)\) is:
\(\[ V = \frac{4}{3} \pi r^3 \]\)
Given the radius of the marble \(\( r = 5 \)\) mm:
\(\[ V = \frac{4}{3} \pi (5)^3 \]\)
\(\[ V = \frac{4}{3} \pi \cdot 125 \]\)
\(\[ V = \frac{500}{3} \pi \]\)
### Step 2: Calculate the cross-sectional area of the cylinder
The cross-sectional area \(\( A \)\) of the cylinder (its base) with radius \(\( r = 9 \)\) mm is:
\(\[ A = \pi r^2 \]\)
Given the radius of the cylinder \(\( r = 9 \)\) mm:
\(\[ A = \pi (9)^2 \]\)
\(\[ A = \pi \cdot 81 \]\)
\(\[ A = 81 \pi \]\)
### Step 3: Calculate the rise in the water level
The rise in the water level \(\( h \)\) 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:
\(\[ h = \frac{V}{A} \]\)
Substituting the values we have calculated:
\(\[ h = \frac{\frac{500}{3} \pi}{81 \pi} \]\)
Simplify the expression:
\(\[ h = \frac{500}{3} \times \frac{1}{81} \]\)
\(\[ h = \frac{500}{243} \]\)
### Step 4: Convert the exact fraction to a decimal and round to the nearest tenth
\(\[ h \approx \frac{500}{243} \approx 2.0576 \]\)
Rounding to the nearest tenth, the rise in the water level is approximately:
\(\[ h \approx 2.1 \text{ mm} \]\)
Thus, the water level in the cylinder rises by approximately 2.1 mm.
