A sphere of radius 2 cm is put into water contained in a cyclinder of base-radius 4 cm. If the sphere is completely immersed in the water, the water level in the cylinder rises by

Options

A. `(1)/(3) cm`
B. `(1)/(2) cm`
C. `(2)/(3) cm`
D. 2 cm

Answer

To find out how much the water level in the cylinder rises when a sphere of radius 2 cm is completely immersed in it, we will follow these steps:

### Step 1: Calculate the Volume of the Sphere
The formula for the volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi r^3 \]
where \( r \) is the radius of the sphere.

Given that the radius of the sphere is 2 cm, we can substitute this value into the formula:
\[ V = \frac{4}{3} \pi (2)^3 \]
Calculating \( (2)^3 \):
\[ (2)^3 = 8 \]
Now substituting back into the volume formula:
\[ V = \frac{4}{3} \pi \times 8 = \frac{32}{3} \pi \text{ cm}^3 \]

### Step 2: Calculate the Volume of Water Displaced
When the sphere is immersed in the water, it displaces a volume of water equal to its own volume. Therefore, the volume of water displaced is:
\[ \text{Volume of water displaced} = \frac{32}{3} \pi \text{ cm}^3 \]

### Step 3: Calculate the Rise in Water Level in the Cylinder
The volume \( V \) of a cylinder is given by:
\[ V = \pi r^2 h \]
where \( r \) is the radius of the base of the cylinder and \( h \) is the height of the water level.

Given that the base radius of the cylinder is 4 cm, we can express the rise in water level \( h \) as:
\[ \frac{32}{3} \pi = \pi (4^2) h \]
This simplifies to:
\[ \frac{32}{3} \pi = \pi (16) h \]
Dividing both sides by \( \pi \):
\[ \frac{32}{3} = 16h \]
Now, solving for \( h \):
\[ h = \frac{32}{3 \times 16} = \frac{32}{48} = \frac{2}{3} \text{ cm} \]

### Conclusion
The water level in the cylinder rises by \( \frac{2}{3} \) cm.