A cylindrical can, whose base is horizontal and of radius 3-5 cm, contains sufficient water so that when a sphere is placed in the can, the water just covers the sphere. Given that the sphere just fits into the can, calculate :
the depth of water in the can before the sphere was put into the can.

Answer

To solve the problem, we need to calculate the depth of water in the cylindrical can before the sphere is placed in it. Here’s a step-by-step solution:

### Step 1: Understand the Problem
We have a cylindrical can with a radius of \( r = 3.5 \) cm. When a sphere is placed in the can, the water level rises to just cover the sphere. We need to find the depth of water in the can before the sphere was inserted.

### Step 2: Identify the Radius of the Sphere
Since the sphere just fits into the can, the radius of the sphere is the same as the radius of the can. Therefore, the radius of the sphere \( r = 3.5 \) cm.

### Step 3: Calculate the Volume of the Sphere
The volume \( V \) of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]
Substituting the radius of the sphere:
\[ V = \frac{4}{3} \pi (3.5)^3 \]

Calculating \( (3.5)^3 \):
\[ (3.5)^3 = 42.875 \]
So, the volume of the sphere becomes:
\[ V = \frac{4}{3} \pi \times 42.875 = \frac{171.5}{3} \pi \approx 57.1667 \pi \text{ cm}^3 \]

### Step 4: Relate the Volume of Water Displaced to the Height of Water
When the sphere is submerged, it displaces an equal volume of water. The volume of water displaced is also given by the formula for the volume of a cylinder:
\[ V = \pi r^2 h \]
Where \( h \) is the height of the water displaced. Since the radius \( r \) of the can is 3.5 cm, we can write:
\[ \pi (3.5)^2 h = \frac{4}{3} \pi (3.5)^3 \]

### Step 5: Cancel Out \( \pi \) and Solve for \( h \)
Dividing both sides by \( \pi \):
\[ (3.5)^2 h = \frac{4}{3} (3.5)^3 \]
Now, substituting \( (3.5)^2 = 12.25 \):
\[ 12.25 h = \frac{4}{3} \times 42.875 \]
Calculating the right side:
\[ \frac{4 \times 42.875}{3} = \frac{171.5}{3} \approx 57.1667 \]
Now, we can solve for \( h \):
\[ h = \frac{57.1667}{12.25} \approx 4.6667 \text{ cm} \]

### Step 6: Calculate the Depth of Water Before the Sphere
The total height of the cylinder is \( 2r \) (since the sphere fits perfectly), which is:
\[ 2r = 2 \times 3.5 = 7 \text{ cm} \]
The depth of water in the can before the sphere was inserted is:
\[ \text{Depth of water} = 2r - h = 7 - 4.6667 \approx 2.3333 \text{ cm} \text{ or } 2 \frac{1}{3} \text{ cm} \]

### Final Answer
The depth of water in the can before the sphere was put into the can is approximately \( 2 \frac{1}{3} \) cm.

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