A cylindrical can whose base is horizontal and is of internal radius 3.5 cm contains sufficient water so that when a solid sphere is placed inside, watre just covers the sphere. The sphere fits in the can exactly. The depth of water in the can before the sphere was put, is

Options

A. a)`(35)/(3) cm`
B. b)`(17)/(3) cm`
C. c)`(7)/(3) cm`
D. d)`(14)/(3) cm`

Answer

To find the depth of water in the cylindrical can before the sphere was placed inside, we can follow these steps:

### Step 1: Understand the dimensions of the sphere
Since the sphere fits exactly into the cylindrical can, the diameter of the sphere is equal to the diameter of the can. Given that the internal radius of the can is 3.5 cm, the diameter of the sphere is:
\[ \text{Diameter of sphere} = 2 \times \text{Radius of can} = 2 \times 3.5 \, \text{cm} = 7 \, \text{cm} \]

### Step 2: Calculate the radius of the sphere
The radius of the sphere is half of its diameter:
\[ \text{Radius of sphere} = \frac{\text{Diameter of sphere}}{2} = \frac{7 \, \text{cm}}{2} = 3.5 \, \text{cm} \]

### 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 \( (3.5)^3 \):
\[ (3.5)^3 = 42.875 \]
Now substituting this value into the volume formula:
\[ V = \frac{4}{3} \pi (42.875) \approx 179.594 \, \text{cm}^3 \, (\text{using } \pi \approx 3.14) \]

### Step 4: Calculate the height of water displaced in the can
The volume of water displaced by the sphere will be equal to the volume of the sphere. The volume of a cylinder is given by:
\[ V = \pi r^2 h \]
Where \( r \) is the radius of the cylinder and \( h \) is the height of the water. We can rearrange this formula to find the height \( h \):
\[ h = \frac{V}{\pi r^2} \]
Substituting the values we have:
\[ h = \frac{179.594}{\pi (3.5)^2} \]
Calculating \( (3.5)^2 \):
\[ (3.5)^2 = 12.25 \]
Now substituting this value:
\[ h = \frac{179.594}{\pi \times 12.25} \approx \frac{179.594}{38.48} \approx 4.67 \, \text{cm} \]

### Conclusion
The depth of water in the can before the sphere was put in is approximately **4.67 cm**.

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