two iron balls each of a diameter 6 cm are immersed in the water contained in a cylindrical vessel of radius 6 cm. The level of the water in the vessel will be raised by

Options

A. 1cm
B. 2 cm
C. 3 cm
D. 6 cm

Answer

To solve the problem of how much the water level in the cylindrical vessel rises when two iron balls are immersed in it, we can follow these steps:

### Step 1: Identify the dimensions of the iron balls and the cylindrical vessel
- **Diameter of each iron ball** = 6 cm
- **Radius of each iron ball (r)** = Diameter / 2 = 6 cm / 2 = 3 cm
- **Radius of the cylindrical vessel (R)** = 6 cm

### Step 2: Calculate the volume of one iron ball
The volume \( V \) of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]
Substituting the radius of the iron ball:
\[ V = \frac{4}{3} \pi (3)^3 = \frac{4}{3} \pi (27) = 36 \pi \, \text{cm}^3 \]

### Step 3: Calculate the volume of two iron balls
Since there are two iron balls, the total volume \( V_{total} \) is:
\[ V_{total} = 2 \times V = 2 \times 36 \pi = 72 \pi \, \text{cm}^3 \]

### Step 4: Relate the volume of the displaced water to the rise in water level
The volume of water displaced by the two iron balls will be equal to the volume of the water raised in the cylindrical vessel. The volume \( V_{cylinder} \) of the cylinder can be expressed as:
\[ V_{cylinder} = \pi R^2 h \]
Where \( h \) is the height the water level rises. Substituting the radius of the cylindrical vessel:
\[ V_{cylinder} = \pi (6)^2 h = 36 \pi h \]

### Step 5: Set the volumes equal to each other
The volume of the displaced water is equal to the volume of the two iron balls:
\[ 72 \pi = 36 \pi h \]

### Step 6: Solve for \( h \)
Dividing both sides by \( 36 \pi \):
\[ h = \frac{72 \pi}{36 \pi} = 2 \, \text{cm} \]

### Conclusion
The level of the water in the vessel will be raised by **2 cm**.

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