A spherical ball of radius 1 cm is dropped into a conical vessel of radius 3 cm and slant height 6 cm. The volume of water (in `cm^(3)`), that can just immerse the ball, is

Options

A. `(5pi)/(3)`
B. `(pi)/(3)`
C. `3pi`
D. `(4pi)/(3)`

Answer

To solve the problem of finding the volume of water that can just immerse a spherical ball of radius 1 cm in a conical vessel of radius 3 cm and slant height 6 cm, we can follow these steps:

### Step 1: Calculate the height of the cone
To find the height (h) of the cone, we can use the Pythagorean theorem. The slant height (l) is 6 cm, and the radius (r) is 3 cm.

Using the formula:
\[ l^2 = r^2 + h^2 \]
Substituting the values:
\[ 6^2 = 3^2 + h^2 \]
\[ 36 = 9 + h^2 \]
\[ h^2 = 36 - 9 = 27 \]
\[ h = \sqrt{27} = 3\sqrt{3} \, \text{cm} \]

### Step 2: Calculate the volume of the cone
The volume (V) of a cone is given by the formula:
\[ V = \frac{1}{3} \pi r^2 h \]
Substituting the values:
\[ V = \frac{1}{3} \pi (3^2)(3\sqrt{3}) \]
\[ V = \frac{1}{3} \pi (9)(3\sqrt{3}) \]
\[ V = \frac{27\sqrt{3}}{3} \pi \]
\[ V = 9\sqrt{3} \pi \, \text{cm}^3 \]

### Step 3: Calculate the volume of the sphere
The volume (V_s) of a sphere is given by the formula:
\[ V_s = \frac{4}{3} \pi r^3 \]
Substituting the radius of the sphere (1 cm):
\[ V_s = \frac{4}{3} \pi (1^3) \]
\[ V_s = \frac{4}{3} \pi \, \text{cm}^3 \]

### Step 4: Calculate the volume of water that can just immerse the ball
The volume of water that can just immerse the ball is equal to the volume of the cone minus the volume of the sphere:
\[ V_{water} = V_{cone} - V_{sphere} \]
Substituting the values:
\[ V_{water} = 9\sqrt{3} \pi - \frac{4}{3} \pi \]

To combine these, we can convert \( 9\sqrt{3} \pi \) into a fraction:
\[ V_{water} = \left( 27\sqrt{3} - 4 \right) \frac{\pi}{3} \]

### Final Answer
Thus, the volume of water that can just immerse the ball is:
\[ V_{water} = \left( 27\sqrt{3} - 4 \right) \frac{\pi}{3} \, \text{cm}^3 \]

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