The volume of the greatest cone obtained by rotating a right - angled triangle of hypotenuse 2 units about a side is `(kpi)/(9sqrt3)` cubic units, then the value of k is equal to

Answer

To solve the problem of finding the volume of the greatest cone obtained by rotating a right-angled triangle with a hypotenuse of 2 units about one of its sides, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Triangle**:
Let the right-angled triangle be \( ABC \) with \( AB \) as one leg, \( BC \) as the other leg, and \( AC \) as the hypotenuse. Given that the hypotenuse \( AC = 2 \) units.

2. **Setting Up the Geometry**:
When we rotate the triangle about one of its legs (say \( AB \)), a cone is formed. The height of the cone will be equal to the length of the leg \( AB \) (let's denote it as \( x \)), and the radius will be equal to the length of the leg \( BC \) (let's denote it as \( r \)).

3. **Using the Pythagorean Theorem**:
According to the Pythagorean theorem:
\[ AB^2 + BC^2 = AC^2 \]
Substituting the values, we have:
\[ x^2 + r^2 = 2^2 = 4 \]
Thus, we can express \( r^2 \) in terms of \( x \):
\[ r^2 = 4 - x^2 \]

4. **Volume of the Cone**:
The volume \( V \) of a cone is given by the formula:
\[ V = \frac{1}{3} \pi r^2 h \]
Substituting \( r^2 \) and \( h = x \):
\[ V = \frac{1}{3} \pi (4 - x^2) x \]
Simplifying this, we get:
\[ V = \frac{1}{3} \pi (4x - x^3) \]

5. **Finding the Maximum Volume**:
To find the maximum volume, we differentiate \( V \) with respect to \( x \) and set the derivative equal to zero:
\[ \frac{dV}{dx} = \frac{1}{3} \pi (4 - 3x^2) \]
Setting the derivative to zero:
\[ 4 - 3x^2 = 0 \implies 3x^2 = 4 \implies x^2 = \frac{4}{3} \implies x = \frac{2}{\sqrt{3}} \]

6. **Calculating the Radius**:
Substitute \( x = \frac{2}{\sqrt{3}} \) back into the equation for \( r^2 \):
\[ r^2 = 4 - \left(\frac{2}{\sqrt{3}}\right)^2 = 4 - \frac{4}{3} = \frac{12}{3} - \frac{4}{3} = \frac{8}{3} \]
Thus, \( r = \frac{2\sqrt{2}}{\sqrt{3}} \).

7. **Calculating the Maximum Volume**:
Substitute \( r \) and \( x \) back into the volume formula:
\[ V = \frac{1}{3} \pi \left(\frac{8}{3}\right) \left(\frac{2}{\sqrt{3}}\right) = \frac{16\pi}{9\sqrt{3}} \]

8. **Finding the Value of \( k \)**:
The volume is given in the form \( \frac{k\pi}{9\sqrt{3}} \). Comparing:
\[ \frac{16\pi}{9\sqrt{3}} = \frac{k\pi}{9\sqrt{3}} \implies k = 16 \]

### Final Answer:
Thus, the value of \( k \) is \( \boxed{16} \).