A rigid body can be hinged about any point on the x-axis. When it is hinged such that the hinge is at `x`, the moment of interia is given by
`I = 2 x^(2) - 12 x + 27`
The x-coordinate of centre of mass is.

Options

A. `x=2`
B. `x=0`
C. `x=1`
D. `x=3`

Answer

To find the x-coordinate of the center of mass of the rigid body, we need to minimize the moment of inertia given by the equation:

\[ I = 2x^2 - 12x + 27 \]

### Step 1: Differentiate the Moment of Inertia
To find the value of \( x \) that minimizes the moment of inertia, we take the derivative of \( I \) with respect to \( x \):

\[ \frac{dI}{dx} = \frac{d}{dx}(2x^2 - 12x + 27) \]

### Step 2: Apply the Derivative
Using the power rule of differentiation:

\[ \frac{dI}{dx} = 2 \cdot 2x - 12 \cdot 1 + 0 = 4x - 12 \]

### Step 3: Set the Derivative to Zero
To find the critical points, we set the derivative equal to zero:

\[ 4x - 12 = 0 \]

### Step 4: Solve for \( x \)
Now, we solve for \( x \):

\[ 4x = 12 \\ x = \frac{12}{4} \\ x = 3 \]

### Conclusion
The x-coordinate of the center of mass is \( x = 3 \).