A thin disc of mass M and radius R has mass per unit area `sigma( r) =kx^(2)` where r is the distance from its centre. Its moment of inertia about an axis going through its centre of mass and perpendicular to its plane is:

Options

A. `(MR^(2))/3`
B. `(2MR^(2))/3`
C. `(MR^(2))/6`
D. `(MR^(2))/2`

Explanation

To find the moment of inertia of a thin disc with a variable mass per unit area, we can follow these steps:

### Step 1: Define the mass per unit area
The mass per unit area (σ) of the disc is given as:
\[ \sigma(r) = k r^2 \]
where \( r \) is the distance from the center of the disc.

### Step 2: Consider an elemental ring
We consider a small elemental ring at a distance \( r \) from the center with a thickness \( dr \). The area of this ring (dA) is:
\[ dA = 2\pi r \, dr \]

### Step 3: Calculate the mass of the elemental ring
The mass \( dm \) of the elemental ring can be expressed as:
\[ dm = \sigma(r) \cdot dA = k r^2 \cdot (2\pi r \, dr) = 2\pi k r^3 \, dr \]

### Step 4: Calculate the moment of inertia of the elemental ring
The moment of inertia \( dI \) of the elemental ring about the axis through the center is given by:
\[ dI = r^2 \, dm = r^2 \cdot (2\pi k r^3 \, dr) = 2\pi k r^5 \, dr \]

### Step 5: Integrate to find the total moment of inertia
To find the total moment of inertia \( I \) of the disc, we integrate \( dI \) from \( r = 0 \) to \( r = R \):
\[ I = \int_0^R 2\pi k r^5 \, dr \]
Calculating the integral:
\[ I = 2\pi k \left[ \frac{r^6}{6} \right]_0^R = 2\pi k \cdot \frac{R^6}{6} = \frac{2\pi k R^6}{6} = \frac{\pi k R^6}{3} \]

### Step 6: Relate k to the total mass M
To find \( k \), we need to relate it to the total mass \( M \) of the disc. The total mass \( M \) can be found by integrating the mass density over the area of the disc:
\[ M = \int_0^R dm = \int_0^R 2\pi k r^3 \, dr = 2\pi k \left[ \frac{r^4}{4} \right]_0^R = 2\pi k \cdot \frac{R^4}{4} = \frac{\pi k R^4}{2} \]
From this, we can solve for \( k \):
\[ k = \frac{2M}{\pi R^4} \]

### Step 7: Substitute k back into the moment of inertia equation
Now we substitute \( k \) back into the equation for \( I \):
\[ I = \frac{\pi \left(\frac{2M}{\pi R^4}\right) R^6}{3} = \frac{2M R^6}{3 R^4} = \frac{2M R^2}{3} \]

### Final Result
Thus, the moment of inertia of the disc about an axis through its center and perpendicular to its plane is:
\[ I = \frac{2}{3} M R^2 \]