A thin disc of mass M and radius R has mass per unit area `sigma( r)=kr^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 : a) `(MR^3)/3` b) `(2MR^2)/3` c) `(MR^2)/6` d) `(MR^2)/2`

Options

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

Answer

To find the moment of inertia of a thin disc with a mass per unit area given by \(\sigma(r) = kr^2\), we can follow these steps:

### Step 1: Define the mass element
Consider a thin ring of radius \(r\) and thickness \(dr\) within the disc. The area of this ring is given by:
\[ dA = 2\pi r \, dr \]
The mass of this ring, \(dm\), can be expressed in terms of the mass per unit area:
\[ dm = \sigma(r) \cdot dA = kr^2 \cdot (2\pi r \, dr) = 2\pi k r^3 \, dr \]

### Step 2: Define the moment of inertia of the ring
The moment of inertia \(dI\) of the thin ring about the axis perpendicular to the plane of the disc and through its center is given by:
\[ dI = r^2 \, dm = r^2 \cdot (2\pi k r^3 \, dr) = 2\pi k r^5 \, dr \]

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

### Step 4: Find the total mass of the disc
Next, we need to find the total mass \(M\) of the disc:
\[ M = \int_0^R dm = \int_0^R 2\pi k r^3 \, dr \]
Calculating this integral:
\[ M = 2\pi k \int_0^R r^3 \, dr = 2\pi k \left[\frac{r^4}{4}\right]_0^R = 2\pi k \cdot \frac{R^4}{4} = \frac{\pi k R^4}{2} \]

### Step 5: Solve for \(k\) in terms of \(M\)
From the equation for total mass, we can express \(k\):
\[ k = \frac{2M}{\pi R^4} \]

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

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

### Answer
The correct option is (b) \(\frac{2MR^2}{3}\).