Surface mass density of disc of mass `m` and radius `R` is `Kr^(2)`. Then its moment of inertia `w.r.t.` axis of rotation passing through center and perpendicular to the plane of disc is :

Options

A. `1=(3)/(2)nR^(2)`
B. `1=(4)/(3)nR^(2)`
C. `1-(2)/(5)nR^(2)`
D. `1-(2)/(3)nR^(2)`

Answer

To find the moment of inertia of a disc with a surface mass density given by \( \sigma = K r^2 \), we can follow these steps:

### Step 1: Define the mass of a ring
Consider a thin ring of radius \( r \) and thickness \( dr \) on the disc. The area of this ring is given by:
\[ dA = 2\pi r \, dr \]
The surface mass density is given as \( \sigma = K r^2 \). Therefore, the mass \( dm \) of the ring can be expressed as:
\[ dm = \sigma \, dA = K r^2 \cdot 2\pi r \, dr = 2\pi K r^3 \, dr \]

### Step 2: Calculate the moment of inertia of the ring
The moment of inertia \( dI \) of the ring about the axis perpendicular to the plane of the disc and passing 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 need to integrate \( dI \) from \( r = 0 \) to \( r = R \):
\[ I = \int_0^R dI = \int_0^R 2\pi K r^5 \, dr \]
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: Relate \( K \) to the total mass \( m \)
The total mass \( m \) of the disc can be found by integrating the mass of the entire 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 express \( K \) in terms of \( m \):
\[ K = \frac{2m}{\pi R^4} \]

### Step 5: Substitute \( K \) back into the moment of inertia equation
Substituting \( K \) into the moment of inertia equation:
\[ I = \frac{\pi K R^6}{3} = \frac{\pi \left(\frac{2m}{\pi R^4}\right) R^6}{3} = \frac{2m R^6}{3 R^4} = \frac{2m R^2}{3} \]

### Final Answer
Thus, the moment of inertia of the disc with respect to the axis of rotation passing through the center and perpendicular to the plane of the disc is:
\[ I = \frac{2}{3} m R^2 \]