A thin circular plate of mass M and radius R has its density varying as `rho(r)=rho_(0)r` with `rho_0` as constant and r is the distance from its center. The moment of Inertia of the circular plate about an axis perpendicular to the plate and passing through its edge is `I = aMR^(2)` The value of the coefficient a is :

Options

A. `1//2`
B. `3//5`
C. `8//5`
D. `3//2`

Answer

To find the moment of inertia of a thin circular plate with a varying density, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Density Variation**:
The density of the circular plate varies as \(\rho(r) = \rho_0 r\), where \(\rho_0\) is a constant and \(r\) is the distance from the center of the plate.

2. **Element of Mass**:
We consider a thin ring of radius \(r\) and thickness \(dr\) within the circular plate. The area of this thin ring is given by:
\[ dA = 2\pi r \, dr \]
The mass of this ring can be expressed as:
\[ dm = \rho(r) \cdot dA = \rho_0 r \cdot (2\pi r \, dr) = 2\pi \rho_0 r^2 \, dr \]

3. **Moment of Inertia of the Ring**:
The moment of inertia \(dI\) of this thin ring about an axis perpendicular to the plate and passing through its center is given by:
\[ dI = r^2 \, dm = r^2 \cdot (2\pi \rho_0 r^2 \, dr) = 2\pi \rho_0 r^4 \, dr \]

4. **Total Moment of Inertia**:
To find the total moment of inertia \(I\), we integrate \(dI\) from \(0\) to \(R\):
\[ I = \int_0^R dI = \int_0^R 2\pi \rho_0 r^4 \, dr \]
Evaluating the integral:
\[ I = 2\pi \rho_0 \left[ \frac{r^5}{5} \right]_0^R = 2\pi \rho_0 \frac{R^5}{5} = \frac{2\pi \rho_0 R^5}{5} \]

5. **Finding Total Mass**:
The total mass \(M\) of the plate can be found by integrating \(dm\) from \(0\) to \(R\):
\[ M = \int_0^R dm = \int_0^R 2\pi \rho_0 r^2 \, dr = 2\pi \rho_0 \left[ \frac{r^3}{3} \right]_0^R = 2\pi \rho_0 \frac{R^3}{3} \]

6. **Expressing \(\rho_0\) in terms of \(M\)**:
Rearranging the equation for mass:
\[ \rho_0 = \frac{3M}{2\pi R^3} \]

7. **Substituting \(\rho_0\) back into the Moment of Inertia**:
Substitute \(\rho_0\) into the expression for \(I\):
\[ I = \frac{2\pi \left(\frac{3M}{2\pi R^3}\right) R^5}{5} = \frac{3M R^2}{5} \]

8. **Comparing with the Given Form**:
The moment of inertia is given in the form \(I = aMR^2\). From our calculation:
\[ I = \frac{3}{5} MR^2 \]
Therefore, we can identify:
\[ a = \frac{3}{5} \]

### Final Answer:
The value of the coefficient \(a\) is \(\frac{3}{5}\).