The surface mass density of a disc of radius a varies with radial distance as `sigma = A + Br` where A & B are positive constants then moment of inertia of the disc about an axis passing through its centre and perpendicular to the plane

Options

A. `2pia^4(A/4+Ba/5)`
B. `2pia^4(Aa/4+B/5)`
C. `pia^4(A/4+Ba/5)`
D. `3pia^4(A/4+Ba/5)`

Explanation

To find the moment of inertia of a disc with a surface mass density that varies with radial distance, we can follow these steps:

### Step 1: Define the surface mass density
The surface mass density of the disc is given by:
\[ \sigma(r) = A + Br \]
where \(A\) and \(B\) are positive constants, and \(r\) is the radial distance from the center of the disc.

### Step 2: Consider a differential ring
We consider a thin ring of radius \(r\) and thickness \(dr\). The area of this ring is given by:
\[ dA = 2\pi r \, dr \]

### Step 3: Calculate the mass of the differential ring
The mass \(dm\) of the differential ring can be calculated by multiplying the surface mass density by the area:
\[ dm = \sigma(r) \cdot dA = (A + Br) \cdot (2\pi r \, dr) \]
Thus,
\[ dm = 2\pi r (A + Br) \, dr \]

### Step 4: Write the expression for the moment of inertia of the ring
The moment of inertia \(dI\) of the differential ring about the axis passing through the center and perpendicular to the plane is given by:
\[ dI = r^2 \, dm \]
Substituting for \(dm\):
\[ dI = r^2 \cdot (2\pi r (A + Br) \, dr) = 2\pi r^3 (A + Br) \, dr \]

### 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 \(r = a\):
\[ I = \int_0^a dI = \int_0^a 2\pi r^3 (A + Br) \, dr \]
This can be split into two integrals:
\[ I = 2\pi \left( \int_0^a A r^3 \, dr + \int_0^a Br^4 \, dr \right) \]

### Step 6: Calculate the integrals
Calculating the first integral:
\[ \int_0^a A r^3 \, dr = A \left[ \frac{r^4}{4} \right]_0^a = \frac{A a^4}{4} \]
Calculating the second integral:
\[ \int_0^a Br^4 \, dr = B \left[ \frac{r^5}{5} \right]_0^a = \frac{B a^5}{5} \]

### Step 7: Combine the results
Substituting these results back into the expression for \(I\):
\[ I = 2\pi \left( \frac{A a^4}{4} + \frac{B a^5}{5} \right) \]

### Step 8: Factor out common terms
Factoring out \(a^4\):
\[ I = 2\pi a^4 \left( \frac{A}{4} + \frac{B a}{5} \right) \]

### Final Result
Thus, the moment of inertia of the disc about an axis passing through its center and perpendicular to the plane is:
\[ I = 2\pi a^4 \left( \frac{A}{4} + \frac{B a}{5} \right) \]