Mass per unit area of a disc of inner radius `'a'` and outer radius `'b'` is given by `((sigma_(0))/(r)),r` distance from center. Its radius of gyration `w.r.t.` axis of rotation passing through center and perpendicular to plane is

Options

A. `sqrt((B^(4)-a^(4))/(3(b^(2)-a^(2))))`
B. `sqrt((b^(3)-a^(3))/(3(b-a)))`
C. `sqrt((b^(3)-a^(3))/(5(b-a)))`
D. `sqrt((b^(4)-a^(4))/(5(b^(2)-a^(2)))`

Answer

To solve the problem, we need to find the radius of gyration of a disc with an inner radius \( a \) and an outer radius \( b \), where the mass per unit area is given by \( \sigma_0 / r \), with \( r \) being the distance from the center.

### Step-by-Step Solution:

1. **Understand the Geometry of the Disc:**
- The disc has an inner radius \( a \) and an outer radius \( b \).
- We will consider a small annular section of the disc at a distance \( r \) from the center with thickness \( dr \).

2. **Determine the Mass Element \( dm \):**
- The mass per unit area is given as \( \sigma(r) = \frac{\sigma_0}{r} \).
- The area of the small annular section is \( dA = 2\pi r \, dr \).
- Therefore, the mass of this section is:
\[ dm = \sigma(r) \cdot dA = \frac{\sigma_0}{r} \cdot (2\pi r \, dr) = 2\pi \sigma_0 \, dr. \]

3. **Calculate the Moment of Inertia \( I \):**
- The moment of inertia of this small section about the axis perpendicular to the plane and passing through the center is:
\[ dI = r^2 \, dm = r^2 \cdot (2\pi \sigma_0 \, dr) = 2\pi \sigma_0 r^2 \, dr. \]
- To find the total moment of inertia \( I \) of the disc, integrate \( dI \) from \( r = a \) to \( r = b \):
\[ I = \int_a^b 2\pi \sigma_0 r^2 \, dr. \]
- Evaluating the integral:
\[ I = 2\pi \sigma_0 \left[ \frac{r^3}{3} \right]_a^b = 2\pi \sigma_0 \left( \frac{b^3}{3} - \frac{a^3}{3} \right) = \frac{2\pi \sigma_0}{3} (b^3 - a^3). \]

4. **Calculate the Total Mass \( M \):**
- The total mass \( M \) of the disc is obtained by integrating \( dm \) from \( r = a \) to \( r = b \):
\[ M = \int_a^b dm = \int_a^b 2\pi \sigma_0 \, dr = 2\pi \sigma_0 (b - a). \]

5. **Find the Radius of Gyration \( k \):**
- The radius of gyration \( k \) is related to the moment of inertia \( I \) and mass \( M \) by the formula:
\[ I = M k^2 \quad \Rightarrow \quad k^2 = \frac{I}{M}. \]
- Substitute the values of \( I \) and \( M \):
\[ k^2 = \frac{\frac{2\pi \sigma_0}{3} (b^3 - a^3)}{2\pi \sigma_0 (b - a)}. \]
- Simplifying this gives:
\[ k^2 = \frac{(b^3 - a^3)}{3(b - a)}. \]
- Therefore, the radius of gyration \( k \) is:
\[ k = \sqrt{\frac{b^3 - a^3}{3(b - a)}}. \]

### Final Answer:
The radius of gyration \( k \) with respect to the axis of rotation passing through the center and perpendicular to the plane is:
\[ k = \sqrt{\frac{b^3 - a^3}{3(b - a)}}. \]