We have a disc of neglligible thickness and whose surface mass density varies as radial disance from centre as `sigma=sigma_(0)(1+r/R)`, where `R` is the radius of the disc. Specific heat of the material of the disc is`C`.
Disc is given an angular velocity `omega_(0)` and placed on a horizontal rough surface such that the plane of the disc is parallel to the surface. Coefficient of friction between disc and suface is `mu`. The temperature of the disc is `T_(0)`.
Answer the following question on the base of information provided in the above paragraph
Magnitude of angular acceleration of disc is

Options

A. `alpha=(7mug)/(27R)`
B. `alpha=(21mug)/(27R)`
C. `alpha=(35mug)/(27R)`
D. `alpha=(42mug)/(27R)`

Explanation

To find the magnitude of the angular acceleration of the disc, we will follow these steps:

### Step 1: Define the mass element of the disc
The surface mass density of the disc varies with the radial distance as:
\[ \sigma = \sigma_0 \left(1 + \frac{r}{R}\right) \]
where \( \sigma_0 \) is a constant, \( r \) is the radial distance from the center, and \( R \) is the radius of the disc.

For a small elemental ring at a distance \( r \) from the center with thickness \( dr \), the area \( dA \) of the ring is:
\[ dA = 2 \pi r \, dr \]
Thus, the mass \( dm \) of the elemental ring is:
\[ dm = \sigma \, dA = \sigma_0 \left(1 + \frac{r}{R}\right) \cdot 2 \pi r \, dr \]

### Step 2: Calculate the torque due to friction
The frictional force \( df \) acting on this mass element is given by:
\[ df = \mu g \, dm \]
The torque \( d\tau \) due to this frictional force about the center of the disc is:
\[ d\tau = df \cdot r = \mu g \, dm \cdot r \]
Substituting for \( dm \):
\[ d\tau = \mu g \cdot \left(2 \pi \sigma_0 \left(1 + \frac{r}{R}\right) \cdot r \, dr\right) \cdot r = \mu g \cdot 2 \pi \sigma_0 \left(1 + \frac{r}{R}\right) \cdot r^2 \, dr \]

### Step 3: Integrate to find total torque
Now, we need to integrate \( d\tau \) from \( r = 0 \) to \( r = R \):
\[ \tau = \int_0^R d\tau = \int_0^R \mu g \cdot 2 \pi \sigma_0 \left(1 + \frac{r}{R}\right) r^2 \, dr \]
This expands to:
\[ \tau = \mu g \cdot 2 \pi \sigma_0 \left( \int_0^R r^2 \, dr + \frac{1}{R} \int_0^R r^3 \, dr \right) \]
Calculating the integrals:
\[ \int_0^R r^2 \, dr = \frac{R^3}{3}, \quad \int_0^R r^3 \, dr = \frac{R^4}{4} \]
Thus:
\[ \tau = \mu g \cdot 2 \pi \sigma_0 \left( \frac{R^3}{3} + \frac{1}{R} \cdot \frac{R^4}{4} \right) = \mu g \cdot 2 \pi \sigma_0 \left( \frac{R^3}{3} + \frac{R^3}{4} \right) \]
Combining the fractions:
\[ \tau = \mu g \cdot 2 \pi \sigma_0 \cdot \frac{7R^3}{12} \]

### Step 4: Calculate the moment of inertia
The moment of inertia \( I \) of the disc can also be calculated by integrating the mass elements:
\[ I = \int_0^R r^2 \, dm = \int_0^R r^2 \cdot 2 \pi \sigma_0 \left(1 + \frac{r}{R}\right) r \, dr \]
This expands to:
\[ I = 2 \pi \sigma_0 \left( \int_0^R r^3 \, dr + \frac{1}{R} \int_0^R r^4 \, dr \right) \]
Calculating the integrals:
\[ I = 2 \pi \sigma_0 \left( \frac{R^4}{4} + \frac{1}{R} \cdot \frac{R^5}{5} \right) = 2 \pi \sigma_0 \left( \frac{R^4}{4} + \frac{R^4}{5} \right) \]
Combining the fractions:
\[ I = 2 \pi \sigma_0 \cdot \frac{9R^4}{20} \]

### Step 5: Relate torque and angular acceleration
Using the relation \( \tau = I \alpha \):
\[ \mu g \cdot 2 \pi \sigma_0 \cdot \frac{7R^3}{12} = \left(2 \pi \sigma_0 \cdot \frac{9R^4}{20}\right) \alpha \]
Canceling \( 2 \pi \sigma_0 \) from both sides:
\[ \mu g \cdot \frac{7R^3}{12} = \frac{9R^4}{20} \alpha \]
Solving for \( \alpha \):
\[ \alpha = \frac{\mu g \cdot 7R^3}{12} \cdot \frac{20}{9R^4} = \frac{35 \mu g}{27 R} \]

### Final Result
The magnitude of the angular acceleration \( \alpha \) of the disc is:
\[ \alpha = \frac{35 \mu g}{27 R} \]