We have a disc of neglligible thickness and whose surface mass density varies as radial distance 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
If the kinetic energy lost due to the friction between disc and surface is absorbed by the disc the moment heat is generated, by the mass lying at the point where the energy is lost. Assume that there is no radial heat flow in betwen disc particles. Rate of change in temperature at a point lying at `r` distance away from centre with time, before disc stops rotating, will be (`alpha`-magnitude of angular acceleration of rod)

Options

A. `(mu gr)/C(omega_(0)-alphat)`
B. `(mugr)/(2C)(omega_(0)-alphat)`
C. `(mugr)/(3C)(omega_(0)-alphat)`
D. `(mugr)/(4C)(omega_(0)-alphat)`

Explanation

To solve the problem, we need to find the rate of change of temperature \( \frac{dT}{dt} \) at a point lying at a distance \( r \) from the center of the disc before it stops rotating. We will follow these steps:

### Step 1: Determine the Force of Friction
The force of friction \( F \) acting on the disc can be expressed as:
\[ F = \mu \cdot N \]
where \( N \) is the normal force. For a disc lying on a horizontal surface, the normal force is equal to the weight of the disc. The weight can be calculated by integrating the mass density over the area of the disc.

### Step 2: Calculate the Mass Element \( dm \)
The mass element \( dm \) at a distance \( r \) from the center can be expressed as:
\[ dm = \sigma \cdot dA = \sigma \cdot 2\pi r \, dr \]
where \( dA \) is the differential area element. Given that \( \sigma = \sigma_0 \left(1 + \frac{r}{R}\right) \), we can substitute this into the equation.

### Step 3: Calculate the Work Done by Friction
The work done \( dW \) by the friction force during a small displacement \( ds \) is given by:
\[ dW = F \cdot ds = \mu \cdot N \cdot ds \]
The displacement \( ds \) can be expressed as \( ds = r \cdot d\theta \), where \( d\theta \) is the angular displacement.

### Step 4: Relate Work Done to Change in Temperature
The energy lost due to friction is absorbed as heat by the mass \( dm \). The relationship between the heat absorbed \( dQ \) and the change in temperature \( dT \) is given by:
\[ dQ = C \cdot dm \cdot dT \]
where \( C \) is the specific heat capacity of the material.

### Step 5: Set Up the Equation
Equating the work done to the heat absorbed, we have:
\[ \mu \cdot N \cdot ds = C \cdot dm \cdot dT \]

### Step 6: Substitute Expressions
Substituting the expressions for \( N \), \( dm \), and \( ds \) into the equation, we can solve for \( \frac{dT}{dt} \).

### Step 7: Use Angular Velocity Relation
The angular velocity \( \omega \) changes with time due to angular acceleration \( \alpha \):
\[ \omega = \omega_0 - \alpha t \]
Substituting this into our equation will give us the final expression for \( \frac{dT}{dt} \).

### Final Expression
After substituting and simplifying, we arrive at:
\[ \frac{dT}{dt} = \frac{\mu g r (\omega_0 - \alpha t)}{C} \]