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)`

Answer

A

Explanation

`omega=omega_(0)-alphat`
`dE=(mu dmgr)(omega_(0)-alphat)dt=dmCdT`
`=(dT)/(dt)=(mugr)/C(omega_(0) -alphat)`