The Red Conducting Carpet.

Consider the line connecting the center of the rolled carpet and the bottom of the rolled carpet. Consider a point at a distance $x$ below the center. Let velocity of center of mass be $v$ and angular velocity $\omega = \frac{v}{r}$

The velocity of this point is : $v - \omega x$

= $v_0 - \frac{v x}{r}$

Now the emf produced(V) would be $\displaystyle \int_{0}^{r} B (v - \frac{v x}{r}) dx$

= $\frac{B v r}{2}$ .

Using conservation of energy, loss in gravitational potential energy equals heat dissipation ,

$\Rightarrow MgR = \displaystyle \int_{0}^{T} \frac{V^2}{m} dt$

$\Rightarrow MgR = \displaystyle \int_{0}^{T} \frac{B^2 }{4m} (vr)^2 dt$

$\Rightarrow \displaystyle \int_{0}^{T} (vr)^2 dt = \frac{4MmgR}{B^2}$

Hence, required rms value = $n = \frac{\frac{4MmgR}{B^2}}{T}$

Hence, $\boxed{\lfloor \frac{n}{4} \rfloor = 29}$

let at some time t the velocity of the center of mass of the rolling part be v and radius be r. let there be p turns in the carpet per unit length.Let the current be then be I Let the total number of turns be n

$\phi=B.A=\int B\pi r^{2}dN=\int_{0}^{r}B\pi r^{2}pdr=\frac{\pi Bpr^{3}}{3}$

$\frac{d\phi}{dt}=\pi pBr^{2}\frac{dr}{dt}=mI$

We know that the mass is constant

let the mass per unit length of carpet be $\rho$

$\rho x+\rho\int_{0}^{r}2\pi rpdr=k$

$v=\frac{dx}{dt}$

$-\frac{v}{2\pi rp}=\frac{dr}{dt}$

$i=-\frac{Bvr}{2m}$

finally all the carpet is rolled out and velocity finally is 0 after t=Ts

applying energy conservation.

$MgR=\int_{0}^{t}I^{2}mdt=\int_{0}^{t}(\frac{-B^{2}v^{2}r^{2}}{2m})^{2}mdt$

$[\frac{\int_{0}^{t}v^{2}r^{2}dt}{4\int_{0}^{t}dt}]=[\frac{MmgR}{B^{2}T}]=29$