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Assuming the initial linear and angular velocities to be ${v}_{0}$ and ${\omega}_{0}$ , such that $2{v}_{0}=R{\omega}_{0}$

Then, initially there is forward slipping. So friction acts backward.

$ma=\mu mg\\mR^2\alpha=\mu mgR$

So,

$v={v}_{0}-\mu gt$

$R\omega=R{\omega}_{0}-\mu gt=2{v}_{0}-\mu gt$

Its evident that becomes zero first. Still due to the rotational component, the slipping is forward. Friction continues to act bacward leading to developement of velocity in the backward direction. So the disc turns back after $t=\frac{{v}_{0}}{\mu g}$ and then

$v=\mu gt'$

$R\omega=R{\omega}_{0}-\mu gt'$

where $t'=t-\frac{{v}_{0}}{g}$

The velocity of point of contact is

${v}_{PC}=v-R\omega=2\mu gt'-R{\omega}_{0}$

Let pure rolling start at , then,

${v}_{PC}(T)=0\Rightarrow T=\frac{{v}_{0}}{\mu g}$

Hence, the disc moves forward till $t=\frac{{v}_{0}}{\mu g}$ andd then it starts moving back and attains pure rolling at $t=\frac{2{v}_{0}}{\mu g}$ (just when the angular velocity becomes zero)

"if sum of n terms in agp is in the form "

In which "form" ??

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Narayana or FIITJEE

The figure isn't clear.

I am assuming the direction of angular velocity ${\omega}_{0}$ to be clockwise

Then the bottommost point slips down the incline and hence friction is up the incline.

Now, as $\mu < tan\theta \Rightarrow\mu mg cos\theta<mgsin\theta$

And hence the acceleration of the CM is down the plane, which will lead to a translational velocity down the plane. And the angular velocity being clockwise, the velocity of point of contact is downward. The cylinder moves downward, and after the angular velocity reduces to zero due to the anticlockwise torque, a counterclockwise angular velocity is created due to friction, that after sometime will lead to pure rolling.

But essentially, the cylinder can't climb the inclined plane if $\mu<tan\theta$

Write the question

Plz write the question fully

Information about a particle in a given energy level (orbit) is contained in a mathematical expression or function. This is called the wave function and is denoted by the Greek letter $\psi$ . The wave function is obtained by solving Schrodinger's Wave Equation. The square of the wave function ( $\psi^2$ ) gives the probability of finding a particle (electron) within a region of space.

Acceptable solutions to the wave equation must have certain properties, and these give some insight into the significance of the wave equation.

$\psi$ must be continuous

$\psi$ must be finite

$\psi$ must be single valued

$\int_{-\infty}^{\infty}\psi^2=1$

Possibly the most important outcome of all this is that we now must have a model of the atom where "probability" is important. What is the probability of finding (the) electron within a certain volume of space ?

Using Schrodingers equation we can calculate the probability density of the electron - i.e. the probability that the electron exists in a given volume of space. For example we can calculate this as a function of distance from the nucleus and we can define the "shape" of an orbital. We shall look at, for example, the radial distribution of the electron in class.

Before going onto look at where we might "place" a particular electron we need to delve a little deeper into quantum mechanics.

There are acceptable solutions to the wave equation. To obtain such solutions, it is necessary to use four quantum numbers to describe the orbital and the electron within that orbital. By defining the values of each of these four quantum numbers we can identify both the electron and the "orbit" of that electron. In some ways this is rather like giving a map reference for a person and indicating which way that person is facing. (Although these quantum numbers are not co-ordinates as such).

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The link is http://www.annauniv.edu/tnea2008/

The drop out setas in AIEEE counselling are alloted to those who have lower ranks ... but then there is a second counselling for it. I suggest that you read up from the AIEEE site.

Good work @srujana. Slight correction, it should be ${v}_{x}={v}_{0}cos(\theta +\alpha)$ if $\theta$ is being measured form horizontal

Refer to the figure provided by srujana. Take origin at the point of projection, X axis downwards along the inclined plane, and Y axis perpendiculato to the plane upwards.

Then,Particle's initial position $\vec{{r}_{0}} = \vec{0}$

Initial velocity, $\vec{{v}_{0}}={v}_{0}(cos\theta\hat{i}+sin\theta\hat{j})$

And the acceleartion, $\vec{a}=g(sin\alpha \hat{i}-cos\alpha \hat{j})$

Then, using $\Delta\vec{r}=\vec{{v}_{0}}\Delta t + \frac{1}{2}\vec{a}(\Delta t)^2$ with the assumption that the particle is thrown at we get

$\vec{r}=x\hat{i}+y\hat{j}=({v}_{0}cos\theta+\frac{1}{2}gsin\alpha t^2)\hat{i}+({v}_{0}sin\theta-\frac{1}{2}gcos\alpha t^2)\hat{i}$

Giving, $x=({v}_{0}cos\theta t+\frac{1}{2}gsin\alpha t^2)$ and $y=({v}_{0}sin\theta t-\frac{1}{2}gcos\alpha t^2)$

Let at

, the particle hit the inclined plane again. Then $y(T)=0\Rightarrow T=\frac{2{v}_{0} sin\theta}{gcos\alpha}$

And at that instant the x coordinate gives the range of the particle on the inclined plane, i.e.,

$R=x(T) = \frac{2{v}_{0}^2sin\theta cos(\theta-\alpha)}{gcos^2\alpha}$