Problem Solving Strategies

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Yusuf Hasan

05:25

@KavinIshwaran That is not due to the JEE. In any other polyhedral geometry there is a possibility for bond distortions (mostly elongation) in response to repulsive bond pairs. That corresponds to an overall bond energy optimization towards a local minima in accordance with Bent's rule. We don't have this luxury in the diatomic case since it has only one bond

As an example: PCl5 distorts axially in response to repulsions. Hence the 5 sp3d hybrid orbitals do not remain equivalent. Since for the same value of principal quantum number, s orbitals have the lowest energy followed by p and d, hence it redistributes the percent s character towards the equatorial plane. You can imagine this as the axial bond pairs shift away from the central P, so it redistributes to bond pairs closer to P. Net result is energetic stabilization

Kavin Ishwaran

06:01

@YusufHasan Oh !

@YusufHasan Thank you for the explanation :-)

@JohnRennie Hi !

John Rennie

Hi :-)

Kavin Ishwaran

A solid cylinder with inner radius a and outer radius b has length l and resistivity rho. find the resistance across its lateral surfaces

@JohnRennie I tried integrating by taking a small cylindrical element. But i couldn't able to frame a correct equation

John Rennie

What does it mean by lateral surfaces?

The inner and outer surface? Or between the ends?

Kavin Ishwaran

The inner and outer surface

John Rennie

Suppose we treat the cylinder as a lot of cylindrical shells of thickness dr.

Then what we have is the shells all in series so their resistances all just add. Yes?

Kavin Ishwaran

06:16

Yes

the dr = rho dr/A

But I am stuck with finding what should be the area for this cylindrical element

John Rennie

If we take dr to be infinitesimally small then the inner and outer areas are the same. Yes?

Kavin Ishwaran

Yes

John Rennie

So that area is 2𝜋𝑟 × 𝓁
Yes?

Kavin Ishwaran

06:46

@JohnRennie Yes

I get it now :-)

I will take it from here

John Rennie

OK :-)

Aayush Sethia

06:58

Hey my friend can you help me

John Rennie

Hi :-) What's the question?

Aayush Sethia

A particle of mass m is initially at rest at the origin. It is subjected to a force and starts moving along the x- axis. Its kinetic energy K changes with time as dK/dt = t, where  is a positive constant of appropriate dimensions. Which of the following statements is (are) true?
(A) The force applied on the particle is constant
(B) The speed of the particle is proportional to time
(C) The distance of the particle from the origin increases linearly with time (D) The force is conservative

Please help only in (D)

PYQ JEE ADVANCED 2018

John Rennie

Forces are conservative as long as energy is not lost to friction, or something similar to friction like aerodynamic drag.

Aayush Sethia

How do I know if a force is conservative just by looking at its formula

John Rennie

It isn't clear from the question whether any form of friction is present, but I assume not since friction isn't mentioned, and in that case the force will be conservative.

@AayushSethia There is no way to know just from the formula for the force.

Aayush Sethia

07:06

Only friction dissipated energy

dissipates

John Rennie

Yes

Aayush Sethia

This means " If in the question it is not written about any type of friction I assume that the force is conservative and it does not dissipated energy "

John Rennie

Yes. If there is no friction all the work done by the force goes into mechanical energy, and then that mechanical energy can be used to produce the same force.

4 hours later…

Kavin Ishwaran

11:38

@JohnRennie Hi !

John Rennie

Hi :-)

Kavin Ishwaran

We just need to find potential of a sphere and equate it to kinetic energy ?

John Rennie

How far have you got with this?

@KavinIshwaran Yes, just use conservation of energy.

Kavin Ishwaran

Ok :-)

I was not sure if we can do as all the three will be moving and the potential will change..

John Rennie

Consider the top sphere. PEs just add, so its PE is the sum of the PEs due to the other two spheres i.e. V = 2Gmm/d.
Yes?

Kavin Ishwaran

11:45

Yes

John Rennie

And when the spheres collide the separation between them is 2R, so now the PE is U = 2Gmm/2a = Gmm/R

Kavin Ishwaran

Yes

John Rennie

So ¹⁄₂mv² = 2Gmm/d - Gmm/R

Kavin Ishwaran

Yes

John Rennie

Oops, other way round:
¹⁄₂mv² = Gmm/R - 2Gmm/d

The answer is (4) isn't it?

Kavin Ishwaran

11:50

Shouldn't it be ¹⁄₂mv² = Gmm/R - 2Gmm/d ?

@JohnRennie Yes :-)

John Rennie

@KavinIshwaran Yes :-)

Kavin Ishwaran

@JohnRennie Yes :-)

John Rennie

Actually you could answer this without doing any work. Suppose d = ∞ so 1/d = 0.

Kavin Ishwaran

Yes

John Rennie

Options 1 and 2 would give the square root of a negative number so they cannot be right.

Kavin Ishwaran

11:52

Yes

John Rennie

Then take d = 2R i.e. the spheres start and end at the same separation. Here the velocity has to be zero because there was no PE change, but option 3 gives a non-zero velocity.

So only option 4 can be correct.

Kavin Ishwaran

Yes

@JohnRennie Thank you for the clarifications :-)

John Rennie

You're welcome :-)

Kavin Ishwaran

Will you be around tomorrow ?

John Rennie

During the day yes, but not in the evening as we are having a big New Year dinner. I will be here this evening.

Kavin Ishwaran

11:55

Ok :-)

@JohnRennie Will see you later :-)

John Rennie

Bye :-)

5 hours later…

Kavin Ishwaran

16:31

@JohnRennie Hi !

John Rennie

Hi :-)

Kavin Ishwaran

I am having a doubt in probability

John Rennie

I'll help if I can ...

Kavin Ishwaran

John Rennie

I have no idea how to do that. Sorry :-(

Kavin Ishwaran

16:33

Its Ok :-)

John Rennie

I'm going to drop out now. I'll be back tomorrow morning as usual.

Kavin Ishwaran

Bye :-)

2 hours later…

Harjot Dhillon

18:31

Happy New Year IST

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Problem Solving Strategies

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