As the roller moves up the step it rotates about the point P, so P is the pivot point, and the sphere will rotate about p when the total clockwise torque about P is positive.
The clockwise torque is the force F times the distance BP. Yes?
Because if we take P as the pivot point, the weight of the roller acts at the centre and is pulling the roller downwards i.e. making it rotate antiiclockwise.
Sir about resistances: using Kirchoff's laws can get us the answer in most cases, but it takes a longer time than those arguments which do not use current, and due to the time constraint, it is better to use the latter. But I do not know many arguments except symmetry, so I am not able to decide what would be the best approach, as frequently, finding equivalent resistance is only a part of the question (e.g, a question about LCR circuits with a resistance combination)
So I want to ask how to analyse a complicated resistor most efficiently
I don't think there is a general answer as each problem is different. Really the only good way is to do lots of problems so you get used to what works.
To be honest a lot of JEE preparation is just doing so many problems that you learn to quickly recognise the problems you see in the exam and know how to answer them.
The strong nuclear force is much stronger that the electrostatic force, so it can easily hold protons together despite the repulsion between the positive charges.
@JohnRennie So, if I understood it correctly, if we are able to bring two protons together, seperated by a distance where the SNF is greater than ESF, they attract each other (?)
- the force between protons and neutrons is called the strong nuclear force
@DevanshBhardwaj oops :-)
The forces are really the same because the SNF comes from the SF acting between quarks in the two protons, but it's useful to consider them separately.
I'm not entirely sure what IISc and IISERs are and how they differ, but yes if you do a physics degree and choose to specialise in quantum mechanics then you'll study this stuff.