Recent content by NTesla

Here's all the steps: ## \Delta\vec{P} = \vec{P_{f}} - \vec{P_{i}}## ##\vec{P_{i}} = \lambda (l - x - h)v\hat{i} - \lambda hv\hat{j}## ##\vec{P_{f}} = \lambda (l - (x + \Delta{x}) - h)(v + \Delta{v}) \hat{i} - \lambda h(v + \Delta{v}) \hat{j}## In the ##\hat{i}## direction: ##\Delta P\hat{i} =...

Yes, and I've already mentioned that in post#22. But there's another problem that I've mentioned in post#22, that is the cause of much problem.

Equations that you wrote are not legible. Kindly see the attached pic.

Appreciate your input. However, I've already understood that mechanical energy will not be conserved in the given situation. That's why I've now tried to solve the question using Newton's 2nd law. Kindly see the post#17 and 22 above, wherein I've written the equations. I don't understand why the...

I'm trying to understand your solution. Could you kindly tell me what does the last term ##o(ds)## mean, and where did that come from. I understood till the term ##N'(x)ds##. Secondly, by simplifying the equation: $$\rho a=N'(x)-g\rho.$$, we get: $$\rho vdv = dN - g\rho dx$$. LHS of this can...

By reverse engineering the correct equation, I've reached a few conclusion: The ##v^{2}## term should not be there in the last equation in post#17. Also, the force on the horizontal part of the chain is not = ##\lambda hg##, but is = ##\lambda h(g - \frac{dv}{dt})##. But for eliminating the...

In your equation, you wrote, x is length of horizontal tail, that led me to think that by "tail" you meant the distance moved by the left end of the horizontal part of the chain. But now, I understand that by tail you wanted to mean length of the horizontal part of the chain. It would have been...

No. That's not the answer. I've posted the correct answer in the original post.

Here's how I'm trying to solve it, using Newton's 2nd law. I'm considering the whole chain as the system. Let at any instant of time, t, the horizontal part of the chain is x distance from the left end. ##v## is the velocity of the chain at the instant t. ##\lambda## is the mass per unit length...

All the equations that I had written have been deleted by the moderator. I don't understand why. Is that just because ##\ln## was not written as it should have been ?

Yes, that's a valid point.

Yes, thanks. Edited. The missing screenshot has been added in the original post.

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## +...

Dude, your teacher seems to be clueless on the given topic, if he's saying that we should answer according to the options given. The answer to a question shouldn't be dependent on the options given. It should stand on it's own, irrespective of whether there are any options given or not. Instead...

I don't understand how did you reach that conclusion. Study material is NOT provided by NTA. As far as I know, only syllabus is provided by NTA in addition to conducting the exam. That link that you've shared is not an authentic source. It's just something written by someone on internet, without...