ICSE − Q & A

https://m.timesofindia.com/home/education/news/history-teachers-allege-mistakes-in-icse-question-paper/amp_articleshow/68231390.cms Grace mark chances???...

Please can someone explain Q 1 (d)
https://www.respaper.com/stpeters1983/1308-pdf.html

I know this is quite late but for the civics question of list 4 writs I only wrote the names and my teacher told me that it was correct. Your thoughts? Only asking because the explanation of the writs was not in the syllabus

A ball of mass 10 g falls from a height of 5 m. It rebounds from the ground to a height of 4 m. (take g=10 m/s)  Find: 
(i) the initial potential energy of the ball. 0.5 Joule
(ii) the kinetic energy of the ball just before striking the ground.  0.5 Joule
(iii) the kinetic energy of the ball after striking the ground.  ???
(iv) the loss in kinetic energy of the ball on striking the ground.   0.1 Joule

+ 1 more questions by teleansh  

https://www.respaper.com/aayush25/3848-pdf.html --> work,power and energy 8(iii)
asked by Ansh Anand (teleansh) 5 years ago
0

CALCULATE the work done by a force of 3 kgf in lifting a load of 1 kgf to a height of 5 m.

Critical angle for diamond is 24 degrees.Explain the statement

in this question
A man having a box in head climbs a slope and another man  having an identical box walks the same distance on a levelled road.Who does more work against the force of gravity and why?
why everyone is concluding that the work in 2 case is zero.
is it not possible that the man is pushing the box on the road for causing displacement ??????

https://www.respaper.com/vsharp/435/432/4224-pdf.html how to solve and what is formula 6 and 7

Can they ask us to draw diagrams from any other chapter other than light?

The UCM acceleration "confusion" :

Honestly, I am baffled. I know for sure that the acceleration is non-uniform, and I'll give you a valid reason for that, but then even @shauns has made a valid argument. If you care to look at Q 5 a(ii) of ICSE 2016, they have asked "Is the acceleration uniform?", and in the Pupil's Analysis marking scheme, they have answered "Yes".

I really don't understand the mysterious logic they have applied here. I think, in this case of confusion, you should go by whatever your Selina says. The only plausible logic behind their strange answer can be that by "uniform" they imply that the magnitude is constant. I really am not sure though.

However, in actual "non-confusing" terms, the acceleration in UCM is variable. Reason is short and precise: the magnitude of the acceleration is constant, the direction is not. 

Now if you think that the direction is also uniformly changing, you're wrong, because we don't talk of "change in magnitude" and "change in direction" separately when dealing with vectors.

For eg, consider a particle in UCM, and note its acceleration vector at one point. Let us assume that it is 4 m/s^2, towards the left. Now, when the particle reaches the diametrically opposite point, its acceleration will be 4 m/s^2, towards the right. Now, if you have any knowledge of vectors, you will understand that when you subtract these two vectors, it would be -8 i cap.

To explain this simply, it is as if the acceleration at the first point that I mentioned is -4 and the acceleration at the diametrically opposite point is +4, so when you subtract +4 from -4 to calculate change, it is -4 - (+4) = -8.

Thus, there is clearly a change in the acceleration vector, so acceleration is not constant. 

But this still doesn't answer the burning question : What the hell do we write in the paper?

I think you should stand with logic, and answer "No, because the magnitude is constant, but the direction is changing as when the particle moves in a circle, the vector towards the center changes in direction."

I think that if you give a correct reason, it will remove any scope for doubt. 

In the end though, it is all up to your prudence. If you want to go with the 2016 answer, your choice. If you don't, it's again your choice.

I don't know how the board thinks, so in any case, I am not in a position to tell you what they will accept as correct. I can only tell you what I know is correct as per logic. If you didn't get what I'm trying to hint at subtly, these last two lines are a sort of disclaimer. Please be advised :P.

ATB, see you later.