🚀 NEW from the xtremepape.rs team: AI exam prep — 120,000+ worked solutions, and it marks your handwritten working from a photo. Sign in with your forum account & try it free → prepare.xtremepape.rs
V21 or 22? And I'm sure there will be ecf so most probably if ur method for the second part is correct u will get all 3 marks but loose 2 for incorrectly counting the wavelengthCan anyone tell me how many marks they will take off if I counted the number of wavelength wrongly (+.5) and carried on using it for the next questions?
The question has 2 marks for the +.5 thing and the next one has 3 marks.
Dude I did 22 and there was no such question like thatAs the thread title mentioned, P22. I thought I could at least get 1 mark for getting the number of wavelength and do the calculation correctly. Anyway, thank you!
yes, some one please tell how much we need to score for an A*?What is the threshold for A*...as in is it like (gt for A+ difference between gt a and gt of b)?
Ah true, I just realised its 21Dude I did 22 and there was no such question like that
Why did u cancel it!! What u did was correctGuys we had to calculate the slit separation so we have to count the no. of wavelengths because in lamda = ax/D, the x is fringe separation...I did it but cancelled it coz it was written not to scale on the diagram....
2:1 tho not entirely sureGuys what was the answer to the ratio question the first was 4:1 what was the second
i think increases the p.d cuz the current decreases but i dont rememberDid the potential difference increase as the resistance increased in that resistance current graph q
ecf will save you brotherCan anyone tell me how many marks they will take off if I counted the number of wavelength wrongly (+.5) and carried on using it for the next questions?
The question has 2 marks for the +.5 thing and the next one has 3 marks.
Did the potential difference increase as the resistance increased in that resistance current graph q
you just had to divide the fringes total distance by number of fringes.. youll get x... then use the formula lambda=ax/d I guess thats how it was supposed to be doneGuys we had to calculate the slit separation so we have to count the no. of wavelengths because in lamda = ax/D, the x is fringe separation...I did it but cancelled it coz it was written not to scale on the diagram....
🚀 NEW from the xtremepape.rs team: AI exam prep — 120,000+ worked solutions, and it marks your handwritten working from a photo. Sign in with your forum account & try it free → prepare.xtremepape.rs