AS PHYS 9702 PAPER 1 | February/March 2023 | Paper 12 | 9702/12/F/M/23 | Full and Detailed Solution

[@yassineadel2181]

goodluck everyone tomorrow :)

[@alevelandapphysics-ali]

All the best for tomorrow exam .

[@TresalynMasheka-dw3nz]

Thank you so much for your videos, your teaching is great cause you dont do the "assumed knowlegde" stuff like other youtubers

[@ZainAli-nk1rq]

Wish I had studied from you sooner! Absolutely clear explanations,breaking down everything to the most understandable thing. I hope Allah gives you a long life for the work you're putting it, your work is not going unnoticed and it's gonna be helping thousands of students to come in the near future! Stay blessed Sir.

[@alevelandapphysics-ali]

Thanks a lot ,It's always a pleasure and blessing to have students like you . Stay blessed !

[@anonnoanupom2053]

life saver fr fr

[@skillyroy3289]

well explained sir

[@hairlesssheep2805]

thank you so much for your explanations

[@alevelandapphysics-ali]

Glad it was helpful

[@sohaibmuhammad-vv9cd]

Thank you so much sir

[@mika3541]

For question 24, why can't we use I = I naught cos square feta So the new intensity is Intensity for 30 degreees × cos²(30)? So the answer is 0.75×Intensity for 30º (Ans C)?

[@alevelandapphysics-ali]

It's ONLY correct if two filters are placed in front of vertically polarized light one makes angle of 30 degrees with vertically polarized light and second filter makes an angle of 30 degrees with first polarizer transmission axis then your calculation is correct but in this case one filter makes angle of 30 degrees with incident vertically polarized light and then second time same filter makes an angle of 60 degrees with vertically polarized light so answer should be B as i already explained in the video . Hope it makes sense . Feel free to ask if you need more detailed explanation . Cheers

[@rhrd33]

For question 14, I thought they said “upthrust of probe and thread is negligible”, since Upthrust = Weight, doesn’t that mean that we don’t have to count in the combined weight of 2N? So shouldn’t it just be Upthrust = 5N?

[@alevelandapphysics-ali]

Total weight of sphere , probe and string =5N+2N=7 N , as sphere is in equilibrium it means that Weight = Upthrust =7 N . As question said upthrust due to probe and thread is negligible , it simply means that MOST of the upthrust is due to sphere only . In this case upthrust due to probe and string is negligible because they have very very small volume as compared with volume of sphere . Hope it makes sense . For better understanding , you can listen explanation again . Let me know if you still have any questions .

[@AmusedAirboat-mc4xm]

Sir pls solve 2024 feb march paper

[@alevelandapphysics-ali]

It's coming on this weekend

[@rhrd33]

Sir, for Q29, since N=1/d, doesn’t that mean N affects too? When N changes, d changes, and since d affects, N must affect?

[@alevelandapphysics-ali]

dsin*angle=n* Wavelength , it's very clear that d affects the angle or fringe spacing . Imagine a diffraction grating of length 1 cm and it has 10 slits so d = 0.1 cm . Now imagine that we have diffraction grating of 2cm and there are 20 slits so again d=2cm/20=0.1 cm . As long as d is same , N does not affect . Hope it makes sense . You can watch solution again for better understanding .

[@alevelandapphysics-ali]

Feel free to ask if you still have any questions .

[@cappybaraeats]

best phy teacher can you just come to our school and teach

[@alevelandapphysics-ali]

Appreciated your comment .

[@eighoreishongsar2336]

Hi, Did you mean height of hill or average diameter of the top of the hill for diffraction to happen in Q . 27?

[@alevelandapphysics-ali]

Height of the hill but it also depends on distance of transmitter and receiver from the hill . In order to understand differaction in a better , look at Huygen's principle :every point on an advancing wavefront is a source of outgoing spherical waves. So diffraction takes place everywhere, all the time. But if all the points on said wavefront are in a plane, those outgoing spherical waves add coherently to make a continuing plane wave. Only when part of the plane wave is blocked off do you see the uncompensated spherical waves out to the side if different directions.

[@eighoreishongsar2336]

Sir, Any update!

[@tasbi21yt]

In question number 13 , why did you divided by area of circle . Why not area of hemisphere

[@alevelandapphysics-ali]

Pressure is the force per *unit perpendicular area* over which the force is applied. In this case , area perpendicular to the force is circular cross-sectional area of hemisphere . Hope it helps . Let me know if you still have any questions

[@leeroy14r60]

26:56 😂😂