HOW WAS PHYSICS PAPER 2 AS!!!!!!!!!!!

[Nisrit Pandey]

NO BUH I ALSO GOT 1000 FREQUENCY and almost all my frnds got the same how can u say it was 500

i still remember
speed was 340
wavelenght was 0.34! and hence 1000

 

[Nisrit Pandey]

since

how can it be half??

the young modulus of copper was half, for the same extension it would require half the force! and the graph always begins from the origin so

 

[Nisrit Pandey]

For copper, Young Modulus is half. Hence, the force should be halved, i.e. 1.5 N.
Same graph but with peak at 1.5 N instead of 3.

yeah i did the same Ahmed! thanks confidence increased

 

[Mattman]

WHat u guys awnsered for the last part of the paper? the mass conservation

 

[Ahmed Tariq]

WHat u guys awnsered for the last part of the paper? the mass conservation

I wrote 'Initially, the neutrons have some kinetic energy.'
Meh.

 

[Nisrit Pandey]

what did you do for experiments people? i used a tuning fork experiment on the water level and we can observe the resonance by adjusting the tuning fork, the first sound heard would denote that a first harmonic stationary wave is created which is 1/4 th of wavelength and hence showed how to caculate it

 

[Nisrit Pandey]

I wrote 'Initially, the neutrons have some kinetic energy.'
Meh.

i wrote a stupid answer that it gives out more particles

 

[Mattman]

:| i wrote that the equation didnt factor in energy released as mass
for the experiment i did the loudspeaker into tube and dust piles thing
distance between dust piles is 1/2lambda

 

[Nimra Asif]

The paper was either TOO long for 1 hour or 1 hour is TOO short for the paper... :/
I cudnt draw the graph

exactly... =( i lost 8 marks =/ 1 hour passed like a ray of light..

 

[Soulgamer]

IMO Oct/Nov was relatively easier than the paper we had today. The gt for Oct/Nov was 30. This paper was very time consuming and tricky. :|

 

[Beaconite]

i still remember
speed was 340
wavelenght was 0.34! and hence 1000

Actually, it was a stationary wave, since it told you to label node and antinodes earlier. In a stationary wave, length b/w 3 nodes is wavelength, not 2. 0.34 cm was distance between 2 nodes, and half wavelength. Trust me, I made the same mistake initially. Wavelength was 68.5 cm, so 340/68.5x10-2 =496 Hz.

 

[xxfarhaxx]

i wrote a stupid answer that it gives out more particles

sme here..i wrote anthr prticle emittd..whch is ryt??

 

[xxfarhaxx]

IMO Oct/Nov was relatively easier than the paper we had today. The gt for Oct/Nov was 30. This paper was very time consuming and tricky. :|

ya..vry trcky indeed..hpe da gt is vv low!!!

 

[Ahmed Tariq]

Actually, it was a stationary wave, since it told you to label node and antinodes earlier. In a stationary wave, length b/w 3 nodes is wavelength, not 2. 0.34 cm was distance between 2 nodes, and half wavelength. Trust me, I made the same mistake initially. Wavelength was 68.5 cm, so 340/68.5x10-2 =496 Hz.

In a stationary wave, the distance between successive nodes and antinodes is lambda/2 while that between a node and an antinode is lambda/4.

 

[Soulgamer]

In a stationary wave, the distance between successive nodes and antinodes is lambda/2 while that between a node and an antinode is lambda/4.

Please refrain from discussing the contents of the paper!

 

[Beaconite]

In a stationary wave, the distance between successive nodes and antinodes is lambda/2 while that between a node and an antinode is lambda/4.

Exactly. So distance b/w 2 nodes i.e 0.34 cm, was half wavelength.

 

[maheenali]

any idea for p34 tomorrow??

 

[WhereAmazingHappens]

Young Modulus is (F/e)(L/A), where F/e was the gradient.

No check your text books

God u guys want me to derive it??

Sheez wait a sec

 

[Ahmed Tariq]

Exactly. So distance b/w 2 nodes i.e 0.34 cm, was half wavelength.

Wasn't the distance between two antinodes on the graph 17cm?

 

[WhereAmazingHappens]

what were the axes btw