Last minute preparation on MCQS 9701 and 9702. Discussion here.

[DarkEclipse]

I dont understand this logic?
How did this clicked to ur mind?

I don't know how else to explain it. Sorry.

 

[The Sarcastic Retard]

Thank.

 

[Shujaat Khan]

I dont understand this logic?
How did this clicked to ur mind?

He might have said it in a bad way, the "period" he found by dividing 0.45/0.3 is the number of cycles in 45cm of line
so 1 cycle= 0.3m
how many cycles in 0.45 m?
you just divide and get 1.5 cycles.
Which equals to 1 complete cycle and 1/2 a cycle

Just draw , you understand it right away

 

[The Sarcastic Retard]

He might have said it in a bad way, the "period" he found by dividing 0.45/0.3 is the number of cycles in 45cm of line
so 1 cycle= 0.3m
how many cycles in 0.45 m?
you just divide and get 1.5 cycles.
Which equals to 1 complete cycle and 1/2 a cycle

Just draw , you understand it right away

I figured it out something like this, There are two nodes formed at transmitter and at plate. So that is lambda/2 = 0.45. Hence lambda = 0.9m. Also I did, lambda = 3*10^8 / 1 * 10 ^9 = 0.3m.

Therefore 1 antinode in 0.3m so how many in 0.9?
Hence answer is 3.

Is it correct?

Thanks.

 

[DarkEclipse]

I figured it out something like this, There are two nodes formed at transmitter and at plate. So that is lambda/2 = 0.45. Hence lambda = 0.9m. Also I did, lambda = 3*10^8 / 1 * 10 ^9 = 0.3m.

Therefore 1 antinode in 0.3m so how many in 0.9?
Hence answer is 3.

Is it correct?

Thanks.

No, lambda/2 is used for adjacent nodes. The two nodes in the question are not adjacent.

 

[The Sarcastic Retard]

No, lambda/2 is used for adjacent nodes. The two nodes in the question are not adjacent.

So its wrong... Nyways, thanks.

 

[The Sarcastic Retard]

 

[The Sarcastic Retard]

 

[The Sarcastic Retard]

my avg score for chem is 31 and for phy is 29
what about u all???

 

[DarkEclipse]

View attachment 57324

Okay, first, you have to realize that X cannot be a cyclic compound. Why? Because the NH group in coniine is the reason the compound is cyclic. In its absence, the compound is a straight chain. Now notice one thing. When 'N' is removed, that leaves two ends, which is originally occupied by the two 'Br' atoms. When you draw X as a straight chain, you'll notice that the 1st Br atom attaches to the first carbon, and the 2nd Br attaches to the fifth carbon. Therefore, the compound is 1,5-dibromooctane.
Here's what I mean (sorry for the crude drawing, I'm not good at Paint):

my avg score for chem is 31 and for phy is 29
what about u all???

26 for Physics. Dunno about Chem.

 

[saturn21]

View attachment 57323

From which paper is this question from?

 

[saturn21]

For q29 i chose D don't remember why but the answer is C
I drew the structures..i just wanna know if they are correct

 

[saturn21]

Ans is C..how?

 

[The Sarcastic Retard]

From which paper is this question from?

I dont remember

 

[saturn21]

Btw guys i found an uncle who solves chem p1 past papers on youtube ... it's Meta Tutor..for me his videos were realllyy helpful

 

[saturn21]

I dont remember

Oh..okey

 

[The Sarcastic Retard]

View attachment 57338

For q29 i chose D don't remember why but the answer is C
I drew the structures..i just wanna know if they are correct

View attachment 57338 View attachment 57339

There is same functional groups, so there would be one same cis trans isomer hence 3

 

[The Sarcastic Retard]

View attachment 57340
Ans is C..how?

I know this question
I will have my dinner and solve it
I am in hostel

 

[saturn21]

I know this question
I will have my dinner and solve it
I am in hostel

Lol okey

 

[The Sarcastic Retard]

Lol okey

I totally forgot.. :3
I will do it now..