Physics: Post your doubts here!

[crazytaylorfanXD]

6)
The question states that "X is measured with a percentage uncertainty of ±1 % of its value at all temperatures." meaning that the percentage uncertainty is unchanged, so A and B is eliminated.
then you need to apply some logic here, will the actual uncertainty get bigger as you approach 100 or 0?.
you can quickly do some math here, 100x1% = ±1 uncertainty while 0x1% = 0
so answer is C, least actual uncertainty when temperature close to 0

11)
The object hits the wall with a speed v, and rebounds with the same speed, but a velocity in the opposite direction.
Kinetic Energy does not depend on the direction of an object's velocity, just the magnitude of that objects velocity, i.e. the speed. Therefore, since the speed remains the same even after the collision, the Kinetic Energy of the object is conserved.

By the same logic applied above, we can say that the speed is conserved, so that option is eliminated.

Lastly, the mass of the object doesn't change, and therefore the mass also remains conserved.
By elimination, we can confidently say that the momentum of the object is the only value that is not conserved - momentum, unlike the other values, is a vector quantity; since the direction of the object's speed changes, the momentum also changes, and is therefore not conserved.

23)
We use F = kx
so we know k for P and Q is k and for R is 3k
--> Therefore total extension of P and Q = k + k = 2k and extension for P,Q and R = 1/2k + 1/3k = (5/6)^-1k = 6/5k
so F = W, k = 6/5k and x = ?
W = 6/5k * x
x = 5w/6k

29)
There should be more points on to which the light strikes so it's either B or C, but hence in C the lines are closer, there is more intensity

36)
Firstly the ammeter is directly connected with the fixed resistor so there is no change in readings.
And its obvious, when there are two resistors, if one resistor's resistance is increasing, so others resistor's resistance would decrease so R is proportional to V hence V decreases.

umm do u mind explaining 23 again??
i didnt understand why u just added the k's and why u inversed them
is there a concept behind this that im missing out on??

 

[ZaqZainab]

Gosh! I like your presentation ! - A LOT

you being sarcastic?
Thanks though

Dushyant, here I took distance XY and XZ and solve them simultaneously.
This was the only way I could think of
ZaqZainab I tried your way now LOL
Anyways here is my method :¬

Use the equation of motion d = ut + 0.5at^2

equation 1 : 40 = 12u + 72a

equation 2 : 80 = 18u + 162a

[I added the distances, so that initial speed of both equations will be same]

now solve these simultaneously, and you will get 0.37 m/s^2 as the answer.

P.S. I solved it in full detail, go some few pages back, and you'll get your answer

thats the same thing i did

 

[Thought blocker]

you being sarcastic?
Thanks though

thats the same thing i did

You took XY and YZ - Not the same

 

[Thought blocker]

umm do u mind explaining 23 again??
i didnt understand why u just added the k's and why u inversed them
is there a concept behind this that im missing out on??

We add them like circuits.
When in series just add 'em.
When in Parallel add them like 1/something and inverse it..

 

[crazytaylorfanXD]

We add them like circuits.
When in series just add 'em.
When in Parallel add them like 1/something and inverse it..

Wait , i think u mean the other way around right?!?

 

[Thought blocker]

Wait , i think u mean the other way around right?!?

Sorry, but I am not able to figure out what you said.. This chemistry had sucked my memory

 

[crazytaylorfanXD]

Sorry, but I am not able to figure out what you said.. This chemistry had sucked my memory

Awww man i know how u feel

 

[Thought blocker]

Awww man i know how u feel

Tell me what exactly you cannot figure out in Q23 ?

 

[crazytaylorfanXD]

i didnt understand why u added K's in one place and inversed them and added them in another place??
where do we add and where do we use the inverse thingy??

 

[Thought blocker]

i didnt understand why u added K's in one place and inversed them and added them in another place??
where do we add and where do we use the inverse thingy??

Circuit Thingy yaar... lemme send you snap!

 

[Thought blocker]

Circuit Thingy yaar... lemme send you snap!

i didnt understand why u added K's in one place and inversed them and added them in another place??
where do we add and where do we use the inverse thingy??

Okay so we add here same as we do in circuits :¬

 

[Asad Moosvi]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w10_qp_13.pdf

question 17

 

[crazytaylorfanXD]

Okay so we add here same as we do in circuits :¬
View attachment 45053

soo the W is going to be equal for all the springs??

 

[Thought blocker]

soo the W is going to be equal for all the springs??

No..
case i)
You had 3 strings with load W
so imagine if you have 3 students and 1 chocolate
What is the ratio of getting chocolate to each child ?
Obviously 1/3
so same over here
3 strings = 3 students
W load = 1 chocolate
so load (Force) on each spring will be W/3

case ii)
You have 2 strings with load 2W
so imagine now you have 2 students and 2 chocolates
What is the ratio of getting chocolate to each child ?
Obviously 1
so same over here
2 strings = 2 students
2W load = 2 chocolates
so load (Force) on each spring will be W

Got it ?

 

[chishtyguy]

View attachment 45039

Answer is A as.... voltage across 15 Ohm resister is 9-7.5= 1.5V By using the formula 15/(R+15)x 9 = 1.5, R= 75 .... Then total resistance is 75+15 = 90 Ohm.... I=V/R so I= 9/90= .1 A

 

[Thought blocker]

Answer is A as.... voltage across 15 Ohm resister is 9-7.5= 1.5V By using the formula 15/(R+15)x 9 = 1.5, R= 75 .... Then total resistance is 75+15 = 90 Ohm.... I=V/R so I= 9/90= .1 A

Answer is B from my side... But let her answer if it is A or B

 

[Thought blocker]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_13.pdf

question 17

17)
The question is asking what is the rate does the energy provided to the system by the motor. So this shows that the motor is actually supplying some of its power. You must also know that the mass of the elevator is greater that the mass of the heavy weight because the question already stated "the elevator is partly counterbalanced by a heavy weight." So (m1 > m2) but if that is the case then how can the elevator move up? Yes. The motor and the heavy weight(m2) are providing the energy required to move the elevator up. So I came up with a formula: Power by m2 + power by motor = Power required to move m1. And we will have to use a formula P=FV where F can be the weight and V can be the velocity/speed. So substitute this formula into the earlier formula and we will get: m2gv + Power by motor = m1gv. Rearrange the formula: Power by motor = m1gv - m2gv. Simplify further and we will get: Power by motor = (m1 - m2)gv

 

[..sacrifice4Revenge..]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s13_qp_12.pdf
q9


http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w13_qp_12.pdf
q11

 

[Asad Moosvi]

17)
The question is asking what is the rate does the energy provided to the system by the motor. So this shows that the motor is actually supplying some of its power. You must also know that the mass of the elevator is greater that the mass of the heavy weight because the question already stated "the elevator is partly counterbalanced by a heavy weight." So (m1 > m2) but if that is the case then how can the elevator move up? Yes. The motor and the heavy weight(m2) are providing the energy required to move the elevator up. So I came up with a formula: Power by m2 + power by motor = Power required to move m1. And we will have to use a formula P=FV where F can be the weight and V can be the velocity/speed. So substitute this formula into the earlier formula and we will get: m2gv + Power by motor = m1gv. Rearrange the formula: Power by motor = m1gv - m2gv. Simplify further and we will get: Power by motor = (m1 - m2)gv

Oh. I see. Thanks.

 

[crazytaylorfanXD]

No..
case i)
You had 3 strings with load W
so imagine if you have 3 students and 1 chocolate
What is the ratio of getting chocolate to each child ?
Obviously 1/3
so same over here
3 strings = 3 students
W load = 1 chocolate
so load (Force) on each spring will be W/3

case ii)
You have 2 strings with load 2W
so imagine now you have 2 students and 2 chocolates
What is the ratio of getting chocolate to each child ?
Obviously 1
so same over here
2 strings = 2 students
2W load = 2 chocolates
so load (Force) on each spring will be W

Got it ?

Ohh i get it now thanks :***