Physics: Post your doubts here!

[Thought blocker]

View attachment 45062

What is the answer ?

 

[ashcull14]

y not c

 

[ashcull14]

What is the answer ?

What is the answer ?

A

 

[Rockstar RK]

Two wires P and Q are made from the same material.
Wire P is initially twice the diameter and twice the length of wire Q. The same force, applied to
each wire, causes the wires to extend elastically.
What is the ratio of the extension in P to that in Q?
(A) 1/2 B 1 C 2 D 4

 

[Thought blocker]

Two wires P and Q are made from the same material.
Wire P is initially twice the diameter and twice the length of wire Q. The same force, applied to
each wire, causes the wires to extend elastically.
What is the ratio of the extension in P to that in Q?
(A) 1/2 B 1 C 2 D 4

Here you go :

 

[Thought blocker]

View attachment 45062

IGNORE ---------------> = in middle

 

[Thought blocker]

Aoa everyone.
Im REALLY stuck on this one specific type of question in all my pastpapers and they are the diffraction grating ones. I havent even gotten a single one right yet and I would EXTREMELY appreciate if someone could just explain 2 of these so I'll understand the rest myself. JazakAllah
Q. 30 from O.N/2012
Q. 27 from O.N/2010

w12_12
30)
Apply d*sin thetha = n*lamida
d = 1/n where n is no. of line/m. Metre not millimetre.
Grating's got 300lines/mm. So this means it has 300*1000 lines/m
(Coz 1m = 1000mm)
d = 1/n ie 1/300,000
U wanna see all the maximas possible, so to get max. no. of maximas on one half of the screen, u put thetha = 90
lamida is given. Convert it into m from nm.

U get n = 4
This means 4 maximas on one half of the screen.
On whole (both halves) of screen, maximas are = 4*2 = 8

BUT, u also have to consider the central maxima.
So total maximas are 8+1 = 9

Central is in middle.
4 maximas above and 4 below it.


27)
Let the distance between the double slit and the screen be 1m initially.
When the distance is increased BY 2m the NEW distance is 3m.
Using the formula:
Wavelength= (fringe seperation x distance between the double slits)/ distance between the screen and the slits.
Wavelenght=600nm=600x10^-9m.
Fringe seperation= 3mm= 3x10^-3m
Distance between screen and slits= 3m (1+2)
Distance between the double slit=?
Put these values in the formula, the final answer is 6x10^-4m =0.6mm
Hence the answer is B.

 

[Thought blocker]

Q1 answer is C
as c= Wavelenght/ time
t=wavelenght/c= 2x10^15
waves in one sec is frequency so f= 1/2x10^15= 5x10^14
Q5
Well I got it wrong at first... But it's just
look at the actual true current .8 and look at faulty ammeter its also .8, so both the readings are closest
Q10
M not completely sure of the formula but in A's We use the formula
initial + final velocity of sphere 1= initial +final velocity of sphere 2
by arranging them we will have
u1+v1= -u2 + v2
rearrange them
u1+u2=v2- v1
About the formula Thought blocker M I correct???

10)
look ... this question deals with relative speeds rather than just momentum.... in a perfectly elastic collision, the relative speed of approach is equal to the relative speed of separation.. but what is meant by relative speed of the 2 objects.... imagine u are moving in a car at a speed of 100 km/h beside another car moving at the same speed that is 100 km/h in the same direction.. u will feel that the other car is not moving.....also if u move at a speed of 100 km/h and u look at another car moving in opposite direction at a speed of 100 km/h u feel that the other car is moving very fast .. that is at a speed of 200 km/h......... so to find relative speed of 2 objects ...add their speeds if they are moving in opposite direction and subtract their speeds when they are moving in the same direction... but only in a perfectly elastic collision.
In this question, the relative speed of approach is equal to u1 + u2 since they are moving in opposite directions.. this is equal to the relative speed of separation which is equal to v2 - v1 since they are moving in the same direction after the collision..

 

[Thought blocker]

Any doubts left ?

 

[crazytaylorfanXD]

Can someone help me with JUN 11 P11 Qs 27?

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s11_qp_11.pdf

 

[chishtyguy]

10)
look ... this question deals with relative speeds rather than just momentum.... in a perfectly elastic collision, the relative speed of approach is equal to the relative speed of separation.. but what is meant by relative speed of the 2 objects.... imagine u are moving in a car at a speed of 100 km/h beside another car moving at the same speed that is 100 km/h in the same direction.. u will feel that the other car is not moving.....also if u move at a speed of 100 km/h and u look at another car moving in opposite direction at a speed of 100 km/h u feel that the other car is moving very fast .. that is at a speed of 200 km/h......... so to find relative speed of 2 objects ...add their speeds if they are moving in opposite direction and subtract their speeds when they are moving in the same direction... but only in a perfectly elastic collision.
In this question, the relative speed of approach is equal to u1 + u2 since they are moving in opposite directions.. this is equal to the relative speed of separation which is equal to v2 - v1 since they are moving in the same direction after the collision..

Ahahaha..... Thanks... I know about relative speeds... Add Maths RV.... Bro u have amazing concepts.... (Y)

 

[Thought blocker]

Can someone help me with JUN 11 P11 Qs 27?

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_11.pdf

Can someone help me with JUN 11 P11 Qs 27?

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_11.pdf

find d which is 1/n -->1/500 * 10^ (-3) = 2 x 10^-6.
then d sin90=n x (600 x 10^-9) = 3 then he asked for the images so it is 3 orders for one side which is 45 degrees so for the 90 degrees it is 3 x 2= 6 + the normal ray = 7 so D

 

[crazytaylorfanXD]

find d which is 1/n -->1/500 * 10^ (-3) = 2 x 10^-6.
then d sin90=n x (600 x 10^-9) = 3 then he asked for the images so it is 3 orders for one side which is 45 degrees so for the 90 degrees it is 3 x 2= 6 + the normal ray = 7 so D

aah now i get it
thanks

 

[Thought blocker]

"there are two vectors, P and P' where P is initial momentum and P' is final momentum.
P=mv
P'=-mv
change in momentum is P'-P which is same as P + (-P)
just add them like this.
-mv + (-mv) = -2mv"

Haris Bin Zahid
This answer is for you from Suchal.. He is unable to post here..

 

[Suchal Riaz]

"there are two vectors, P and P' where P is initial momentum and P' is final momentum.
P=mv
P'=-mv
change in momentum is P'-P which is same as P + (-P)
just add them like this.
-mv + (-mv) = -2mv"

Haris Bin Zahid
This answer is for you from Suchal.. He is unable to post here..

thanks my problem is now solved and the planet master has received enough lectures from me.

 

[Thought blocker]

thanks my problem is now solved and the planet master has received enough lectures from me.

haha

 

[sagar65265]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s13_qp_12.pdf
q9


http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w13_qp_12.pdf
q11

Q9)

Either ways, the working for this question can be very easy if you work in a particular manner; first point is that Force = Δ(mv)/Δt, and if we take Δt to be 1 second, the equation is reduced to it's simplest form - Force = Δ(mv) in 1 second, i.e. force is equal to change in momentum in 1 second.

So, in 1 second, the velocity of some mass of air goes from 33 ms^-1 to 0 ms^-1. Therefore, the change in momentum = (mass of air) * (0-33) = -33(mass of air)
Okay. Now we know the change in momentum, but there still a few pieces of the puzzle left - how much mass is there in that "some mass of air"?

We have taken the time period over which this change in momentum occurs to be 1 second. Therefore, the total change in momentum of the air in 1 second is what we need.

What we can do is write m = ρV (by the definition of density) and calculate the volume of air that comes to a stop in 1 second. Multiplying this by ρ which they have given is 1.2 kgm^-3, we can find out how much air has lost it's momentum.

In that one second, air traveling at 33ms^-1 slows down and comes to a stop. This happens to all the air within 33 meters of the wall - if a small cross section of air is 33 meters from the wall, in one second it will travel 33 meters and come to a stop, right? Air that is 40 meters from the wall will not stop in that one second. Air that is 25 meters from the wall will stop before that one second is over. But totally, 33 meters length of air will come to a stop.

We have the length of the air column that stops in 1 second. But that doesn't give us the volume; we need the area as well.
Luckily, the question tells us that the area of the wall is 12m^2. Correspondingly, the volume of air that comes to a stop is 12m^2 * 33 m = 396 m^3 (which is the units for volume; we are on the right track, then!).

So now we can find out the mass. This mass = 1.2 kgm^-3 * 396 m^3 = 475.2 kg (which is the unit of mass; still on the right track!)
And since we have change in momentum in 1 second= -33(mass of air), change in momentum in 1 second = Force exerted on air BY wall = -15,681.6 Newtons.

By Newton's Third Law, the force exerted by the air on the wall acts in the opposite direction, with an equal magnitude. Therefore, the force is approximately 16,000 Newtons = D.

Q11)

In 1 second, 5.0 × 10⁴ α-particles collide into the lead sample. Let's take these 50,000 particles as our system and work from there.

Imagine the collisions happening; initially, the particles each have a certain amount of momentum, by virtue of their mass and velocity. They keep moving closer and closer to the lead sample over time, and eventually they smash into the lead sheet. Their momentum seems to vanish since they come to a stop, but something does happen here, something very important - there is a force on our system (the particles) from the lead sheet, and that force is large enough to bring all these particles to a halt in one second.

So how large is this force that acts on our system? By Newton's Second Law, this force is equal to the rate of change of momentum of our system (again, our system consists of the particles).
So they go from moving at a high speed to coming to a stop. Their speed changes, and that is why their momentum changes!

Let's find out this change in momentum. In 1 second, 50,000 α particles, each with a mass of 6.6 × 10⁻²⁷ kg and a speed of 1.5 × 10⁷ ms⁻¹ collide with the lead sheet over an area of 1.0 cm². After the collision, their velocity becomes zero. Therefore, we can write that the initial momentum is:

Initial total momentum (of all particles) = 50,000 particles * (6.6 × 10⁻²⁷ kg)(1.5 × 10⁷ ms⁻¹) = 4.95 * 10⁻¹⁵ kg ms⁻¹.
Final total momentum (of all particles) = 50,000 particles * (6.6 × 10⁻²⁷ kg)(0 ms⁻¹) = 0 kg ms⁻¹
Change in momentum (of system) = 0 - 4.95 * 10⁻¹⁵ = -4.95 * 10⁻¹⁵ kg ms⁻¹

This change in momentum takes 1 full second to occur, therefore the force on the system is

Force = (Change in Momentum)/(Time taken for change in momentum) = (Change in momentum)/1 = Change in momentum/second = -4.95 * 10⁻¹⁵ kg ms⁻².

By Newton's Third Law, this is the same force exerted by the system (all the particles) on the lead. Therefore, the pressure on the lead per second is equal to

Pressure = (Magnitude of Force)/(Area) = (4.95 * 10⁻¹⁵ kg ms⁻²)/(1.0/10000) = 4.95 * 10⁻¹¹ kg m⁻¹s⁻² = 5.00 * 10⁻¹¹ Pascals = C.

Hope this helped!
Good Luck for all your exams!

 

[Hadi Murtaza]

Best ov luck for tomorrow every1 <3 may Allah be wid u all

 

[..sacrifice4Revenge..]

Best ov luck for tomorrow every1 <3 may Allah be wid u all

but its not tomorrow. o_o

 

[Asad Moosvi]

Best ov luck for tomorrow every1 <3 may Allah be wid u all

Yo, it ain't tomorrow.