• We need your support!

    We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
    Click here to Donate Now (View Announcement)

Physics: Post your doubts here!

A

_Ahmad

Messages
186
Reaction score
332
Points
73
Thought blocker said:
8) I will solve it later.

20)
Simple yaar, take a look on book, that middle thing is work done..

24)
The amplitude increases 2 times so intensity increases by 2^2 = 4 times
The frequency decreases by 1/2 so intensity decreases by (1/2)^2 = 1/4 times
So net change in intensity= 4 * (1/4) = 1
So intensity is unchanged and is equal to I0
So the answer is B

25)
In air, the frequency would remain the same. Now do it yourself
Why frequency is same :
when waves travel, they carry energy. When passing from one media to another the energy content of the waves has to stay the same due to the law of conservation of energy.
we know that E=hf where E=Energy of wave, h=planck's constant, f=frequency of wave.
so h being a constant would not change, and f could not change either.
So, the only characteristic of the wave which can change is the velocity and wavelength, which adjust themselves so as to give the same frequence before and after passing through different media.

Hope you got it :)
Click to expand...

THANKS ALOT
 
Thought blocker

Thought blocker

Messages
8,477
Reaction score
34,837
Points
698
Ahmed H. Al-Neel said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_12.pdf
questions 8,12,13,14,20 please
Click to expand...
8)
Pi = Pf
Pi = 2mu - mu =mu
So check for each condition Pi = Pf ?
so in A,C and D Pi = Pf
but C and D are inelastic collision so answer is A :)

12)
the torque between the pulley Q and P in the upper wire is the torque at Q.
as it is causing the driving force for Q.
i.e T=F*d
F= T/d
F=3/100*10^-3
F=30*2 (since, the formula suggests force into perpendicular distance between the two forces)
F=60 N.
Torque on P,
T=F*d
T=30*150*10^-2 (Force remains the same, i.e 30N)
T=4.5 N m

13)
assuming it is projected with velocity v, so the initial KE, E= (mv^2)/2
at the highest point, the vertical component of the veolcity = 0, so its left with v cos 45,
so final KE = 1/2 x m x (v cos 45)^2 = 0.5E

14)
By conservation of momentum, 2 x 2 = 1 x v, v =4
so the 1kg trolley moves 4ms-1 to the right,
calculating the total KE of the both trolleys, which is 1/2 x 2 x 2^2 + 1/2 x 1 x 4^2 = 12J
assuming no energy is lost, the energy stored in the spring is converted to KE gained by the trolleys, so its 12J

20)
Use 0.5*f*x formula = 0.5*100*(2*10^-3)
= 0.1 But the actual strain energy is bit more than this, so 0.11 J
 
Thought blocker

Thought blocker

Messages
8,477
Reaction score
34,837
Points
698
Ahmed H. Al-Neel

Ahmed H. Al-Neel

Messages
45
Reaction score
138
Points
8
Thought blocker said:
8)
Pi = Pf
Pi = 2mu - mu =mu
So check for each condition Pi = Pf ?
so in A,C and D Pi = Pf
but C and D are inelastic collision so answer is A :)

12)
the torque between the pulley Q and P in the upper wire is the torque at Q.
as it is causing the driving force for Q.
i.e T=F*d
F= T/d
F=3/100*10^-3
F=30*2 (since, the formula suggests force into perpendicular distance between the two forces)
F=60 N.
Torque on P,
T=F*d
T=30*150*10^-2 (Force remains the same, i.e 30N)
T=4.5 N m

13)
assuming it is projected with velocity v, so the initial KE, E= (mv^2)/2
at the highest point, the vertical component of the veolcity = 0, so its left with v cos 45,
so final KE = 1/2 x m x (v cos 45)^2 = 0.5E

14)
By conservation of momentum, 2 x 2 = 1 x v, v =4
so the 1kg trolley moves 4ms-1 to the right,
calculating the total KE of the both trolleys, which is 1/2 x 2 x 2^2 + 1/2 x 1 x 4^2 = 12J
assuming no energy is lost, the energy stored in the spring is converted to KE gained by the trolleys, so its 12J

20)
Use 0.5*f*x formula = 0.5*100*(2*10^-3)
= 0.1 But the actual strain energy is bit more than this, so 0.11 J
Click to expand...
thank you sooo sooo much again but can u also explain q.)21 also? i can w8 till 2morro if ur really busy! sorry for any troubles and thnx again
 
Thought blocker

Thought blocker

Messages
8,477
Reaction score
34,837
Points
698
Ahmed H. Al-Neel said:
thank you sooo sooo much again but can u also explain q.)21 also? i can w8 till 2morro if ur really busy! sorry for any troubles and thnx again
Click to expand...
21)
use the formula for E
E = F L / Ax
rearrange to get the ratio x / L on one side (change in length / original length)
you'll get is x / L = F / E A : Where - (A = pi r ^2 )
= 20 / 2 * 10^(11) * pi x (2.5 * 10^(-4))^2
= 5.1 * 10^(-4)
multiply this by a 100 to get the percentage
5.1 x 10^-4 x 100 = 5.1 x 10^-2 %
So answer is B.
 
S

sagar65265

Messages
227
Reaction score
571
Points
103
_Ahmad said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_12.pdf

Q12,28,31
Click to expand...

Q12)

Vectors.JPG

Q28)

Wow, this is a toughie!

Suppose you have two sources of light, beside each other and projecting light onto a screen equidistant from them. This is an alternative set-up to a double slit experiment, where two slits are small enough to act as light sources on their own, and transfer light from a single source onto a screen in from of them.

Every second, thousands of wavelengths of light pour out through the light sources (in both experiments), all the time interfering with other light waves, combining constructively and destructively, traveling through the medium around them, and finally striking the screen, millions at a time. How can a steady diffraction fringe be formed? How is the image so steady?

The answer is because the situation outside the screen remains the same all the time. Every second, the same number of photons are found between the slits and the screen (those that enter through the source replace those that collide into the screen), every second the same number of waves interact (the wavelength of the light concerned has to be similar; simultaneously, the frequency has to be similar and the color has to be similar. Red light cannot from a fringe with blue light, for example) and every second the situation is exactly the same as it is the previous second. How?

This is because the sources are behave in exactly the same manner. The sources both have the same intensity, they produce the same number of wavelengths per second, the waves they produce are in phase, and only because of that do all those waves interact the same way, all the time. In other words, the steady fringe is formed only because the sources are coherent. Therefore, we can rule out B.

Suppose we polarize the light. That still doesn't make any difference.

If one source has light vibrating in the vertical plane (just imagine that) and the other source is polarized "at right angles to light from the other source", i.e. polarized in the horizontal plane, that simply means that one source will have horizontally oriented vibrations and the other will have vertically oriented vibrations.

No matter how we put them together, it is impossible to get zero resultant intensity from that! E.g. it's like trying to add two perpendicular vectors (say the sides of the triangle) and getting a zero resultant!

It just won't happen. Sadly, though, for a fringe to be created, there are places where zero intensity of light has to be found, and places where double intensity of source light has to be found. Therefore, since polarizing the light as they have mentioned it will not create points of zero intensity, we won't be able to create a fringe like that. Therefore, C is also wrong.

Lastly, let's take D: suppose the light from the two sources do not even overlap, they cannot possibly interfere with each other! And if they can't interfere, they can't form points of zero intensity and maximum intensity, and so they can never form a fringe at all - this means that D is also not the answer, and the only remainder is A, our final answer.

Q31)

At any point in the electric field, the force on a charged particle will be along the tangent to the field line. Therefore, the force on any particle Q at a point P in some electric field can never act at an angle (which is not 0 or 180 degrees) to the tangent of the field lines at point P.

To put an example, in this question, the force at A is along the tangent to the field line at that point - suppose you draw the tangent to the field lines at that point, the force they show at A will probably be along that tangent. So A might be the answer.

At B, suppose you draw a tangent there, the force shown will again be mostly along the field lines. So B might be the answer.

At C, we have a problem. If you draw a tangent at C, then the force shown will definitely not be along that tangent. So C is not the answer.

At D, the force shown might lie along the tangent. So D might be the answer.

The last thing we need to know is that:

".....at any point on an electric field line, the arrow on the field line will point in the direction of the force on a positive charge, placed at that point on the field line. The force on a negative charge placed at the point will be in the opposite direction."

So, positive charge = force along the field line direction. Negative charge = force opposite the field line direction.
Since we are talking about a negative charge here, we are therefore looking for a force that points in the opposite direction of the field line arrow.

At D, the field line arrow points to the right. The force also points in this direction, so this option is wrong.

At B (we have eliminated C) the field line points upward. The force is also pointing in this direction, so B is wrong.

Lastly, at A, the field line moves downwards, and then to the right. The force shown is upwards and to the left. Therefore, A is our answer.

Hope this helped!
Good Luck for all your exams!
 
ShreeyaBeatz

ShreeyaBeatz

Messages
84
Reaction score
178
Points
43
Which statement about electrical resistivity is correct?
A The resistivity of a material is numerically equal to the resistance in ohms of a cube of that
material, the cube being of side length one metre and the resistance being measured
between opposite faces.
B The resistivity of a material is numerically equal to the resistance in ohms of a one metre
length of wire of that material, the area of cross-section of the wire being one square
millimetre and the resistance being measured between the ends of the wire.
C The resistivity of a material is proportional to the cross-sectional area of the sample of the
material used in the measurement.
D The resistivity of a material is proportional to the length of the sample of the material used in
the measurement.


WHY IS A right but not C??
 
Thought blocker

Thought blocker

Messages
8,477
Reaction score
34,837
Points
698
sagar65265 said:
Q12)

View attachment 44428

Q28)

Wow, this is a toughie!

Suppose you have two sources of light, beside each other and projecting light onto a screen equidistant from them. This is an alternative set-up to a double slit experiment, where two slits are small enough to act as light sources on their own, and transfer light from a single source onto a screen in from of them.

Every second, thousands of wavelengths of light pour out through the light sources (in both experiments), all the time interfering with other light waves, combining constructively and destructively, traveling through the medium around them, and finally striking the screen, millions at a time. How can a steady diffraction fringe be formed? How is the image so steady?

The answer is because the situation outside the screen remains the same all the time. Every second, the same number of photons are found between the slits and the screen (those that enter through the source replace those that collide into the screen), every second the same number of waves interact (the wavelength of the light concerned has to be similar; simultaneously, the frequency has to be similar and the color has to be similar. Red light cannot from a fringe with blue light, for example) and every second the situation is exactly the same as it is the previous second. How?

This is because the sources are behave in exactly the same manner. The sources both have the same intensity, they produce the same number of wavelengths per second, the waves they produce are in phase, and only because of that do all those waves interact the same way, all the time. In other words, the steady fringe is formed only because the sources are coherent. Therefore, we can rule out B.

Suppose we polarize the light. That still doesn't make any difference.

If one source has light vibrating in the vertical plane (just imagine that) and the other source is polarized "at right angles to light from the other source", i.e. polarized in the horizontal plane, that simply means that one source will have horizontally oriented vibrations and the other will have vertically oriented vibrations.

No matter how we put them together, it is impossible to get zero resultant intensity from that! E.g. it's like trying to add two perpendicular vectors (say the sides of the triangle) and getting a zero resultant!

It just won't happen. Sadly, though, for a fringe to be created, there are places where zero intensity of light has to be found, and places where double intensity of source light has to be found. Therefore, since polarizing the light as they have mentioned it will not create points of zero intensity, we won't be able to create a fringe like that. Therefore, C is also wrong.

Lastly, let's take D: suppose the light from the two sources do not even overlap, they cannot possibly interfere with each other! And if they can't interfere, they can't form points of zero intensity and maximum intensity, and so they can never form a fringe at all - this means that D is also not the answer, and the only remainder is A, our final answer.

Q31)

At any point in the electric field, the force on a charged particle will be along the tangent to the field line. Therefore, the force on any particle Q at a point P in some electric field can never act at an angle (which is not 0 or 180 degrees) to the tangent of the field lines at point P.

To put an example, in this question, the force at A is along the tangent to the field line at that point - suppose you draw the tangent to the field lines at that point, the force they show at A will probably be along that tangent. So A might be the answer.

At B, suppose you draw a tangent there, the force shown will again be mostly along the field lines. So B might be the answer.

At C, we have a problem. If you draw a tangent at C, then the force shown will definitely not be along that tangent. So C is not the answer.

At D, the force shown might lie along the tangent. So D might be the answer.

The last thing we need to know is that:

".....at any point on an electric field line, the arrow on the field line will point in the direction of the force on a positive charge, placed at that point on the field line. The force on a negative charge placed at the point will be in the opposite direction."

So, positive charge = force along the field line direction. Negative charge = force opposite the field line direction.
Since we are talking about a negative charge here, we are therefore looking for a force that points in the opposite direction of the field line arrow.

At D, the field line arrow points to the right. The force also points in this direction, so this option is wrong.

At B (we have eliminated C) the field line points upward. The force is also pointing in this direction, so B is wrong.

Lastly, at A, the field line moves downwards, and then to the right. The force shown is upwards and to the left. Therefore, A is our answer.

Hope this helped!
Good Luck for all your exams!
Click to expand...
Were thins question though ? -_-
 
Thought blocker

Thought blocker

Messages
8,477
Reaction score
34,837
Points
698
ShreeyaBeatz said:
Which statement about electrical resistivity is correct?
A The resistivity of a material is numerically equal to the resistance in ohms of a cube of that
material, the cube being of side length one metre and the resistance being measured
between opposite faces.
B The resistivity of a material is numerically equal to the resistance in ohms of a one metre
length of wire of that material, the area of cross-section of the wire being one square
millimetre and the resistance being measured between the ends of the wire.
C The resistivity of a material is proportional to the cross-sectional area of the sample of the
material used in the measurement.
D The resistivity of a material is proportional to the length of the sample of the material used in
the measurement.


WHY IS A right but not C??
Click to expand...
Ugh! Confused -_-
 
ShreeyaBeatz

ShreeyaBeatz

Messages
84
Reaction score
178
Points
43
sagar65265 said:
Q12)

View attachment 44428

Q28)

Wow, this is a toughie!

Suppose you have two sources of light, beside each other and projecting light onto a screen equidistant from them. This is an alternative set-up to a double slit experiment, where two slits are small enough to act as light sources on their own, and transfer light from a single source onto a screen in from of them.

Every second, thousands of wavelengths of light pour out through the light sources (in both experiments), all the time interfering with other light waves, combining constructively and destructively, traveling through the medium around them, and finally striking the screen, millions at a time. How can a steady diffraction fringe be formed? How is the image so steady?

The answer is because the situation outside the screen remains the same all the time. Every second, the same number of photons are found between the slits and the screen (those that enter through the source replace those that collide into the screen), every second the same number of waves interact (the wavelength of the light concerned has to be similar; simultaneously, the frequency has to be similar and the color has to be similar. Red light cannot from a fringe with blue light, for example) and every second the situation is exactly the same as it is the previous second. How?

This is because the sources are behave in exactly the same manner. The sources both have the same intensity, they produce the same number of wavelengths per second, the waves they produce are in phase, and only because of that do all those waves interact the same way, all the time. In other words, the steady fringe is formed only because the sources are coherent. Therefore, we can rule out B.

Suppose we polarize the light. That still doesn't make any difference.

If one source has light vibrating in the vertical plane (just imagine that) and the other source is polarized "at right angles to light from the other source", i.e. polarized in the horizontal plane, that simply means that one source will have horizontally oriented vibrations and the other will have vertically oriented vibrations.

No matter how we put them together, it is impossible to get zero resultant intensity from that! E.g. it's like trying to add two perpendicular vectors (say the sides of the triangle) and getting a zero resultant!

It just won't happen. Sadly, though, for a fringe to be created, there are places where zero intensity of light has to be found, and places where double intensity of source light has to be found. Therefore, since polarizing the light as they have mentioned it will not create points of zero intensity, we won't be able to create a fringe like that. Therefore, C is also wrong.

Lastly, let's take D: suppose the light from the two sources do not even overlap, they cannot possibly interfere with each other! And if they can't interfere, they can't form points of zero intensity and maximum intensity, and so they can never form a fringe at all - this means that D is also not the answer, and the only remainder is A, our final answer.

Q31)

At any point in the electric field, the force on a charged particle will be along the tangent to the field line. Therefore, the force on any particle Q at a point P in some electric field can never act at an angle (which is not 0 or 180 degrees) to the tangent of the field lines at point P.

To put an example, in this question, the force at A is along the tangent to the field line at that point - suppose you draw the tangent to the field lines at that point, the force they show at A will probably be along that tangent. So A might be the answer.

At B, suppose you draw a tangent there, the force shown will again be mostly along the field lines. So B might be the answer.

At C, we have a problem. If you draw a tangent at C, then the force shown will definitely not be along that tangent. So C is not the answer.

At D, the force shown might lie along the tangent. So D might be the answer.

The last thing we need to know is that:

".....at any point on an electric field line, the arrow on the field line will point in the direction of the force on a positive charge, placed at that point on the field line. The force on a negative charge placed at the point will be in the opposite direction."

So, positive charge = force along the field line direction. Negative charge = force opposite the field line direction.
Since we are talking about a negative charge here, we are therefore looking for a force that points in the opposite direction of the field line arrow.

At D, the field line arrow points to the right. The force also points in this direction, so this option is wrong.

At B (we have eliminated C) the field line points upward. The force is also pointing in this direction, so B is wrong.

Lastly, at A, the field line moves downwards, and then to the right. The force shown is upwards and to the left. Therefore, A is our answer.

Hope this helped!
Good Luck for all your exams!
Click to expand...

Sagar could you like help a bit on my post? Thank you :)
 
usama321

usama321

Messages
1,258
Reaction score
1,397
Points
173
ShreeyaBeatz said:
Which statement about electrical resistivity is correct?
A The resistivity of a material is numerically equal to the resistance in ohms of a cube of that
material, the cube being of side length one metre and the resistance being measured
between opposite faces.
B The resistivity of a material is numerically equal to the resistance in ohms of a one metre
length of wire of that material, the area of cross-section of the wire being one square
millimetre and the resistance being measured between the ends of the wire.
C The resistivity of a material is proportional to the cross-sectional area of the sample of the
material used in the measurement.
D The resistivity of a material is proportional to the length of the sample of the material used in
the measurement.


WHY IS A right but not C??
Click to expand...

Resistivity of a material does not change with length or cross sectional area. Each material has its own resistivity constant, so option C and D are out of the question.
from what i can deduce, It can not be B because there is inconsistency in the units ie one square millimeter. However, I am not sure about what it has to do with the opposite sides of a cube or a wire.
 
You must log in or register to reply here.
Top