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Physics: Post your doubts here!

Thought blocker

Thought blocker

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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

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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

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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.
 
The Godfather

The Godfather

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

Q2. Q9. and Q12.
Click to expand...
for 2 :
3000 revolutions per minute means 50 revolutions in one second. And therefor this the frequency.
Frequency = 50Hz
Now, using the frequency, we find the time taken for one revolution -----> T = 1/f = 1/50 -----> 0.02 seconds.
Time is o.o2 seconds and so, in other words, 20 milliseconds (0.02 x 1000).
So you see, out of all the options given, 10 ms cm^-1 is the closed. So, the answer is B.
If you use 1 s cm^-1 as time base, there would be too many oscillations on the screen and wouldn't give you a good display.
Similarly, 100 microseconds or 1 microsecond cm^-1 will give you a very extended display, you wouldn't be able to see proper oscillations.
for 9 :
Use the momentum equation m1v1+m2v2=m1u1+m2u2
The initial velocity is 0 while the final direction of the final velocities is opposite to each other as the masses move away so we will use a negative sign.
M1V1-M2V2=0
M1V1=M2V2
V1/V2=M2/V2
for 12 :
use the formula v^2 = u^2 + 2as
where
positive direction = direction of train
u = original velocity (speed of train)
v = final velocity (zero)
a = acceleration of train (in this case, acceleration is negative)
s = distance (from point where velocity = u to where velocity = v)

The problem uses x for distance, not s, so the equation we'll use is v^2 = u^2 + 2ax

because v = 0, we can write
u^2 + 2ax = 0, rearranging as:
x = -(u^2) / 2a . . . . . . . . Note - a has a negative value, thus making x positive
x = Ku^2 where K = -1 / (2a)
ie. x varies as the square of u
so if u increases by 20% (ie changes by a factor of 1.2), then x will change by a factor of (1.2)^2
which is a factor of 1.44
thus the minimum distance between yellow and red must now be 1.44x
Source:
 
S

sagar65265

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ShreeyaBeatz said:
Sagar could you like help a bit on my post? Thank you :)
Click to expand...

Sorry for the late reply, wasn't online for quite some time.

Let's take this option by option, with the following formula in focus:

ρ = RA/l

Where ρ = resistivity of the material through which current is being passed,
R = resistance of the sample through which current is being passed,
A = Area of the cross section of the sample through which current is being passed (perpendicular to the direction of current), and
l = length of cross section of sample through which current flows (i.e. the distance through which the current flows in the sample used).

A neat way of finding out the area A is by imagining the way the current passes through the sample (say left to right) and imagining using a large blade/ knife to cut it off in between (grotesque, but bear with me for this bit. Alternatively, you can put up a plane perpendicular to the direction of the current and use that). The blade is perpendicular to the current, and the area that the sample occupies on the face of the blade is your area.

So, for option A, let's see what the formula tells us:

If the cube has a one meter side and the current is passed from one face to another, we can say that the current travels 1 meter from one end of the cube to the other; therefore, l = 1 meter. Alternatively, the current first enters the cube on one side, travels 1 meter, and exits the cube, giving us the same value.

Further, the cross section through which the current travels, perpendicular to the direction of current - suppose you place the cube on the table, and cut it parallel to the edges, the area you get after the cut is an area of 1 meter x 1 meter = 1 m^2.
Another way of getting this is to imagine there is no current passing through the cube. You close the circuit, and current starts flowing. Soon, it gets to the beginning of the cube, and starts moving through it. Suppose no charge carrier (electron/proton, either is fine) travels faster than another, you have a "wall" of such charge carriers advancing along the cube. Now ask yourself. How large is that wall? What is the area of that wall? In this case, the area of that wall is 1 m^2, which is our result.

Putting these in the equation, we get

ρ = R * 1 m^2/1 m = R
So ρ = R
Therefore, A is right - the value of ρ is numerically equal to the value of the resistance of the cube in this situation. But for the others, what can we say?

Option B: Doing the maths here again, we can say that the length through which the current passes is l = 1 meter and the cross sectional area through which the current passes is 1 mm^2 = (1/1000 meters)^2 = (10^-3)^2 = 10^-6.
Putting it in the equation,

ρ = R * 10^-6/1 = 10^-6 * R
Which is not right.

Option C: This has already been answered before by usama321 and Asad rehman - the option says that resistivity is dependent on the cross sectional area of the sample used. Even though the formula says that ρ = R/l * A, the wording of the option is something very precise; when it says proportional, what it means is

"...a change in A results in a change in ρ, such that the change in A can be equated to the corresponding change in ρ if a suitable constant k is introduced as a factor of change (i.e. ΔA = k * Δρ)"

This is, of course, false. At a fixed temperature, given no other external electric or magnetic interference (literally and figuratively), the resistivity of a material will not change with any change in area. Instead, the resistance will change to ensure that the value of ρ remains the same. Since resistivity does not depend on the dimensions of a sample and only on the innate nature of the material, C cannot be right.

Changing the cross-sectional area of a sample may change the resistance, but it cannot change the resistivity.

Option D: The same argument as option C can be supplied here, replacing Cross Sectional Area A with length l.

Hope this helped!

Good Luck for all your exams!
(BTW, Thought blocker, thanks for the symbols, bro! Credit to you!)
 
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