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Physics: Post your doubts here!

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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