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Physics: Post your doubts here!

Catherine7532

Catherine7532

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25 A wire stretches 8 mm under a load of 60 N.
A second wire of the same material, with half the diameter and a quarter of the original length of
the first wire, is stretched by the same load.
Assuming that Hooke’s law is obeyed, what is the extension of this wire?
A 1 mm B 4 mm C 8 mm D 16 mm
 
Venkata Gurrala

Venkata Gurrala

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Catherine7532 said:
How is the answer D?

5 A quantity x is to be determined from the equation
x = P – Q.
P is measured as 1.27 ± 0.02 m.
Q is measured as 0.83 ± 0.01 m.
What is the percentage uncertainty in x to one significant figure?
A 0.4 % B 2 % C 3 % D 7 %
Click to expand...
well, when you subtract 1.27 and 0.83, you get 0.44. However, you always add the uncertainties. you never subtract them just because the values are being subtracted. So when you add 0.02 and 0.01 and get 0.03, divide 0.03 by 0.44 *100 and you should get around 7%
 
Zepudee

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Thought blocker said:
I already asked sagar :p
36)
Any two conductors connected to each other by another conductor will have the same potential.

This is like having two separate tubes of water (each one represents one conductor), with different amounts and different heights in each (the height represents the potential, the amount represents the charge) - if they are connected by a tube or any other medium that allows free flow of water (this medium represents the conductor) then the heights will become equal (the potential will become equal).

Note that this does not mean "potential" flows from one to another - just like "height" does not flow, potential won't. The charges on an object make up it's electric potential, and these charges will flow (since all materials involved are conductors).
In the analogy, water (the charges) flows to equalize the levels.
In the end, the volume of water in each tube may be different (i.e. the charge on each conductor may be different) but the height (potential) will be the same.

However, this potential only becomes equal in a static situation.

So we can say the potential at point X is +24 Volts because it is connected to the positive terminal of the source, which is at a potential of +24 Volts itself.
Suppose current is flowing through the motor, energy is lost in the motor until the potential of the current becomes equal to the potential of the negative terminal. So the potential at Y is 0 Volts.

The problem arises when either wire is cut - suppose you cut the positive cable somewhere along it's length. What is the potential of X then? It is not connected to the 24 Volt terminal, so what else could the potential there be?

The answer is that the potential of X is 0 Volts, since it is connected to the negative terminal of the battery, which we have assigned to have a zero potential. When the wire is cut no current can flow through the circuit, so the resistance inside the motor stops acting like a resistance and acts like a normal conductor, which means the point X will be at 0 Volts.

Suppose we cut the negative cable. Following the same logic, since the motor no longer does anything (no current because circuit is incomplete), it simply acts as a conductor and equalizes the potential at X and Y, so that both have a potential of +24 Volts.

Suppose the connection in the motor breaks, again no current will flow through the circuit, and X will be at a potential of +24 Volts (since it is still connected by an unbroken connector to the positive terminal of the source) with Y remaining at 0 Volts (since it is connected to the negative terminal of the source).

So having seen all this, we can say that D is the only right answer among all the options.
Let me know if you have any doubts, since it was a little difficult to explain this one, and the concept is not so straightforward anyways.

38)
This is a neat little question, with the only law needed being Kirchoff's Second Law.

There is a very, very quick way of solving it, but first the long way:

i) Suppose you start at the bottom right corner of the circuit.
ii) You go left, across the battery, and find an increase in potential of 20 Volts.
iii) Okay, now you go up (no resistance), go right till the junction (again, no resistance) and take the upper branch.
iv) As you pass the resistor L, you see a drop of 7 Volts.
v) You continue across M where there is some unknown change in potential and then
vi) return to the bottom right corner.

So, the total change in potential should be zero. Therefore,

+20 Volts + (-7)Volts + (Change in potential across M) = 0
13 + (change in potential across M) = 0
Therefore, the change in potential across M = -13 Volts. The change is -13 Volts, the drop is 13 Volts. So B or C.

Let's narrow it down using Q.

i) Again, start at the bottom right corner of the circuit.
ii) Go left, cross the battery, and see an increase in 20 Volts.
iii) Continue until you reach the junction, and pick the upper branch.
iv) Cross P, seeing a drop of 7 Volts, and take the bridge between the branches.
v) Cross N, seeing a drop of 4 Volts, and go towards Q, i.e. to the right on the diagram..
vi) There is some unknown change in potential across Q, and
vii)you then return to the bottom right corner.

Again, the total change in potential should be zero. So,

+20 Volts + (-7) Volts + (-4) Volts + (Change in potential across Q) = 0
9 Volts + (Change in potential across Q) = 0
Change in potential across Q = -9 Volts. Therefore drop = 9 Volts, so the only option that has that is C.

The thing is, that you could have gotten the answer directly without the first part if you had just gone past Q - you would get 9 Volts, and the only option with a 9 Volts drop for Q would be C!

But be careful here - all these changes are so simple because we are going in the direction of current. Suppose you go across a resistor with this method and you are going in the direction opposite to the current, then you will have anincrease in potential, not a decrease.

This is because current flows from a region of high potential to low potential - suppose you are going along the current, you are also going from the region of high potential to the region of lower potential. But if you go in the direction opposite to the current you are going from a region of low potential to a region of high potential, which is an increase in potential - you have to make sure you adjust the signs correctly.
Click to expand...

But be careful here - all these changes are so simple because we are going in the direction of current. Suppose you go across a resistor with this method and you are going in the direction opposite to the current, then you will have an increase in potential, not a decrease.
About this, can you give me an example? :)
sagar65265 ZaqZainab
 
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..sacrifice4Revenge..

..sacrifice4Revenge..

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Catherine7532 said:
25 A wire stretches 8 mm under a load of 60 N.
A second wire of the same material, with half the diameter and a quarter of the original length of
the first wire, is stretched by the same load.
Assuming that Hooke’s law is obeyed, what is the extension of this wire?
A 1 mm B 4 mm C 8 mm D 16 mm
Click to expand...
Second wire would have same youngs modulus ad force
Now extension= force x length/area
=1/4length / area over 4
Than gives us simple length over area whichbis 1
Extension wld be same, i.e 8
 
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Zepudee

Zepudee

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Find the total resistance, so it is equal to 10/17
V=IR
V=5 x 10/17
=3
Then, the resistance is 2 ohms, so to find the current
is 3 / 2 = 1.5
 
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sagar65265

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Venkata Gurrala said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_11.pdf
Need help with Q9,14,15,19,26, and 28. Its pretty frustrating.
Click to expand...

About your earlier query, I have no idea what you said "yeah" for, so here's an answer anyways ;).
Either ways, the book shows the graphs for different materials, and since different materials have different behavioral patterns under tension and compression, the graphs for different materials vary. So, the graph for an elastic band cannot be expected to conform to those patterns.

However, experience and assumption says that as you stretch a rubber band, it starts becoming more and more difficult to extend - in other words, after some time, a small extension will require a large force. In other words, after some time, the gradient of the graph (gradient = [change in applied force]/[change in extension]) increases.
The only graph that shows this kind of pattern (where after some extension the gradient of this graph increases) is A.

Q9)

First concept here - both energy and momentum are conserved in an elastic collision. So the kinetic energy and momentum and the colliding bodies before the collision are the same as the kinetic energy and momentum after the collision.

One more important characteristic of an elastic collision - the relative velocity between the interacting bodies is the same before the interaction as after. In other words, the rate of change of the distance between them is the same.

So, let's take D first, since it seems to be an outlier - it is an odd one out here, since it is the only one where both colliding objects move off together, so this is one we can easily eliminate as follows:

Initially, the ball moving right has kinetic energy 1/2 * (2m) * u² = mu².
The ball moving to the left has kinetic energy 1/2 * (m) * u² = 0.5 * mu².
Adding these up, the initial kinetic energy is equal to mu² + 0.5 * mu² = 1.5 * mu².

Now consider the situation after the collision. The mass of the moving object is 2m + m (since the velocities are the same and the objects stick together - you can take each separately and use the same speed to calculate their K.E., but it gives the same answer) = 3m. The option shows us a final velocity of (u/3).

So, the final K.E. = 1/2 * (3m) * (u/3)² = 1/2 * 3m * u²/9 = mu²/6.
This is not the value of the initial Kinetic Energy, so we can assume energy is lost somewhere or the other - so, this is not an elastic collision.

Let's do C next. The Initial Kinetic Energy, as calculated above, is 1.5 * mu².

In the final situation, the ball of mass (2m) has a velocity of (u/6), so it's K.E. = 0.5 * 2m * (u/6)² = m * u²/36 = mu²/36.
Also, the ball of mass (m) has a velocity of (2u/3), so it's K.E. = 0.5 * m * (2u/3)² = 0.5m * (4u²/9) = 2mu²/9 = 8mu²/36.
Adding these up, we get mu²/36 + 8mu²/36 = 9mu²/36 = mu²/4.

This is, again, not the same as before. So C is also out.

B is next - you can see immediately that this option is wrong, since it has the same speeds and masses as C - the (2m) mass is moving at (u/6) and the (m) mass is moving at (2u/3), which is the same combination we proved wrong before. Therefore, the only remaining option and our final answer is A.

Q14) When the projectile reaches it's highest point, it has a purely horizontal velocity and no vertical component of velocity.
Additionally, since there is no force on the projectile in the horizontal direction, it has the same horizontal velocity at it's highest point as it did when it was shot.

A little more on this - suppose the initial velocity has a magnitude of "v", the initial kinetic energy E = 1/2 * m * v² = 0.5mv².

Now, onto the focus of the question: since the initial velocity points at an angle 45° to the horizontal, the component of velocity in the horizontal direction should be

v(horizontal) = v * cos(45) = v/(√2)

This is, by our discussion above, the velocity of the projectile at it's maximum height. So, the kinetic energy at it's greatest height is

E(f) = (1/2) * m * (v/(√2))² = 0.5m * v²/2 = 0.25mv².

Let's divide E(f) by E, giving us

E(f)/E = (0.25mv²)/(0.5mv²) = 1/2
Therefore, E(f) = (1/2)E = 0.5E = A.

I'll post the rest after a while, let me know if you have any doubts.
 
U

unique111

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Venkata Gurrala said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_11.pdf
Need help with Q9,14,15,19,26, and 28. Its pretty frustrating.
Click to expand...
19) use the formula P=hdg where g=9.81 ... D comes out to be 1936.8=1900.
26) use dsin(theta)=n lambda. Where d=1/N and N=300,000 lines/m. Solve the equation 1=nlambda/d and n=7. This is the number of order, total maxima is double this plus 1=15 ( there is one more maxima in the zero order).
28) f=mg and f=Eq equate these two equations and charge/mass will be g/E. Don't know its polarity though.

* how did you do q.13? I got the torque, but what about the tension?
 
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sagar65265

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Zepudee said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_12.pdf

Q5 (D) WHY?
Click to expand...

upload_2014-6-11_13-27-16.png

Since the pattern displayed as λ above repeats itself, that length is equivalent to the time period of the wave (since the horizontal direction is the time-base axis).
This is less that 1 centimeter, and actually measuring the length on the paper would give you an answer somewhere close to 0.6 cm, 0.7 cm or 0.8 cm - whichever one it is, we can use that for the calculation.

If 1 cm represents 10 ms = 10/1000 = 1/100 seconds, the 0.6, 0.7 or 0.8 centimeters represent 0.6/100, 0.7/100 or 0.8/100 seconds respectively.
So, if the time period of the wave is one of the above durations, then the frequency must be equal to 1/T, which turns out to be:

(100/0.6) = 166.6 Hertz, if the time period is 0.6/100 seconds.
(100/0.7) = 142.8 Hertz, if the time period is 0.7/100 seconds.
(100/0.8) = 125.0 Hertz, if the time period is 0.8/100 seconds.

So, whatever the answer time period is, the closest option appears to be 140 Hz, which is equal to B, our final answer.

Hope this helped!
Good Luck for all your exams!
 
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