As Physics MCQs Help

While solving some past papers, i faced these problems which i couldn't solve. Any help would be greatly appreciated.
First Question:

If we stretch a wire to half its diameter, what would happen to its length? would it increase by 4 times?
Second question:

Third:

(edited 10 years ago)

Reply 1

hello calum

Original post by phenomenon.ahmad

While solving some past papers, i faced these problems which i couldn't solve. Any help would be greatly appreciated.

(I) calculate the volume of the unstretched wire. This should be equal to the volume of the stretched wire. So you can now find the length. V=A*L

(II) It must be (c) because if you resolve the forces horizontally, there is no way it can be in equilibrium because the is a larger component to the left. If you resolve forces on all the other diagrams then you will see that they can all be in equilibrium.
[To be in equilibrium you must be able to make a complete triangle with the forces]

(III) At 18 seconds they are in anti-phase (1/2 a cycle out). So at (1/8) of a cycle out of phase the time much be 18(1/4) which is 4.5 seconds. So the answer is (B).
[they]

(edited 10 years ago)

Reply 2

b.k

the answer to first MCQ is D. The volume of wire will remain same. Since the diameter is halved, for volume to be the same the new length will four times the initial. Using the formula Resistance=length/area (since resistivity does not change with dimensions we can ignore it in this calculation) find out the original resistance in terms of length and diametre.

Original will turn out to be 4lenght/diametre.

The new resistance will be 64length/diametre.

Since the new resistance is 16 times the original,
the answer will be
0.2*16=3.2.

Reply 3

phenomenon.ahmad

Thanks for the replies. Answers according to marks scheme:
1) D
2) A
3) B

Original post by hello calum

(II) It must be (c) because if you resolve the forces horizontally, there is no way it can be in equilibrium because the is a larger component to the left. If you resolve forces on all the other diagrams then you will see that they can all be in equilibrium.
[To be in equilibrium you must be able to make a complete triangle with the forces]

Ok i understand q1 and q3 now but what about q2. The question seems ridiculous to me. All of them can or cannot be in equilibrium depending on magnitudes of forces... any thought? :confused:

Reply 4

Bones:

Original post by phenomenon.ahmad

Ok i understand q1 and q3 now but what about q2. The question seems ridiculous to me. All of them can or cannot be in equilibrium depending on magnitudes of forces... any thought? :confused:

I asked this question here before: http://www.thestudentroom.co.uk/showthread.php?t=2374997.

Reply 5

Stonebridge

Original post by Bones:

I asked this question here before: http://www.thestudentroom.co.uk/showthread.php?t=2374997.

This forces question is flawed on a number of levels.
It should never have got passed the initial "shredding" stage.
The question talks about a "body", not a system.
Any system of three forces on a body could be in equilibrium.
All diagrams have 3 forces.
It's not clear from the question if the lengths of the "forces" shown represent the vectors and have a length proportional to the magnitude, or if they represent physical strings, as is obviously the case in diagram D, where the length is clearly not proportional.
Diagram A
If the two upper forces are acting on the object then is the third force also doing so, as it is also shown acting on the surface of the object. The question says the forces shown are acting on the bodies. It actually looks like it represents the object's weight.
If it is meant to be the force (a possibility) of the object on the slope (which I doubt), then the two other forces, which are on the object, cannot be in equilibrium. So what are we to believe here? See also my comment below regarding A*.
B and C are fairly conventional 3-force diagrams which could both be in equilibrium.
Diagram D shows a system of forces that seems to imply strings attached to the object and to each other.
This system could be in equilibrium.
The diagram only shows one force acting directly on the object in a downwards direction.
There would naturally be another force of tension in the upper part of the vertical string acting upwards on the lamp. These two forces would produce the equilibrium in the forces on the body. But only one is shown??
So are we to assume there is a force there even though it isn't shown? The other two forces are acting on the lower string, not the body directly.

*Or are we to assume that in diagram A the forces do not all act through the (implied) centre of mass of the object, and would therefore not produce rotational equilibrium? I must say on second thoughts this could be what they mean, but it still leaves the question of what to make of D.

I must say I find this question poorly presented.
But maybe by dissecting it in this way you can learn the important physics and put this one down to poor question construction.

(edited 10 years ago)

Reply 6

Bones:

Original post by Stonebridge

*Or are we to assume that in diagram A the forces do not all act through the (implied) centre of mass of the object, and would therefore not produce rotational equilibrium? I must say on second thoughts this could be what they mean, but it still leaves the question of what to make of D.

I must say I find this question poorly presented.
But maybe by dissecting it in this way you can learn the important physics and put this one down to poor question construction.

That is, in fact, what they meant. You can see it in the examiner's report here: http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Physics%20(9702)/9702_w12_er.pdf.
'Three forces can only be in equilibrium if they all pass through the same point, otherwise there will be a turning force on the object. The correct answer A perhaps looked too familiar to candidates so many chose B or C.'