AS physics MCQs help

leosco1995

I need help with this one. I don't understand why the trend is decreasing because V = IR, and R = ρL/A. So if ρ is increasing, then shouldn't R also increase and hence V also increase? I'm confused.

(edited 11 years ago)

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Reply 1

SalBhatti

Original post by leosco1995

R = \rho \frac{l}{A}

and

V = IR

V = I \rho \frac{l}{A}

And as we're considering the distance x, I think we can take it as length of wire, so gradient :-

\frac{V}{l} = \frac{I\rho}{A}

As the resistivity increases and the cross-sectional area and current remain constant, the gradient also increases. The answer should be B.

Do you get why the gradient is negative?

(edited 11 years ago)

Reply 2

leosco1995

Oh, I see what you did there. But I still don't get why the gradient is negative.. maybe I'm missing something obvious..

Reply 3

Stonebridge

Original post by leosco1995

Oh, I see what you did there. But I still don't get why the gradient is negative.. maybe I'm missing something obvious..

Potential drop.
The potential drops as you move along the wire. The drop is biggest for the wire with the biggest resistance. (V=IR and same current in all 3)
The wire with the biggest resistance is the one with the biggest resistivity. (Same length and cross section)

Reply 4

SalBhatti

Original post by leosco1995

Oh, I see what you did there. But I still don't get why the gradient is negative.. maybe I'm missing something obvious..

Stonebridge has already answered it qualitatively, but if you want a mathematical proof, here you go:
the voltage at a particular point on the wire is

V_l

. This will be the voltage at the start of the wire

V_0

less the voltage lost by resistance

V

V_l=V_0-V

the voltage lost is given by

V=IR

byt we know that

R=\rho\dfrac{l}{A}

so we can write

V_l=V_0-I\rho\dfrac{l}{A}

since

V_0

A

and

I

are constant, we now have

V_l \propto -\rho l

so plotting

V_l

against

l

gives a straight line of gradient

-\rho

Reply 5

leosco1995

Oh, I see. I never took into consideration that the voltage drops across the wire. Thanks for the explanation guys.

Reply 6

leosco1995

I hate these type of questions.. I don't know where to start and I invariably get them wrong. I just made an equation:

Fnet = ma
x - 6 = 8 * a

Where x is the forward force. But I think I'm wrong and don't know what else to do.

Reply 7

SalBhatti

Original post by leosco1995

This might be helpful:

Reply 8

leosco1995

That was really helpful indeed, thanks. :smile:

Reply 9

leosco1995

I don't know how to solve this.. I know that tension is like when you pull something and compression is when you push it.. but I'm still confused because for me it's hard to tell from just looking at a point on the diagram.

I thought A was the answer, BTW.

(edited 11 years ago)

Reply 10

Stonebridge

It's common sense really.
If you pull down on that weight, which parts will be stretched (tension) and which will be compressed.
The key is to remember that if you try to bend a bar, the outside edge stretches and the inside edge is compressed.
If you pull down on that weight, which way would you expect the two bars to bend?

Reply 11

leosco1995

Original post by Stonebridge

Is that always the case? I mean, when you bend something the outside edge will stretch and inside edge compress? It is kind of hard for me to imagine it on a diagram.

If you don't mind, could you answer this question too? For this one, I've eliminated options B and C (B because it isn't a stationary wave and C because there is some displacement at point R), but I can't tell if it's A or D. How can you figure out the velocity and acceleration from a displacement/distance graph?

Once again, thanks a lot for all of your help.

(edited 11 years ago)

Reply 12

Stonebridge

The points on the string are performing SHM.
You should know that for SHM (think of a pendulum) the mass has maximum velocity at the centre and zero velocity at its greatest distance from the centre.
It's acceleration is greatest when furthest from the centre.
Look back at your SHM notes for this.

Reply 13

SalBhatti

Original post by leosco1995

Is that always the case? I mean, when you bend something the outside edge will stretch and inside edge compress? It is kind of hard for me to imagine it on a diagram.

Do you have CIE Physics MCQs exam in coming days? I just miss doing all these questions. . . :cry2:

Reply 14

leosco1995

Original post by Stonebridge

This is interesting.. although I don't think we did SHM in class. I actually thought it was in the A2 syllabus but after going through the AS syllabus I guess it is there. I'll look up some more stuff on it. Thanks for the help.

Reply 15

leosco1995

Original post by Zishi

Do you have CIE Physics MCQs exam in coming days? I just miss doing all these questions. . . :cry2:

Yeah, it's on 14th June. :P And haha, I can sort of relate to that, I miss doing the MCQs I used to do in my O-levels. I'll probably miss these too.

Reply 16

leosco1995

I hope you guys don't mind me posting all these questions..

I don't understand this one properly. If half of the water gets transferred to the other vessel, wouldn't only the height be halved? The total energy then lost by the water would just be mg * (h/2)? But the answer is mgh/4.

Reply 17

Stonebridge

Original post by leosco1995

Is that always the case? I mean, when you bend something the outside edge will stretch and inside edge compress? It is kind of hard for me to imagine it on a diagram.

Reply 18

Stonebridge

Original post by leosco1995

I hope you guys don't mind me posting all these questions..

Think where the centre of gravity is for the water in the LHS before opening the tap. The PE of the water in the LHS depends on the height of the c of g. Not the total height of the water.
Think where the c of g is after opening, with half the water now there.
How much PE was lost?
Think where the c of g of the water now is in the RHS
How much PE was gained?

(edited 11 years ago)

Reply 19

leosco1995

Hmm, so if I'm understanding you correctly,

Column X went from mgh to mgh/2.
Column Y went from 0 to mgh/4.

Total loss in P.E = loss in column X + gain in column Y
= -mgh/2 + mgh/4?

I never realized the P.E changes because of the c.o.g and not the height itself. :redface: