B Is this mark scheme correct, poorly worded, or wrong?

[Elbow_Patches]

Hello there,

I'd like some help with this question from a sample A level Physics paper (please see the question and mark scheme attached). The issue is with 16 c). If the zero error means that the voltages used are less than they should have been, that makes the resistances calculated less than they should be, so the gradient is less than it should be, so the calculated resistivity is less than it should be. So that means the 'actual' voltages should be more, so the actual resistances are more, so the gradient is more, so the actual resistivity is more. The mark scheme makes two rather ambiguous statements followed by a third that clarifies what they mean "Hence resistivity of the metal will be smaller than the value in (b)"

So I say the direct opposite of what the mark scheme says. Please could anyone here give an opinion? Am I wrong and have just misread the question? Is the mark scheme wrong? Or, as several people in my department hold, I've got the Physics right but am reading the mark scheme wrong or that the mark scheme is poorly worded and ambiguous. I don't think that third statement is ambiguous, I think it makes a very definite statement.

Thanks for any help! :)

Elbow

 

[mjc123]

I think you are right and the mark scheme is wrong. If you compare the wording with the question, "actual" means the true value as distinct from the measured value, and the former should be greater than the latter if the voltmeter is reading low.

 

[Elbow_Patches]

I think you are right and the mark scheme is wrong. If you compare the wording with the question, "actual" means the true value as distinct from the measured value, and the former should be greater than the latter if the voltmeter is reading low.

Thanks for the reply! Good to know I'm not going insane. Does anyone else out there agree/disagree?

Elbow

 

[DLeuPel]

Hello there,

I'd like some help with this question from a sample A level Physics paper (please see the question and mark scheme attached). The issue is with 16 c). If the zero error means that the voltages used are less than they should have been, that makes the resistances calculated less than they should be, so the gradient is less than it should be, so the calculated resistivity is less than it should be. So that means the 'actual' voltages should be more, so the actual resistances are more, so the gradient is more, so the actual resistivity is more. The mark scheme makes two rather ambiguous statements followed by a third that clarifies what they mean "Hence resistivity of the metal will be smaller than the value in (b)"

So I say the direct opposite of what the mark scheme says. Please could anyone here give an opinion? Am I wrong and have just misread the question? Is the mark scheme wrong? Or, as several people in my department hold, I've got the Physics right but am reading the mark scheme wrong or that the mark scheme is poorly worded and ambiguous. I don't think that third statement is ambiguous, I think it makes a very definite statement.

Thanks for any help! :)

Elbow

I imagine that the question refers to the effects of temperature in the value of the resistance, affecting therefore the potential difference. When temperature increases resistance increases because more particles collide with the flow of electrons. All of this taking into account that intensity doesn’t vary. I’m not sure how increasing the resistance increases temperature, even after searching a little I haven’t found anything yet. Yet resistance does vary with temperature so maybe the exercise wanted you to respond that.

 

[sophiecentaur]

Hi and welcome to PF.
As far as I can see, the mark scheme seems to be wrong. tut tut!
@mjc123 :The wording is really not very helpful for students and that could throw some of them. "A zero error" could be read as "zero error". The question is quite a hard one for students who are not doing that sort of thing every day. They could waste a lot of time over two alternative answers.
The straight line part of the graph should pass through the origin. The straight portion of the line, produced, will pass below the origin so each calculated R value must be low - meaning that the actual value would be higher than calculated. If the measured R is lower than it should be, the measured V must be lower.
The gradient would be higher with the actual values because it goes from a higher value, for a given length to the origin.

 

[sophiecentaur]

I imagine that the question refers to the effects of temperature in the value of the resistance,

Yes. The information is there but it doesn't seem to be require a verbal answer about it - except that the student would be required to ignore the lost two (hot) points on the graph. "Null points" (blame Eurovision Song Contest for the pun) for doing a best fit line through all the points

 

[Tom.G]

that makes the resistances calculated less than they should be, so the gradient is less than it should be, so the calculated resistivity is less than it should be. So that means the 'actual' voltages should be more, so the actual resistances are more, so the gradient is more, so the actual resistivity is more.

I think you are right and the mark scheme is wrong. If you compare the wording with the question, "actual" means the true value as distinct from the measured value, and the former should be greater than the latter if the voltmeter is reading low.

What you said... and what he said... (and whoever did that mark scheme needs to find a competent proof reader!)

Cheers,
Tom

p.s. Good catch. Keep up the good work.

 

[sophiecentaur]

Not just a competent proof reader. You need someone competent to be very aware of the Science / Maths for every question. If it was intended as a ‘trick’ question then the appropriate way of thinking should be a central part of the course. I can’t say I ever found that in A Level Specifications.
There’s a choice here between filling the course with sexy new stuff (Quarks etc.) or developing thinking skills. Bums on seats always wins out with choice of course content.

 

[mjc123]

@sophiecentaur, I don't quite get what you're saying. The straight-line portion of the graph seems to me to pass through the origin pretty closely. If there is a constant error in the voltage, this will translate into a proportional error in the calculated R, i.e. a change in slope in the graph, but not a change of intercept. I agree that both the question and the answer could be better expressed.
@DLeuPel, the temperature has nothing to do with this part of the question. It is mentioned in part b to explain why the lowest point is above the line, and a clued-up student should leave it out, but the only error in part c is due to the voltmeter.

 

[sophiecentaur]

The straight-line portion of the graph seems to me to pass through the origin pretty closely.

Not on my screen. It's aiming to intercept the x-axis at around 0.02. The 0.2 value is actually deviating from a straight line afaics.
With a zero error, there will be 'impossible' values, calculated for the resistance, where there is zero current and a finite voltage, so the graph need not pass through the origin.
Yes - once the bottom point has been ignored, the temperature effect has been dealt with.

 

[hutchphd]

Not on my screen. It's aiming to intercept the x-axis at around 0.02. The 0.2 value is actually deviating from a straight line afaics.
With a zero error, there will be 'impossible' values, calculated for the resistance, where there is zero current and a finite voltage, so the graph need not pass through the origin.
Yes - once the bottom point has been ignored, the temperature effect has been dealt with.

The voltage is not being changed during the experiment. The zero offset provides a fixed percentage error.

 

[sophiecentaur]

The voltage is not being changed during the experiment. The zero offset provides a fixed percentage error.

Let me think... Ah yes - the 4V will always be 4+ΔV so the Resistance will measure as (4+ΔV)/I. That's inversely proportional to the Current. So there wouldn't be an offset - right, thanks just a difference in slope of the graph. There will be an Offset from the origin on a V/I graph - but that's not the picture!

 

[hutchphd]

Let me think... Ah yes - the 4V will always be 4+ΔV so the Resistance will measure as (4+ΔV)/I. That's inversely proportional to the Current. So there wouldn't be an offset - right, thanks just a difference in slope of the graph. There will be an Offset from the origin on a V/I graph - but that's not the picture!

I very nearly went down that road myself..

 

[Elbow_Patches]

Many thanks for the replies guys.

One other little question if I may. I will typically give an answer to the minimum number of significant figures given by the information in the question, to a minimum of two, and this is what I understand to be the rule for mark schemes in A level exams. However, in this case, when filling out the data table, each voltage is given to 2sf, and each current given to 2 or 3sf, so I would have given the calculated resistances to 2sf. They give each to 3sf. I understand that we want to give the values for resistance to the same significant figures because we have calculated them (whereas if we measured them with an ohmmeter, we'd give the data to the same precision, or decimal places), but why is it given to 3sf instead of 2sf?

Many thanks,

Elbow

 

[hutchphd]

Many thanks for the replies guys.

One other little question if I may. I will typically give an answer to the minimum number of significant figures given by the information in the question, to a minimum of two, and this is what I understand to be the rule for mark schemes in A level exams. However, in this case, when filling out the data table, each voltage is given to 2sf, and each current given to 2 or 3sf, so I would have given the calculated resistances to 2sf. They give each to 3sf. I understand that we want to give the values for resistance to the same significant figures because we have calculated them (whereas if we measured them with an ohmmeter, we'd give the data to the same precision, or decimal places), but why is it given to 3sf instead of 2sf?

Many thanks,

Elbow

I don't know what the exam convention is, but you should be aware of one experimental reality: In the presence of random noise, multiple readings of a good 2sf device can produce 3sf data when averaged over multiple readings. This also applies when creating a least squares linear fits.
In general I use your method (it certainly doesn't overestimate precision)

 

[sophiecentaur]

Also, the last sig fig in 0.98 is no less 'accurate' than the last sig fig in 1.02 (only marginally at least) so the result from calculations using either of those are of equal accuracy.
The attention given to sig figs is important for students who have calculators with perhaps 9 sig figs and they all need to be educated into rounding appropriately when presenting their final answers but they should also use the full resolution for intermediate stages if they want to minimise errors.

 

[cosmik debris]

I don't know what the exam convention is, but you should be aware of one experimental reality: In the presence of random noise, multiple readings of a good 2sf device can produce 3sf data when averaged over multiple readings. This also applies when creating a least squares linear fits.

This is the idea behind dithering isn't it?

Cheers

 

[hutchphd]

Yes...a little bit counter-intuitive until you give it some thought.

 

[Herman Trivilino]

I agree that the question is poorly authored. Unfortunately there is far too much of this kind of crap in lots of these tests. It's not a surprise to me that they got the grading rubric wrong, too.