C for the last question ...any one plz tell hw much i ll lose approx? I found the reciprocal for each and added to the column head 10^-3 with the unit...then i drew the graph without any conversion of anything! and found the gradient and then the h...but all my way was correct...like large triangle and substituting etc besides tht conversion...so i may lose 3 ryt? and the last part i wrote zero error in the voltmeter
Is it just me, or wasn't the May 2012 paper much easier than the jan 2013 paper? Maybe grade boundary is lower because more people sit for the May exam?
Is it just me, or wasn't the May 2012 paper much easier than the jan 2013 paper? Maybe grade boundary is lower because more people sit for the May exam?
You sure? I thought the Jan 13 paper was really easy, especially the last question where the graph was incredibly simple to draw.
I'm pretty sure we'll have lower boundaries this time, because this paper was quite tricky at some points, especially compared to this January.
I'm not sure about whether what you're saying is right or not about the prefix, but it doesn't matter, we'll see what they'll accept and what not.
As for the adv+disadv question, I mentioned that the metre rule had greater uncertainty but they just said include uncertainty in your answer, so the whole answer shouldn't just be about that.
What did you write for it? And could you please tell me how many marks you think I'll lose for that silly gradient mistake? Thanks.
The unit didn't make any difference to the value of the gradient when U tried both ways because at last both units would turn our to be the same thing if you have misunderstood the prefix thing. So no marks lost I guess.
So even if you don't break the graph, it doesn't cut the origin? Because of errors? We had to assume all that and draw a graph? Damn I only thought no y intercept, so cuts through origin
I'm not too sure but in the past papers there was lot of graphs that didn't pass through the origin that the mark schemes accepted as correct line of best fit even when there was no c. I'm not too sure. As long as the line looks like a line of best fit its fine.
I used the cal when in comes to finding the line of best fit, by deriving the equation for the least square regression line we learnt in S1. There's no eye guessing there.
I'm guessing a grade boundary of about 30/31, but May 2012 was 29! That means this might go even lower
What are your predictions?
I'd say 30 or 31 as well... 29 for May 2012 seems incredibly low, I was quite surprised considering the fact that Unit 3B in general isn't a very challenging paper compared to the actual main exams (units 1, 2, 4, 5).
I'm not too sure but in the past papers there was lot of graphs that didn't pass through the origin that the mark schemes accepted as correct line of best fit even when there was no c. I'm not too sure. As long as the line looks like a line of best fit its fine.
I used the cal when in comes to finding the line of best fit, by deriving the equation for the least square regression line we learnt in S1. There's no eye guessing there.
I wasted a lot of time trying to make the damn graph go through the damn origin while still being a line of best fit
My gradient ended up getting screwed, wish I went through the mark schemes properly, or failed to notice that there was no y intercept...
My line cut through 3 points, with one anomaly down, that was very close to the graph and 2 anomalies up, that were further away from the line. Do you think it sounds like a suitable line of best fit?
I wasted a lot of time trying to make the damn graph go through the damn origin while still being a line of best fit
My gradient ended up getting screwed, wish I went through the mark schemes properly, or failed to notice that there was no y intercept...
My line cut through 3 points, with one anomaly down, that was very close to the graph and 2 anomalies up, that were further away from the line. Do you think it sounds like a suitable line of best fit?
Yeah, so did I. Wasted a huge amount of time trying to get it to go through the origin. Plotting the points took a long time too. I probably took as much time to draw the graph alone as I did to finish everything else on the paper
I didn't really go through the markschemes properly either, but I'm not too bothered about this paper as I only need 10/60 for an A in Physics AS.
Yeah, so did I. Wasted a huge amount of time trying to get it to go through the origin. Plotting the points took a long time too. I probably took as much time to draw the graph alone as I did to finish everything else on the paper
I didn't really go through the markschemes properly either, but I'm not too bothered about this paper as I only need 10/60 for an A in Physics AS.
Haha yea, I too finished the rest of the paper in about 30 mts. Spent a lot of time erasing and plotting, then erasing again, then checking if my points are right, then wondering what happened to my gradient
Haha yea, I too finished the rest of the paper in about 30 mts. Spent a lot of time erasing and plotting, then erasing again, then checking if my points are right, then wondering what happened to my gradient
Yeah, I had to be really careful with plotting the points! Probably one of the hardest graphs I've ever had the displeasure of plotting and drawing.
The unit didn't make any difference to the value of the gradient when U tried both ways because at last both units would turn our to be the same thing if you have misunderstood the prefix thing. So no marks lost I guess.
When I calculated the gradient, I used y-0/0-x but the graph didn't start from zero. That's a mistake I did due to time constraints as I was short on time due to wasting 20 minutes drawing the wrong gradient then realizing that was wrong.
Anyway, I got h to be 4.1x10^-34 but the gradient calculation had an error. How much do you think that costs out of 6?
well i doubt that, finding manufacture's fault is not a very gud way, we used two measurement they were wave length and V so the only measurement we take is Voltmeter's so it should had some prob.
well i doubt that, finding manufacture's fault is not a very gud way, we used two measurement they were wave length and V so the only measurement we take is Voltmeter's so it should had some prob.
well i'm not certain either
I wrote microvoltmeter could be used for more accuracy and the manufacturer's error. What do you think about it?
for the last question ...any one plz tell hw much i ll lose approx? I found the reciprocal for each and added to the column head 10^-3 with the unit...then i drew the graph without any conversion of anything! and found the gradient and then the h...but all my way was correct...like large triangle and substituting etc besides tht conversion...so i may lose 3 ryt? and the last part i wrote zero error in the voltmeter
I don't know if i am correct or not...but i ran out of ideas and wrote that we considered speed of light in vacuum in our calculations which isn't the same in air
Well if i remember correctly the question asked, the reason for the calculated value to be lower than the actual value, they didn't ask for how to improve accuracy so i'm not sure about ur answer too.
I don't know if i am correct or not...but i ran out of ideas and wrote that we considered speed of light in vacuum in our calculations which isn't the same in air
well u might get a mark for tht i assume, because wht u stated is correct. i don't remeber if it was a one mark or for two mark ? if it was for two u sholud have stated about the actual value of s.of light being either low or high to make the calculated value low.
well i'm saying all this and all other post from my experiance in pastpaper ans scheme and Exm.report only so don't rely on my comments
For the last question i wrote lost volts due to internal heating in the wire since no wire wound resistor was used and also volts reading intervals where inconsistent. Correct?