Posted April 27, 2016 (edited) Hey guys, I can't solve these correctly. Need explanations, thanks! Edited April 27, 2016 by IBdoingnothing 1 person likes this Share this post Link to post Share on other sites

Posted April 27, 2016 (edited) For 18, you can easily cross off C and D since the slit length won't affect the intensity so it's between A and B, but we also know the single slit diffraction formula which goes by s = lamda D/ d, where s is the length of diffraction, D is length between screen and slit, and d the size of the slit. so when you reduce d (slit), the diffraction pattern increases, so it should be A. For 25, period is 5ms and average power is half of max so I would go with A again (idk though, might be wrong) For the data analysis question, they want the power supplied TO the mains so I would just draw a horizontal line at y = 2kW to show the cap off since anything after this will be fed back into the main. Then, I'll draw an mean line between all the data plots till 12:00 and find the area of the triangle formed (0.5 x b x h), but be sure to change the 3 hours (between 9:00 and 12:00) from hours to seconds since power is j/s. Then the area should give the energy provided to the main supply. Obviously I can't give the exact answer since I can't work out the data plots without paper, but my estimate would be around 22Mj (0.5 x 4000 x 10800) I hope I'm right and I also hope this helped! EDIT: @Viohgot it right Edited April 27, 2016 by kevG 1 person likes this Share this post Link to post Share on other sites

Posted April 27, 2016 @kevG I'm afraid your reasoning is a bit wrong for the multiple choice questions. For question 18, reducing the slit width does reduce the intensity, because if you have a smaller slit, less light will be able to pass through the slit to the observer (another way to think about this is that if you have a smaller window in your room, the room will obviously be less bright). So the choice here is between C and D. And since the first minimum of diffraction pattern can be estimated using the formula θ ≈ λ/b (where theta is the angle of diffraction and 'b' is the width size), then for smaller 'b', theta gets bigger. Hence, the graph must look wider than the original graph, which implies the answer to be D. To help us clearly understand question 25 (correct answer should be B), I've attached a short derivation (taken from my own study guide) of the power/emf/current here: From the picture above, we see that the average EMF and average current are deduced using the root-mean-square method. But since the power is equal to the EMF multiplied by the current, so the square root of 2 in the denominator becomes 2, which means that the average power should be half of the maximum power. In this case, average power is 60kW. To calculate the period, we have to be a little more careful. The question asks about the period of the rotation of the coil, not the period of the graph itself. Normally, the graph that you would get for the EMF is a sinusoidal graph fluctuating around the x-axis. However, as you can see from the picture above, the formula for the instantaneous power is equal the square of the EMF graph divided by the resistance, such that the power is never a negative number. This means that all the parts below the x-axis are inverted around the x-axis (it's like taking the absolute value). So the period must be 10 ms. Another way to think about this question is to compare the graph of sin(x) and sin^{2}(x). Both of them have the same period, even though the distance between 2 maximums in the sin^{2}(x) graph is only half the distance between 2 maximums in the sin(x) graph. For the data analysis question, kevG has done everything correctly. @IBdoingnothing, in case you don't have access to the markscheme, here is a screenshot of it: 6 people like this Share this post Link to post Share on other sites

Posted April 27, 2016 thanks so much guys! @Vioh @kevG Share this post Link to post Share on other sites

Posted April 28, 2016 Hey, I don't get the next sub question from the same past paper either. Since V = sqrt(PR), shouldn't the fractional/percentage uncertainties be added and then divided by 2? In the mark scheme they are subtracted. PLS HELP! Share this post Link to post Share on other sites

Posted April 28, 2016 @ibstudent321 I agree with what you're saying. The fractional uncertainties should be added, instead of being subtracted. So I think the markscheme is wrong in this case. 2 people like this Share this post Link to post Share on other sites

Posted April 28, 2016 Guys, for that question I have a feeling the markscheme might be correct. As the question mentioned that it is the "Power is calculated" which means that the uncertainty of power is the one that has been processed. On the other hand, for example, in this question, the volume was the measured variable, and radius was the processed one. Here, after making the radius the subject and add the uncertainty % we get the answer, which is 8% in the markscheme. I'm not so sure if this logic is right, what do you guys think? @Vioh @ibstudent321 3 people like this Share this post Link to post Share on other sites

Posted April 28, 2016 Hey, Does it make a difference which variable is measured and which isn't? Can you please explain how that works? Considering the mark scheme agrees with you in both questions I think you are right. @IBdoingnothing @Vioh Share this post Link to post Share on other sites

Posted April 30, 2016 On 4/28/2016 at 5:06 AM, ibstudent321 said: Hey, Does it make a difference which variable is measured and which isn't? Can you please explain how that works? Considering the mark scheme agrees with you in both questions I think you are right. @IBdoingnothing @Vioh We only "propagate errors" for values which we did not directly measure. That means in @IBdoingnothing example, we essentially solve R as sqrt (V / (H * pi)), and the error is R was propagated using this formula. Note that if we actually go measure the radius then regardless of what the uncertainty is, we don't need to propagate error or even calculate anything since we can measure everything. Share this post Link to post Share on other sites

Posted April 30, 2016 On 4/27/2016 at 2:16 AM, Vioh said: And since the first minimum of diffraction pattern can be estimated using the formula θ ≈ λ/b (where theta is the angle of diffraction and 'b' is the width size), then for smaller 'b', theta gets bigger. Hence, the graph must look wider than the original graph, which implies the answer to be D. Wouldn't the answer be A if the graph must be wider than the original graph? The dotted graph represents the original graph. Share this post Link to post Share on other sites

Posted April 30, 2016 13 minutes ago, iblearner said: Wouldn't the answer be A if the graph must be wider than the original graph? The dotted graph represents the original graph. Here are the corresponding examiner's comments in the subject review As a result this is the second most difficult question on the exam. (The most difficult question is question 25, see above) I concur with the markscheme and@Vioh, though I only learned about diffraction in the SL Option last year. To be fair, the question did not make it very clear that the diffraction pattern is more spread out in choice D. It takes a second glance to notice that s has in fact increased. I assume that had choice D resembled choice A in making the new s look distinctively bigger, then the difficulty index and discrimination index would have gotten up. 2 people like this Share this post Link to post Share on other sites

Posted April 30, 2016 On 4/27/2016 at 2:16 AM, Vioh said: 6 minutes ago, kw0573 said: Here are the corresponding examiner's comments in the subject review As a result this is the second most difficult question on the exam. (The most difficult question is question 25, see above) I concur with the markscheme and@Vioh, though I only learned about diffraction in the SL Option last year. To be fair, the question did not make it very clear that the diffraction pattern is more spread out in choice D. It takes a second glance to notice that s has in fact increased. I assume that had choice D resembled choice A in making the new s look distinctively bigger, then the difficulty index and discrimination index would have gotten up. Thanks for clearing that for me ^^ Share this post Link to post Share on other sites

Posted April 30, 2016 4 hours ago, kw0573 said: We only "propagate errors" for values which we did not directly measure. That means in @IBdoingnothing example, we essentially solve R as sqrt (V / (H * pi)), and the error is R was propagated using this formula. Note that if we actually go measure the radius then regardless of what the uncertainty is, we don't need to propagate error or even calculate anything since we can measure everything. If R = sqrt ( V / pi*H ) then shouldn't the % uncertainty in R = (%uncertainty of V - the %uncertainty of H) / 2 ? When dividing values you subtract the % uncertainties right? Share this post Link to post Share on other sites

Posted April 30, 2016 7 minutes ago, ibstudent321 said: If R = sqrt ( V / pi*H ) then shouldn't the % uncertainty in R = (%uncertainty of V - the %uncertainty of H) / 2 ? When dividing values you subtract the % uncertainties right? No we still add the % uncertainties. This is to account when the numerator and denominator are off in opposite directions. That is, if the actual numerator is numerator + 10% and actual denominator is denominator - 10%, then we can't just say that is no uncertainty. There is now 20% uncertainty. note that 1.1/0.9 gives 1.22 or 22% uncertainty, but 20% is a better estimate then 0%. Share this post Link to post Share on other sites

Posted May 1, 2016 @kw0573 OK I get it now, thanks a lot! Share this post Link to post Share on other sites

Posted May 2, 2016 @IBdoingnothing @ibstudent321 Yes it was my mistake. I misread the question. The markscheme is actually correct for question (c) in the data analysis part. This is because the question explicitly states that the power P is calculated from the voltage V and the resistance R, which means that the fractional uncertainty of P is equal to the fractional uncertainty of V^{2} PLUS the fractional uncertainty of R. Hence, to get the uncertainty of V, we need to use the minus sign instead of the plus sign. Share this post Link to post Share on other sites

Posted May 2, 2016 If you guys dont mind, could you help me solve these as well? hehe @Vioh @kw0573 @ibstudent321 Share this post Link to post Share on other sites

Posted May 2, 2016 Yo yo For 19 I would go with B cause I know that the average power is half of max which is P/2 (IV/2). You can even check by multiply and check using Irms and Vrms and you would get the Average P. However they're asking for max power which is simply just P = IV For 25, I know that the charges cannot be opposite so I can cross out III so it's between A and B and I've seen two equal masses having equipotential lines like that somewhere so my best guess would be A? For 37, I would go with B, cause work function is defined as the minimum work done to remove an electron from a solid's surface 17, I found it to be a little tricky, but I think I got it. I just converted the parallel circuit into a series one, so will have a 1 ohm internal resistance and 3 ohms on the outside. Then I just used V = IR to find I. I attached a picture of my workings if you didn't understand what I did. I hope I'm right for some of these haha 2 people like this Share this post Link to post Share on other sites

Posted May 2, 2016 (edited) Ah I see, for 19 they asked MAX power so its P = VI If they asked average, then its VI/2 For 25, I'm very confused For 37, I answered B as well, but its actually D For 17, so if two EMFs/voltages are in parallel, we can just add them up along with their internal resistance to make one new unit? I thought this only works if they are in series, only then we can add them up to make one new voltage/EMF @kevG Hahahaha, you nearly got all of them right! Edited May 2, 2016 by IBdoingnothing Share this post Link to post Share on other sites

Posted May 2, 2016 (edited) 4 minutes ago, IBdoingnothing said: Ah I see, for 19 they asked MAX power so its P = VI If they asked average, then its VI/2 For 25, I'm very confused For 37, I answered B as well, but its actually D For 17, so if two EMFs/voltages are in parallel, we can just add them up along with their internal resistance to make one new unit? @kevG Hahahaha, you nearly got all of them right! Is the answer for Q25 B? Edited May 2, 2016 by iblearner Share this post Link to post Share on other sites

Posted May 2, 2016 @IBdoingnothing oh hahaha for 37, I really don't know why its D then and for 17 yeah you can just convert it into series cause remember, the voltage is the same for a parallel circuit Share this post Link to post Share on other sites

Posted May 2, 2016 2 minutes ago, kevG said: @IBdoingnothing oh hahaha for 37, I really don't know why its D then and for 17 yeah you can just convert it into series cause remember, the voltage is the same for a parallel circuit In series the current is same and in parallel the voltage is same, right? Share this post Link to post Share on other sites

Posted May 2, 2016 8 minutes ago, kevG said: @IBdoingnothing oh hahaha for 37, I really don't know why its D then The work function Ф is related by the following equation, which is in the data booklet E_{kmax}= h*f - Ф where f is the frequency of light, h is Planck's constant and E_{kmax }is the maximum possible kinetic energy of the emitted electron from the metal. By rearranging the formula we get that Ф= h*f - E_{kmax }and that tells us that the work function is the difference between the energy of the absorbed photon and the maximum kinetic energy of the emitted photon. The answer is clearly not B since the electron can be emitted with an energy ranging from zero to any value. Thus it must be D. Hope this helps. 2 people like this Share this post Link to post Share on other sites

Posted May 2, 2016 @kevG @IBdoingnothing Even though your answer is correct for Question 19, but your reasoning is incorrect. If you look at the picture (from my study guide) attached in my previous post, you can see that: P_{average} = V_{average} * I_{average} = V_{max}/sqrt(2) * I_{max}/sqrt(2) = V_{max} * I_{max} / 2 = P_{max} / 2 This means that: P_{max} = V_{average} * I_{average} * 2 However, note that this is the maximum power for the whole circuit (which consists of 2 resistors connected in series). But the question only asks for one of the resistors, so you need to divided the result above by 2. This is why the answer is B (where P = VI). For question 25, I think the answer is A. The way that I do this question is to first draw the field lines, and then draw the equipotential lines such that the equipotential lines are always perpendicular to the field lines. If you draw all of these correctly, then you should get a vertical equipotential line in the middle of the 2 opposite charges (see the picture below). Thus the picture in the question must be for 2 equal charges of same sign. As for why choice 1 (two equal masses) is also correct, you should think of contour lines. Besides, gravitational field between 2 equal masses is very similar to the electric field between 2 equal charges of the same sign, so the answer must be A. For question 37 & 17, I completely agree with Haitham Wahid and kevG . 1 person likes this Share this post Link to post Share on other sites