Questions Q1. Select one answer from A to D and put a cross in the box ( )

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1 Questions Q1. Select one answer from A to D and put a cross in the box ( ) The resistance of a length of copper wire is 6 Ω. A second piece of copper wire has twice the length and twice the cross-sectional area. The resistance of the second piece of copper wire is A 3Ω B 6Ω C 12 Ω D 24 Ω (Total for Question = 1 mark) Q2. (a) Explain the difference between resistance and resistivity. (2) (b) The resistivity of copper is Ω m. A copper wire is 0.50 m long and has a cross sectional area of m2. Calculate its resistance. (2) Resistance =...

(Total for Question = 4 marks) Q3. The photograph shows a typical hairdryer. (a) The hairdryer contains a heating element which consists of a long nichrome wire wound around an insulator.

2 (Total for Question = 4 marks) Q3. The photograph shows a typical hairdryer. (a) The hairdryer contains a heating element which consists of a long nichrome wire wound around an insulator. The heating element operates at 230 V and has a power rating of 1 kw. Show that the resistance of the heating element is about 50 Ω. (3) (b) The nichrome wire has a cross-sectional area of m2. Calculate the length of the wire. resistivity of nichrome = Ω m (2) Length =...

3 (c) The nichrome wire has a diameter of 0.40 mm. A manufacturer wishes to make a hairdryer of the same resistance but using half the length of wire. Calculate the diameter of nichrome wire that must be used. (3) Diameter =... (Total for Question = 8 marks) Q4. Select one answer from A to D and put a cross in the box ( ) The resistance of a negative temperature coefficient thermistor A becomes zero above a certain temperature. B decreases as the temperature decreases. C increases as the temperature decreases. D is constant at temperatures below 0 C. (Total for Question = 1 mark) Q5. When a semiconductor has its temperature increased from room temperature, its resistance usually decreases because A the electrons are moving faster. B the lattice atoms vibrate with greater amplitude. C the lattice atoms vibrate with smaller amplitude. D the number of charge carriers per unit volume increases.

4 (Total for Question = 1 mark) Q6. When tidying a prep room, a teacher discovers a tray of resistance wires that have lost their labels. She decides to ask her students to carry out experiments to determine the material that each wire is made of by measuring the resistivity of the wires. (a) Explain why the teacher asks the students to measure the resistivity and not the resistance of the wires. (2) *(b) You are to describe a method to determine accurately the resistivity of one of the metal wires. Your description should include: the circuit diagram you would use the quantities you would measure the graph you would plot how you would determine the resistivity (9)

5 (Total for question = 11 marks) Q7. A strain gauge measures changes in the resistance of a metal under strain to find the applied force. The kitchen balance in the photograph uses strain gauges to measure the weight of cooking ingredients. A student tests this method by measuring the resistance of a wire before a force is applied and while it is under tension. (a) Calculate the initial resistance of the wire. length of wire = 1.0 m cross sectional area of wire = m2 resistivity of wire = Ω m (2) Resistance of wire =...

6 (b) The student applies a force to the wire and measures the new length. He calculates the increase in the resistance to be Ω. He measures the increase in resistance and finds it to be Ω. The student suggests that the difference between these two values is because the cross-sectional area of the wire changes under strain. Explain why a change in cross-sectional area would cause this difference. (3) (Total for Question = 5 marks) Q8. A student carried out a series of measurements to determine how the resistance of a wire varies with its length. The student obtained the following results. The student plotted the results on a graph.

(a) Calculate the resistivity of the wire used. cross-sectional area of wire = 1.06 10 7 m2 (4) Resistivity =.

7 (a) Calculate the resistivity of the wire used. cross-sectional area of wire = m2 (4) Resistivity =... (b) One precaution taken by the student was to keep the current small. Explain why this precaution was necessary. (2) (c) Explain one other precaution which should be taken by the student to ensure the accuracy of the results in the table. (2)

8 (Total for Question = 8 marks) Q9. The instruction booklet for an electric garden shredder includes the following advice. (a) Describe the relationship between area and length in the table. (1) (b) The cable for the shredder contains two conductors in series, the live wire and the neutral wire. A cable of length 40 m has a total conductor length of 80 m. (i) Show that the resistance of a copper conductor of length 80 m and cross-sectional area 1.00 mm2 is about 1.3 Ω. resistivity of copper = Ω m (2)

9 (ii) When in use the current for the shredder is 11 A. Calculate the rate of energy dissipation by the 40 m, 1.00 mm2 cable when it is used with the shredder. (2) Rate of energy dissipation =... (iii) Calculate the total potential difference across the conductors in the 40 m cable when it is used with the shredder. (2) Potential difference =... (c) Suggest why the advice in the instruction booklet is included. (2) (Total for question = 9 marks)

10 Q1. No Examiner's Report available for this question Q2. (a) Majority of candidates scored one or zero for this question, demonstrating a lack of understanding about resistance and resistivity. Some candidates who stated that resistivity was a constant for a material, then went on to say that it also depended on dimensions. Most candidates defined resistance for the equation R = V/I. Although this equation defines the unit of resistance, current is determined by the potential difference and resistance. This candidate shows an understanding of the physics, but omits any reference to resistance depending on physical dimensions. This scores 1 mark An example where candidate incorrectly relates resistivity to length and area and defines resistance in terms of p.d. and current. Results Plus: Examiner Tip Equations give you a mathematical relationship between variables but do not tell you which variable is dependant on the others. This needs to be learnt.

11 (b) This question required a straightforward substitution into a formula to find a resistance and not surprisingly the majority of candidates scored both marks. Where errors were made, it was because of the omission of a unit or confusion between resistance and resistivity. Some candidates chose to rearrange the equation and substitute resistivity as resistance. A significant number of candidates gave this answer, where they confused resistance and resistivity. Results Plus: Examiner Tip Make sure that you know what all the symbols in the equation stand for. Q3. (a-b) The most common outcome in each part of question 13 was a full score, with the great majority gaining five marks for parts (a) and (b) and at least two marks for part (c). There were some problems with applying powers of 10, occasionally with kw but more frequently, if at all, with mm in the last part. In part (b) candidates occasionally lost a mark by failing to include the unit with their answers. A minority of candidates chose the correct formula but got resistance and resistivity confused and some could not rearrange it. Some did not understand what was represented by the symbol rho and calculated the product of resistivity and length rather than length, apparently satisfied with the product of two variables rather than a single value.

12 Resistance and resistivity have been reversed.

13 Scores 3, 0. This answer treats the product of resistivity and length as a single quantity equivalent to length.

14 Results Plus: Examiner Tip Be sure to learn what all the symbols in the supplied formulae stand for. (c) The normal approach in part (c) was to use the full resistivity equation rather than applying the ratio of area to length and a large majority got this far. Nearly half completed it successfully and about a sixth failed to apply the factor of two correctly to get from the radius to the diameter. Correct working, but missing the unit. Results Plus: Examiner Tip Physical quantities must have a magnitude and a unit.

15 The radius has been correctly calculated, but the instruction to find the diameter has not been followed. Q4. 78% of candidates answered this question correctly. Q5. B was the favoured incorrect choice. This is a true statement, but not one that explains negative temperature coefficient behaviour. Q6. (a) The majority of candidates linked resistivity with a material, but only a quarter got both marks. Some said resistivity was a property of a wire instead of a material. Some said resistance depended on temperature, but so does resistivity, and some said that resistance depends on potential difference or on current. Some seemed to reverse resistance and resistivity, and some may have thought resistivity depends on length and area because of experiments such as that from (b).

16 The candidate has some understanding in saying that resistance can be calculated from resistivity, but the reverse is also true and it does not have the required detail. 0 marks This example gets the resistivity mark, but not resistance because it is linked to p.d. and current, rather than dimensions. (b) The great majority of candidates achieved at least 3 marks, over half at least 6, a third at least 7, but very few, 9. The question clearly identified the required points, for those who read it properly. The diagrams were usually satisfactory, although the wire was not always identified correctly. Some included power supplies with resistance meters. The need to measure length, current and potential difference was usually mentioned, although candidates sometimes only said 'take readings from the voltmeter and ammeter'. The diameter measurement was not always included, with some seeking to measure crosssectional area directly. Many candidates chose to plot graphs of V against I, using these to find resistance and effectively using only one length. A variety of possible graphs was used, including RA vs l, with the gradient giving the resistivity. The equation was usually rearranged correctly, but the use of the gradient was not always stated.

17 This response receives marks for an acceptable diagram, 2 for measuring V, I and length and 1 for the final equation. A graph is mentioned but not identified, the gradient is not referred to and

18 it suggests that area can be measured directly. 5 marks Results Plus: Examiner Tip Be sure that you know the difference between what you can measure directly and what you can calculate from your measurements. If you are using a graph, state the axes clearly, and identify the gradient and how you will use it.

19 This is a good answer, but gets 7 out of 9 in the end. because the calculation of cross-sectional area is not mentioned. An interesting graph is proposed, and the gradient is identified, but the final step to go from gradient to resistivity is omitted. Q7. (a) The great majority completed this calculation successfully, although a few applied the unit Ωm to resistance. Some went wrong with powers of 10 and some reversed resistance and resistivity in the formula. The candidate has applied the correct formula, but is treating resistivity as resistance and vice versa. This may be because the symbols have not been learned correctly, so ρ is being treated as resistance and R as resistivity. This may in turn be because of the way the formula appears under the heading Resistivity in the list of formulae.. Results Plus: Examiner Tip The list of data etc gives formulae in symbol form. Be sure to learn what each symbol represents.

20 The formula and substitution are correct, but an error has been made in applying powers of ten so the answer is incorrect. It is so small that a candidate might have been expected to notice the error and check the calculation. Questions are not set to test mathematics alone and will always give realistic answers. Results Plus: Examiner Tip Check that answers are realistic and attempt calculations again when they are not. The working and numerical answer are fine, but the unit given is the unit of resistivity, not

21 resistance. (b) The great majority managed to get at least one mark, often for identifying the relationship between cross-sectional area and resistance. Those who invoked the formula were at a significant advantage. Many described in detail the effect of both an increase and a decrease in cross-sectional area, but failed to state which they thought had occurred, hence the common award of the second mark only. This is another example of failure to answer the specific question asked, which here was about 'this difference'. Some just wrote about change of area and change of resistance, with no reference to increase or decrease at all. Others got two marks for identifying the decrease in cross-sectional area and making a vague, descriptive link to increased resistance in terms of how hard it is for electrons to get through or crowding of electrons. The candidate has demonstrated an understanding of the link between cross-sectional area and resistance, but has not stated which applies in this case, as required by the question where it says 'this difference' and not 'a difference'.

22 This candidate has successfully identified the decrease in cross-sectional area and has linked it to an increase in resistance for two marks. The explanation itself, in terms of less space for electrons, does not clearly establish the link. When there is a straightforward formula linking the variables, as in this case, it is better to use the formula in explaining the outcome, such as in the previous example. Q8. (a) While well over three quarters gained three marks for completing the calculation, the proportion gaining full marks was under half because the line was often not drawn on the graph. The question didn't state 'use the graph', but candidates should recognise from it that different points would give different values of resistivity, so the best way to find 'the' resistivity would be to use the gradient of a best fit line. Errors occasionally seen included omitting the unit, calculating from a pair of values that gave a result outside the accepted range, converting 100 cm to m incorrectly and getting R and ρ mixed up in the formula. (b) Candidates again lost marks by not giving answers to the specific situation described. They often said that the temperature would be affected and resistance would change without specifying an increase in either. Overall, half of them got the mark for temperature increase, often saying it gets hot, and a half of those the second mark. Some tried to interpret the precaution in terms of personal safety, saying it might cause a shock

23 or a burn. (c) The question referred to the table, so precautions needed to be related to accuracy in the measurement of length, current or potential difference. Only about a sixth chose a suitable precaution. One of the most frequent suggestions was repeating the measurement of diameter and calculating the average, but diameter, radius and cross-sectional area were not variables in the table. Common accepted precautions were ensuring that the wire was straight, avoiding parallax errors in reading analogue meters and avoiding zero errors. Q9. (a) About half of candidates were credited for the answer to this question, most of these for quoting direct proportionality. A large minority just made a statement like 'as area increases, length increases', which was true but is not a relationship as such. This is just a very general statement that as one quantity increases so does the other. Candidates are expected to be able to identify direct proportionality.

24 In this question, credit was given to answers establishing the relationship between the quantities in terms such as length (in m) = 40 x area (in mm2), but this answer involves too many steps. (b) (i-iii) The majority of candidates achieved at least five marks for this six mark section. (i) The most common error in this part was using the wrong power of ten for the area in m2, starting with As this was a 'show that' question, most could see that this gave them the wrong answer so they could go back to their working and identify their mistake. (ii) There was not generally a problem with this calculation, although it was apparent that some candidates did not realise that rate of energy transfer required a power equation. A significant minority applied a length of 40 m for this part, even though they used 80 m in part (i), so they recalculated resistance or halved the value from part (i). (iii) This part was tackled very well using one of the three available methods, although some candidates again worked with a length of 40 m.

25 (i) The method is correct, for one mark, but there has been an error in converting the units of area. The question states that the value should be about 1.3 Ω, so a candidate would be expected to try to identify the error when the answer is not as expected.(ii) Correct answer. Note that the quoted 'show that' value of resistance has sensibly been used instead of the incorrect answer from part (i). (iii) Correct answer. Results Plus: Examiner Tip Remember that 1 m2 is (1000 mm)2, i.e mm2. Also, when an answer does not mention the 'show that' value, go over the working to identify the error.

27 (i) Correct answer. (ii) The candidate has recalculated the resistance using a length of only 40 m, so the answer is only half what it should be. (iii) The power from the previous part has been used here and processed correctly so the answer, although incorrect, is awarded both marks applying 'error carried forward'. (c) Only a fifth of the entry gained any credit here, usually for suggesting that the resistance would be too large. Candidates most commonly stated that the cable would overheat in this case, sometimes adding that this was because a larger current would be needed to provide the shredder with a high enough potential difference. They did not think of the fixed mains supply or about the current required by the shredder remaining at 11 A. This includes an odd reference to voltage travelling along a wire. There is a suggestion that a greater current would be required but it doesn't make it clear that this couldn't happen because the potential difference from the supply couldn't be increased.

28 This response gets a mark for stating that a longer cable would have a very large resistance, but, when it refers to heating, does not take into account the fixed potential difference which would mean a reduced current. Mark Scheme Q1. Q2.

29 Q3. Q4. Q5.

30 Q6. Q7.

31 Q8.

32 Q9. Powered by TCPDF (

Which one of the following graphs correctly shows the relationship between potential difference (V) and current (I) for a filament lamp?

Questions Q1. Select one answer from A to D and put a cross in the box ( ) Which one of the following graphs correctly shows the relationship between potential difference (V) and current (I) for a filament