· web viewis the potential difference across the external resistor. (1) (ii) add a voltmeter to...

12 – Practical skills and synoptic – 05-06-17

Q1.(a) (i) Describe how you would make a direct measurement of the emf ɛ of a cell, stating the type of meter you would use.

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(ii) Explain why this meter must have a very high resistance.

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(b) A student is provided with the circuit shown in the diagram below.

The student wishes to determine the efficiency of this circuit.

In this circuit, useful power is dissipated in the external resistor. The total power input is the power produced by the battery.

Efficiency =

The efficiency can be determined using two readings from a voltmeter.

(i) Show that the efficiency = where ɛ is the emf of the cell

and V is the potential difference across the external resistor.

(1)


(ii) Add a voltmeter to the diagram and explain how you would use this new circuit to take readings of ɛ and V.

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(c) Describe how you would obtain a set of readings to investigate the relationship between efficiency and the resistance of the external resistor. State any precautions you would take to ensure your readings were reliable.

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(d) State and explain how you would expect the efficiency to vary as the value of R is increased.

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(Total 9 marks)


Q2.(a)

Figure 1

In a laboratory experiment, monochromatic light of wavelength 633 nm from a laser is incident normal to a diffraction grating. The diffracted waves are received on a white screen which is parallel to the plane of the grating and 2.0 m from it. Figure 1 shows the positions of the diffraction maxima with distances measured from the central maximum.

By means of a graphical method, use all these measurements to determine a mean value for the number of rulings per unit length of the grating.

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(6)


(b) Describe and explain the effect, if any, on the appearance of the diffraction pattern of

(i) using a grating which has more rulings per unit length.

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(ii) using a laser source which has a shorter wavelength.

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(iii) increasing the distance between the grating and the screen.

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(c) Figure 2, below, shows the diffracted waves from four narrow slits of a diffraction grating similar to the one described in part (a). The slit separation AB = BC = CD = DE = d and EQ is a line drawn at a tangent to several wavefronts and which makes an angle θ with the grating.


Figure 2

(i) Explain why the waves advancing perpendicular to EQ will reinforce if superposed.

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(ii) Show that this will happen when sin θ =

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(Total 15 marks)

Q3.A strain gauge is made from a constantan wire of original length 25 mm. If the wire stretches its resistance changes. The gauge is attached to an object that is then placed under tension, which causes the length of the constantan wire to increase. The resistance, R, was measured for various lengths, l, and the following results were obtained:

R / Ω 99.96 100.64 101.76 102.80 103.85 104.71

l / 10–2 m 2.500 2.508 2.523 2.536 2.548 2.557

When the wire is stretched, it may be assumed that for small extensions:

R ∝ l2

(a) Complete the table showing the value of l2 for each value of R.(2)

(b) Plot a graph of R on the y-axis against l2 on the x-axis.(4)

(c) Use your graph and the value for the resistivity of constantan given below to find the diameter of the wire when its resistance is 103.40 Ω.

resistivity of constantan = 4.7 × 10–7 Ωm

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(5)

(d) Define tensile strain. Use your graph to determine the strain when the resistance of the wire is 103.40 Ω.

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(Total 13 marks)

Q4. A climber falls 2.3 m before being stopped by his climbing rope that is secured above him.The weight of the climber is 840 N.

(a) Calculate the loss in gravitational potential energy of the climber.

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loss in potential energy .......................................................... J(2)

(b) The figure below shows a force-extension graph for the rope being used.


(b) (i) Use the figure above to find the stiffness of the rope when it is being used with forces up to 350 N. Give the appropriate unit.

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stiffness .................................................................................

unit .................................................................................(4)

(ii) Use the figure above to determine the energy stored in the rope when it is stretched by 0.25 m.

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energy ................................................................................. J(3)

(Total 9 marks)

Q5.A laser is used with a double slit positioned 5.0 m from a white screen. The separation of


the slits is 0.25 mm and the wavelength of the laser light is 630 mm.

The screen is removed and a linear air track positioned so that a glider on the air track moves in the same plane that was occupied by the screen.

A light dependent resistor, LDR, is attached to a glider and connected by loosely hanging leads to a datalogger. When the glider moves at a constant speed, the datalogger records the output voltage from a circuit containing the LDR. Output from the datalogger, plotted against time, is shown below.

(a) (i) Explain why the LDR output voltage varies with the position of the glider.

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(ii) Calculate the separation between two adjacent positions of the glider when the LDR is under maximum illumination.

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(iii) Use your answer to (ii) and the graph to calculate a speed for the glider consistent with these results.

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(b) A potential divider current is used to derive an output from the LDR.

In this experiment the resistance of the LDR is 10 kΩ when under maximum illumination and 100 kΩ when under minimum illumination. The value of R is 50 Ω. Calculate the values of V1 and V2 shown on the graph and state which corresponds to maximum illumination.

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(c) If the experiment were to be repeated how could you ensure that the glider is launched with the same speed each time?

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(Total 10 marks)


Q6.The diagram below shows a motor lifting a small mass. The energy required comes from a charged capacitor

The capacitor was charged to a potential difference of 4.5 V and then discharged through the motor.

(a) (i) The motor only operates when the voltage at its terminals is at least 2.5 V.

Calculate the energy delivered to the motor when the potential difference across the capacitor falls from 4.5 V to 2.5 V.

(3)

(ii) The motor lifted the mass through a distance of 0.35 m. Calculate the efficiency of the transfer of energy from the capacitor to gravitational potential energy of the mass. Give your answer as a percentage.

gravitational field strength = 9.8 N kg–1

(2)

(iii) Give two reasons why the transfer is inefficient.

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(b) The motor operated for 1.3 s as the capacitor discharged from 4.5 V to 2.5 V.

Calculate:

(i) the average useful power developed in lifting the mass;

(2)

(ii) the effective resistance of the motor, assuming that it remained constant.

(3)(Total 12 marks)


M1.(a) (i) Voltmeter across terminals with nothing else connected to battery / no additional load.

1

(ii) This will give zero / virtually no current 1

(b) (i)

Answer must clearly show power: εI and VI, with I cancelling out to give formula stated in the question

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(ii) Voltmeter connected across cell terminals

Switch open, voltmeter records εSwitch closed, voltmeter records VBoth statements required for mark

Candidates who put the voltmeter in the wrong place can still achieve the second mark providing they give a detailed description which makes it clear that:To measure emf, the voltmeter should be placed across the cell with the external resistor disconnectedAndTo measure V, the voltmeter should be connected across the external resistor when a current is being supplied by the cell

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(c) Vary external resistor and measure new value of V, for at least 7 different values of external resistor

Precautions - switch off between readings / take repeat readings (to check that emf or internal resistance not changed significantly)

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(d) Efficiency increases as external resistance increases

ExplanationEfficiency = Power in R / total power generatedI2R / I2(R + r) = R / (R + r)So as R increases the ratio becomes larger or ratio of power in load to power in internal resistance increases

Explanation in terms of V and ε is acceptable


[9]

M2.(a)

x / m sin θ

1 0.173 0.086

2 0.316 0.156

3 0.499 0.242

4 0.687 0.325

5 0.860 0.395

1

If angles only calculated 1 / 2

at least 4 points plotted correctly (1)best straight line (1)gradient calculated from suitable triangle, 50% of each axis (1)correct value from readings (1)appropriate use of d sin θ = nλ (1)hence N (rulings per metre) = 1.25 × 105 m–1 (1.1 to 1.4 ok) (1)

max 2 / 6 if no graph and more than one data set used correctly, 1 / 6 only one setif tan calc but plotted as sin, mark as schemetan or distance plotted, 0 / 6

max 6

(b) (i) maxima wider spaced [or pattern brighter] (1)sin θ or θ increases with N [or light more concentrated] (1)


(ii) maxima spacing less (1)sin θ or θ decreases with λ [or statement] (1)

(iii) maxima wider spaced [or pattern less bright] (1)same θ but larger D [or light more spread out] (1)

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(c) (i) waves in phase from (1)any sensible ref to coherence (1)whole number of wavelengths path difference (1)

(ii) use of geometry to show that sin θ = max 3

[15]

M3.(a)

R / Ω 99.96 100.64 101.76 102.80 103.85 104.71

l / 10–2 m 2.500 2.508 2.523 2.536 2.548 2.557

l2 / 10–4 m2 6.250 6.290 6.366 6.432 6.492 6.538

correct values above (1)(1) (deduct one mark for each error) (If two significant figures only – deduct one mark)

(2)

(b) both axes, with units, correctly labelled (1) six points correctly plotted (1) best straight line through plotted points (1) sensible scale (1)

(4)


(c) at R = 103.40 Ω, l2 = 6.46 (1) × 10–4 m2 (1) thus l = 2.542 × 10–2 (1) (m) constantan resistivity = 47 × 10–8 (Ωm) (1) = 1.2(1) × 10–5 m (1)

(5)

(d) tensile strain = (1)

(at R = 103.40 Ω, l = 2.542 × 10–2 m)

thus strain = (1) = 0.017 (1)(max 2)

[13]

M4. (a) 840 × 2.3

C1

1900 (J)/1930 (J)

A12

(b) (i) uses gradient

C1

data extraction correct –350 N, 0.3 m

C1

1170

A1

N m–1

B14


(ii) uses area

B1

6.5 to 7.0 squares or 1 square is equivalent to5 J/area is ½ base × height

B1

32.5 to 35 (J)

B13

[9]

M5.(a) (i) Because the bright fringes and dark fringes (1) produced by double slit interference (1)

(ii) w = (1) = 12.6 (13) mm (1)

(iii) from graph, time to travel between adjacent bright fringes = 0.25 s (1)

use w = 13 mm and υ = (1)

to give υ = = 52 mm s–1 or 5.2 × 10–2 m s–1 (1)(max 5)

(b) V2 = value corresp to low resist ∴ bright fringe (1)


= 6 × = 2.0 V (1)

V2 = 6 × = 0.29 V (1)(3)

(c) e.g. use spring between glider & track end (1) compress to same length each time, release (1)

(2)[10]

M6.(a) (i) Energy = ½CV2 or ½QV and Q = VCC1

Calculation initial or final energy correctly (0.202 J or 0.0625 J)or energy = ½ (20 000 × 10−6 (4.52 − 2.52)condone no or incorrect power of 10

C1

0.137 to 0.140 JA1

3

(ii) PE = 0.015 × 9.8 × 0.35 (0.0515) (0.052 J)or arrives at 0.368 or 0.371

C1

36 to 38 (36.8 (37) % is correct)(ecf 0.052 × 100 / their (a)(i)) (penalise use of 0.05 J)

A12

(iii) heating / energy loss due to resistance of wiresB1

work done against friction allow energy / heat loss due to frictionB1

work done against air resistance due to motion of the massB1

sound energy due to vibrations of the motorB1

Max2


(b) (i) Power = work done / time (W / t or E / t)

or work done = their PE from (a)(ii) / 1.3

or power = 0.14 / 1.3(i.e. use of input energy from (a)(i))

C1

40 (39.6) mWA1

2

(ii) V = VOe−t / RC or Q = QOe−t / RC and Q = VCC1

2.5 = 4.5 e−1.3 / 0.02R (ignore incorrect power of 10 for C in substitution)C1

111 (110) ΩAllow B1 for realising 0.69CR ≈ 1.3 leading to 94 Ω

A13

[12]


E1.(a) (i) Students had to make it clear that the voltmeter ‘alone’ should be connected across the cell.

(ii) A correct explanation was given by a large proportion of students.

(b) (i) Answered well by the more able students.

(ii) A proportion of students seemed to understand how to use the voltmeter but failed to show the correct position on the circuit diagram.

(c) This question discriminated well. Many students failed to give sufficient detail as required by the mark scheme for the first marking point. The second marking point proved to be more accessible, with a greater proportion of students able to suggest an appropriate precaution.

(d) As anticipated this proved to be very demanding, with only the more able students successfully stating and explaining why efficiency would increase as external resistance increases.

E2.This question attracted the smallest proportion of answers. For those candidates who attempted the question, the answers to part (a) fell mainly into one of two categories. In the first group were candidates who had obviously either seen or done the experiment or who at least were able to interpret the diagram correctly. This group of candidates could calculate the diffraction angles, plot an appropriate graph and use it to determine the final answer, scoring full marks. The second group consisted of those candidates who confused the situation with a Young double slit experiment. Although these candidates could plot a graph, there was no way that it could be related to any sensible means of determining the number of rulings per metre on a grating. It was impossible to award any marks to answers of this kind. There were some candidates who made some progress with the question but whose answers fell a long way short of the ideal. For example, some calculated the tangent of the diffraction angle but interpreted this as sine. Others did not plot a graph at all but calculated at least one value from the data. There was some merit in answers of this kind but, at best, they were worth one or two marks out of six.

Success with part (b) hinged upon the recognition of the relevance of the diffraction grating equation. Many, but not all, of those candidates who had interpreted the question as relating to the double slit experiment, continued with the same theme and few qualified for any marks. There was a minority of very good answers in which the grating equation


was clearly stated and the effect of changing the appropriate variables was analysed. Some otherwise quite good answers were spoiled by imprecise use of language such as “more (or less) diffraction” or “the pattern becomes less”. Some candidates thought that, in answer to part (b)(iii), the pattern remained unchanged because the diffraction angle did not alter.

Answers to part (c) were generally weak and many candidates left the answer space blank. It had been hoped that the diagram would provide a useful prompt, but this was not the case. Of those candidates who answered the question, many scored a mark for some reference to either “waves in phase” or “coherence”. Few candidates were able to put together the logical steps stating the waves from each slit to have a whole number of wavelengths path difference, so being in phase and reinforcing when superposed. In answer to part (c)(iii), most candidates put n = 1 in the grating equation and did not use the diagram to show, for example, that a line drawn from D perpendicular to EQ has a length equal to the wavelength. The equation then follows naturally from this diagram.

E4. Most candidates correctly answered the calculation in (a), with the most common error being the introduction of a spurious factor of g. Nearly all candidates successfully completed the graph work in part (b).

E6.(a) (i) Most candidates knew which formula to use and there were many correct solutions. A common error was using (4.5 – 2.0)2 instead of subtracting the energy for a p.d. of 2.0 V from the energy for a p.d of 4.5 V. Many of those who used ½ QV and Q = CV failed to take into account the fact that at 2.0 V the charge is different from that at 4.5 V.

(ii) This was well done by most candidates allowing the error carried forward. However, there was a significant proportion who rounded off the output energy to 1 sf early in the calculation hence producing an inaccurate answer.

(iii) Most candidates were able to gain some credit but weak answers failed to address the context or gave vague statements such as ‘due to heat produced in the wires’.

(b) (i) This was generally well done. The majority knew that power is ‘E / t’ but a significant proportion used the energy input rather than the useful energy.

(ii) Those who appreciated that this was a capacitor discharge problem usually coped well with the mathematical demands of the question. Many tried to


solve it using equations such as V = IR, Q = It and Q = VC. Some incorrectly thought 1.3 s was equal to the time constant RC.

· web viewis the potential difference across the external resistor. (1) (ii) add a voltmeter to...

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