Topic 5 Revision [142 marks] - Peda.net · Topic 5 Revision [142 marks] 1. Three resistors are...

Topic 5 Revision [142 marks]

1. Three resistors are connected as shown. What is the value of the total resistancebetween X and Y?

A. 1.5 ΩB. 1.9 ΩC. 6.0 ΩD. 8.0 Ω

MarkschemeA

Examiners report[N/A]

2. A liquid that contains negative charge carriers is flowing through a square pipe withsides A, B, C and D. A magnetic field acts in the direction shown across the pipe.

On which side of the pipe does negative charge accumulate?

MarkschemeA

[1 mark]

[1 mark]


3. Five resistors of equal resistance are connected to a cell as shown.

What is correct about the power dissipated in the resistors?

A. The power dissipated is greatest in resistor X.

B. The power dissipated is greatest in resistor Y.

C. The power dissipated is greatest in resistor Z.

D. The power dissipated is the same in all resistors.

MarkschemeC


4. Two resistors X and Y are made of uniform cylinders of the same material. X and Y areconnected in series. X and Y are of equal length and the diameter of Y is twice thediameter of X.

The resistance of Y is R.

What is the resistance of this series combination?

A.

B.

C. 3R

D. 5R

5R4

3R2

[1 mark]

[1 mark]

MarkschemeD


5. A cell with negligible internal resistance is connected as shown. The ammeter and thevoltmeter are both ideal.

What changes occur in the ammeter reading and in the voltmeter reading when the resistanceof the variable resistor is increased?

MarkschemeC


An electron enters the region between two charged parallel plates initially moving

[1 mark]

6. An electron enters the region between two charged parallel plates initially movingparallel to the plates.

The electromagnetic force acting on the electron

A. causes the electron to decrease its horizontal speed.

B. causes the electron to increase its horizontal speed.

C. is parallel to the field lines and in the opposite direction to them.

D. is perpendicular to the field direction.

MarkschemeC


7. A beam of electrons moves between the poles of a magnet.

What is the direction in which the electrons will be deflected?

A. Downwards

B. Towards the N pole of the magnet

C. Towards the S pole of the magnet

D. Upwards

MarkschemeD


[1 mark]

[1 mark]

8. A cell has an emf of 4.0 V and an internal resistance of 2.0 Ω. The ideal voltmeter reads3.2 V.

What is the resistance of R?

A. 0.8 ΩB. 2.0 ΩC. 4.0 ΩD. 8.0 Ω

MarkschemeD


9. An ion of charge +Q moves vertically upwards through a small distance s in a uniformvertical electric field. The electric field has a strength E and its direction is shown in thediagram.

What is the electric potential difference between the initial and final position of the ion?

A.

B. EQs

C. Es

D.

MarkschemeC

EQ

s

Es

[1 mark]

[1 mark]


10. When an electric cell of negligible internal resistance is connected to a resistor ofresistance 4R, the power dissipated in the resistor is P.

What is the power dissipated in a resistor of resistance value R when it is connected to thesame cell?

A.

B. P

C. 4P

D. 16P

MarkschemeC


P4

11. A cell of emf 6.0 V and negligible internal resistance is connected to three resistors asshown.

The resistors have resistance of 3.0 Ω and 6.0 Ω as shown.

What is the current in resistor X?

A. 0.40 A

B. 0.50 A

C. 1.0 A

D. 2.0 A

MarkschemeC

[1 mark]

[1 mark]


12a.

An ohmic conductor is connected to an ideal ammeter and to a power supply of output voltageV.

The following data are available for the conductor:

density of free electrons = 8.5 × 10 cm

resistivity ρ = 1.7 × 10 Ωm

dimensions w × h × l = 0.020 cm × 0.020 cm × 10 cm.

The ammeter reading is 2.0 A.

Calculate the resistance of the conductor.

Markscheme1.7 × 10 ×

0.043 «Ω»

[2 marks]


22 −3

−8

–8 0.10

(0.02×10−2)2

[2 marks]

12b. Calculate the drift speed v of the electrons in the conductor in cm s . State youranswer to an appropriate number of significant figures.

Markschemev «= » =

0.368 «cms »

0.37 «cms »

Award [2 max] if answer is not expressed to 2 sf.

[3 marks]


–1

I

neA

28.5×1022×1.60×10−19×0.022

–1

–1

13a.

An electron moves in circular motion in a uniform magnetic field.

The velocity of the electron at point P is 6.8 × 10 m s in the direction shown.

The magnitude of the magnetic field is 8.5 T.

State the direction of the magnetic field.

5 –1

[3 marks]

[1 mark]

Markschemeout of the page plane / ⊙

Do not accept just “up” or “outwards”.

[1 mark]


13b. Calculate, in N, the magnitude of the magnetic force acting on the electron.

Markscheme1.60 × 10 × 6.8 × 10 × 8.5 = 9.2 × 10 «N»

[1 mark]


–19 5 –13

13c. Explain why the electron moves at constant speed.

[1 mark]

[1 mark]

Markschemethe magnetic force does not do work on the electron hence does not change the electron’skinetic energy

OR

the magnetic force/acceleration is at right angles to velocity

[1 mark]


13d. Explain why the electron moves on a circular path.

Markschemethe velocity of the electron is at right angles to the magnetic field

(therefore) there is a centripetal acceleration / force acting on the charge

OWTTE

[2 marks]


The diagram shows a potential divider circuit used to measure the emf E of a cell X. Both cells

[2 marks]

14a.

The diagram shows a potential divider circuit used to measure the emf E of a cell X. Both cellshave negligible internal resistance.

State what is meant by the emf of a cell.

Markschemethe work done per unit charge

in moving charge from one terminal of a cell to the other / all the way round the circuit

Award [1] for “energy per unit charge provided by the cell”/“power per unit current”

Award [1] for “potential difference across the terminals of the cell when no current isflowing”

Do not accept “potential difference across terminals of cell”

[2 marks]


AB is a wire of uniform cross-section and length 1.0 m. The resistance of wire AB is 80 Ω. When

[2 marks]

14b.

AB is a wire of uniform cross-section and length 1.0 m. The resistance of wire AB is 80 Ω. Whenthe length of AC is 0.35 m the current in cell X is zero.

Show that the resistance of the wire AC is 28 Ω.

Markschemethe resistance is proportional to length / see 0.35 AND 1«.00»

so it equals 0.35 × 80

«= 28 Ω»

[2 marks]


14c. Determine E.

Markschemecurrent leaving 12 V cell is = 0.15 «A»

OR

E = × 28

E = «0.15 × 28 =» 4.2 «V»

Award [2] for a bald correct answer

Allow a 1sf answer of 4 if it comes from a calculation.

Do not allow a bald answer of 4 «V»

Allow ECF from incorrect current

[2 marks]

1280

1280

[2 marks]

[2 marks]


15a. Calculate the resistance of the conductor.

Markscheme1.7 × 10 ×

0.043 «Ω»

[2 marks]


–8 0.10

(0.02×10−2)2

An ohmic conductor is connected to an ideal ammeter and to a power supply of output voltage

[2 marks]

15b.

An ohmic conductor is connected to an ideal ammeter and to a power supply of output voltageV.

The following data are available for the conductor:

density of free electrons = 8.5 × 10 cm

resistivity ρ = 1.7 × 10 Ωm

dimensions w × h × l = 0.020 cm × 0.020 cm × 10 cm.

The ammeter reading is 2.0 A.

Calculate the drift speed v of the electrons in the conductor in cm s .

Markschemev «= » =

0.37 «cms »

[2 marks]


22 −3

−8

–1

I

neA

28.5×1022×1.60×10−19×0.022

–1

The electric field E inside the sample can be approximated as the uniform electric field between

[2 marks]

15c.

The electric field E inside the sample can be approximated as the uniform electric field betweentwo parallel plates.

Determine the electric field strength E.

MarkschemeV = RI = 0.086 «V»

«» 0.86 «V m »

Allow ECF from 4(a).

Allow ECF from MP1.

[2 marks]


= =Vd

0.0860.10

–1

15d. Show that .=v

E1

neρ

[2 marks]

[3 marks]

MarkschemeALTERNATIVE 1

clear use of Ohm’s Law ( V = IR)

clear use of R =

combining with I = nAve and V = EL to reach result.

ALTERNATIVE 2

attempts to substitute values into equation.

correctly calculates LHS as 4.3 × 10 .

correctly calculates RHS as 4.3 × 10 .

For ALTERNATIVE 1 look for:

V = IR

R =

V = EL

I = nAve

V = I

EL = I

E = I

E = nAve = nveρ

[3 marks]


ρL

A

9

9

ρL

A

ρL

A

ρL

A

ρ

A

ρ

A

=vE

1neρ

Hydrogen atoms in an ultraviolet (UV) lamp make transitions from the first excited state to the

16a.

Hydrogen atoms in an ultraviolet (UV) lamp make transitions from the first excited state to theground state. Photons are emitted and are incident on a photoelectric surface as shown.

Show that the energy of photons from the UV lamp is about 10 eV.

MarkschemeE = –13.6 «eV» E = – = –3.4 «eV»

energy of photon is difference E – E = 10.2 «≈ 10 eV»

Must see at least 10.2 eV.

[2 marks]


1 213.6

4

2 1

The photons cause the emission of electrons from the photoelectric surface. The work function

[2 marks]

16b.

The photons cause the emission of electrons from the photoelectric surface. The work functionof the photoelectric surface is 5.1 eV.

Calculate, in J, the maximum kinetic energy of the emitted electrons.

Markscheme10 – 5.1 = 4.9 «eV»

4.9 × 1.6 × 10 = 7.8 × 10 «J»

Allow 5.1 if 10.2 is used to give 8.2×10 «J».


–19 –19

−19

16c. Suggest, with reference to conservation of energy, how the variable voltage sourcecan be used to stop all emitted electrons from reaching the collecting plate.

[2 marks]

[2 marks]

MarkschemeEPE produced by battery

exceeds maximum KE of electrons / electrons don’t have enough KE

For first mark, accept explanation in terms of electric potential energy difference ofelectrons between surface and plate.

[2 marks]


16d. The variable voltage can be adjusted so that no electrons reach the collecting plate.Write down the minimum value of the voltage for which no electrons reach thecollecting plate.

Markscheme4.9 «V»

Allow 5.1 if 10.2 is used in (b)(i).

Ignore sign on answer.

[1 mark]


The electric potential of the photoelectric surface is 0 V. The variable voltage is adjusted so that

[1 mark]

16e.

The electric potential of the photoelectric surface is 0 V. The variable voltage is adjusted so thatthe collecting plate is at –1.2 V.

On the diagram, draw and label the equipotential lines at –0.4 V and –0.8 V.

Markschemetwo equally spaced vertical lines (judge by eye) at approximately 1/3 and 2/3

labelled correctly

[2 marks]


16f. An electron is emitted from the photoelectric surface with kinetic energy 2.1 eV.Calculate the speed of the electron at the collecting plate.

[2 marks]

[2 marks]

Markschemekinetic energy at collecting plate = 0.9 «eV»

speed = «

» = 5.6 × 10 «ms »

Allow ECF from MP1

[2 marks]


√ 2×0.9×1.6×10−19

9.11×10−315 –1

17a. State what is meant by the emf of a cell.

Markschemethe work done per unit charge

in moving charge from one terminal of a cell to the other / all the way round the circuit

Award [1] for “energy per unit charge provided by the cell”/“power per unit current”

Award [1] for “potential difference across the terminals of the cell when no current isflowing”

Do not accept “potential difference across terminals of cell”

[2 marks]


Show that the resistance of the wire AC is 28 Ω.

[2 marks]

17b. Show that the resistance of the wire AC is 28 Ω.

Markschemethe resistance is proportional to length / see 0.35 AND 1«.00»

so it equals 0.35 × 80

«= 28 Ω»

[2 marks]


17c. Determine E.

Markschemecurrent leaving 12 V cell is = 0.15 «A»

OR

E = × 28

E = «0.15 × 28 =» 4.2 «V»


Allow a 1sf answer of 4 if it comes from a calculation.

Do not allow a bald answer of 4 «V»

Allow ECF from incorrect current

[2 marks]


1280

1280

[2 marks]

[2 marks]

17d. Cell X is replaced by a second cell of identical emf E but with internal resistance 2.0Ω. Comment on the length of AC for which the current in the second cell is zero.

Markschemesince the current in the cell is still zero there is no potential drop across the internalresistance

and so the length would be the same

OWTTE

[2 marks]


A negatively charged thundercloud above the Earth’s surface may be modelled by a parallel

[2 marks]

18a.

A negatively charged thundercloud above the Earth’s surface may be modelled by a parallelplate capacitor.

The lower plate of the capacitor is the Earth’s surface and the upper plate is the base of thethundercloud.

The following data are available.

Show that the capacitance of this arrangement is C = 6.6 × 10 F.

MarkschemeC = «ε

=» 8.8 × 10 ×

«C = 6.60 × 10 F»

[1 mark]


Area of thundercloud base = 1.2 × 108 m2

Charge on thundercloud base = −25 CDistance of thundercloud base from Earth's surface = 1600 mPermittivity of air = 8.8 × 10−12 F m−1

–7

A

d–12 1.2×108

1600

–7

Calculate in V, the potential difference between the thundercloud and the Earth’s

[1 mark]

18b. Calculate in V, the potential difference between the thundercloud and the Earth’ssurface.

MarkschemeV = «

=»

V = 3.8 × 10 «V»


[2 marks]


Q

C

256.6×10−7

7

18c. Calculate in J, the energy stored in the system.

[2 marks]

[2 marks]

MarkschemeALTERNATIVE 1

E = «QV =» × 25 × 3.8 × 10

E = 4.7 × 10 «J»

ALTERNATIVE 2

E = «CV =» × 6.60 × 10 × (3.8 × 10 )

E = 4.7 × 10 «J» / 4.8 × 10 «J» if rounded value of V used


Allow ECF from (b)(i)

[2 marks]


12

12

7

8

12

2 12

–7 7 2

8 8

18d.

Lightning takes place when the capacitor discharges through the air between the thundercloudand the Earth’s surface. The time constant of the system is 32 ms. A lightning strike lasts for 18ms.

Show that about –11 C of charge is delivered to the Earth’s surface. [3 marks]

MarkschemeQ = « =» 25 ×

Q = 14.2 «C»

charge delivered = Q = 25 – 14.2 = 10.8 «C»

«≈ –11 C»

Final answer must be given to at least 3 significant figures

[3 marks]


Q0e− tτ e− 18

32

18e. Calculate, in A, the average current during the discharge.

MarkschemeI «=

» ≈ 610 «A»

Accept an answer in the range 597 − 611 «A»

[1 mark]


=ΔQ

Δt

1118×10−3

State one assumption that needs to be made so that the Earth-thundercloud

[1 mark]

18f. State one assumption that needs to be made so that the Earth-thundercloudsystem may be modelled by a parallel plate capacitor.

Markschemethe base of the thundercloud must be parallel to the Earth surface

OR

the base of the thundercloud must be flat

OR

the base of the cloud must be very long «compared with the distance from the surface »

[1 mark]


19a. Bohr modified the Rutherford model by introducing the condition mvr = n . Outlinethe reason for this modification.

h2π

[1 mark]

[3 marks]

Markschemethe electrons accelerate and so radiate energy

they would therefore spiral into the nucleus/atoms would be unstable

electrons have discrete/only certain energy levels

the only orbits where electrons do not radiate are those that satisfy the Bohr condition «mvr= n

»

[3 marks]


h2π

19b. Show that the speed v of an electron in the hydrogen atom is related to the radius r ofthe orbit by the expression

where k is the Coulomb constant.

Markscheme

OR

KE = PE hence m v =

«solving for v to get answer»

Answer given – look for correct working

[1 mark]


v = √ ke2

mer

=mev2

rke2

r2

12

12 e

2 12

ke2

r

[1 mark]

19c. Using the answer in (b) and (c)(i), deduce that the radius r of the electron’s orbit in theground state of hydrogen is given by the following expression.

Markschemecombining v = with m vr = using correct substitution

«eg»

correct algebraic manipulation to gain the answer

Answer given – look for correct working

Do not allow a bald statement of the answer for MP2. Some further working eg cancellationof m or r must be shown

[2 marks]


r =h2

4π2kmee2

√ ke2

mer eh

2π

me2 r2 =ke2

merh2

4π2

19d. Calculate the electron’s orbital radius in (c)(ii).

[2 marks]

[1 mark]

Markscheme« r =

»

r = 5.3 × 10 «m»

[1 mark]


(6.63×10−34)2

4π2×8.99×109×9.11×10−31×(1.6×10−19)2

–11

19e.

Rhodium-106 ( ) decays into palladium-106 ( ) by beta minus (β ) decay. Thediagram shows some of the nuclear energy levels of rhodium-106 and palladium-106. The arrowrepresents the β decay.

Explain what may be deduced about the energy of the electron in the β decay.

Markschemethe energy released is 3.54 – 0.48 = 3.06 «MeV»

this is shared by the electron and the antineutrino

so the electron’s energy varies from 0 to 3.06 «MeV»

[3 marks]

10645Rh 106

46Pd –

–

– [3 marks]


19f. Suggest why the β decay is followed by the emission of a gamma ray photon.

Markschemethe palladium nucleus emits the photon when it decays into the ground state «from theexcited state»

[1 mark]


–

19g. Calculate the wavelength of the gamma ray photon in (d)(ii).

MarkschemePhoton energy

E = 0.48 × 10 × 1.6 × 10 = «7.68 × 10 J»

λ = « =» 2.6 × 10 «m»


Allow ECF from incorrect energy

[2 marks]

6 –19 –14

=hcE

6.63×10−34×3×108

7.68×10−14–12

[1 mark]

[2 marks]


20. In the circuit shown, the fixed resistor has a value of 3 Ω and the variable resistor canbe varied between 0 Ω and 9 Ω.

The power supply has an emf of 12 V and negligible internal resistance. What is thedifference between the maximum and minimum values of voltage V across the 3 Ω resistor?

A. 3 V

B. 6 V

C. 9 V

D. 12 V

MarkschemeC


Kirchhoff’s laws are applied to the circuit shown.

[1 mark]

21. Kirchhoff’s laws are applied to the circuit shown.

What is the equation for the dotted loop?

A. 0 = 3 I + 4I

B. 0 = 4 I − 3I

C. 6 = 2I + 3I + 4I

D. 6 = 3I + 4I

MarkschemeB


2 3

3 2

1 2 3

2 3

22. With reference to internal energy conversion and ability to be recharged, what arethe characteristics of a primary cell?

MarkschemeB


The diagram shows two current-carrying wires, P and Q, that both lie in the plane of the

[1 mark]

[1 mark]

23. The diagram shows two current-carrying wires, P and Q, that both lie in the plane of thepaper. The arrows show the conventional current direction in the wires.

The electromagnetic force on Q is in the same plane as that of the wires. What is the directionof the electromagnetic force acting on Q?

MarkschemeA


24. Two wires, X and Y, are made from the same metal. The wires are connected inseries. The radius of X is twice that of Y. The carrier drift speed in X is v and in Y it isv .

What is the value of the ratio ?

A. 0.25

B. 0.50

C. 2.00

D. 4.00

MarkschemeA


X

Y

vX

vY

[1 mark]

[1 mark]

25. The diagram shows the magnetic field surrounding two current-carrying metal wires Pand Q. The wires are parallel to each other and at right angles to the plane of the page.

What is the direction of the electron flow in P and the direction of the electron flow in Q?

MarkschemeB


Electrical resistors can be made by forming a thin film of carbon on a layer of an insulating

[1 mark]

26a.

Electrical resistors can be made by forming a thin film of carbon on a layer of an insulatingmaterial.

A carbon film resistor is made from a film of width 8.0 mm and of thickness 2.0 µm. The diagramshows the direction of charge flow through the resistor.

The resistance of the carbon film is 82 Ω. The resistivity of carbon is 4.1 x 10 Ω m.Calculate the length l of the film.

Markscheme«l = »

0.032 «m»


–5

=RAρ

82×8×10−3×2×10−6

4.1×10−5

The film must dissipate a power less than 1500 W from each square metre of its

[1 mark]

26b. The film must dissipate a power less than 1500 W from each square metre of itssurface to avoid damage. Calculate the maximum allowable current for the resistor.

Markschemepower = 1500 × 8 × 10 × 0.032 «= 0.384»

«current ≤ »

0.068 «A»

Be aware of ECF from (a)(i)

Award [1] for 4.3 «A» where candidate has not calculated area


–3

√ = √powerresistance

0.38482

26c. State why knowledge of quantities such as resistivity is useful to scientists.

[2 marks]

[1 mark]

Markschemequantities such as resistivity depend on the material

OR

they allow the selection of the correct material

OR

they allow scientists to compare properties of materials


26d. The current direction is now changed so that charge flows vertically through the film.

Deduce, without calculation, the change in the resistance.

Markschemeas area is larger and length is smaller

resistance is «very much» smaller

Award [1 max] for answers that involve a calculation

[2 marks]


26e. Draw a circuit diagram to show how you could measure the resistance of the carbon-film resistor using a potential divider arrangement to limit the potential differenceacross the resistor.

Markschemecomplete functional circuit with ammeter in series with resistor and voltmeter across it

potential divider arrangement correct

eg:


There is a proposal to power a space satellite X as it orbits the Earth. In this model, X is

[2 marks]

27a.

There is a proposal to power a space satellite X as it orbits the Earth. In this model, X isconnected by an electronically-conducting cable to another smaller satellite Y.

Satellite X orbits 6600 km from the centre of the Earth.

Mass of the Earth = 6.0 x 10 kg

Show that the orbital speed of satellite X is about 8 km s .

Markscheme« » =

7800 «m s »

Full substitution required

Must see 2+ significant figures.


24

–1

v = √ GME

r√ 6.67×10−11×6.0×1024

6600×103

–1

27b.

Satellite Y orbits closer to the centre of Earth than satellite X. Outline why

the orbital times for X and Y are different.

[2 marks]

[1 mark]

MarkschemeY has smaller orbit/orbital speed is greater so time period is less

Allow answer from appropriate equation

Allow converse argument for X


27c. satellite Y requires a propulsion system.

Markschemeto stop Y from getting ahead

to remain stationary with respect to X

otherwise will add tension to cable/damage satellite/pull X out of its orbit


The cable between the satellites cuts the magnetic field lines of the Earth at right

[2 marks]

27d. The cable between the satellites cuts the magnetic field lines of the Earth at rightangles.

Explain why satellite X becomes positively charged.

Markschemecable is a conductor and contains electrons

electrons/charges experience a force when moving in a magnetic field

use of a suitable hand rule to show that satellite Y becomes negative «so X becomespositive»

Alternative 2

cable is a conductor

so current will flow by induction flow when it moves through a B field

use of a suitable hand rule to show current to right so «X becomes positive»

Marks should be awarded from either one alternative or the other.

Do not allow discussion of positive charges moving towards X


Satellite X must release ions into the space between the satellites. Explain why

[3 marks]

27e. Satellite X must release ions into the space between the satellites. Explain whythe current in the cable will become zero unless there is a method for transferringcharge from X to Y.

Markschemeelectrons would build up at satellite Y/positive charge at X

preventing further charge flow

by electrostatic repulsion

unless a complete circuit exists


27f. The magnetic field strength of the Earth is 31 µT at the orbital radius of thesatellites. The cable is 15 km in length. Calculate the emf induced in the cable.

Markscheme«ε = Blv =» 31 x 10 x 7990 x 15000

3600 «V»

Allow 3700 «V» from v = 8000 m s .

–6

–1

[3 marks]

[2 marks]


27g.

The cable acts as a spring. Satellite Y has a mass m of 3.5 x 10 kg. Under certaincircumstances, satellite Y will perform simple harmonic motion (SHM) with a period T of 5.2 s.

Estimate the value of k in the following expression.

T =

Give an appropriate unit for your answer. Ignore the mass of the cable and any oscillation ofsatellite X.

Markschemeuse of k = « »

510

N m or kg s

Allow MP1 and MP2 for a bald correct answer

Allow 500

Allow N/m etc.


2

2π√ mk

=4π2m

T 24×π2×350

5.22

–1 –2

Describe the energy changes in the satellite Y-cable system during one cycle of the

[3 marks]

27h. Describe the energy changes in the satellite Y-cable system during one cycle of theoscillation.

MarkschemeE in the cable/system transfers to E of Y

and back again twice in each cycle

Exclusive use of gravitational potential energy negates MP1


p k

A non-uniform electric field, with field lines as shown, exists in a region where there is no

[2 marks]

28a.

A non-uniform electric field, with field lines as shown, exists in a region where there is nogravitational field. X is a point in the electric field. The field lines and X lie in the plane of thepaper.

Outline what is meant by electric field strength.

Markschemeforce per unit charge

acting on a small/test positive charge


28b. An electron is placed at X and released from rest. Draw, on the diagram, thedirection of the force acting on the electron due to the field.

Markschemehorizontally to the left

Arrow does not need to touch X

[2 marks]

[1 mark]


28c. The electron is replaced by a proton which is also released from rest at X. Compare,without calculation, the motion of the electron with the motion of the proton afterrelease. You may assume that no frictional forces act on the electron or the proton.

Markschemeproton moves to the right/they move in opposite directions

force on each is initially the same

proton accelerates less than electron initially «because mass is greater»

field is stronger on right than left «as lines closer»

proton acceleration increases «as it is moving into stronger field»

OR

electron acceleration decreases «as it is moving into weaker field»

Allow ECF from (b)

Accept converse argument for electron


An electrical circuit is used during an experiment to measure the current I in a variable resistor

[4 marks]

29a.

An electrical circuit is used during an experiment to measure the current I in a variable resistorof resistance R. The emf of the cell is e and the cell has an internal resistance r.

A graph shows the variation of with R.

Show that the gradient of the graph is equal to .

Markscheme«ε = IR + Ir»

identifies equation with y = mx + c

«hence m = »

No mark for stating data booklet equation

Do not accept working where r is ignored or ε = IR is used

OWTTE

1I

1e

= +1I

Rε

rε

1ε

[2 marks]


29b. State the value of the intercept on the R axis.

Markscheme«–» r

Allow answer in words


30. Two pulses are travelling towards each other.

What is a possible pulse shape when the pulses overlap?

[1 mark]

[1 mark]

MarkschemeA


31. The graph shows the variation of current with potential difference for a filament lamp.

What is the resistance of the filament when the potential difference across it is 6.0 V?A. 0.5 mΩB. 1.5 mΩC. 670 ΩD. 2000 Ω

MarkschemeC


32. An electron is accelerated through a potential difference of 2.5 MV. What is the changein kinetic energy of the electron?

A. 0.4µJ

B. 0.4 nJ

C. 0.4 pJ

D. 0.4 fJ

[1 mark]

[1 mark]

MarkschemeC


33. A cell is connected in series with a resistor and supplies a current of 4.0 A for a time of500 s. During this time, 1.5 kJ of energy is dissipated in the cell and 2.5 kJ of energy isdissipated in the resistor.

What is the emf of the cell?

A. 0.50 V

B. 0.75 V

C. 1.5 V

D. 2.0 V

MarkschemeD


34. An electron travelling at speed v perpendicular to a magnetic field of strength Bexperiences a force F.

What is the force acting on an alpha particle travelling at 2 v parallel to a magnetic field ofstrength 2B?

A. 0

B. 2F

C. 4F

D. 8F

MarkschemeA

[1 mark]

[1 mark]


35. The diagram shows two equal and opposite charges that are fixed in place.

At which points is the net electric field directed to the right?

A. X and Y only

B. Z and Y only

C. X and Z only

D. X, Y and Z

MarkschemeC


36. A wire has variable cross-sectional area. The cross-sectional area at Y is double that atX.

At X, the current in the wire is I and the electron drift speed is v. What is the current and theelectron drift speed at Y?

[1 mark]

[1 mark]

MarkschemeB


37. A circuit contains a cell of electromotive force (emf) 9.0 V and internal resistance 1.0 Ωtogether with a resistor of resistance 4.0 Ω as shown. The ammeter is ideal. XY is aconnecting wire.

What is the reading of the ammeter?

A. 0 A

B. 1.8 A

C. 9.0 A

D. 11 A

MarkschemeC


A positively-charged particle moves parallel to a wire that carries a current upwards.

[1 mark]

38. A positively-charged particle moves parallel to a wire that carries a current upwards.

What is the direction of the magnetic force on the particle?

A. To the left

B. To the right

C. Into the page

D. Out of the page

MarkschemeA


39. Electrons, each with a charge e, move with speed v along a metal wire. The electriccurrent in the wire is I.

Plane P is perpendicular to the wire. How many electrons pass through plane P in eachsecond?

A.

B.

C.

D.

MarkschemeD

eI

veI

Ive

Ie

[1 mark]

[1 mark]


40. Positive charge is uniformly distributed on a semi-circular plastic rod. What is thedirection of the electric field strength at point S?

MarkschemeB


41. The diagram shows the path of a particle in a region of uniform magnetic field. The fieldis directed into the plane of the page.

This particle could be

A. an alpha particle.

B. a beta particle.

C. a photon.

D. a neutron.

MarkschemeA

[1 mark]

[1 mark]


42a.

A heater in an electric shower has a power of 8.5 kW when connected to a 240 V electricalsupply. It is connected to the electrical supply by a copper cable.

The following data are available:

Length of cable = 10 mCross-sectional area of cable = 6.0 mmResistivity of copper = 1.7 × 10 Ω m

Calculate the current in the copper cable.

MarkschemeI «= » =35«A»


2

–8

8.5×103

240

42b. Calculate the resistance of the cable.

[1 mark]

[2 marks]

MarkschemeR =

= 0.028 «Ω»

Allow missed powers of 10 for MP1.


1.7×10−8×106.0×10−6

42c. Explain, in terms of electrons, what happens to the resistance of the cable as thetemperature of the cable increases.

Markscheme«as temperature increases» there is greater vibration of the metal atoms/lattice/lattice ions

OR

increased collisions of electrons

drift velocity decreases «so current decreases»

«as V constant so» R increases

Award [0] for suggestions that the speed of electrons increases so resistance decreases.


The heater changes the temperature of the water by 35 K. The specific heat capacity

[3 marks]

Printed for Jyvaskylan Lyseon lukio

© International Baccalaureate Organization 2019 International Baccalaureate® - Baccalauréat International® - Bachillerato Internacional®

42d. The heater changes the temperature of the water by 35 K. The specific heat capacityof water is 4200 J kg K .

Determine the rate at which water flows through the shower. State an appropriate unit for youranswer.

Markschemerecognition that power = flow rate × cΔ T

flow rate « »

= 0.058 «kg s »

kg s / g s / l s / ml s / m s

Allow MP4 if a bald flow rate unit is stated. Do not allow imperial units.


–1 –1

= powercΔT

= 8.5×103

4200×35

–1

−1 −1 −1 −1 3 −1

[4 marks]

Topic 5 Revision [142 marks] - Peda.net · Topic 5 Revision [142 marks] 1. Three resistors are...

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