Question 1
A metal wheel consists of an axle A, eight spokes and a rim, as shown in Fig. 1.1.
Point X is on the rim at the end of one of the spokes.
The rim has a radius of \(0.85\,\mathrm{m}\).
The wheel is rotating clockwise with an angular speed of \(140\,\mathrm{rad\,s^{-1}}\).
(a) For point X, determine:
(i) the speed. (2 marks)
speed = __________________________ \( \mathrm{m\,s^{-1}} \)
(ii) the centripetal acceleration. (2 marks)
acceleration = __________________________ \( \mathrm{m\,s^{-2}} \)
(b) There is a uniform magnetic field of flux density \(0.18\,\mathrm{T}\) into the plane of the page.
(i) State Lenz’s law of electromagnetic induction. (2 marks)
______________________________________________
______________________________________________
(ii) Show that the time taken for point X to complete one revolution is \(45\,\mathrm{ms}\). (1 mark)
______________________________________________
(iii) Calculate the magnetic flux cut by spoke AX during one revolution of the wheel. Give a unit with your answer. (3 marks)
magnetic flux = __________________________ unit __________
(iv) Determine the magnitude of the electromotive force (e.m.f.) induced across spoke AX. (2 marks)
induced e.m.f. = __________________________ \( \mathrm{V} \)
(v) Use Lenz’s law to explain whether the potential is higher at end A or end X of the spoke. (1 mark)
______________________________________________
______________________________________________
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702 Paper 4):
• 12.2: Centripetal acceleration — part (a)(ii)
• 20.5: Electromagnetic induction — parts (b)(i), (b)(iii), (b)(iv) and (b)(v)
▶️ Answer/Explanation
(a)(i)
For uniform circular motion, the linear speed is related to angular speed by
\( v=r\omega \)
Substituting \(r=0.85\,\mathrm{m}\) and \(\omega=140\,\mathrm{rad\,s^{-1}}\),
\( v=(0.85)(140) \)
\( v=119\,\mathrm{m\,s^{-1}} \)
To an appropriate number of significant figures,
Answer: \( \boxed{v=120\,\mathrm{m\,s^{-1}}} \)
(a)(ii)
The centripetal acceleration is
\( a=r\omega^2 \)
Therefore,
\( a=(0.85)(140)^2 \)
\( a=1.666\times10^4\,\mathrm{m\,s^{-2}} \)
Answer: \( \boxed{a=1.7\times10^4\,\mathrm{m\,s^{-2}}} \)
(b)(i)
Lenz’s law states that the direction of the induced e.m.f. and induced current is such that the magnetic effect produced opposes the change that causes the induction.
Answer: \( \boxed{\text{The induced e.m.f. is in a direction that opposes the change producing it.}} \)
(b)(ii)
The period of rotation is
\( T=\frac{2\pi}{\omega} \)
\( T=\frac{2\pi}{140} \)
\( T=0.0449\,\mathrm{s} \)
Thus,
Answer: \( \boxed{T=45\,\mathrm{ms}} \)
(b)(iii)
The magnetic flux is
\( \Phi=BA \)
The area swept out by spoke AX in one revolution is
\( A=\pi r^2 \)
Therefore,
\( \Phi=(0.18)\pi(0.85)^2 \)
\( \Phi=0.408\,\mathrm{Wb} \)
Answer: \( \boxed{\Phi=0.41\,\mathrm{Wb}} \)
(b)(iv)
The magnitude of the induced e.m.f. is given by Faraday’s law:
\( E=\frac{\Delta\Phi}{\Delta t} \)
Using the magnetic flux cut during one revolution and \(T=0.045\,\mathrm{s}\),
\( E=\frac{0.41}{0.045} \)
\( E=9.1\,\mathrm{V} \)
Answer: \( \boxed{E=9.1\,\mathrm{V}} \)
(b)(v)
As the spoke rotates clockwise in the magnetic field, the magnetic force on the charges in the spoke must act anticlockwise so as to oppose the change in magnetic flux.
This causes conventional current to be driven from A towards X.
Therefore, X is at the higher potential.
Answer: \( \boxed{\text{X is at the higher potential than A.}} \)
Question 2
The Sun may be considered as a uniform sphere with a mass of \(1.99\times10^{30}\,\mathrm{kg}\) and a surface temperature of \(5780\,\mathrm{K}\).
A probe with a mass of \(2.63\,\mathrm{kg}\) moves in a straight line towards the Sun.
When it is at a distance \(x\) from the centre of the Sun, the probe measures the gravitational field strength \(g\) and the solar radiant flux \(F\) of radiation from the Sun.
(a) Define gravitational field. (1 mark)
______________________________________________
______________________________________________
(b) For the position of the probe where \(x=1.47\times10^{11}\,\mathrm{m}\):
(i) calculate \(g\). (2 marks)
\(g=\) __________________________ \( \mathrm{N\,kg^{-1}} \)
(ii) determine the gravitational potential energy \(E_{\mathrm{P}}\) of the probe. (2 marks)
\(E_{\mathrm{P}}=\) __________________________ \( \mathrm{J} \)
(c) (i) Show that, for any particular value of \(x\), the numerical values of \(g\) and \(F\) are related by
\( g=\frac{4\pi GM}{L}F \)
where \(M\) is the mass of the Sun, \(L\) is the luminosity of the Sun and \(G\) is the gravitational constant. (3 marks)
______________________________________________
______________________________________________
______________________________________________
(ii) Fig. 2.1 shows the variation of \(g\) with \(F\).
Determine a value for the luminosity \(L\) of the Sun. Give a unit with your answer. (2 marks)
\(L=\) __________________________ unit __________
(iii) Use your answer in (c)(ii) to determine the radius \(r\) of the Sun. (2 marks)
\(r=\) __________________________ \( \mathrm{m} \)
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702 Paper 4):
• 13.3: Gravitational field of a point mass — parts (b)(i), (c)(i), (c)(ii) and (c)(iii)
• 13.4: Gravitational potential — part (b)(ii)
• 14.1: Thermal radiation — part (c)(iii)
▶️ Answer/Explanation
(a)
A gravitational field is a region in which a mass experiences a gravitational force.
Gravitational field strength is the gravitational force per unit mass:
\( g=\frac{F_{\mathrm{G}}}{m} \)
Answer: \( \boxed{\text{gravitational force per unit mass}} \)
(b)(i)
For a point outside a spherical mass,
\( g=\frac{GM}{x^2} \)
Substituting \(G=6.67\times10^{-11}\,\mathrm{N\,m^2\,kg^{-2}}\), \(M=1.99\times10^{30}\,\mathrm{kg}\) and \(x=1.47\times10^{11}\,\mathrm{m}\),
\( g=\frac{(6.67\times10^{-11})(1.99\times10^{30})}{(1.47\times10^{11})^2} \)
\( g=6.14\times10^{-3}\,\mathrm{N\,kg^{-1}} \)
Answer: \( \boxed{g=6.14\times10^{-3}\,\mathrm{N\,kg^{-1}}} \)
(b)(ii)
The gravitational potential energy of a mass \(m\) at distance \(x\) from the centre of the Sun is
\( E_{\mathrm{P}}=-\frac{GMm}{x} \)
Therefore,
\( E_{\mathrm{P}}=-\frac{(6.67\times10^{-11})(1.99\times10^{30})(2.63)}{1.47\times10^{11}} \)
\( E_{\mathrm{P}}=-2.37\times10^9\,\mathrm{J} \)
Answer: \( \boxed{E_{\mathrm{P}}=-2.37\times10^9\,\mathrm{J}} \)
(c)(i)
The gravitational field strength at distance \(x\) is
\( g=\frac{GM}{x^2} \)
The radiant flux from the Sun is
\( F=\frac{L}{4\pi x^2} \)
Rearranging the second equation gives
\( x^2=\frac{L}{4\pi F} \)
From the gravitational field equation,
\( x^2=\frac{GM}{g} \)
Equating the two expressions for \(x^2\),
\( \frac{GM}{g}=\frac{L}{4\pi F} \)
Rearranging,
\( \boxed{g=\frac{4\pi GM}{L}F} \)
Answer: \( \boxed{g=\frac{4\pi GM}{L}F} \)
(c)(ii)
From the graph, a suitable pair of values is approximately
\( g=8.0\times10^{-3}\,\mathrm{N\,kg^{-1}} \)
and
\( F=1.83\times10^3\,\mathrm{W\,m^{-2}} \)
Using
\( g=\frac{4\pi GM}{L}F \)
gives
\( L=\frac{4\pi GMF}{g} \)
\( L=\frac{4\pi(6.67\times10^{-11})(1.99\times10^{30})(1.83\times10^3)}{8.0\times10^{-3}} \)
\( L=3.8\times10^{26}\,\mathrm{W} \)
Answer: \( \boxed{L=3.8\times10^{26}\,\mathrm{W}} \)
(c)(iii)
The luminosity of a spherical black-body source is given by the Stefan-Boltzmann law:
\( L=4\pi r^2\sigma T^4 \)
Hence,
\( 3.8\times10^{26}=4\pi(5.67\times10^{-8})(5780)^4r^2 \)
Rearranging,
\( r=\sqrt{\frac{3.8\times10^{26}}{4\pi(5.67\times10^{-8})(5780)^4}} \)
\( r=6.9\times10^8\,\mathrm{m} \)
Answer: \( \boxed{r=6.9\times10^8\,\mathrm{m}} \)
Question 3
(a) Define specific latent heat. (2 marks)
______________________
(b) A dish containing \(7.2\times10^{-5}\,\mathrm{m^3}\) of a substance rests on a laboratory bench. The substance is initially a liquid of density \(710\,\mathrm{kg\,m^{-3}}\). Atmospheric pressure is \(1.0\times10^5\,\mathrm{Pa}\).
The liquid is heated at its boiling point so that it completely vaporises. The increase in the internal energy of the substance during this process is \(17.6\,\mathrm{kJ}\). The final volume of the vapour is \(0.017\,\mathrm{m^3}\).
(i) Show that the magnitude of the work done on the substance when it vaporises is \(1.7\,\mathrm{kJ}\). (2 marks)
______________________________________________
______________________________________________
(ii) Use the information in (b)(i) to calculate the thermal energy \(Q\), in \(\mathrm{kJ}\), supplied to the substance to cause it to vaporise. (2 marks)
\(Q=\) __________________________ \( \mathrm{kJ} \)
(iii) Use your answer in (b)(ii) to determine a value for the specific latent heat of vaporisation \(L_{\mathrm{V}}\), in \(\mathrm{kJ\,kg^{-1}}\), of the substance. (2 marks)
\(L_{\mathrm{V}}=\) __________________________ \( \mathrm{kJ\,kg^{-1}} \)
(c) The substance in (b) has a specific latent heat of fusion \(L_{\mathrm{F}}\).
Suggest and explain whether \(L_{\mathrm{F}}\) is likely to be less than, the same as, or greater than the answer in (b)(iii). (3 marks)
_____________________
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702 Paper 4):
• 16.1: Internal energy — parts (b)(i), (b)(ii) and (c)
• 16.2: The first law of thermodynamics — parts (b)(i), (b)(ii) and (c)
▶️ Answer/Explanation
(a)
Specific latent heat is the thermal energy required per unit mass to change the state of a substance without changing its temperature.
It can be expressed by
\( L=\frac{Q}{m} \)
Answer: \( \boxed{\text{thermal energy required per unit mass to change state at constant temperature}} \)
(b)(i)
The magnitude of the work done when the substance expands against atmospheric pressure is
\( |W|=p\Delta V \)
The initial volume is \(7.2\times10^{-5}\,\mathrm{m^3}\) and the final volume is \(0.017\,\mathrm{m^3}\).
Hence,
\( \Delta V=0.017-7.2\times10^{-5} \)
\( \Delta V\approx0.017\,\mathrm{m^3} \)
Therefore,
\( |W|=(1.0\times10^5)(0.017) \)
\( |W|=1700\,\mathrm{J} \)
Thus,
Answer: \( \boxed{|W|=1.7\,\mathrm{kJ}} \)
(b)(ii)
Using the first law of thermodynamics, with \(W\) defined as the work done on the substance,
\( \Delta U=Q+W \)
The substance expands, so the work done on it is negative:
\( W=-1.7\,\mathrm{kJ} \)
Therefore,
\( 17.6=Q-1.7 \)
\( Q=17.6+1.7 \)
\( Q=19.3\,\mathrm{kJ} \)
Answer: \( \boxed{Q=19.3\,\mathrm{kJ}} \)
(b)(iii)
First determine the mass of the substance:
\( m=\rho V \)
\( m=(710)(7.2\times10^{-5}) \)
\( m=0.0511\,\mathrm{kg} \)
The specific latent heat of vaporisation is
\( L_{\mathrm{V}}=\frac{Q}{m} \)
\( L_{\mathrm{V}}=\frac{19.3}{0.0511} \)
\( L_{\mathrm{V}}\approx378\,\mathrm{kJ\,kg^{-1}} \)
To an appropriate number of significant figures,
Answer: \( \boxed{L_{\mathrm{V}}=380\,\mathrm{kJ\,kg^{-1}}} \)
(c)
The specific latent heat of fusion is likely to be less than the specific latent heat of vaporisation.
Fusion involves a much smaller change in volume than vaporisation, so there is a much smaller change in the spacing between the particles.
Therefore, the increase in internal energy associated with the change of state is smaller. Also, the work done by the substance during fusion is negligible compared with that during vaporisation.
Hence, less thermal energy is required per unit mass for fusion than for vaporisation.
Answer: \( \boxed{L_{\mathrm{F}}<L_{\mathrm{V}}} \)
Question 4
(a) State three of the basic assumptions of the kinetic theory of gases. (3 marks)
1. ______________________________________________
2. ______________________________________________
3. ______________________________________________
(b) Explain how molecular movement causes the pressure exerted by a gas. (3 marks)
______________________________________________
______________________________________________
______________________________________________
______________________________________________
(c) Fig. 4.1 shows the variation with thermodynamic temperature \(T\) of the mean-square speeds \(\langle c^2\rangle\) for two gases X and Y.
Fig. 4.2 shows the variation with \(T\) of the product \(pV\) for samples of the two gases, where \(p\) is the pressure of the gas and \(V\) is the volume of the gas.
State three conclusions about the gases and their samples that may be drawn from Fig. 4.1 and Fig. 4.2. The conclusions may be qualitative or quantitative. Use the space below for any working that you need. (3 marks)
1. ______________________________________________
______________________________________________
2. ______________________________________________
______________________________________________
3. ______________________________________________
______________________________________________
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702 Paper 4):
• 15.2: Equation of state — part (c), Fig. 4.2
▶️ Answer/Explanation
(a)
Any three of the following basic assumptions are acceptable:
• Gas molecules are in constant random motion.
• All collisions between molecules, and between molecules and the walls, are perfectly elastic.
• There are no forces between molecules except during collisions.
• The volume of the molecules is negligible compared with the volume of the gas.
• Collisions involving molecules are instantaneous.
Answer: Any three of the above assumptions.
(b)
Gas molecules are in continuous random motion and collide with the walls of the container.
When a molecule collides with a wall, its momentum changes because its velocity changes.
The change in momentum occurs because the wall exerts a force on the molecule, so by Newton’s third law the molecule exerts a force on the wall.
A very large number of molecules collide with the wall, producing a resultant force over the area of the wall. Hence, a pressure is exerted on the wall.
Answer: \( \boxed{\text{Molecular collisions with the walls cause a change in momentum and hence a force; the force per unit area produces pressure.}} \)
(c)
From Fig. 4.1, the mean-square speed of both gases increases linearly with thermodynamic temperature.
For an ideal gas,
\( \frac{1}{2}m\langle c^2\rangle=\frac{3}{2}kT \)
so
\( \langle c^2\rangle=\frac{3kT}{m} \)
Therefore, the gradient of the \(\langle c^2\rangle\)-against-\(T\) graph is inversely proportional to the mass of one molecule.
Since gas X has the greater gradient, molecules of gas X have a smaller mass than molecules of gas Y.
From the graphs, the gradient for X is approximately twice that for Y, so the mass of one molecule of gas Y is approximately twice the mass of one molecule of gas X.
From Fig. 4.2, \(pV\) is directly proportional to \(T\) for both samples, so both gases behave as ideal gases.
The \(pV\)-against-\(T\) graph for sample Y has a gradient approximately three times that for sample X. Since
\( pV=NkT \)
the number of molecules in sample Y is approximately three times the number of molecules in sample X.
Possible conclusions:
1. Both gases behave as ideal gases.
2. The mass of one molecule of gas Y is approximately twice the mass of one molecule of gas X.
3. Sample Y contains approximately three times the amount of gas, or number of molecules, as sample X.
Answer: Any three valid conclusions supported by the two graphs.
Question 5
Fig. 5.1 shows a pendulum consisting of a metal sphere suspended by a thin string.
The sphere undergoes small oscillations about its equilibrium position. The oscillations may be considered to be simple harmonic.
Fig. 5.2 shows the variation with time \(t\) of the displacement \(x\) of the sphere from its equilibrium position.
(a) On Fig. 5.1, draw an arrow, from the centre of the sphere, to represent the direction of the resultant force on the sphere when it is in the position shown. (1 mark)
(b) The mass of the sphere is \(0.15\,\mathrm{kg}\).
(i) State the amplitude of the oscillations. (1 mark)
amplitude = __________________________________________ \( \mathrm{m} \)
(ii) Determine the angular frequency of the oscillations. (2 marks)
angular frequency = ______________________________ \( \mathrm{rad\,s^{-1}} \)
(iii) Calculate the total energy of the oscillations. (2 marks)
total energy = __________________________________ \( \mathrm{J} \)
(c) On Fig. 5.3, sketch the variation with \(x\) of the kinetic energy \(E_{\mathrm{k}}\) of the sphere. (3 marks)
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702 Paper 4):
• 17.2: Energy in simple harmonic motion — parts (b)(iii) and (c)
▶️ Answer/Explanation
(a)
The resultant force is the restoring force. It acts towards the equilibrium position and is perpendicular to the string at the position shown.
The arrow should therefore be drawn from the centre of the sphere, perpendicular to the string and pointing left and down.
Answer: \( \boxed{\text{Arrow perpendicular to the string, pointing left and down}} \)
(b)(i)
Amplitude is the maximum displacement from the equilibrium position.
From Fig. 5.2,
\( A=0.016\,\mathrm{m} \)
Answer: \( \boxed{0.016\,\mathrm{m}} \)
(b)(ii)
From Fig. 5.2, the time between two successive maxima is
\( T=0.40\,\mathrm{s} \)
The relationship between angular frequency and period is
\( \omega=\frac{2\pi}{T} \)
\( \omega=\frac{2\pi}{0.40} \)
\( \omega=15.7\,\mathrm{rad\,s^{-1}} \)
Therefore, to an appropriate number of significant figures,
Answer: \( \boxed{16\,\mathrm{rad\,s^{-1}}} \)
(b)(iii)
For a system undergoing simple harmonic motion, the total energy is
\( E=\frac{1}{2}m\omega^2A^2 \)
Substituting \(m=0.15\,\mathrm{kg}\), \(\omega=15.7\,\mathrm{rad\,s^{-1}}\) and \(A=0.016\,\mathrm{m}\),
\( E=\frac{1}{2}(0.15)(15.7)^2(0.016)^2 \)
\( E=4.7\times10^{-3}\,\mathrm{J} \)
Answer: \( \boxed{4.7\times10^{-3}\,\mathrm{J}} \)
(c)
For SHM, the total energy is constant and is shared between kinetic and potential energy.
The kinetic energy is greatest at the equilibrium position, where \(x=0\), because the speed is maximum.
At the extreme positions, \(x=-A\) and \(x=+A\), the speed is zero. Therefore, the kinetic energy is zero.
The kinetic energy is given by
\( E_{\mathrm{k}}=E-\frac{1}{2}m\omega^2x^2 \)
Hence, the graph is a symmetrical inverted parabola.
It should:
• start at \(E_{\mathrm{k}}=0\) when \(x=-0.016\,\mathrm{m}\)
• reach a maximum of \(4.7\times10^{-3}\,\mathrm{J}\) at \(x=0\)
• return to \(E_{\mathrm{k}}=0\) when \(x=+0.016\,\mathrm{m}\)
Answer: \( \boxed{\text{Symmetrical inverted-parabola curve with maximum }E_{\mathrm{k}}=4.7\times10^{-3}\,\mathrm{J}\text{ at }x=0} \)
Question 6
(a) State Coulomb’s law. (2 marks)
______________________________________________
______________________________________________
______________________________________________
(b) Fig. 6.1 shows an isolated hollow conducting sphere that is positively charged.
On Fig. 6.1, draw field lines to represent the electric field outside the sphere. (3 marks)
______________________________________________
(c) Fig. 6.2 shows the variation of the electric field strength \(E\) with distance \(x\) from the centre of the sphere in (b).
(i) Determine the radius, in \(\mathrm{cm}\), of the sphere. (1 mark)
radius = __________________________ \( \mathrm{cm} \)
(ii) Calculate the charge on the sphere. (3 marks)
charge = __________________________ \( \mathrm{C} \)
(iii) Suggest an explanation for the fact that the electric field inside the sphere is zero. (1 mark)
______________________________________________
______________________________________________
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702 Paper 4):
• 18.2: Uniform electric fields — part (c)(iii)
• 18.3: Electric force between point charges — parts (a) and (c)(ii)
▶️ Answer/Explanation
(a)
Coulomb’s law states that the electric force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of their separation.
Thus,
\( F\propto\frac{Q_1Q_2}{r^2} \)
or, in free space,
\( F=\frac{Q_1Q_2}{4\pi\varepsilon_0r^2} \)
Answer: \( \boxed{\text{Force is proportional to the product of the charges and inversely proportional to the square of their separation.}} \)
(b)
The electric field outside a positively charged spherical conductor is radial.
The diagram should show:
• at least four straight radial field lines from the surface of the sphere
• the field lines approximately equally spaced around the sphere
• arrows on the field lines pointing away from the positively charged sphere
Answer: \( \boxed{\text{Straight radial field lines directed outwards from the sphere.}} \)
(c)(i)
Inside the conducting sphere, the electric field is zero. From Fig. 6.2, the electric field begins at the surface when
\( x=3.2\,\mathrm{cm} \)
Hence, the radius of the sphere is
Answer: \( \boxed{r=3.2\,\mathrm{cm}} \)
(c)(ii)
Outside a charged spherical conductor, its electric field is the same as that of a point charge at its centre:
\( E=\frac{Q}{4\pi\varepsilon_0x^2} \)
Rearranging,
\( Q=4\pi\varepsilon_0Ex^2 \)
From Fig. 6.2, at the surface approximately
\( E=2.2\times10^5\,\mathrm{N\,C^{-1}} \)
and
\( x=3.2\,\mathrm{cm}=0.032\,\mathrm{m} \)
Therefore,
\( Q=4\pi(8.85\times10^{-12})(2.2\times10^5)(0.032)^2 \)
\( Q=2.5\times10^{-8}\,\mathrm{C} \)
Answer: \( \boxed{Q=2.5\times10^{-8}\,\mathrm{C}} \)
(c)(iii)
The positive charge is distributed over the outer surface of the conducting sphere.
If an electric field existed inside the conductor, the free charges would experience a force and move. They redistribute themselves until electrostatic equilibrium is reached and the resultant electric field inside is zero.
Answer: \( \boxed{\text{Charges redistribute over the surface until the resultant electric field inside the conductor is zero.}} \)
Question 7
(a) Define the capacitance of a parallel-plate capacitor. (2 marks)
______________________________________________
______________________________________________
(b) An initially uncharged capacitor X, of capacitance \(C\), is gradually charged so that the final potential difference (p.d.) between its plates is \(V\) and the final charge is \(Q\).
(i) On Fig. 7.1, sketch the variation of charge with p.d. for capacitor X as the p.d. increases from \(0\) to \(V\). (2 marks)
(ii) Determine an expression, in terms of \(Q\) and \(V\), for the work \(W\) done on capacitor X during charging. Explain your reasoning. (2 marks)
\(W=\) __________________________________________
(c) Another capacitor Y is initially uncharged. The fully charged capacitor X in (b) is now connected to capacitor Y, as shown in Fig. 7.2.
The capacitance of capacitor Y is \(3C\).
(i) Complete Table 7.1 to show expressions, in terms of \(Q\) and \(V\), for the final p.d.s across, and the final charges on, the two capacitors. Use the space below for any working that you need. (3 marks)
| X | Y | |
| final p.d. | ________________ | ________________ |
| final charge | ________________ | ________________ |
(ii) State whether the total energy stored in the two capacitors is less than, the same as, or greater than the energy initially stored in capacitor X. (1 mark)
______________________________________________
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702 Paper 4):
• 19.2: Energy stored in a capacitor — parts (b)(ii) and (c)(ii)
▶️ Answer/Explanation
(a)
Capacitance is the ratio of the charge \(Q\) stored on one plate of a capacitor to the potential difference \(V\) between its plates.
\( C=\frac{Q}{V} \)
Answer: \( \boxed{\text{capacitance}=\frac{\text{charge}}{\text{potential difference}}} \)
(b)(i)
For a capacitor of constant capacitance,
\( Q=CV \)
Therefore, charge is directly proportional to p.d.
The graph is a straight line through the origin, ending at the point \((V,Q)\).
Answer: \( \boxed{\text{Straight line through the origin with positive gradient, ending at }(V,Q).} \)
(b)(ii)
The work done in charging the capacitor is equal to the area under the \(Q\)-against-\(V\) graph.
The graph is a triangle with base \(V\) and height \(Q\).
Hence,
\( W=\frac{1}{2}QV \)
Using \(Q=CV\), this is also
\( W=\frac{1}{2}CV^2 \)
Answer: \( \boxed{W=\frac{1}{2}QV} \)
(c)(i)
When the fully charged capacitor X is connected to the initially uncharged capacitor Y, the capacitors are connected in parallel.
Therefore, they have the same final p.d. Let this final p.d. be \(V_{\mathrm{f}}\).
The total charge is conserved:
\( Q_X+Q_Y=Q \)
Since \(C_X=C\) and \(C_Y=3C\),
\( Q=C V_{\mathrm{f}}+3C V_{\mathrm{f}} \)
\( Q=4CV_{\mathrm{f}} \)
Hence,
\( V_{\mathrm{f}}=\frac{Q}{4C} \)
Initially, \(Q=CV\), so
\( V_{\mathrm{f}}=\frac{V}{4} \)
For capacitor X,
\( Q_X=C\left(\frac{V}{4}\right)=\frac{Q}{4} \)
For capacitor Y,
\( Q_Y=3C\left(\frac{V}{4}\right)=\frac{3Q}{4} \)
| X | Y | |
| final p.d. | \( \frac{V}{4} \) | \( \frac{V}{4} \) |
| final charge | \( \frac{Q}{4} \) | \( \frac{3Q}{4} \) |
Answer: \( \boxed{V_X=V_Y=\frac{V}{4},\quad Q_X=\frac{Q}{4},\quad Q_Y=\frac{3Q}{4}} \)
(c)(ii)
Initially, the energy stored in capacitor X is
\( E_{\mathrm{i}}=\frac{1}{2}CV^2 \)
After connection, the total capacitance is
\( C_{\mathrm{total}}=C+3C=4C \)
and the final p.d. is \(V/4\).
Therefore,
\( E_{\mathrm{f}}=\frac{1}{2}(4C)\left(\frac{V}{4}\right)^2 \)
\( E_{\mathrm{f}}=\frac{1}{8}CV^2 \)
Since
\( \frac{1}{8}CV^2<\frac{1}{2}CV^2 \)
the total energy stored after connection is less than the initial energy.
Answer: \( \boxed{\text{less than}} \)
Question 8
(a) State what is meant by the frequency of an alternating current. [1]
…………………………………………………………………………………………………………….
(b) An alternating current \(I\) in a resistor of resistance \(680\,\Omega\) varies with time \(t\) according to
\( I=3.5\sin(40\pi t) \)
where \(I\) is in A and \(t\) is in s.
(i) Show that the period of the alternating current is \(50\,\mathrm{ms}\). [1]
…………………………………………………………………………………………………………….
(ii) On Fig. 8.1, sketch the variation of \(I\) with \(t\) between \(t=0\) and \(t=100\,\mathrm{ms}\). [3]
(iii) Determine the root-mean-square (r.m.s.) current in the resistor. [1]
r.m.s. current \(=\) …………………………………. A
(c) Use data from (b), including your answer in (b)(iii), to show by calculation that the mean power in the \(680\,\Omega\) resistor is half the peak power. [3]
…………………………………………………………………………………………………………….
…………………………………………………………………………………………………………….
…………………………………………………………………………………………………………….
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702 Paper 4):
• 21.1 Characteristics of alternating currents
▶️ Answer/Explanation
(a)
Frequency is the number of complete cycles of the alternating current per unit time.
Answer: \( \boxed{\text{number of cycles per unit time}} \)
(b)(i)
Compare
\( I=3.5\sin(40\pi t) \)
with the general form
\( I=I_0\sin(\omega t) \)
Therefore,
\( \omega=40\pi\,\mathrm{rad\,s^{-1}} \)
Using \(T=\frac{2\pi}{\omega}\),
\( T=\frac{2\pi}{40\pi}=0.050\,\mathrm{s} \)
\( T=50\,\mathrm{ms} \)
Answer: \( \boxed{T=50\,\mathrm{ms}} \)
(b)(ii)
The current is
\( I=3.5\sin(40\pi t) \)
At \(t=0\), \(I=0\), and the current initially increases.
The peak current is \(+3.5\,\mathrm{A}\) and the minimum current is \(-3.5\,\mathrm{A}\).
Since the period is \(50\,\mathrm{ms}\), the interval from \(0\) to \(100\,\mathrm{ms}\) contains two complete cycles.
Graph requirements: sinusoidal curve starting at \((0,0)\), initially increasing, showing two complete cycles, with peaks at \(+3.5\,\mathrm{A}\) and troughs at \(-3.5\,\mathrm{A}\).
(b)(iii)
For a sinusoidal alternating current,
\( I_{\mathrm{r.m.s.}}=\frac{I_0}{\sqrt{2}} \)
\( I_{\mathrm{r.m.s.}}=\frac{3.5}{\sqrt{2}} \)
\( I_{\mathrm{r.m.s.}}=2.47\,\mathrm{A}\approx2.5\,\mathrm{A} \)
Answer: \( \boxed{2.5\,\mathrm{A}} \)
(c)
The power dissipated in the resistor is
\( P=I^2R \)
Peak power:
\( P_{\mathrm{peak}}=I_0^2R \)
\( P_{\mathrm{peak}}=(3.5)^2(680) \)
\( P_{\mathrm{peak}}=8330\,\mathrm{W} \)
Mean power:
\( P_{\mathrm{mean}}=I_{\mathrm{r.m.s.}}^2R \)
\( P_{\mathrm{mean}}=(2.47)^2(680) \)
\( P_{\mathrm{mean}}\approx4170\,\mathrm{W} \)
Now,
\( \frac{P_{\mathrm{peak}}}{2}=\frac{8330}{2}=4165\,\mathrm{W} \)
This agrees with the calculated mean power, allowing for rounding.
Therefore, \( \boxed{P_{\mathrm{mean}}=\frac{1}{2}P_{\mathrm{peak}}} \).
Question 9
Electrons in a vacuum are accelerated from rest through a potential difference (p.d.) \(V\) to form a beam. The electrons each have mass \(m\) and charge \(q\).
The beam is incident on a graphite crystal that acts as a diffraction grating. After passing through the crystal, the beam reaches a fluorescent screen. An interference pattern is observed on this screen.
(a) Explain what this observation shows about the nature of electrons. (1 mark)
______________________________________________
______________________________________________
(b) Determine an expression, in terms of \(m\), \(q\) and \(V\), for the momentum \(p\) of an electron in the beam. (3 marks)
\(p=\) __________________________________________
(c) The p.d. through which the electrons are accelerated is now increased to a greater value.
Describe and explain the effect of this change on the interference pattern observed. (2 marks)
______________________________________________
______________________________________________
______________________________________________
(d) The electrons are now accelerated through different values of \(V\), resulting in pairs of corresponding values for \(p\) and the de Broglie wavelength \(\lambda\).
(i) On Fig. 9.1, sketch the variation of \(p\) with \(\frac{1}{\lambda}\). (2 marks)
(ii) State the name of the quantity represented by the gradient of the line in Fig. 9.1. (1 mark)
______________________________________________
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702 Paper 4):
• 22.3: Wave-particle duality — de Broglie wavelength, momentum and Planck constant in parts (b) and (d)
▶️ Answer/Explanation
(a)
Diffraction and interference are characteristic of wave behaviour.
Answer: \( \boxed{\text{Electrons can behave like waves.}} \)
(b)
The electrons are accelerated from rest through a p.d. \(V\), so the electrical energy gained is equal to the kinetic energy:
\( qV=\frac{1}{2}mv^2 \)
Hence,
\( v=\sqrt{\frac{2qV}{m}} \)
Since \(p=mv\),
\( p=m\sqrt{\frac{2qV}{m}} \)
Therefore,
Answer: \( \boxed{p=\sqrt{2mqV}} \)
(c)
Increasing \(V\) increases the kinetic energy and therefore increases the momentum \(p\) of the electrons.
Using the de Broglie relation,
\( \lambda=\frac{h}{p} \)
so the greater momentum gives a smaller de Broglie wavelength.
Therefore, the interference fringes become closer together.
Answer: \( \boxed{\text{The fringes become closer together because the de Broglie wavelength decreases.}} \)
(d)(i)
The de Broglie equation is
\( \lambda=\frac{h}{p} \)
Rearranging gives
\( p=h\left(\frac{1}{\lambda}\right) \)
Therefore, \(p\) is directly proportional to \(\frac{1}{\lambda}\).
Graph requirements: a straight line with positive gradient passing through the origin.
(d)(ii)
From
\( p=h\left(\frac{1}{\lambda}\right) \)
the gradient of the graph is \(h\).
Answer: \( \boxed{\text{Planck constant}} \)
Question 10
(a) Radioactive decay is both random and spontaneous.
(i) State what is meant by random. (1 mark)
______________________________________________
(ii) State what is meant by spontaneous. (1 mark)
______________________________________________
______________________________________________
(iii) State one piece of evidence for the random nature of decay. (1 mark)
______________________________________________
______________________________________________
(b)
(i) Describe the differences between nuclear fission and nuclear fusion. (3 marks)
______________________________________________
______________________________________________
(ii) Explain, with reference to the variation of binding energy per nucleon with nucleon number, why the processes of nuclear fission and nuclear fusion both result in a release of energy. (2 marks)
______________________________________________
______________________________________________
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702 Paper 4):
• 23.1: Mass defect and nuclear binding energy — parts (b)(i) and (b)(ii)
▶️ Answer/Explanation
(a)(i)
Radioactive decay is random because it is not possible to predict when a particular nucleus will decay or which nucleus will decay next.
Answer: \( \boxed{\text{It is impossible to predict when a particular nucleus will decay.}} \)
(a)(ii)
Spontaneous decay occurs without being triggered by external or environmental factors.
Answer: \( \boxed{\text{Radioactive decay is not affected by external factors.}} \)
(a)(iii)
A measured count rate from a radioactive source shows random fluctuations.
Answer: \( \boxed{\text{Fluctuations in measured count rate.}} \)
(b)(i)
Nuclear fission involves a large nucleus splitting into two or more smaller nuclei, whereas nuclear fusion involves two smaller nuclei joining together to form one larger nucleus.
Fission is usually initiated by neutron bombardment, while fusion requires very high temperatures so that the nuclei can overcome their electrostatic repulsion.
Thus, fission involves the splitting of a heavy nucleus, whereas fusion involves the joining of light nuclei.
Answer: \( \boxed{\text{Fission: large nucleus splits; fusion: small nuclei combine.}} \)
(b)(ii)
The binding energy per nucleon is greatest for nuclei with intermediate nucleon numbers.
In both fission and fusion, the products are closer to this region of maximum binding energy per nucleon.
Therefore, the products have a greater binding energy per nucleon than the original nuclei. The total binding energy increases, so energy is released.
Answer: \( \boxed{\text{Both fission and fusion produce nuclei with greater binding energy per nucleon, so energy is released.}} \)