Question 1
(a) State Newton’s law of gravitation. [2 marks]
………………………………………………………………………………………………………………………….
(b) A satellite is in a circular orbit around a planet. The radius of the orbit is \(R\) and the period of the orbit is \(T\). The planet is a uniform sphere.
Use Newton’s law of gravitation to show that \(R\) and \(T\) are related by
\(4\pi^2R^3=GMT^2\)
where \(M\) is the mass of the planet and \(G\) is the gravitational constant. [2 marks]
(c) The Earth may be considered to be a uniform sphere of mass \(5.98\times10^{24}\,\mathrm{kg}\) and radius \(6.37\times10^6\,\mathrm{m}\).
A geostationary satellite is in orbit around the Earth.
Use the expression in (b) to determine the height of the satellite above the Earth’s surface.
height = …………………………………………….. \(\mathrm{m}\) [3 marks]
(d) Another satellite is in a circular orbit around the Earth with the same orbital radius and period as the satellite in (c).
(i) Calculate the angular speed of the satellite in this orbit. Give a unit with your answer. [2 marks]
angular speed = ………………………………………. unit …………….
(ii) Despite having the same orbital period, the orbit of this satellite is not geostationary.
Suggest two ways in which the orbit of this satellite could be different from the orbit of the satellite in (c). [2 marks]
1 ……………………………………………………………………………………………………………………….
2 ……………………………………………………………………………………………………………………….
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702 Paper 4):
• 12.1: Kinematics of uniform circular motion parts (b), (c), (d)(i) and (d)(ii)
▶️ Answer/Explanation
(a) [2 marks]
Newton’s law of gravitation states that the gravitational force between two point masses is directly proportional to the product of their masses and inversely proportional to the square of their separation.
Therefore,
\(F\propto\frac{m_1m_2}{r^2}\)
Answer: \( \boxed{F=G\frac{m_1m_2}{r^2}} \)
(b) [2 marks]
The gravitational force provides the centripetal force required to keep the satellite in circular motion.
Gravitational force:
\(F=\frac{GMm}{R^2}\)
Centripetal force:
\(F=mR\omega^2\)
Equating the two forces,
\(\frac{GMm}{R^2}=mR\omega^2\)
Cancel \(m\):
\(\frac{GM}{R^2}=R\omega^2\)
Since
\(\omega=\frac{2\pi}{T}\)
then
\(\frac{GM}{R^2}=R\left(\frac{2\pi}{T}\right)^2\)
\(\frac{GM}{R^2}=\frac{4\pi^2R}{T^2}\)
Therefore,
\(GMT^2=4\pi^2R^3\)
Answer: \( \boxed{4\pi^2R^3=GMT^2} \)
(c) [3 marks]
For a geostationary satellite, the orbital period is \(24\,\mathrm{h}\).
Convert the period into seconds:
\(T=24\times60\times60=86400\,\mathrm{s}\)
Using
\(4\pi^2R^3=GMT^2\)
\[ \text{Note: all mathematical expressions are kept inline as requested.} \]
Hence,
\(R^3=\frac{GMT^2}{4\pi^2}\)
\(R=\left(\frac{(6.67\times10^{-11})(5.98\times10^{24})(86400)^2}{4\pi^2}\right)^{1/3}\)
\(R=4.22\times10^7\,\mathrm{m}\)
The height above the Earth’s surface is
\(h=R-R_{\mathrm{Earth}}\)
\(h=(4.22\times10^7)-(6.37\times10^6)\)
\(h=3.58\times10^7\,\mathrm{m}\)
Answer: \( \boxed{3.6\times10^7\,\mathrm{m}} \)
(d)(i) Angular speed [2 marks]
The angular speed is related to the period by
\(\omega=\frac{2\pi}{T}\)
Using \(T=24\times60\times60\,\mathrm{s}\),
\(\omega=\frac{2\pi}{24\times60\times60}\)
\(\omega=7.27\times10^{-5}\,\mathrm{rad\,s^{-1}}\)
Answer: \( \boxed{7.3\times10^{-5}\,\mathrm{rad\,s^{-1}}} \)
(d)(ii) [2 marks]
For an orbit to be geostationary, it must have the same period as the Earth’s rotation and must orbit in the equatorial plane in the same direction as the Earth’s rotation.
Two possible differences are:
1. The satellite could orbit from west to east in the opposite direction to the required geostationary direction.
2. The satellite could have an orbit that is not in the equatorial plane, for example, a polar or inclined orbit.
Answer: \( \boxed{\text{opposite direction of orbit and/or non-equatorial orbit}} \)
Question 2
(a) (i) State what is meant by an ideal gas. [2 marks]
………………………………………………………………………………………………………………………….
(ii) State the temperature, in degrees Celsius, of absolute zero.
temperature = ……………………………………………. \(\mathrm{^\circ C}\) [1 mark]
(b) A sealed vessel contains a mass of \(0.0424\,\mathrm{kg}\) of an ideal gas at \(227^\circ\mathrm{C}\). The pressure of the gas is \(1.37\times10^5\,\mathrm{Pa}\) and the volume of the gas is \(0.640\,\mathrm{m^3}\).
Calculate:
(i) the number of molecules of the gas in the vessel
number of molecules = ………………………………………………… [3 marks]
(ii) the mass of one molecule of the gas
mass = ……………………………………………. \(\mathrm{kg}\) [1 mark]
(iii) the root-mean-square (r.m.s.) speed \(v\) of the molecules of the gas.
\(v\) = ………………………………………… \(\mathrm{m\,s^{-1}}\) [3 marks]
(c) The gas in (b) is now cooled gradually to absolute zero.
On Fig. 2.1, sketch the variation with thermodynamic temperature \(T\) of the r.m.s. speed of the molecules of the gas.
[2 marks]
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702 Paper 4):
• 14.2: Temperature scales – part (a)(ii)
• 15.2: Equation of state – part (b)(i)
• 15.3: Kinetic theory of gases – part (b)(ii)
▶️ Answer/Explanation
(a)(i) [2 marks]
An ideal gas is a gas that obeys the equation
\(pV\propto T\)
for all values of \(p\), \(V\) and \(T\), where \(T\) is the thermodynamic temperature.
Answer: \( \boxed{\text{A gas that obeys }pV\propto T\text{ for all values of }p,V\text{ and }T} \)
(a)(ii) [1 mark]
Absolute zero corresponds to \(0\,\mathrm{K}\).
Using the relationship between Celsius and thermodynamic temperature,
\(T/^\circ\mathrm{C}=T/\mathrm{K}-273.15\)
Therefore,
\(T=-273.15^\circ\mathrm{C}\)
Answer: \( \boxed{-273.15^\circ\mathrm{C}} \)
(b)(i) Number of molecules [3 marks]
For an ideal gas,
\(pV=NkT\)
The temperature must be converted to kelvin:
\(T=227+273=500\,\mathrm{K}\)
Rearranging,
\(N=\frac{pV}{kT}\)
\(N=\frac{(1.37\times10^5)(0.640)}{(1.38\times10^{-23})(500)}\)
\(N=1.27\times10^{25}\)
Answer: \( \boxed{1.27\times10^{25}\text{ molecules}} \)
(b)(ii) Mass of one molecule [1 mark]
The mass of one molecule is the total mass divided by the number of molecules:
\(m=\frac{0.0424}{1.27\times10^{25}}\)
\(m=3.34\times10^{-27}\,\mathrm{kg}\)
Answer: \( \boxed{3.34\times10^{-27}\,\mathrm{kg}} \)
(b)(iii) r.m.s. speed [3 marks]
From the kinetic theory of gases,
\(\frac{1}{2}m\langle c^2\rangle=\frac{3}{2}kT\)
For the r.m.s. speed \(v\), \(v^2=\langle c^2\rangle\), so
\(mv^2=3kT\)
Substituting \(m=3.34\times10^{-27}\,\mathrm{kg}\) and \(T=500\,\mathrm{K}\):
\((3.34\times10^{-27})v^2=3(1.38\times10^{-23})(500)\)
\(v=2.49\times10^3\,\mathrm{m\,s^{-1}}\)
Answer: \( \boxed{2.49\times10^3\,\mathrm{m\,s^{-1}}} \)
(c) [2 marks]
From kinetic theory,
\(v_{\mathrm{rms}}=\sqrt{\frac{3kT}{m}}\)
Therefore,
\(v_{\mathrm{rms}}\propto\sqrt{T}\)
The graph must therefore start at the origin and have a positive gradient that decreases as \(T\) increases. It is a curve with decreasing positive gradient throughout.
Required sketch: a curve passing through \((0,0)\), increasing throughout with decreasing positive gradient.
Answer: \( \boxed{v_{\mathrm{rms}}\propto\sqrt{T}} \)
Question 3
(a) State the first law of thermodynamics. Identify the meaning of any symbols that you use. [2 marks]
………………………………………………………………………………………………………………………….
(b) The state of an ideal gas is continuously changed according to the cycle ABCDA shown in Fig. 3.1.
(i) Complete Table 3.1 for the changes A to B and B to C by placing two ticks (\(\checkmark\)) in each row. [4 marks]
| change | change in internal energy | work done on gas | ||||
|---|---|---|---|---|---|---|
| decrease | no change | increase | negative | zero | positive | |
| A to B | ||||||
| B to C | ||||||
(ii) Use the first law of thermodynamics to describe and explain the energy transfers associated with one complete cycle ABCDA. [3 marks]
………………………………………………………………………………………………………………………….
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702 Paper 4):
• 16.1: Internal energy – part (b)(i)
▶️ Answer/Explanation
(a) [2 marks]
The first law of thermodynamics states that the change in internal energy of a system is equal to the energy transferred to the system by heating plus the work done on the system.
Using \(W\) for work done on the system and \(Q\) for energy transferred to the system by heating:
\(\Delta U=W+Q\)
where \(\Delta U\) is the change in internal energy, \(W\) is the work done on the system and \(Q\) is the energy transferred to the system by heating.
Answer: \( \boxed{\Delta U=W+Q} \)
(b)(i) [4 marks]
For an ideal gas, internal energy depends only on temperature.
A to B: The volume decreases at constant pressure. Therefore the temperature decreases, so the internal energy decreases. Since the gas is compressed, work is done on the gas and this work is positive.
Required ticks for A to B:
Change in internal energy: decrease \(\checkmark\)
Work done on gas: positive \(\checkmark\)
B to C: The volume remains constant while the pressure increases. Therefore the temperature increases, so the internal energy increases. Since there is no change in volume, no work is done on the gas.
Required ticks for B to C:
Change in internal energy: increase \(\checkmark\)
Work done on gas: zero \(\checkmark\)
Completed Table 3.1:
| change | change in internal energy | work done on gas | ||||
|---|---|---|---|---|---|---|
| decrease | no change | increase | negative | zero | positive | |
| A to B | ✓ | ✓ | ||||
| B to C | ✓ | ✓ | ||||
(b)(ii) [3 marks]
Over one complete cycle, the gas returns to its original state. Therefore its temperature and hence its internal energy return to their original values.
Thus, for the complete cycle,
\(\Delta U=0\)
From the first law,
\(\Delta U=W+Q\)
Since \(\Delta U=0\), the net energy transferred by heating must balance the net work done.
The work done by the gas during expansion from C to D is greater than the work done on the gas during compression from A to B because the pressure is higher during C to D.
There is no work done during B to C or D to A because the volume is constant.
Therefore, the gas does net work over one complete cycle and there must be an overall input of thermal energy.
Answer: \( \boxed{\text{The internal energy returns to its original value, so }\Delta U=0.\text{ The gas does net work and there is an overall input of thermal energy.}} \)
Question 4
A small steel sphere is oscillating vertically on the end of a spring, as shown in Fig. 4.1.
The velocity \(v\) of the sphere varies with displacement \(x\) from its equilibrium position according to
\(v=\pm9.7\sqrt{11.6-x^2}\)
where \(v\) is in \(\mathrm{cm\,s^{-1}}\) and \(x\) is in \(\mathrm{cm}\).
(a) (i) Calculate the frequency of the oscillations. [2 marks]
frequency = ……………………………………… \(\mathrm{Hz}\)
(ii) Show that the amplitude of the oscillations is \(3.4\,\mathrm{cm}\). [1 mark]
………………………………………………………………………………………………………………………….
(iii) Calculate the maximum acceleration \(a_0\) of the sphere. [2 marks]
\(a_0\) = ……………………………………… \(\mathrm{m\,s^{-2}}\)
(b) On Fig. 4.2, sketch the variation with \(x\) of the acceleration \(a\) of the sphere. [3 marks]
(c) Describe, without calculation, the interchange between the potential energy and the kinetic energy during the oscillations. [3 marks]
………………………………………………………………………………………………………………………….
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702 Paper 4):
• 17.2: Energy in simple harmonic motion – part (c)
▶️ Answer/Explanation
(a)(i) Frequency [2 marks]
For simple harmonic motion, the velocity is given by
\(v=\pm\omega\sqrt{A^2-x^2}\)
Comparing this with
\(v=\pm9.7\sqrt{11.6-x^2}\)
gives
\(\omega=9.7\,\mathrm{s^{-1}}\)
The angular frequency is related to frequency by
\(\omega=2\pi f\)
Therefore,
\(f=\frac{9.7}{2\pi}\)
\(f=1.54\,\mathrm{Hz}\)
Answer: \( \boxed{1.5\,\mathrm{Hz}} \)
(a)(ii) Amplitude [1 mark]
Comparing
\(v=\pm\omega\sqrt{A^2-x^2}\)
with
\(v=\pm9.7\sqrt{11.6-x^2}\)
gives
\(A^2=11.6\)
\(A=\sqrt{11.6}=3.41\,\mathrm{cm}\)
Answer: \( \boxed{3.4\,\mathrm{cm}} \)
(a)(iii) Maximum acceleration [2 marks]
For simple harmonic motion,
\(a=-\omega^2x\)
The maximum acceleration occurs when \(x=A\), so
\(a_0=\omega^2A\)
\(a_0=(9.7)^2(3.41)\,\mathrm{cm\,s^{-2}}\)
\(a_0=320\,\mathrm{cm\,s^{-2}}\)
Converting to \(\mathrm{m\,s^{-2}}\),
\(a_0=3.20\,\mathrm{m\,s^{-2}}\)
Answer: \( \boxed{3.2\,\mathrm{m\,s^{-2}}} \)
(b) Acceleration-displacement graph [3 marks]
For simple harmonic motion,
\(a=-\omega^2x\)
Therefore, acceleration is directly proportional to displacement and is always in the opposite direction to the displacement.
The graph is therefore a straight line with a negative gradient passing through the origin.
At \(x=+3.4\,\mathrm{cm}\), \(a=-a_0=-3.2\,\mathrm{m\,s^{-2}}\).
At \(x=-3.4\,\mathrm{cm}\), \(a=+a_0=+3.2\,\mathrm{m\,s^{-2}}\).
Required sketch: a straight line through the origin, extending from \((-3.4\,\mathrm{cm},+a_0)\) to \((+3.4\,\mathrm{cm},-a_0)\).
(c) Energy interchange [3 marks]
At the maximum displacement, the sphere has zero velocity, so its kinetic energy is zero and its potential energy is maximum.
As the sphere moves towards the equilibrium position, potential energy is transferred into kinetic energy. At the equilibrium position, the speed and kinetic energy are maximum while the potential energy is minimum.
As the sphere moves from the equilibrium position towards the opposite extreme, kinetic energy is transferred back into potential energy.
The total mechanical energy remains constant, assuming negligible energy losses.
Answer: \( \boxed{\text{PE is maximum at the extremes and KE is maximum at equilibrium; energy continually transfers between PE and KE while total energy remains constant.}} \)
Question 5
Two capacitors A and B are connected into the circuit shown in Fig. 5.1.
Capacitor A has capacitance \(C\) and capacitor B has capacitance \(3C\).
The electromotive force (e.m.f.) of the cell is \(V\).
The two-way switch \(S\) is initially at position X, and capacitor B is initially uncharged.
(a) State, in terms of \(V\) and \(C\), expressions for:
(i) the initial charge \(Q_A\) on the plates of capacitor A
\(Q_A\) = ………………………………………………… [1 mark]
(ii) the initial energy \(E_A\) stored in capacitor A.
\(E_A\) = …………………………………………………. [1 mark]
(b) The two-way switch \(S\) is now moved to position Y.
(i) State and explain what happens to the charge that was initially on the plates of capacitor A. [2 marks]
………………………………………………………………………………………………………………………….
(ii) Show that the final potential difference (p.d.) \(V_B\) across capacitor B is given by
\(V_B=\frac{V}{4}\)
Explain your reasoning. [3 marks]
(iii) Determine an expression, in terms of \(V\) and \(C\), for the decrease \(\Delta E\) in the total energy that is stored in the capacitors as a result of the change of the position of the switch.
\(\Delta E\) = ………………………………………………… [2 marks]
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702 Paper 4):
• 19.2: Energy stored in a capacitor – parts (a)(ii) and (b)(iii)
▶️ Answer/Explanation
(a)(i) Initial charge [1 mark]
The charge stored on a capacitor is given by
\(Q=CV\)
Capacitor A has capacitance \(C\) and is connected directly across the cell of e.m.f. \(V\). Therefore,
\(Q_A=CV\)
Answer: \( \boxed{Q_A=CV} \)
(a)(ii) Initial energy [1 mark]
The energy stored in a capacitor is
\(E=\frac{1}{2}CV^2\)
Therefore, for capacitor A,
\(E_A=\frac{1}{2}CV^2\)
Answer: \( \boxed{E_A=\frac{1}{2}CV^2} \)
(b)(i) Transfer of charge [2 marks]
When the switch is moved to position Y, capacitor A and capacitor B become connected together.
The initial potential differences across the two capacitors are not equal because capacitor B is initially uncharged.
Therefore, some of the charge initially stored on capacitor A transfers to capacitor B.
The transfer stops when the potential differences across the two capacitors become equal.
Answer: \( \boxed{\text{Charge transfers from A to B until the p.d.s across the capacitors are equal.}} \)
(b)(ii) Final potential difference [3 marks]
At the final state, the two capacitors are connected in parallel, so their potential differences are equal:
\(V_A=V_B\)
The total charge is conserved. Initially, the total charge on the capacitors is
\(Q_{\mathrm{total}}=CV\)
After connection, the total capacitance is
\(C_{\mathrm{total}}=C+3C=4C\)
Therefore,
\(Q_{\mathrm{total}}=C_{\mathrm{total}}V_B\)
\(CV=4CV_B\)
Hence,
\(V_B=\frac{V}{4}\)
Answer: \( \boxed{V_B=\frac{V}{4}} \)
(b)(iii) Decrease in total stored energy [2 marks]
The initial energy is the energy stored in capacitor A:
\(E_{\mathrm{initial}}=\frac{1}{2}CV^2\)
After the capacitors are connected, their combined capacitance is \(4C\) and their common p.d. is \(V/4\).
Thus,
\(E_{\mathrm{final}}=\frac{1}{2}(4C)\left(\frac{V}{4}\right)^2\)
\(E_{\mathrm{final}}=\frac{1}{8}CV^2\)
The decrease in energy is
\(\Delta E=E_{\mathrm{initial}}-E_{\mathrm{final}}\)
\(\Delta E=\frac{1}{2}CV^2-\frac{1}{8}CV^2\)
\(\Delta E=\frac{3}{8}CV^2\)
Answer: \( \boxed{\Delta E=\frac{3}{8}CV^2} \)
Question 6
A heavy aluminium disc has a radius of \(0.36\,\mathrm{m}\). The disc rotates with the wheels of a vehicle and forms part of an electromagnetic braking system on the vehicle.
In order to activate the braking system, a uniform magnetic field of flux density \(0.17\,\mathrm{T}\) is switched on. This magnetic field is perpendicular to the plane of rotation of the disc, as shown in Fig. 6.1.
(a) (i) Define magnetic flux. [2 marks]
………………………………………………………………………………………………………………………….
(ii) Calculate the magnetic flux through the disc. Give a unit with your answer. [2 marks]
magnetic flux = ……………………………………….. unit ……………….
(b) The disc is rotating at a rate of \(25\) revolutions per second.
Calculate the magnitude of the electromotive force (e.m.f.) induced between the axle and the rim of the disc. [3 marks]
e.m.f. = ……………………………………………… \(\mathrm{V}\)
(c) The axle and the rim are connected into an external circuit that enables the energy of the rotation of the disc to be stored for future use. The direction of rotation is shown in Fig. 6.1.
Use Lenz’s law of electromagnetic induction to determine whether the current in the disc is from the rim to the axle or from the axle to the rim. Explain your reasoning. [3 marks]
………………………………………………………………………………………………………………………….
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702 Paper 4):
• 20.1: Concept of a magnetic field – part (a)(i)
▶️ Answer/Explanation
(a)(i) Magnetic flux [2 marks]
Magnetic flux is the product of the magnetic flux density and the area perpendicular to the magnetic field.
For a uniform magnetic field,
\(\Phi=BA\)
Answer: \( \boxed{\text{Magnetic flux is the product of magnetic flux density and perpendicular area.}} \)
(a)(ii) Magnetic flux through the disc [2 marks]
The magnetic field is perpendicular to the plane of the disc, so the whole area of the disc is perpendicular to the field.
The area of the disc is
\(A=\pi r^2\)
Therefore,
\(\Phi=B\pi r^2\)
\(\Phi=(0.17)\pi(0.36)^2\)
\(\Phi=6.92\times10^{-2}\,\mathrm{Wb}\)
Answer: \( \boxed{6.9\times10^{-2}\,\mathrm{Wb}} \)
(b) Induced e.m.f. [3 marks]
The disc completes \(25\) revolutions per second, so the rate at which the disc cuts magnetic flux is \(25\) times the magnetic flux through the disc.
Hence,
\(\mathcal{E}=\frac{\Delta\Phi}{\Delta t}\)
\(\mathcal{E}=(6.92\times10^{-2})(25)\)
\(\mathcal{E}=1.73\,\mathrm{V}\)
Alternatively, using the motional e.m.f. expression for a rotating disc,
\(\mathcal{E}=\frac{1}{2}B\omega r^2\)
where \(\omega=2\pi f\).
Answer: \( \boxed{1.7\,\mathrm{V}} \)
(c) Direction of current [3 marks]
When the current flows through the disc in the magnetic field, a magnetic force acts on the current-carrying charges.
By Lenz’s law, the induced effect must oppose the change that produces it. Therefore, the magnetic force on the disc must oppose its rotation.
Using the direction of the magnetic field and the direction of rotation shown in Fig. 6.1, the left-hand rule gives the required current direction as from the rim towards the axle.
Answer: \( \boxed{\text{The current flows from the rim to the axle.}} \)
Question 7
Four diodes are used in a bridge rectifier circuit to produce rectification of a sinusoidal a.c. input voltage \(V_{\mathrm{IN}}\). Fig. 7.1 shows part of the circuit, but three of the diodes are missing.
The p.d. across the load resistor \(R\) is the output p.d. \(V_{\mathrm{OUT}}\) of the bridge rectifier.
(a) (i) State the name of the type of rectification produced by a bridge rectifier. [1 mark]
(ii) Complete Fig. 7.1 by drawing the three missing diodes, correctly connected. [2 marks]
(iii) On Fig. 7.1, draw an arrow to indicate the direction of the current in resistor \(R\). [1 mark]
(b) \(V_{\mathrm{IN}}\) has amplitude \(V_0\) and period \(T\). Fig. 7.2 shows the variation with time \(t\) of \(V_{\mathrm{IN}}\).
(i) On Fig. 7.3, sketch the variation of \(V_{\mathrm{OUT}}\) with \(t\) between \(t=0\) and \(t=2.0T\). [3 marks]
(ii) The power dissipated in the resistor is \(P\).
On Fig. 7.4, sketch the variation of \(P\) with \(t\) between \(t=0\) and \(t=2.0T\). [2 marks]
(iii) Suggest, with a reason, how the root-mean-square (r.m.s.) value of \(V_{\mathrm{OUT}}\) compares with the r.m.s. value of \(V_{\mathrm{IN}}\). [1 mark]
………………………………………………………………………………………………………………………….
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702 Paper 4):
▶️ Answer/Explanation
(a)(i) Type of rectification [1 mark]
A bridge rectifier converts both half-cycles of an alternating input into an output of the same polarity.
Answer: \( \boxed{\text{full-wave rectification}} \)
(a)(ii) Missing diodes [2 marks]
The three missing diodes must be connected so that current through the load resistor \(R\) flows in the same direction during both half-cycles of \(V_{\mathrm{IN}}\).
The required diode orientations are:
• lower-left diode pointing towards the left,
• lower-right diode pointing towards the left,
• upper-left diode pointing towards the left.
Answer: \(\boxed{\text{Three diodes connected in the required bridge orientation so both half-cycles give the same load-current direction.}}\)
(a)(iii) Direction of current in \(R\) [1 mark]
The current through the load resistor must have the same direction for both half-cycles.
Answer: \( \boxed{\text{current through }R\text{ is from left to right}} \)
(b)(i) Output voltage [3 marks]
A bridge rectifier produces full-wave rectification. Therefore, the negative half-cycles of \(V_{\mathrm{IN}}\) are inverted to become positive.
Hence,
\(V_{\mathrm{OUT}}=|V_{\mathrm{IN}}|\)
The graph therefore has:
• minimum value \(V_{\mathrm{OUT}}=0\),
• maximum value \(V_{\mathrm{OUT}}=V_0\),
• peaks at \(t=0\), \(0.5T\), \(1.0T\), \(1.5T\) and \(2.0T\),
• zero values halfway between successive peaks.
Required sketch: a full-wave rectified sine curve entirely above the time axis, with successive positive peaks of \(V_0\).
(b)(ii) Power variation [2 marks]
The power dissipated in a resistor is
\(P=\frac{V_{\mathrm{OUT}}^2}{R}\)
Since \(V_{\mathrm{OUT}}=|V_{\mathrm{IN}}|\), squaring removes the effect of the sign.
Therefore, the power is always positive and has peaks whenever \(V_{\mathrm{IN}}\) has a positive or negative peak.
The graph has zero values when \(V_{\mathrm{IN}}=0\), and maximum values at \(t=0\), \(0.5T\), \(1.0T\), \(1.5T\) and \(2.0T\).
Required sketch: a sinusoidal-squared curve entirely above the time axis, with troughs on the time axis and peaks at \(0\), \(0.5T\), \(1.0T\), \(1.5T\) and \(2.0T\).
(b)(iii) Comparison of r.m.s. values [1 mark]
The r.m.s. value depends on the mean of \(V^2\), not on the sign of \(V\).
Rectification changes the negative half-cycles to positive half-cycles, but \(V^2\) remains unchanged.
Answer: \( \boxed{V_{\mathrm{OUT,rms}}=V_{\mathrm{IN,rms}}} \)
Question 8
Fig. 8.1 shows the lowest four energy levels of an electron in an isolated atom.
Fig. 8.2 shows the lines in the emission spectrum of the atom that correspond to the transitions of the electron from \(n=3\) to \(n=1\) and from \(n=4\) to \(n=1\).
(a) Explain, with reference to photons, why there is a single frequency of electromagnetic radiation that corresponds to each of these transitions. [2 marks]
………………………………………………………………………………………………………………………….
(b) (i) On Fig. 8.2, draw a line that corresponds to the transition of the electron from \(n=2\) to \(n=1\).
Label this line A. [2 marks]
(ii) On Fig. 8.2, draw a line that corresponds to the transition of the electron from \(n=3\) to \(n=2\).
Label this line B. [2 marks]
(c) The frequency of radiation represented by line A is \(f_A\).
The frequency of radiation represented by line B is \(f_B\).
The energy of the ground state (\(n=1\)) is \(E_1\).
Determine an expression, in terms of \(f_A\), \(f_B\), \(E_1\) and the Planck constant \(h\), for the energy \(E_3\) of the energy level \(n=3\). [2 marks]
\(E_3\) = …………………………………………………
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702 Paper 4):
• 22.1: Energy and momentum of a photon – parts (a) and (c)
▶️ Answer/Explanation
(a) [2 marks]
When an electron makes a transition between two energy levels, it emits a photon whose energy is equal to the difference between the two energy levels.
The energy of a photon is related to its frequency by
\(E=hf\)
Since the two energy levels have fixed energies, their energy difference is fixed. Therefore, the photon has a fixed energy and hence a single corresponding frequency.
Answer: \( \boxed{\text{Each transition emits a photon with a fixed energy difference, so }E=hf\text{ gives a single frequency.}} \)
(b)(i) Transition \(n=2\) to \(n=1\) [2 marks]
The transition from \(n=2\) to \(n=1\) has a smaller energy difference than the transitions from \(n=3\) to \(n=1\) and from \(n=4\) to \(n=1\).
Therefore, it has a lower frequency and must appear to the left of the two existing lines.
The line is labelled A.
Answer: \( \boxed{\text{Line A is to the left of the existing pair of lines.}} \)
(b)(ii) Transition \(n=3\) to \(n=2\) [2 marks]
The transition from \(n=3\) to \(n=2\) has a smaller energy difference than the transition from \(n=2\) to \(n=1\).
Therefore, it has an even lower frequency and must appear to the left of line A.
The line is labelled B.
Answer: \( \boxed{\text{Line B is to the left of line A.}} \)
(c) Energy of level \(n=3\) [2 marks]
For line A, the transition is from \(n=2\) to \(n=1\), so
\(E_2-E_1=hf_A\)
Therefore,
\(E_2=E_1+hf_A\)
For line B, the transition is from \(n=3\) to \(n=2\), so
\(E_3-E_2=hf_B\)
Hence,
\(E_3=E_2+hf_B\)
Substituting \(E_2=E_1+hf_A\),
\(E_3=E_1+hf_A+hf_B\)
Therefore,
Answer: \( \boxed{E_3=E_1+h(f_A+f_B)} \)
Question 9
(a) Define mass defect. [2 marks]
………………………………………………………………………………………………………………………….
(b) Table 9.1 shows the mass defects of three nuclei.
The nuclear fusion process in a particular star is described by
\({}^{2}_{1}\mathrm{H}+{}^{3}_{1}\mathrm{H}\rightarrow{}^{4}_{2}\mathrm{He}+X\)
where \(X\) is a particle that has no mass defect.
(i) State the name of particle \(X\). [1 mark]
………………………………………………………………………………………………………………………….
(ii) Show that the energy released when one nucleus of \({}^{4}_{2}\mathrm{He}\) is formed in this fusion reaction is \(2.8\times10^{-12}\,\mathrm{J}\). [3 marks]
………………………………………………………………………………………………………………………….
(c) The star in (b) has a radius of \(2.3\times10^9\,\mathrm{m}\) and a luminosity of \(1.4\times10^{28}\,\mathrm{W}\).
All the energy released from the formation of \({}^{4}_{2}\mathrm{He}\) is radiated away from the star.
All the energy that is radiated from the star has been released in the formation of \({}^{4}_{2}\mathrm{He}\).
Determine:
(i) the mass of \({}^{4}_{2}\mathrm{He}\) produced per unit time by the fusion process
mass per unit time = ……………………………………….. \(\mathrm{kg\,s^{-1}}\) [3 marks]
(ii) the surface temperature of the star.
temperature = ……………………………………………… \(\mathrm{K}\) [2 marks]
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702 Paper 4):
• 25.2: Stellar radii – part (c)(ii)
▶️ Answer/Explanation
(a) Mass defect [2 marks]
Mass defect is the difference between the total mass of the individual nucleons when they are infinitely separated and the mass of the nucleus formed from those nucleons.
Answer: \( \boxed{\text{Mass defect is the difference between the mass of the separated nucleons and the mass of the nucleus.}} \)
(b)(i) Particle \(X\) [1 mark]
The reaction is
\({}^{2}_{1}\mathrm{H}+{}^{3}_{1}\mathrm{H}\rightarrow{}^{4}_{2}\mathrm{He}+X\)
Conservation of nucleon number gives \(2+3=4+A_X\), so \(A_X=1\).
Conservation of proton number gives \(1+1=2+Z_X\), so \(Z_X=0\).
Answer: \( \boxed{\text{neutron}} \)
(b)(ii) Energy released [3 marks]
The mass defect for the fusion reaction is
\(\Delta m=0.030377-0.002388-0.009105\)
\(\Delta m=0.018884\,\mathrm{u}\)
Using
\(E=\Delta mc^2\)
and \(1\,\mathrm{u}=1.66\times10^{-27}\,\mathrm{kg}\),
\(E=(0.018884)(1.66\times10^{-27})(3.00\times10^8)^2\)
\(E=2.82\times10^{-12}\,\mathrm{J}\)
Answer: \( \boxed{2.8\times10^{-12}\,\mathrm{J}} \)
(c)(i) Mass of helium produced per unit time [3 marks]
The luminosity is the energy released per unit time, so the number of helium nuclei produced per unit time is
\(N=\frac{1.4\times10^{28}}{2.8\times10^{-12}}\)
\(N=5.0\times10^{39}\,\mathrm{s^{-1}}\)
The mass of one \({}^{4}_{2}\mathrm{He}\) nucleus is approximately
\(m=4(1.66\times10^{-27})\)
\(m=6.64\times10^{-27}\,\mathrm{kg}\)
Therefore, the mass produced per unit time is
\(\frac{m}{t}=(6.64\times10^{-27})(5.0\times10^{39})\)
\(\frac{m}{t}=3.32\times10^{13}\,\mathrm{kg\,s^{-1}}\)
Answer: \( \boxed{3.3\times10^{13}\,\mathrm{kg\,s^{-1}}} \)
(c)(ii) Surface temperature [2 marks]
Using the Stefan-Boltzmann law,
\(L=4\pi\sigma r^2T^4\)
Substituting \(L=1.4\times10^{28}\,\mathrm{W}\), \(r=2.3\times10^9\,\mathrm{m}\) and \(\sigma=5.67\times10^{-8}\,\mathrm{W\,m^{-2}\,K^{-4}}\):
\(1.4\times10^{28}=4\pi(5.67\times10^{-8})(2.3\times10^9)^2T^4\)
Rearranging and solving gives
\(T=7.8\times10^3\,\mathrm{K}\)
Answer: \( \boxed{7800\,\mathrm{K}} \)
Question 10
(a) X-rays for use in medical diagnosis are produced in an X-ray tube. In the X-ray tube, charged particles are accelerated towards a metal target by an applied potential difference (p.d.).
(i) State the name of the charged particles that are accelerated by the applied p.d. [1 mark]
………………………………………………………………………………………………………………………….
(ii) Explain how X-rays are produced at the metal target. [2 marks]
………………………………………………………………………………………………………………………….
(iii) Calculate the minimum wavelength of X-rays produced when the applied p.d. is \(5.80\,\mathrm{kV}\). [3 marks]
wavelength = …………………………………………….. \(\mathrm{m}\)
(b) X-rays pass through a medium that has an attenuation coefficient of \(1.4\,\mathrm{cm^{-1}}\).
Calculate the percentage of the X-ray energy that is absorbed by a \(2.8\,\mathrm{cm}\) thickness of this medium. [3 marks]
percentage absorbed = …………………………………………….. \(\%\)
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702 Paper 4):
• 22.1: Energy and momentum of a photon – part (a)(iii)
▶️ Answer/Explanation
(a)(i) [1 mark]
The charged particles accelerated by the applied p.d. are electrons.
Answer: \( \boxed{\text{electrons}} \)
(a)(ii) [2 marks]
The electrons are accelerated to high speeds and are then rapidly decelerated or stopped when they strike the metal target.
Some of the kinetic energy lost by the electrons is emitted as X-ray photons.
Answer: \( \boxed{\text{Electrons are rapidly decelerated at the target and their lost kinetic energy is emitted as X-ray photons.}} \)
(a)(iii) Minimum wavelength [3 marks]
The maximum energy of an X-ray photon occurs when all the kinetic energy gained by an electron is converted into photon energy.
Therefore,
\(eV=\frac{hc}{\lambda_{\min}}\)
Rearranging,
\(\lambda_{\min}=\frac{hc}{eV}\)
The applied p.d. is
\(V=5.80\times10^3\,\mathrm{V}\)
Hence,
\(\lambda_{\min}=\frac{(6.63\times10^{-34})(3.00\times10^8)}{(1.60\times10^{-19})(5.80\times10^3)}\)
\(\lambda_{\min}=2.14\times10^{-10}\,\mathrm{m}\)
Answer: \( \boxed{2.14\times10^{-10}\,\mathrm{m}} \)
(b) Percentage absorbed [3 marks]
The attenuation equation is
\(I=I_0e^{-\mu x}\)
Therefore,
\(\frac{I}{I_0}=e^{-\mu x}\)
Using \(\mu=1.4\,\mathrm{cm^{-1}}\) and \(x=2.8\,\mathrm{cm}\),
\(\frac{I}{I_0}=e^{-(1.4)(2.8)}\)
\(\frac{I}{I_0}=0.0198\)
Thus, \(1.98\%\) of the original X-ray energy is transmitted.
Percentage absorbed is
\(\%\,\mathrm{absorbed}=(1-0.0198)\times100\)
\(\%\,\mathrm{absorbed}=98.0\%\)
Answer: \( \boxed{98\%} \)