Question 1
(a) Define gravitational potential at a point. [2 marks]
(b) Artemis is a spherical planet that may be assumed to be isolated in space. The variation with distance \(x\) from the centre of Artemis of the gravitational potential \(\phi\) is shown in Fig. 1.1.
(i) The radius of Artemis is \(4800\,\mathrm{km}\).
Determine the value of \(\phi\) on the surface of Artemis.
\(\phi\) = ……………………………………….. \( \mathrm{J\,kg^{-1}} \) [1 mark]
(ii) Show that the mass of Artemis is \(2.55\times10^{24}\,\mathrm{kg}\). [1 mark]
________________________________
(iii) Calculate the gravitational field strength \(g\) on the surface of Artemis.
\(g\) = ………………………………………. \( \mathrm{N\,kg^{-1}} \) [2 marks]
(iv) A satellite is in an orbit at a fixed position above a point on the surface of Artemis. The satellite is located above the equator of Artemis at a height above the surface where the gravitational potential is \(-0.65\times10^7\,\mathrm{J\,kg^{-1}}\). Calculate the period, in hours, of rotation of Artemis.
period = ……………………………………….. hours [4 marks]
(c) State one similarity and one difference between gravitational potential due to a point mass and electric potential due to a point charge.
similarity…………………………………………………………..
difference…………………………………………………………… [2 marks]
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702 Paper 4):
• 13.3: Gravitational field of a point mass — part (b)(iii)
• 13.2: Gravitational force between point masses — part (b)(iv)
▶️ Answer/Explanation
(a)
Gravitational potential at a point is the work done per unit mass in bringing a small test mass from infinity to that point.
Answer: \( \boxed{\text{work done per unit mass in moving a mass from infinity to the point}} \)
(b)(i)
The radius of Artemis is \(4800\,\mathrm{km}\), so the surface corresponds to \(x=4800\,\mathrm{km}\) on the graph.
Reading the graph gives
\( \phi=-3.55\times10^7\,\mathrm{J\,kg^{-1}} \)
Answer: \( \boxed{-3.55\times10^7\,\mathrm{J\,kg^{-1}}} \)
(b)(ii)
For a spherical mass, the gravitational potential is
\( \phi=-\frac{GM}{r} \)
Therefore,
\( M=-\frac{\phi r}{G} \)
\( M=-\frac{(-3.55\times10^7)(4.80\times10^6)}{6.67\times10^{-11}} \)
\( M=2.55\times10^{24}\,\mathrm{kg} \)
Answer: \( \boxed{2.55\times10^{24}\,\mathrm{kg}} \)
(b)(iii)
The gravitational field strength due to a point mass is
\( g=\frac{GM}{r^2} \)
Using \(M=2.55\times10^{24}\,\mathrm{kg}\) and \(r=4.80\times10^6\,\mathrm{m}\):
\( g=\frac{(6.67\times10^{-11})(2.55\times10^{24})}{(4.80\times10^6)^2} \)
\( g=7.4\,\mathrm{N\,kg^{-1}} \)
Answer: \( \boxed{7.4\,\mathrm{N\,kg^{-1}}} \)
(b)(iv)
At the satellite’s orbit,
\( \phi=-\frac{GM}{r} \)
Hence
\( r=-\frac{GM}{\phi} \)
\( r=\frac{(6.67\times10^{-11})(2.55\times10^{24})}{0.65\times10^7} \)
\( r\approx2.62\times10^7\,\mathrm{m} \)
For a circular orbit, the gravitational force provides the centripetal force:
\( \frac{mv^2}{r}=\frac{GMm}{r^2} \)
The orbital speed is
\( v=\frac{2\pi r}{T} \)
Therefore,
\( T^2=\frac{4\pi^2r^3}{GM} \)
\( T^2=\frac{4\pi^2(2.62\times10^7)^3}{(6.67\times10^{-11})(2.55\times10^{24})} \)
\( T\approx6.48\times10^4\,\mathrm{s} \)
Converting to hours:
\( T=\frac{6.48\times10^4}{3600}\approx18\,\mathrm{h} \)
Answer: \( \boxed{18\,\mathrm{hours}} \)
(c)
Similarity: Both potentials are inversely proportional to distance from the point source, and both are zero at infinity.
Difference: Gravitational potential is always negative, whereas electric potential may be positive or negative depending on the sign of the charge.
Answer: Any one valid similarity and one valid difference.
Question 2
(a) State what is meant by an ideal gas. [2 marks]
(b) A fixed amount of helium gas is sealed in a container. The helium gas has a pressure of \(1.10\times10^5\,\mathrm{Pa}\), and a volume of \(540\,\mathrm{cm^3}\) at a temperature of \(27^{\circ}\mathrm{C}\).
The volume of the container is rapidly decreased to \(30.0\,\mathrm{cm^3}\). The pressure of the helium gas increases to \(6.70\times10^6\,\mathrm{Pa}\) and its temperature increases to \(742^{\circ}\mathrm{C}\), as illustrated in Fig. 2.1.
No thermal energy enters or leaves the helium gas during this process.
(i) Show that the helium gas behaves as an ideal gas. [2 marks]
________________________________
(ii) The first law of thermodynamics may be expressed as
\( \Delta U=q+W \)
Use the first law of thermodynamics to explain why the temperature of the helium gas increases. [2 marks]
________________________________
(iii) The average translational kinetic energy \(E_{\mathrm{K}}\) of a molecule of an ideal gas is given by
\( E_{\mathrm{K}}=\frac{3}{2}kT \)
where \(k\) is the Boltzmann constant and \(T\) is the thermodynamic temperature.
Calculate the change in the total kinetic energy of the molecules of the helium gas.
change in kinetic energy = ……………………………………………. \( \mathrm{J} \) [3 marks]
(c) The mass of nitrogen gas in another container is \(24.0\,\mathrm{g}\) at a temperature of \(27^{\circ}\mathrm{C}\). The gas is cooled to its boiling point of \(-196^{\circ}\mathrm{C}\). Assume all the gas condenses to a liquid.
For this change the specific heat capacity of nitrogen gas is \(1.04\,\mathrm{kJ\,kg^{-1}\,K^{-1}}\).
The specific latent heat of vaporisation of nitrogen is \(199\,\mathrm{kJ\,kg^{-1}}\).
Determine the thermal energy, in \(\mathrm{kJ}\), removed from the nitrogen gas.
energy = ……………………………………………. \( \mathrm{kJ} \) [3 marks]
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702 Paper 4):
• 16.2: The first law of thermodynamics — part (b)(ii)
• 15.3: Kinetic theory of gases — part (b)(iii)
• 14.3: Specific heat capacity and specific latent heat — part (c)
▶️ Answer/Explanation
(a)
An ideal gas is a gas for which \(pV\) is proportional to \(T\), where \(T\) is the thermodynamic temperature.
Answer: \( \boxed{pV\propto T} \)
(b)(i)
For a fixed amount of an ideal gas, \( \frac{pV}{T} \) should remain constant.
Initial state:
\( T_1=27+273=300\,\mathrm{K} \)
\( \frac{p_1V_1}{T_1}=\frac{(1.10\times10^5)(540\times10^{-6})}{300} \)
\( \frac{p_1V_1}{T_1}=0.198\,\mathrm{Pa\,m^3\,K^{-1}} \)
Final state:
\( T_2=742+273=1015\,\mathrm{K} \)
\( \frac{p_2V_2}{T_2}=\frac{(6.70\times10^6)(30.0\times10^{-6})}{1015} \)
\( \frac{p_2V_2}{T_2}=0.198\,\mathrm{Pa\,m^3\,K^{-1}} \)
Since \( \frac{pV}{T} \) has the same value in both states, the helium gas behaves as an ideal gas.
Answer: \( \boxed{\frac{pV}{T}=\text{constant}} \)
(b)(ii)
No thermal energy enters or leaves the gas, so \(q=0\).
The volume of the gas decreases, so work is done on the gas. Therefore, \(W\) is positive.
From the first law of thermodynamics,
\( \Delta U=q+W \)
Since \(q=0\), \( \Delta U=W>0 \).
The internal energy therefore increases. For an ideal gas, an increase in internal energy corresponds to an increase in temperature.
Answer: \( \boxed{\text{work is done on the gas, so internal energy and temperature increase}} \)
(b)(iii)
For an ideal gas,
\( pV=NkT \)
Hence the number of molecules is
\( N=\frac{pV}{kT} \)
Using \(k=1.38\times10^{-23}\,\mathrm{J\,K^{-1}}\):
\( N=\frac{(1.10\times10^5)(540\times10^{-6})}{(1.38\times10^{-23})(300)} \)
\( N=1.435\times10^{22} \)
The change in total kinetic energy is
\( \Delta E_{\mathrm{K}}=\frac{3}{2}Nk\Delta T \)
The temperature change is
\( \Delta T=742-27=715\,\mathrm{K} \)
Therefore,
\( \Delta E_{\mathrm{K}}=\frac{3}{2}(1.435\times10^{22})(1.38\times10^{-23})(715) \)
\( \Delta E_{\mathrm{K}}\approx212\,\mathrm{J} \)
Answer: \( \boxed{212\,\mathrm{J}} \)
(c)
The mass of nitrogen is
\( m=24.0\,\mathrm{g}=0.0240\,\mathrm{kg} \)
The temperature decrease is
\( \Delta T=27-(-196)=223\,\mathrm{K} \)
The thermal energy removed while cooling the gas to its boiling point is
\( E_1=mc\Delta T \)
\( E_1=(0.0240)(1.04)(223) \)
\( E_1=5.57\,\mathrm{kJ} \)
The energy removed during condensation is
\( E_2=mL \)
\( E_2=(0.0240)(199)=4.78\,\mathrm{kJ} \)
Therefore, the total thermal energy removed is
\( E=E_1+E_2 \)
\( E=5.57+4.78=10.3\,\mathrm{kJ} \)
Answer: \( \boxed{10.3\,\mathrm{kJ}} \)
Question 3
An object is suspended from a vertical spring as shown in Fig. 3.1.
The object is displaced vertically and then released so that it oscillates, undergoing simple harmonic motion.
Fig. 3.2 shows the variation with displacement \(x\) of the energy \(E\) of the oscillations.
The kinetic energy, the potential energy and the total energy of the oscillations are each represented by one of the lines \(P\), \(Q\) and \(R\).
(a) State the energy that is represented by each of the lines \(P\), \(Q\) and \(R\).
P ……………………………………………………………………………………………………………………………..
Q ……………………………………………………………………………………………………………………………..
R …………………………………………………………………………………………………………………………….. [2 marks]
(b) The object has a mass of \(130\,\mathrm{g}\). Determine the period of the oscillations.
period = ……………………………………………… \( \mathrm{s} \) [4 marks]
(c)(i) State the cause of damping.
…………………………………………………………………………………………………………………………. [1 mark]
(ii) A light card is attached to the object. The object is displaced with the same initial amplitude and then released. During each complete oscillation the total energy of the system decreases by \(8.0\%\) of the total energy at the start of that oscillation.
Determine the decrease in total energy, in \(\mathrm{mJ}\), of the system by the end of the first 6 complete oscillations.
energy lost = …………………………………………… \( \mathrm{mJ} \) [2 marks]
(iii) State, with a reason, the type of damping that the card introduces into the system. [1 mark]
………………………………………………………………………………………………………………………….
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702 Paper 4):
• 17.2: Energy in simple harmonic motion — parts (a) and (b)
• 17.3: Damped and forced oscillations, resonance — parts (c)(i), (c)(ii) and (c)(iii)
▶️ Answer/Explanation
(a)
Line \(P\) is constant for all values of displacement, so it represents the total energy.
Line \(Q\) is zero at maximum displacement and maximum at \(x=0\), so it represents the kinetic energy.
Line \(R\) is maximum at maximum displacement and zero at \(x=0\), so it represents the potential energy.
Answer: \( \boxed{P:\text{ total energy},\quad Q:\text{ kinetic energy},\quad R:\text{ potential energy}} \)
(b)
From the graph, the total energy is
\( E=6.4\times10^{-3}\,\mathrm{J} \)
The amplitude is
\( x_0=1.5\,\mathrm{cm}=0.015\,\mathrm{m} \)
For simple harmonic motion,
\( E=\frac{1}{2}m\omega^2x_0^2 \)
Using \(m=130\,\mathrm{g}=0.130\,\mathrm{kg}\):
\( 6.4\times10^{-3}=\frac{1}{2}(0.130)\omega^2(0.015)^2 \)
\( \omega^2\approx438\,\mathrm{rad^2\,s^{-2}} \)
\( \omega\approx20.9\,\mathrm{rad\,s^{-1}} \)
The period is
\( T=\frac{2\pi}{\omega} \)
\( T=\frac{2\pi}{20.9} \)
\( T\approx0.30\,\mathrm{s} \)
Answer: \( \boxed{0.30\,\mathrm{s}} \)
(c)(i)
Damping is caused by resistive forces, such as air resistance or friction.
Answer: \( \boxed{\text{resistive forces}} \)
(c)(ii)
During each oscillation, \(92\%\) of the energy remains.
After 6 oscillations, the remaining energy is
\( E_6=6.4(0.92)^6\,\mathrm{mJ} \)
Therefore, the energy lost is
\( E_{\mathrm{lost}}=6.4-6.4(0.92)^6 \)
\( E_{\mathrm{lost}}\approx2.5\,\mathrm{mJ} \)
Answer: \( \boxed{2.5\,\mathrm{mJ}} \)
(c)(iii)
The card introduces light damping because the system continues to oscillate while its amplitude gradually decreases.
Answer: \( \boxed{\text{light damping}} \)
Question 4
(a) State Coulomb’s law. [2 marks]
…………………………………………
(b) A charged sphere X is supported on an insulating stand. A second charged sphere Y is suspended by an insulating thread so that sphere Y is in equilibrium at the position shown in Fig. 4.1.
The charge on sphere X is \(+96\,\mathrm{nC}\) and the charge on sphere Y is \(+64\,\mathrm{nC}\). Assume that the spheres behave as point charges.
The length of the thread is \(1.2\,\mathrm{m}\) and the centres of sphere X and sphere Y are separated horizontally by a distance of \(0.080\,\mathrm{m}\).
(i) On Fig. 4.2, draw and label all the forces acting on sphere Y. [1 mark]
(ii) Determine the mass of sphere Y.
mass = ……………………………………………. \( \mathrm{kg} \) [4 marks]
(iii) Calculate the total electric potential energy stored between X and Y.
energy = ……………………………………………. \( \mathrm{J} \) [1 mark]
(c) An electron enters the region between two parallel plates P and Q, that are separated by a distance of \(18\,\mathrm{mm}\), as shown in Fig. 4.3.
The space between the plates is a vacuum.
The potential difference between the plates is \(250\,\mathrm{V}\). The electric field may be assumed to be uniform in the region between the plates and zero outside this region.
(i) State the direction of the electric force on the electron when between the plates. [1 mark]
………………………………………………………………………………………………………………………….
(ii) Determine the magnitude of the force acting on the electron due to the electric field.
force = ……………………………………………. \( \mathrm{N} \) [2 marks]
(iii) Explain why the electron does not follow a circular path. [1 mark]
………………………………………………………………………………………………………………………….
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702 Paper 4):
• 18.5: Electric potential — part (b)(iii)
• 18.2: Uniform electric fields — parts (c)(i), (c)(ii) and (c)(iii)
▶️ Answer/Explanation
(a)
Coulomb’s law states that the electrostatic force between two point charges is proportional to the product of their charges and inversely proportional to the square of their separation.
The force acts along the line joining the two charges.
\( F=\frac{1}{4\pi\varepsilon_0}\frac{Q_1Q_2}{r^2} \)
Answer: \( \boxed{F\propto\frac{Q_1Q_2}{r^2}} \)
(b)(i)
Three forces act on sphere Y:
• weight \(mg\), vertically downwards
• tension \(T\), along the thread towards the point of suspension
• electrostatic repulsive force \(F\), horizontally away from sphere X
Answer: \( \boxed{\text{tension, weight and electrostatic repulsive force}} \)
(b)(ii)
The electrostatic force between X and Y is
\( F=\frac{1}{4\pi\varepsilon_0}\frac{Q_XQ_Y}{r^2} \)
\( F=\frac{(8.99\times10^9)(96\times10^{-9})(64\times10^{-9})}{(0.080)^2} \)
\( F=8.63\times10^{-3}\,\mathrm{N} \)
From the geometry,
\( \sin\theta=\frac{0.080}{1.2} \)
\( \theta=3.82^{\circ} \)
For equilibrium, resolving the tension horizontally and vertically gives
\( T\sin\theta=F \)
\( T\cos\theta=mg \)
Dividing these equations:
\( \tan\theta=\frac{F}{mg} \)
Therefore,
\( m=\frac{F}{g\tan\theta} \)
\( m=\frac{8.63\times10^{-3}}{9.81\tan3.82^{\circ}} \)
\( m=1.32\times10^{-2}\,\mathrm{kg} \)
Answer: \( \boxed{1.32\times10^{-2}\,\mathrm{kg}} \)
(b)(iii)
The electric potential energy between two point charges is
\( E_{\mathrm{P}}=\frac{1}{4\pi\varepsilon_0}\frac{Q_XQ_Y}{r} \)
\( E_{\mathrm{P}}=\frac{(8.99\times10^9)(96\times10^{-9})(64\times10^{-9})}{0.080} \)
\( E_{\mathrm{P}}=6.90\times10^{-4}\,\mathrm{J} \)
Answer: \( \boxed{6.9\times10^{-4}\,\mathrm{J}} \)
(c)(i)
Plate P is at \(+250\,\mathrm{V}\) and plate Q is at lower potential, so the electric field is directed downwards.
An electron has negative charge, so the electric force acts opposite to the electric field.
Answer: \( \boxed{\text{upwards, towards plate P}} \)
(c)(ii)
The electric field strength is
\( E=\frac{V}{d} \)
\( E=\frac{250}{18\times10^{-3}} \)
\( E=1.39\times10^4\,\mathrm{V\,m^{-1}} \)
The magnitude of the electric force is
\( F=eE \)
\( F=(1.60\times10^{-19})(1.39\times10^4) \)
\( F=2.22\times10^{-15}\,\mathrm{N} \)
Answer: \( \boxed{2.2\times10^{-15}\,\mathrm{N}} \)
(c)(iii)
The electric force on the electron has a constant direction while it is between the plates. For circular motion, the resultant force must continuously change direction and always act towards the centre of the circle.
Answer: \( \boxed{\text{the force does not continuously act towards a fixed centre}} \)
Question 5
A capacitor, a battery of electromotive force (e.m.f.) \(12\,\mathrm{V}\), a resistor \(R\) and a two-way switch are connected in the circuit shown in Fig. 5.1.
The switch is initially in position S. When the capacitor is fully charged, the switch is moved to position T so that the capacitor discharges. At time \(t\) after the switch is moved the charge on the capacitor is \(Q\).
The variation with \(t\) of \(\ln(Q/\mu\mathrm{C})\) is shown in Fig. 5.2.
(a) Show that the capacitance of the capacitor is \(1.5\,\mu\mathrm{F}\). [3 marks]
________________________________
(b) Determine the resistance of \(R\).
resistance = …………………………………………….. \( \Omega \) [3 marks]
(c) Calculate the energy stored in the capacitor at time \(t=0\).
energy = ……………………………………………… \( \mathrm{J} \) [2 marks]
(d) A second identical resistor is now connected in parallel with \(R\).
The switch is initially in position S. When the capacitor is fully charged, the switch is moved to position \(T\) so that the capacitor discharges. At time \(t\) after the switch is moved the charge on the capacitor is \(Q\).
On Fig. 5.2, sketch a line to show the variation of \(\ln(Q/\mu\mathrm{C})\) with \(t\) between time \(t=0\) and time \(t=5.0\,\mathrm{s}\). [2 marks]
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702 Paper 4):
• 19.3: Discharging a capacitor — parts (b) and (d)
• 19.2: Energy stored in a capacitor — part (c)
▶️ Answer/Explanation
(a)
From the graph, at \(t=0\),
\( \ln(Q/\mu\mathrm{C})\approx2.9 \)
Therefore,
\( Q=e^{2.9}\,\mu\mathrm{C} \)
\( Q\approx18.2\,\mu\mathrm{C} \)
When fully charged, the capacitor has the same potential difference as the battery, so
\( Q=CV \)
Hence,
\( C=\frac{Q}{V} \)
\( C=\frac{18.2}{12}\,\mu\mathrm{F} \)
\( C=1.5\,\mu\mathrm{F} \)
Answer: \( \boxed{1.5\,\mu\mathrm{F}} \)
(b)
For a discharging capacitor,
\( Q=Q_0e^{-t/RC} \)
Taking logarithms:
\( \ln Q=\ln Q_0-\frac{t}{RC} \)
Therefore, the gradient of the graph is
\( \text{gradient}=-\frac{1}{RC} \)
From Fig. 5.2,
\( \text{gradient}\approx-0.25\,\mathrm{s^{-1}} \)
Hence,
\( R=\frac{1}{0.25C} \)
\( R=\frac{1}{(0.25)(1.5\times10^{-6})} \)
\( R=2.7\times10^6\,\Omega \)
Answer: \( \boxed{2.7\times10^6\,\Omega} \)
(c)
The energy stored in a charged capacitor is
\( W=\frac{1}{2}QV \)
At \(t=0\), \(Q=18.2\,\mu\mathrm{C}\) and \(V=12\,\mathrm{V}\).
\( W=\frac{1}{2}(18.2\times10^{-6})(12) \)
\( W=1.1\times10^{-4}\,\mathrm{J} \)
Answer: \( \boxed{1.1\times10^{-4}\,\mathrm{J}} \)
(d)
With a second identical resistor connected in parallel with \(R\), the equivalent resistance is
\( R_{\mathrm{eq}}=\frac{R}{2} \)
For capacitor discharge, the gradient of the \(\ln(Q/\mu\mathrm{C})\) against \(t\) graph is
\( -\frac{1}{R_{\mathrm{eq}}C} \)
Since \(R_{\mathrm{eq}}=R/2\), the magnitude of the gradient becomes twice as large.
The new line therefore starts at the same point \((0,2.9)\) but has twice the negative gradient of the original line.
Answer: \( \boxed{\text{straight line starting at }(0,2.9)\text{ with twice the negative gradient}} \)
Question 6
(a) A Hall probe is placed in a magnetic field. The Hall voltage is zero. The Hall probe is rotated to a new position in the magnetic field. The Hall voltage is now maximum.
Explain these observations. [2 marks]
………………………………………………………………………………………………………………………….
(b) The formula for calculating the Hall voltage \(V_{\mathrm{H}}\) as measured by a Hall probe is
\( V_{\mathrm{H}}=\frac{BI}{ntq} \)
Table 6.1 shows the value of \(n\) for two materials.
(i) State the meaning of \(n\).
…………………………………………………………………………………………………………………….. [1 mark]
(ii) Explain why a Hall probe is made from silicon rather than copper.
…………………………………………………………………………………………………………………….. [1 mark]
(c) A Hall probe gives a maximum reading of \(24\,\mathrm{mV}\) when placed in a uniform magnetic field of flux density \(32\,\mathrm{mT}\).
The same Hall probe is then placed in a magnetic field of fixed direction and varying flux density. The Hall probe is in a fixed position so that the angle between the Hall probe and the magnetic field is the same as when the Hall voltage was \(24\,\mathrm{mV}\).
The variation of the reading \(V_{\mathrm{H}}\) on the Hall probe with time \(t\) from time \(t=0\) to time \(t=8.6\,\mathrm{s}\) is shown in Fig. 6.1.
A coil with 780 turns and a diameter of \(3.6\,\mathrm{cm}\) is placed in this varying magnetic field. The plane of the coil is perpendicular to the field lines.
Calculate the magnitude of the maximum electromotive force (e.m.f.) induced in the coil in the time between \(t=0\) and \(t=8.6\,\mathrm{s}\).
e.m.f. = ……………………………………………… \( \mathrm{V} \) [4 marks]
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702 Paper 4):
• 20.5: Electromagnetic induction — part (c)
▶️ Answer/Explanation
(a)
The Hall voltage is zero when the plane of the Hall probe is parallel to the magnetic field lines.
When the probe is rotated so that its plane is perpendicular to the magnetic field lines, the Hall voltage becomes maximum.
Answer: \( \boxed{\text{zero when the probe is parallel to the field; maximum when perpendicular}} \)
(b)(i)
\(n\) is the number density of charge carriers.
Answer: \( \boxed{\text{number density of charge carriers}} \)
(b)(ii)
From \( V_{\mathrm{H}}=\frac{BI}{ntq} \), the Hall voltage is inversely proportional to \(n\).
Silicon has a much smaller value of \(n\) than copper, so it produces a greater Hall voltage for the same magnetic field and current.
Answer: \( \boxed{\text{silicon has a smaller }n\text{, giving a greater Hall voltage}} \)
(c)
The Hall voltage is proportional to the magnetic flux density \(B\), since the Hall probe is kept at the same orientation.
A Hall voltage of \(24\,\mathrm{mV}\) corresponds to \(32\,\mathrm{mT}\).
From Fig. 6.1, the maximum Hall voltage is \(36\,\mathrm{mV}\).
Therefore, the maximum magnetic flux density is
\( B_{\max}=32\times\frac{36}{24}\,\mathrm{mT} \)
\( B_{\max}=48\,\mathrm{mT} \)
The greatest rate of change of magnetic flux occurs when the magnetic flux density falls from \(48\,\mathrm{mT}\) to zero between \(t=7.2\,\mathrm{s}\) and \(t=8.6\,\mathrm{s}\).
Thus,
\( \Delta t=8.6-7.2=1.4\,\mathrm{s} \)
The radius of the coil is
\( r=\frac{3.6}{2}\,\mathrm{cm}=1.8\times10^{-2}\,\mathrm{m} \)
The area of the coil is
\( A=\pi r^2=\pi(1.8\times10^{-2})^2 \)
Using Faraday’s law,
\( \mathcal{E}=N\frac{\Delta\Phi}{\Delta t} \)
Since the plane of the coil is perpendicular to the field lines,
\( \Delta\Phi=A\Delta B \)
Therefore,
\( \mathcal{E}=\frac{N A\Delta B}{\Delta t} \)
\( \mathcal{E}=\frac{780\times\pi(1.8\times10^{-2})^2\times48\times10^{-3}}{1.4} \)
\( \mathcal{E}=2.7\times10^{-2}\,\mathrm{V} \)
Answer: \( \boxed{0.027\,\mathrm{V}} \)
Question 7
(a) A beam of white light passes through a cloud of cool gas. The spectrum of the transmitted light is viewed and contains a number of dark lines.
Explain why these dark lines occur. [4 marks]
………………………………………………………………………………………………………………………….
(b) Some energy levels for the electron in an isolated hydrogen atom are illustrated in Fig. 7.1.
Table 7.1 shows the wavelengths of photons that are emitted in the transitions to \(n=2\) from the other energy levels shown in Fig. 7.1.
The energy associated with the energy level \(n=2\) is \(-3.40\,\mathrm{eV}\).
Calculate the energy, in \(\mathrm{J}\), of energy level \(n=3\).
energy = ……………………………………………… \( \mathrm{J} \) [3 marks]
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702 Paper 4):
▶️ Answer/Explanation
(a)
Photons from the white light are absorbed by electrons in the atoms of the cool gas.
An electron absorbs a photon and is excited to a higher energy level when the photon energy is equal to the difference between two energy levels.
The absorbed photon has a particular energy and therefore a particular frequency or wavelength.
These particular wavelengths are removed from the continuous white-light spectrum, producing dark absorption lines. The excited electrons later de-excite and emit photons in different directions.
Answer: \( \boxed{\text{specific wavelengths are absorbed by electrons during transitions between energy levels}} \)
(b)
For a photon emitted during a transition,
\( \frac{hc}{\lambda}=\Delta E \)
For the transition from \(n=3\) to \(n=2\), the wavelength from Table 7.1 is \(658\,\mathrm{nm}\).
The photon energy is therefore
\( E_{\mathrm{photon}}=\frac{(6.63\times10^{-34})(3.00\times10^8)}{658\times10^{-9}} \)
\( E_{\mathrm{photon}}=3.02\times10^{-19}\,\mathrm{J} \)
The energy of the \(n=2\) level is
\( E_2=-3.40\,\mathrm{eV} \)
Converting to joules:
\( E_2=(-3.40)(1.60\times10^{-19}) \)
\( E_2=-5.44\times10^{-19}\,\mathrm{J} \)
For the transition from \(n=3\) to \(n=2\),
\( E_{\mathrm{photon}}=E_3-E_2 \)
Therefore,
\( E_3=E_{\mathrm{photon}}+E_2 \)
\( E_3=(3.02\times10^{-19})+(-5.44\times10^{-19}) \)
\( E_3=-2.42\times10^{-19}\,\mathrm{J} \)
Answer: \( \boxed{-2.42\times10^{-19}\,\mathrm{J}} \)
Question 8
Plutonium-238 \(\left({}^{238}_{94}\mathrm{Pu}\right)\) is unstable and undergoes alpha decay.
(a) Complete the equation to show the decay of plutonium-238. [2 marks]
\({}^{238}_{94}\mathrm{Pu}\rightarrow{}^{\phantom{238}}_{\phantom{94}}\mathrm{U}+{}^{4}_{2}\alpha\)
(b) The power source in a space probe contains \(0.874\,\mathrm{kg}\) of plutonium-238. Each nucleus of plutonium-238 that decays emits \(5.59\,\mathrm{MeV}\) of energy. The half-life of plutonium-238 is \(87.7\,\mathrm{years}\).
(i) Calculate the initial number \(N_0\) of nuclei of plutonium-238 in the power source.
\(N_0=\) ……………………………………………… [1 mark]
(ii) Determine the initial activity of the source. Give a unit with your answer.
activity = ……………………………. unit ……………… [2 marks]
(iii) Use your answer in (b)(ii) to determine the initial power output from the source due to the decay of plutonium-238.
power output = …………………………………………….. \( \mathrm{W} \) [2 marks]
(iv) The space probe will continue to function until the power output from the plutonium in the source decreases to \(65.3\%\) of its initial value.
Calculate the time, in years, for which the space probe will function.
time = ……………………………………………… years [2 marks]
(c) An alternative power source uses energy generated from the radioactive decay of polonium-210. This isotope has a half-life of \(0.378\,\mathrm{years}\). The mass of the isotope needed for the same initial power output as in (b) is \(3.37\,\mathrm{g}\).
Suggest one advantage and one disadvantage of using polonium-210 as the source of energy.
advantage ………………………………………………………………………………………………………………..
disadvantage …………………………………………………………………………………………………………….. [2 marks]
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702 Paper 4):
• 23.2: Radioactive decay — parts (b)(i), (b)(ii), (b)(iii), (b)(iv) and (c)
▶️ Answer/Explanation
(a)
In alpha decay, the mass number decreases by \(4\) and the proton number decreases by \(2\).
Therefore, plutonium-238 decays to uranium-234 and an alpha particle:
\({}^{238}_{94}\mathrm{Pu}\rightarrow{}^{234}_{92}\mathrm{U}+{}^{4}_{2}\alpha\)
Answer: \( \boxed{{}^{238}_{94}\mathrm{Pu}\rightarrow{}^{234}_{92}\mathrm{U}+{}^{4}_{2}\alpha} \)
(b)(i)
The mass of one plutonium-238 nucleus is approximately
\(m_{\mathrm{nucleus}}=238(1.66\times10^{-27})\,\mathrm{kg}\)
Therefore,
\(N_0=\frac{0.874}{238(1.66\times10^{-27})}\)
\(N_0=2.21\times10^{24}\)
Answer: \( \boxed{2.21\times10^{24}} \)
(b)(ii)
Activity is given by
\(A=\lambda N\)
The decay constant is
\(\lambda=\frac{\ln2}{t_{1/2}}\)
Converting the half-life into seconds:
\(t_{1/2}=87.7\times365\times24\times3600\,\mathrm{s}\)
Hence,
\(A=\frac{\ln2}{87.7\times365\times24\times3600}(2.21\times10^{24})\)
\(A=5.54\times10^{14}\,\mathrm{Bq}\)
Answer: \( \boxed{5.54\times10^{14}\,\mathrm{Bq}} \)
(b)(iii)
Each decay releases \(5.59\,\mathrm{MeV}\).
Converting this energy into joules:
\(E=5.59\times10^6\times1.60\times10^{-19}\)
\(E=8.94\times10^{-13}\,\mathrm{J}\)
Power is the energy released per second:
\(P=AE\)
\(P=(5.54\times10^{14})(5.59\times10^6)(1.60\times10^{-19})\)
\(P=496\,\mathrm{W}\)
Answer: \( \boxed{496\,\mathrm{W}} \)
(b)(iv)
The power output is proportional to the activity, and hence to the number of undecayed nuclei.
Using the exponential decay equation:
\(P=P_0e^{-\lambda t}\)
Since \(P/P_0=0.653\),
\(0.653=e^{-\frac{\ln2}{87.7}t}\)
Taking logarithms:
\(\ln(0.653)=-\frac{\ln2}{87.7}t\)
Therefore,
\(t=\frac{-87.7\ln(0.653)}{\ln2}\)
\(t=53.9\,\mathrm{years}\)
Answer: \( \boxed{53.9\,\mathrm{years}} \)
(c)
Advantage: Polonium-210 requires a much smaller mass, \(3.37\,\mathrm{g}\), for the same initial power output, so less mass would need to be launched into space.
Disadvantage: Polonium-210 has a much shorter half-life, so its power output decreases more rapidly and it will not provide useful power for as long.
Answer: \( \boxed{\text{smaller mass required; shorter operating lifetime}} \)
Question 9
Ultrasound is used to produce diagnostic information about internal body structures.
(a) Explain how ultrasound waves are detected. [3 marks]
………………………………………………………………………………………………………………………….
(b) An alternating voltage \(V\) varies with time \(t\) according to
\( V=V_0\sin\omega t \)
The voltage is applied to an ultrasound probe.
The root-mean-square (r.m.s.) voltage is \(66\,\mathrm{V}\). The frequency of the ultrasound generated by the probe is \(4.3\,\mathrm{MHz}\).
Determine the values of
(i) \(V_0\).
\(V_0=\) ……………………………………………… \( \mathrm{V} \) [1 mark]
(ii) \(\omega\).
\(\omega=\) ……………………………………………… \( \mathrm{rad\,s^{-1}} \) [1 mark]
(c) Table 9.1 contains information about air and soft tissue.
(i) Determine the unit for the specific acoustic impedance values shown in Table 9.1. [1 mark]
………………………………………………………………………………………………………………………….
(ii) Calculate the density of soft tissue.
density = ………………………………………. \( \mathrm{kg\,m^{-3}} \) [1 mark]
(iii) Use data from Table 9.1 to explain why ultrasound cannot be used to produce an image inside an air-filled cavity such as the lungs.
………………………………………………………………………………………………………………………….
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702 Paper 4):
• 21.1: Characteristics of alternating currents — parts (b)(i) and (b)(ii)
▶️ Answer/Explanation
(a)
Ultrasound waves are detected using a piezo-electric crystal in the transducer.
The ultrasound wave causes the crystal to change shape and vibrate.
The changing shape of the crystal produces an e.m.f., which is detected and converted into an electrical signal.
Answer: \( \boxed{\text{piezo-electric crystal; ultrasound causes vibrations; vibrations produce an e.m.f.}} \)
(b)(i)
For a sinusoidal alternating voltage,
\( V_{\mathrm{rms}}=\frac{V_0}{\sqrt{2}} \)
Therefore,
\( V_0=V_{\mathrm{rms}}\sqrt{2} \)
\( V_0=66\sqrt{2} \)
\( V_0=93\,\mathrm{V} \)
Answer: \( \boxed{93\,\mathrm{V}} \)
(b)(ii)
Angular frequency is related to frequency by
\( \omega=2\pi f \)
The frequency is
\( f=4.3\times10^6\,\mathrm{Hz} \)
Hence,
\( \omega=2\pi(4.3\times10^6) \)
\( \omega=2.7\times10^7\,\mathrm{rad\,s^{-1}} \)
Answer: \( \boxed{2.7\times10^7\,\mathrm{rad\,s^{-1}}} \)
(c)(i)
Specific acoustic impedance is
\( Z=\rho c \)
The units are
\( \mathrm{kg\,m^{-3}}\times\mathrm{m\,s^{-1}}=\mathrm{kg\,m^{-2}\,s^{-1}} \)
Answer: \( \boxed{\mathrm{kg\,m^{-2}\,s^{-1}}} \)
(c)(ii)
Using
\( Z=\rho c \)
Therefore,
\( \rho=\frac{Z}{c} \)
For soft tissue, \(Z=1.7\times10^6\,\mathrm{kg\,m^{-2}\,s^{-1}}\) and \(c=1600\,\mathrm{m\,s^{-1}}\).
\( \rho=\frac{1.7\times10^6}{1600} \)
\( \rho=1.1\times10^3\,\mathrm{kg\,m^{-3}} \)
Answer: \( \boxed{1.1\times10^3\,\mathrm{kg\,m^{-3}}} \)
(c)(iii)
The specific acoustic impedance of air and soft tissue are very different:
\( Z_{\mathrm{air}}=4.3\times10^2\,\mathrm{kg\,m^{-2}\,s^{-1}} \)
\( Z_{\mathrm{tissue}}=1.7\times10^6\,\mathrm{kg\,m^{-2}\,s^{-1}} \)
This gives an intensity reflection coefficient close to \(1\), so almost all of the ultrasound is reflected at the tissue-air boundary.
Very little or no ultrasound is transmitted into the air-filled cavity, so an ultrasound image cannot be formed effectively inside it.
Answer: \( \boxed{\text{large impedance difference causes almost complete reflection, so little ultrasound enters the cavity}} \)
Question 10
(a) A student observes different stars from the Earth. Give two reasons why some stars appear brighter than others.
1 ………………………………………………………………………………………………………………………………
2 ……………………………………………………………………………………………………………………………… [2 marks]
(b) State what is meant by a standard candle.
……………………………………………………………………………………………………………………………. [1 mark]
(c) A spectral line from a star within a galaxy is observed to have a wavelength of \(660.9\,\mathrm{nm}\). The same spectral line measured in the laboratory is observed to have a wavelength of \(656.3\,\mathrm{nm}\).
(i) Show that the speed of the star relative to the Earth is \(2.1\times10^6\,\mathrm{m\,s^{-1}}\). [1 mark]
________________________________
(ii) Calculate the distance to the star.
The Hubble constant is \(2.3\times10^{-18}\,\mathrm{s^{-1}}\).
distance = ……………………………………………… \( \mathrm{m} \) [2 marks]
(iii) State and explain what can be concluded about the Universe based on this change in observed wavelength.
………………………………………………………………………………………………………………………….
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702 Paper 4):
• 25.3: Hubble’s law and the Big Bang theory — parts (c)(i), (c)(ii) and (c)(iii)
▶️ Answer/Explanation
(a)
A star may appear brighter because it is closer to the Earth.
A star may also appear brighter because it has a greater luminosity, meaning it emits more power as radiation.
Answer: \( \boxed{\text{different distances and different luminosities}} \)
(b)
A standard candle is an object with a known luminosity.
Answer: \( \boxed{\text{an object of known luminosity}} \)
(c)(i)
The change in wavelength is
\( \Delta\lambda=660.9-656.3=4.6\,\mathrm{nm} \)
For redshift,
\( \frac{\Delta\lambda}{\lambda}\approx\frac{v}{c} \)
Therefore,
\( \frac{660.9-656.3}{656.3}=\frac{v}{3.00\times10^8} \)
\( v=2.1\times10^6\,\mathrm{m\,s^{-1}} \)
Answer: \( \boxed{2.1\times10^6\,\mathrm{m\,s^{-1}}} \)
(c)(ii)
Using Hubble’s law,
\( v=H_0d \)
Therefore,
\( d=\frac{v}{H_0} \)
\( d=\frac{2.1\times10^6}{2.3\times10^{-18}} \)
\( d=9.1\times10^{23}\,\mathrm{m} \)
Answer: \( \boxed{9.1\times10^{23}\,\mathrm{m}} \)
(c)(iii)
The observed wavelength is greater than the laboratory wavelength, so the spectral line has been redshifted.
The redshift indicates that the star is moving away from the Earth.
The observation of redshift from distant objects provides evidence that the Universe is expanding.
Answer: \( \boxed{\text{the Universe is expanding}} \)