▶️ Answer/Explanation
(a)(i) Radius of Cambridge’s circular path [1 mark]
The radius of the circle is the horizontal component of the Earth’s radius:
\(r=R\cos52.2^\circ\)
\(r=(6.37\times10^6)\cos52.2^\circ\)
\(r=3.90\times10^6\,\mathrm{m}\)
Answer: \(r=3.90\times10^6\,\mathrm{m}\)
(a)(ii) Speed of Cambridge [3 marks]
The Earth completes one rotation in \(24\,\mathrm{h}\), so the period is \(T=24\times60\times60\,\mathrm{s}\).
For uniform circular motion, \(v=\dfrac{2\pi r}{T}\).
Therefore, \(v=\dfrac{2\pi(3.90\times10^6)}{24\times60\times60}\)
\(v=280\,\mathrm{m\,s^{-1}}\)
Answer: \(v=280\,\mathrm{m\,s^{-1}}\)
(b)(i) Resultant force on the student [2 marks]
The student undergoes circular motion, so the resultant force is the centripetal force:
\(F=\dfrac{mv^2}{r}\)
\(F=\dfrac{(58.6)(280)^2}{3.90\times10^6}\)
\(F=1.2\,\mathrm{N}\)
Answer: \(F=1.2\,\mathrm{N}\)
(b)(ii) Direction of resultant force [1 mark]
The centripetal resultant force acts horizontally towards the centre of the circular path.
Answer: Arrow pointing horizontally to the left.
(b)(iii) Forces acting on the student [2 marks]
The student’s weight acts vertically downwards, along the dotted radial line towards the centre of the Earth.
The contact force from the ground acts upwards, perpendicular to the Earth’s surface. It is directed to the left of the normal shown in the diagram.
These two forces have a resultant directed horizontally towards the centre of the student’s circular path.
Answer: Draw a downward arrow labelled weight along the dotted radial line, and an upward contact-force arrow directed to the left of the normal and above the tangent to the Earth’s surface.
▶️ Answer/Explanation
(a) Newton’s law of gravitation [2 marks]
The gravitational force between two point masses is directly proportional to the product of their masses.
The force is inversely proportional to the square of their separation.
\(F=\dfrac{Gm_1m_2}{r^2}\)
Answer: \(F\propto\dfrac{m_1m_2}{r^2}\).
(b) Assumptions of the kinetic theory of gases [2 marks]
Any two of the following:
1. The molecules are in continuous random motion.
2. The molecules have negligible volume compared with the volume of the gas.
3. Collisions involving molecules are perfectly elastic.
4. Collisions between molecules are instantaneous.
(c)(i) Pressure of the gas [2 marks]
Using the ideal gas equation, \(pV=nRT\).
Therefore, \(p=\dfrac{nRT}{V}\).
\(p=\dfrac{(0.0160)(8.31)(282)}{1.87\times10^{-4}}\)
\(p=2.01\times10^5\,\mathrm{Pa}\)
Answer: \(p=2.01\times10^5\,\mathrm{Pa}\)
(c)(ii) Average separation of the molecules [2 marks]
First determine the number of molecules:
\(N=nN_{\mathrm A}\)
\(N=(0.0160)(6.02\times10^{23})\)
\(N=9.63\times10^{21}\)
If each molecule occupies an average volume \(s^3\), then \(s^3=\dfrac{V}{N}\).
\(s=\sqrt[3]{\dfrac{1.87\times10^{-4}}{0.0160(6.02\times10^{23})}}\)
\(s=2.7\times10^{-9}\,\mathrm{m}\)
Answer: \(s=2.7\times10^{-9}\,\mathrm{m}\)
(d)(i) Average gravitational force [2 marks]
Using Newton’s law of gravitation, \(F=\dfrac{Gm^2}{r^2}\).
Here, \(m=3.34\times10^{-27}\,\mathrm{kg}\) and \(r=2.7\times10^{-9}\,\mathrm{m}\).
\(F=\dfrac{(6.67\times10^{-11})(3.34\times10^{-27})^2}{(2.7\times10^{-9})^2}\)
\(F=1.0\times10^{-46}\,\mathrm{N}\)
Answer: \(F=1.0\times10^{-46}\,\mathrm{N}\)
(d)(ii) Comparison with the weight of a molecule [1 mark]
The weight of one molecule is approximately \(W=mg\approx10^{-26}\,\mathrm{N}\).
The gravitational force between adjacent molecules is only about \(10^{-46}\,\mathrm{N}\), which is enormously smaller than the weight.
Answer: The gravitational force between molecules is negligible, so the result is consistent with the kinetic theory assumption that intermolecular forces can be ignored except during collisions.
▶️ Answer/Explanation
(a) Thermal equilibrium [2 marks]
Two objects are in thermal equilibrium when they are at the same temperature.
There is no net transfer of thermal energy between them.
(b)(i) Property of the liquid [1 mark]
The property used is the density of the liquid.
The pressure difference is related to the height difference by \(\Delta p=\rho g\Delta h\).
(b)(ii) Disadvantages of a constant volume gas thermometer [2 marks]
Any two valid disadvantages:
1. It has a large response time because the gas must reach thermal equilibrium with the environment, so it cannot measure rapidly changing temperatures.
2. A significant transfer of thermal energy may be required to reach thermal equilibrium, which can change the temperature of the environment being measured and makes it unsuitable for very small objects.
Another valid disadvantage is that the apparatus is bulky and difficult to set up or take readings from.
(b)(iii) Appropriate situation [1 mark]
It would be appropriate for measuring the temperature of a substance with a large mass or a temperature that remains constant over time.
It may also be used to calibrate other thermometers in a laboratory.
(b)(iv) Determination of \(\theta\) [3 marks]
The level of \(X\) remains fixed at \(2.31\,\mathrm{cm}\), so the height difference is \(\Delta h=Y-X\).
At \(0^\circ\mathrm{C}\):
\(\Delta h_0=8.69-2.31=6.38\,\mathrm{cm}\)
At temperature \(\theta\):
\(\Delta h_\theta=7.83-2.31=5.52\,\mathrm{cm}\)
Since the gas is at constant volume, its thermodynamic temperature is proportional to its pressure, and the pressure is proportional to \(\Delta h\):
\(\dfrac{T_\theta}{273}=\dfrac{5.52}{6.38}\)
\(T_\theta=273\left(\dfrac{5.52}{6.38}\right)\)
\(T_\theta\approx236\,\mathrm{K}\)
Therefore, \(\theta=T_\theta-273\)
\(\theta\approx-36.8^\circ\mathrm{C}\)
Answer: \(\boxed{\theta\approx-37^\circ\mathrm{C}}\)
▶️ Answer/Explanation
(a) First law of thermodynamics [2 marks]
The change in internal energy of a system is equal to the work done on the system plus the thermal energy transferred to the system by heating.
\(\Delta U=W+Q\)
Here, \(W\) is the work done on the system and \(Q\) is the thermal energy supplied to the system.
(b)(i) General expression for internal energy [1 mark]
From Fig. 4.1 and Fig. 4.2, the internal energy is proportional to the product of pressure and volume:
\(U=\dfrac{3}{2}pV\)
Answer: \(U=\dfrac{3}{2}pV\)
(b)(ii) Internal energy in terms of \(N\) and \(T\) [2 marks]
For an ideal gas, \(pV=NkT\), where \(k\) is the Boltzmann constant.
Substituting into \(U=\dfrac{3}{2}pV\):
\(U=\dfrac{3}{2}NkT\)
Answer: \(U=\dfrac{3}{2}NkT\), where \(k\) is the Boltzmann constant.
(c)(i) Work done during AB [1 mark]
During AB, the pressure is constant at \(2Y\), while the volume decreases from \(6X\) to \(2X\).
The work done on the gas is positive because the gas is compressed:
\(W=(2Y)(6X-2X)=8XY\)
Answer: \(W=+8XY\)
(c)(ii) Work done during CD [1 mark]
During CD, the pressure is constant at \(5Y\), while the volume increases from \(2X\) to \(6X\).
\(W=-(5Y)(6X-2X)=-20XY\)
Answer: \(W=-20XY\)
(d) Net thermal energy supplied during one cycle [3 marks]
During the vertical stages BC and DA, the volume is constant, so no work is done:
\(W_{\mathrm{BC}}=W_{\mathrm{DA}}=0\)
Over one complete cycle, the gas returns to its initial state, so its change in internal energy is zero:
\(\Delta U=0\)
From the first law, \(\Delta U=W+Q\).
Therefore, for the complete cycle, \(Q=-W\).
The net work done on the gas is \(W=8XY-20XY=-12XY\).
Hence, \(Q=-(-12XY)=12XY\).
Answer: \(Q=+12XY\)
▶️ Answer/Explanation
(a)(i) Simple harmonic motion [2 marks]
The graph is a straight line through the origin, showing that the acceleration is proportional to the displacement:
\(a\propto x\)
The negative gradient shows that the acceleration is always in the opposite direction to the displacement.
Therefore, \(a\propto -x\), which is the condition for simple harmonic motion.
(a)(ii) Period of oscillations [3 marks]
For simple harmonic motion, \(a_0=\omega^2x_0\).
From the graph, \(x_0=1.2\,\mathrm{cm}\) and \(a_0=13\,\mathrm{cm\,s^{-2}}\).
Hence, \(\omega=\sqrt{\dfrac{a_0}{x_0}}\).
Also, \(\omega=\dfrac{2\pi}{T}\).
Therefore, \(T=2\pi\sqrt{\dfrac{x_0}{a_0}}\).
\(T=2\pi\sqrt{\dfrac{1.2}{13}}\)
\(T=1.9\,\mathrm{s}\)
Answer: \(T=1.9\,\mathrm{s}\)
(b)(i) Damping [2 marks]
Damping is the loss of energy from an oscillating system.
The energy is lost due to resistive force(s).
(b)(ii) Displacement-time graph for heavily damped motion [3 marks]
At \(t=0\), the displacement has its maximum value, so the graph starts at a non-zero positive displacement of approximately \(+1.2\,\mathrm{cm}\).
The oscillations are heavily damped, so the magnitude of the displacement decreases continuously with time.
A suitable sketch remains on one side of the \(t\)-axis, with both the magnitude of \(x\) and the magnitude of the gradient continuously decreasing.
Answer: A decaying curve starting from \(x\approx1.2\,\mathrm{cm}\) at \(t=0\), remaining above the \(t\)-axis and approaching \(x=0\) by \(t=2T\).
▶️ Answer/Explanation
(a) Electric field [1 mark]
Electric field at a point is the force per unit positive charge placed at that point.
\(E=\dfrac{F}{Q}\)
(b)(i) Electric field lines [2 marks]
The sphere has a positive charge, so the electric field lines are radial and directed away from the sphere.
Answer: Draw radial field lines perpendicular to the surface, with arrows pointing away from the positively charged sphere.
(b)(ii) Electric potential at the surface [2 marks]
The capacitance is related to charge and potential by \(C=\dfrac{Q}{V}\).
Therefore, \(V=\dfrac{Q}{C}\).
\(V=\dfrac{83}{69}\)
\(V=1.2\,\mathrm{V}\)
Answer: \(V=+1.2\,\mathrm{V}\)
(b)(iii) Radius of the sphere [2 marks]
For an isolated conducting sphere, \(V=\dfrac{Q}{4\pi\varepsilon_0r}\).
Rearranging, \(r=\dfrac{Q}{4\pi\varepsilon_0V}\).
\(r=\dfrac{83\times10^{-12}}{4\pi(8.85\times10^{-12})(1.2)}\)
\(r=0.62\,\mathrm{m}\)
Answer: \(r=0.62\,\mathrm{m}\)
(b)(iv) Electric field strength at the surface [2 marks]
For a point charge, \(E=\dfrac{Q}{4\pi\varepsilon_0r^2}\).
\(E=\dfrac{83\times10^{-12}}{4\pi(8.85\times10^{-12})(0.62)^2}\)
\(E=1.9\,\mathrm{N\,C^{-1}}\)
Answer: \(E=1.9\,\mathrm{N\,C^{-1}}\)
(c) Discharge of the sphere [2 marks]
The charge on a discharging capacitor follows \(Q=Q_0e^{-t/RC}\).
Here, \(Q_0=83\,\mathrm{pC}\), \(Q=26\,\mathrm{pC}\), \(R=120\times10^6\,\Omega\), and \(C=69\times10^{-12}\,\mathrm{F}\).
\(26=83\exp\left[-\dfrac{t}{(120\times10^6)(69\times10^{-12})}\right]\)
\(t=9.6\times10^{-3}\,\mathrm{s}\)
Answer: \(t=9.6\times10^{-3}\,\mathrm{s}\)
▶️ Answer/Explanation
(a)(i) Period of the alternating voltage [1 mark]
Comparing \(V=18\cos40\pi t\) with \(V=V_0\cos\omega t\):
\(\omega=40\pi\,\mathrm{rad\,s^{-1}}\)
Using \(\omega=\dfrac{2\pi}{T}\):
\(T=\dfrac{2\pi}{40\pi}\)
\(T=0.050\,\mathrm{s}\)
Answer: \(T=0.050\,\mathrm{s}\)
(a)(ii) r.m.s. voltage [1 mark]
For a sinusoidal alternating voltage, \(V_{\mathrm{r.m.s.}}=\dfrac{V_0}{\sqrt{2}}\).
\(V_{\mathrm{r.m.s.}}=\dfrac{18}{\sqrt{2}}\)
\(V_{\mathrm{r.m.s.}}\approx13\,\mathrm{V}\)
Answer: \(13\,\mathrm{V}\)
(b) Sketch of alternating voltage [3 marks]
The graph is a sinusoidal curve with period \(50\,\mathrm{ms}\).
At \(t=0\), \(V\) has its maximum value of \(+18\,\mathrm{V}\). The voltage reaches \(-18\,\mathrm{V}\) at \(t=25\,\mathrm{ms}\), returns to \(+18\,\mathrm{V}\) at \(t=50\,\mathrm{ms}\), and repeats this pattern.
The sketch therefore shows two complete sinusoidal cycles between \(0\) and \(100\,\mathrm{ms}\), with maximum and minimum voltages of \(\pm18\,\mathrm{V}\).
(c) Conclusions from Fig. 7.3 [3 marks]
Any three valid conclusions:
• The rectification is full-wave.
• The mean power is approximately \(14\,\mathrm{W}\).
• The resistance of the load resistor is approximately \(12\,\Omega\).
• The peak current in \(R\) is approximately \(1.6\,\mathrm{A}\), or the r.m.s. current is approximately \(1.1\,\mathrm{A}\).
• The period of the output voltage or power is \(25\,\mathrm{ms}\).
Therefore, the frequency of the output voltage or power is \(40\,\mathrm{Hz}\), or its angular frequency is approximately \(250\,\mathrm{rad\,s^{-1}}\).
▶️ Answer/Explanation
(a) Photon model and discrete energy levels [3 marks]
When an electron moves between two energy levels, it emits a single photon.
The energy of the photon is equal to the difference between the two electron energy levels: \(\Delta E=hf\).
Since the spectrum contains only discrete frequencies, the possible differences between electron energies must also be discrete. Therefore, the electron energy levels are discrete.
(b)(i) Energy of the ground state [1 mark]
\(E=-(13.6)(1.60\times10^{-19})\)
\(E=-2.18\times10^{-18}\,\mathrm{J}\)
Answer: \(-2.18\times10^{-18}\,\mathrm{J}\)
(b)(ii) Energy difference between \(n=1\) and \(n=2\) [2 marks]
The lowest-frequency spectral line shown is \(2.47\times10^{15}\,\mathrm{Hz}\), corresponding to the transition between \(n=2\) and \(n=1\).
Using \(\Delta E=hf\):
\(\Delta E=(6.63\times10^{-34})(2.47\times10^{15})\)
Converting to eV: \[ \Delta E=\dfrac{(6.63\times10^{-34})(2.47\times10^{15})}{1.60\times10^{-19}} \]
\(\Delta E=10.2\,\mathrm{eV}\)
Answer: \(\Delta E=10.2\,\mathrm{eV}\)
(b)(iii) Hydrogen energy levels [4 marks]
For \(n=2\), the energy difference from the ground state is \(10.2\,\mathrm{eV}\), so
\(E_2=-13.6+10.2=-3.4\,\mathrm{eV}\)
The spectral line at \(2.92\times10^{15}\,\mathrm{Hz}\) corresponds to the transition from \(n=3\) to \(n=1\):
\(\Delta E=\dfrac{(6.63\times10^{-34})(2.92\times10^{15})}{1.60\times10^{-19}}\)
\(\Delta E=12.1\,\mathrm{eV}\)
Therefore, \(E_3=-13.6+12.1=-1.5\,\mathrm{eV}\).
The spectral line at \(3.09\times10^{15}\,\mathrm{Hz}\) corresponds to the transition from \(n=4\) to \(n=1\):
\(\Delta E=\dfrac{(6.63\times10^{-34})(3.09\times10^{15})}{1.60\times10^{-19}}\)
\(\Delta E=12.8\,\mathrm{eV}\)
Therefore, \(E_4=-13.6+12.8=-0.8\,\mathrm{eV}\).
Completed values:
| Level | Energy difference from \(n=1\) / eV | Energy / eV |
|---|---|---|
| \(n=4\) | \(12.8\) | \(-0.8\) |
| \(n=3\) | \(12.1\) | \(-1.5\) |
| \(n=2\) | \(10.2\) | \(-3.4\) |
| \(n=1\) | \(0.0\) | \(-13.6\) |
▶️ Answer/Explanation
(a) Mass defect [2 marks]
Mass defect is the difference between the mass of a nucleus and the total mass of its constituent nucleons when the nucleons are separated to infinity.
(b) Energy released in nuclear fusion [4 marks]
The mass defect for the formation of one helium-4 nucleus is
\(\Delta m=(2\times2.013553)-4.001505\)
\(\Delta m=0.025601\,\mathrm{u}\)
Using \(E=c^2\Delta m\):
\(E=0.025601\times1.66\times10^{-27}\times(3.00\times10^8)^2\)
\(E=3.82\times10^{-12}\,\mathrm{J}\)
One mole contains \(6.02\times10^{23}\) nuclei, so
\(E=(3.82\times10^{-12})(6.02\times10^{23})\)
\(E=2.30\times10^{12}\,\mathrm{J}\)
Answer: \(2.30\times10^{12}\,\mathrm{J}\)
(c)(i) Luminosity of Sirius [2 marks]
The energy released per kilogram of mass converted is \(c^2\). Therefore, the power released is
\(L=\dot{m}c^2\)
\(L=(1.09\times10^{11})(3.00\times10^8)^2\)
\(L=9.81\times10^{27}\,\mathrm{W}\)
Answer: \(L=9.81\times10^{27}\,\mathrm{W}\)
(c)(ii) Surface temperature of Sirius [2 marks]
Using the Stefan-Boltzmann law, \(L=4\pi\sigma r^2T^4\).
\(9.81\times10^{27}=4\pi(5.67\times10^{-8})(1.19\times10^9)^2T^4\)
Rearranging, \(T=\left[\dfrac{9.81\times10^{27}}{4\pi(5.67\times10^{-8})(1.19\times10^9)^2}\right]^{1/4}\)
\(T=9930\,\mathrm{K}\)
Answer: \(T=9930\,\mathrm{K}\)
(d) Standard candles [3 marks]
Standard candles are objects whose luminosity is known.
The radiant flux intensity from the object is measured on Earth.
The distance can then be calculated using the inverse-square relationship:
\(F=\dfrac{L}{4\pi d^2}\)
Hence, \(d=\sqrt{\dfrac{L}{4\pi F}}\).
▶️ Answer/Explanation
(a) Contrast [1 mark]
Contrast is the difference in degrees of blackening between different parts of an X-ray image.
(b)(i) Intensity in region B [1 mark]
The X-rays in region B pass through \(5.8\,\mathrm{cm}\) of material P.
Using the attenuation equation \(I=I_0\exp(-\mu x)\):
\(I=I_0\exp[-(0.35)(5.8)]\)
\(I=0.13I_0\)
Answer: \(I=0.13I_0\)
(b)(ii) Linear attenuation coefficient of Q [3 marks]
The thickness of material P in region A is
\(5.8-2.1=3.7\,\mathrm{cm}\)
Therefore,
\(0.053I_0=I_0\exp[-(0.35\times3.7+2.1\mu)]\)
Dividing by \(I_0\):
\(0.053=\exp[-(0.35\times3.7+2.1\mu)]\)
Solving for \(\mu\):
\(\mu\approx0.78\,\mathrm{cm^{-1}}\)
Answer: \(\mu=0.78\,\mathrm{cm^{-1}}\)
(b)(iii) Contrast [1 mark]
The detected intensities are only different by a factor of \(\dfrac{0.13}{0.053}\approx2.5\), so there is only a small difference in blackening and hence poor contrast.
(c) CT scanning [3 marks]
The structure is scanned in thin sections.
Many X-ray scans of each section are taken from different angles.
The scanning is repeated for all sections and the data are processed and combined to form a three-dimensional image.