Question
The Earth may be assumed to be an isolated uniform sphere with its mass of \(6.0 × 10^{24}\) kg concentrated at its centre. A satellite of mass 1200kg is in a circular orbit about the Earth in the Earth’s gravitational field.
The period of the orbit is 94 minutes.
(a) Define gravitational field strength.
…………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………
(b) Calculate the radius of the orbit of the satellite.
radius = …………………………………………….. m
(c) Rockets on the satellite are fired so that the satellite enters a different circular orbit that has a period of 150 minutes. The change in the mass of the satellite may be assumed to be negligible.
(i) Show that the radius of the new orbit is 9.4 × 106m.
(ii) State, with a reason, whether the gravitational potential energy of the satellite increases or decreases.
………………………………………………………………………………………………………………………….
…………………………………………………………………………………………………………………….
(iii) Determine the magnitude of the change in the gravitational potential energy of the satellite.
change in potential energy = ……………………………………………… J
Answer/Explanation
Ans:
(a) force per unit mass
(b) \(GMm / r ^2 = mrω^2\) and \(ω = 2π/T\)
or
\(GMm / r ^2 = mv^2 / r\) and \(v = 2πr / T\)
\(6.67× 10^{–11}× 6.0× 10^{24} = r^3 × [2π / (94× 60)]^2\)
\(r = 6.9× 10^6 m\)
(c)(i) \(r^3ω^2 = constant\) or \(r^3 / T^2 = constant\)
\(r^3 / (6.9× 10^6)^3 = (150 / 94)^2\) so \(r = 9.4× 10^6 m\)
or
\(GMT^2/4 π^2 = r^3\) and clear that M is \(6.0× 10^{24}\)
\(6.67× 10^{–11}× 6.0× 10^{24} = r^3 × [2π / (150× 60)]^2\)
so \(r = 9.4× 10^6 m\)
(c)(ii) separation increases so (potential energy) increases
or
movement is against gravitational force so (potential energy) increases
(c)(iii) potential energy \(= (–)GMm / r\)
\(ΔE_P = 6.67× 10^{–11}× 6.0× 10^{24}× 1200× [(6.9× 10^6)^{–1} – (9.4× 10^6)^{–1}] = 1.9× 10^{10}\) J
Question
An ideal gas is contained in a cylinder by means of a movable frictionless piston, as illustrated in Fig. 2.1.
Initially, the gas has a volume of \(1.8 × 10^{−3}m^3\) at a pressure of \(3.3 × 10^5Pa\) and a temperature of 310K.
(a) Show that the number of gas molecules in the cylinder is \(1.4 × 10^{23}\).
(b) Use kinetic theory to explain why, when the piston is moved so that the gas expands, this causes a decrease in the temperature of the gas.
…………………………………………………………………………………………………………………………………
(c) The gas expands so that its volume increases to \(2.4 × 10^{−3}m^3\) at a pressure of \(2.3 × 10^5Pa\) and a temperature of 288K, as shown in Fig. 2.2.
(i) The average translational kinetic energy \(E_K\) of a molecule of an ideal gas is given by
E3=\(\frac{3}{2}\)KT
where k is the Boltzmann constant and T is the thermodynamic temperature.
Calculate the increase in internal energy ΔU of the gas during the expansion.
ΔU = ……………………………………………… J
(ii) The work done by the gas during the expansion is 76J.
Use your answer in (i) to explain whether thermal energy is transferred to or from the gas during the expansion.
………………………………………………………………………………………………………………………….
Answer/Explanation
Ans:
(a) \(pV = NkT \)
\(N = (1.8× 10^{–3}× 3.3× 10^5) / (1.38× 10^{–23}× 310) = 1.4× 10^{23}\)
or
\(pV = nRT\) and \(nNA = N \)
\(N = (1.8× 10^{–3}× 3.3× 10^5 × 6.02× 10^{23}) / (8.31× 310) = 1.4× 10^{23}\)
(b) speed of molecule decreases on impact with moving piston
mean square speed (directly) proportional to (thermodynamic) temperature
or
mean square speed (directly) proportional to kinetic energy (of molecules)
or
kinetic energy (of molecules) (directly) proportional to (thermodynamic) temperature
kinetic energy (of molecules) decreases (so temperature decreases)
(c)(i) \(ΔU = 3/2× k×ΔT×N\)
\(= 3/2× 1.38× 10^{–23}× (288 – 310)× 1.4× 10^{23}\)
\(=– 64 J \)
(c)(ii) decrease in internal energy is less than work done by gas
(thermal energy is) transferred to the gas (during the expansion)
Question
(a) State what is meant by simple harmonic motion.
…………………………………………………………………………………………………………………………………
(b) A trolley of mass m is held on a horizontal surface by means of two springs. One spring is attached to a fixed point P. The other spring is connected to an oscillator, as shown in Fig. 3.1.
The springs, each having spring constant k of 130Nm−1, are always extended. The oscillator is switched off. The trolley is displaced along the line of the springs and then released. The resulting oscillations of the trolley are simple harmonic. The acceleration a of the trolley is given by the expression
f = ……………………………………………. Hz
(c) The oscillator in (b) is switched on. The frequency of oscillation of the oscillator is varied, keeping its amplitude of oscillation constant. The amplitude of oscillation of the trolley is seen to vary. The amplitude is a maximum at the frequency calculated in (b).
(i) State the name of the effect giving rise to this maximum.
…………………………………………………………………………………………………………………….
(ii) At any given frequency, the amplitude of oscillation of the trolley is constant. Explain how this indicates that there are resistive forces opposing the motion of the trolley.
………………………………………………………………………………………………………………………….
Answer/Explanation
Ans:
(a) acceleration (directly) proportional to displacement
acceleration is in opposite direction to displacement
(b)\(ω^2 = 2k / m \)and \(ω = 2πf\)
\((2πf)^2 = (2× 130) / 0.84 \)
\(f = 2.8 Hz\)
(c)(i) resonance
(c)(ii) oscillator supplies energy (continuously)
energy of trolley constant so energy must be dissipated
or
without loss of energy the amplitude would continuously increase
Question
Outline the use of ultrasound to obtain diagnostic information about internal body structures.
……………………………………………………………………………………………………………………………………….
……………………………………………………………………………………………………………………………………….
Answer/Explanation
Ans:
(ultrasound) pulse
reflected at boundaries
gel is used to minimise reflection at skin
or
generated and detected by quartz crystal
time delay between generation and detection gives information about depth
intensity (of reflected wave) gives information about nature of boundary
Question
(a) State what is meant by the amplitude modulation (AM) of a radio wave.
…………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………………
(b) A radio wave is modulated by an audio signal.
The variation with frequency f of the amplitude of the modulated wave is shown in Fig. 5.1.
Determine:
(i) the wavelength of the carrier wave
wavelength = …………………………………………….. m [1]
(ii) the bandwidth of the modulated wave
bandwidth = ………………………………………….. kHz [1]
(iii) the maximum frequency of the audio signal.
maximum frequency = ………………………………………….. kHz [1]
(c) The power of a radio signal at a transmitter is PT.
At a receiver, the received power PR is given by the expression
where x is the distance, in metres, between the transmitter and the receiver. For the transmission of this signal, the attenuation is 73dB. Determine the distance x.
x = …………………………………………….. m
Answer/Explanation
Ans:
(a) amplitude of the carrier wave varies
in synchrony with the displacement of the (information) signal
(b)(i) wavelength \(= (3.0× 10^8) / (300× 10^3)= 1000 m\)
(b)(ii) bandwidth \(= 16 kHz \)
(b)(iii) frequency \(= 8 kHz \)
(c) attenuation \(= 10 lg (P_1 / P_2)\)
\(73 = 10 lg (P_T / P_R)\)
\(73 = 10 lg (\frac{P_T (x^2) }{ 0.082 P_T})\) or \(x^2 / 0.082 = 10^{7.3}\)
\(x = 1300 m\)
Question
(a) An isolated metal sphere of radius r is charged so that the electric field strength at its surface is E0. On Fig. 6.1, sketch the variation of the electric field strength E with distance x from the centre of the sphere. Your sketch should extend from x = 0 to x = 3r.
(c) A radioactive isotope decays with a half-life of 15s to form a stable product.
A fresh sample of the radioactive isotope at time t = 0 contains N0 nuclei and no nuclei of the stable product. On Fig. 6.3, sketch the variation with t of the number n of nuclei of the stable product for time t = 0 to time t = 45s.
Answer/Explanation
Ans:
(a) from x = 0 to x = r: E = 0
from x = r to x = 3r: curve with negative gradient of decreasing magnitude passing through (r, E0)
line passing through (2r, E0 / 4) and (3r, E0 / 9)
(b) from p =p0 / 2 to p =p0: curve with negative gradient of decreasing magnitude passing through (p0, λ0)
line passing through (1⁄2p0, 2λ0)
(c) from t = 0 to t = 45 s: curve with positive gradient of decreasing magnitude starting at (0, 0)
line passing through (15, 1⁄2N0)
line passing through (30, 0.75N0) and (45, 0.88N0)
Question
(a) State what is meant by the capacitance of a parallel plate capacitor.
…………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………
(b) A capacitor of capacitance C is connected into the circuit shown in Fig. 7.1.
When the two-way switch is in position A, the capacitor is charged so that the potential difference across it is V. The switch moves to position B and the capacitor fully discharges through the sensitive ammeter. The switch moves repeatedly between A and B so that the capacitor charges and then discharges with frequency f.
(i) Show that the average current I in the ammeter is given by the expression
\(I = fCV\).
(ii) For a potential difference V of 150V and a frequency f of 60Hz, the average current in the ammeter is 4.8μA.
Calculate the capacitance, in pF, of the capacitor.
capacitance = ……………………………………………. pF [2]
(c) A second capacitor, having the same capacitance as the capacitor in (b), is connected into the circuit of Fig. 7.1. The two capacitors are connected in series.
State and explain the new reading on the ammeter.
new reading = ………………………………………………… μA
…………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………
Answer/Explanation
Ans:
(a) charge / potential
charge is on one plate, potential is p.d. between the plates
(b)(i) \(I = Q / t \)
\(charge = CV\) and \(time = 1 / f\) leading to \(I = fCV\)
(b)(ii) \(4.8 × 10^{–6} = 150 × 60 × C \)
\(C = 530 pF\)
(c) (total) capacitance is halved
charge (for each cycle/discharge) is halved
or
since f and V are constant, current is proportional to capacitance
current \(= 2.4 μA\)
Question
The variation with temperature of the resistance of a thermistor is shown in Fig. 8.1.
(a) Calculate the potential \(V^+\) at the non-inverting input of the op-amp.
\(V^+\) = ……………………………………………… V [2]
(b) At 10°C, the resistance of the thermistor is 2.5kΩ.
State and explain whether the light-emitting diode (LED) is emitting light.
…………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………… [2]
(c) Explain why the student’s circuit will not indicate any change in temperature above 0°C.
…………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………… [2]
(d) The resistor of resistance 5.0kΩ is changed to a resistor of resistance R so that the LED
switches on or off at a temperature of 20°C.
Determine R in kΩ.
Answer/Explanation
Ans:
(a) \(V^+ = 3.0× 3.0 / (2.5 + 3.0) = 1.6V\)
(b)\(V^–\) is \(+2.0 V\) or \(V^– > V^+\)
output is negative so (LED) does not emit light
(c) at 0 °C, \(V– = 1.7 V\)
or
for all temperatures above 0 °C, resistance of thermistor \(< 4.2kΩ\)
\(V^–\) always greater than \(V^+\) (so no switching)
(d) (at 20 °C,) \(RT = 1.8 kΩ\)
\(2.5 / 3.0 = 1.8 / R\)
or
\([R / (R + 1.8)]× 3.0 = 1.6\)
\(R = 2.2kΩ\)
Question
(a) State what is meant by a magnetic field.
…………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………… [2]
(b) A rectangular piece of aluminium foil is situated in a uniform magnetic field of flux density B, as shown in Fig. 9.1.
The magnetic field is normal to the face PQRS of the foil. Electrons, each of charge −q, enter the foil at right angles to the face PQTV.
(i) On Fig. 9.1, shade the face of the foil on which electrons initially accumulate. [1]
(ii) Explain why electrons do not continuously accumulate on the face you have shaded.
………………………………………………………………………………………………………………………….
………………………………………………………………………………………………………………………….
………………………………………………………………………………………………………………………….
……………………………………………………………………………………………………………………. [3]
(c) The Hall voltage \(V_H\) developed across the foil in (b) is given by the expression
\(V_H =\frac{ BI}{ ntq}\)
where I is the current in the foil.
(i) State the meaning of the quantity n.
………………………………………………………………………………………………………………………….
……………………………………………………………………………………………………………………. [1]
(ii) Using the letters on Fig. 9.1, identify the distance t.
……………………………………………………………………………………………………………………. [1]
(d) Suggest why, in practice, Hall probes are usually made using a semiconductor material rather than a metal.
…………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………… [1]
[Total: 9]
Answer/Explanation
Ans:
(a) region where there is a force exerted on a current-carrying conductor
or
a moving charge
or
a magnetic material/magnetic pole
(b)(i) face PSWV shaded
(b)(ii) accumulating electrons cause an electric field (between the faces)
force due to electric field opposes force due to magnetic field
accumulation stops when magnetic force equals electric force
(c)(i) number density of charge carriers
(c)(ii) PV or QT or SW
(d) (for semiconductor,) n is (much) smaller so \(V_H\) (much) larger
Question
(a) State Lenz’s law.
…………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………… [2]
(b) A metal ring is suspended from a fixed point P by means of a thread, as shown in Fig. 10.1.
The ring is displaced a distance d and then released. The ring completes many oscillations before coming to rest.
The poles of a magnet are now placed near to the ring so that the ring hangs midway between the poles of the magnet, as shown in Fig. 10.2.
The ring is again displaced a distance d and then released.
Explain why the ring completes fewer oscillations before coming to rest.
…………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………………
(c) The ring in (b) is now cut so that it has the shape shown in Fig. 10.3.
Explain why, when the procedure in (b) is repeated, the cut ring completes more oscillations
than the complete ring when oscillating between the poles of the magnet.
…………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………… [3]
[Total: 9]
Answer/Explanation
Ans:
(a) direction of (induced) e.m.f.
is such as to oppose the change causing it
(b) ring cuts (magnetic) flux and causes induced e.m.f. in ring
(induced) e.m.f. causes (eddy/induced) currents (in ring)
currents (in ring) cause magnetic field (around ring)
two fields interact to cause resistive/opposing force
or
current (in ring) is in a magnetic field
which causes resistive force
or
currents (in ring) dissipate thermal energy
(thermal) energy comes from energy of oscillations
(c) current cannot pass all the way around the ring
(induced) currents smaller
smaller resistive force (so more oscillations)
or
smaller rate of dissipation of energy (so more oscillations)
Question
(a) State how, in a modern X-ray tube, the intensity of the X-ray beam and its hardness are
controlled.
intensity: …………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………………………
hardness: ………………………………………………………………………………………………………………….
…………………………………………………………………………………………………………………………………
[2]
(b) A model of a limb consists of soft tissue and bone, as illustrated in Fig. 11.1.
A parallel beam of X-rays of intensity \(I_0\) is incident normally on the model.
Calculate, in terms of \(I_0\):
(i) the transmitted intensity \(I_S\) through soft tissue alone
\(I_S\) = …………………………………………….. \(I_0\) [2]
(ii) the transmitted intensity IC through soft tissue and bone.
\(I_C\) = …………………………………………….. \(I_0\) [2]
(c) By reference to your answers in (b), suggest, with a reason, whether good contrast on an X-ray image would be obtained.
…………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………
Answer/Explanation
Ans:
(a) intensity: vary filament current/p.d. across filament
hardness: vary accelerating potential difference
(b)(i) \(I =I_0e^{ –μx}\)
\(I_S =I_0 exp(–0.92× 9.0)= 2.5× 10^{–4}I_0\)
(b)(ii) \(IC = [exp(–0.92× 6.0)× exp(–2.9× 3.0)]I_0\)
\(= 6.7× 10^{–7}I_0\)
(c) conclusion consistent with values in (b)(i) and (b)(ii)
e.g.
\(I_S ≫ I_C\) so good contrast
Question
(a) Electromagnetic radiation of a single constant frequency is incident on a metal surface. This causes an electron to be emitted.
Explain why the maximum kinetic energy of the electron is independent of the intensity of the incident radiation.
…………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………… [3]
(b) Ultraviolet radiation of wavelength 250nm is incident on the surface of a sheet of zinc. The maximum kinetic energy of the emitted electrons is 1.4eV.
Determine, in eV:
(i) the energy of a photon of the ultraviolet radiation
energy = ……………………………………………. eV [3]
(ii) the work function energy of the surface of the zinc.
energy = …………………………………………….
Answer/Explanation
Ans:
(a)
- frequency determines energy of photon
- intensity determines number of photons (per unit time)
- intensity does not determine energy of a photon
Any two points, 1 mark each
kinetic energy (of the electron) depends on the energy of one photon
(b)(i) \(E = hc / λ\)
or
\(E = hf\) and \(c = fλ\)
\(E = (6.63× 10^{–34}× 3.00× 10^8) / (250× 10^{–9})\)
\((= 7.96× 10^{–19} J)= 5.0 eV\)
(b)(ii) \(E_{MAX}\) = photon energy – work function
work function \(= 5.0 – 1.4 = 3.6 eV\)