June 2017

SECTION A
Answer ALL questions.
All multiple-choice questions must be answered with a cross in the box for the correct answer from A to D. If you change your mind about an answer, put a line through the box and then mark your new answer with a cross.
1. A spring is hung vertically and masses are added to the lower end. The graph shows how the extension Δx of the spring varies with the mass m added.
The work done in extending the spring can be expressed as
Answer: C
Explain:
The work done in extending a spring is equal to the elastic potential energy stored in the spring. The force required to extend a spring is not constant; it increases linearly with the extension. The work done can be calculated by finding the area under the force-extension graph.
[math]F = mg [/math]
The extension is [math] \Delta x[/math]
[math]\text{Work} = \frac{1}{2} \times \text{base} \times \text{height}[/math]
In this case, the base of the triangle is the extension and the height is the force
[math]\text{Work} = \frac{1}{2}(F)(\Delta x) \\
\text{Work} = \frac{1}{2}(mg)(\Delta x) \\
\text{Work} = \frac{1}{2}mg\Delta x[/math]
2. The velocity of a wave on a string is given by [math]v = \sqrt{\frac{T}{\mu}}[/math] Where T is the tension in the string and μ is a constant.
Which of the following pairs of variables, if plotted, would produce a straight line graph?
Answer: D
Explain:
The given equation is:
[math]v = \sqrt{\frac{T}{\mu}}[/math]
To produce a straight line graph, the equation must be in the form of a linear equation,
[math]y = mx + c[/math]
We can rearrange the given equation to match the linear equation form.
[math]v^2 = \left(\sqrt{\frac{T}{\mu}}\right)^2 \\
v^2 = \frac{1}{\mu} T \\
y = mx \\
y = v^2 \\
x = T[/math]
The slop is
[math]m = \frac{1}{\mu} \\
v = \frac{1}{\sqrt{\mu}} \sqrt{T} \\
m = \frac{1}{\sqrt{\mu}}[/math]
Therefore, plotting v on the y – axis and [math]\sqrt{T}[/math] on the x – axis will produce a straight line graph.
3. In an experiment to determine the wavelength of light, a diffraction grating is illuminated with light from a monochromatic source. A series of bright spots is observed.
The experiment is repeated and the distance between consecutive bright spots increases.
Select the row of the table that gives two changes to the experimental set up which would both cause the distance between consecutive bright spots to increase.
Answer: A
Explain:
Diffraction Grating Experiment
The relationship between the distance between consecutive bright spots (fringe separation, x) the distance from the grating to the screen (D) the wavelength of the light source (λ), and the slit separation (d) is given by the equation:
[math]x = \frac{\lambda D}{d}[/math]
The slit separation d is the inverse of the number of slits per unit length (e.g., per mm). So, if the number of slits per mm increases, the slit separation d decreases.
To increase the distance between consecutive bright spots (x) we need to either increase the wavelength [math]\lambda[/math] or decrease the slit separation (d). Decreasing the slit separation (d) is equivalent to increasing the number of slits per mm.
Therefore, the two changes that would cause the distance between consecutive bright spots to increase are:
1. Increasing the wavelength of the light source.
2. Increasing the number of slits per mm (which decreases the slit separation).
Based on the table, option C matches these conditions. The number of slits per mm is decreased, and the wavelength is increased. Wait, let’s re-read the question. The question asks for two changes that would both cause the distance to increase.
4. Displacement-time graphs are shown for two waves, each of frequency f and period T.
The phase difference in radians between the two waves is given by
Answer: A
Explain:
Analyze the graph
The graph shows two waves with the same frequency f and period T. The time difference between the waves is given as t. This is the time it takes for one wave to reach a point that the other wave has already passed.
Calculate the phase difference:
The phase difference (∅) is the ratio of the time difference (t) to the period (T), multiplied by 2 radians,

5. The diagram shows the lowest energy levels for a certain atom.
A photon with energy 3.2 eV is absorbed.
An electron could move from.
Answer: B
Explain:
To determine the possible transition, we must calculate the energy difference between the given levels. An electron can only move to a higher energy level if the energy of the absorbed photon matches the exact energy difference between the initial and final states.
The energy difference is
[math]\Delta E = E_2 – E_{\text{ground}} \\
\Delta E = (-2.3\,\text{eV}) – (-5.5\,\text{eV}) \\
\Delta E = 3.2\,\text{eV}[/math]
This matches the photon energy.
6. The rate at which energy is transferred to a solar cell of radius r is P.
In the same conditions, the rate at which energy is transferred to a solar cell of radius 2r is given by
Answer: D
Explain:
The rate at which energy is transferred to a solar cell is proportional to the area of the solar cell. The area of a circular solar cell is given by the formula
[math] A = \pi r^2[/math]
Where r is the radius
Let [math]P_1[/math] be the rate of energy transfer for a solar cell with radius r.
[math]P_1 \propto A_1 \\
P_1 = \pi r^2[/math]
Let [math]P_2[/math] be the rate of energy transfer for a solar cell with radius 2r.
[math]P_2 \propto A_2 \\
P_2 = \pi(2r)^2 \\
P_2 = 4\pi r^2 \\
P_2 = 4P_1[/math]
7. For two waves of light to be coherent the waves must.
Answer: D
Explain:
For two waves of light to be coherent, they must have the same frequency and a constant phase difference. While option (a) states a specific phase difference of zero, any constant phase difference allows for coherence. The most common way to achieve a constant phase difference is to have the waves originate from a single source and then split them, as this ensures they start with a fixed phase relationship.
8. Prenatal ultrasound scans are performed to produce an image of a fetus.
The ultrasound wave is transmitted in pulses so that
Answer: C
Explain
Ultrasound imaging works by sending out a pulse of sound waves and then listening for the echoes that bounce back from different tissues and organs. The time it takes for an echo to return is used to calculate the distance to the reflecting surface. If a new pulse were sent out before the previous echoes had a chance to return, the machine would not be able to distinguish between the echoes from the first pulse and the second, leading to a confused image. Therefore, the ultrasound is transmitted in pulses to allow the machine to receive and process the reflected echoes from one pulse before sending out the next one.
9. In the 1920s Louis de Broglie proposed that an electron could behave as a wave.
Calculate the wavelength of an electron that is travelling at a speed of [math]2.2 \times 10^7 \text{m·s}^{-1}[/math] .
Explain:
– Speed of the electron, v = [math]2.2 \times 10^7 \text{m·s}^{-1}
[/math]
– Planck’s constant, h = [math]6.626 \times 10^{-34} \text{m·s}^{-1}
[/math]
– Mas of electron = [math]m_e = 9.1 \times 10^{-31} \text{m·s}^{-1}
[/math]
– Find the wavelength = [math]\lambda[/math]
[math]\lambda = h / p \\
\lambda = h / (m_e v) \\
\lambda = (6.626 \times 10^{-34}) / \bigl( (9.1 \times 10^{-31})(2.2 \times 10^{7}) \bigr) \\
\lambda = (6.626 \times 10^{-34}) / (2.004 \times 10^{-23}) \\
\lambda = 3.3 \times 10^{-11}\ \text{m}[/math]
Wavelength = [math]3.3 \times 10^{-11} \text{m}[/math]

10. Details supplied with a school microwave transmitter and receiver include the following information:

Transmitter supplies a 10 GHz polarised EM wave. Receiver detects EM waves in a single plane containing the direction of propagation, producing an audible output proportional to the microwave intensity

a) The receiver and transmitter are initially set up, as shown in the diagram, so that a maximum audible output is produced.
The receiver is rotated through 180° relative to the transmitter.
Describe what is heard as the receiver is rotated.
Explain:
The sound will decrease to a minimum (or become silent) and then increase to a maximum again.
The transmitter emits a polarized EM wave. The receiver detects this wave in a single plane. When the receiver is rotated, the alignment of its detection plane with the polarization of the EM wave changes.
– At the start, the receiver is aligned with the transmitter’s polarization, resulting in a maximum audible output.
– As the receiver is rotated, the component of the polarized wave detected by the receiver decreases, so the sound decreases.
– At a rotation of [math]90^0[/math] the receiver’s plane is perpendicular to the wave’s polarization, and no wave is detected, resulting in a minimum (or zero) audible output.
– As the rotation continues from [math]90^0[/math] to [math]90^0[/math] the component of the wave detected by the receiver increases again, so the sound increases.
– At [math]180^0[/math] the receiver is again aligned with the transmitter’s polarization, and the sound returns to a maximum.
b) A student uses the microwave transmitter and receiver to investigate interference using the set-up shown.
As the receiver is moved along the dotted line, alternate points of maximum and minimum intensity are detected.
Explain why points of maximum and minimum intensity are detected.
Explain:
Understand the setup:
The setup described is analogous to a double-slit experiment, but with microwaves instead of light. The microwave transmitter acts as a source of waves, and the two gaps in the metal plates act as two coherent sources of waves.
Explain interference
Interference is the superposition of two or more waves. When the waves from the two slits reach the receiver, they combine. The outcome of this combination depends on the phase difference between the waves.
Explain constructive and destructive interference
Maximum intensity:
– This occurs when the two waves arrive at the receiver in phase. This is known as constructive interference. When the waves are in phase, their amplitudes add up, resulting in a larger resultant amplitude and thus a higher intensity. This happens when the path difference between the two waves is an integer multiple of the wavelength [math]n \lambda[/math]
Minimum intensity:
– This occurs when the two waves arrive at the receiver out of phase. This is known as destructive interference. When the waves are out of phase, their amplitudes cancel out, resulting in a smaller resultant amplitude and thus a lower intensity. This happens when the path difference between the two waves is a half-integer multiple of the wavelength [math]((n + \tfrac{1}{2})\lambda)[/math].
Result:
The points of maximum and minimum intensity are detected due to the interference of the microwave waves passing through the two slits in the metal plates. The two slits act as coherent sources of waves.
When the receiver is moved, the path difference between the waves from the two slits changes. At locations where the path difference is an integer multiple of the wavelength, the waves arrive in phase, leading to constructive interference and a maximum intensity. At locations where the path difference is a half-integer multiple of the wavelength, the waves arrive out of phase, leading to destructive interference and a minimum intensity.
11. Converging and diverging lenses may be used in glasses to correct problems with eyesight.
a) The diagram shows three parallel rays of light incident on a diverging lens and the path of the rays after passing through the lens. The diagram is drawn to actual size.
Add to the diagram to determine the focal length of the lens. (2)
Explain:
Extend the refracted rays backward
For a diverging lens, the focal point is a virtual focal point. The refracted rays appear to originate from this point. To find the focal point, extend the three refracted rays backward as dashed lines.
Locate the focal point
The point where the backward extensions of the refracted rays intersect is the principal focal point (F) of the lens.
Measure the focal length
The focal length (f) is the distance from the optical center of the lens to the principal focal point. Measure this distance using a ruler. Based on the provided diagram, the focal length is approximately 2.5 cm.
Focal length = 2.5 cm
b) The eye acts as a converging lens system.
The diagram shows light rays from an object 25 cm in front of an eye converging to a point on the retina at the back of the eye. The eye has a depth of 2.4 cm.
Calculate the optical power of the eye. (3)
Explain:
– The image distance (v) is the depth of the eye, v = 2.4 cm = 0.024 m
– The image distance (u) = 25 cm = 0.25 m
– Focal length = f = ?
– Optical Power P = ?
Use
[math]\frac{1}{u} + \frac{1}{v} = \frac{1}{f} \\[6pt]
\frac{1}{0.25} + \frac{1}{0.024} = \frac{1}{f} \\[6pt]
\frac{1}{f} = 4 + 41.7 \\[6pt]
\frac{1}{f} = 45.7 \\[6pt]
f = \frac{1}{45.7} \\[6pt]
f \approx 0.022\,\text{m}[/math]
Calculate the optical power (P) of the eye:
[math]p = \frac{1}{f} \\[6pt]
p = \frac{1}{0.022} \\[6pt]
p \approx 45.5\,\text{D}[/math]
Power = 46 D/Dioptre
c) A person who is long-sighted cannot clearly see objects that are close to the eye.
Rays of light from an object 25 cm in front of the eye would converge to a point behind the retina as shown in the diagram.
This may be corrected by using an additional converging lens.
State how an additional converging lens would enable the light rays from an object 25 cm in front of the eye to converge at a point on the retina.
Explain:
A converging lens would increase the overall converging power of the eye’s lens system. By placing an additional converging lens in front of the eye, the light rays from the object will be converged more strongly before they enter the eye. This will cause the light rays to converge at a point on the retina, rather than behind it, allowing the person to see the object clearly.

12. On sunny days a mirage can sometimes be observed when a virtual image of the sky is seen on the surface of a road.
The Sun’s rays heat up the surface of the road. The heated road then heats the surrounding air so that the layer of air just above the road is at a higher temperature than the air above it.
Warm air has a lower refractive index than cool air.
a) State what is meant by a virtual image. (1)
Explain:
A virtual image is an image that is formed when light rays appear to diverge from a point but do not actually meet at that point. This type of image cannot be projected onto a screen.
b) The diagram represents a simple model which is sometimes used to explain how a mirage is formed. The three layers, each with a different refractive index, represent air at three different temperatures. Layer 3 represents the air at the highest temperature closest to the road. A light ray is shown incident at the interface between layer 1 and layer 2.
Use the information to discuss whether the observer sees a mirage on the road in the position shown. (6)
Explain:
– [math]n_1 = 1.00032[/math]
– [math]\theta_1 = 89.59^o[/math]
– [math]n_2 = 1.00030[/math]
– Calculate the angle of refraction at the interface between layer 1 and layer 2 = [math]\theta_2[/math] ?
By using the Snell’s Law
[math] n_1 \sin \theta_1 = n_2 \sin \theta_2 \\
n_2 \sin \theta_2 = n_1 \sin \theta_1 \\
\sin \theta_2 = \frac{n_1 \sin \theta_1}{n_2} \\
\theta_2 = \sin^{-1}\!\left(\frac{n_1 \sin \theta_1}{n_2}\right) \\
\theta_2 = \sin^{-1}\!\left(\frac{(1.00032)\sin(89.59^\circ)}{1.00030}\right) \\
\theta_2 = \sin^{-1}\!\left(\frac{1.00032 \times 0.999974}{1.00030}\right) \\
\theta_2 = 89.81^\circ
[/math]
13. A student carries out an experiment to determine the viscosity of glycerol. She does this by determining the terminal velocity of a steel sphere falling through glycerol.
a) The equation shows how the terminal velocity of a solid sphere falling through a liquid depends on the density of both the solid and the liquid.
[math]v = \frac{V g (\rho_s – \rho_t)}{6 \pi \eta}[/math]
Where:
– [math]p_s[/math] = density of solid
– [math]p_t[/math] = density of liquid
– r = radius of sphere
– [math]\eta[/math] = viscosity of liquid
– v = terminal velocity
The derivation of the equation for terminal velocity has been started below. Complete the derivation. (3)
At terminal velocity: weight of solid sphere = drag + uptrust
Explain:
Write the equations for the forces
The problem states that at terminal velocity, the weight of the solid sphere is equal to the sum of the drag force and the upthrust (buoyancy force).
Weight of solid sphere:
The weight of an object is its mass (m) multiplied by the acceleration due to gravity (g). The mass of a sphere can be expressed as its density [math]p_s[/math], multiplied by its volume (V). The volume of a sphere is given by
[math]V = \frac{4}{3}\pi r^3 \\
W = mg \\
W = \rho_s V g \\
W = \rho_s \left(\frac{4}{3}\pi r^3\right) g[/math]
Upthrust (buoyancy force):
The upthrust is equal to the weight of the fluid displaced by the sphere. The weight of the fluid is its mass [math](m_f) [/math]multiplied by g. The mass of the fluid is its density [math](p_l) [/math] multiplied by the volume of the displaced fluid (V).
[math]U = m_f g \\
U = \rho_l V g \\
U = \rho_l \left(\frac{4}{3}\pi r^3\right) g[/math]
Drag force:
For a small sphere moving slowly through a viscous fluid, the drag force [math] (F_d) [/math] is given by Stoke’s Law:
[math]F_d = 6 \pi \eta rv[/math]
Set up the terminal velocity equation
At terminal velocity, the forces are in equilibrium.
[math]\text{Weight} = \text{Drag} + \text{Upthrust} \\
\rho_s \left(\frac{4}{3}\pi r^3\right) g = 6\pi\eta r v + \rho_l \left(\frac{4}{3}\pi r^3\right) g \\
6\pi\eta r v = \rho_s \left(\frac{4}{3}\pi r^3\right) g – \rho_l \left(\frac{4}{3}\pi r^3\right) g \\
6\pi\eta r v = \left(\frac{4}{3}\pi r^3\right)(\rho_s – \rho_l) g \\
v = \frac{\left(\frac{4}{3}\pi r^3\right)(\rho_s – \rho_l) g}{6\pi\eta r} \\ v = \frac{4 r^3 g (\rho_s – \rho_l)}{18 \eta} \\
v = \frac{2 r^2 g (\rho_s – \rho_l)}{9\eta}[/math]
The derivation is as follows:
[math]\text{Weight} = \text{Drag} + \text{Upthrust} \\
\rho_s V g = 6\pi\eta r v + \rho_l V g \\
6\pi\eta r v = \rho_s V g – \rho_l V g \\
6 \pi \eta r v = \frac{(\rho_s – \rho_l) V g}{6 \pi \eta r}[/math]
b)
i) The student drops a steel sphere with a radius of 4.0 mm into a cylinder of glycerol. The sphere reaches terminal velocity and takes 3.9 s to fall 0.50 m.
Calculate the viscosity of glycerol.
– Density of steel = 7800 kg.m^-3
– Density of glycerol = 1300 kg. m^-3 (4)
Explain:
– Distance = d = 0.50 m
– Time = t = 3.9 s
Calculate the terminal velocity = [math]v_t[/math] = ?
[math]v_t = \frac{d}{t} \\
v_t = \frac{0.50}{3.9} \\
v_t = 0.128\ \text{m /s}[/math]
State the formula for viscosity:
The formula for the viscosity of a fluid, derived from Stokes’ law, is given by:
[math]\eta = \frac{2 r^2 g (\rho_s – \rho_l)}{9 v_t}[/math]
Radius = r = 4.0mm = [math]4 \times 10^{-3}[/math]
[math]\eta = \frac{2 r^2 g (\rho_s – \rho_l)}{9 v_t} \\
\eta = \frac{2(4 \times 10^{-3})^2 (9.8)(7800 – 1300)}{9(0.128)} \\
\eta = \frac{2(1.6 \times 10^{-5})(9.8)(6500)}{1.1538} \\
\eta = \frac{2.0384}{1.1538} \\
\eta = 1.7667\ \text{Pa·s} \\
\eta \approx 1.8\ \text{Pa·s}[/math]
Viscosity of glycerol = 1.8 Pa s
ii) There are two cylinders available for the student to use. One cylinder has a diameter of 1.5 cm and the other has a diameter of 5.0 cm.
State and justify which cylinder the student should use in order to gain a more accurate value for the viscosity of glycerol. (2)
Explain:
The student should use the cylinder with a diameter of 5.0 cm.
The justification is that the wider cylinder allows for laminar flowand reduces the effects of the cylinder walls. This ensures that Stokes’ law applies, allowing the sphere to fall at a more constant terminal velocity, which leads to a more accurate value for the viscosity of glycerol.