9 June 2016

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. Which statement about sound is correct?
Answer: A
Explain:
Sound waves are mechanical waves, meaning they require a medium to travel through. They can travel through solids, liquids, and gases. Sound travels fastest in solids because the particles are more closely packed, allowing vibrations to be transferred more efficiently.
2. Which of the following is a correct unit for intensity of light?
Answer: C
Explain:
The intensity of light is defined as the power per unit area. Power is measured in watts (W) and area is measured in square meters [math]m^2[/math]. Therefore, the unit for intensity of light is watts per square meter, or [math]W.m^{-2}[/math]. This is a unit for irradiance, which is a measure of the intensity of electromagnetic radiation, including light.
3. The diagram shows a force-extension graph for a wire.
L is the elastic limit.
Which point represents the yield point?
Answer: B
Explain:
The yield point is the point on a stress-strain or force-extension graph where plastic deformation begins to occur. It is the point just after the elastic limit, where the material starts to deform permanently. In the given graph, point L is identified as the elastic limit. Point B is the first point after the elastic limit where the material begins to deform plastically, making it the yield point.
4. A bubble of air is rising through a vertical column of water.
Which statement, about the motion of the bubble, is correct to a good approximation?
Answer: B
Explain:
An air bubble rising through water is subject to three main forces:
1. Gravity (Weight):
– This force acts downward.
– The weight of the air bubble is very small due to the low density of air.
Upthrust (Buoyancy):
– This force acts upward. It is equal to the weight of the water displaced by the bubble.
2. Viscous Drag:
– This force opposes the motion of the bubble and acts downward.
– It increases with the bubble’s velocity.
As the bubble starts to rise, the upward upthrust force is initially much greater than the combined downward forces of weight and viscous drag. This causes the bubble to accelerate. As the bubble’s velocity increases, the viscous drag force also increases. Eventually, the viscous drag force becomes large enough that the total downward force (weight + viscous drag) equals the upward upthrust force.
At this point, the net force on the bubble is zero, and according to Newton’s First Law of Motion, the bubble will stop accelerating and continue to rise at a constant velocity, known as the terminal velocity. Since the weight of the air bubble is negligible compared to the upthrust and viscous drag, the condition for constant velocity can be approximated as the upthrust being equal to the viscous drag.
5. In an experiment a student is taking a measurement of a time interval in seconds.
He repeats the measurement and collects the following three readings:
3.2 s 3.2 s 3.3 s
Which one of the following is a correct statement?
Answer: C
Explain:
The three readings are 3.2 s, 3.2 s, and 3.3 s. The average value is
[math]\frac{3.2 + 3.2 + 3.3}{3} = 3.233[/math]
However, the measurements are given to one decimal place, so the average should also be rounded to one decimal place, which is 3.2 s.
The uncertainty is the range of the measurements. The range of the given readings is from 3.2 s to 3.3 s. The uncertainty is half of the range, which is
[math]\frac{3.3 – 3.2}{2} = 0.05[/math]
However, a more common way to state the uncertainty is to use the smallest division of the measuring instrument, which in this case is 0.1 s. The readings themselves have a precision of 0.1 s.
– The reading of 3.3 s is not necessarily an error; it’s a valid measurement within the expected range.
– The average should be recorded as 3.2 s, not 3.23 s, to reflect the precision of the measurements.
– The error is not 0.1 s. Error is the difference between the measured value and the true value, which is not known.
Therefore, the most correct statement is that the uncertainty is 0.1 s, as it represents the precision of the measurements taken.

6.The graph shows the variation of displacement with time for two waves.
What is the phase difference between these two waves?
Answer: D
Explain:
From the graph, the solid wave completes one cycle = T = 20 ms
From the graph, the dotted wave is shifted to the right by 5 ms relative to the solid wave.
Therefore [math]\Delta t = 5 ms[/math]
Calculate the phase difference in radians
[math]\Phi = \frac{5}{20} \times 2\pi \\
\Phi = \frac{1}{4} \times 2\pi \\
\Phi = \frac{\pi}{2} \text{ radians}[/math]
The phase difference is [math]\frac{\pi}{2}[/math]. This is equals to[math]90^o[/math].
7. A source of unpolarised light is viewed through two crossed polarising filters X and Y.
Which row in the table correctly describes the light emerging from the two filters?
Answer: D
Explain:
Light emerging from filter X:An unpolarized light wave oscillates in all directions. When it passes through a polarizing filter (filter X), the filter only allows light waves that oscillate in a specific direction to pass through. Therefore, the light emerging from filter X will oscillate in only one direction.
Light emerging from filter Y:The second filter (filter Y) is “crossed” with respect to the first filter (filter X). This means their polarization axes are perpendicular to each other. The light that emerges from filter X is already polarized, oscillating in a single direction. When this light reaches filter Y, its oscillations are perpendicular to the polarization axis of filter Y. As a result, filter Y blocks all the light, and no light emerges.
8. Sound waves are produced by a vibrating guitar string.
Which row in the table correctly describes the waves produced?
Answer: D
Explain:
A vibrating guitar string produces a transverse wave. In a transverse wave, the particles of the medium vibrate perpendicular to the direction of the wave’s propagation.
As the guitar string vibrates, it displaces the surrounding air, creating compressions and rarefactions that travel away from the string.
This disturbance in the air is a sound wave, which is a longitudinal wave. In a longitudinal wave, the particles of the medium vibrate parallel to the direction of the wave’s propagation.
9. A simple optical fibre consists of a core surrounded by cladding. The refractive index of the core is 1.56 and the refractive index of the cladding is 1.20.
a) Show that the critical angle for light between these two media is about [math]5^o[/math].
Explain:
– The refractive index of the core [math]n_1[/math] = 1.56
– The refractive index of the cladding [math]n_2[/math] = 1.20
– Find the critical angle = [math]\theta_c = ?[/math]
[math]\sin \theta_c = \frac{n_2}{n_1} \\
\sin \theta_c = \frac{1.20}{1.56} \\
\theta_c = \sin^{-1}(0.7692) \\
\theta_c = 50.30^\circ[/math]
b) The diagram shows a diverging beam of light incident on the boundary between the core and the cladding. One side of the beam strikes the boundary at [math]60^o[/math] and the other side at [math]30^o[/math] as shown.
Three students each suggest a different outcome for the beam of light at the boundary.
Student A says “all the beam will totally internally reflect”.
Student B says “all the beam will refract”.
Student C says “some of the beam will totally internally reflect and some will refract”.
State which student is correct, adding to the diagram to illustrate your answer.
Explain:
Student C is correct.
Total internal reflection occurs when a light ray traveling from a denser medium to a less dense medium strikes the boundary at an angle greater than the critical angle. Refraction occurs when the angle of incidence is less than the critical angle.
In this problem, the beam of light is diverging, which means it consists of multiple rays incident on the boundary at different angles. The diagram shows that the angles of incidence range from
The critical angle for the core-cladding boundary is not given, but based on the diagram, the range of angles of incidence spans both sides of the critical angle.
– The rays with an angle of incidence greater than the critical angle will undergo total internal reflection.
– The rays with an angle of incidence less than the critical angle will refract into the cladding.
Therefore, some parts of the beam will totally internally reflect, and some parts will refract. This makes Student C’s statement correct.

10. Shrilk is a new material made from discarded shrimp shells. It is biodegradable and is easily moulded into different shapes. Shrilk is an alternative to polythene and could be used to make waste bags in the future.
The graph shows a stress-strain curve for a 25.0 cm length of shrilk and for a similar length of polythene, up to breaking point.
a) (i) Calculate the force applied to the shrilk at a strain of 0.02
Cross – sectional area = [math]1.2 \times 10^{-6}\,\text{m}^2[/math]
Explain:
– The stress at a strain of 0.02 for shrilk is approximately 55 MPa
[math]\sigma = 55\,\text{MPa} = 55 \times 10^{6}\,\text{Pa}[/math]
– Cross – sectional area = A = [math]1.2 \times 10^{-6}\,\text{m}^2[/math]
– Find the force applied to the shrilk = F = ?
[math]\sigma = \frac{F}{A} \\
F = \sigma A \\
F = (55 \times 10^{6})(1.2 \times 10^{-6}) \\
F = 55 \times 1.2 \\
F = 66\,\text{N}[/math]
Force = 66 N
(ii) Determine the extension of the shrilk at a strain of 0.04 (2)
Explain:
– The original length of the shrilk = 25.0 cm
– The strain = [math]\varepsilon[/math] = 0.04
– To find the extension = [math]\Delta x[/math]
[math]\varepsilon = \frac{\Delta x}{x} \\
x \varepsilon = \Delta x \\
\Delta x = (25)(0.04) \\
\Delta x = 1.0\,\text{cm}[/math]
Extension = 1.0 cm
b) Deduce whether shrilk or polythene is better for making waste bags. (3)
Explain:
Analyze the stress-strain curves
The stress-strain curve shows the relationship between the stress applied to a material and the resulting strain (deformation). The area under the curve represents the toughness of the material, which is its ability to absorb energy before fracturing. The breaking point is the end of the curve, where the material fractures.
Compare the breaking points of shrilk and polythene
From the graph, the breaking point for shrilk occurs at a higher stress and a higher strain compared to polythene. This means shrilk can withstand more force and stretch more before breaking.
Compare the toughness of shrilk and polythene
The area under the stress-strain curve for shrilk is significantly larger than the area under the curve for polythene. This indicates that shrilk is a tougher material and can absorb more energy before breaking.
Conclude which material is better for making waste bags
A good material for waste bags needs to be tough and resistant to breaking under stress. Since shrilk has a higher breaking point and is tougher (larger area under the curve) than polythene, it is a better material for making waste bags.
11. The diagram shows a coherent beam of light incident on a metal ball bearing.
A dark shadow is seen on a screen behind the ball bearing. There is a small spot of light in the centre of the shadow. This spot of light is known as the Arago spot.
a) Use Huygens’ construction to explain the behaviour of light as it travels past the edge of the ball bearing. (2)
Explain:
According to Huygens’ principle, every point on a wavefront can be considered a source of secondary spherical wavelets that spread out in the forward direction. When a coherent beam of light encounters the edge of the ball bearing, the wavefront is obstructed, but the portions of the wavefront that pass the edge act as new sources of these secondary wavelets.
These secondary wavelets from the unobstructed parts of the wavefront interfere with each other. Due to the circular symmetry of the ball bearing, the wavelets that travel along the central axis behind the ball bearing all arrive in phase and interfere constructively, creating a bright spot of light at the center of the shadow. This phenomenon is known as diffraction.
b) Explain why a spot of light is produced at the centre of the shadow.
Explain:
The Arago spot, also known as the Poisson spot, is a bright point that appears at the center of the shadow of a circular opaque object when illuminated by a point source or a coherent beam of light. This phenomenon is a direct consequence of the wave nature of light and can be explained by the principles of Fresnel diffraction and Huygens-Fresnel principle.
Diffraction:
As the coherent light waves travel past the edges of the ball bearing, they diffract, or bend, into the geometric shadow region.
Constructive Interference:
According to the Huygens-Fresnel principle, every point on the wavefront that passes the object’s edge acts as a secondary source of light. The light waves from all these points on the circular edge of the ball bearing travel the same distance to reach the exact center of the shadow on the screen.
Path Difference:
Because the path length is the same for all the diffracted waves arriving at the center, they arrive in phase. This results in constructive interference, creating a bright spot of light.
c) François Arago first demonstrated this experiment in 1818 for a group of eminent scientists, to show the behaviour of light.
State the model for the behaviour of light that this experiment demonstrated and explain why the scientific community accepted this model.
Explain:
The experiment demonstrated the wave model of light.
The scientific community accepted the wave model of light because the Arago spot, a bright spot at the center of the shadow of an opaque disk, could not be explained by the particle model of light. According to the particle model, light travels in straight lines, and therefore a solid object like a ball bearing should cast a completely dark shadow.
The presence of a bright spot within the shadow provides evidence that light waves bend around the edges of the object and interfere constructively at the center of the shadow, a phenomenon known as diffraction. This observation was a key piece of evidence that supported the wave nature of light.
d) The ball bearing shown in the experimental set-up has a diameter of about 1cm.
Describe how the diameter could be measured accurately.
Explain:
Select the appropriate measuring tool
A micrometer screw gauge is the most suitable tool for this measurement as it provides a high degree of accuracy for small diameters.
Calibrate the micrometer
Check for zero error by closing the jaws of the micrometer. If the zero mark on the thimble scale does not align with the main scale, record the zero error. This error will be subtracted from or added to the final reading.
Take multiple measurements
Place the ball bearing between the jaws of the micrometer. Tighten the ratchet until it clicks to ensure the correct pressure is applied. Record the reading. Repeat this process at several different positions around the ball bearing to account for any non-uniformity.
Calculate the average diameter and final result
Calculate the average of all the measurements taken. Then, apply the zero error correction to the average value to get the final, accurate diameter of the ball bearing.
12. A student carries out an experiment using a guitar string. She investigates the effect of varying the tension in the guitar string on the frequency of sound produced when the string is plucked.
a) Describe a method of varying and measuring the tension in the string. (2)
Explain:
To vary and measure the tension in a guitar string, a student can set up a pulley system.
Varying the tension:
The string can be stretched between a fixed point and a pulley. A mass hanger can be attached to the end of the string that passes over the pulley. By adding or removing slotted masses from the hanger, the tension in the string can be varied.
Measuring the tension:
The tension in the string is equal to the weight of the masses attached to the hanger. The weight (W) can be calculated using the formula
[math]W = mg[/math]
Where m is the total mass on the hanger and g is the acceleration due to gravity . Therefore, the tension can be measured by calculating the total weight of the masses.
b) The student records the following data and plots a graph.
Complete the table and graph. (3)
Explain:
The table shows the relationship between tension and frequency. The third row requires the square of the frequency.
For a tension of 35 N, the frequency is 95 Hz,
The squared frequency is 9025 [math]Hz^2[/math]