Physics (9-1) 0972 | Paper 5 Mock 01 | Practical Test
Question 1 Report
In this experiment, you will investigate total internal reflection using a semicircular glass block. You directs a narrow ray of light from a ray box into the curved surface of the block so that the ray passes through the glass and arrives at the centre of the flat surface, point O. The apparatus is shown in Fig. 5.1.
You gradually increases the angle of incidence at the flat surface by rotating the ray box. At a certain angle, the refracted ray disappears and the light is totally internally reflected. You records this angle as the critical angle c. The ray-trace diagram for the critical angle is shown in Fig. 5.2.
(a) Using Fig. 5.2, measure and record the critical angle c. [1]
(b) Calculate sin c. [1]
(c) Calculate the refractive index n of the glass using the equation n = 1 / sin c. [2]
(d) State why the ray of light must enter through the curved surface of the block, directed towards the centre of the flat surface. [1]
(e) Describe what happens to the ray at the flat surface when the angle of incidence is less than the critical angle. [2]
(f) Describe what happens to the brightness of the refracted ray as the angle of incidence is increased towards the critical angle. [1]
(g) State one practical use of total internal reflection. [1]
(h) The critical angle for diamond is 24°. Calculate the refractive index of diamond. [1]
(a) From Fig. 5.2, the critical angle c is measured between the incident ray and the normal at the flat surface at point O. Using a protractor on the diagram:
\[ c = 42^\circ \] [1]
(Accept 41° to 43°.)
(b) Calculating the sine of the critical angle:
\[ \sin c = \sin 42^\circ = 0.669 \] [1]
(Accept a value consistent with the measured angle.)
(c) The refractive index of the glass is found using \( n = 1 / \sin c \):
\[ n = \frac{1}{\sin c} \] [1]
\[ n = \frac{1}{0.669} = 1.49 \] [1]
(Accept 1.4 to 1.6 if consistent with the measured critical angle.) This value is typical for glass, confirming the measurement is reasonable.
(d) The ray must enter through the curved surface because at the curved surface, the ray travels along a radius of the semicircle, which means it hits the curved surface at right angles (perpendicular to the surface). A ray striking a surface perpendicularly does not refract, so it passes straight through to the flat surface without changing direction. [1]
This ensures that refraction occurs only at the flat surface, so the measured angle of incidence at the flat surface is the true angle.
(e) When the angle of incidence at the flat surface is less than the critical angle:
- The ray is partly refracted as it passes from the glass into the air. The refracted ray bends away from the normal (since glass is optically denser than air). [1]
- At the same time, a weak reflected ray is also produced inside the glass block. This partial reflection always accompanies refraction. [1]
(f) As the angle of incidence increases towards the critical angle, the refracted ray becomes dimmer (less bright). [1]
More and more of the light energy is reflected rather than refracted. At the critical angle, the refracted ray just grazes along the surface (at 90° to the normal), and beyond this angle, all the light is reflected.
(g) One practical use of total internal reflection (any one): [1]
- Optical fibres for telecommunications or medical endoscopes: light is guided along the fibre by repeated total internal reflection.
- Reflecting prisms in binoculars, periscopes, or SLR cameras.
- Bicycle reflectors and road reflectors.
(h) For diamond, the critical angle is 24°:
\[ n = \frac{1}{\sin 24^\circ} = \frac{1}{0.407} = 2.46 \] [1]
Diamond has a much higher refractive index than glass (2.46 compared to about 1.5), which means it has a much smaller critical angle. This causes extensive internal reflection, producing the characteristic sparkle of a cut diamond.
Answer Details
(a) From Fig. 5.2, the critical angle c is measured between the incident ray and the normal at the flat surface at point O. Using a protractor on the diagram:
\[ c = 42^\circ \] [1]
(Accept 41° to 43°.)
(b) Calculating the sine of the critical angle:
\[ \sin c = \sin 42^\circ = 0.669 \] [1]
(Accept a value consistent with the measured angle.)
(c) The refractive index of the glass is found using \( n = 1 / \sin c \):
\[ n = \frac{1}{\sin c} \] [1]
\[ n = \frac{1}{0.669} = 1.49 \] [1]
(Accept 1.4 to 1.6 if consistent with the measured critical angle.) This value is typical for glass, confirming the measurement is reasonable.
(d) The ray must enter through the curved surface because at the curved surface, the ray travels along a radius of the semicircle, which means it hits the curved surface at right angles (perpendicular to the surface). A ray striking a surface perpendicularly does not refract, so it passes straight through to the flat surface without changing direction. [1]
This ensures that refraction occurs only at the flat surface, so the measured angle of incidence at the flat surface is the true angle.
(e) When the angle of incidence at the flat surface is less than the critical angle:
- The ray is partly refracted as it passes from the glass into the air. The refracted ray bends away from the normal (since glass is optically denser than air). [1]
- At the same time, a weak reflected ray is also produced inside the glass block. This partial reflection always accompanies refraction. [1]
(f) As the angle of incidence increases towards the critical angle, the refracted ray becomes dimmer (less bright). [1]
More and more of the light energy is reflected rather than refracted. At the critical angle, the refracted ray just grazes along the surface (at 90° to the normal), and beyond this angle, all the light is reflected.
(g) One practical use of total internal reflection (any one): [1]
- Optical fibres for telecommunications or medical endoscopes: light is guided along the fibre by repeated total internal reflection.
- Reflecting prisms in binoculars, periscopes, or SLR cameras.
- Bicycle reflectors and road reflectors.
(h) For diamond, the critical angle is 24°:
\[ n = \frac{1}{\sin 24^\circ} = \frac{1}{0.407} = 2.46 \] [1]
Diamond has a much higher refractive index than glass (2.46 compared to about 1.5), which means it has a much smaller critical angle. This causes extensive internal reflection, producing the characteristic sparkle of a cut diamond.
Question 2 Report
In this experiment, you will use a metre bridge to find the resistance of an unknown resistor X. You connects X and a standard 10.0 Ω resistor S in the gaps of the metre bridge as shown in Fig. 31.1. A cell and a galvanometer are also connected. You moves the sliding contact (jockey) along the bridge wire until the galvanometer reads zero. You records the balance length l1 from end A. You then swaps X and S and finds a new balance length.
You obtains the following results:
First reading: balance length l1 = 38.5 cm from end A.
After swapping X and S: balance length l2 = 61.0 cm from end A.
(a) Record the two balance lengths. [1]
(b) For the first reading, calculate X using X/S = l1/(100 − l1). Show your working. [2]
(c) For the second reading, calculate X using S/X = l2/(100 − l2). Show your working. [2]
(d) Calculate the average value of X from your two results. [1]
(e) State why you takes readings with X and S swapped. [1]
(f) State what the galvanometer reading of zero means. [1]
(g) Describe one precaution when using the jockey on the bridge wire. [1]
(h) State one source of error in this experiment. [1]
(a) The two balance lengths are: \( l_1 = 38.5 \; \text{cm} \) (first reading) and \( l_2 = 61.0 \; \text{cm} \) (after swapping X and S). [1]
(b) Using the balance condition for the first reading:
\[ \frac{X}{S} = \frac{l_1}{100 - l_1} = \frac{38.5}{100 - 38.5} = \frac{38.5}{61.5} = 0.626 \][1]
\[ X = 0.626 \times S = 0.626 \times 10.0 = 6.3 \; \Omega \][1]
(c) After swapping X and S, the balance condition becomes:
\[ \frac{S}{X} = \frac{l_2}{100 - l_2} = \frac{61.0}{100 - 61.0} = \frac{61.0}{39.0} = 1.564 \][1]
\[ X = \frac{S}{1.564} = \frac{10.0}{1.564} = 6.4 \; \Omega \][1]
(d) Average value of X:
\[ X_{\text{avg}} = \frac{6.3 + 6.4}{2} = 6.35 \; \Omega \][1]
(e) Swapping X and S and taking a second reading helps to eliminate systematic errors caused by non-uniform resistance along the bridge wire or by end corrections (extra resistance at the terminal connections). Averaging the two results gives a more reliable value. [1]
(f) A galvanometer reading of zero means there is no current flowing through the galvanometer. This occurs when the bridge is balanced: the potential difference across the two halves of the bridge are equal, so there is no driving force to push current through the galvanometer. [1]
(g) Do not press the jockey too hard onto the wire and do not drag it along the surface. Pressing too firmly can deform or scratch the wire, locally changing its cross-sectional area and resistance. Instead, tap the jockey briefly at each test position. [1]
(h) Sources of error include: contact resistance at the connections between the wire and the resistors, non-uniformity of the bridge wire (its resistance per unit length may vary along its length), or temperature changes in the wire caused by the current flowing through it. [1]
Why this matters: The metre bridge is a practical form of the Wheatstone bridge. At the balance point, the ratio of resistances in one arm equals the ratio in the other: \( \frac{X}{S} = \frac{l_1}{100 - l_1} \). Because the bridge wire has (ideally) uniform resistance per unit length, the resistance of each section is proportional to its length. The null method (zero galvanometer current) is powerful because at balance the measurement does not depend on the e.m.f. of the cell or the resistance of the galvanometer, eliminating two common sources of error.
Answer Details
(a) The two balance lengths are: \( l_1 = 38.5 \; \text{cm} \) (first reading) and \( l_2 = 61.0 \; \text{cm} \) (after swapping X and S). [1]
(b) Using the balance condition for the first reading:
\[ \frac{X}{S} = \frac{l_1}{100 - l_1} = \frac{38.5}{100 - 38.5} = \frac{38.5}{61.5} = 0.626 \][1]
\[ X = 0.626 \times S = 0.626 \times 10.0 = 6.3 \; \Omega \][1]
(c) After swapping X and S, the balance condition becomes:
\[ \frac{S}{X} = \frac{l_2}{100 - l_2} = \frac{61.0}{100 - 61.0} = \frac{61.0}{39.0} = 1.564 \][1]
\[ X = \frac{S}{1.564} = \frac{10.0}{1.564} = 6.4 \; \Omega \][1]
(d) Average value of X:
\[ X_{\text{avg}} = \frac{6.3 + 6.4}{2} = 6.35 \; \Omega \][1]
(e) Swapping X and S and taking a second reading helps to eliminate systematic errors caused by non-uniform resistance along the bridge wire or by end corrections (extra resistance at the terminal connections). Averaging the two results gives a more reliable value. [1]
(f) A galvanometer reading of zero means there is no current flowing through the galvanometer. This occurs when the bridge is balanced: the potential difference across the two halves of the bridge are equal, so there is no driving force to push current through the galvanometer. [1]
(g) Do not press the jockey too hard onto the wire and do not drag it along the surface. Pressing too firmly can deform or scratch the wire, locally changing its cross-sectional area and resistance. Instead, tap the jockey briefly at each test position. [1]
(h) Sources of error include: contact resistance at the connections between the wire and the resistors, non-uniformity of the bridge wire (its resistance per unit length may vary along its length), or temperature changes in the wire caused by the current flowing through it. [1]
Why this matters: The metre bridge is a practical form of the Wheatstone bridge. At the balance point, the ratio of resistances in one arm equals the ratio in the other: \( \frac{X}{S} = \frac{l_1}{100 - l_1} \). Because the bridge wire has (ideally) uniform resistance per unit length, the resistance of each section is proportional to its length. The null method (zero galvanometer current) is powerful because at balance the measurement does not depend on the e.m.f. of the cell or the resistance of the galvanometer, eliminating two common sources of error.
Question 3 Report
In this experiment, you will connect the output of a signal generator to a cathode-ray oscilloscope (CRO). You adjusts the signal generator frequency and amplitude and records the traces displayed on the CRO screen. Fig. 4.1 shows the first trace. The Y-gain is set to 2.0 V/div and the time base is set to 5.0 ms/div. Fig. 4.2 shows a second trace from a different signal. The Y-gain for Fig. 4.2 is 5.0 V/div and the time base is 20 ms/div.
(a) Measure the peak voltage of the waveform shown in Fig. 4.1. Show your working. [2]
(b) Measure the time period of one complete cycle of the waveform in Fig. 4.1. Show your working. [2]
(c) Use your answer to (b) to calculate the frequency of the signal. Show your working. [1]
(d) You increases the frequency of the signal generator without changing the CRO settings. Describe how the trace on the CRO screen changes. [1]
(e) Record the number of complete cycles shown on the screen in Fig. 4.2. [1]
(f) Measure the peak-to-peak voltage of the waveform in Fig. 4.2. Show your working. [1]
(g) State what you should do to the time base setting to display more complete cycles on the screen. [1]
(h) Explain why the CRO is more suitable than a moving-coil voltmeter for examining a.c. signals. [1]
(a) In Fig. 4.1, the waveform peaks at 3.0 divisions above the centre line (the horizontal axis through the middle of the screen). With the Y-gain set to 2.0 V/div:
\[ \text{peak height} = 3.0 \text{ divisions} \] [1]
\[ V_{\text{peak}} = 3.0 \times 2.0 = 6.0 \text{ V} \] [1]
(b) One complete cycle of the waveform in Fig. 4.1 spans from one peak to the next (or any equivalent full wavelength). Counting the grid squares, one cycle covers 4.0 divisions. With the time base set to 5.0 ms/div:
\[ \text{divisions per cycle} = 4.0 \] [1]
\[ T = 4.0 \times 5.0 = 20 \text{ ms} = 0.020 \text{ s} \] [1]
(c) Frequency is the reciprocal of the time period:
\[ f = \frac{1}{T} = \frac{1}{0.020} = 50 \text{ Hz} \] [1]
This is the standard mains frequency in many countries.
(d) If the frequency increases while the CRO settings remain unchanged, the time period of each cycle becomes shorter. Since the time base still sweeps at the same rate, each cycle occupies fewer divisions on the screen. The waves appear closer together (compressed horizontally) and more complete cycles are visible on the screen. [1]
(e) Counting the complete cycles in Fig. 4.2 (from one point where the wave crosses the centre line going upwards to the next equivalent point, repeated):
\[ \text{Number of complete cycles} = 2 \] [1]
(f) In Fig. 4.2, the waveform extends 2.0 divisions above and 2.0 divisions below the centre line, giving a peak-to-peak height of 4.0 divisions. With the Y-gain set to 5.0 V/div:
\[ V_{\text{peak-to-peak}} = 4.0 \times 5.0 = 20 \text{ V} \] [1]
The peak voltage (amplitude) is half this value: 10 V.
(g) To display more complete cycles on the screen, you should decrease the time base setting (reduce the number of ms per division). [1]
A lower time base means the beam sweeps faster across the screen, so more cycles fit into the same screen width.
(h) A CRO displays the waveform shape, showing how the voltage varies with time. This allows you to see the amplitude, frequency, and shape of the signal. A moving-coil voltmeter can only display a single steady reading (and reads zero for a.c. with zero mean, or shows r.m.s. values), so it cannot reveal the frequency or waveform shape. [1]
Answer Details
(a) In Fig. 4.1, the waveform peaks at 3.0 divisions above the centre line (the horizontal axis through the middle of the screen). With the Y-gain set to 2.0 V/div:
\[ \text{peak height} = 3.0 \text{ divisions} \] [1]
\[ V_{\text{peak}} = 3.0 \times 2.0 = 6.0 \text{ V} \] [1]
(b) One complete cycle of the waveform in Fig. 4.1 spans from one peak to the next (or any equivalent full wavelength). Counting the grid squares, one cycle covers 4.0 divisions. With the time base set to 5.0 ms/div:
\[ \text{divisions per cycle} = 4.0 \] [1]
\[ T = 4.0 \times 5.0 = 20 \text{ ms} = 0.020 \text{ s} \] [1]
(c) Frequency is the reciprocal of the time period:
\[ f = \frac{1}{T} = \frac{1}{0.020} = 50 \text{ Hz} \] [1]
This is the standard mains frequency in many countries.
(d) If the frequency increases while the CRO settings remain unchanged, the time period of each cycle becomes shorter. Since the time base still sweeps at the same rate, each cycle occupies fewer divisions on the screen. The waves appear closer together (compressed horizontally) and more complete cycles are visible on the screen. [1]
(e) Counting the complete cycles in Fig. 4.2 (from one point where the wave crosses the centre line going upwards to the next equivalent point, repeated):
\[ \text{Number of complete cycles} = 2 \] [1]
(f) In Fig. 4.2, the waveform extends 2.0 divisions above and 2.0 divisions below the centre line, giving a peak-to-peak height of 4.0 divisions. With the Y-gain set to 5.0 V/div:
\[ V_{\text{peak-to-peak}} = 4.0 \times 5.0 = 20 \text{ V} \] [1]
The peak voltage (amplitude) is half this value: 10 V.
(g) To display more complete cycles on the screen, you should decrease the time base setting (reduce the number of ms per division). [1]
A lower time base means the beam sweeps faster across the screen, so more cycles fit into the same screen width.
(h) A CRO displays the waveform shape, showing how the voltage varies with time. This allows you to see the amplitude, frequency, and shape of the signal. A moving-coil voltmeter can only display a single steady reading (and reads zero for a.c. with zero mean, or shows r.m.s. values), so it cannot reveal the frequency or waveform shape. [1]
Question 4 Report
In this experiment, you will measure the current through and the potential difference across a resistor to determine its resistance. You connects the resistor in a circuit with a battery, a switch, an ammeter and a voltmeter. The ammeter is connected in series with the resistor and the voltmeter is connected in parallel across it. You closes the switch and reads both meters. The ammeter and voltmeter scales are shown in Fig. 15.1 and Fig. 15.2. The ammeter has a range of 0 to 1.0 A with ten divisions between each numbered mark. The voltmeter has a range of 0 to 10 V with four divisions between each numbered mark.
(a) Measure and record the ammeter reading shown in Fig. 15.1. [1]
(b) Measure and record the voltmeter reading shown in Fig. 15.2. [1]
(c) Calculate the resistance of the resistor using these readings. Show your working. [2]
(d) State the resolution of the ammeter. [1]
(e) State the resolution of the voltmeter. [1]
(f) You wants to find out whether the resistance changes with voltage. Describe the measurements you should take and the calculations you should do. [3]
(g) State one precaution you should take when reading an analogue meter to reduce parallax error. [1]
(a) The ammeter reading is 0.63 A. The pointer lies between the 0.6 A and 0.7 A division marks, approximately three-tenths of the way from 0.6 to 0.7. [1]
(b) The voltmeter reading is 6.5 V. The pointer lies between the 6 V and 7 V marks, approximately halfway between them. [1]
(c) Using \( R = \frac{V}{I} \): [1]
\[ R = \frac{6.5}{0.63} = 10.3 \; \Omega \]The resistance of the resistor is approximately 10 Ω. [1]
(d) The ammeter scale runs from 0 to 1.0 A. The smallest division on the ammeter represents 0.1 A, so the resolution is 0.1 A. [1]
(e) The voltmeter scale runs from 0 to 10 V with divisions every 1 V. The resolution is 1 V (accept 0.5 V if half-divisions are counted). [1]
(f) To investigate whether resistance changes with voltage:
- Change the voltage using a variable power supply or add a variable resistor to the circuit. [1]
- Record several pairs of V and I readings at different voltages. [1]
- Calculate \( R = \frac{V}{I} \) for each pair and compare the values. If R is constant, the resistor obeys Ohm's law. Alternatively, plot V against I: a straight line through the origin confirms ohmic behaviour. [1]
(g) Read the meter with your eye positioned directly in front of the pointer, perpendicular to the scale, to avoid parallax error. If the meter has a mirror strip behind the scale, align the pointer with its reflection to confirm you are viewing from directly in front. [1]
Why this matters: Reading analogue instruments correctly is a core practical skill. Parallax error occurs when your line of sight is at an angle to the scale, making the pointer appear to be at a different position than it truly is. The resolution of an instrument determines the smallest measurable change. A measurement can never be more precise than the resolution of the instrument used. When calculating resistance from V and I, the uncertainties in both readings combine to affect the reliability of the final value.
Answer Details
(a) The ammeter reading is 0.63 A. The pointer lies between the 0.6 A and 0.7 A division marks, approximately three-tenths of the way from 0.6 to 0.7. [1]
(b) The voltmeter reading is 6.5 V. The pointer lies between the 6 V and 7 V marks, approximately halfway between them. [1]
(c) Using \( R = \frac{V}{I} \): [1]
\[ R = \frac{6.5}{0.63} = 10.3 \; \Omega \]The resistance of the resistor is approximately 10 Ω. [1]
(d) The ammeter scale runs from 0 to 1.0 A. The smallest division on the ammeter represents 0.1 A, so the resolution is 0.1 A. [1]
(e) The voltmeter scale runs from 0 to 10 V with divisions every 1 V. The resolution is 1 V (accept 0.5 V if half-divisions are counted). [1]
(f) To investigate whether resistance changes with voltage:
- Change the voltage using a variable power supply or add a variable resistor to the circuit. [1]
- Record several pairs of V and I readings at different voltages. [1]
- Calculate \( R = \frac{V}{I} \) for each pair and compare the values. If R is constant, the resistor obeys Ohm's law. Alternatively, plot V against I: a straight line through the origin confirms ohmic behaviour. [1]
(g) Read the meter with your eye positioned directly in front of the pointer, perpendicular to the scale, to avoid parallax error. If the meter has a mirror strip behind the scale, align the pointer with its reflection to confirm you are viewing from directly in front. [1]
Why this matters: Reading analogue instruments correctly is a core practical skill. Parallax error occurs when your line of sight is at an angle to the scale, making the pointer appear to be at a different position than it truly is. The resolution of an instrument determines the smallest measurable change. A measurement can never be more precise than the resolution of the instrument used. When calculating resistance from V and I, the uncertainties in both readings combine to affect the reliability of the final value.
