Question 1
Fig. 1.1 shows a balloon filled with helium gas. (Sub-topic – 1.3)
The mass of the balloon is 120 kg.
(a) Calculate the weight of the balloon. Show your working.
▶️Answer/Explanation
Solution:
Weight is calculated using the formula:
\[ \text{Weight} = \text{mass} \times \text{acceleration due to gravity} \]
Given:
\[ \text{mass} = 120 \, \text{kg}, \quad \text{acceleration due to gravity} = 9.8 \, \text{m/s}^2 \]
Therefore:
\[ \text{Weight} = 120 \times 9.8 = 1176 \, \text{N} \]
The weight of the balloon is 1176 N.
(b) The resultant force on the balloon is 54 N. Show that the acceleration of the balloon is \(0.45 \, \text{m/s}^2\). (Sub-topic – 1.5)
▶️Answer/Explanation
Solution:
Newton’s second law states:
\[ F = ma \]
Rearranging for acceleration:
\[ a = \frac{F}{m} \]
Given:
\[ F = 54 \, \text{N}, \quad m = 120 \, \text{kg} \]
Therefore:
\[ a = \frac{54}{120} = 0.45 \, \text{m/s}^2 \]
The acceleration of the balloon is \(0.45 \, \text{m/s}^2\).
(c) The balloon accelerates upwards from rest at \(0.45 \, \text{m/s}^2\) for 8.0 s. Calculate the velocity of the balloon after 8.0 s. (Sub-topic – 1.2)
▶️Answer/Explanation
Solution:
The final velocity can be calculated using the formula:
\[ v = u + at \]
Given:
\[ u = 0 \, \text{m/s}, \quad a = 0.45 \, \text{m/s}^2, \quad t = 8.0 \, \text{s} \]
Therefore:
\[ v = 0 + (0.45 \times 8.0) = 3.6 \, \text{m/s} \]
The velocity of the balloon after 8.0 s is 3.6 m/s.
(d) Calculate the distance travelled by the balloon in the first 8.0 s. (Sub-topic – 1.2)
▶️Answer/Explanation
Solution:
The distance travelled can be calculated using the formula:
\[ s = ut + \frac{1}{2}at^2 \]
Given:
\[ u = 0 \, \text{m/s}, \quad a = 0.45 \, \text{m/s}^2, \quad t = 8.0 \, \text{s} \]
Therefore:
\[ s = 0 \times 8.0 + \frac{1}{2} \times 0.45 \times (8.0)^2 = 0 + 0.225 \times 64 = 14.4 \, \text{m} \]
The distance travelled by the balloon in the first 8.0 s is 14.4 m.
Question 2
(a) (i) Define pressure. (Sub-topic – 1.8)
▶️Answer/Explanation
Pressure is defined as force per unit area. Mathematically, it is expressed as:
\[ P = \frac{F}{A} \]
where \( P \) is pressure, \( F \) is force, and \( A \) is area.
(ii) Describe how pressure in a liquid varies with its depth and with its density.
▶️Answer/Explanation
Variation with depth: Pressure in a liquid increases with depth. This is because the weight of the liquid above exerts a force on the liquid below, increasing the pressure as you go deeper.
Variation with density: Pressure in a liquid also increases with its density. A denser liquid has more mass per unit volume, which means it exerts a greater force (and thus pressure) at a given depth compared to a less dense liquid.
(b) State two energy resources for which the Sun is not the main source. (Sub-topic – 1.7.3)
▶️Answer/Explanation
Two energy resources for which the Sun is not the main source are:
- Geothermal energy: This energy is derived from the heat within the Earth’s core, which is generated by radioactive decay and residual heat from the planet’s formation.
- Nuclear energy: This energy is produced from the fission or fusion of atomic nuclei, which releases energy stored in the nucleus of atoms, not from the Sun.
(c) State and explain whether each of the following methods of electrical power generation is renewable. (Sub-topic – 1.7.3)
(i) Power generation in a nuclear power station.
▶️Answer/Explanation
Statement: Non-renewable.
Explanation: Nuclear power generation relies on uranium or other radioactive materials, which are finite resources. Once these materials are used, they cannot be replenished on a human timescale, making nuclear power non-renewable.
(ii) Power generation from waves in the sea.
▶️Answer/Explanation
Statement: Renewable.
Explanation: Wave energy is generated by the movement of ocean waves, which are driven by wind and tidal forces. Since waves are continuously generated by natural processes, wave energy is considered renewable.
Question 3
(a)(i) State which state of matter, solid, liquid or gas, has the greatest thermal expansion and which has the least. (Sub-topic – 2.2.1)
▶️Answer/Explanation
Answer:
Greatest expansion: Gas
Least expansion: Solid
Explanation:
Gases have the greatest thermal expansion because their particles are far apart and move freely, allowing them to expand significantly when heated. Solids have the least thermal expansion because their particles are closely packed and vibrate in fixed positions, limiting their ability to expand.
(ii) Describe, in terms of the motion and arrangement of particles, the structures of solids and gases. (Sub-topic – 2.1.2)
▶️Answer/Explanation
Answer:
Solids: Particles vibrate in fixed positions and are closely packed in a regular arrangement.
Gases: Particles move freely and are randomly arranged with large spaces between them.
Explanation:
In solids, particles are held together by strong forces, causing them to vibrate in fixed positions. This results in a rigid structure. In gases, particles have weak forces between them, allowing them to move freely and fill the container they are in, leading to a random arrangement.
(b)(i) Define specific heat capacity. (Sub-topic – 2.2.2)
▶️Answer/Explanation
Answer:
Specific heat capacity is the amount of energy required to raise the temperature of 1 kg of a substance by 1°C (or 1 K).
Explanation:
Specific heat capacity is a property of a material that indicates how much energy is needed to change its temperature. It is an important concept in thermodynamics and is used to calculate the energy required for heating or cooling substances.
(b)(ii) A student carries out an experiment to determine the specific heat capacity of a metal. A cylinder of the metal is heated by a 12W electrical heater. State the readings that the student takes.
(Sub-topic – 2.2.2)
▶️Answer/Explanation
Answer:
The student takes the following readings:
1. Time (t) for which the heater is on.
2. Mass (m) of the metal cylinder.
3. Initial temperature (T1) of the metal.
4. Final temperature (T2) of the metal.
Explanation:
To calculate the specific heat capacity, the student needs to measure the time the heater is on to determine the energy supplied (E = P × t), the mass of the metal, and the temperature change (ΔT = T2 – T1). These readings are essential for applying the formula c = E / (m × ΔT).
Question 4
(a) Fig. 4.1 is an incomplete ray diagram showing an object O, a converging lens, and the principal axis. The focal points of the lens are each labelled F. (Sub-topic – 3.2.3)
(i) Complete the ray diagram to draw the image formed by the lens. Label your image I.
(ii) Circle three descriptions in the list which describe the image formed in (i).
▶️Answer/Explanation
(i) To complete the ray diagram:
- Draw a ray from the top of the object parallel to the principal axis. After passing through the lens, it will refract through the focal point on the opposite side.
- Draw a ray from the top of the object through the focal point on the same side as the object. After passing through the lens, it will refract parallel to the principal axis.
- The point where these two rays intersect (or appear to intersect) is the top of the image. Draw the image from this point to the principal axis.
(ii) The correct descriptions are: enlarged, virtual, and upright.
(b) (i) State the name for the defect of vision that can be corrected by a converging lens. (Sub-topic – 3.2.3)
(ii) Describe how a converging lens corrects the defect in (i). You may find it helpful to sketch a ray diagram.
▶️Answer/Explanation
(i) The defect of vision that can be corrected by a converging lens is long-sightedness (hypermetropia).
(ii) A converging lens corrects long-sightedness by refracting the light rays so that they converge earlier and focus on the retina. In a long-sighted eye, the light rays focus behind the retina, causing blurred vision for nearby objects. The converging lens brings the focal point forward onto the retina, allowing clear vision.
Question 5
(a) Two types of electromagnetic radiation are used in glass optical fibres for high-speed broadband. (Sub-topic – 3.3)
(i) State the type of electromagnetic radiation, other than visible light, which is used in glass optical fibres.
▶️Answer/Explanation
Infrared
(ii) Give two reasons why these two types of electromagnetic radiation are used in glass optical fibres for high-speed broadband.
▶️Answer/Explanation
1. Glass is transparent to visible light and (some) infrared.
2. Visible light and some infrared can carry high rates of data/information.
(b) (i) The critical angle of the glass in an optical fibre is 45°. (Sub-topic – 3.2.2)
Calculate the refractive index of the glass.
▶️Answer/Explanation
Refractive index \( n = \frac{1}{\sin c} \)
Given \( c = 45^\circ \),
\( n = \frac{1}{\sin 45^\circ} = \frac{1}{0.7071} \approx 1.4 \)
(ii) Fig. 5.1 shows an optical fibre made of the glass described in (i).
On Fig. 5.1, draw carefully a ray of light in the fibre undergoing total internal reflection.
▶️Answer/Explanation
Diagram should show that total internal reflection is taking place inside the optical fibre with one or more correct reflections where the angle of incidence \( i > 45^\circ \).
Question 6
An electric heater uses a resistance wire of resistance 26 Ω. The power dissipated in the resistance wire is 2500 W. (Sub-topic – 4.2.4)
(a) Calculate the current in the resistance wire.
▶️Answer/Explanation
Solution:
We know that power \( P \) is related to current \( I \) and resistance \( R \) by the formula: \[ P = I^2 R \] Rearranging the formula to solve for \( I \): \[ I = \sqrt{\frac{P}{R}} \] Substituting the given values: \[ I = \sqrt{\frac{2500}{26}} = \sqrt{96.15} \approx 9.8 \, \text{A} \] Therefore, the current in the resistance wire is approximately 9.8 A.
(b) The resistance wire of the heater has a length of 1.2 m and a cross-sectional area of \(7.9 \times 10^{-7} \, \text{m}^2\). A new heater is designed using wire of the same material with length 1.8 m and cross-sectional area \(5.8 \times 10^{-7} \, \text{m}^2\). (Sub-topic – 4.2.4)
Calculate the resistance of this wire.
▶️Answer/Explanation
Solution:
The resistance \( R \) of a wire is given by: \[ R = \rho \frac{L}{A} \] where \( \rho \) is the resistivity of the material, \( L \) is the length of the wire, and \( A \) is the cross-sectional area.
Since the material is the same, the resistivity \( \rho \) remains constant. We can use the ratio of the resistances: \[ \frac{R_2}{R_1} = \frac{L_2}{L_1} \times \frac{A_1}{A_2} \] Substituting the given values: \[ \frac{R_2}{26} = \frac{1.8}{1.2} \times \frac{7.9 \times 10^{-7}}{5.8 \times 10^{-7}} \] Simplifying: \[ \frac{R_2}{26} = 1.5 \times 1.362 \approx 2.043 \] Therefore: \[ R_2 = 26 \times 2.043 \approx 53 \, \Omega \] The resistance of the new wire is approximately 53 Ω.
(c) The 2500 W heater is used in a country where electricity costs 0.30 dollars per kilowatt-hour. (Sub-topic – 4.2.5)
Calculate the cost of using the heater continuously for two days.
▶️Answer/Explanation
Solution:
First, calculate the total energy consumed by the heater in two days: \[ \text{Energy} = \text{Power} \times \text{Time} \] The power is 2500 W, which is 2.5 kW. The time is 2 days, which is 48 hours. \[ \text{Energy} = 2.5 \, \text{kW} \times 48 \, \text{h} = 120 \, \text{kWh} \] Next, calculate the cost: \[ \text{Cost} = \text{Energy} \times \text{Cost per kWh} \] Substituting the given values: \[ \text{Cost} = 120 \, \text{kWh} \times 0.30 \, \text{dollars/kWh} = 36 \, \text{dollars} \] Therefore, the cost of using the heater continuously for two days is 36 dollars.
Question 7
The voltage across the primary coil of a 100% efficient transformer is 220 V and the voltage across the secondary coil is 12 V. (Sub-topic – 4.5.6)
(a) The current in the secondary coil is 2.5 A. Calculate the current in the primary coil.
▶️Answer/Explanation
Answer: 0.14 A
Explanation:
For a 100% efficient transformer, the power in the primary coil is equal to the power in the secondary coil. Therefore:
\( V_p \times I_p = V_s \times I_s \)
Given:
\( V_p = 220 \, \text{V} \), \( V_s = 12 \, \text{V} \), \( I_s = 2.5 \, \text{A} \)
Substituting the values:
\( 220 \times I_p = 12 \times 2.5 \)
\( 220 \times I_p = 30 \)
\( I_p = \frac{30}{220} \)
\( I_p = 0.136 \, \text{A} \approx 0.14 \, \text{A} \)
(b) Calculate the ratio of the number of turns on the primary coil to the number of turns on the secondary coil of the transformer. (Sub-topic – 4.5.6)
▶️Answer/Explanation
Answer: 18:1
Explanation:
The ratio of the number of turns on the primary coil (\( N_p \)) to the number of turns on the secondary coil (\( N_s \)) is given by:
\( \frac{N_p}{N_s} = \frac{V_p}{V_s} \)
Given:
\( V_p = 220 \, \text{V} \), \( V_s = 12 \, \text{V} \)
Substituting the values:
\( \frac{N_p}{N_s} = \frac{220}{12} \)
\( \frac{N_p}{N_s} = 18.33 \)
Therefore, the ratio is approximately 18:1.
Question 8
(a) During β-decay, one of the neutrons in the nucleus changes. (Sub-topic – 5.2.3)
(i) State what happens to this neutron.
(ii) Explain how charge is conserved during this change.
▶️Answer/Explanation
(i) The neutron changes into a proton and an electron (β-particle).
(ii) The neutron has a charge of 0. When it decays into a proton and an electron, the proton has a charge of +1 and the electron has a charge of -1. The total charge remains 0, so charge is conserved.
(b) Complete the nuclide equation for the α-decay of radon-212 to form an isotope of polonium, symbol Po. (Sub-topic – 5.2.3)
\[ \frac{212}{86} Rn \rightarrow \]
▶️Answer/Explanation
The nuclide equation for the α-decay of radon-212 is: \[ \frac{212}{86} Rn \rightarrow \frac{208}{84} Po + \frac{4}{2} He \] Explanation: In α-decay, the radon-212 nucleus loses an alpha particle, which consists of 2 protons and 2 neutrons. This reduces the atomic number by 2 (from 86 to 84) and the mass number by 4 (from 212 to 208), resulting in the formation of polonium-208.
Question 9
Fig. 9.1 shows the Sun as the central dot and the planets Saturn, Jupiter, and Earth labelled \( S_0 \), \( J_0 \), and \( E_0 \). The planets orbit the Sun anticlockwise. From the Earth’s orbit, the planets appear aligned. (Sub-topic – 6.1.2)
Assume that Saturn takes 30 years to orbit the Sun and that Jupiter takes 12 years to orbit the Sun.
(a) On Fig. 9.1, mark the positions of Saturn and Jupiter 5.0 years after the original positions shown. Label these positions \( S_1 \) and \( J_1 \). Show your working.
▶️Answer/Explanation
Solution:
To determine the positions of Saturn and Jupiter after 5.0 years:
- Saturn completes one orbit in 30 years, so in 5 years, it will have moved \( \frac{5}{30} \times 360^\circ = 60^\circ \) from its original position.
- Jupiter completes one orbit in 12 years, so in 5 years, it will have moved \( \frac{5}{12} \times 360^\circ = 150^\circ \) from its original position.
Therefore, mark Saturn at \( 60^\circ \) and Jupiter at \( 150^\circ \) from their original positions.
(b) (i) On Fig. 9.1, mark the positions of Saturn and Jupiter 20 years after the original positions shown in Fig. 9.1. Label these positions \( S_2 \) and \( J_2 \). (Sub-topic – 6.1.2)
(ii) State what is observed from the Earth’s orbit after 20 years.
▶️Answer/Explanation
Solution:
(i) To determine the positions of Saturn and Jupiter after 20 years:
- Saturn completes one orbit in 30 years, so in 20 years, it will have moved \( \frac{20}{30} \times 360^\circ = 240^\circ \) from its original position.
- Jupiter completes one orbit in 12 years, so in 20 years, it will have moved \( \frac{20}{12} \times 360^\circ = 600^\circ \). Since 600° is equivalent to 240° (600° – 360°), Jupiter will also be at 240° from its original position.
Therefore, mark both Saturn and Jupiter at \( 240^\circ \) from their original positions.
(ii) After 20 years, Saturn and Jupiter will be aligned again, with Jupiter exactly in front of Saturn as observed from Earth.
(c) (i) Choose two words from the list to describe each planet. (Sub-topic – 6.1.2)
| gaseous | large | rocky | small | |
|---|---|---|---|---|
| Jupiter | ||||
| Earth |
(ii) The average density of Jupiter is much less than that of the Earth. The gravitational field strength at the surface of Jupiter is greater than that at the surface of the Earth. Explain how these differences in density and in gravitational field strength are consistent with your answers to (c)(i).
▶️Answer/Explanation
Solution:
(i) Jupiter: gaseous, large
Earth: rocky, small
(ii) The differences in density and gravitational field strength are consistent with the descriptions:
- Jupiter is a gas giant, so it has a low density despite its large size. Its large mass results in a strong gravitational field at its surface.
- Earth is a rocky planet with a high density and smaller size, but its gravitational field strength is less than Jupiter’s due to its smaller mass.
(d) The average density of Jupiter is 1300 kg/m\(^3\) and its volume is \(1.4 \times 10^{15}\) km\(^3\). Calculate the mass of Jupiter. (Sub-topic – 1.4)
▶️Answer/Explanation
Solution:
To calculate the mass of Jupiter, use the formula:
\[ \text{mass} = \text{density} \times \text{volume} \] Given:
\[ \text{density} = 1300 \, \text{kg/m}^3 \] \[ \text{volume} = 1.4 \times 10^{15} \, \text{km}^3 = 1.4 \times 10^{15} \times 10^9 \, \text{m}^3 = 1.4 \times 10^{24} \, \text{m}^3 \] Therefore:
\[ \text{mass} = 1300 \times 1.4 \times 10^{24} = 1.82 \times 10^{27} \, \text{kg} \]
Question 10
(a) Show that 1 light-year = 9.5 × 1015 m. (Sub-topic – 6.2.2)
▶️Answer/Explanation
Solution:
To show that 1 light-year = 9.5 × 1015 m, we need to calculate the distance that light travels in one year.
The speed of light, \( c \), is approximately \( 3.0 \times 10^8 \) m/s.
First, calculate the number of seconds in one year:
\[ 1 \text{ year} = 365 \text{ days} \times 24 \text{ hours/day} \times 60 \text{ minutes/hour} \times 60 \text{ seconds/minute} \] \[ 1 \text{ year} = 365 \times 24 \times 60 \times 60 = 31,536,000 \text{ seconds} \]
Now, calculate the distance light travels in one year:
\[ \text{Distance} = \text{Speed} \times \text{Time} = c \times \text{Time} \] \[ \text{Distance} = 3.0 \times 10^8 \, \text{m/s} \times 31,536,000 \, \text{s} \] \[ \text{Distance} = 9.4608 \times 10^{15} \, \text{m} \]
Rounding to two significant figures, we get:
\[ 1 \text{ light-year} \approx 9.5 \times 10^{15} \, \text{m} \]
(b) (i) State one measurement that is taken when determining the speed \( v \) at which a galaxy is moving away from the Earth. (Sub-topic – 6.2.3)
▶️Answer/Explanation
Answer:
One measurement that is taken is the change in wavelength of the galaxy’s starlight due to redshift.
(ii) Write down an equation relating \( v \) and the distance \( d \) of a far galaxy. (Sub-topic – 6.2.3)
▶️Answer/Explanation
Answer:
The equation relating \( v \) and \( d \) is:
\[ H_0 = \frac{v}{d} \] where \( H_0 \) is the Hubble constant.
(iii) State how the distance \( d \) of a far galaxy can be determined other than by using the equation in (ii). (Sub-topic – 6.2.3)
▶️Answer/Explanation
Answer:
The distance \( d \) of a far galaxy can also be determined by measuring the brightness of a supernova in that galaxy.