Question 1
(i) Describe a method to measure the volume of the irregularly shaped solid object.
(ii) The volume of the object is $83\text{ cm}^3$. Calculate the mass of the object.
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $1.4$ — Density (Parts $\mathrm{(a)}$, $\mathrm{(b)(i)}$, $\mathrm{(b)(ii)}$)
▶️ Answer/Explanation
(a)
The oil will float on the surface of the water because its density ($\rho_{\text{oil}} = 0.80\text{ g/cm}^3$) is lower than that of water ($\rho_{\text{water}} = 1.0\text{ g/cm}^3$). In a mixture of non-mixing liquids, the substance with the lower density always displaces the more dense liquid and resides on top.
(b)(i)
To find the volume, use the displacement method: fill a measuring cylinder with a known volume of water, $V_1$, then fully submerge the object and record the new volume, $V_2$. The volume of the object is calculated as $V = V_2 – V_1$, ensuring no air bubbles are trapped and the object is completely immersed.
(b)(ii)
For the correct answer:
$224.1\text{ g}$
The mass is calculated using the density formula $\rho = \frac{m}{V}$, rearranged as $m = \rho \times V$. Substituting the given values: $m = 2.7\text{ g/cm}^3 \times 83\text{ cm}^3 = 224.1\text{ g}$. This product directly yields the mass since the units for volume and density are consistent.
Question 2
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 1.5.1 — Effects of forces (Parts (a)(i), (a)(ii), (b)(i), (b)(ii))
▶️ Answer/Explanation
(a)(i)
For the correct answer:
350 N/m
The spring constant k is defined as the force per unit extension, expressed by the equation k= x F . From the linear portion of the graph, a load of 14 N corresponds to an extension of 4.0 cm, which is 0.04 m. Calculating the ratio gives k= 0.04 m 14 N =350 N/m, representing the stiffness of the spring.
(a)(ii)
For the correct answer:
Point L marked where the graph ceases to be a straight line.
The limit of proportionality is the specific point on a load-extension graph beyond which Hooke’s Law no longer applies and the extension is no longer directly proportional to the force. On Fig 2.1, this is identified as the exact coordinate where the linear diagonal line begins to curve. Labeling this point L indicates the maximum load the spring can support while still maintaining a constant spring constant.
(b)(i)
For the correct answer:
Arrow pointing towards the center of the circular path.
When an object moves in a circular path at a constant speed, it undergoes a change in velocity due to the continuous change in its direction. This acceleration requires a resultant force, known as centripetal force, which must act perpendicular to the motion. Therefore, the arrow on Fig. 2.2 should be drawn starting from the car and pointing directly toward the center of the curve A.
(b)(ii)
For the correct answer:
The force increases.
For motion in a circular path, the required centripetal force is related to the radius by the relationship F∝ r 1 when mass and speed remain constant. As the car moves from corner A to a sharper corner B with a smaller radius, a larger resultant force is necessary to maintain the same speed. Consequently, the force acting on the car increases as the radius of the turn decreases.
Question 3
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 1.6 — Momentum (Parts (a), (b))
• Topic 1.7.1 — Energy (Part (c))
▶️ Answer/Explanation
(a)
For the correct answer:
1.3 kg m/s
Momentum is defined as the product of mass and velocity, expressed by the equation p=mv. First, convert the mass from grams to kilograms: 190 g=0.19 kg. Substituting the values, p=0.19 kg×6.9 m/s=1.311 kg m/s, which rounds to 1.3 kg m/s when considering significant figures.
(b)
For the correct answer:
0.89 m/s
According to the principle of conservation of momentum, the total initial momentum equals the total final momentum: m 1 u 1 +m 2 u 2 =m 1 v 1 +m 2 v 2 . Taking the initial direction as positive, the ball’s final velocity is −1.5 m/s. The equation becomes 1.311=(0.19×−1.5)+(1.8×v), leading to 1.311=−0.285+1.8v. Solving for v, we get v= 1.8 1.596 =0.886… m/s, rounded to 0.89 m/s.
(c)
For the correct answer:
3.6 J
First, calculate the final kinetic energy of the object using E k = 2 1 mv 2 , which gives 2 1 ×1.8×0.886 2 ≈0.71 J. The total final kinetic energy of the system is 0.2 J+0.71 J=0.91 J. The change in total kinetic energy is the difference between the initial energy of the ball and this total final energy: ΔE k =4.5 J−0.91 J=3.59 J, which rounds to 3.6 J.
Question 4
(a) The lowest possible temperature is zero kelvin (0 K).
(i) State the name of this lowest possible temperature.
(ii) Nitrogen boils at 77 K. Calculate the boiling point of nitrogen on the Celsius scale.
(b) The temperature of a fixed mass of gas at constant volume changes from 300 K to 400 K.
State and explain, in terms of particles, the effect on the pressure of the gas.
(c) A sample of gas is at a pressure of 120 kPa. The volume of the gas is doubled at constant temperature.
Calculate the new pressure of the gas.
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 2.2.1 — Thermal expansion of solid, liquid and gases (Parts (a), (b), (c))
▶️ Answer/Explanation
(a)(i)
For the correct answer:
absolute zero
The temperature at which the particles of a substance have the least possible kinetic energy is known as absolute zero. On the thermodynamic (Kelvin) scale, this point is defined as 0 K, which corresponds to −273.15 ∘ C on the Celsius scale.
(a)(ii)
For the correct answer:
−196 ∘ C
To convert a temperature from the Kelvin scale to the Celsius scale, subtract 273 from the Kelvin value using the formula T( ∘ C)=θ(K)−273. Substituting the given value, T=77−273=−196 ∘ C.
(b)
For the correct answer:
pressure increases; particles move faster/have more kinetic energy and collide more frequently/with more force.
As temperature increases, gas particles gain kinetic energy and move at higher average speeds. Consequently, they strike the container walls more frequently and with greater impact force. Since pressure is defined as force per unit area (P= A F ), these more energetic and frequent collisions result in an overall increase in gas pressure.
(c)
For the correct answer:
60 kPa
According to Boyle’s Law, for a fixed mass of gas at a constant temperature, pressure is inversely proportional to volume, expressed as p 1 V 1 =p 2 V 2 . If the volume doubles (V 2 =2V 1 ), the pressure must halve to maintain the constant product: 120 kPa×V 1 =p 2 ×2V 1 , which solves to p 2 = 2 120 =60 kPa.
Question 5
On Fig. 5.2:
- draw three wavefronts of this wave before it reaches the barrier
- draw three wavefronts after the wave passes through the gap.
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 3.1 — General properties of waves (Parts (a), (b))
▶️ Answer/Explanation
(a)
For the correct answer:
Three semi-circular wavefronts centered on the gap with constant wavelength λ.
When waves pass through a gap comparable to their wavelength, they undergo diffraction, spreading into the region behind the barrier. The resulting wavefronts are circular arcs centered at the center of the gap. Crucially, the speed v and frequency f of the light remain constant, so the wavelength λ in the equation v=fλ does not change; thus, the spacing between the drawn arcs must exactly match the spacing of the incident wavefronts.
(b)
For the correct answer:
Three straight parallel lines before the gap and three tighter semi-circular arcs after the gap.
Blue light has a higher frequency and thus a shorter wavelength than red light (\lambda_{blue} < \lambda_{red}). Before the barrier, three straight lines should be drawn parallel to the barrier with a smaller separation than in Fig 5.1. After passing through the gap, the wavefronts diffract as semi-circles; however, because the gap width is now larger relative to the shorter blue wavelength, the diffraction (spreading) is less pronounced than for red light, though they remain centered on the gap with their original shorter wavelength λ.
Question 6
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 4.2.4 — Resistance (Part (a))
• Topic 4.3.2 — Series and parallel circuits (Parts (b)(i), (b)(ii), (b)(iii), (b)(iv))
• Topic 4.2.5 — Electrical energy and electrical power (Part (b)(v))
▶️ Answer/Explanation
(a)
For the correct answer:
The graph is a curve with a decreasing gradient as voltage increases.
As current passes through the filament, the temperature of the wire increases, leading to more frequent collisions between electrons and lattice ions. This increase in temperature causes the resistance R to rise, as described by R= I V . Consequently, the current I increases at a slower rate than the potential difference V, resulting in a non-linear I−V characteristic curve that flattens out.
(b)(i)
For the correct answer:
8.0 V
First, find the total resistance R total =4.2Ω+2.1Ω=6.3Ω. The circuit current is I= R total V total = 6.3Ω 12 V ≈1.905 A. The voltmeter reading across the 4.2Ω resistor is then calculated using V=IR=1.905 A×4.2Ω=8.0 V.
(b)(ii)
For the correct answer:
1.9 A
The current I in the 4.2Ω resistor is the same as the total current in this series circuit. Using Ohm’s Law, I= R V = (4.2Ω+2.1Ω) 12 V = 6.3 12 ≈1.9 A. This value represents the flow of charge through the component per unit time.
(b)(iii)
For the correct answer:
1.9 A
In a series circuit, the current at every point is identical because there is only one path for the electrons to follow. Therefore, the current flowing through the 2.1Ω resistor must be equal to the current flowing through the 4.2Ω resistor, which is approximately 1.9 A.
(b)(iv)
For the correct answer:
1.9 A
The ammeter is connected in series with the resistors and the power supply, meaning it measures the total current flowing through the entire loop. Since the current is uniform throughout a series arrangement, the ammeter reading will be the same as the current calculated for the individual resistors, which is 1.9 A.
(b)(v)
For the correct answer:
15 W
Electrical power P can be calculated using the formula P=I 2 R or P=VI. Using the current I≈1.905 A and resistance R=4.2Ω, the power is P=(1.905 A) 2 ×4.2Ω≈15.2 W, which rounds to 15 W. This represents the rate at which energy is transferred to the resistor.
Question 7
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 4.2.1 — Electric charge (Parts (a), (b), (c))
▶️ Answer/Explanation
(a)(i)
For the correct answer:
An arrow at Y pointing directly towards X (radially inward).
The electric field lines represent the direction of force on a positive test charge. Since a negative charge experiences a force opposite to the field direction, and the field lines at Y are directed radially outward from X, the force on a negative charge at Y will be directed toward X. This is governed by the attraction between unlike charges.
(a)(ii)
For the correct answer:
A positive point charge.
The electric field pattern shown consists of straight lines radiating symmetrically outward from a single point. By convention, electric field lines always point away from positive charges and toward negative charges. Therefore, the source at point X must be a positive point charge to produce this specific outward radial pattern.
(b)
For the correct answer:
Electrons are transferred from the plastic to the other material.
Charging by friction in solids involves only the transfer of negative charge carriers, specifically electrons. When plastic becomes positively charged, it means it has a deficit of electrons. This occurs because the friction process provides energy for electrons to move from the plastic onto the material it is being rubbed against, leaving the plastic with a net positive charge.
(c)
For the correct answer:
Conductors contain free (delocalised) electrons while insulators do not.
In electrical conductors, such as metals, some electrons are not bound to specific atoms and are free to move throughout the lattice structure as delocalised electrons. In contrast, all electrons in an electrical insulator are tightly bound within atoms or molecules and are not free to migrate. This structural difference allows conductors to facilitate charge flow while insulators resist it.
Question 8
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 4.5.1 — Electromagnetic induction (Parts (a), (b))
▶️ Answer/Explanation
(a)(i)
For the correct answer:
Clockwise
Using Fleming’s Left-Hand Rule, the magnetic field acts from North to South (left to right) and the current flows from the positive terminal. This creates an upward force on the left side and a downward force on the right side. The resulting moment causes the coil to rotate clockwise from the perspective of observer O.
(a)(ii)
For the correct answer:
The turning effect is greater (three times larger).
The total force F acting on the coil is proportional to the number of turns N in the coil, as shown by the relationship F∝N. Increasing the turns from one to three while keeping the current constant triples the magnetic force on each side of the coil, thereby increasing the total turning effect proportionally.
(a)(iii)
For the correct answer:
The turning effect is less (one-third as large).
According to Ohm’s Law, I= R V , so increasing the resistance by three times at a constant potential difference reduces the current to one-third of its original value. Since the magnetic force is directly proportional to the current (F∝I), the reduction in current results in a weaker turning effect.
(b)(i)-(iii)
For the correct answer:
(i) Vertical (A at top, B at bottom); (ii) Horizontal (A at left, B at right); (iii) Vertical (A at bottom, B at top)
At t 0 , the voltage is zero because the coil is horizontal and moving parallel to field lines. At t 1 , the voltage is maximum positive, meaning the coil is vertical (rotated 90 ∘ ). At t 2 , the voltage returns to zero as the coil completes a 180 ∘ turn to a horizontal position. At t 3 , the negative peak indicates a vertical position but with the sides A and B inverted relative to t 1 .
Question 9
(a) For each application of radioactive isotopes, state and explain which type of radioactive emission is suitable and suggest an appropriate half-life for the isotope.
(i) Household smoke alarm
(ii) Measuring the thickness of aluminium strips produced in a factory
(b) Lead-208 ({82}^{208}\textrm{Pb}) has the highest nucleon number of the stable isotopes of lead. Explain why lead-214 ({82}^{214}\textrm{Pb}) is radioactive.
(c) State two different sources of background radiation.
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 5.2.4 — Half-life (Part (a))
• Topic 5.2.3 — Radioactive decay (Part (b))
• Topic 5.2.1 — Detection of radioactivity (Part (c))
▶️ Answer/Explanation
(a)(i)
Type of radioactive emission: Alpha (α)
Explanation: Alpha particles have the highest ionizing power, allowing them to ionize air molecules inside the detector chamber to create a current. When smoke particles enter, they absorb these alpha particles, reducing the current and triggering the alarm. Their low penetration (stopped by a few cm of air) ensures safety outside the device. A long half-life (e.g., 432 years for Americium-241) is required so the source activity remains nearly constant over many years, avoiding frequent replacements.
(a)(ii)
Type of radioactive emission: Beta (β)
Explanation: Beta particles are ideal for monitoring metal thickness because they can penetrate thin aluminium but are partially absorbed; the count rate changes significantly if the thickness varies. Alpha particles would be completely stopped, while gamma rays would pass through with almost no absorption. A long half-life is necessary to ensure that a drop in the recorded count rate is due to increased material thickness rather than the natural decay of the radioactive source over time.
(b)
For the correct answer:
The nucleus contains an excess of neutrons, making it unstable.
Radioactive decay is a spontaneous and random process that occurs in unstable nuclei to reach a more stable state. While Lead-208 is stable, Lead-214 has 214−82=132 neutrons, compared to 126 in Lead-208. This excess of neutrons disrupts the balance between the strong nuclear force and electrostatic repulsion. To increase stability, the nucleus undergoes β − decay, where a neutron transforms into a proton and an electron: n→p+e − .
(c)
For the correct answer:
Radon gas (in the air) and Rocks/buildings
Background radiation refers to the ionizing radiation present in the environment from both natural and artificial sources. Natural contributors include radon gas, which seeps from the ground, and radioactive isotopes like Carbon-14 or Potassium-40 found in rocks, building materials, and food. Cosmic rays from outer space also contribute to this constant level of radiation. Artificial sources, though typically a smaller percentage, include medical X-rays and fallout from past nuclear testing or accidents.
Question 10
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 6.2.2 & 6.2.3 — Stars and the Universe (Parts (a), (b))
▶️ Answer/Explanation
(a)(i) 1.
For the correct answer:
Hubble constant
The symbol H 0 represents the Hubble constant, which describes the current expansion rate of the Universe. It provides a proportional relationship between the recession speed of a distant galaxy and its distance from an observer on Earth, acting as a fundamental parameter in the Big Bang theory.
(a)(i) 2.
For the correct answer:
H 0 = d v
The Hubble constant is defined as the ratio of the speed v at which a galaxy is moving away (receding) from Earth to its distance d from Earth. This linear relationship shows that galaxies further away move faster, expressed by the equation H 0 = d v , where v is recession speed and d is distance.
(a)(ii)
For the correct answer:
s −1 OR per second
Based on the defining equation H 0 = d v , the units of speed (m/s) are divided by the units of distance (m). This results in the units of length cancelling out, leaving the reciprocal of time. Therefore, the standard SI unit for the Hubble constant is s −1 (per second).
(a)(iii)
For the correct answer:
4.5×10 17 s
The age of the Universe can be estimated as the reciprocal of the Hubble constant, using the formula t= H 0 1 . By substituting the given value, the calculation becomes t= 2.2×10 −18 s −1 1 , which yields approximately 4.55×10 17 seconds. This calculation assumes a constant expansion rate since the Big Bang.
(b)
For the correct answer:
Produced shortly after the Universe was formed AND detected from all points in space (all directions)
Cosmic microwave background radiation (CMBR) was produced shortly after the Big Bang when the early Universe cooled sufficiently for atoms to form, allowing radiation to travel freely. This ancient light has since been redshifted into the microwave region and is now observed as a uniform background signal coming from every direction in the sky.