Question 1

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To reliably measure the volume of a liquid, a measuring cylinder is the most appropriate standard laboratory device, especially for a quantity like 200 ml which doesn’t require the extreme precision of a titration pipette. When evaluating the diameter of a thin wire, a simple ruler is entirely too imprecise because the wire’s diameter is exceptionally small. A micrometer screw gauge, however, is specifically designed to measure these very small distances with high precision, typically down to $0.01\text{ mm}$. Therefore, pairing the measuring cylinder with the micrometer screw gauge is the correct and practical approach.
Question 2

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The distance travelled by any moving body can be found by calculating the area under its speed-time graph. Let’s compute the areas for each given graph over the $6.0\text{ s}$ interval. Graph A forms a rectangle, giving an area of $6.0 \times 6.0 = 36\text{ m}$. Graph B forms a triangle, so its area is $\frac{1}{2} \times 6.0 \times 4.0 = 12\text{ m}$. Graphs C and D are also triangles, giving areas of $\frac{1}{2} \times 6.0 \times 6.0 = 18\text{ m}$ and $\frac{1}{2} \times 6.0 \times 8.0 = 24\text{ m}$, respectively. Comparing these calculated areas, $12\text{ m}$ is the smallest value, meaning body B travelled the least distance.
Question 3
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Mass is an intrinsic property that measures the amount of matter in an object, remaining completely constant regardless of its location in the universe. Weight, on the other hand, is not matter; it is the downward force experienced by that object due to gravity. This physical relationship is perfectly summarized by the mathematical equation $W = mg$, where $W$ is weight, $m$ is mass, and $g$ is the gravitational field strength. Therefore, weight is truly defined as the effect of a gravitational field acting upon a specific mass.
Question 4

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To correctly determine the density of any substance, you must measure both its mass and its volume, as governed by the fundamental density equation $\rho = \frac{m}{V}$. When dealing specifically with a liquid, the volume is most accurately and easily measured using a measuring cylinder. The mass of that exact liquid volume can be found by placing the cylinder on a standard laboratory balance (weighing it empty and then full). Because a stop-watch measures time and a ruler measures linear distance, only the balance and measuring cylinder are the functional tools for this task.
Question 5


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The crucial piece of information given in the question is that the metre rule “rests steadily” on the supports, which tells us that the entire system is perfectly in a state of rotational and translational equilibrium. According to the principle of moments, for any object in static equilibrium, the sum of clockwise moments about any chosen pivot point must exactly equal the sum of anticlockwise moments. Because the rule is not actively rotating around any axis, the net (or total) moment calculated about point M must be zero, and equally, the net moment about point N must also be zero.
Question 6

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When an object is traveling in a circular path, even if its speed is constant, its direction is continuously changing, which means it is constantly accelerating. According to Newton’s second law, this acceleration requires a resultant force. For circular motion, this requisite force is known as the centripetal force. The very definition of centripetal force dictates that it must always point directly toward the center of the circular path. In the provided diagram, the arrow pointing strictly toward the center of the circle is labeled D.
Question 7

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This problem is a classic application of the principle of conservation of momentum, which states that total initial momentum equals total final momentum in a closed system. First, calculate the total momentum before the collision: $(4.0\text{ kg} \times 4.0\text{ m/s}) + (2.0\text{ kg} \times 2.0\text{ m/s}) = 16 + 4 = 20\text{ kg m/s}$. After the collision, the momentum equation is $(4.0\text{ kg} \times 3.0\text{ m/s}) + (2.0\text{ kg} \times v) = 20$. Simplifying this yields $12 + 2v = 20$. Subtracting 12 from both sides gives $2v = 8$, which means the final velocity $v$ is exactly $4.0\text{ m/s}$.
Question 8

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As an object falls freely towards the Earth, its height above the ground continuously decreases. Because gravitational potential energy is directly proportional to height ($E_p = mgh$), a loss of height inevitably means the gravitational potential energy is decreasing. At the exact same time, the object accelerates downward due to the force of gravity, meaning its velocity is increasing. Since kinetic energy is tied to the square of velocity ($E_k = \frac{1}{2}mv^2$), its kinetic energy must be simultaneously increasing, effectively transforming potential energy into kinetic energy.
Question 9

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To find the power developed by the boy, we must first calculate the work done, which equals the change in gravitational potential energy. The vertical distance the book was lifted is $1.7\text{ m} – 0.80\text{ m} = 0.90\text{ m}$. Using the formula for potential energy change, $\Delta E_p = mgh$, we get $0.60\text{ kg} \times 10\text{ N/kg} \times 0.90\text{ m} = 5.4\text{ J}$. Power is defined as the rate at which work is done, formulated as $P = \frac{W}{t}$. Plugging in our values gives $P = \frac{5.4\text{ J}}{0.60\text{ s}} = 9.0\text{ W}$, leading us to Option C.
Question 10

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The law of conservation of energy establishes that work done on a system transfers energy into that system. In this scenario, the mechanical work actively lifts the mass against gravity while simultaneously accelerating it from rest. Because there is no energy lost to air resistance, all the mechanical work done must exactly equal the total energy gained by the mass. This total energy encompasses both the increase in its gravitational potential energy (P) and its newly acquired kinetic energy (Q). Therefore, the total work done is expressed simply as $P + Q$.
Question 11

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To find the gauge pressure (pressure relative to atmospheric pressure) at point P, we use the fluid pressure formula $P = \rho g h$, where $\rho$ is the density of the fluid, $g$ is the gravitational field strength ($10\text{ N/kg}$), and $h$ is the depth from the surface. Based on the provided diagram measurements, the vertical depth $h$ of point P below the surface is approximately $7.4\text{ m}$ (or similar derived from the total column). Plugging in the values gives $P = 1020\text{ kg/m}^3 \times 10\text{ N/kg} \times 7.35\text{ m} \approx 75000\text{ Pa}$. Therefore, the pressure at P is 75000 Pa above atmospheric pressure.
Question 12

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This scenario relies on Boyle’s Law, which states that for a fixed mass of gas at a constant temperature, the pressure is inversely proportional to its volume ($P \propto 1/V$). As the volume of the gas in the cylinder is slowly decreased over time, the gas molecules become more crowded and hit the cylinder walls more frequently, leading to a steady increase in pressure. The graph must start at a non-zero initial pressure because a fixed mass of gas always exerts some baseline pressure. Therefore, the graph showing a steady upward slope from a positive starting value correctly represents this relationship.
Question 13

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Drying clothes is a practical example of evaporation, the process where liquid water turns into water vapor. The rate of evaporation is significantly influenced by several environmental factors. A higher temperature gives the water molecules more kinetic energy, allowing more of them to escape the liquid’s surface. A high wind speed continuously removes the newly formed water vapor from the immediate vicinity of the clothes, preventing the local air from becoming saturated and thus maintaining a steep concentration gradient. Combining both high temperature and high wind speed provides the optimal conditions for rapid drying.
Question 14
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Sensitivity in a liquid-in-glass thermometer refers to how much the liquid thread moves along the scale for a given change in temperature. The volume expansion of the liquid depends entirely on the initial volume of liquid present. By incorporating a larger liquid reservoir (bulb), there is a greater total volume of liquid to expand when heated. Because the expansion is constrained to move up a narrow capillary tube, this larger total expansion translates into a much longer movement of the thread per degree, thus significantly increasing the thermometer’s sensitivity.
Question 15
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We can solve this by calculating the total energy supplied and the mass of liquid vaporised. The energy supplied by the heater is $E = \text{Power} \times \text{time} = 100\text{ W} \times (12 \times 60)\text{ s} = 72000\text{ J}$. The mass of the liquid that actually boiled away into a gas is $m = 300\text{ g} – 100\text{ g} = 200\text{ g}$, which converts to $0.2\text{ kg}$. Using the specific latent heat formula $E = mL$, we can rearrange it to find $L = \frac{E}{m}$. Substituting our calculated values gives $L = \frac{72000\text{ J}}{0.2\text{ kg}} = 360000\text{ J/kg}$.
Question 16
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Detailed solution:
An electric fire operates by converting electrical energy into thermal energy, heavily radiating this heat into the surroundings as infrared radiation. Infrared radiation is a distinct section of the electromagnetic spectrum, meaning the primary functional output of an electric fire is indeed electromagnetic waves. In contrast, an electric generator produces alternating electrical current, an electric motor produces mechanical kinetic energy, and an electromagnet simply produces a static magnetic field, none of which are primarily radiating electromagnetic waves for their main purpose.
Question 17

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When any wave transitions from one medium to another, its frequency remains absolutely constant because frequency is solely determined by the initial source of the vibration. However, the speed of a sound wave drastically increases when it moves from a gas (air) into a denser liquid (water) due to the closer proximity of particles propagating the mechanical energy. According to the wave equation $v = f\lambda$, if the wave speed ($v$) increases while the frequency ($f$) stays the same, the wavelength ($\lambda$) must proportionally increase to maintain the mathematical balance.
Question 18

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The physical phenomenon where plane wavefronts spread out radially into the shadow regions after passing through a narrow gap or moving past an edge is scientifically termed diffraction. This is a fundamental property of all wave types, whether they are water waves, sound, or light. Reflection involves waves bouncing back, while refraction involves waves changing direction and speed as they enter a distinctly different medium. Here, the water waves remain in the same medium but naturally flare outwards around the barriers, which perfectly describes diffraction.
Question 19

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To calculate the refractive index using Snell’s Law ($n = \frac{\sin i}{\sin r}$), you must ensure all angles are measured accurately from the normal line, not the boundary surface. The diagram provides the angle between the incident ray and the surface as $40^\circ$, so the true angle of incidence is $i = 90^\circ – 40^\circ = 50^\circ$. The angle of refraction $r$ is already correctly given relative to the normal as $27^\circ$. Substituting these into the formula yields $n = \frac{\sin(50^\circ)}{\sin(27^\circ)} \approx \frac{0.7660}{0.4540} \approx 1.687$. Rounding to one decimal place gives 1.7.
Question 20

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A fundamental property of reflection in a plane mirror is that the virtual image is always formed perfectly perpendicularly behind the mirror surface. Furthermore, the distance of the image behind the mirror is exactly equal to the distance of the physical object in front of the mirror. Looking at the provided diagram, position B is the only point located along the direct perpendicular line from the object into the mirror, and it matches the object’s original distance perfectly. The position of the observing eye does not alter where the image is intrinsically formed.
Question 21
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It’s a common trap to mistake the angle given with the mirror’s surface for the angle of incidence. In optics, all critical angles—like incidence and reflection—are measured directly from the “normal”, which is an imaginary line drawn exactly at $90^{\circ}$ (perpendicular) to the mirror’s surface. If the incident light ray makes a $70^{\circ}$ angle with the flat surface of the mirror, we simply subtract this from $90^{\circ}$ to find the angle it makes with the normal line. Calculating $90^{\circ} – 70^{\circ}$ gives us $20^{\circ}$, which is the true angle of incidence.
Question 22

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Sound waves are mechanical longitudinal waves, meaning the particles of the medium vibrate back and forth parallel to the exact direction that the wave itself is travelling. The diagram shows alternating vertical bands of compressions and rarefactions arranged side-by-side, which indicates the wave energy is propagating horizontally across the page. Because the wave is moving horizontally, it is physically impossible for the wave to be travelling towards the top of the page (which would imply a transverse wave motion in this context). Therefore, statement D is completely incorrect.
Question 23
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When you look into a standard flat (plane) mirror, the image you see has several fixed characteristics defined by the laws of reflection. First, the image appears to be exactly the same size as the real object; it does not magnify or shrink what it reflects. Second, the image is “virtual”, which means the light rays only appear to diverge from a point behind the mirror, and the image cannot be projected onto a physical screen. Finally, it is upright and laterally inverted. Option D perfectly captures the core properties of being the same size and virtual.
Question 24


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This scenario highlights the massive difference between the speed of light and the speed of sound. Light travels at an incredible $3 \times 10^8\text{ m/s}$, so the visual signal of the smoke crossing the $100\text{ m}$ distance is essentially instantaneous to human perception. Sound, however, travels much slower through air, at roughly $330\text{ m/s}$. To calculate the time delay for the sound, we use $t = \frac{d}{v}$. Plugging in the distance gives $t = \frac{100\text{ m}}{330\text{ m/s}}$, which calculates to approximately $0.3\text{ seconds}$. Thus, the judge sees the smoke immediately but hears the bang about $0.3\text{ s}$ later.
Question 25


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By analyzing the magnetic field line pattern shown between the two magnets, we see continuous lines stretching directly from the pole of magnet X to the pole of magnet Y. Field lines bridging the gap between two poles in this manner indicate that the poles are opposite (one North and one South), creating an attractive magnetic field. Because opposite poles attract, magnet X will experience a pulling force directed to the right (towards Y), and magnet Y will experience an equal pulling force directed to the left (towards X). Option B correctly illustrates these attractive inward forces.
Question 26

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To calculate the electrical resistance of the lamp, we rely on Ohm’s Law, which defines resistance as the ratio of the potential difference across a component to the current flowing through it. The mathematical formula is $R = \frac{V}{I}$. The problem explicitly gives us the potential difference (voltage) across the lamp as $V = 6.0\text{ V}$ and the current flowing through it as $I = 0.5\text{ A}$. Substituting these given values into the equation yields $R = \frac{6.0\text{ V}}{0.5\text{ A}}$, which calculates to exactly $12.0\text{ }\Omega$.
Question 27
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Electric current is fundamentally defined as the rate at which electrical charge flows past a specific point in a circuit over a given period. It does not directly depend on the resistance in its defining equation, though resistance affects its magnitude in a full circuit. The formal mathematical definition states that Current equals the total Charge divided by the Time taken. Using the standard physics symbols where $I$ represents current, $Q$ is charge, and $t$ is time, the relationship is expressed perfectly by the equation $I = \frac{Q}{t}$.
Question 28
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When insulating materials are rubbed together, charging by friction occurs entirely due to the transfer of electrons. Protons are locked tightly inside the atomic nuclei and never move during simple electrostatic charging, which immediately rules out options C and D. Since electrons carry a negative charge, gaining them makes an object negative, while losing them leaves behind an uncancelled positive charge. For the plastic rod to become positively charged, it must have lost negative particles. Therefore, electrons physically transferred from the rod over to the woollen cloth.
Question 29

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A filament lamp is a non-ohmic conductor. As the voltage across the lamp increases, the current forced through it also increases, which causes the thin metal filament to heat up significantly. As the temperature of the metal rises, its internal lattice atoms vibrate more vigorously, actively hindering electron flow and increasing the electrical resistance. On an $I-V$ graph, a higher resistance translates to a flatter slope ($I/V$ decreases). Thus, the graph curves over, showing that current increases at a progressively slower rate as voltage increases, matching curve D.
Question 30

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In a potential divider circuit, the supply voltage is shared between components based on their individual resistance; the component with the highest resistance gets the largest share of the voltage. A Light-Dependent Resistor (LDR) works such that as light intensity increases, its resistance sharply drops. If component X is the LDR, increasing light lowers its resistance, so it takes a smaller share of the total voltage. Because the sum of voltages must equal the supply, component Y (the fixed resistor) must subsequently take a larger share, meaning its p.d. increases.
Question 31
Diagram 2 shows the output voltage from the power supply.


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The circuit shown is a classic bridge rectifier consisting of a network of four diodes. Its primary function is to convert the incoming alternating current (a.c.), which swings positive and negative, into direct current (d.c.) that flows in only one direction. When the a.c. supply is in its positive half-cycle, two opposite diodes conduct, driving current down through resistor R. Crucially, during the negative half-cycle, the other pair of diodes takes over, but they route the current through the resistor in the exact same downward direction. This means the negative halves of the wave are flipped completely positive, creating a continuous series of positive peaks known as full-wave rectification, matching graph B.
Question 32

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In standard international circuit diagrams, very specific symbols are used so anyone can read a schematic without confusion. A plain, empty rectangle always represents a standard fixed resistor. However, a fuse is specifically drawn as a rectangular box with the main circuit wire passing continuously straight through its center from end to end. This is a very literal visual representation of the thin sacrificial wire inside the fuse casing that is designed to melt and snap if the current dangerously exceeds safety limits. Since diagram D shows this exact rectangle with a line right through the middle, it is the universally accepted symbol for a fuse.
Question 33


Which logic gate is used?

▶️ Answer/Explanation
Detailed solution:
To solve this, we need to translate the real-world condition into binary logic. The problem explicitly states that the alarm siren will sound (outputting a 1) if there is “any indication of a fire.” This means that if either the temperature sensor detects dangerous heat (input A = 1), OR the smoke detector triggers on smoke (input B = 1), OR if both happen at the same time, the alarm must trigger (output = 1). The only time it stays quiet is if both sensors are 0. This behavior is the exact definition of an OR gate’s truth table. In standard logic gate symbols, the OR gate has a curved input side and a pointed, shield-like output side, matching diagram B.
Question 34


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When a direct current flows through a tightly wound coil of wire called a solenoid, it acts exactly like a classic bar magnet, generating its own strong magnetic field. Using the right-hand grip rule, one end of the solenoid acts as a North pole and the opposite end acts as a South pole. The fundamental rule of magnetic fields is that the invisible field lines always emerge out of the North pole, curve smoothly around the outside space, and enter directly into the South pole. The tiny compass needles will naturally align themselves exactly along these looping field lines. Therefore, diagram D correctly illustrates this distinct, continuous closed-loop pattern around the electromagnet.
Question 35

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Transmitting massive amounts of electrical power over very long distances through national grid cables inevitably causes energy losses due to the heating effect of the current, mathematically defined as $P = I^2R$. To drastically minimize this wasted heat energy, the current must be kept as low as physically possible. Therefore, directly at the power station, a step-up transformer is used to push the transmission voltage to extremely high levels, which proportionally slashes the current. However, these massive transmission voltages are incredibly dangerous for domestic use. Before the electricity is routed into homes, a step-down transformer is necessary to safely drop the voltage back down to standard household levels.
Question 36


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The diagram neatly details the internal components of a simple alternating current (a.c.) generator. For an a.c. generator to function efficiently, the fast-rotating coil must maintain an unbroken electrical connection with the stationary external circuit without the wires twisting and snapping. Part 2 points to one of the two solid, completely continuous metal rings securely attached to the rotating axle; these are universally called slip-rings. Part 1 clearly points to the small, stationary block (usually made of carbon) that lightly presses and slides against the rotating slip-ring to conduct the current outwards. This static conductive block is known as a brush.
Question 37
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This perfectly describes the historic Rutherford alpha-particle scattering experiment that revolutionized atomic theory. The initial observation that the vast majority of alpha particles punch straight through the gold foil undeflected indicates that atoms are overwhelmingly composed of empty space. However, the shocking discovery that a very tiny fraction of these dense, highly positive alpha particles are violently deflected at massive angles proves they collided with a repulsive force of immense strength. This provides the definitive evidence that almost all of an atom’s mass and its entire positive charge is densely packed into a tiny central core, establishing the existence of the atomic nucleus.
Question 38
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When an unstable radioactive nucleus decays specifically by emitting an alpha particle, it is forcefully ejecting a helium nucleus from its core. This alpha particle is composed of exactly two protons and two neutrons. Because the parent nucleus loses two protons, its atomic number (the bottom sub-number) must decrease by exactly 2. Furthermore, because it loses a total of four nucleons altogether, its mass number (the top super-number) must drop by exactly 4. Starting with Thorium-230 (${}_{90}^{230}\text{Th}$), the new mass number calculates to $230 – 4 = 226$, and the new atomic number becomes $90 – 2 = 88$. This perfectly yields the radium nuclide ${}_{88}^{226}\text{Ra}$.
Question 39


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We can systematically identify the exact types of radiation present by analyzing how they interact with specific absorbing materials. When thin paper is introduced, the count rate remains completely unaffected at 350 counts per second. Since alpha particles are very weakly penetrating and are easily stopped by just a sheet of paper, this lack of a drop proves conclusively that there are no alpha particles present. However, when 1.0 mm of aluminium is added, the count rate crashes significantly to 180, proving the definite presence of beta particles, which are blocked by thin metals. Finally, placing thick lead drops the count even further down to 23. Since only highly penetrating gamma rays can punch through the aluminium and require thick lead shielding to be stopped, gamma rays must also be actively present.
Question 40

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The decay graph starts at a high initial count rate of $X$ but notably doesn’t drop to zero; it eventually flattens out steadily at $Y$. This constant baseline count rate $Y$ is the ever-present background radiation of the room, which does not decay away. Therefore, the pure radiation coming specifically from the isotope source at the very beginning is actually $X – Y$. After exactly one half-life duration, the radiation emitting from the source itself will halve to $\frac{X – Y}{2}$. However, the question asks for the total measured count rate, which means the un-decayed background radiation $Y$ is still being detected on top of it. Adding them together gives $\frac{X – Y}{2} + Y$. To combine these, finding a common denominator results in $\frac{X – Y + 2Y}{2}$, which simplifies cleanly down to $\frac{X + Y}{2}$.
