Question 1

In fluid mechanics, if there were no gravitational field acting on a fluid, which one of the following effects would no longer exist for the fluid?

Accepted Answer

Introduction / Context: Fluids such as liquids and gases exhibit several important properties, including pressure, viscosity, surface tension, and buoyant force. Many of these are familiar from everyday experiences, such as floating objects, flowing water, and droplets forming on leaves. This question asks you to think conceptually about which effect would disappear if gravity were completely absent. It tests your understanding of which fluid properties fundamentally depend on the gravitational field and which are independent of it. Given Data / Assumptions: We consider an idealised situation where gravitational acceleration is zero. The fluid still exists with molecules interacting through intermolecular forces. Properties listed include viscosity, surface tension, hydrostatic pressure due to weight, and upward buoyant thrust. We assume no other external fields create equivalent effects. Concept / Approach: Buoyant thrust or upthrust on a body immersed in a fluid is described by Archimedes principle. It states that the upward force equals the weight of the fluid displaced. Weight itself is the product of mass and gravitational acceleration. Without gravity, there would be no weight and hence no buoyant force supporting floating or rising bodies. In contrast, viscosity arises from internal friction between layers of fluid and depends on molecular interactions, not gravity. Surface tension arises due to cohesive forces at the liquid surface and also does not vanish without gravity. Pressure in a fluid has two components: microscopic kinetic pressure due to molecular motion, which would still exist, and hydrostatic pressure variation with depth due to weight, which would vanish, but the general concept of pressure would not disappear completely. Step-by-Step Solution: Step 1: Recall Archimedes principle: buoyant force equals the weight of displaced fluid.Step 2: Weight is given by mass multiplied by gravitational acceleration g.Step 3: If g becomes zero, the weight of any volume of fluid becomes zero, so the buoyant force also becomes zero.Step 4: Recognise that viscosity is caused by molecular friction and surface tension by cohesive forces, both of which do not require gravity.Step 5: Conclude that in the absence of gravity, upward buoyant thrust on immersed bodies would no longer be present, even though other properties remain. Verification / Alternative check: Consider astronauts aboard a spacecraft in orbit, where they experience effective weightlessness. Liquids behave strangely, forming spherical blobs that float rather than staying at the bottom of containers. Objects do not float or sink because there is no meaningful up or down direction defined by gravity, and the usual buoyant force relationships break down. However, the liquid still has viscosity, and its surface still behaves with surface tension, forming droplets. These examples from microgravity environments confirm that buoyant thrust is the effect that depends directly on gravity and disappears when gravity is absent. Why Other Options Are Wrong: Viscosity, option A, is an internal property of the fluid due to intermolecular forces and relative motion of layers; it exists even without gravity. Surface tension, option B, arises from cohesive forces at the surface, and droplets still form in space, so it does not vanish. Hydrostatic pressure due to weight, part of the idea behind option C, would lose its depth variation without gravity, but the general concept of pressure from molecular motion still exists. The question specifically points to an effect on a fluid that relies totally on gravity, which is the upward buoyant thrust on immersed bodies, covered by option D. Common Pitfalls: Many learners confuse pressure in general with pressure due to weight. While hydrostatic pressure gradients vanish without gravity, fluids still exert pressure through molecular collisions. Another pitfall is to think that all fluid properties disappear if gravity is removed, which is not true. Viscosity and surface tension are still very important in microgravity environments. To answer such questions correctly, always ask whether the definition of the property explicitly involves weight or gravitational acceleration. If it does, that property is likely to vanish when gravity is absent. Final Answer: Upward buoyant thrust on immersed bodies

Question 2

In a simple electric circuit, if you want to decrease the overall or effective resistance of the circuit by adding more resistors, in which way should these additional resistances be connected?

Accepted Answer

Introduction / Context: In basic electrical circuits, resistors can be combined in series or in parallel to obtain different effective resistance values. Understanding how these combinations change the total resistance is crucial for designing circuits that work safely and efficiently. This question asks you to identify the arrangement that decreases the overall resistance when additional resistors are added to a circuit. It tests your grasp of the formulas for series and parallel combinations of resistances. Given Data / Assumptions: We have more than one resistor that can be connected in a circuit. The goal is specifically to decrease the effective or total resistance. Possible arrangements are series, parallel, mixed, or unspecified. We assume all resistors have finite, positive resistance values. Concept / Approach: When resistors are connected in series, their resistances simply add, so the total resistance increases as more resistors are added. The formula is R_total = R1 + R2 + R3 and so on. In contrast, when resistors are connected in parallel, the reciprocal of the total resistance equals the sum of the reciprocals of individual resistances: 1 / R_total = 1 / R1 + 1 / R2 + 1 / R3 and so on. For any two or more positive resistances in parallel, the effective resistance is always less than the smallest individual resistance. Therefore, to reduce the overall resistance of a circuit, you should connect additional resistors in parallel across the same two points of the circuit. Step-by-Step Solution: Step 1: Recall the formula for series combination: R_series = R1 + R2 + R3, which always makes the total resistance larger than any single resistance.Step 2: Recall the formula for parallel combination: 1 / R_parallel = 1 / R1 + 1 / R2 + 1 / R3.Step 3: Observe that for positive resistances, adding more terms to the right side increases the sum of reciprocals, making R_parallel smaller.Step 4: Conclude that the effective resistance in parallel is always less than each individual resistor in that branch.Step 5: Therefore, to decrease overall resistance, you must connect additional resistors in parallel, not in series. Verification / Alternative check: Take a simple numerical example. Suppose you have a single 10 ohm resistor. Its resistance is 10 ohm. If you add another 10 ohm resistor in series, the total becomes 20 ohm, which is larger. But if you add the same second resistor in parallel, the effective resistance is given by 1 / R_total = 1 / 10 + 1 / 10 = 2 / 10, so R_total = 5 ohm, which is smaller. This direct numerical check confirms that series connections increase resistance, while parallel connections reduce it. Mixed arrangements can produce intermediate values but are not guaranteed to reduce the resistance below the smallest component. Why Other Options Are Wrong: Connecting resistors in series, option A, always increases the total resistance and therefore does the opposite of what the question requires. A mixed series parallel arrangement, option C, may either increase or decrease resistance depending on the details but does not guarantee a reduction and is not the simplest answer. Claiming that the connection method does not matter, option D, is incorrect because the formulas for series and parallel combinations give clearly different results. Only option B, connecting resistors in parallel across the same two points, reliably decreases the total resistance. Common Pitfalls: Students sometimes memorise the formulas but forget their physical meaning. It is easy to think that adding more resistors must always increase resistance, like adding more obstacles in a path. However, a parallel branch offers extra pathways for current, which effectively reduces the opposition to current flow. To avoid confusion, remember the guiding idea: series narrows the path and increases resistance, while parallel opens new paths and decreases resistance. Combining this intuition with the formulas helps in solving many circuit questions quickly. Final Answer: Connected in parallel across the same two points

Question 3

When a small air bubble rises up through water, it appears to glitter or shine. Which of the following optical phenomena is mainly responsible for this sparkling appearance of the bubble?

Accepted Answer

Introduction / Context: When you observe small air bubbles rising through water in a glass or aquarium, you may notice that they often appear bright and glittering. This sparkling effect is not just a matter of simple reflection but involves deeper optical behaviour at the boundary between air and water. The question tests your understanding of total internal reflection, refraction, and scattering in the context of everyday phenomena, which is a common topic in optics based general knowledge questions. Given Data / Assumptions: There is a small air bubble inside water, rising upward. Light from the surroundings falls on the bubble and its surface. The refractive index of water is higher than that of air. The bubble surface forms a curved air water interface. Concept / Approach: The boundary between water and the air inside the bubble acts like a spherical interface between a denser medium (water) and a rarer medium (air). When light travels from water (higher refractive index) toward air (lower refractive index) and hits the boundary at angles greater than a certain critical angle, it undergoes total internal reflection. Instead of passing into the air inside the bubble, the light is reflected back into the water side, but the curved geometry and multiple reflections make portions of the bubble appear bright to an outside observer. Simple reflection or refraction alone cannot fully account for the intense, sparkling effect seen at certain angles, whereas total internal reflection explains the high brightness and directional nature of the glittering. Step-by-Step Solution: Step 1: Recognise that the air bubble has a curved surface separating water and air.Step 2: Light travelling in water toward the bubble meets the boundary from the denser side going to a rarer medium.Step 3: For incidence angles greater than the critical angle, the light rays cannot escape into the air; instead, they are totally internally reflected.Step 4: These internally reflected rays are redirected within the water and bubble interface, making parts of the bubble appear very bright from specific viewing directions.Step 5: Therefore, the glittering appearance is mainly due to total internal reflection of light inside and at the surface of the bubble. Verification / Alternative check: Similar shimmering effects are seen in many situations where total internal reflection occurs, such as the sparkling of cut gemstones, the bright edges of glass, and the guiding of light in optical fibres. In all these cases, light is trapped and reflected inside a medium with higher refractive index, emerging in intensified directions. The air bubble in water is a small, natural example of this effect. If the phenomenon were due only to simple reflection or refraction, the brightness would be more uniform and less dependent on viewing angle. The concentration of brightness at particular positions supports the idea of total internal reflection. Why Other Options Are Wrong: Simple reflection at the flat water surface, option A, does not explain why the bubble itself appears to glitter as it rises below the surface. Refraction at the curved interface, option B, does bend light rays but typically reduces brightness by transmitting light into the air rather than reflecting it strongly. Scattering by water molecules, option D, creates diffuse effects like the bluish colour of water but does not produce sharp sparkling points on the bubble. Only option C, total internal reflection of light inside the bubble interface, accounts for the intense, directional glittering observed. Common Pitfalls: Students often select refraction because they know that light bends at a curved interface. However, refraction alone does not usually lead to strong sparkling effects. Another pitfall is to associate any bright appearance in a medium with scattering, while ignoring the role of internal reflections. To avoid such errors, remember that total internal reflection occurs only when light travels from a denser to a rarer medium and strikes the boundary at angles greater than the critical angle, producing bright and often shimmering effects. This key concept explains many optical phenomena, including the glittering of rising air bubbles in water. Final Answer: Total internal reflection of light inside the bubble

Question 4

In an optical fibre communication system, the light signal is able to travel along the curved fibre over long distances mainly due to which physical phenomenon?

Accepted Answer

Introduction / Context: Optical fibres are widely used to transmit data as light signals over long distances in telecommunication and internet networks. They are thin strands of glass or plastic that can guide light even when the fibre is bent. The question asks which physical principle allows light to remain confined inside the fibre core instead of leaking out. Understanding this principle is central in optics and modern communication technology. Given Data / Assumptions: Light is transmitted through an optical fibre, which may be curved. The fibre has a core surrounded by cladding with lower refractive index. The signal must stay inside the fibre without significant loss. Options mention straight line motion, refraction, total internal reflection, and absorption related processes. Concept / Approach: In an optical fibre, the core has a slightly higher refractive index than the cladding. When light enters the core at a suitable angle, it strikes the core cladding boundary from the denser side. If the angle of incidence is greater than the critical angle for that boundary, the light is totally internally reflected. Because this happens repeatedly along the length of the fibre, the light is effectively trapped and guided along the fibre, even if the fibre bends gently. This repeated total internal reflection is the key guiding mechanism. Simple straight line travel cannot explain guidance along curves, and ordinary refraction alone would cause the light to escape into the cladding instead of staying confined. Step-by-Step Solution: Step 1: Recognise that the core of the optical fibre has a higher refractive index than the surrounding cladding.Step 2: Light entering the core at a suitable angle strikes the core cladding interface from the denser medium toward the rarer medium.Step 3: If the incidence angle exceeds the critical angle, total internal reflection occurs, and the light is reflected back into the core.Step 4: As the fibre extends and even bends, the light continues to strike the boundary at angles above the critical angle, undergoing repeated total internal reflections.Step 5: This process allows the light signal to stay within the core and travel long distances with low loss, so repeated total internal reflection is the correct phenomenon. Verification / Alternative check: Applications like endoscopes in medicine and undersea fibre optic cables in communication all rely on optical fibres. Textbooks and technical references clearly describe total internal reflection as the operating principle of these fibres. If light simply travelled in a straight line, it would escape whenever the fibre curves. If refraction were dominant, light would gradually leak out through the cladding. Experimental demonstrations with glass rods and water filled tanks also show that light can be guided around bends when conditions for total internal reflection are met. These observations confirm the explanation based on total internal reflection. Why Other Options Are Wrong: Option A, straight line motion, is only valid in a homogeneous medium with no boundaries and cannot account for guiding around bends. Option B, guidance by refraction, would cause light to bend but ultimately escape rather than being perfectly confined. Option D, absorption and re emission due to heating, would greatly reduce signal strength and speed and is not how optical fibres operate. Only option C, repeated total internal reflection of light within the core, accurately describes the mechanism that keeps light trapped inside an optical fibre. Common Pitfalls: Some learners think of refraction whenever light crosses boundaries with different refractive indices and may incorrectly select it in this context. Others assume that light must always travel in straight lines and forget that internal reflections can redirect it. To avoid confusion, remember the specific conditions for total internal reflection: the ray must travel from denser to rarer medium and the angle of incidence must exceed the critical angle. Optical fibres are carefully designed to satisfy these conditions, making total internal reflection the central phenomenon in their operation. Final Answer: Repeated total internal reflection of light within the core

Question 5

Using Ohm's law, what is the electrical resistance (in ohms) of a component if a current of 0.1 A flows through it when a potential difference of 5 V is applied across its ends?

Accepted Answer

Introduction / Context: Ohm's law is a fundamental relationship in electric circuits that connects voltage, current, and resistance. It is commonly written as V = I * R, where V is potential difference, I is current, and R is resistance. This question gives you the voltage and current for an electrical component and asks you to calculate its resistance. Such numerical applications of Ohm's law are common in school level physics and competitive exams. Given Data / Assumptions: Potential difference V across the component is 5 V. Current I flowing through it is 0.1 A. The relationship between V, I, and R follows Ohm's law. We assume the component behaves like an ohmic resistor with constant resistance. Concept / Approach: Ohm's law states that the potential difference across a conductor is directly proportional to the current flowing through it, provided temperature and other physical conditions remain constant. Mathematically, V = I * R. If you know any two of the three quantities, you can rearrange the formula to find the third. In this problem, we need to find R, so we rearrange to R = V / I. Substituting the given values of V and I will give the resistance in ohms. This is a straightforward single step substitution problem, typical of easy numerical questions in electricity. Step-by-Step Solution: Step 1: Write down Ohm's law in the form V = I * R.Step 2: Rearrange the formula to solve for resistance: R = V / I.Step 3: Substitute the given values, V = 5 V and I = 0.1 A.Step 4: Compute R = 5 / 0.1 = 5 / (1 / 10) = 5 * 10 = 50 ohm.Step 5: Conclude that the resistance of the component is 50 ohm. Verification / Alternative check: You can verify the result by reversing the calculation. If the resistance is 50 ohm and the current is 0.1 A, then according to Ohm's law, the voltage should be V = I * R = 0.1 * 50 = 5 V, which matches the given potential difference. This consistency check confirms that the calculation is correct. Additionally, thinking about the numbers, a small current of 0.1 A flowing under a modest voltage of 5 V suggests a resistance significantly larger than 1 ohm, making 50 ohm a reasonable value. Why Other Options Are Wrong: A resistance of 0.5 ohm, option A, would give V = I * R = 0.1 * 0.5 = 0.05 V, which is much smaller than 5 V and therefore incorrect. A resistance of 100 ohm, option B, would give V = 0.1 * 100 = 10 V, which does not match the given 5 V. A resistance of 1 ohm, option D, would produce V = 0.1 * 1 = 0.1 V, again inconsistent with the data. Only 50 ohm leads back to the given voltage when combined with the stated current. Common Pitfalls: One common mistake is to divide I by V instead of V by I, leading to an incorrect value of resistance. Another error is mishandling decimal numbers such as 0.1, where students might incorrectly compute 5 / 0.1 as 0.5 instead of 50. To avoid such errors, remember that dividing by 0.1 is the same as multiplying by 10. Writing out the rearranged formula carefully before substituting the numbers helps ensure you are performing the correct operation in Ohm's law problems. Final Answer: 50 ohm

Question 6

In kinematics, the motion of a freely falling body near the earth (neglecting air resistance) is an example of which type of accelerated motion?

Accepted Answer

Introduction / Context: When an object is released from rest and allowed to fall freely under gravity near the earth's surface, its speed increases as it falls. This is a classic example in kinematics and dynamics used to illustrate the concept of acceleration. The question asks you to classify this motion as uniformly or non uniformly accelerated. Understanding this example helps in solving many numerical problems related to free fall, projectile motion, and gravitational effects. Given Data / Assumptions: The body is freely falling near the earth's surface. Air resistance is neglected to simplify the analysis. The acceleration due to gravity is treated as constant and equal to g. We focus on motion in one dimension, vertically downward. Concept / Approach: Acceleration is defined as the rate of change of velocity with respect to time. If this rate is constant, the motion is called uniformly accelerated. Near the earth's surface, the acceleration due to gravity g has a nearly constant value, approximately 9.8 metres per second squared, directed downward. When air resistance is negligible, this constant acceleration acts on the falling body throughout its motion, so its velocity increases linearly with time. Therefore, the motion of a freely falling body is uniformly accelerated motion, not non uniform or special variable acceleration. Step-by-Step Solution: Step 1: Recall that for free fall near the earth, the only significant force on the body is its weight, equal to mass multiplied by gravitational acceleration g.Step 2: This constant force produces a constant acceleration g, independent of the body's mass.Step 3: In uniformly accelerated motion, acceleration remains the same at all instants of time.Step 4: Since the acceleration due to gravity is taken as constant in this idealised situation, the falling body experiences uniform acceleration.Step 5: Conclude that the motion is uniformly accelerated motion, matching standard kinematics equations such as v = u + g * t and s = u * t + (1 / 2) * g * t^2. Verification / Alternative check: You can use experimental or textbook graphs to verify this. A velocity time graph for a freely falling body, neglecting air resistance, is a straight line with slope equal to g, indicating constant acceleration. The displacement time relation forms a parabola, consistent with uniform acceleration. Additionally, many standard kinematics formulas for free fall assume constant acceleration and are derived under that assumption. If acceleration were non uniform, those equations would not hold, and the graphs would look different, which they do not in the ideal case. Why Other Options Are Wrong: Non uniformly accelerated motion, option A, refers to situations where acceleration varies with time or position, such as motion with air resistance or changing forces, which is not assumed here. Option C mentions unique accelerated motion with variable direction, which might apply to circular or more complex motion but not to straight downward free fall. Option D, special motion without acceleration, is incorrect because the body clearly speeds up, indicating non zero acceleration. Only option B correctly identifies free fall (without air resistance) as uniformly accelerated motion. Common Pitfalls: Students sometimes think that because the speed is changing, the acceleration must also be changing, which is not true. A constant acceleration can still produce changing velocity; in fact, that is its defining feature. Another pitfall is to confuse real life falling, where air resistance plays a role and acceleration decreases over time, with the idealised free fall model used in basic physics. Always pay attention to phrases like neglecting air resistance or in vacuum, which indicate that you should use the constant acceleration model for gravity. Final Answer: Uniformly accelerated motion

Question 7

A sensitive galvanometer can be converted into a voltmeter suitable for measuring potential difference by connecting which type of resistance with it?

Accepted Answer

Introduction / Context: A galvanometer is a delicate instrument that detects and measures small currents. With suitable modifications, it can be used either as an ammeter or as a voltmeter. The question asks how to convert a galvanometer into a voltmeter, which measures potential difference across a circuit element. Knowing which resistance arrangement achieves this conversion is an important application of basic circuit principles and is frequently asked in electricity and electronics questions. Given Data / Assumptions: We start with a sensitive galvanometer that can measure small currents. The goal is to modify it so that it measures voltage (potential difference). We may connect either high or low resistance, in series or parallel, with the galvanometer. We assume the galvanometer has a known internal resistance and full scale deflection current. Concept / Approach: A voltmeter is designed to measure potential difference across a component without drawing significant current from the circuit. To achieve this, it must have very high resistance so that only a small current flows through it even at the maximum voltage it is supposed to measure. To convert a galvanometer into a voltmeter, a suitable high resistance, called a multiplier, is connected in series with the galvanometer. The series combination then has a large total resistance. When a voltage is applied across this combination, the current is limited, and the galvanometer deflects proportionally to the applied voltage. Connecting high resistance in parallel would lower the total resistance, which is undesirable for a voltmeter. Step-by-Step Solution: Step 1: Recall that a good voltmeter should have very high resistance to minimise current drawn from the circuit.Step 2: The galvanometer by itself has relatively low resistance and is designed for small currents.Step 3: To increase the overall resistance, we connect a high resistance in series with the galvanometer, creating a large combined resistance.Step 4: When a voltage is applied across this series combination, the current is limited and proportional to the voltage due to Ohm's law, allowing calibration in volts.Step 5: Therefore, the correct modification is to connect a high resistance in series with the galvanometer. Verification / Alternative check: Compare this with the method of converting a galvanometer into an ammeter. For an ammeter, we want very low resistance so that it does not change the current in the circuit. A low resistance shunt is connected in parallel with the galvanometer in that case. This contrast helps you remember that voltmeters require high resistance in series, while ammeters require low resistance in parallel. Textbooks on basic electronics and circuit theory consistently explain voltmeter conversion using a high series resistance, confirming the correctness of this arrangement. Why Other Options Are Wrong: High resistance in parallel, option A, would create a low equivalent resistance path and is used for ammeter shunts, not for voltmeters. Low resistance in parallel, option B, again lowers the overall resistance and is associated with ammeter conversion, not voltmeter conversion. Low resistance in series, option D, would not sufficiently increase the total resistance and would allow too much current to flow, disturbing the circuit and possibly damaging the galvanometer. Only option C, high resistance connected in series with the galvanometer, satisfies the requirement for a voltmeter. Common Pitfalls: Students often mix up the modifications needed for ammeters and voltmeters, especially under exam pressure. A useful memory aid is to associate voltmeters with "very high resistance" and the word series, while associating ammeters with "very low resistance shunt" and the word parallel. Also, always think about the effect on the circuit: voltmeters must not significantly change the circuit conditions, so they must draw minimal current, which requires high resistance. Keeping this physical reasoning in mind helps you choose the correct configuration even if you momentarily forget the standard rule. Final Answer: High resistance connected in series with the galvanometer

Question 8

During which operating condition of an automobile engine is the carbon monoxide (CO) content in the exhaust gas generally at its maximum level?

Accepted Answer

Introduction / Context: Carbon monoxide is a toxic gas produced by incomplete combustion of fuel in internal combustion engines. The amount of CO in exhaust gases depends on how efficiently the fuel air mixture burns under different operating conditions. The question asks you to identify when, in typical automobile operation, the carbon monoxide content of the exhaust is generally highest. This is important in environmental science, automotive engineering, and pollution control studies. Given Data / Assumptions: The engine is a conventional petrol automobile engine. Operating conditions mentioned are acceleration, cruising, idle running, and deceleration. Carbon monoxide is formed when there is incomplete combustion, especially with rich mixtures. We consider typical tuning and behaviour of standard engines without advanced emission controls. Concept / Approach: Incomplete combustion, which produces more carbon monoxide, occurs when the fuel air mixture is richer than ideal, meaning there is more fuel relative to air than needed for complete combustion. At idle, the engine is running with the throttle nearly closed and airflow is low. Many engines tend to run on a slightly rich mixture at idle to maintain smooth operation and prevent stalling. The low airflow and rich mixture conditions promote incomplete combustion and higher CO formation. During steady cruising, mixtures are closer to stoichiometric or slightly lean for fuel economy, leading to lower CO. Acceleration can momentarily enrich the mixture, but the engine also receives more air and there is strong mixing. Deceleration may even cut off fuel in modern engines, reducing CO. Step-by-Step Solution: Step 1: Recall that high CO emissions are associated with rich fuel mixtures and incomplete combustion.Step 2: At idle running, the engine operates at low speed with the vehicle stationary, and the airflow through the intake is minimal.Step 3: To maintain stable combustion at idle, the carburettor or injection system often supplies a slightly rich mixture.Step 4: This combination of low air supply and relatively high fuel per unit air results in incomplete combustion and high CO formation.Step 5: Therefore, the exhaust gas typically has maximum carbon monoxide content during idle running. Verification / Alternative check: Emission testing data for older petrol engines and technical literature often show idle CO measurements as a key indicator of mixture richness and engine tuning. Regulations may set limits for CO levels at idle because they can be significantly higher than during normal cruising. Mechanics adjust the idle mixture screw to bring CO within acceptable limits. While acceleration can produce short bursts of higher CO, the sustained condition with consistently high CO is idle running. This practical evidence supports the selection of idle running as the condition with maximum CO content in the exhaust. Why Other Options Are Wrong: During rapid acceleration, option A, the mixture may briefly become richer, but the increased airflow and changing conditions make this a transient state, not typically where sustained maximum CO is measured. Steady cruising at constant speed, option B, is usually tuned for efficiency with near stoichiometric or slightly lean mixtures, resulting in lower CO. During deceleration, option D, the throttle is largely closed and fuel supply may be reduced, sometimes even cut off, lowering CO emissions. Only option C, idle running of the engine at standstill, matches the condition where engines often produce their highest sustained carbon monoxide levels. Common Pitfalls: Students may intuitively pick acceleration because it feels like the engine is working hardest and burning more fuel. However, emissions are not just about total fuel burned but about how completely it burns. Another pitfall is to assume that cruising produces maximum emissions because the vehicle is moving, but engines are usually most efficient in that state. To avoid such mistakes, focus on mixture richness and completeness of combustion rather than speed alone. Remember that idle operation with low airflow and slightly rich mixtures is typically associated with higher CO in exhaust gases. Final Answer: Idle running of the engine at standstill

Question 9

Even after sunset, the air near the earth's surface continues to receive heat. This continued heating is mainly due to which of the following?

Accepted Answer

Introduction / Context: During the day, the earth's surface is heated by solar radiation. After the sun sets, however, the air near the ground does not become cold immediately. Instead, it continues to be warmed for some time. The question asks you to identify the main mechanism that supplies this heat to the lower atmosphere after sunset. This concept is central in physical geography and climatology and helps explain daily temperature patterns. Given Data / Assumptions: Solar radiation stops reaching the surface directly after sunset. The earth's surface has stored heat obtained during the day. The air near the surface is still in contact with the warmed ground. Several processes are possible: direct insolation, terrestrial radiation, conduction, and convection. Concept / Approach: Insolation refers to incoming solar radiation, which effectively stops at the surface when the sun is below the horizon. The earth's surface, having absorbed solar energy during the day, becomes a secondary source of heat. It emits long wave infrared radiation back into the atmosphere; this is called terrestrial radiation. The lower layers of the atmosphere absorb this long wave radiation and are warmed. Conduction and convection also play roles in redistributing heat near the surface, but the primary ongoing source of energy after sunset is the terrestrial radiation emitted by the warmed land and water surfaces rather than new energy arriving from the sun. Step-by-Step Solution: Step 1: Recognise that after sunset, there is no direct solar insolation reaching the earth's surface at that location.Step 2: Recall that the earth's surface absorbs solar energy during the day and becomes warmer.Step 3: After sunset, this warmed surface begins to emit long wave infrared radiation back toward the atmosphere, known as terrestrial radiation.Step 4: The air layers just above the surface absorb some of this terrestrial radiation and are heated from below.Step 5: Conclude that this process of terrestrial radiation is the main reason the air continues to receive heat after sunset. Verification / Alternative check: Daily temperature curves show that the maximum temperature often occurs in the afternoon, not at noon, and temperatures do not drop sharply at sunset. Instead, there is a gradual cooling as the surface loses heat by radiation. Meteorology and climatology texts emphasise that the earth acts as a heat source for the lower atmosphere by emitting terrestrial radiation, especially during the evening and night. While conduction and convection distribute this heat within the air, their energy ultimately comes from the surface radiation, confirming that terrestrial radiation is the fundamental mechanism indicated in the question. Why Other Options Are Wrong: Direct insolation from the sun, option A, is absent after sunset simply because the sun is no longer above the horizon. Conduction from the earth's core, option C, operates on geological timescales and contributes negligibly to daily atmospheric temperature changes. Convection currents from upper layers, option D, redistribute heat but are not the basic source of new energy after sunset; they depend on the heat supplied by the surface and its radiation. Only option B, terrestrial radiation emitted by the warmed earth, correctly identifies the main process that continues to heat the air near the surface after sunset. Common Pitfalls: Students sometimes overemphasise conduction and convection because these words are commonly associated with heat transfer in school experiments. However, in the atmosphere, radiation plays a dominant role, particularly in interactions between the surface and the air. Another pitfall is assuming that once the sun sets, no heating process occurs, which contradicts observed evening temperatures. To understand climate and weather, remember this sequence: the sun heats the surface by insolation, the surface heats the air mainly by terrestrial radiation, and conduction and convection then redistribute that heat within the atmosphere. Final Answer: Terrestrial radiation emitted by the warmed earth

Question 10

In motion graphs, the slope of a velocity time graph for an object moving in a straight line represents which physical quantity?

Accepted Answer

Introduction / Context: Graphs are a powerful tool in kinematics for visualising how physical quantities change with time. A velocity time graph shows how the velocity of an object changes as time passes. The question asks what the slope of such a graph represents. Recognising the meaning of slopes and areas in motion graphs is crucial for interpreting experimental data and solving many physics problems involving speed, velocity, acceleration, and displacement. Given Data / Assumptions: The graph under discussion has velocity on the vertical axis and time on the horizontal axis. The motion considered is along a straight line. The slope refers to the change in velocity divided by the change in time. Possible interpretations include acceleration, distance, speed, and momentum. Concept / Approach: The slope of any graph is defined as the change in the quantity on the vertical axis divided by the change in the quantity on the horizontal axis. For a velocity time graph, the vertical axis represents velocity and the horizontal axis represents time. Therefore, the slope equals change in velocity divided by change in time, which is the definition of acceleration. Distance or displacement is represented by the area under the velocity time graph, not by the slope. Speed and momentum are related to velocity and mass but are not given by the slope in this type of graph. Step-by-Step Solution: Step 1: Write the general formula for slope in a graph: slope = change in vertical quantity / change in horizontal quantity.Step 2: In a velocity time graph, the vertical axis quantity is velocity v and the horizontal axis quantity is time t.Step 3: The slope becomes change in velocity divided by change in time, or delta v / delta t.Step 4: This ratio delta v / delta t is precisely the definition of acceleration.Step 5: Conclude that the slope of a velocity time graph represents the acceleration of the object. Verification / Alternative check: Consider some special cases. If the velocity time graph is a horizontal line, its slope is zero, meaning velocity does not change, so acceleration is zero. This matches the idea of uniform motion. If the graph is a straight line with positive slope, velocity increases linearly with time, indicating constant positive acceleration. These interpretations align with the concept of acceleration and confirm that the slope corresponds to acceleration. In contrast, distance travelled can be obtained from the area under the graph, not the slope, further validating the distinction. Why Other Options Are Wrong: Distance, option B, is associated with the area under the velocity time curve and not with the slope. Instantaneous speed, option C, is simply the value of velocity at a specific point on the graph, not the slope between two points. Linear momentum, option D, is mass multiplied by velocity; it depends on the value of velocity and mass, not on the rate of change of velocity, and therefore is not represented by the slope. Only option A correctly identifies acceleration as the physical quantity represented by the slope of a velocity time graph. Common Pitfalls: Students sometimes confuse what the slope and area represent in different kinds of graphs. For example, in a displacement time graph, the slope represents velocity, while in a velocity time graph, the slope represents acceleration. Mixing these interpretations leads to incorrect answers. To avoid confusion, always check which quantity is on which axis and apply the general rule: slope is change in vertical quantity over change in horizontal quantity, and area under the curve often represents a cumulative effect, such as distance covered. Final Answer: Acceleration of the object

Question 11

In mechanics, the physical quantity known as jerk is defined as the rate of change of acceleration with respect to time. What is the SI unit of jerk?

Accepted Answer

Introduction / Context: In kinematics, we commonly study displacement, velocity, and acceleration. Jerk is a higher order quantity that describes how acceleration itself changes with time. It is particularly important in engineering applications where sudden changes in acceleration can cause discomfort or structural stress, such as in elevators, vehicles, and roller coasters. The question asks for the correct SI unit of jerk, which you can determine by analysing its definition in terms of more familiar quantities. Given Data / Assumptions: Jerk is defined as the rate of change of acceleration with respect to time. Acceleration has SI unit metre per second square (m/s^2). Time has SI unit second (s). We apply basic dimensional analysis to find the unit of jerk. Concept / Approach: If a quantity is the rate of change of another quantity with respect to time, its unit is the unit of the original quantity divided by the unit of time. For example, velocity is the rate of change of displacement, so its unit is metres per second. Acceleration is the rate of change of velocity, so its unit is metres per second square. Jerk, being the rate of change of acceleration, has unit acceleration per unit time. Since acceleration has unit m/s^2, dividing by time (s) gives m/s^2 per s, which simplifies to m/s^3. Thus, the SI unit of jerk is metre per second cube (m/s^3). Step-by-Step Solution: Step 1: Write the definition of jerk: jerk = change in acceleration / change in time.Step 2: Note that the SI unit of acceleration is metre per second square, written as m/s^2.Step 3: Time has SI unit second, written as s.Step 4: The unit of jerk becomes (m/s^2) / s = m / (s^2 * s) = m/s^3.Step 5: Therefore, the SI unit of jerk is metre per second cube (m/s^3). Verification / Alternative check: Dimensional analysis provides a quick cross check. Displacement has dimension L. Velocity has dimension L T^-1. Acceleration has dimension L T^-2. Jerk, as the time derivative of acceleration, has dimension L T^-3. The SI combination that matches L T^-3 is metre per second cube, confirming that m/s^3 is the correct unit. No other combination among the options matches this dimension. This systematic approach avoids guessing based on appearance alone and ensures consistency with fundamental definitions. Why Other Options Are Wrong: Metre second, option A, would have dimension L T and is not associated with any standard kinematic quantity. Metre per second square, option C, is the unit of acceleration, not jerk. Metre per second, option D, is the unit of velocity. None of these represent the rate of change of acceleration. Only option B, metre per second cube (m/s^3), directly follows from the definition of jerk as acceleration per unit time. Common Pitfalls: Students often confuse jerk with acceleration and may mistakenly select metre per second square. Another common error is overlooking the extra division by time when moving from acceleration to jerk. To avoid such mistakes, remember the sequence: displacement (m), velocity (m/s), acceleration (m/s^2), jerk (m/s^3). Each step introduces an additional division by time. Writing out the units carefully during problem solving rather than relying on memory alone can greatly reduce errors in questions involving derived quantities like jerk. Final Answer: Metre per second cube (m/s^3)

Question 12

A man standing very close to a platform at a railway station feels a pulling force towards a fast moving train as it passes by. This effect is mainly due to which of the following?

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Introduction / Context: When a fast train passes close to a person standing on a platform, the person sometimes feels as if they are being pulled towards the train. This can be dangerous and is often mentioned in safety warnings. The question asks you to identify the main physical reason for this effect. Understanding it involves applying basic fluid dynamics and Bernoulli's principle to the air flow around the moving train and the stationary environment. Given Data / Assumptions: The train is moving at high speed past a stationary man on the platform. Air flows rapidly in the narrow space between the train and the man. The man is close enough to feel a force towards the train. We neglect any direct contact or mechanical pulling; only air flow effects are considered. Concept / Approach: As the train moves quickly, it drags air along with it, creating a region of fast moving air in the narrow gap between the train and the person. According to Bernoulli's principle, where a fluid (including air) moves faster, its static pressure is lower compared to slower moving regions. The air on the far side of the person, away from the train, is more or less still and thus at higher pressure. This pressure difference creates a net force pushing the man towards the region of lower pressure, that is, towards the train. Gravity between the train and man is negligible, and centripetal force is not relevant because the man is not moving in a circular path. Step-by-Step Solution: Step 1: Recognise that a fast moving train creates strong air currents along its sides due to friction and drag.Step 2: In the narrow gap between the train and the man, the air flows rapidly in the direction of the train's motion.Step 3: Apply Bernoulli's principle, which states that in a streamline flow, increased speed of the fluid is accompanied by decreased pressure.Step 4: The air pressure in the fast moving region between the train and man is therefore lower than the relatively still air on the other side of the man.Step 5: This pressure difference results in a net force pushing the man towards the train, making him feel a pulling effect. Verification / Alternative check: Similar effects can be observed in other situations: two ships moving close to each other may be drawn together, and two nearby trains can experience lateral forces due to pressure differences. Aerodynamics studies show that fast moving bodies through air disturb the pressure distribution around them. Safety guidelines at stations often advise standing well behind the yellow line to reduce the effect of these air currents. These practical observations agree with the explanation based on pressure differences caused by fast moving air rather than gravitational or imaginary forces. Why Other Options Are Wrong: Gravitational attraction between the train and the man, option A, is extremely small compared to the gravitational pull of the earth and cannot account for the noticeable sideways pulling force. The idea that the effect is only an illusion, option B, ignores the real aerodynamic forces that can even cause accidents. Centripetal force, option C, is associated with circular motion and does not apply to a stationary man and a linearly moving train. Only option D, pressure difference due to fast moving air between the train and the man, correctly explains the phenomenon using Bernoulli's principle. Common Pitfalls: Students may attribute any unusual force to gravity out of habit, forgetting that gravity acts vertically rather than sideways in this situation. Others may think of centripetal force whenever a force towards something is mentioned, even when circular motion is absent. To avoid such mistakes, carefully consider the geometry and the actual motion involved. Remember that moving vehicles can create strong air currents and pressure differences, and Bernoulli's principle provides the correct framework to understand these effects near fast moving trains and vehicles. Final Answer: Pressure difference caused by fast moving air between train and man

Question 13

Two resistors of 20 ohm and 30 ohm are connected in parallel across the same potential difference. What is the effective resistance of the combination?

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Introduction / Context: Combining resistors in series and parallel is a key skill in circuit analysis. Many real circuits use combinations rather than single resistors, so being able to calculate the effective or equivalent resistance helps determine current and voltage distributions. This question asks for the effective resistance of two resistors, 20 ohm and 30 ohm, connected in parallel. It tests your knowledge of the parallel resistance formula and your ability to apply it correctly to a numerical example. Given Data / Assumptions: Two resistors have resistances R1 = 20 ohm and R2 = 30 ohm. They are connected in parallel across the same potential difference. We need to find the effective resistance R_eq of the parallel combination. We assume ideal resistors with no additional circuit elements. Concept / Approach: For resistors connected in parallel, the reciprocal of the equivalent resistance equals the sum of the reciprocals of the individual resistances. The formula is 1 / R_eq = 1 / R1 + 1 / R2. After calculating this sum, we invert it to obtain R_eq. An important property is that the equivalent resistance of resistors in parallel is always less than the smallest individual resistance in the group. Using these principles, we can compute the effective resistance for 20 ohm and 30 ohm connected in parallel. Step-by-Step Solution: Step 1: Write the parallel resistance formula for two resistors: 1 / R_eq = 1 / R1 + 1 / R2.Step 2: Substitute R1 = 20 ohm and R2 = 30 ohm into the formula.Step 3: Compute 1 / R_eq = 1 / 20 + 1 / 30.Step 4: Find a common denominator, which is 60: 1 / 20 = 3 / 60 and 1 / 30 = 2 / 60, so 1 / R_eq = 3 / 60 + 2 / 60 = 5 / 60.Step 5: Invert the fraction to get R_eq = 60 / 5 = 12 ohm. Verification / Alternative check: You can verify that the answer makes sense by checking the property that the equivalent resistance of a parallel combination is less than the smallest individual resistor. Here, the smallest resistor is 20 ohm, and our result is 12 ohm, which is indeed smaller. Another way to verify is to consider an applied voltage, for example 60 V. The current through the 20 ohm resistor would be 3 A, and through the 30 ohm resistor 2 A, giving a total current of 5 A. The equivalent resistance would then be R_eq = V / I_total = 60 / 5 = 12 ohm, confirming the calculation. Why Other Options Are Wrong: Option A, 50 ohm, is simply the sum of the resistances and would be the correct answer for series connection, not parallel. Option C, 24 ohm, might result from incorrectly averaging the two resistances, which is not the correct method. Option D, 36 ohm, does not follow from any standard formula and is larger than the smaller resistor, which cannot be correct for a parallel combination. Only option B, 12 ohm, matches the result obtained from the proper parallel resistance formula. Common Pitfalls: A frequent error is to add the resistances directly for parallel combinations, confusing them with series combinations. Another mistake is to compute a simple arithmetic mean instead of using the reciprocal formula. To avoid these errors, remember the key rule: in series, resistances add directly; in parallel, their reciprocals add. Also, use the useful check that the equivalent resistance in parallel must be less than the smallest individual resistance. This provides a quick sanity check on your numerical answer. Final Answer: 12 ohm

Question 14

If the price of an article decreases from an initial price P1 to Rs 25, and as a result the quantity demanded increases from 900 units to 1200 units, and the (absolute) point elasticity of demand is 2, what is the initial price P1?

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Introduction / Context: Elasticity of demand is an important concept in economics that measures how responsive the quantity demanded of a good is to changes in its price. Point elasticity uses a specific formula based on small changes in price and quantity around a particular point. This question gives a change in price and quantity demanded, along with the value of point elasticity, and asks you to find the original price P1. It tests your ability to apply the point elasticity formula correctly and interpret percentage changes in quantity and price. Given Data / Assumptions: Initial price is P1 (unknown), final price is 25 rupees. Initial quantity demanded is 900 units, final quantity demanded is 1200 units. Point elasticity of demand (in absolute value) is given as 2. We use the usual point elasticity formula based on the initial price and quantity. Concept / Approach: For small changes, the point price elasticity of demand (ignoring the minus sign for inverse relation) can be written as E = (ΔQ / ΔP) * (P1 / Q1), where ΔQ is the change in quantity demanded, ΔP is the change in price, P1 is the initial price, and Q1 is the initial quantity demanded. Here, ΔQ = 1200 - 900 = 300 units. The price decreases from P1 to 25, so ΔP = 25 - P1, which is negative, but we use the absolute value when equating to the given elasticity of 2. Solving this equation for P1 will give the required initial price. Step-by-Step Solution: Step 1: Compute the change in quantity: ΔQ = 1200 - 900 = 300 units.Step 2: Write the point elasticity formula in absolute value form: E = (ΔQ / ΔP) * (P1 / Q1).Step 3: Substitute E = 2, Q1 = 900, and ΔQ = 300. We use ΔP in magnitude as P1 - 25.Step 4: The equation becomes 2 = (300 / (P1 - 25)) * (P1 / 900).Step 5: Simplify: 2 = (300 * P1) / (900 * (P1 - 25)) = (P1) / (3 * (P1 - 25)).Step 6: Cross multiply: 2 * 3 * (P1 - 25) = P1, so 6 * (P1 - 25) = P1.Step 7: Expand: 6P1 - 150 = P1, so 5P1 = 150, giving P1 = 30 rupees. Verification / Alternative check: Check that this P1 value is consistent with the data. The price falls from 30 to 25, a change of 5 rupees. The quantity rises from 900 to 1200, a change of 300 units. Using the point elasticity formula again: E = (ΔQ / ΔP) * (P1 / Q1) in magnitude = (300 / 5) * (30 / 900) = 60 * (30 / 900) = 60 * (1 / 30) = 2. This matches the given elasticity of 2, confirming that P1 = 30 rupees is correct. Why Other Options Are Wrong: If P1 were 20 rupees, option A, the price would have increased to 25, which contradicts the statement that price decreases. P1 = 25, option C, would mean no change in price, making elasticity undefined in this context. P1 = 35, option D, would produce a different elasticity value; substituting into the formula would not give 2. Only P1 = 30 rupees yields the correct elasticity when combined with the given quantity changes. Common Pitfalls: Students sometimes confuse arc elasticity formulas with point elasticity formulas or forget to use the initial price and quantity in the fraction P1 / Q1. Another common error is mishandling the sign of ΔP and ΔQ, but since the question gives elasticity as a positive number, you can safely work with absolute values. To avoid mistakes, write down the formula carefully, substitute known values step by step, and simplify systematically, as shown above. Final Answer: Rs 30

Question 15

The absolute refractive index of any transparent medium (defined relative to vacuum) is always:

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Introduction / Context: The refractive index of a medium describes how much the speed of light is reduced when it passes from vacuum into that medium. The absolute refractive index is defined as the ratio of the speed of light in vacuum to the speed of light in the medium. This question tests your understanding of this definition and asks you to identify a general property of the absolute refractive index for any real transparent medium compared with vacuum. Given Data / Assumptions: Absolute refractive index n is defined as n = c / v. c is the speed of light in vacuum. v is the speed of light in the medium. We consider ordinary transparent media such as air, water, and glass. Concept / Approach: Light always travels fastest in vacuum. When it enters any material medium, interactions with atoms and molecules cause it to travel more slowly on average. Therefore, v, the speed of light in the medium, is always less than c, the speed of light in vacuum, for any transparent material. Because the refractive index is defined as n = c / v, and c > v, the ratio c / v must be greater than 1. If n were exactly 1, it would mean that v equals c and the medium behaves like vacuum. A value less than 1 would require light to travel faster in the medium than in vacuum, which contradicts established physics. Step-by-Step Solution: Step 1: Write the definition of absolute refractive index: n = c / v.Step 2: Recall that c, the speed of light in vacuum, is the maximum possible speed of light.Step 3: In any real transparent medium, the speed of light v is reduced due to interactions with the medium, so v < c.Step 4: Since c / v is a ratio with numerator larger than the denominator, n = c / v must be greater than 1.Step 5: Conclude that the absolute refractive index of any transparent medium is always more than one. Verification / Alternative check: Check typical refractive index values from tables: air has n approximately 1.0003, water about 1.33, and common glass about 1.5. All these values are greater than 1, consistent with the definition. A medium with n exactly 1 would behave like vacuum for light, and none with n less than 1 exist in normal circumstances, as that would imply superluminal speeds. Advanced physics may describe some unusual effective indices in special materials, but in standard optics for real transparent media, n is always greater than 1. This confirms the general statement required by the question. Why Other Options Are Wrong: Zero, option B, would imply c / v = 0, which is impossible because c and v are both positive speeds. Exactly one, option C, would mean the medium does not slow light at all and is equivalent to vacuum, which is not true for ordinary transparent materials. Less than one, option D, would suggest light travels faster in the medium than in vacuum, contradicting basic principles of relativity and optics. Only option A, more than one, is consistent with the definition and observed values of absolute refractive index. Common Pitfalls: Some students misinterpret absolute refractive index as a relative measure between two media and forget that the definition here is specifically with respect to vacuum. Others may have seen approximate values close to one, such as for air, and mistakenly think that values less than or equal to one are common. To avoid confusion, remember that in this context absolute means relative to vacuum, and vacuum provides the reference speed c. Because no medium allows light to exceed this speed, the ratio c / v is always greater than one. Final Answer: More than one

Question 16

An electric iron rated at 750 W is used for 8 hours per day. How much electrical energy (in units, where 1 unit = 1 kWh) is consumed by the iron in one day?

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Introduction / Context: Electricity bills for households and industries are usually based on energy consumption measured in kilowatt hour units, often simply called units. Electrical appliances such as irons, heaters, and fans have a rated power in watts and are used for certain durations each day. This question asks you to calculate the daily energy consumption of an electric iron with a known power rating and usage time. It tests your understanding of the relationship between power, time, and electrical energy in practical units. Given Data / Assumptions: Power rating of the iron P = 750 W. Usage time per day t = 8 hours. 1 kilowatt hour (kWh) is defined as the energy used by a 1 kW appliance running for 1 hour. 1 unit of electrical energy on the bill is equal to 1 kWh. Concept / Approach: Electrical energy consumed is equal to power multiplied by time. To express energy in kWh, we convert power into kilowatts and time into hours, then multiply. The formula in billing units is Energy (in kWh) = Power (in kW) * Time (in hours). Here, 750 W equals 0.75 kW, and the iron is used for 8 hours. So the daily energy consumption is 0.75 * 8 kWh. This gives a straightforward numerical answer that can be compared with the options provided. Step-by-Step Solution: Step 1: Convert the power from watts to kilowatts: 750 W = 0.75 kW.Step 2: Note the usage time per day: t = 8 hours.Step 3: Use the formula Energy (kWh) = Power (kW) * Time (h).Step 4: Substitute the values: Energy = 0.75 * 8 = 6 kWh.Step 5: Recognise that 1 kWh is one electrical unit, so the iron consumes 6 units of energy in one day. Verification / Alternative check: You can verify the calculation using joules if desired. One kilowatt hour equals 1000 W * 3600 s = 3.6 * 10^6 J. The iron uses 750 W for 8 hours, which is 8 * 3600 = 28800 seconds. The energy in joules is E = 750 * 28800 = 21.6 * 10^6 J. Dividing by 3.6 * 10^6 J per kWh gives E = 6 kWh. This matches the previous result, confirming that 6 units is correct. This cross check using fundamental units ensures that no conversion errors have occurred. Why Other Options Are Wrong: Option B, 600 units, would correspond to 600 kWh, which is unrealistic for a single 750 W appliance used only 8 hours per day; it would require much higher power or usage time. Option C, 0.6 units, is too small and would come from treating 750 W as 0.075 kW or using 0.8 hours instead of 8 hours. Option D, 60 units, is ten times too large and would require either 10 times the usage time or power. Only option A, 6 units, fits the correct calculation based on the given power and time. Common Pitfalls: Common errors include forgetting to convert watts to kilowatts, mixing up hours and minutes, or misplacing decimal points when multiplying. Some students also confuse power and energy, thinking that the wattage itself is the energy consumed. To avoid mistakes, always convert to kilowatts, use time in hours, and apply the simple formula Energy = Power * Time for kWh calculations. Writing each step clearly, as shown here, helps prevent unit and decimal errors in exam situations. Final Answer: 6 units

Question 17

In basic physics, the energy that is obtained from changes within the nuclei of atoms, such as fission or fusion processes, is known as which type of energy?

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Introduction / Context: Atoms consist of a nucleus made of protons and neutrons, surrounded by electrons. Most ordinary energy transformations in chemistry involve electrons and chemical bonds. However, there is a much larger store of energy bound up in the nucleus itself. When nuclear reactions occur, such as fission in nuclear reactors or fusion in stars, this nuclear binding energy can be released. The question asks you to identify the correct name for energy obtained from atoms in this nuclear sense, which is a foundational concept in modern physics and energy studies. Given Data / Assumptions: The energy comes from processes occurring within atomic nuclei. Examples include nuclear fission and nuclear fusion. We distinguish this from energy coming from electrons or chemical bonds. Options include solar, thermal, nuclear, and chemical energy. Concept / Approach: Nuclear energy is defined as the energy released during nuclear reactions, which involve changes in the composition of the nucleus. In nuclear fission, a heavy nucleus splits into smaller nuclei, releasing energy. In nuclear fusion, light nuclei combine to form a heavier nucleus, also releasing energy. This energy comes from a small loss of mass, converted into energy according to Einstein's equation E = m * c^2. Solar energy, on the other hand, is the energy received from the sun at the earth's surface and is itself produced by nuclear fusion in the sun's core. Chemical energy arises from rearrangements of electrons in atoms and molecules, not from changes in the nucleus. Step-by-Step Solution: Step 1: Identify that the question specifically mentions energy obtained from atoms, particularly from processes like fission and fusion in their nuclei.Step 2: Recall that such processes are nuclear reactions, involving protons and neutrons in the nucleus rather than electrons.Step 3: Recognise that the term used for energy released in nuclear reactions is nuclear energy.Step 4: Differentiate this from solar energy, which is nuclear energy from the sun observed as radiation, and from chemical energy, which is associated with bonds between atoms.Step 5: Conclude that nuclear energy is the correct term for energy obtained from atoms via nuclear processes. Verification / Alternative check: Examples help verify this conclusion. Nuclear power plants use controlled nuclear fission of uranium 235 or plutonium to generate heat, which is then converted to electricity. This is called nuclear energy. Atomic bombs and hydrogen bombs also release massive amounts of nuclear energy in an uncontrolled manner. In contrast, burning coal or gasoline releases chemical energy stored in molecular bonds. Solar panels capture solar energy, which originally comes from nuclear fusion reactions in the sun but is not directly referred to as nuclear energy at the point of use. These distinctions in terminology confirm that energy obtained from changes in atomic nuclei is properly called nuclear energy. Why Other Options Are Wrong: Solar energy, option A, refers to the electromagnetic radiation from the sun reaching the earth. While it originates from nuclear fusion, the term used at the earth's surface is solar, not nuclear, energy. Heat or thermal energy, option B, describes energy associated with random motion of particles and can come from many sources, not specifically from nuclear processes. Chemical energy, option D, arises from the arrangement and bonds of electrons in atoms and molecules, not from the nucleus. Only option C, nuclear energy, correctly names the energy released directly from atomic nuclei in fission and fusion. Common Pitfalls: Students sometimes confuse the source with the manifestation of energy. For example, the heat produced in a nuclear reactor is thermal energy, but its origin is nuclear. Another pitfall is associating atomic energy solely with bombs, forgetting peaceful applications in power generation and medicine. To answer conceptual questions correctly, focus on where the energy originates at the microscopic level: if it comes from changing nuclei, it is nuclear energy; if it comes from changing electron configurations, it is chemical energy. Final Answer: Nuclear energy

Question 18

The potential difference between the terminals of an electric heater is 60 V when it draws a current of 4 A from the source. Assuming the resistance of the heater remains constant, what current will it draw if the potential difference is increased t…

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Introduction / Context: Many electrical appliances such as heaters can be treated as resistive loads over a certain operating range. For such devices, Ohm's law applies, and the current drawn depends on the applied voltage and the fixed resistance of the appliance. This question gives you the current at an initial voltage and asks for the new current when the voltage is increased, assuming the resistance does not change. It tests your ability to calculate resistance first and then apply Ohm's law again at the new voltage. Given Data / Assumptions: Initial potential difference V1 = 60 V. Initial current I1 = 4 A. Heater resistance R is assumed constant. New potential difference V2 = 127.5 V. We are asked to find the new current I2 at V2. Concept / Approach: First, we use Ohm's law V = I * R with the initial data to find the resistance of the heater. The formula rearranged gives R = V / I. Once we know R, we treat it as constant and use Ohm's law again with the new voltage to solve for the new current I2 = V2 / R. This two step approach is standard when you are given one set of voltage current data and asked to predict current at another voltage for an ohmic device. Step-by-Step Solution: Step 1: Use Ohm's law with the initial values to find the resistance: R = V1 / I1.Step 2: Substitute V1 = 60 V and I1 = 4 A to get R = 60 / 4 = 15 ohm.Step 3: Assume the heater's resistance remains at R = 15 ohm when the voltage changes.Step 4: Apply Ohm's law with the new voltage: I2 = V2 / R.Step 5: Substitute V2 = 127.5 V and R = 15 ohm to get I2 = 127.5 / 15 = 8.5 A. Verification / Alternative check: We can verify the arithmetic by checking the division: 15 * 8 = 120 and 15 * 0.5 = 7.5, so 15 * 8.5 = 127.5, confirming that 127.5 / 15 is indeed 8.5. Also, since the voltage a little more than doubles from 60 V to 127.5 V, we expect the current to a little more than double as well, because current is directly proportional to voltage for a fixed resistance. The initial current was 4 A, and a little more than twice that is around 8 A, so 8.5 A is a reasonable and consistent value. Why Other Options Are Wrong: Option B, 24 A, would imply a resistance far smaller than 15 ohm and would not match the initial V I data. Option C, 10 A, would correspond to a resistance of 12.75 ohm at the new voltage, which contradicts the constant resistance assumption. Option D, 12 A, implies an even smaller resistance when combined with 127.5 V. Only option A, 8.5 A, is consistent with both the original and new conditions under the assumption of constant resistance. Common Pitfalls: Some students try to directly proportion the currents and voltages without first finding the resistance, which can lead to mistakes if they misinterpret the proportionality. Another pitfall is miscalculating the initial resistance or failing to keep it constant in the second step. Always remember the correct order: use the first set of data to determine resistance, then apply Ohm's law with the second voltage. Writing out the formulas and numerical values step by step, as shown, helps avoid arithmetic and conceptual errors in these types of questions. Final Answer: 8.5 A

Question 19

According to Archimedes’ principle, the magnitude of the buoyant force acting on a body immersed in a fluid depends primarily on which physical property of the fluid?

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Introduction / Context: This general physics question is about Archimedes’ principle and the concept of buoyant force. When an object is partly or completely immersed in a fluid such as water or air, it experiences an upward force called the buoyant force. Understanding which property of the fluid this buoyant force depends on helps students apply the principle correctly in problems related to floating, sinking, and fluid mechanics. Given Data / Assumptions: An object is immersed in a fluid (for example, water or oil). The fluid has a definite density, mass, volume, and temperature. We are interested only in which property of the fluid determines the magnitude of the buoyant force. Standard form of Archimedes’ principle is assumed: buoyant force equals weight of displaced fluid. Concept / Approach: According to Archimedes’ principle, the buoyant force on an immersed body is equal to the weight of the fluid displaced by that body. Weight of the displaced fluid is given by the product of fluid density, displaced volume, and acceleration due to gravity. Therefore, the magnitude of the buoyant force depends directly on the density of the fluid and the volume displaced, along with the constant g. The question specifically asks about a property of the fluid itself, so density is the relevant choice among the given options. Step-by-Step Solution: Step 1: Write the relation for buoyant force: F_b = rho * V_displaced * g, where rho is fluid density, V_displaced is volume of displaced fluid, and g is acceleration due to gravity. Step 2: Identify which quantities are properties of the fluid alone. Among these, rho, the density, is a fluid property, whereas V_displaced depends on the object shape and immersion. Step 3: Recognise that mass of the fluid is related to density and volume, but mass is not as fundamental for Archimedes’ formula as density. Step 4: Note that temperature may influence density, but the expression for buoyant force is written directly in terms of density, not temperature. Step 5: Conclude that the magnitude of buoyant force depends primarily on the density of the fluid for a given displaced volume and g. Verification / Alternative check: Consider placing the same object in water and in a lighter fluid such as kerosene. The object may float higher or lower depending on the fluid. Water has higher density than kerosene, so it provides a greater buoyant force for the same displaced volume. This practical observation matches the formula, confirming that density is the key fluid property that affects the buoyant force magnitude. Why Other Options Are Wrong: Mass: The total mass of the fluid in the container does not by itself determine buoyant force on the object; local displaced volume and density matter more. Temperature: Temperature can change density, but buoyant force is not expressed directly in terms of temperature. It is an indirect factor, not the primary property in the formula. Volume: Fluid volume in the container is not the controlling factor. The volume of fluid displaced by the body matters, but that is not a property of the fluid itself. Common Pitfalls: Students sometimes confuse the volume of the object or the volume of the container with a fluid property. Others focus on temperature because they know density varies with temperature, but they overlook that the formula is written directly in terms of density. Remember that in Archimedes’ principle, the buoyant force is directly proportional to fluid density for a given displaced volume. Final Answer: The magnitude of the buoyant force depends on the density of the fluid.

Question 20

A constant force of 200 N acts on an object of mass 100 kg. According to Newton’s second law of motion, what is the acceleration produced in the object?

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Introduction / Context: This numerical physics question tests direct application of Newton’s second law of motion, which relates the net force acting on an object to its mass and acceleration. Being able to compute acceleration correctly from a known force and mass is a fundamental skill used in mechanics, engineering, and everyday reasoning about how objects move when forces act on them. Given Data / Assumptions: Net force on the object, F = 200 N. Mass of the object, m = 100 kg. Motion is assumed to be in a straight line, and the force is constant. We neglect friction and other opposing forces, so 200 N is the net force. Concept / Approach: Newton’s second law states that the net force acting on a body equals the product of its mass and acceleration, written as F = m * a. Rearranging to find acceleration gives a = F / m. Substituting the given values of force and mass will directly yield the acceleration. Unit consistency is important: force in newtons, mass in kilograms, and acceleration in metres per second squared (m/s^2). Step-by-Step Solution: Step 1: Write Newton’s second law: F = m * a. Step 2: Rearrange to express acceleration: a = F / m. Step 3: Substitute the given values: F = 200 N and m = 100 kg. Step 4: Compute the value: a = 200 / 100 = 2. Step 5: Attach the appropriate unit. In SI, acceleration is measured in m/s^2, so a = 2 m/s^2. Verification / Alternative check: As a quick check, notice that if the mass were 100 kg and the acceleration 1 m/s^2, the required force would be 100 N. Doubling the force to 200 N should logically double the acceleration to 2 m/s^2 if mass stays the same. This proportional reasoning is consistent with the formula F = m * a and confirms that 2 m/s^2 is the correct result. Why Other Options Are Wrong: 2 m/s: This has the wrong unit. Metres per second describes speed, not acceleration. 0.5 m/s^2: This would correspond to a smaller force of 50 N on 100 kg, not the given 200 N. 4 m/s^2: This would require a net force of 400 N on a 100 kg mass, which is double the given force. Common Pitfalls: A common error is to ignore units and choose an answer that looks numerically simple but is dimensionally incorrect. Others may mistakenly divide mass by force instead of force by mass, which would give a nonsensical value with wrong units. Some students also forget that acceleration must be in m/s^2 and confuse it with velocity. Always start from F = m * a, rearrange carefully, and check the units of your final answer. Final Answer: The acceleration of the object is 2 m/s^2.

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