Question 1

The audible Doppler effect for sound waves is observed when a source of sound moves in what way relative to the listener?

Accepted Answer

Introduction / Context: The Doppler effect is a well known phenomenon in wave physics and everyday life. You hear it when an ambulance or train passes by and the pitch of the siren seems to change as it approaches and then moves away. This question focuses on when the Doppler effect occurs for sound waves, specifically in terms of the motion of the sound source relative to the listener. Given Data / Assumptions: We are considering sound waves propagating through air. The listener is at rest in the medium, while the source of sound may move. We assume standard conditions: uniform medium and speeds much less than the speed of sound. We want to know for which directions of source motion the Doppler effect is observed. Concept / Approach: The Doppler effect arises whenever there is relative motion between the source and the observer along the line joining them. This relative motion changes the frequency of wave fronts that reach the observer compared with the frequency emitted. If the source moves towards the observer, wave fronts are compressed and the observed frequency increases, producing a higher pitch. If the source moves away, wave fronts are stretched and the observed frequency decreases, producing a lower pitch. Thus, the effect occurs for both approaching and receding motion, not just one direction. Step-by-Step Solution: Step 1: Imagine a sound source emitting waves at a constant frequency while stationary. The listener hears the original frequency without shift. Step 2: Now let the source move towards the listener. Successive wave fronts are emitted from positions closer to the listener, effectively reducing the spacing between them in front of the source. Step 3: Reduced spacing between wave fronts corresponds to a shorter wavelength, and because the speed of sound in air is nearly constant, the frequency heard by the listener increases. Step 4: If the source moves away from the listener, the opposite happens. Wave fronts are emitted from positions farther away, increasing the spacing and thus the wavelength, so the observed frequency decreases. Step 5: In both cases, approaching and receding, there is a change between emitted and observed frequencies, meaning the Doppler effect occurs whenever there is relative motion along the line of sight. Verification / Alternative check: Real life examples confirm this explanation. When a siren approaches, you hear a higher pitched sound, and as it passes and moves away, the pitch drops suddenly. Similar effects occur with moving trains or vehicles. Physics formulas for Doppler shift for a moving source provide separate expressions for motion towards and away from the observer, but both show a non zero frequency shift whenever relative motion is present. No shift occurs only when source and observer are at rest relative to each other along the line of propagation. Why Other Options Are Wrong: Away from the listener only: This includes only the case of receding motion and ignores the increase in pitch when the source approaches. Towards the listener only: This covers only the approaching case and ignores the frequency decrease when the source recedes. When there is no relative motion between source and listener: In that case, there is no Doppler shift, so this does not describe when the effect occurs. Common Pitfalls: Some learners incorrectly think that the Doppler effect only applies when the source moves towards the observer, because the higher pitch is more dramatic. Others forget that the listener can also move and create a similar effect. The key idea is relative motion along the line of sight. As long as there is such motion, whether the source moves, the observer moves, or both move, there will be a Doppler shift in the observed frequency. Final Answer: The audible Doppler effect for sound is observed when the source moves either towards or away from the listener so that there is relative motion along the line joining them.

Question 2

Kilohertz (kHz) is a unit commonly used in electronics and communication to measure which physical quantity?

Accepted Answer

Introduction / Context: Modern electronic and communication systems rely heavily on signals that vary with time. The rate at which these signals repeat is described by their frequency. Units such as hertz, kilohertz, megahertz, and gigahertz are widely used to specify operating ranges of radio stations, microprocessors, and communication links. This question asks what physical quantity is measured in kilohertz (kHz). Given Data / Assumptions: Kilohertz is written as kHz, where kilo indicates one thousand. We are dealing with electromagnetic and radio wave applications. Options include electromagnetic radio wave frequencies, resistance, power for one ampere, and voltage. We assume SI based definitions of electrical and wave quantities. Concept / Approach: The hertz (Hz) is the SI unit of frequency and is defined as one cycle per second. Kilohertz is one thousand hertz, meaning one thousand cycles per second. Frequency describes how many times a periodic phenomenon repeats every second. In radio and communication engineering, frequency in kilohertz or megahertz is used to label radio stations, carrier signals, and clock signals. Electric resistance is measured in ohms, power in watts, and voltage in volts, not in kilohertz. Step-by-Step Solution: Step 1: Recognise that the base unit hertz (Hz) measures frequency, which is number of cycles per second. Step 2: Understand that kilohertz (kHz) means one thousand hertz, so 1 kHz = 1000 Hz. Step 3: In radio and electromagnetic applications, broadcast bands are labelled, for example, as 1000 kHz or 100 MHz, indicating how many times the electromagnetic field oscillates each second. Step 4: Electric resistance is measured in ohms (symbol ohm), power is measured in watts (W), and voltage is measured in volts (V); none of these use kilohertz as their unit. Step 5: Therefore, kilohertz is a unit that measures electromagnetic and radio wave frequencies, that is, frequency of periodic signals. Verification / Alternative check: When tuning an AM radio, station frequencies are often displayed in kilohertz, such as 810 kHz or 1200 kHz. When examining a microcontroller data sheet, the clock speed may be given in kilohertz or megahertz, for example 16 MHz, which is 16000 kHz. In both cases, the quantity described is how many cycles occur each second. Ohm, watt, and volt appear elsewhere in such data sheets associated with resistance, power rating, and supply voltage, confirming the different roles of these units. Why Other Options Are Wrong: Electric resistance: Measured in ohms, representing opposition to current flow, not in kilohertz. Power used by a current of one ampere: Power is measured in watts and involves both current and voltage, not frequency units. Electric potential difference (voltage): Measured in volts; it can be associated with alternating signals that have a frequency, but the unit kilohertz applies to the rate of variation, not to the voltage itself. Common Pitfalls: Students may confuse frequency with speed or think that kilohertz measures how strong a signal is. It is important to remember that frequency only tells how often a signal repeats, not how large it is. Another common confusion is mixing units seen together in circuit descriptions. For example, a waveform may be described as 5 V amplitude at 1 kHz, but here volt refers to amplitude and kilohertz to repetition rate. Keeping these distinct helps avoid mistakes. Final Answer: Kilohertz (kHz) is a unit that measures electromagnetic and radio wave frequencies, that is, frequency in thousands of cycles per second.

Question 3

In communication systems, the bandwidth of a channel is primarily defined as what?

Accepted Answer

Introduction / Context: The term bandwidth is widely used in physics, electronics, and computer networking. In strict communication theory, it refers to the range of frequencies that a system or channel can handle. In everyday language, people often use bandwidth to mean data speed or capacity of an internet connection. This question checks whether you know the primary technical definition of bandwidth in communication systems. Given Data / Assumptions: The context is signals and communication channels in electronics and networking. Bandwidth may be used in frequency domain or data communications contexts. Options mention frequency range, data rate, both, or none. We assume the formal physics and engineering definition as primary. Concept / Approach: In communication theory and signal processing, bandwidth is defined as the difference between the upper and lower frequencies of a band of interest. For example, if a low pass filter passes frequencies up to 3 kHz, its bandwidth is 3 kHz. If a radio channel passes from 88 MHz to 108 MHz, the bandwidth is 20 MHz. While data rate is related to bandwidth through information theory, it is not identical to bandwidth in definition. Therefore the primary meaning of bandwidth is the width of a frequency range, not the data rate itself. Step-by-Step Solution: Step 1: Consider a band pass filter that allows signals between f1 and f2 hertz to pass. The bandwidth of the filter is defined as f2 minus f1. Step 2: For a typical voice telephone channel with frequencies from about 300 Hz to 3400 Hz, the bandwidth is approximately 3100 Hz. Step 3: Data transmission capacity over a channel is influenced by this bandwidth and by the signal to noise ratio, but it is not numerically identical to bandwidth; it is measured in bits per second, not in hertz. Step 4: In everyday language, people sometimes say a connection has more bandwidth when it has a higher data rate, but this is a derived usage based on the underlying frequency domain concept. Step 5: Therefore, the core technical definition of bandwidth is a measure of the width of a range of frequencies that a channel or signal occupies. Verification / Alternative check: Textbooks on signal processing and communication systems define bandwidth explicitly as a frequency interval. The Shannon capacity formula for a channel, C = B * log2 (1 + S / N), shows that capacity C in bits per second depends on B, the bandwidth in hertz. This equation further confirms that bandwidth and data rate are related but distinct quantities. Unit analysis also helps: bandwidth is in hertz, whereas data rate is in bits per second, so they cannot be the same by definition. Why Other Options Are Wrong: The maximum rate of data transfer of a digital link only: This describes data rate or throughput, usually measured in bits per second, not bandwidth in hertz, though the two are linked. Both the width of the frequency range and the data transfer rate by definition: This confuses a fundamental physical property (frequency range) with a performance measure (data rate). They are related but not both part of the formal definition. None of the above: Incorrect because the first option gives the correct primary definition of bandwidth. Common Pitfalls: Many learners and even practitioners casually use the word bandwidth to mean download speed. While this is acceptable in informal conversation, exam questions usually expect the precise technical meaning. To avoid errors, remember that bandwidth lives in the frequency domain (measured in hertz), whereas data rate lives in the time domain (bits per second). Connecting both via formulas is important, but not the same as equating them in definition. Final Answer: In communication systems, bandwidth is primarily defined as a measure of the width of a range of frequencies carried by a signal or channel.

Question 4

In which type of mechanical wave do particles of the medium vibrate in a direction perpendicular to the direction of wave propagation?

Accepted Answer

Introduction / Context: Mechanical waves can be classified based on how particles of the medium move relative to the direction in which the wave travels. Understanding this distinction helps in analysing waves on strings, water surfaces, and sound in air. This question focuses on identifying the type of wave for which particle vibration is perpendicular to the direction of wave propagation. Given Data / Assumptions: The wave is a mechanical wave requiring a material medium. Particles of the medium are said to vibrate at right angles to the direction in which the wave travels. Options include longitudinal, rolling, transverse, and field waves. We assume standard textbook definitions of these wave types. Concept / Approach: In a transverse wave, particles of the medium move perpendicular to the direction of energy transfer or wave propagation. A classic example is a wave on a stretched string, where the string moves up and down while the disturbance travels horizontally. In longitudinal waves, particles oscillate parallel to the direction of propagation, as in sound waves in air. Rolling and field waves are not standard basic categories in this context. Therefore, the correct classification for a wave with perpendicular particle motion is transverse wave. Step-by-Step Solution: Step 1: Recall that mechanical waves involve oscillations of particles about their mean positions. Step 2: Identify that the question specifies particle motion perpendicular to the wave direction, which is the defining feature of transverse waves. Step 3: Visualise a rope tied at one end; if you jerk the free end up and down, loops move along the rope, but each point on the rope moves vertically while the disturbance travels horizontally. Step 4: Compare this with a longitudinal wave like sound in air, where compressions and rarefactions move along the direction of propagation, and particles move back and forth along that same line. Step 5: Conclude that the correct type of wave for the described motion is a transverse wave. Verification / Alternative check: Check standard diagrams of transverse and longitudinal waves. In transverse wave diagrams, displacement is drawn vertically while the wave moves horizontally, emphasising the perpendicular relationship. Texts explicitly state that in transverse waves, particle displacement is perpendicular to the direction of propagation. No such description is given for longitudinal waves. Rolling and field waves are used in other contexts but are not the basic categories for simple mechanical waves in school level physics. Why Other Options Are Wrong: Longitudinal wave: Particles move parallel to the direction of propagation, as in sound waves, so this does not match perpendicular motion. Rolling wave: This term may appear in discussions of water surface motion but is not a standard basic category used in introductory definitions. Field wave: This is not a commonly defined mechanical wave type at this level and does not specifically describe particle motion. Common Pitfalls: Students often confuse the names because they associate longitudinal with long distances and transverse with crossing. A clearer strategy is to remember specific examples: light on a string is transverse, sound in air is longitudinal. Visualising particle motion with arrows helps fix the concept: perpendicular arrows for transverse waves and parallel arrows for longitudinal waves. Final Answer: The described wave, where particles vibrate perpendicular to the direction of propagation, is a transverse wave.

Question 5

A body of mass 20 kg is at a height of 8 m above the ground. Taking g = 9.8 m/s^2, what is the gravitational potential energy possessed by the body relative to the ground?

Accepted Answer

Introduction / Context: Gravitational potential energy is the energy stored in a body due to its position in a gravitational field. Near the Earth surface, this energy is directly proportional to the mass of the body, the height above a reference level, and the acceleration due to gravity. This question asks you to apply the basic formula for gravitational potential energy to a specific set of values for mass, height, and g. Given Data / Assumptions: Mass of the body, m = 20 kg. Height above the ground, h = 8 m. Acceleration due to gravity, g = 9.8 m/s^2. The reference level for potential energy is the ground, where potential energy is taken as zero. We use the formula for gravitational potential energy near Earth surface. Concept / Approach: The gravitational potential energy U of a body of mass m at height h above a reference level in a uniform gravitational field is given by the formula U = m * g * h. Here m is in kilograms, g in m/s^2, and h in metres. The resulting potential energy is measured in joules (J), where 1 J = 1 kg * m^2 / s^2. Once we substitute the given values into this formula, we can compute the numeric value and identify the correct unit among the options. Step-by-Step Solution: Step 1: Write the formula for gravitational potential energy: U = m * g * h. Step 2: Substitute the given values: m = 20 kg, g = 9.8 m/s^2, and h = 8 m. Step 3: Compute the product g * h: 9.8 * 8 = 78.4. Step 4: Multiply by the mass: U = 20 * 78.4 = 1568. Step 5: Attach the correct unit. Since we used kg, m, and m/s^2, the unit is kg * m^2 / s^2, which is called the joule (J). So U = 1568 J. Verification / Alternative check: Check the order of magnitude. A 10 kg mass at 10 m height with g taken as roughly 10 m/s^2 would have U approximately 10 * 10 * 10 = 1000 J. In this problem, the mass is 20 kg and height is 8 m with g slightly less than 10, which should give a value somewhat above 1500 J. The computed value 1568 J fits this expectation. Also, asking whether coulomb, watt, or newton could be the unit of energy immediately confirms that joule is appropriate for potential energy in mechanics. Why Other Options Are Wrong: 1568 C: Coulomb (C) is the unit of electric charge, not mechanical energy. 1568 W: Watt (W) is the unit of power, which is energy per unit time, not energy itself. 1568 N: Newton (N) is the unit of force, equal to kg * m / s^2, not kg * m^2 / s^2. Common Pitfalls: Students sometimes forget to include the height or the acceleration due to gravity in the formula, or they choose the wrong unit at the end. Mixing up joule, watt, and newton is especially common, because all three appear in mechanics problems. To avoid this, remember that energy and work are always in joules, power is in watts, and force is in newtons. Also be careful with g; using 10 m/s^2 as an approximation is acceptable for rapid calculations, but here the value 9.8 m/s^2 leads to the more precise result 1568 J. Final Answer: The gravitational potential energy of the 20 kg mass at a height of 8 m is 1568 J.

Question 6

In air at normal conditions, what is the minimum distance in metres between a sound source and a reflecting surface that is required to hear a distinct echo?

Accepted Answer

Introduction / Context: An echo is a separate repetition of a sound due to reflection from a distant surface such as a wall, building, or hill. For the human ear to perceive an echo as distinct from the original sound, there must be a sufficient time gap between the original sound and the reflected sound. This time gap translates to a minimum distance for the reflecting surface, based on the speed of sound in air. This question asks you to recall or derive that minimum distance under normal conditions. Given Data / Assumptions: The medium is air at ordinary temperature and pressure. The speed of sound in air is approximately 340 m/s (or 343 m/s; we use 340 m/s for easy calculation). The human ear can distinguish two sounds as separate if they reach the ear at least about 0.1 second apart. Sound must travel from the source to the reflecting surface and back to the listener. Concept / Approach: Let t be the minimum time interval required for the ear to perceive two sounds separately, and v be the speed of sound. For an echo, the sound travels a total distance of 2 * d, where d is the distance between the source and the reflecting surface. The total time taken is t = (2 * d) / v. Rearranging this gives d = (v * t) / 2. Substituting the typical values v approximately 340 m/s and t approximately 0.1 s gives a minimum distance of about 17 m. Step-by-Step Solution: Step 1: Use the relation time = distance / speed. For the echo path, total distance travelled by sound is 2 * d. Step 2: Write t = (2 * d) / v, where t is the time interval between original and reflected sound reaching the listener. Step 3: Rearrange for distance: d = (v * t) / 2. Step 4: Substitute v = 340 m/s and t = 0.1 s: d = (340 * 0.1) / 2. Step 5: Compute: 340 * 0.1 = 34; then 34 / 2 = 17 m. So the minimum distance needed is about 17 metres. Verification / Alternative check: If the reflecting surface were only 10 m away, the total distance for the sound to travel out and back would be 20 m. With v = 340 m/s, the time taken is 20 / 340, which is about 0.059 s, less than 0.1 s, so the reflected sound would blend with the original. At 17 m, the round trip distance is 34 m, giving a travel time of 34 / 340 = 0.1 s, exactly the minimum separation required. This confirms that the standard textbook value of 17 m is consistent with human hearing limits and speed of sound in air. Why Other Options Are Wrong: 10 m: Too small; the time difference between original and reflected sound is less than 0.1 s, so no distinct echo is heard. 13 m: Also too small; round trip distance is 26 m, giving a time of about 0.076 s, still below the 0.1 s threshold. 21 m: Greater than the minimum required distance, so an echo would be heard, but the question asks for the minimum distance, which is smaller and equal to 17 m. Common Pitfalls: Students often forget that the sound travels to the wall and back, so they mistakenly use d instead of 2 * d in the time calculation. Others may use an approximate speed of sound but forget to include the ear resolution time of about 0.1 s. To avoid mistakes, always remember that for an echo you must consider the round trip path and the minimum time separation that the human ear can distinguish. Final Answer: The minimum distance required between the sound source and the reflecting surface to hear a distinct echo in air is 17 m.

Question 7

Natural radioactivity, the spontaneous emission of radiation from certain unstable nuclei, was first discovered by which scientist?

Accepted Answer

Introduction / Context: Radioactivity is the spontaneous emission of particles or electromagnetic radiation from the unstable nucleus of an atom. The discovery of natural radioactivity at the end of the nineteenth century opened a new era in physics, chemistry, and medicine. Several famous scientists worked in this area, but one physicist is credited with the first discovery of natural radioactivity. This question asks you to identify that scientist. Given Data / Assumptions: The phenomenon in question is natural radioactivity, not artificial transmutation. The scientists listed are Marie Curie, Ernest Rutherford, Henri Becquerel, and Enrico Fermi. We are interested in who first observed and reported spontaneous emission from uranium salts. We assume standard historical accounts of the discovery. Concept / Approach: Henri Becquerel, a French physicist, discovered natural radioactivity in 1896 while studying phosphorescence in uranium salts. He found that certain salts emitted penetrating radiation that could fog photographic plates even in the absence of sunlight. Marie and Pierre Curie later extended this work by isolating new radioactive elements such as polonium and radium, and Rutherford made major contributions to understanding radioactive decay and the structure of the atom. Enrico Fermi worked much later on neutron induced radioactivity and nuclear reactors. Therefore, the credit for the first discovery of natural radioactivity goes to Henri Becquerel. Step-by-Step Solution: Step 1: Recall that Becquerel was investigating potential connections between X rays and phosphorescence using uranium salts placed on photographic plates wrapped in black paper. Step 2: He observed that even when the salts were kept in the dark, the photographic plates became fogged, indicating that the salts themselves were emitting some invisible penetrating radiation. Step 3: This spontaneous emission did not depend on external stimulation like sunlight, proving that it was a property of the uranium atoms. Step 4: This phenomenon was later named radioactivity, and similar properties were found in other elements by the Curies. Step 5: Considering the timeline, Becquerel discovery in 1896 predates the detailed work of Marie Curie, Rutherford, and Fermi, so he is recognised as the discoverer of natural radioactivity. Verification / Alternative check: Historical timelines confirm that in 1895, Roentgen discovered X rays, and in 1896, Becquerel discovered uranium rays, now known as radioactivity. In the late 1890s and early 1900s, Marie Curie and Pierre Curie isolated radium and polonium and coined the term radioactivity. Rutherford later distinguished between alpha, beta, and gamma radiation and proposed the nuclear model of the atom. Enrico Fermi work on artificial radioactivity and chain reactions took place in the 1930s and 1940s. This chronology supports the conclusion that Becquerel made the first discovery of natural radioactivity. Why Other Options Are Wrong: Marie Curie: She greatly advanced the study of radioactivity and discovered new radioactive elements but did not make the initial discovery. Ernest Rutherford: He made fundamental contributions to understanding radioactive decay and nuclear structure but came after Becquerel. Enrico Fermi: Known for his work on neutron induced radioactivity and nuclear reactors, he worked decades after the discovery of natural radioactivity. Common Pitfalls: Because Marie Curie is strongly associated with radioactivity and even gave the phenomenon its name, students sometimes mistakenly credit her with the original discovery. The key to avoiding this error is to separate the initial observation of the effect, made by Becquerel, from the later detailed investigations and applications developed by the Curies, Rutherford, and others. Final Answer: Natural radioactivity was first discovered by Henri Becquerel.

Question 8

In power units, 1 kilowatt (1 kW) is equal to how many watts (W)?

Accepted Answer

Introduction / Context: Electrical and mechanical power are often expressed in watts and kilowatts. Understanding the relationship between these units is essential when reading power ratings of household appliances, motors, and power plants. This question asks you to convert one kilowatt into its equivalent number of watts. Given Data / Assumptions: The base SI unit of power is the watt (W). The prefix kilo means one thousand times the base unit. We want to know how many watts are contained in 1 kilowatt. We assume standard SI prefixes and definitions. Concept / Approach: The SI system uses prefixes to denote multiples and submultiples of base units. The prefix kilo, with symbol k, denotes a factor of 10^3, which equals 1000. This rule applies uniformly across units: one kilometre is 1000 metres, one kilogram is 1000 grams in older usage, and one kilowatt is 1000 watts. Therefore, converting from kilowatts to watts simply means multiplying the numerical value by 1000. Step-by-Step Solution: Step 1: Recall the SI prefix rule: 1 kilo unit = 1000 base units. Step 2: Apply this rule to power: 1 kilowatt = 1 kilo watt. Step 3: Replace kilo with 1000: 1 kilowatt = 1000 watts. Step 4: Express this result in compact form: 1 kW = 1000 W. Step 5: Select the option that matches this numeric equivalence, which is 1000 W. Verification / Alternative check: Consider a household appliance rated at 0.5 kW. Converting to watts gives 0.5 * 1000 = 500 W. This matches typical power values for devices like microwave ovens or small heaters. Power plants are often rated in megawatts, where 1 MW = 1000 kW = 1,000,000 W, again showing the consistent factor of 1000 between kilowatt and watt. These examples reinforce the simple conversion factor. Why Other Options Are Wrong: 10 W: This would imply that kilo represents a factor of 10, which conflicts with the SI definition of kilo. 100 W: This would treat kilo as 100, again inconsistent with the standard prefix. 10000 W: This corresponds to a factor of 10^4 and would be appropriate for 10 kW, not 1 kW. Common Pitfalls: Some learners mix up kilo and mega or forget that SI prefixes are powers of ten. Confusion can also arise from using non SI units such as horsepower alongside watts. To avoid mistakes, remember that 1 kW = 1000 W and that 1 MW = 1000 kW. Writing out the factors in powers of ten, such as 10^3 and 10^6, can help maintain clarity when converting between units. Final Answer: One kilowatt is equal to 1000 W (1000 watts).

Question 9

When a body experiences negative acceleration (retardation), the acceleration vector acts in a direction opposite to which of the following vectors?

Accepted Answer

Introduction / Context: In kinematics, acceleration describes how the velocity of an object changes with time. When acceleration acts in the same direction as velocity, the object speeds up. When it acts in the opposite direction, the object slows down, and this is often called negative acceleration or retardation. This question asks you to identify which vector negative acceleration opposes. Given Data / Assumptions: We describe motion using vectors: displacement, velocity, acceleration, force, and momentum. Negative acceleration means that the acceleration is directed opposite to the chosen positive direction of velocity. We assume straight line motion for simplicity. The options include force, displacement, velocity, and momentum. Concept / Approach: Velocity is the rate of change of displacement with time and points in the direction of motion. Acceleration is the rate of change of velocity with time. If acceleration has a component opposite to the velocity vector, it reduces the magnitude of the velocity, slowing the body down. This is the situation referred to as negative acceleration or retardation in elementary physics. While force is related to acceleration through Newton second law, the phrase negative acceleration is best understood as acceleration opposite to velocity, not necessarily opposite to displacement or momentum in every context. Step-by-Step Solution: Step 1: Consider an object moving to the right. We take motion to the right as the positive direction, so its velocity vector points to the right. Step 2: If the object is slowing down, its speed is decreasing even though it is still moving to the right for some time. Step 3: For speed to decrease, the acceleration must have a component opposite to the direction of velocity, which in this case is to the left. Step 4: This acceleration directed opposite to velocity is called negative acceleration or retardation when we have chosen the direction of velocity as positive. Step 5: Therefore, negative acceleration is in the direction opposite to the velocity vector. Verification / Alternative check: Mathematically, if velocity v is positive and decreasing, then dv/dt is negative, indicating acceleration a is negative with respect to the chosen coordinate axis. Graphs of velocity versus time for a braking car show a straight line sloping downwards, representing negative acceleration while velocity is still positive. This graphical interpretation again highlights that acceleration points opposite to velocity during slowing down motion, matching the conceptual description of negative acceleration. Why Other Options Are Wrong: Net force: Although net force and acceleration are in the same direction by Newton second law, the question asks what negative acceleration is opposite to, which is best expressed in terms of velocity, not force. Displacement: Displacement points from the initial to the final position and may not coincide with the instantaneous velocity or acceleration directions at all times. Momentum: Momentum is mass times velocity, so its direction is the same as velocity, but the standard explanation of negative acceleration is given directly in relation to velocity, not momentum. Common Pitfalls: Students may confuse negative acceleration with simply changing direction of motion. A body may have positive acceleration and yet change direction, as in circular motion, if the coordinate system is chosen differently. The key is to think of negative acceleration as acceleration that reduces the magnitude of velocity along its current direction of motion, which means it acts opposite to the velocity vector in that frame of reference. Final Answer: Negative acceleration, or retardation, acts in a direction opposite to the velocity of the body.

Question 10

The speed of light is maximum when it travels in which of the following media?

Accepted Answer

Introduction / Context: Light and other electromagnetic waves can travel through different media, such as vacuum, air, water, and glass. The speed of light is not the same in all media; it depends on the optical properties of the material. This question asks you to identify the medium in which the speed of light attains its maximum value. Given Data / Assumptions: We are comparing the speed of light in vacuum, solids, liquids, and gases. The fundamental constant c denotes the speed of light in vacuum. When light enters a material medium, its speed decreases compared with c. We are not considering exotic media with special properties, only typical physical states. Concept / Approach: The speed of light in vacuum is a fundamental constant of nature, usually denoted by c and approximately equal to 3 * 10^8 m/s. When light passes through a material medium, its speed is reduced. The ratio of c to the speed of light in the medium is called the refractive index. For any transparent material, the refractive index is greater than or equal to 1, so the speed of light in that material is less than or equal to c. Among the listed options, only vacuum has refractive index exactly 1 and allows light to travel at the maximum speed c. Step-by-Step Solution: Step 1: Recall that in vacuum, light travels at speed c, which is approximately 3 * 10^8 m/s. Step 2: In air, which is a gas, the speed of light is slightly less than c because the refractive index of air is just above 1, though very close to 1. Step 3: In liquids such as water and solids such as glass, the refractive index is greater than that of air, so the speed of light is further reduced. Step 4: Since speed decreases as refractive index increases, the maximum speed is achieved in the medium with the lowest refractive index, which is vacuum. Step 5: Therefore, the speed of light is maximum in vacuum, compared with solids, liquids, or gases. Verification / Alternative check: Typical refractive index values are about 1 for vacuum, approximately 1.0003 for air, about 1.33 for water, and around 1.5 for common glass. The speed of light in any medium v is given by v = c / n, where n is the refractive index. Substituting these values shows that v is very slightly less than c in air and significantly less in water and glass. This confirms that light travels fastest in vacuum and progressively slower in gases, liquids, and solids. Why Other Options Are Wrong: Solids: Have refractive indices greater than 1, so light travels slower than in vacuum. Liquids: Also have refractive indices greater than 1 and slow light more than air does. Gases: Most gases have refractive indices slightly greater than 1; light speed is close to but still less than c. Common Pitfalls: Some learners assume that light travels fastest in air because it seems to move very freely there, but they forget that air is still matter, not vacuum. Others confuse sound and light, thinking that because sound travels fastest in solids, light might also. In reality, sound is a mechanical wave requiring a medium, while light is an electromagnetic wave that needs no medium and achieves its maximum speed in vacuum. Final Answer: The speed of light is maximum when it travels in a vacuum.

Question 11

For the photoelectric effect to occur efficiently, the metal used as the cathode should have which of the following properties?

Accepted Answer

Introduction / Context: The photoelectric effect is the emission of electrons from a metal surface when it is exposed to electromagnetic radiation of sufficient frequency, such as ultraviolet light. This effect provided crucial evidence for the particle nature of light and the concept of photons. The ease with which electrons are emitted depends strongly on the nature of the metal used as the cathode. This question asks what property the metal should have to make photoemission easier. Given Data / Assumptions: We are considering metals used as cathodes in photoelectric experiments or devices such as photo cells. Important properties include melting point, work function, and electrical resistance. Work function is the minimum energy needed to remove an electron from the metal surface. We assume light of fixed frequency is incident on the metal and we want efficient electron emission. Concept / Approach: According to Einstein photoelectric equation, the maximum kinetic energy of emitted electrons is given by K max = h * f minus phi, where h is Planck constant, f is the frequency of incident light, and phi is the work function of the metal. The work function represents the minimum photon energy required to liberate an electron from the surface. A lower work function means electrons can be emitted with lower frequency or lower energy light. Therefore, to obtain photoelectric emission easily, the cathode metal should have a low work function. Melting point and resistance may be important for other design reasons but are not the primary factor governing photoelectric emission threshold. Step-by-Step Solution: Step 1: Recall that work function phi is the energy needed to remove an electron from the metal surface to a point just outside the metal. Step 2: The condition for photoelectric emission is that the photon energy h * f must be greater than or equal to the work function phi. Step 3: If phi is small, then even light of moderate frequency can satisfy h * f greater than or equal to phi and cause emission. Step 4: If phi is large, much higher frequency light is required, which may not be available or practical in an experiment or device. Step 5: Therefore, to ensure that the photoelectric effect occurs readily, one chooses metals with low work function as cathode materials. Verification / Alternative check: Common photoelectric materials include alkali metals such as cesium and potassium, which have relatively low work functions and readily emit electrons under visible or near ultraviolet light. In contrast, metals with high work function require ultraviolet or even higher energy photons to show significant photoelectric effect. Device designers pick low work function materials to build sensitive photo detectors and light meters. This practical engineering choice confirms the theoretical reasoning that low work function is desirable. Why Other Options Are Wrong: High melting point: While useful for high temperature applications, it does not directly determine the threshold for photoelectric emission. Low melting point: This could make the device unstable at moderate temperatures and again does not control electron emission threshold. Low electrical resistance: Good electrical conductivity is helpful, but even highly conductive metals may have high work functions and thus be poor photoemitters. Common Pitfalls: A frequent confusion is to associate photoelectric sensitivity solely with good conductivity. While the metal must conduct, the crucial quantity is the work function. Another mistake is to think that any strong light will cause emission from any metal, ignoring the threshold frequency requirement. Remember that for efficient photoelectric emission, the cathode metal should combine adequate conductivity with a low work function so that incident photons can easily liberate electrons. Final Answer: For the photoelectric effect to occur easily, the cathode metal should have a low work function.

Question 12

A photon of X ray has energy of 1 keV, while a photon of visible light has energy of about 3 eV. In this context, which one of the following statements is not correct?

Accepted Answer

Introduction / Context: Photons are quanta of electromagnetic radiation, and their properties such as energy, wavelength, and frequency are related. X rays are high energy electromagnetic waves, while visible light has lower photon energy. Although their energies and wavelengths differ, all electromagnetic waves share certain fundamental properties when they travel in vacuum. This question asks you to identify which of the listed statements about an X ray photon and a visible light photon is not correct. Given Data / Assumptions: Energy of the X ray photon E x = 1 keV (1000 eV). Energy of the visible light photon E v = 3 eV. Both photons travel in vacuum when we compare their speeds. Standard relations E = h * f and c = lambda * f apply, where h is Planck constant, f is frequency, lambda is wavelength, and c is speed of light in vacuum. Concept / Approach: Photon energy E is proportional to frequency f through the relation E = h * f. Therefore, a larger energy photon has higher frequency. Since the speed of light in vacuum c is constant for all electromagnetic waves, wavelength lambda is related by c = lambda * f, so higher frequency implies shorter wavelength. Thus, a 1 keV X ray photon has much higher frequency and shorter wavelength than a 3 eV visible photon. However, the speed in vacuum is the same c for both. Any statement claiming different speeds in vacuum is therefore incorrect. Step-by-Step Solution: Step 1: Compare energies: 1 keV equals 1000 eV, while the visible photon has 3 eV. So E x is much greater than E v. Step 2: Use E = h * f. Higher energy means higher frequency. Therefore, the X ray photon has higher frequency than the visible photon. Step 3: Use c = lambda * f. For a given constant c, a higher frequency implies a shorter wavelength. So the X ray photon has a shorter wavelength than the visible photon. Step 4: Recall that all electromagnetic waves in vacuum travel at the same speed c, regardless of their frequency or wavelength. This applies to X rays, visible light, radio waves, and gamma rays alike. Step 5: Therefore, any statement that claims that the speeds of the two photons in vacuum are different is not correct. Verification / Alternative check: Standard physics references define c, the speed of light in vacuum, as a universal constant approximately equal to 3 * 10^8 m/s. This value applies to all electromagnetic waves in vacuum. Differences in speed occur only when waves travel through media with different refractive indices, and even then the speed depends on the medium, not on photon energy alone. Since the question explicitly states speeds in vacuum, c must be the same. On the other hand, X ray wavelengths are well known to be much shorter (for example, nanometres to picometres) than visible wavelengths (hundreds of nanometres), consistent with the first statement being correct. Why Other Options Are Wrong (meaning they are actually correct statements): The wavelength of the X ray photon is less than the wavelength of the visible light photon: This is correct because higher energy and higher frequency correspond to shorter wavelength for the X ray. The two photons have different energies: This is obviously correct, since 1 keV is much greater than 3 eV. The frequency of the X ray photon is higher than the frequency of the visible light photon: This follows directly from E = h * f and the fact that the X ray photon has higher energy. Common Pitfalls: Students sometimes mistakenly think that higher energy electromagnetic waves must travel faster in vacuum. This confusion arises from everyday experience with mechanical objects where more energy can mean higher speed. However, for photons in vacuum, speed is fixed at c regardless of energy. Energy differences change frequency and wavelength, not speed. Remember that all electromagnetic waves share the same speed in vacuum, and differences appear only in media where refractive index varies with wavelength. Final Answer: The incorrect statement is that the speeds of the two photons in vacuum are different.

Question 13

Optical fibre communication cables guide light over long distances based mainly on which fundamental principle of light?

Accepted Answer

Introduction / Context: Optical fibres are thin strands of glass or plastic that can transmit light signals over long distances with very low loss. They are the backbone of modern high speed communication networks. The ability of optical fibres to confine light within their cores even when bent is based on a specific principle of geometrical optics. This question asks you to identify that principle. Given Data / Assumptions: An optical fibre has a higher refractive index core surrounded by a lower refractive index cladding. Light is launched into the fibre within a certain range of angles. We assume clear, lossless fibre and neglect absorption and scattering for conceptual understanding. Options include reflection, refraction, diffraction, and total internal reflection. Concept / Approach: When light travelling in a denser medium reaches the boundary with a rarer medium at an angle greater than the critical angle, it is completely reflected back into the denser medium. This phenomenon is called total internal reflection. Optical fibres are designed so that light rays inside the core always meet the core cladding boundary at angles greater than the critical angle, causing repeated total internal reflection. As a result, the light is trapped and guided along the fibre, even if the fibre is curved. Simple reflection at a mirror or ordinary refraction are not sufficient to explain this efficient light guiding mechanism. Step-by-Step Solution: Step 1: Recall that total internal reflection occurs when light travels from a medium with higher refractive index to one with lower refractive index and the angle of incidence exceeds the critical angle. Step 2: In an optical fibre, the core has a slightly higher refractive index than the cladding, satisfying the condition for potential total internal reflection at the core cladding interface. Step 3: Light is injected into the core at suitable angles so that when it reaches the boundary, the angle of incidence is greater than the critical angle. Step 4: At each such encounter, light is totally reflected back into the core, with virtually no loss into the cladding, allowing it to travel long distances through multiple reflections. Step 5: This repeated total internal reflection is the key mechanism that keeps light confined within the optical fibre. Verification / Alternative check: Diagrams of optical fibres in physics and communication textbooks show zigzag paths of light bouncing inside the core, with reflection angles above the critical angle. Mathematical analysis using Snell law and critical angle formulas further confirms that for typical core and cladding indices, total internal reflection occurs for a range of entry angles. Practical fibre optic systems rely on this principle to maintain signal strength over many kilometres, while simple reflection or refraction at a single interface would not provide the same level of confinement. Why Other Options Are Wrong: Reflection at a plane mirror surface: Mirrors reflect light, but optical fibres do not contain mirror coatings inside; they rely on internal reflections at glass interfaces instead. Simple refraction at a single interface: Refraction changes the direction of light as it enters a new medium, but by itself does not trap light in a waveguide. Diffraction through narrow slits: Diffraction describes spreading of waves around obstacles and apertures; it is important in optics but is not the main guiding principle in standard step index fibres. Common Pitfalls: Students sometimes confuse total internal reflection with ordinary reflection or think that refraction is enough to explain light guidance. While there is initial refraction as light enters the fibre, the crucial ongoing mechanism is repeated total internal reflection. Another confusion is to associate total internal reflection only with prisms and diamonds; in fact, optical fibres are a major practical application of the same phenomenon. Final Answer: Optical fibres work primarily on the principle of total internal reflection inside the higher refractive index core.

Question 14

The fact that planets move around the Sun in elliptical orbits rather than perfect circles was first clearly demonstrated by which astronomer?

Accepted Answer

Introduction / Context: The motion of planets in the Solar System has been studied for centuries. Early models often assumed circular orbits, but precise observations eventually showed that planetary paths are actually ellipses with the Sun at one focus. The formulation of this fact as a clear law of planetary motion was a major milestone in astronomy. This question asks which astronomer first demonstrated that planetary orbits are elliptical rather than circular. Given Data / Assumptions: We are considering historical contributions to the understanding of planetary orbits. Names given include Galileo Galilei, Martin Luther, Johannes Kepler, and Nicolaus Copernicus. The focus is on who formulated the law about elliptical orbits around the Sun. We assume standard historical accounts of the scientific revolution. Concept / Approach: Johannes Kepler, using the extremely accurate observational data collected by Tycho Brahe, studied the orbit of Mars and other planets. He found that circular orbits with epicycles could not fully explain the data. After years of analysis, he concluded that the planets move in ellipses with the Sun at one focus. This became Kepler first law of planetary motion. Copernicus introduced the heliocentric model but still used circular orbits. Galileo supported heliocentrism and made many telescopic discoveries but did not derive the elliptical nature of orbits. Martin Luther was a religious reformer, not an astronomer. Thus, the correct answer is Johannes Kepler. Step-by-Step Solution: Step 1: Recall that Copernicus proposed that the Sun, not the Earth, is at the centre of the planetary system, but he retained circular orbits and epicycles. Step 2: Galileo observations of the phases of Venus and moons of Jupiter provided evidence for heliocentrism but did not involve detailed orbital shapes. Step 3: Kepler inherited Brahe precise measurements of planetary positions and used them to model Mars orbit. Step 4: After testing and rejecting circular paths, Kepler discovered that an ellipse with the Sun at one focus described the motion accurately. Step 5: This result became Kepler first law: planets move in elliptical orbits with the Sun at one focus, clearly demonstrating the non circular nature of planetary orbits. Verification / Alternative check: Modern celestial mechanics confirms that, in Newtonian gravity, the natural closed orbit for a body bound to another by an inverse square force is an ellipse. Kepler empirical law was later derived from Newton gravitational theory. Historical records attribute this discovery of elliptical orbits to Kepler, whose three laws are named in his honour. Copernicus and Galileo are celebrated for other contributions, but not specifically for the elliptical orbit law, and Martin Luther is not associated with astronomy at all. Why Other Options Are Wrong: Galileo Galilei: He made important observations supporting heliocentrism but did not formulate the law of elliptical orbits. Martin Luther: A key figure in the Protestant Reformation, he was a theologian, not an astronomer. Nicolaus Copernicus: Proposed the Sun centred model but still used circular orbits and epicycles rather than ellipses. Common Pitfalls: Students often conflate the roles of Copernicus, Galileo, and Kepler, remembering only that all supported heliocentrism. To avoid confusion, remember the sequence: Copernicus proposed the heliocentric model, Galileo provided observational support, and Kepler refined the model by introducing elliptical orbits and quantitative laws of planetary motion. This clear division of contributions makes it easier to answer historical questions correctly. Final Answer: The demonstration that planets move around the Sun in elliptical orbits was first given by Johannes Kepler.

Question 15

For which of the following adventure sports is an understanding of Pascal law and pressure in liquids especially important for safety and equipment design?

Accepted Answer

Introduction / Context: Pascal law is a fundamental principle in fluid mechanics. It states that pressure applied to an enclosed fluid is transmitted undiminished in all directions. Closely related ideas describe how pressure in a fluid increases with depth according to p = p0 + rho * g * h. These concepts are critical for understanding how pressure changes affect the human body and equipment underwater. This question asks which adventure sport particularly requires awareness of such fluid pressure principles. Given Data / Assumptions: Options include climbing, paragliding, rafting, and scuba diving. Pascal law deals with pressure transmission in liquids. Hydrostatic pressure increases with depth in water. We consider which sport involves significant immersion and pressure changes in water. Concept / Approach: Scuba diving involves a person moving and breathing underwater at different depths. Water pressure on the diver body and on the air inside the scuba tank and breathing equipment is much higher than atmospheric pressure at the surface and increases with depth. Understanding how pressure changes are transmitted through liquids and how they affect the ears, lungs, and other air spaces is essential for safe diving. While climbing, paragliding, and rafting involve air and water flows, they do not require the same detailed knowledge of how pressure in an enclosed liquid environment is transmitted and increases with depth as scuba diving does. Step-by-Step Solution: Step 1: Recall the hydrostatic pressure formula in a liquid: p = p0 + rho * g * h, where h is depth, rho is density, and g is gravitational acceleration. Step 2: As a scuba diver descends, h increases, so the water pressure on the body and equipment rises linearly with depth. Step 3: The compressed air system must balance this external pressure so that the diver can breathe, and pressure changes are transmitted through the liquid according to Pascal related principles. Step 4: Climbing and paragliding mainly involve atmospheric pressure variations with altitude, which are relatively small over typical sport ranges and do not rely heavily on Pascal law in the same way. Step 5: Rafting involves moving on the water surface and dealing with currents and waves, but participants are not usually subjected to large changes in water pressure at depth like scuba divers are. Verification / Alternative check: Diving manuals emphasise the importance of understanding pressure doubling approximately every 10 metres of depth in seawater, explaining how pressure is transmitted through the body and equipment. Concepts of equalisation, decompression sickness, and buoyancy control devices all depend on accurate knowledge of fluid pressure. Adventure sports literature for climbing or paragliding does not highlight Pascal law in the same direct way, focusing instead on friction, aerodynamics, or strength of materials. This confirms that scuba diving is the sport where knowledge of fluid pressure is most critical. Why Other Options Are Wrong: Rock climbing: Involves friction, rope tension, and gravity, but not significant changes in liquid pressure. Paragliding: Deals more with aerodynamics and wind currents than with pressure transmission in liquids. River rafting: Involves surface water flow and turbulence rather than deep immersion and hydrostatic pressure changes. Common Pitfalls: Students may associate Pascal law strictly with hydraulic machines like lifts and brakes and not see its relation to diving. However, the same principles of pressure transmission apply inside divers bodies and equipment. Another pitfall is to choose rafting simply because it involves water. To avoid mistakes, focus on which sport actually involves being surrounded by liquid at depth, where pressure changes with depth are crucial for safety. Final Answer: An understanding of Pascal law and pressure in liquids is especially important in scuba diving.

Question 16

At what temperature (in degree Celsius) do the numerical values on the Celsius and Fahrenheit temperature scales become equal?

Accepted Answer

Introduction / Context: The Celsius and Fahrenheit scales are two commonly used temperature scales. They have different zero points and degree sizes, so the same physical temperature has different numerical values on each scale. Interestingly, there is one temperature at which the Celsius and Fahrenheit readings are numerically equal. This question asks you to determine that temperature in degree Celsius. Given Data / Assumptions: Celsius and Fahrenheit scales are related by a linear conversion formula. The general relation is C = (5 / 9) * (F minus 32), where C is in degree Celsius and F is in degree Fahrenheit. We want the temperature where C and F have the same numerical value. We assume standard definitions of both scales. Concept / Approach: To find the temperature where the two scales coincide numerically, we can set C equal to F in the conversion equation and solve for that common value. This reduces the problem to a simple algebraic equation. The solution will give a temperature that is negative, reflecting the fact that the intersection of the two linear scales occurs below the freezing point of water. Step-by-Step Solution: Step 1: Start with the conversion formula: C = (5 / 9) * (F minus 32). Step 2: At the temperature where the scales have equal numerical values, set C = F. Let this common value be T. Step 3: Substitute C = T and F = T into the formula: T = (5 / 9) * (T minus 32). Step 4: Multiply both sides by 9 to remove the denominator: 9 * T = 5 * (T minus 32). Step 5: Expand the right side: 9 * T = 5 * T minus 160. Bring terms involving T to one side: 9 * T minus 5 * T = minus 160, so 4 * T = minus 160. Divide both sides by 4 to get T = minus 40. Verification / Alternative check: Check the result with direct conversion. If C = minus 40, then F = (9 / 5) * C + 32 = (9 / 5) * (minus 40) + 32 = minus 72 + 32 = minus 40. This confirms that at minus 40 degree Celsius, the Fahrenheit reading is also minus 40 degree. This is the only point where the linear conversion graphs of the two scales intersect. Temperature values such as 0, 100, or 273 would give very different values on the other scale, so minus 40 is unique. Why Other Options Are Wrong: 40: At 40 degree Celsius, Fahrenheit is (9 / 5) * 40 + 32 = 72 + 32 = 104 degree Fahrenheit, not equal. 273: 273 degree Celsius corresponds to a very high Fahrenheit value and is associated with absolute temperature in kelvin, not with equality of the two scales. -273: This is close to absolute zero in degree Celsius, but the Fahrenheit value at this point is not numerically equal to minus 273; instead it is even lower. Common Pitfalls: Some learners try to guess the answer using approximate values or by confusing Celsius with kelvin. Others may incorrectly rearrange the conversion formula or forget to set C equal to F. The safest method is to treat the common temperature as an unknown T and solve the simple linear equation carefully. Doing this step by step leads straightforwardly to T = minus 40 degree. Final Answer: The Celsius and Fahrenheit scales give equal numerical readings at -40 degree Celsius (which is also -40 degree Fahrenheit).

Question 17

Which of the following unit–physical quantity pairs is incorrect?

Accepted Answer

Introduction / Context: In physics, many special units are used for measuring length, volume, and other quantities. Examples include parsec for astronomical distances, barrel for measuring volumes of liquids such as oil, and light year for very large distances in space. This question presents three unit–quantity pairs and asks you to identify which pairing is incorrect. Given Data / Assumptions: Pair I: Parsec – Distance. Pair II: Barrel – Liquid (volume). Pair III: Light year – Time. We assume standard definitions used in astronomy and trade. Concept / Approach: A parsec is an astronomical unit of distance defined based on parallax measurements. A barrel is a unit of volume commonly used in the petroleum industry to measure liquid quantities. A light year, despite having the word year in its name, is defined as the distance light travels in one year in vacuum, not a unit of time. Therefore, the first two pairs correctly associate units with distance and volume, while the third pair incorrectly associates a distance unit with time. So only pair III is incorrect. Step-by-Step Solution: Step 1: Consider pair I: Parsec – Distance. A parsec is defined as the distance at which 1 astronomical unit subtends an angle of 1 arcsecond. This is clearly a measure of distance, so pair I is correct. Step 2: Consider pair II: Barrel – Liquid. A barrel is a traditional unit of volume used for liquids such as crude oil, typically equal to about 159 litres. Although the word liquid is used informally, the underlying quantity is volume, so the pairing is effectively correct. Step 3: Consider pair III: Light year – Time. A light year is defined as the distance that light travels in vacuum in one year. It is therefore a unit of distance, not time. This makes the pairing with time incorrect. Step 4: Since I and II are correct and III is incorrect, only pair III is wrong. Step 5: Among the options given, the statement that only III is incorrect matches this analysis. Verification / Alternative check: Astronomy sources state that 1 parsec is approximately 3.26 light years, emphasizing that both parsec and light year are distance units. Oil industry standards use barrels as a unit of volume for crude oil, again confirming that the second pair relates a volume unit to liquids. No serious scientific reference treats light year as a time unit; instead, time is measured in seconds, years, and other direct time units, while light year is always quoted as a very large length scale. Why Other Options Are Wrong: Only I and III: This would imply parsec is incorrectly matched with distance, which is false because parsec is indeed a distance unit. Only II: This would imply barrel does not relate to measuring liquids, which contradicts common usage in trade and industry. All are correct: This would incorrectly treat light year as a time unit, which is not accepted in physics. Common Pitfalls: Students often get misled by the word year in light year and assume it must measure time. However, the correct interpretation is distance covered by light in one year. A helpful way to remember this is to note that the speed of light is in units of distance per time, so multiplying by a time (one year) gives a distance, not another time. Always check whether the name of a unit reflects its actual definition or is just historical terminology. Final Answer: The incorrect unit–physical quantity pair is Only III, because a light year is a unit of distance, not time.

Question 18

The motion of a freely falling body near the Earth surface (neglecting air resistance) is an example of which type of motion?

Accepted Answer

Introduction / Context: Freely falling bodies, such as an object dropped from a height near the Earth surface, are standard examples used in kinematics and dynamics. When air resistance is neglected, their motion is governed almost entirely by gravity, which produces a constant acceleration. Recognising the type of motion helps in applying the correct equations of motion. This question asks you to classify the motion of such a freely falling object. Given Data / Assumptions: The body is falling freely under gravity near the Earth surface. Air resistance and other forces are neglected. The acceleration due to gravity g is taken as constant in magnitude and direction. We are considering motion along a straight vertical line. Concept / Approach: Uniformly accelerated motion is defined as motion in which the acceleration remains constant in magnitude and direction. For a freely falling body near the Earth surface and over moderate height ranges, the gravitational field can be treated as uniform, giving a constant acceleration g downward. Therefore, the velocity of the body changes at a constant rate, making the motion uniformly accelerated. Non uniform acceleration would involve changing acceleration, and constant velocity motion requires zero acceleration, which does not match the situation of a falling object. Step-by-Step Solution: Step 1: Recognise that the only significant force acting on a freely falling body, in the absence of air resistance, is its weight, given by m * g. Step 2: By Newton second law, the acceleration a of the body is F / m = (m * g) / m = g, which is constant in magnitude and directed downward. Step 3: Since acceleration is constant, the velocity of the body changes linearly with time, and displacement follows the standard equations of uniformly accelerated motion, such as v = u + a * t and s = u * t + (1 / 2) * a * t^2. Step 4: Because acceleration is constant and non zero, the motion is not with constant velocity or constant speed. Step 5: Therefore, the motion of the freely falling body is an example of uniformly accelerated motion. Verification / Alternative check: Experimental measurements of falling objects over small height ranges show that the time of fall and distance satisfy the relation s proportional to t^2, which is characteristic of constant acceleration motion. If acceleration were non uniform, this relation would not hold. Also, standard kinematics problems treat free fall with a single constant g value, confirming the use of uniformly accelerated motion equations. Why Other Options Are Wrong: Non-uniformly accelerated motion: This would mean acceleration changes with time or position, which is not assumed for simple free fall near the Earth surface. Motion with constant velocity: A constant velocity requires zero acceleration, but free fall clearly involves a non zero downward acceleration g. Motion with constant speed in any direction: The speed of a freely falling body increases as it falls, so the speed is not constant. Common Pitfalls: Students sometimes confuse uniformly accelerated motion with uniform motion. Uniform motion has constant velocity, while uniformly accelerated motion has constant acceleration but changing velocity. Another source of confusion is the effect of air resistance; in realistic situations, acceleration may decrease as speed increases, but in introductory physics, air resistance is neglected to highlight the ideal case of constant acceleration g. Final Answer: The motion of a freely falling body near the Earth surface (neglecting air resistance) is an example of uniformly accelerated motion.

Question 19

The contact force that resists the relative sliding motion between two surfaces in contact is commonly known as which of the following?

Accepted Answer

Introduction / Context: Forces in mechanics can be broadly classified as contact forces, which act when bodies are in physical contact, and non contact forces, which act at a distance. When two surfaces touch and there is a tendency for one to slide over the other, a particular contact force arises that opposes this relative motion. This force plays a crucial role in walking, braking, and the functioning of machines. This question asks you to identify that contact force by name. Given Data / Assumptions: Two solid surfaces are in contact with each other. There is either actual relative sliding or a tendency to slide between the surfaces. The force we are interested in acts along the surfaces and opposes this relative motion. Options include friction, magnetic force, electrostatic force, and muscular force. Concept / Approach: Friction is defined as the contact force that opposes relative motion or the tendency of relative motion between two surfaces in contact. It acts along the surface of contact and is responsible for resistance to sliding. Magnetic and electrostatic forces are non contact forces that act at a distance through fields and do not require surfaces to touch. Muscular force is the force generated by muscles in the human body or animals and can act through contact or via tools but is not specifically the force between surfaces resisting sliding. Therefore, the correct name for the contact force described is friction. Step-by-Step Solution: Step 1: Identify that the force in question arises only when two surfaces are in contact. Step 2: Note that the force acts along the interface between the surfaces and opposes sliding or the tendency to slide. Step 3: Recall that this is exactly how friction is defined in mechanics textbooks. Step 4: Compare with magnetic and electrostatic forces, which can act between objects that are not touching and are therefore not purely contact forces. Step 5: Recognise that muscular force refers to the active force generated by muscles, not the passive resisting force between surfaces. Verification / Alternative check: Everyday examples support this identification. When you push a box along the floor, friction between the box and the floor opposes the motion. When a car brakes, friction between the tyres and the road surface is what slows the car down. In both cases, friction is clearly a contact force between surfaces that resists sliding. Magnetic or electrostatic forces might act in some situations but are not responsible for these typical examples of resistance to motion. Why Other Options Are Wrong: Magnetic force: A non contact force that can act across air or vacuum and does not require surfaces to touch. It is not specifically the force resisting sliding between surfaces. Electrostatic force: Also a non contact force between charged bodies and does not describe the general resistance to sliding between neutral surfaces. Muscular force: Describes the force produced by muscles, such as when you push or pull an object. It is a cause of motion, not the resisting contact force between surfaces. Common Pitfalls: Students may confuse the broad term contact force with friction and think they are always interchangeable. In fact, contact forces include normal reaction, friction, tension, and others. However, when the question specifically mentions resisting sliding motion between surfaces, it is clearly referring to friction. Another misconception is to think of friction as always undesirable, but in many cases such as walking and driving, friction is essential for control and safety. Final Answer: The contact force that resists relative sliding between two surfaces in contact is called friction.

Question 20

In basic units conversion for speed, 1 metre per second (1 m/s) is equal to how many kilometres per hour (km/h)?

Accepted Answer

Introduction / Context: This question tests your understanding of basic unit conversion for speed, a very common topic in physics and aptitude exams. Speed is often given in metre per second (m/s) in physics problems, while in real life and competitive exams it is frequently expressed in kilometre per hour (km/h). Knowing how to convert correctly between these units helps you switch comfortably between theoretical and practical contexts. Given Data / Assumptions: We are given a speed of 1 metre per second (1 m/s). We want to express this same speed in kilometres per hour (km/h). Standard metric relationships are used: 1 kilometre = 1000 metres and 1 hour = 3600 seconds. No approximations are made beyond these exact relationships. Concept / Approach: To convert a speed from m/s to km/h, we must convert metres to kilometres and seconds to hours. Because 1 kilometre has 1000 metres, we divide by 1000 when going from metres to kilometres. Because 1 hour has 3600 seconds, we multiply by 3600 when going from seconds to hours. Combining both effects gives a constant conversion factor between m/s and km/h. This factor is 3.6, so any speed in m/s can be changed to km/h by multiplying by 3.6. We then express 3.6 as a fraction to match the options. Step-by-Step Solution: Step 1: Start with the given speed: v = 1 m/s. Step 2: Convert metres to kilometres. Since 1 kilometre = 1000 metres, 1 metre = 1/1000 kilometre. Step 3: Convert seconds to hours. Since 1 hour = 3600 seconds, 1 second = 1/3600 hour. Step 4: Write 1 m/s in kilometre per hour form: 1 m/s = (1/1000) kilometre / (1/3600) hour. Step 5: Divide by a fraction: (1/1000) / (1/3600) = (1/1000) * (3600/1) = 3600/1000 km/h. Step 6: Simplify 3600/1000 by dividing numerator and denominator by 200: 3600/1000 = 18/5. Step 7: Therefore, 1 m/s equals 18/5 km/h. Verification / Alternative check: A quick mental check is to remember the standard conversion 1 m/s = 3.6 km/h. Now 3.6 can be written as 36/10, which simplifies to 18/5. This matches the detailed calculation above, confirming that 18/5 km/h is correct. Also, approximate values support this: 1 m/s is slightly more than 3 km/h, and 18/5 = 3.6, which is reasonable for walking and running speeds when compared with everyday experience. Why Other Options Are Wrong: 5/16 is much less than 1 and would imply that 1 m/s is slower than 1 km/h, which is not reasonable. 16/5 equals 3.2 km/h, which is close but does not match the exact conversion factor of 3.6. 5/18 is about 0.277 km/h, which is far too small and clearly inconsistent with known speeds. Common Pitfalls: Students often mix up the direction of conversion and either multiply or divide incorrectly by 3.6. A useful memory trick is that km/h is a larger unit for distance but a larger unit for time as well, leading to the factor 3.6 between m/s and km/h. Remember that m/s to km/h always multiplies by 3.6, while km/h to m/s divides by 3.6. Writing out the unit fractions carefully, as shown, helps avoid mistakes in exams. Final Answer: 1 metre per second is equal to 18/5 km/h.

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