Question 1

A free soap bubble in air naturally attains a nearly perfect spherical shape mainly because of which physical effect?

Accepted Answer

Introduction / Context: Soap bubbles are beautiful everyday examples of surface tension at work. When a thin film of soap solution encloses air, the bubble almost always forms a spherical shape if it is free in air. This question asks you to identify the main physical effect responsible for that spherical shape. Given Data / Assumptions: A thin soap film encloses air and forms a free bubble in still air. No strong external disturbances such as wind or collisions are present. Gravity effects on the bubble shape are small for small bubbles. The liquid film has surface tension and some viscosity, and the air exerts pressure. Concept / Approach: Surface tension is the property of a liquid surface that tends to minimise its area due to cohesive forces between molecules. For a given volume of air enclosed by a thin liquid film, the shape that minimises surface area is a sphere. Therefore, surface tension pulls the film into a spherical shape. Inertia refers to resistance to change in motion, not to static shape. Viscosity describes internal friction in the fluid and affects how quickly the shape changes, but not the final equilibrium shape. Pressure differences between inside and outside are related to curvature but alone do not explain why a spherical shape is selected; they are a consequence of surface tension. Step-by-Step Solution: Step 1: Recognise that the bubble encloses a fixed volume of air. Step 2: For that fixed volume, different shapes correspond to different surface areas. Step 3: A sphere is the geometrical shape that has the minimum surface area for a given volume. Step 4: Surface tension tends to pull the liquid film into a shape of minimum surface area to minimise surface energy. Step 5: Thus, the bubble naturally adopts a spherical shape to reduce surface energy. Step 6: Conclude that surface tension is the primary reason for the spherical shape of the soap bubble. Verification / Alternative check: If surface tension is reduced, for example by adding substances that disrupt the liquid surface, bubbles become unstable and may not form properly. Experiments with different liquids show that liquids with higher surface tension form more stable spherical drops and bubbles. Rain droplets in air, oil drops in water and small mercury droplets all tend to spherical shapes when gravity is not dominant, reinforcing the idea that surface tension prefers minimal surface area shapes, which are spheres. Why Other Options Are Wrong: Inertia of the liquid film: Inertia resists changes in motion but does not determine the static equilibrium shape of a bubble. Air pressure inside the bubble only: Pressure difference across the curved surface is related to surface tension and curvature; it is not an independent cause of the spherical shape. Viscosity of the liquid film: Viscosity affects how quickly the film flows and deforms but does not set the final equilibrium shape for a given volume. Common Pitfalls: Students sometimes confuse pressure and surface tension because both appear in the formula for pressure difference across a curved surface. It is important to remember that surface tension is the driving effect that tends to minimise surface area, and the pressure difference is a consequence of this curvature and tension balance. Thinking in terms of energy minimisation makes it easier to associate spherical shapes with surface tension. Final Answer: A soap bubble attains a spherical shape mainly because of the surface tension of the soap solution.

Question 2

Electromagnetic radiations with wavelengths 200 nm, 500 nm and 1000 nm are given. Which of these wavelengths can produce a visual sensation (visible light) in the human eye?

Accepted Answer

Introduction / Context: The electromagnetic spectrum covers a wide range of wavelengths, from very short gamma rays to very long radio waves. The human eye can detect only a narrow band of this spectrum, known as visible light, and radiations outside this band are invisible to us even though they carry energy. This question asks you to determine which of the given wavelengths lie within the visible region and therefore cause visual sensation. Given Data / Assumptions: Three wavelengths are given: 200 nm, 500 nm and 1000 nm. The human eye is sensitive to the visible range of approximately 400 nm to 700 nm. Wavelengths shorter than about 400 nm are ultraviolet (UV). Wavelengths longer than about 700 nm are infrared (IR). Concept / Approach: To solve this, we simply compare each of the given wavelengths with the standard visible range. If a wavelength falls within about 400 nm to 700 nm, it will stimulate the retina and be perceived as light of a particular colour. Wavelengths outside that band will not be seen, even though they may have physical effects such as heating or chemical action. We do not need complicated formulas; only a clear understanding of the approximate visible window. Step-by-Step Solution: Step 1: Recall that visible light spans roughly from 400 nm (violet) to 700 nm (red). Step 2: Compare 200 nm with this range: 200 nm is less than 400 nm, so it lies in the ultraviolet region and is invisible. Step 3: Compare 500 nm with this range: 500 nm lies between 400 nm and 700 nm, so it is visible, typically perceived as greenish light. Step 4: Compare 1000 nm with this range: 1000 nm is greater than 700 nm, so it lies in the infrared region and is invisible. Step 5: Conclude that only 500 nm is within the visible band and can cause visual sensation. Verification / Alternative check: Spectral charts used in physics and biology consistently mark the approximate visible range at 400 to 700 nm, with violet at the short wavelength end and red at the long wavelength end. Ultraviolet radiation, starting below 400 nm, is known for causing sunburn and fluorescence but is invisible. Infrared radiation, starting above about 700 nm, is associated with heat. The given wavelengths of 200 nm and 1000 nm clearly fall into these invisible regions, reinforcing that 500 nm is the only visible wavelength among the options. Why Other Options Are Wrong: 200 nm and 500 nm: Includes 200 nm, which is ultraviolet and does not produce a visual sensation in the eye. 500 nm and 1000 nm: Includes 1000 nm, which is infrared and invisible, so this pair is not correct. 200 nm and 1000 nm: Both wavelengths are outside the visible range, so neither produces a visual sensation. Common Pitfalls: Students sometimes memorise only one end of the visible range and misclassify wavelengths near the boundaries. Others may confuse frequency and wavelength, forgetting that smaller wavelength means higher frequency. A simple way to remember is to associate visible light roughly with 0.4 to 0.7 micrometres (400 to 700 nm). Anything much smaller is UV; anything much larger is IR. Using that rule of thumb quickly identifies 500 nm as visible and the other two as non visible. Final Answer: Among the given wavelengths, 500 nm only lies in the visible range and can produce a visual sensation in the human eye.

Question 3

A copper wire of radius r and length l has resistance R. A second copper wire has radius 2r and the same length l. The two wires are connected in parallel. What is the resultant resistance of this parallel combination in terms of R?

Accepted Answer

Introduction / Context: This problem tests your understanding of how resistance depends on length, area and resistivity of a conductor, and how resistances combine when connected in parallel. It uses two copper wires of different radii but the same length and material. You need to find the equivalent resistance of their parallel combination, expressed in terms of the known resistance R of the thinner wire. Given Data / Assumptions: First copper wire: radius r, length l, resistance R. Second copper wire: radius 2r, same length l, same material (same resistivity). The two wires are connected in parallel across the same potential difference. Copper is assumed to be homogeneous with constant resistivity. Concept / Approach: The resistance of a uniform wire is given by R = rho * l / A, where rho is resistivity, l is length and A is cross sectional area. For cylindrical wires, A = pi * r^2. If radius changes, area and therefore resistance change. For parallel combination of two resistances R1 and R2, the equivalent resistance R_eq is given by 1 / R_eq = 1 / R1 + 1 / R2, or R_eq = (R1 * R2) / (R1 + R2). We first find the resistance of the thicker wire in terms of R, then apply the parallel formula. Step-by-Step Solution: Step 1: Write the resistance of the first wire: R1 = R = rho * l / (pi * r^2). Step 2: For the second wire, the radius is 2r, so its area is A2 = pi * (2r)^2 = 4 * pi * r^2. Step 3: Its resistance R2 is R2 = rho * l / A2 = rho * l / (4 * pi * r^2) = (1/4) * rho * l / (pi * r^2) = R/4. Step 4: Now use the parallel resistance formula: R_eq = (R1 * R2) / (R1 + R2). Step 5: Substitute R1 = R and R2 = R/4 to get R_eq = (R * R/4) / (R + R/4). Step 6: Simplify: numerator = R^2 / 4; denominator = (5R/4). So R_eq = (R^2 / 4) * (4 / 5R) = R / 5. Verification / Alternative check: As a consistency check, the equivalent resistance of two resistors in parallel is always less than the smaller of the two. Here, the smaller resistance is R2 = R/4. Our result R_eq = R/5 is indeed smaller than R/4, so it passes this important sanity check. Additionally, if both wires had the same resistance R, the parallel result would be R/2, so getting a value smaller than R/4 is reasonable because one branch is R and the other is R/4, providing more overall conductance. Why Other Options Are Wrong: 5R: This is much larger than either individual resistance and cannot be the result of a parallel combination, which must be smaller than the smallest branch resistance. 5R/4 or 4R/5: Both are greater than R/4, again impossible for a parallel combination of R and R/4. Such values might arise from incorrect algebra or using a series formula instead of the parallel formula. Common Pitfalls: A common mistake is to confuse series and parallel formulas, accidentally adding resistances instead of using the reciprocal relationship. Another pitfall is failing to square the radius when computing the area, leading to an incorrect factor for the resistance of the thicker wire. Always remember that area is proportional to r^2, so doubling the radius makes the area four times larger and the resistance four times smaller, not just twice smaller. Final Answer: The resultant resistance of the parallel combination of the two copper wires is R/5.

Question 4

On a rainy day, a rainbow is observed in the sky. Which optical phenomenon is primarily responsible for the formation of the different colours in a rainbow?

Accepted Answer

Introduction / Context: Rainbows are among the most striking natural displays of optics. They appear as multicoloured arcs in the sky when sunlight interacts with water droplets in the atmosphere. While several optical processes are involved, one key phenomenon is responsible for splitting white sunlight into its colours and arranging them in order across the arc. This question asks you to identify that primary phenomenon. Given Data / Assumptions: Sunlight is essentially white light, containing a spectrum of wavelengths. Raindrops in the atmosphere act as tiny transparent prisms or lenses. Light enters, travels through and may reflect internally in each raindrop. The observer sees an arc of colours from red to violet. Concept / Approach: White light is a mixture of different wavelengths. When it passes from one medium to another, such as from air into water, different wavelengths refract by slightly different amounts because the refractive index of the medium depends on wavelength. This separation of colours due to wavelength dependent refraction is called dispersion. In a rainbow, dispersion of sunlight by raindrops, combined with refraction and internal reflection, spreads the colours out into a continuous spectrum. Diffraction and thin film interference create other colourful patterns but are not the main mechanism of a rainbow. Step-by-Step Solution: Step 1: Recognise that sunlight entering a raindrop is refracted at the air water boundary. Step 2: Different wavelengths of light experience slightly different refractive indices in water. Step 3: Because of this, red, green and violet rays bend by different angles as they enter and exit the drop. Step 4: This separation of white light into its constituent colours is called dispersion. Step 5: Internal reflection inside the drop sends the dispersed rays back toward the observer. Step 6: The observer perceives an arc of colours, with red on the outside and violet on the inside, due primarily to dispersion. Verification / Alternative check: Prism experiments in school demonstrate dispersion clearly: when a narrow beam of white light passes through a glass prism, it spreads into a spectrum of colours. The behaviour of water droplets is similar in principle. Detailed ray tracing of light through a spherical drop shows that dispersion is necessary to separate colours into distinct angles, while refraction and reflection alone, without wavelength dependence, would not produce a colourful spectrum. Why Other Options Are Wrong: Diffraction of sunlight by raindrops: Diffraction effects occur when light encounters edges or apertures comparable to its wavelength, but they do not primarily create the ordered colour bands of a rainbow. Interference in thin films: This phenomenon creates colour patterns in soap bubbles or oil films, where multiple reflections interfere, but it is not the main cause of a large scale rainbow in the sky. Simple reflection from raindrops: Reflection alone would produce bright spots or glints but would not separate white light into a full spectrum of distinct colours. Common Pitfalls: Learners sometimes lump many optical phenomena together under the general idea of bending or spreading of light and mistakenly pick diffraction or interference. To avoid confusion, remember the classic association: prism plus white light gives a spectrum via dispersion. Rainbows are natural, large scale versions of the prism experiment, so dispersion is the key term to choose. Final Answer: The primary optical phenomenon responsible for a rainbow is dispersion of sunlight into its component colours.

Question 5

An electric motor in a fan or machine converts which form of energy primarily into mechanical energy of rotation?

Accepted Answer

Introduction / Context: Energy conversion is a central idea in physics and everyday technology. Electric motors are devices found in fans, pumps, mixers, washing machines and many other appliances. They take energy in one form and convert it into mechanical energy, usually rotational motion of a shaft. This question asks which initial energy form is converted by an electric motor into mechanical energy. Given Data / Assumptions: An electric motor is used in common appliances such as fans and mixers. The output of the motor is mechanical energy in the form of rotation. The motor is powered by a standard electrical supply or battery. Other forms of energy such as chemical and sound may appear as side effects but are not the primary input. Concept / Approach: Electric motors operate on the principle that a current carrying conductor placed in a magnetic field experiences a mechanical force. When multiple coils are arranged correctly and supplied with electric current, they produce continuous rotation. The energy that drives this process is electrical energy supplied from a power source. Inside the motor, some of this electrical energy is converted into mechanical energy of rotation, while some is lost as heat and a little as sound. Chemical energy is stored in fuels or batteries and must first be converted to electrical energy (in batteries) or mechanical energy (in engines) before use. Step-by-Step Solution: Step 1: Identify what goes into an electric motor: a supply of electrical energy, either from mains or a battery. Step 2: Identify what comes out: mechanical rotation of the shaft, used to drive fans, pumps or wheels. Step 3: Recognise that the key physical principle is the force on a current carrying conductor in a magnetic field. Step 4: As current flows through the motor windings, magnetic forces act on them and produce torque on the rotor. Step 5: This torque performs work and delivers mechanical energy to any connected load. Step 6: Therefore, the primary energy conversion is from electrical energy to mechanical energy. Verification / Alternative check: If the electrical supply to a motor is cut off, the motor quickly slows and stops, showing that it depends on continuous electrical input to sustain mechanical output. Technical specifications for motors always rate them in terms of electrical input power (watts or kilowatts) and mechanical output power. This confirms that the primary energy form entering the device is electrical. Why Other Options Are Wrong: Sound energy: Motors make some sound as a by product of vibration and air movement, but sound is not the energy input; it is a minor output. Mechanical energy: Mechanical energy is the output, not the input, of a motor. An engine driven generator, by contrast, converts mechanical energy into electrical energy. Chemical energy: Batteries store chemical energy and convert it to electrical energy, which then powers the motor; the motor itself takes electrical energy as input. Common Pitfalls: A frequent confusion occurs between batteries and motors; students sometimes say motors convert chemical to mechanical energy, skipping the intermediate electrical step inside the battery. It helps to think in terms of separate devices in an energy chain: chemical energy in fuel or battery to electrical energy in circuits, and then electrical energy to mechanical energy in the motor. Final Answer: An electric motor primarily converts electrical energy into mechanical energy.

Question 6

In acoustics, what is the SI-derived unit commonly used to express sound intensity level (loudness) on a logarithmic scale?

Accepted Answer

Introduction / Context: Sound can be described physically by its intensity (power per unit area) and perceptually by how loud it seems to the human ear. Because the range of sound intensities we can hear is extremely large, acousticians usually work with a logarithmic scale called sound intensity level or sound pressure level. This question asks for the commonly used SI derived unit for this sound level quantity. Given Data / Assumptions: We are concerned with sound intensity level or loudness as normally measured in practice. Sound intensity level is defined using a logarithmic ratio of intensity to a reference value. The question refers to the SI context but focuses on the practical unit. Candidate units include decibel, newton, hertz and tesla. Concept / Approach: Pure physical intensity of sound is measured in watt per square metre (W/m^2). However, to match human hearing and handle the large dynamic range, sound level is defined as L = 10 * log10(I / I0) for intensity or L = 20 * log10(p / p0) for pressure, where I0 and p0 are reference values. The resulting quantity is expressed in bels, and more commonly in decibels (dB), where 1 bel = 10 decibels. Decibel is thus the standard practical unit for sound level in the SI system. Newton is a unit of force, hertz is a unit of frequency and tesla is a unit of magnetic flux density. Step-by-Step Solution: Step 1: Recall that sound level is often quoted as, for example, 60 dB, 90 dB or 120 dB. Step 2: Understand that dB stands for decibel, a logarithmic unit derived from the SI system. Step 3: Recognise that decibel is dimensionless but is tied to SI base units through the intensity or pressure ratio definition. Step 4: Note that newton (N) is the SI unit of force, not of sound level. Step 5: Note that hertz (Hz) measures frequency, such as pitch of a sound, not its loudness. Step 6: Note that tesla (T) is the unit of magnetic field strength, unrelated to acoustics. Verification / Alternative check: Any standard noise regulation document or sound level meter specification will list readings in decibels, often with weightings such as dB(A) to approximate the ear frequency response. Environmental noise limits, workplace safety recommendations and headphone volume guidelines are all specified in dB. This widespread usage confirms that decibel is the conventional unit for sound intensity level. Why Other Options Are Wrong: Newton (N): Measures force (kg·m/s^2), used in mechanics, not for sound loudness. Hertz (Hz): Measures frequency (cycles per second), relevant to pitch, not to how loud a sound is. Tesla (T): Measures magnetic flux density in electromagnetism, not an acoustic quantity. Common Pitfalls: Students sometimes confuse the physical quantity intensity (W/m^2) with the practical measure of loudness (dB). It is important to distinguish between the base SI unit used in formulas and the logarithmic unit used in real world sound measurement. When a question mentions sound intensity level, loudness ratings or decibel values, decibel (dB) is the correct unit to choose. Final Answer: The SI derived unit commonly used for sound intensity level is the decibel (dB).

Question 7

A ball is thrown vertically upwards in the Earth's gravitational field. Neglecting air resistance, which of the following quantities remains constant during its entire upward and downward motion?

Accepted Answer

Introduction / Context: Vertical motion under gravity is a classic topic in kinematics. When a ball is thrown straight up and then falls back down, several physical quantities change with time, such as velocity and height. However, some quantities remain constant under ideal conditions. This question asks you to identify which of the listed quantities stays constant throughout the motion when air resistance is neglected. Given Data / Assumptions: A ball is projected vertically upward from near the Earth surface. Only the gravitational force acts on the ball; air resistance is neglected. The acceleration due to gravity g is taken as constant near the Earth surface. We consider the entire journey: going up, momentary rest at the top and coming down. Concept / Approach: Under these assumptions, the only force acting on the ball is its weight, mg, directed downward. According to Newton second law, this leads to a constant downward acceleration of magnitude g (about 9.8 m/s^2). The velocity of the ball changes continuously because the acceleration is non zero: it decreases on the way up, is zero at the highest point and increases downward on the way down. The displacement from the starting point clearly changes as the ball moves. Total mechanical energy is conserved if we ignore air resistance, but its distribution between kinetic and potential changes. Among the options, acceleration due to gravity is the simplest quantity that remains strictly constant. Step-by-Step Solution: Step 1: Identify forces: only weight mg acts on the ball (downward). Step 2: Apply Newton second law: net force = m * a, so m * a = m * g downward, giving constant acceleration a = g downward. Step 3: On the way up, velocity decreases linearly with time because of this constant downward acceleration. Step 4: At the highest point, velocity becomes zero momentarily, but acceleration remains g downward. Step 5: On the way down, velocity increases in magnitude downward, again under the same constant acceleration. Step 6: Therefore acceleration due to gravity is constant throughout the motion, while velocity and displacement change. Verification / Alternative check: Equations of motion under constant acceleration, such as v = u + a * t and s = u * t + (1/2) * a * t^2, are derived assuming a is constant. These equations accurately describe vertical projectile motion near the Earth, confirming that g is treated as constant over moderate heights. Graphs of velocity versus time for such motion are straight lines with slope equal to g, again indicating constant acceleration. Why Other Options Are Wrong: Total mechanical energy: Although total mechanical energy is conserved if air resistance is neglected, its value remains constant, but the question at this level usually highlights acceleration as the explicitly constant quantity; many curricula place emphasis on g as constant. Displacement: The displacement from the starting point changes continuously as the ball moves upward and then downward. Velocity: Velocity changes in both magnitude and direction; it is positive (upward), zero at the top and negative (downward). Common Pitfalls: Students often think velocity is constant because they confuse it with uniform motion. Another confusion is between speed and acceleration; some imagine that if acceleration is constant, speed must be constant, which is not true. Constant acceleration means velocity changes at a uniform rate, not that it stays the same. Final Answer: The quantity that remains constant during the motion is the acceleration due to gravity.

Question 8

A ball rolling along the ground gradually slows down and finally comes to rest. Which force is mainly responsible for this loss of motion?

Accepted Answer

Introduction / Context: When an object such as a ball is set rolling on the ground, it does not continue forever. Instead, it gradually slows down and eventually comes to rest. Understanding why motion dies out in everyday situations is an important part of learning about forces, friction and Newton laws. This question asks which force is primarily responsible for the rolling ball losing its motion. Given Data / Assumptions: A ball is rolling along a horizontal ground surface. The ball was set in motion earlier, perhaps by a kick or push. No one continues to push the ball while it is rolling. We neglect exotic forces and focus on common forces like friction and contact forces. Concept / Approach: According to Newton first law, an object in motion continues in a straight line at constant speed unless acted on by a net external force. In ideal physics problems, a frictionless surface would allow the ball to roll indefinitely. In reality, contact between the ball and the ground is not perfectly smooth, and there is rolling friction and possibly some sliding friction. These frictional forces oppose the motion of the ball and convert its mechanical energy into heat and a little sound, causing it to slow down and eventually stop. Magnetic and electrostatic forces are usually negligible in this everyday context, and the muscular force from the person acts only briefly at the start. Step-by-Step Solution: Step 1: Recognise that once the ball has left the player foot or hand, no continuous muscular force is applied. Step 2: Identify forces acting during rolling: weight of the ball downward, normal reaction upward and friction at the contact with the ground. Step 3: The vertical forces weight and normal mostly balance each other, so they do not slow the ball horizontally. Step 4: The frictional force acts opposite to the direction of motion at the point of contact. Step 5: This frictional force does negative work on the ball, reducing its kinetic energy. Step 6: As kinetic energy decreases, the ball speed reduces until it eventually comes to rest. Verification / Alternative check: On very smooth surfaces like ice, a puck or ball can slide or roll much farther before stopping, showing that when friction is reduced, motion lasts longer. On rough surfaces like carpet or grass, the ball stops more quickly, which confirms that friction is the key factor controlling how fast the ball slows down. There is no noticeable magnetic or electrostatic interaction in these everyday rolling situations. Why Other Options Are Wrong: Magnetic force from the Earth: Earth magnetic field is too weak to significantly affect an ordinary non magnetised ball rolling on the ground. Electrostatic force: Any static charges on the ball or ground are negligible compared with mechanical contact forces in this context. Muscular force: The muscular force acts only during the initial push or kick; after that, it no longer acts on the ball. Common Pitfalls: Some learners incorrectly think that a continuous force is necessary to keep an object moving, contradicting Newton first law. This misconception leads them to believe that the ball slows because the push runs out. In reality, an ideal force free ball would go on forever; it is friction that removes kinetic energy and stops the ball. Remember that in everyday life, friction is almost always present and is the main cause of objects coming to rest. Final Answer: The ball slows down and stops mainly because of the frictional force between the ball and the ground.

Question 9

Friction between two solid surfaces in contact arises mainly because of what microscopic property of those surfaces?

Accepted Answer

Introduction / Context: Friction is a common force that opposes motion between surfaces that are in contact. Although surfaces may look smooth to the naked eye, at the microscopic level they are not perfectly flat. Understanding the microscopic origin of friction helps explain why rough surfaces have higher friction and why lubrication reduces it. This question asks which microscopic property of surfaces is mainly responsible for frictional forces. Given Data / Assumptions: Two solid surfaces are in contact and one tends to move or slides over the other. We are dealing with ordinary materials like wood, metal, rubber or glass. There is no special magnetic or electrostatic effect being considered. We focus on the physical nature of the contact between surfaces. Concept / Approach: Even polished surfaces are made of atoms and molecules and therefore have microscopic peaks and valleys known as asperities. When two surfaces are pressed together, the asperities on one surface interlock with those on the other, creating many tiny contact points. As one surface tries to move, these interlocking irregularities resist motion, producing friction. Additional contributions can come from molecular adhesion and deformation, but irregularities provide the basic intuitive picture. If surfaces were perfectly smooth and there were no other interactions, friction would be much smaller. Step-by-Step Solution: Step 1: Consider a supposedly smooth surface under a microscope; it reveals tiny bumps and hollows. Step 2: When two such surfaces are pressed together, only the highest points asperities actually make contact. Step 3: These contact points may interlock or deform slightly, resisting relative motion. Step 4: As one surface attempts to slide, it must continually break these micro contacts and form new ones, which requires force. Step 5: The cumulative effect of many such microscopic interactions is the macroscopic friction force. Step 6: Therefore, friction is mainly due to irregularities and roughness of the surfaces in contact. Verification / Alternative check: Experiments show that polishing a surface and making it smoother generally reduces friction, which is consistent with the idea of fewer and smaller asperities. Lubricating surfaces with oil or grease fills the microscopic gaps and separates the solid surfaces with a thin liquid layer, greatly reducing direct contact of irregularities and thus reducing friction. This behaviour supports the explanation based on roughness and irregularities. Why Other Options Are Wrong: Extreme smoothness of the surfaces: Greater smoothness tends to reduce friction rather than cause it, except in special cases where adhesion dominates. High densities of the materials only: Density affects mass and inertia but does not directly determine the friction force between surfaces. Large gaps with no contact: If there were large gaps and no actual contact, friction would be nearly zero; real friction requires contact at asperities. Common Pitfalls: Some students imagine friction as a mysterious force that arises simply whenever two objects touch, without considering microscopic structure. Others think friction depends only on roughness and forget that normal force and material properties also matter. While the full theory of friction is complex, at the school level it is reasonable and helpful to say that friction arises mainly from interlocking irregularities on surfaces. Final Answer: Friction is mainly caused by irregularities and roughness on the two surfaces in contact.

Question 10

During which type of motion of an object along a straight line does its velocity remain constant with time?

Accepted Answer

Introduction / Context: Kinematics distinguishes between different types of motion based on how position, velocity and acceleration change with time. When we say that velocity remains constant, we mean both speed and direction do not change. For motion along a straight line, direction is fixed, so constant velocity means constant speed in that direction. This question asks which named type of motion along a straight line corresponds to constant velocity. Given Data / Assumptions: The object moves along a straight line path. We compare various descriptive terms for motion. Velocity is considered as a vector quantity, including both magnitude and direction. Constant velocity means no change in speed or direction with time. Concept / Approach: Uniform motion along a straight line is defined as motion in which the object covers equal displacements in equal intervals of time, no matter how small those intervals are. This is equivalent to saying that its velocity is constant. Linear motion alone merely indicates straight line motion, not whether speed is constant or changing. Translational motion is a general term for any motion in which all parts of a body move in the same way and can still be accelerated. Equilibrium refers to net force being zero; in a straight line, that condition implies constant velocity, but the safest and most precise term among the options is uniform motion. Step-by-Step Solution: Step 1: Recall that velocity remains constant only when there is no acceleration acting on the object. Step 2: For straight line motion, this means the object covers equal distances in equal time intervals. Step 3: This behaviour matches the definition of uniform motion along a straight line. Step 4: Linear motion simply describes the path as straight but does not specify whether speed is uniform. Step 5: Translational motion is broader and may include accelerating or decelerating motion. Step 6: Therefore, the correct term for straight line motion with constant velocity is uniform motion. Verification / Alternative check: If you plot position versus time for uniform motion, you obtain a straight line graph with constant slope. The slope represents velocity, and its constancy shows that velocity is unchanging. In contrast, non uniform motion yields curved position time graphs, indicating changing velocity. This graphical perspective confirms that uniform motion is the appropriate description for constant velocity along a straight line. Why Other Options Are Wrong: Linear motion of any kind: Includes both uniform and accelerated motion; it does not guarantee constant velocity. Translational motion with changing speed: By definition involves changing velocity, so it contradicts the idea of constant velocity. Motion in equilibrium only: While dynamic equilibrium can imply constant velocity, the question asks for a standard term for the motion itself, which is best described as uniform motion. Common Pitfalls: Students sometimes confuse the words linear and uniform, thinking that a straight path automatically means constant speed. In reality, a car can speed up or slow down while still moving straight. The key word that signals constant velocity is uniform. Always link uniform motion with the idea of equal distances in equal times and zero acceleration. Final Answer: Velocity remains constant during uniform motion along a straight line.

Question 11

Which law states that internal energy is a state function and that the increase in internal energy of a system equals the heat supplied to the system plus the work done on (or equivalently, minus the work done by) the system?

Accepted Answer

Introduction / Context: Thermodynamics deals with heat, work and internal energy. One of its foundational principles connects these quantities and expresses the law of conservation of energy for thermodynamic systems. This principle states that any change in the internal energy of a system results from the heat added to the system and the work done on or by the system. This question asks you to identify the named law that encapsulates this relationship. Given Data / Assumptions: A thermodynamic system exchanges heat and work with its surroundings. Internal energy is treated as a state function depending on variables like temperature and volume. The change in internal energy is related to heat supplied and work done. We use the sign convention that increase in internal energy equals heat supplied plus work done on the system (or heat supplied minus work done by the system). Concept / Approach: The first law of thermodynamics is essentially a statement of energy conservation in thermodynamic processes. In differential form, it can be expressed as dU = dQ - dW, where dU is the change in internal energy, dQ is the heat supplied to the system and dW is the work done by the system. Alternatively, with a different sign convention for work, it may be written as dU = dQ + dW_on. The crucial idea is that internal energy is a state function and its change depends only on initial and final states, while heat and work are path dependent. Step-by-Step Solution: Step 1: Identify the key statement: increase in internal energy equals heat supplied plus or minus work term, depending on sign convention. Step 2: Recognise that this is a direct expression of energy conservation applied to thermodynamic systems. Step 3: Recall that the law saying stress is proportional to strain within elastic limit is Hooke's law, not related to heat and work. Step 4: Recall that Coulomb's law describes the electrostatic force between point charges, again unrelated to internal energy changes. Step 5: Faraday's law deals with electromagnetic induction, relating induced emf to the rate of change of magnetic flux. Step 6: Therefore, the only option that matches the described heat work energy relationship is the first law of thermodynamics. Verification / Alternative check: Examples from thermodynamics courses show that for an isochoric process (constant volume) where no work is done, the change in internal energy equals the heat supplied. For isothermal processes of ideal gases, internal energy change is zero, so heat supplied equals work done by the gas. These applications all follow directly from the first law and confirm its role as the energy conservation rule in thermodynamic contexts. Why Other Options Are Wrong: Hooke's law: States that within the elastic limit of a material, stress is proportional to strain; it does not involve internal energy, heat or work. Coulomb's law: Gives the magnitude of electrostatic force between two charges and depends on distance and charge values, not on thermodynamic variables. Faraday's law: Relates induced emf in a circuit to the time rate of change of magnetic flux through it, unrelated to heat and internal energy in a thermodynamic sense. Common Pitfalls: Students sometimes mix up the numerical forms of gas laws and the first law. It is useful to remember that the first law always links three central quantities internal energy, heat and work and is a qualitative and quantitative statement of energy conservation. The presence of these three key words in the statement is a strong clue that the answer is the first law of thermodynamics. Final Answer: The statement describes the first law of thermodynamics.

Question 12

In vector form, torque τ on a particle is defined as τ = r × F, where r is the position vector and F is the applied force. In which direction does the torque vector point relative to r and F?

Accepted Answer

Introduction / Context: Torque (or moment of force) is a vector quantity that measures the turning effect of a force about a point or axis. In three dimensional mechanics, torque is defined using a vector cross product, τ = r × F, where r is the position vector from the axis to the point of application of the force, and F is the force vector. This question asks you to identify the direction of the torque vector relative to r and F. Given Data / Assumptions: The torque vector τ is defined by the cross product τ = r × F. r and F lie in some plane through the origin or pivot point. We use the standard right-hand rule convention for cross products. We are interested in the direction of τ, not its magnitude. Concept / Approach: The cross product of two vectors a × b is a vector that is perpendicular to the plane containing a and b. Its direction is given by the right-hand rule: if you point the fingers of your right hand along a and curl them toward b, your thumb points in the direction of a × b. Thus, for τ = r × F, the torque vector is perpendicular to both r and F and points along the axis about which rotation tends to occur. It is not parallel to the force, nor opposite to it, and is not generally parallel to the position vector r. Step-by-Step Solution: Step 1: Write the vector definition τ = r × F. Step 2: Recall the geometric rule for a cross product: the result is perpendicular to the plane containing the two vectors. Step 3: Visualise r and F lying in a plane; the torque vector must then come out of or go into this plane. Step 4: Apply the right-hand rule: align the fingers of your right hand with r and curl them toward F. Step 5: The thumb then points in the direction of τ, perpendicular to both r and F. Step 6: Conclude that the correct description is that τ is perpendicular to the plane containing r and F, given by the right-hand rule. Verification / Alternative check: In rotational dynamics around a fixed axis, we often treat torque as pointing along the axis of rotation. For example, when you turn a screw, the force is applied tangentially, the radius extends from the axis to the point of application and the torque vector points along the screw axis. This is consistent with the cross product direction being perpendicular to the plane of r and F. Problems in textbooks that use right-handed coordinate systems also show torque vectors along axes, further confirming this relationship. Why Other Options Are Wrong: Parallel to the applied force F: If torque were parallel to F, it would not correctly represent the sense of rotation relative to the axis defined by r and F; the cross product definition contradicts this. Exactly opposite to F: Torque is not defined as a negative of the force vector; its direction is determined by the perpendicular to the plane of r and F. Always parallel to r: In many simple problems r and F are perpendicular, but τ is still perpendicular to both; it does not align with r except in special degenerate cases. Common Pitfalls: Students sometimes confuse the direction of torque with that of force because both are involved in rotational motion. Another confusion is between the scalar magnitude of torque r * F * sin(θ) and the vector direction. To avoid these mistakes, always remember that the cross product provides both magnitude and a direction perpendicular to the plane of the vectors, and the right-hand rule is the standard tool to find that direction. Final Answer: The torque vector points perpendicular to the plane containing r and F, as given by the right-hand rule.

Question 13

Find the resistance, in megaohms (MΩ), of a wire of length 20 m and cross-sectional area 1 cm² made of a material whose resistivity is 200 Ω·m.

Accepted Answer

Introduction / Context: This question checks your understanding of the relationship between resistance, resistivity, length and cross sectional area of a conductor. It also tests your ability to handle unit conversions, particularly from square centimetres to square metres and from ohms to megaohms. The goal is to compute the resistance of a wire using the standard formula R = rho * l / A. Given Data / Assumptions: Resistivity of the material, rho = 200 Ω·m. Length of the wire, l = 20 m. Cross sectional area, A = 1 cm². We are asked for resistance in megaohms (MΩ), where 1 MΩ = 10^6 Ω. The wire is uniform, with constant cross sectional area and resistivity. Concept / Approach: The resistance R of a uniform conductor is given by R = rho * l / A. To use this formula consistently in SI units, resistivity should be in Ω·m, length in metres and area in square metres. Therefore, we must convert 1 cm² to m². Once we compute R in ohms, we then convert that value into megaohms by dividing by 10^6. Step-by-Step Solution: Step 1: Convert the area from cm² to m². Since 1 cm = 10^-2 m, 1 cm² = (10^-2 m)² = 10^-4 m². So A = 1 * 10^-4 m². Step 2: Write the resistance formula: R = rho * l / A. Step 3: Substitute the values: R = 200 Ω·m * 20 m / (1 * 10^-4 m²). Step 4: Multiply the numerator: 200 * 20 = 4000 Ω·m². Step 5: Divide by 10^-4 m²: R = 4000 / 10^-4 Ω = 4000 * 10^4 Ω = 4 * 10^7 Ω. Step 6: Convert ohms to megaohms: R = 4 * 10^7 Ω / 10^6 = 40 MΩ. Verification / Alternative check: You can check the order of magnitude: a long wire (20 m) of relatively high resistivity (200 Ω·m) and very small cross sectional area (1 cm² is 10^-4 m²) should have a large resistance. The formula R = rho * l / A shows that if the area is tiny and resistivity and length are large, R should be very large, in the tens of megaohms range. Our answer 40 MΩ is consistent with this expectation and matches option 40. Why Other Options Are Wrong: 4000: This would correspond to 4000 MΩ if read directly or 4000 Ω if interpreted wrongly, both inconsistent with the calculation based on correct units. 80 and 2000: These values do not match the computed 40 MΩ and likely result from incorrect handling of the 10^-4 factor or from forgetting to convert to megaohms. Common Pitfalls: A frequent mistake is to treat 1 cm² as 10^-2 m² instead of 10^-4 m², forgetting that the square applies to the whole conversion factor. Another common error is to forget the final conversion to megaohms and leave the answer in plain ohms, or to incorrectly multiply instead of divide by 10^6. Always keep track of units at each step and remember that 1 MΩ = 10^6 Ω. Final Answer: The resistance of the wire is 40 megaohms (40 MΩ).

Question 14

Air movement (wind) occurs as air flows from one region to another. In general, air moves from the region where the air ______ to the region where the air ______.

Accepted Answer

Introduction / Context: Atmospheric motion, including winds and breezes, is largely driven by differences in air pressure between regions. Understanding the basic rule for how air moves helps explain weather patterns, sea breezes, land breezes and the formation of storms. This question asks you to identify the correct direction of air flow in terms of high and low pressure regions. Given Data / Assumptions: We compare regions of the atmosphere with different air pressures. Other factors like Earth rotation and terrain can modify wind patterns, but the basic driving force is pressure difference. We assume air can move freely from one region to another. Humidity differences may also exist but are not the primary driver of bulk air flow. Concept / Approach: According to basic fluid dynamics, air tends to move from regions of higher pressure to regions of lower pressure. The difference in pressure creates a net force on the air, pushing it from high pressure towards low pressure, in an attempt to equalise pressure differences. This is analogous to water flowing downhill from high potential to low potential. Humidity differences influence density and can indirectly affect pressure, but the direct cause of wind is the pressure gradient. Step-by-Step Solution: Step 1: Recall that pressure is force per unit area exerted by air molecules in random motion. Step 2: A region of higher pressure has more air molecules or higher temperature compared with a low pressure region. Step 3: The pressure difference across regions creates a net pushing force on air at the boundary. Step 4: Air is accelerated from the region of higher pressure toward the region of lower pressure. Step 5: This movement continues until pressure differences are reduced or balanced by other forces, such as the Coriolis effect. Step 6: Therefore, the correct general rule is that air moves from high pressure regions to low pressure regions. Verification / Alternative check: Weather maps showing isobars lines of equal pressure often illustrate winds crossing from areas of high pressure toward low pressure areas, though deflected by Earth rotation. Sea breeze and land breeze phenomena are explained by pressure differences created by unequal heating of land and sea. In every case, air movement is traced back to flows from higher pressure zones toward lower pressure zones. Why Other Options Are Wrong: Humidity is high to low humidity: While humidity affects density and can influence pressure, air does not directly move simply from high humidity to low humidity; pressure gradients remain the primary driver. Pressure is low to high pressure: This is opposite to what is observed; air cannot spontaneously move from low pressure into high pressure without an external force. Humidity is low to high humidity: Again, humidity gradients alone do not control the basic direction of wind; pressure gradients do. Common Pitfalls: Some students confuse pressure with density or temperature and attempt to reason from hot to cold or humid to dry regions. While temperature and humidity differences often create pressure differences, the simplest rule is that air responds directly to pressure gradients. Always focus on which region has higher pressure when deciding the basic direction of air movement. Final Answer: Air generally moves from the region where pressure is high to the region where pressure is low.

Question 15

The demand curve for Alphonso mangoes is D = 25000 - 70P and the supply curve is S = 10000 + 80P, where P is price and D and S are quantities. What is the equilibrium quantity (in units) where demand equals supply?

Accepted Answer

Introduction / Context: Even though this question appears under physics, it is actually a simple microeconomics problem about market equilibrium. The demand and supply of Alphonso mangoes are given as linear functions of price. The equilibrium point in a competitive market occurs where the quantity demanded equals the quantity supplied. This question asks you to find the corresponding equilibrium quantity. Given Data / Assumptions: Demand function: D = 25000 - 70P. Supply function: S = 10000 + 80P. P represents the price per unit (exact currency units are not specified). Equilibrium occurs when D = S. We assume all quantities are in the same units (for example, kilograms or number of units). Concept / Approach: In a simple linear demand-supply model, equilibrium is found by solving the equation D = S for the price P and then substituting this price back into either the demand or supply function to obtain the equilibrium quantity. Algebraically, this becomes solving a single linear equation in one variable. Careful handling of signs is important because demand typically slopes downward with price (negative coefficient for P), while supply slopes upward (positive coefficient for P). Step-by-Step Solution: Step 1: Set demand equal to supply for equilibrium: 25000 - 70P = 10000 + 80P. Step 2: Bring like terms together: 25000 - 10000 = 80P + 70P. Step 3: Simplify the left side: 15000 = 150P. Step 4: Solve for price: P = 15000 / 150 = 100. Step 5: Substitute P = 100 into the demand function to get equilibrium quantity: D = 25000 - 70 * 100. Step 6: Compute D: 70 * 100 = 7000, so D = 25000 - 7000 = 18000 units. This equals the supply at equilibrium as well. Verification / Alternative check: Check with the supply equation: S = 10000 + 80 * 100 = 10000 + 8000 = 18000 units, which matches the demand at price 100. Since both demand and supply give 18000 units at this price, the equilibrium quantity is confirmed as 18000 units. Additionally, the demand curve slope is negative (coefficient -70) and the supply curve slope is positive (coefficient +80), so they intersect exactly once, giving a unique equilibrium. Why Other Options Are Wrong: 2,000 units, 1,800 units and 200 units: These values do not satisfy both the demand and supply equations at any common price. They likely result from algebraic errors such as misplacing zeros, ignoring the minus sign in the demand equation or miscomputing the product 70 * 100. Common Pitfalls: A common mistake is to forget the minus sign in the demand equation and treat it as D = 25000 + 70P, which would give an incorrect equilibrium. Another frequent error is to solve for quantity directly without first solving for price, leading to confusion between variables. To avoid this, always set D = S, solve for P carefully and then use that value to compute the equilibrium quantity. Final Answer: The equilibrium quantity of Alphonso mangoes is 18,000 units.

Question 16

In electricity and electronics, the farad (F) is the SI unit of which physical quantity?

Accepted Answer

Introduction / Context: Many physical quantities in electricity have specific SI units, such as coulomb for charge, ohm for resistance and farad for another important quantity. Recognising these units is essential for understanding circuit diagrams, component ratings and formulas. This question focuses on the farad, denoted by the symbol F, and asks which quantity it measures. Given Data / Assumptions: The farad (F) is an SI derived unit used in circuit analysis. Other related quantities include electric charge, conductance and reactance. We assume conventional definitions from basic physics and electronics. Capacitors in circuits are often marked with values such as microfarads (µF) or picofarads (pF). Concept / Approach: Capacitance is the property of a system that describes its ability to store electric charge per unit potential difference. For a simple capacitor, capacitance C is defined as C = Q / V, where Q is the magnitude of charge stored and V is the potential difference across the plates. The SI unit of charge is coulomb (C), and the SI unit of potential difference is volt (V), so the unit of capacitance is coulomb per volt, which is named the farad. Reactance, electric charge and conductance each have their own distinct units. Step-by-Step Solution: Step 1: Recall the definition of capacitance: C = Q / V. Step 2: Identify that Q is measured in coulombs and V in volts, so C has units of coulomb per volt. Step 3: Recognise that this derived unit is named the farad (F). Step 4: Compare with electric charge, whose unit is coulomb (C), not farad. Step 5: Recall that conductance, the reciprocal of resistance, has the unit siemens (S). Step 6: Note that reactance, like resistance, is measured in ohms (Ω), not in farads. Verification / Alternative check: Look at any capacitor data sheet or the markings on a small capacitor. You will typically see values like 100 nF, 10 µF or 220 pF. All of these abbreviations refer to nanofarads, microfarads and picofarads, confirming that capacitance is measured in farads. Circuit analysis formulas such as Q = C * V and energy stored in a capacitor E = (1/2) * C * V^2 further confirm this relationship between capacitance, charge and voltage. Why Other Options Are Wrong: Reactance: This is the opposition offered by a capacitor or inductor to alternating current and, like resistance, is measured in ohms (Ω). Electric charge: Charge is measured in coulombs (C), not in farads. Electric conductance: Conductance has the SI unit siemens (S), which is the reciprocal of ohm. Common Pitfalls: New learners sometimes mix up farad with coulomb because both appear in the formula C = Q / V. It helps to remember that C is the symbol for both charge and capacitance in different contexts, so you must pay attention to the formula and units. The farad is often a large unit in practice, which is why actual capacitors are rated in microfarads or nanofarads rather than whole farads. Final Answer: The farad (F) is the SI unit of capacitance.

Question 17

In electromagnetism, what is the SI unit of the physical quantity "inductance"?

Accepted Answer

Introduction / Context: Inductance is an important concept in electromagnetism and circuit theory, especially in circuits containing coils and inductors. It measures the ability of a conductor or coil to oppose changes in current by inducing an electromotive force (emf). The SI system assigns a specific unit to inductance, named after a famous scientist. This question asks you to identify that unit. Given Data / Assumptions: Inductance appears in formulas relating magnetic flux, current and induced emf. The symbol for inductance is usually L. Other magnetic quantities include flux (measured in weber) and flux density (measured in tesla). Capacitance has the unit farad. Concept / Approach: For a coil, inductance L is defined by the relationship Φ = L * I, where Φ is the magnetic flux linked with the coil and I is the current through it. Inductance also appears in Faraday law of induction, where the magnitude of induced emf is proportional to the rate of change of current multiplied by inductance. The SI unit of inductance is the henry (H). One henry is defined as the inductance of a circuit in which an emf of one volt is induced when the current changes at the rate of one ampere per second. Step-by-Step Solution: Step 1: Recall the definition Φ = L * I, so L = Φ / I. Step 2: The unit of magnetic flux Φ is the weber (Wb), and the unit of current is ampere (A). Step 3: Therefore, the unit of L is Wb/A. Step 4: This derived unit is given the special name henry (H). Step 5: Compare to farad (F), which is the unit of capacitance, not inductance. Step 6: Tesla (T) measures magnetic flux density (Wb/m²), and weber alone measures magnetic flux, not inductance. Verification / Alternative check: Inductor specifications in circuit diagrams and data sheets use values like 10 mH, 100 µH or 1 H, where mH and µH stand for millihenry and microhenry respectively. Formulas such as energy stored in an inductor E = (1/2) * L * I^2 explicitly show L carrying units of henry when current is in amperes and energy in joules, which is consistent with the definition. Why Other Options Are Wrong: Weber (Wb): This is the unit of magnetic flux, not of inductance. Farad (F): This is the unit of capacitance, used for capacitors, not for inductors. Tesla (T): This is the unit of magnetic flux density (Wb/m²), not of inductance. Common Pitfalls: Students sometimes confuse magnetic flux (weber) with inductance (henry) because both occur in similar formulas, or they mix up henry and farad since inductors and capacitors often appear together in AC circuits. A good way to remember is to associate henry with inductors and coils, and farad with capacitors and stored charge. Final Answer: The SI unit of inductance is the henry (H).

Question 18

A body of mass 5 kg accelerates uniformly from 12 m/s to 20 m/s in 4 seconds due to a constant net force. What is the magnitude of this force (in newtons)?

Accepted Answer

Introduction / Context: This problem uses Newton second law of motion, which relates the net force acting on a body to the product of its mass and acceleration. The body speed changes from 12 m/s to 20 m/s over a specified time interval, so we can find its acceleration and then determine the force causing this acceleration. It is a straightforward application of kinematics and dynamics. Given Data / Assumptions: Mass of the body, m = 5 kg. Initial velocity, u = 12 m/s. Final velocity, v = 20 m/s. Time interval, t = 4 s. The net force is constant, producing uniform acceleration. Motion is along a straight line. Concept / Approach: Acceleration a is defined as the rate of change of velocity, a = (v - u) / t. Once we compute acceleration, we apply Newton second law, F = m * a, to find the net force. Careful substitution and correct units are crucial. Since all values are small and neat, the arithmetic should be manageable, but attention must still be paid to the correct formula and order of operations. Step-by-Step Solution: Step 1: Compute the change in velocity: v - u = 20 m/s - 12 m/s = 8 m/s. Step 2: Use the definition of acceleration: a = (v - u) / t = 8 m/s / 4 s = 2 m/s^2. Step 3: Apply Newton second law: F = m * a. Step 4: Substitute the mass and acceleration: F = 5 kg * 2 m/s^2. Step 5: Multiply to get F = 10 kg·m/s^2. Step 6: Recognise that 1 kg·m/s^2 is 1 newton, so F = 10 N. Verification / Alternative check: As a quick check, note that the acceleration is modest (2 m/s^2), and the mass is 5 kg, so a force of 10 N is reasonable. If you recompute acceleration with the numbers swapped or misread the time as 2 s, you would get different accelerations and forces, which would not match both the given velocities and time. Rechecking the arithmetic confirms the correctness of 2 m/s^2 and 10 N. Why Other Options Are Wrong: 40 N: This would correspond to an acceleration of 8 m/s^2 for a 5 kg mass, which is inconsistent with the given change in velocity over 4 seconds. 20 N: This would correspond to an acceleration of 4 m/s^2, which would change velocity by 16 m/s in 4 seconds, not by 8 m/s. 80 N: This is far too large for the given rate of change of velocity and would imply an acceleration of 16 m/s^2, again inconsistent with the data. Common Pitfalls: Students sometimes plug values into the wrong kinematic formula or forget whether to subtract u from v or vice versa. Always remember that acceleration is (final velocity minus initial velocity) divided by time, and then apply F = m * a. Keeping units visible at each step helps avoid mixing up time and velocity or forgetting to square units where needed. Final Answer: The magnitude of the force acting on the body is 10 N.

Question 19

According to Newton's Second Law of Motion, for a given constant force, the acceleration of an object is inversely proportional to which property of the object?

Accepted Answer

Introduction / Context: Newton second law of motion provides a quantitative link between force, mass and acceleration. It tells us not only that force causes acceleration but also how the magnitude of acceleration depends on the applied force and the mass of the object. This question focuses on the relationship between acceleration and an object properties when the applied force is kept constant. Given Data / Assumptions: We consider a constant net force acting on an object. Newton second law is written in the form F = m * a. We investigate how acceleration a changes when mass m changes, while F remains fixed. Other properties like density and volume may vary but are not included explicitly in the law. Concept / Approach: From F = m * a, we can write acceleration as a = F / m. This expression directly shows that for a fixed F, acceleration is inversely proportional to mass. That is, if the mass doubles, acceleration halves; if the mass is reduced, acceleration increases. Density and volume may influence mass, but the fundamental inverse proportionality is between acceleration and mass, not between acceleration and density or volume directly. Step-by-Step Solution: Step 1: Start with Newton second law: F = m * a. Step 2: Rearrange to express acceleration: a = F / m. Step 3: For a given constant force F, this becomes a ∝ 1 / m. Step 4: This means that as mass increases, acceleration decreases proportionally, and vice versa. Step 5: Note that density is mass per unit volume and does not appear directly in this formula. Step 6: Therefore, the quantity with which acceleration is inversely proportional, for given force, is the mass of the object. Verification / Alternative check: Everyday experience supports this idea. It is harder to accelerate a heavy object than a light one with the same push. For example, pushing an empty trolley is easier than pushing a loaded trolley, because the loaded trolley has more mass and thus smaller acceleration for the same applied force. In physics labs, experiments with carts and known masses confirm that acceleration varies inversely with the total mass. Why Other Options Are Wrong: Density: While density relates mass and volume, acceleration depends directly on total mass, not on density alone. Two objects with different densities but the same mass will have the same acceleration for a given force. Volume: An object volume does not directly appear in F = m * a. Large but light objects and small but heavy objects can have very different accelerations depending on their mass. Magnitude of the applied force: For a given force, acceleration is directly proportional to F, not inversely. The question explicitly states that force is fixed, so we focus on mass. Common Pitfalls: Students sometimes think that heavier objects fall faster or accelerate less only because of gravity, mixing everyday impressions with the precise law. To avoid confusion, always go back to the formula a = F / m. When F is fixed, any change in m must be reflected inversely in a. This logic works for all types of net forces, not just gravity. Final Answer: For a given force, acceleration is inversely proportional to the mass of the object.

Question 20

In an electric room heater or electric cooking heater, the high resistance coil of wire that becomes red hot and produces heat is commonly called the ________.

Accepted Answer

Introduction / Context: Electric room heaters and electric cooking heaters convert electrical energy into heat energy using a specially designed coil of wire. This coil has high resistance and is mounted in a safe way so that it can become very hot without melting immediately. Knowing the common name of this part is an important piece of basic physics and general knowledge about household electric appliances. Given Data / Assumptions: We are talking about an electric room heater or electric cooking heater used in homes. Inside such a heater there is a visible or hidden coil made of high resistance wire, often nichrome. When current passes through this coil it becomes red hot and produces heat. We only need the common practical name used by electricians and users for this coil part. Concept / Approach: In an electric heater, electrical energy is converted into heat energy by the Joule heating effect. A wire of high resistance is chosen so that even a moderate current produces significant heating. This wire is usually coiled and mounted on an insulating support. In everyday language and in service manuals this coil is called the heating element or simply the element of the heater. The word circuit refers to the complete path of current, not the coil itself. Filament is usually used for the thin wire inside an electric bulb. Cells are sources of electrical energy, not heating parts. Therefore the correct everyday technical term here is heating element. Step-by-Step Solution: Step 1: Identify the part of the heater that gets hot. It is the high resistance coiled wire inside the appliance.Step 2: Recall the term commonly used by technicians. They talk about replacing a burned out heater element.Step 3: Compare with the term circuit, which means the entire path of current, including wires, switch and plug.Step 4: Compare with filament, which is the glowing wire in a bulb, not in a typical room heater.Step 5: Recognise that cells are individual electrochemical units that supply electricity, not heat by themselves.Step 6: Conclude that heating element is the correct description of the coil wire in an electric heater. Verification / Alternative check: User manuals and repair guides for electric heaters usually list the main spare part as heater element or heating element. When this part burns out due to overheating or overvoltage, technicians say that the heater element is gone and needs replacement. In electric cooking heaters, the spiral or coil is also described as the element. This consistent real world usage confirms that element or heating element is the proper term rather than circuit, filament or cells. Why Other Options Are Wrong: Circuit refers to the complete closed path of electric current and not to any single part such as the heater coil. Filament is the fine wire that glows in an incandescent bulb and is not the standard term for heater coils. Cells are devices that produce electricity by chemical reactions, so they are sources of energy, not heating components that convert electric current into heat. These terms do not correctly name the heater coil. Common Pitfalls: Sometimes students confuse the filament of a bulb with the heating coil of a heater because both become hot and may glow. However, in standard terminology, the word filament is reserved for lamps, and element is used for heaters, geysers and cooking coils. Another common mistake is to think that any part carrying current can be called a circuit, but circuit is a broader concept, not a single component. Remember that in electric heaters the correct practical term is heating element. Final Answer: In an electric room heater or electric cooking heater, the high resistance coil of wire is called the heating element.

Physics Questions