Question 1

An object is dropped from a certain height in a region where there is no air resistance (that is, it is in free fall). What happens to its speed and acceleration as it falls?

Accepted Answer

Introduction / Context: This conceptual physics question is about free fall motion near the Earth surface. When an object is dropped from rest in the absence of air resistance, its motion is governed only by the gravitational force. Understanding how speed and acceleration behave in this idealised situation is essential for solving many kinematics and dynamics problems in mechanics. Given Data / Assumptions: An object is released from rest at some height above the ground. No air resistance or other forces act, only the gravitational force. Acceleration due to gravity is taken as g, approximately 9.8 m/s^2, directed downward. We are interested in how both speed and acceleration change during the fall. Concept / Approach: In free fall, the only force acting on the object is its weight, mg, downwards. Newton’s second law then gives a constant downward acceleration equal to g. Because acceleration is constant, the velocity of the object increases linearly with time from its initial value of zero. Speed, which is the magnitude of velocity, therefore increases steadily, while the acceleration remains constant in magnitude and direction throughout the fall. Step-by-Step Solution: Step 1: Use Newton’s second law: F_net = m * a. Here, F_net = mg downward, so a = g downward, which is constant. Step 2: Recognise that since a is constant and non zero, the velocity of the object cannot remain constant. Step 3: For motion starting from rest, the velocity at time t is v = g * t downward, so its magnitude (speed) increases as time passes. Step 4: Since g does not change during the fall in this idealised model, the acceleration remains constant in magnitude and direction. Step 5: Therefore, the correct description is that the object speed increases while its acceleration remains constant. Verification / Alternative check: The standard equations of uniformly accelerated motion support this. For example, displacement s = 0.5 * g * t^2, velocity v = g * t, and acceleration a = g, all derived from constant acceleration. Laboratory experiments using vacuum tubes or drop towers confirm that bodies in free fall without air resistance all accelerate at the same constant rate, regardless of mass. These facts are consistent only with increasing speed and constant acceleration. Why Other Options Are Wrong: It will fall with constant speed and constant acceleration: Constant acceleration cannot coexist with constant speed; if acceleration is non zero, speed must change. Its acceleration will keep on increasing as it falls: In the free fall model, gravitational acceleration near the Earth surface is taken as constant g, not increasing with time. Both its speed and its acceleration will keep increasing together: Speed increases, but acceleration remains constant, so this statement is incorrect. Common Pitfalls: Many learners confuse acceleration with speed and think that if an object is speeding up, its acceleration must also be increasing. Another common mistake is to imagine that gravity somehow becomes stronger as the object approaches the ground in simple school level problems; however, we treat g as approximately constant over the small height changes involved. Keeping the distinction clear between velocity, speed, and acceleration helps to avoid these misunderstandings. Final Answer: In free fall without air resistance, the object speed increases while its acceleration remains constant.

Question 2

When we see an object with the human eye, what is the nature of the image formed on the retina inside the eye?

Accepted Answer

Introduction / Context: This question relates to basic optics and human vision. The human eye acts like a small camera with a convex lens and a light sensitive screen called the retina. Understanding the nature of the image formed on the retina helps students link geometric optics with biology and understand how the brain interprets visual information from the external world. Given Data / Assumptions: The eye contains a convex (converging) lens system. The retina is located behind the lens and is the surface where the image is formed. Objects we see are generally at distances much greater than the focal length of the eye lens. The question asks about whether the image is real or virtual, erect or inverted. Concept / Approach: A convex lens forms real and inverted images of objects placed beyond its focal length on the side opposite to the object. In the human eye, the lens focuses light rays from an external object onto the retina. Because the rays actually converge onto the retina, the image is real. Due to the geometry of refraction, this real image is inverted (upside down) with respect to the object. The brain processes the signals from the retina and interprets them so that we perceive the world as upright, even though the physical image on the retina is inverted. Step-by-Step Solution: Step 1: Recall the image formation by a convex lens for an object located beyond the focal length: the image is real and inverted. Step 2: Recognise that the eye lens adjusts its focal length slightly (accommodation) but still forms the image on the retina using converging rays. Step 3: Since light rays actually meet at the retina, the image must be a real image, not virtual. Step 4: Due to the lens geometry, the top of the object is focused at the lower part of the retina and the bottom of the object at the upper part, which means the image is inverted. Step 5: Therefore, the image on the retina is real and inverted. Verification / Alternative check: Optics diagrams in textbooks that trace rays through the eye lens clearly show a small inverted image on the retina. This is similar to the image produced by a camera lens on photographic film or a digital sensor. Experiments with simple lenses also show that real images form on screens, and those images are inverted. Since the retina functions as a biological screen, the same behaviour occurs in the eye. Why Other Options Are Wrong: Real and erect: A convex lens does not form a real and erect image for distant objects; that combination does not match the eye real image. Virtual and erect: Virtual images are formed when rays appear to diverge from a point but do not actually meet there, which is not the case on the retina. Virtual and inverted: This combination is not produced by a simple convex lens for usual object positions and does not describe the retinal image. Common Pitfalls: Some students think that because we see objects upright, the image on the retina must also be upright. They forget that perception involves brain processing. Others confuse real images (formed on a screen) with virtual images (seen through an optical instrument but not projected on a screen). Remember that the retina is a physical surface where light rays converge, so the image must be real and, due to lens geometry, inverted. Final Answer: The image formed on the retina is real and inverted.

Question 3

In mechanics, the kinetic energy of a moving body depends on which of the following quantities?

Accepted Answer

Introduction / Context: This question tests understanding of kinetic energy, a fundamental concept in mechanics. Kinetic energy represents the energy a body possesses due to its motion. Knowing how kinetic energy depends on mass and speed is important in physics, engineering, vehicle safety, sports science, and many everyday situations where moving objects do work or cause damage. Given Data / Assumptions: The body has a certain mass m and is moving with speed v. Kinetic energy is defined for translational motion (straight line motion) in this context. We use the standard expression for kinetic energy in classical mechanics. Concept / Approach: In classical physics, the kinetic energy (KE) of a body of mass m moving with speed v is given by KE = (1/2) * m * v^2. This formula shows that kinetic energy is directly proportional to mass and proportional to the square of speed. Therefore, both mass and speed influence kinetic energy. Pressure is not part of this basic formula for the kinetic energy of a rigid body in translational motion. Step-by-Step Solution: Step 1: Recall the formula for kinetic energy: KE = (1/2) * m * v^2. Step 2: Identify the variables in the formula. Mass m appears as a direct factor, and speed v appears squared. Step 3: Since KE is directly proportional to m, doubling the mass doubles the kinetic energy if speed remains the same. Step 4: Since KE is proportional to v^2, doubling the speed increases kinetic energy by a factor of 4, even if mass remains unchanged. Step 5: Conclude that kinetic energy depends on both mass and speed, not on pressure. Verification / Alternative check: Consider two vehicles: one light and one heavy, both moving at the same speed. The heavier vehicle has more kinetic energy and can cause more damage in a collision, reflecting the dependence on mass. Now imagine the same vehicle at two different speeds. At higher speed, stopping distance and potential damage are much greater, consistent with the v^2 dependence. Practical road safety campaigns often emphasise that small increases in speed lead to much larger increases in kinetic energy, matching the formula. Why Other Options Are Wrong: The velocity or speed of the moving body only: This ignores the role of mass; a heavier object at the same speed has more kinetic energy. The mass of the moving body only: This ignores the fact that a stationary object has zero kinetic energy; motion (speed) is essential. The pressure of the moving body: Pressure is not a factor in the simple particle kinetic energy formula and is irrelevant here. Common Pitfalls: Students sometimes mistakenly think that kinetic energy is proportional to speed, not speed squared, leading them to underestimate the effect of speed. Others may focus only on mass and assume that a heavy object always has high kinetic energy regardless of motion, forgetting that kinetic energy becomes zero if the object is at rest. Keeping the full formula KE = (1/2) * m * v^2 in mind helps to avoid these misunderstandings. Final Answer: Kinetic energy depends on both the mass and the speed of the moving body.

Question 4

During a change in temperature of a substance without any change of state, in which microscopic form is the supplied heat energy mainly stored inside the substance?

Accepted Answer

Introduction / Context: This question deals with the microscopic interpretation of heat and temperature. When a substance is heated and its temperature rises, there is an increase in the internal energy of the substance. Understanding whether this increase is mainly in kinetic energy, potential energy, or some other form helps clarify the difference between temperature change and change of state in basic thermodynamics. Given Data / Assumptions: The substance is being heated and its temperature is changing. There is no change of state (for example, no melting or boiling during the process described). We are interested in what happens at the microscopic particle level. The behavior is considered in the usual school level model of solids, liquids, and gases. Concept / Approach: Temperature is directly related to the average kinetic energy of the particles in a substance. When the temperature of a body increases without a change of state, the speed of its atoms or molecules increases. In solids they vibrate more vigorously, and in liquids and gases they move faster. This means that the heat supplied in such a process is stored mainly as an increase in kinetic energy of the particles. In contrast, during a change of state at constant temperature, the supplied heat is mainly used to change potential energy associated with intermolecular separation, not kinetic energy. Step-by-Step Solution: Step 1: Recall that temperature is a measure of the average kinetic energy of the particles in a substance. Step 2: When heat is supplied and temperature rises, the average kinetic energy of the particles must increase. Step 3: During this process, the arrangement of particles does not change phase; for example, the solid remains solid, or the liquid remains liquid. Step 4: Since no phase change occurs, most of the energy does not go into breaking or forming intermolecular bonds, so the main increase is in kinetic energy rather than potential energy. Step 5: Therefore, during a pure temperature change, the supplied heat energy is stored mainly as increased kinetic energy of the particles. Verification / Alternative check: A common classroom experiment heats water from room temperature to just below boiling. The temperature steadily rises as heat is supplied, indicating that the particles are moving faster on average. During this period, no phase change occurs. When water starts to boil at constant temperature, further heat does not increase temperature but instead changes liquid to steam, reflecting a change in potential energy. This clear difference supports the idea that heat during a temperature rise mainly increases kinetic energy, while heat during a phase change mainly alters potential energy. Why Other Options Are Wrong: As heat energy separate from internal energy: Heat is energy in transfer, not a separate stored form inside the substance. Internal energy is stored as kinetic and potential energy of particles. As potential energy only: Potential energy changes are more important during phase changes at constant temperature, not primarily during simple temperature rises. Equally as kinetic and potential energy for every temperature change: The simple school level model emphasises kinetic energy increase for temperature rise; equal sharing is not a general rule. Common Pitfalls: Learners often confuse heat, temperature, and internal energy, treating heat as a substance that is stored. It is also easy to mix up processes with and without phase changes. Remember that when temperature changes, average kinetic energy changes. When state changes at constant temperature, potential energy changes. Keeping this distinction clear makes thermodynamics problems much easier to understand. Final Answer: During a temperature change without change of state, the supplied heat energy is mainly stored as kinetic energy of the particles.

Question 5

In SI units, 1 kilowatt hour (1 kWh) of electrical energy is equal to how many joules?

Accepted Answer

Introduction / Context: This numerical question is about converting a practical energy unit used by electricity companies into the standard SI unit, the joule. Electricity bills often show energy consumption in kilowatt hours (kWh), but physics equations typically use joules (J). Knowing the exact conversion between these units is essential for solving power and energy problems correctly. Given Data / Assumptions: 1 watt (W) = 1 joule per second (1 J/s). 1 kilowatt (kW) = 1000 W. 1 hour (h) = 3600 seconds. We want to find the energy in joules corresponding to 1 kWh. Concept / Approach: Energy is related to power and time by the relation Energy = Power * Time. A kilowatt hour means that a device with a power of 1 kilowatt operates for one hour. To convert to joules, we convert kilowatts to watts and hours to seconds, then multiply. Careful handling of powers of 10 is needed to avoid errors in the final exponent. Step-by-Step Solution: Step 1: Write the basic relation: Energy (in joules) = Power (in watts) * Time (in seconds). Step 2: Convert 1 kilowatt to watts: 1 kW = 1000 W. Step 3: Convert 1 hour to seconds: 1 h = 3600 s. Step 4: Multiply to get energy: Energy = 1000 W * 3600 s = 3,600,000 J. Step 5: Express this number in scientific notation: 3,600,000 J = 3.6 x 10^6 J. Verification / Alternative check: You can think of a 1000 W heater running for one hour. Each second it uses 1000 J of energy. In 3600 seconds, it uses 1000 * 3600 J, which is 3.6 million joules. Many textbooks and electricity company references confirm the standard conversion 1 kWh = 3.6 x 10^6 J, so the result matches widely accepted data. Why Other Options Are Wrong: 3.6 x 10^5 J: This is smaller by a factor of 10 and results from missing a zero in the calculation. 3.6 x 10^-5 J and 3.6 x 10^-6 J: These represent extremely tiny amounts of energy and clearly cannot correspond to one kilowatt hour, which is a large everyday energy unit. Common Pitfalls: Typical mistakes include forgetting that 1 kilowatt is 1000 watts, miscalculating the number of seconds in an hour, or misplacing the decimal when writing the final answer in scientific notation. Some students also confuse kilowatt hours with kilojoules. Carefully doing the unit conversions step by step and then rewriting the final number with the correct power of 10 helps avoid these errors. Final Answer: 1 kilowatt hour is equal to 3.6 x 10^6 joules.

Question 6

When several resistors are connected in series in an electric circuit, how does the electric current compare at different points along that series branch?

Accepted Answer

Introduction / Context: This question is from basic electric circuits. Understanding the behavior of current in series and parallel connections is essential for analysing household wiring, electronic circuits, and many practical devices. In a series connection, components are connected end to end so that the same current flows through each component in the chain. Given Data / Assumptions: Several resistors are connected in series in a single branch of the circuit. The circuit is powered by a source such as a battery or power supply. We focus on how the current varies along that series branch, not on its overall magnitude compared to a different circuit. Concept / Approach: In a series circuit, there is only one path for charge to flow. Because charge cannot pile up at any point in a steady state direct current circuit, the rate of flow of charge, that is, the electric current, must be the same at every cross section along the series path. The total resistance in series is the sum of individual resistances, and this affects the overall magnitude of the current, but the current value itself is identical through each resistor and through the connecting wires in that branch. Step-by-Step Solution: Step 1: Recall that current is the rate of flow of charge, measured in coulombs per second. Step 2: In a series connection, there is only one continuous path for this charge to move through each resistor. Step 3: Since charge cannot accumulate at intermediate points in a steady state, the same amount of charge that enters a resistor per second must leave it per second. Step 4: Therefore, the current entering each resistor equals the current leaving it, and this value is the same at all points in the series branch. Step 5: Conclude that in a series circuit, the current remains the same through all the resistors. Verification / Alternative check: Using Ohm’s law, for a series combination the total resistance R_total equals R1 + R2 + R3 and so on. The circuit current is I = V / R_total, where V is the applied voltage. Because there is only one branch, the same current I flows through each resistor. Measurements with an ammeter placed at different positions in a simple series circuit confirm that the current reading does not change, provided the circuit remains unchanged and connections are ideal. Why Other Options Are Wrong: The current increases as it passes through each resistor: Resistors oppose current; they do not create additional current. The current becomes exactly half after the first resistor: There is no rule that halves current at each resistor in a series circuit. The current decreases after each resistor: In an ideal series circuit, current is equal at all points; only voltage drops across each resistor. Common Pitfalls: Students sometimes confuse current with voltage. They may think that because voltage drops across each resistor, current must also drop, which is incorrect. In a series circuit, the current is constant, while voltage is divided among components. In parallel circuits, the voltage across each branch is the same, while currents can differ. Keeping these two cases clearly separated helps avoid many basic circuit mistakes. Final Answer: In a series circuit, the current remains the same through all the resistors.

Question 7

A 4 kg object is moving horizontally with a speed of 5 m/s. To increase its speed to 10 m/s, how much net work must be done on the object?

Accepted Answer

Introduction / Context: This question is a straightforward application of the work energy theorem in mechanics. The theorem states that the net work done on an object equals the change in its kinetic energy. By comparing the initial and final kinetic energies of a moving object, we can determine how much work is required to increase its speed from one value to another. This concept is very useful in physics, engineering, and real life calculations, such as estimating the energy needed for acceleration in vehicles. Given Data / Assumptions: Mass of the object, m = 4 kg. Initial speed, v1 = 5 m/s. Final speed, v2 = 10 m/s. The motion is horizontal and friction or other energy losses are neglected; only net work is considered. Concept / Approach: The kinetic energy KE of a body of mass m moving with speed v is KE = (1/2) * m * v^2. The work energy theorem states that net work done, W_net, equals the change in kinetic energy: W_net = KE_final - KE_initial. To solve the problem, we first compute initial and final kinetic energies, then subtract to find the work done. Careful substitution and correct squaring of speeds are essential to arrive at the correct numerical answer. Step-by-Step Solution: Step 1: Write the expression for kinetic energy: KE = (1/2) * m * v^2. Step 2: Calculate initial kinetic energy with v1 = 5 m/s: KE_initial = (1/2) * 4 * (5^2) = 2 * 25 = 50 J. Step 3: Calculate final kinetic energy with v2 = 10 m/s: KE_final = (1/2) * 4 * (10^2) = 2 * 100 = 200 J. Step 4: Use the work energy theorem: W_net = KE_final - KE_initial = 200 J - 50 J = 150 J. Step 5: Conclude that 150 J of net work must be done on the object to increase its speed from 5 m/s to 10 m/s. Verification / Alternative check: We can also reason qualitatively: kinetic energy depends on the square of the speed. Doubling the speed from 5 m/s to 10 m/s increases v^2 by a factor of 4 (from 25 to 100). Therefore, kinetic energy should also increase by a factor of 4, assuming mass is constant. Initial KE is 50 J, and four times that is 200 J, which matches the calculated final KE. The difference, 200 - 50 = 150 J, is the needed work, confirming the result. Why Other Options Are Wrong: 100 J: This might arise from incorrectly using linear proportionality with speed instead of speed squared, or from calculation mistakes. 50 J: This equals the initial kinetic energy and would only raise the energy from 0 to 50 J, not from 50 J to 200 J. 75 J: This is half the correct answer and reflects a miscalculation or partial change in kinetic energy. Common Pitfalls: A typical error is to treat kinetic energy as proportional to speed rather than speed squared, leading to incorrect scaling when speed is doubled. Another common mistake is mishandling the square during calculation, such as squaring the mass instead of the speed. Finally, students sometimes forget to subtract initial kinetic energy and incorrectly equate work done with final kinetic energy only. Carefully applying the formula KE = (1/2) * m * v^2 and the work energy theorem helps avoid these issues. Final Answer: The required net work is 150 J.

Question 8

The reflector used in a powerful search light to produce a strong parallel beam of light is which type of mirror?

Accepted Answer

Introduction / Context: This question tests basic knowledge of spherical mirrors and their applications. Search lights, car headlights, and some torches use specially shaped mirrors to produce strong, roughly parallel beams of light. Understanding which type of mirror is used helps reinforce concepts of focal point, converging and diverging reflection, and practical uses of concave and convex mirrors in optics. Given Data / Assumptions: We are dealing with a search light that needs to project a strong, almost parallel beam of light over long distances. A small electric bulb or lamp is placed near the mirror. The options include convex, concave, plane, and cylindrical mirrors. Concept / Approach: A concave mirror is a converging mirror for rays incident near its principal axis. If a point source of light is placed at its focus, the reflected rays travel nearly parallel to the principal axis, creating a parallel beam. This principle is used in search lights and car headlights. A convex mirror spreads rays and cannot form a strong parallel beam in this way. A plane mirror simply reflects light without focusing, and a cylindrical mirror focuses light differently in one direction only, which is not the standard choice for search lights. Step-by-Step Solution: Step 1: Recall that a concave mirror can form parallel rays when a point source is placed at its principal focus. Step 2: In a search light, the bulb is placed close to the focal point of a large concave mirror. Step 3: Light rays from the bulb spread out, strike the concave mirror, and are reflected so that they emerge approximately parallel to the principal axis. Step 4: This produces a bright, narrow beam that can travel a long distance, which is exactly the function of a search light. Step 5: Therefore, the reflector of a search light is a concave mirror. Verification / Alternative check: Diagrams in optics textbooks for car headlights and search lights consistently show a concave reflector behind the bulb. You can also test the idea with a small concave shaving mirror: placing a light source near its focus produces a relatively collimated beam. Convex mirrors, in contrast, are used where a wide field of view is required, such as rear view mirrors, not for focused beams. This practical evidence supports the choice of concave mirror for search lights. Why Other Options Are Wrong: Convex mirror: Diverges incident rays and is used for wide angle viewing, not for forming a strong parallel beam. Plane mirror: Reflects light without focusing it into a parallel beam; the reflected rays remain as divergent as the incident rays. Cylindrical mirror: Can focus light in one dimension but is not typically used as the reflector in search lights, which require axisymmetric focusing. Common Pitfalls: Some students confuse concave and convex mirrors because both are curved and their names can be mixed up. Remember that concave mirrors converge light and are used where focusing is needed, such as in solar furnaces, shaving mirrors, and search lights. Convex mirrors diverge light and are used for rear view mirrors and security mirrors. Keeping these roles clearly in mind helps prevent mistakes in questions like this. Final Answer: The reflector of a search light is a concave mirror.

Question 9

In the International System of Units (SI), what is the unit used to measure electrical resistance?

Accepted Answer

Introduction / Context: This general physics question is about SI units for basic electrical quantities. Electrical resistance describes how strongly a material opposes the flow of electric current. Using the correct unit symbol is important for interpreting circuit diagrams, formulas, and specifications of electrical components such as resistors and wires. Given Data / Assumptions: We are dealing with electrical resistance R in Ohm’s law, where V = I * R. The question asks specifically for the SI unit of resistance. Other SI units mentioned are those of force, charge, and energy. Concept / Approach: Ohm’s law relates voltage V (in volts), current I (in amperes), and resistance R by V = I * R. From this, resistance can be expressed as R = V / I. In SI units, the unit of resistance is defined as one volt per ampere and is given the special name ohm, with symbol capital Greek letter omega. Other units such as newton, coulomb, and joule correspond to force, electric charge, and energy respectively, not resistance. Step-by-Step Solution: Step 1: Recall Ohm’s law: V = I * R. Step 2: Rearrange to express resistance: R = V / I. Step 3: In SI units, voltage V is measured in volts (V) and current I is measured in amperes (A). Step 4: Therefore, the unit of resistance is volt per ampere, given the special name ohm. Step 5: Conclude that the SI unit of resistance is ohm. Verification / Alternative check: Any resistor component is labelled with a value such as 10 ohm, 1 kiloohm, or 1 megaohm, using the symbol omega to indicate resistance. Circuit analysis textbooks always list the SI units for basic quantities as volt for voltage, ampere for current, and ohm for resistance. This consistent usage across physics and electrical engineering confirms that ohm is the correct unit for resistance. Why Other Options Are Wrong: Newton: SI unit of force, defined from F = m * a. Coulomb: SI unit of electric charge, related to current by Q = I * t. Joule: SI unit of energy or work, used in equations such as W = F * s or E = P * t. Common Pitfalls: Students sometimes mix up Coulomb and ohm because both are linked to electricity. However, Coulomb refers to the quantity of charge, while ohm refers to how much a component resists current flow. Others confuse ohm with volt or ampere. Remembering the Ohm’s law triangle (V, I, R) and associating ohm with resistance, volt with potential difference, and ampere with current helps avoid these errors. Final Answer: The SI unit of electrical resistance is the ohm.

Question 10

A convex mirror has a focal length f when it is in air. If this mirror is completely immersed in a liquid of refractive index 4/3, what will be its focal length in the liquid?

Accepted Answer

Introduction / Context: This question is about spherical mirrors and the effect of surrounding medium on focal length. Students often learn that the focal length of a lens depends on refractive indices of the lens material and the surrounding medium. However, mirrors form images by reflection, not refraction, and therefore behave differently. Understanding this distinction helps avoid common conceptual mistakes. Given Data / Assumptions: A convex (diverging) spherical mirror has focal length f in air. The mirror is then immersed in a liquid of refractive index 4/3 relative to air. The mirror material and its radius of curvature remain unchanged. We are asked for the new focal length of the mirror in the liquid. Concept / Approach: The focal length of a spherical mirror depends only on its radius of curvature and the geometry of reflection. For a mirror, the approximate relation is f = R / 2, where R is the radius of curvature. This relation arises purely from reflection laws (angle of incidence equals angle of reflection) and the mirror shape. Unlike lenses, mirrors do not depend on the refractive index of the surrounding medium because light does not refract through the mirror; it reflects from the mirror surface. Therefore, immersing a mirror in a different medium does not change its focal length. Step-by-Step Solution: Step 1: Recall the mirror formula for focal length in terms of radius of curvature: f = R / 2 for spherical mirrors. Step 2: Note that this formula is derived assuming reflection at the mirror surface, using laws of reflection, which do not involve refractive index. Step 3: When the mirror is immersed in a liquid, neither the radius of curvature R nor the reflection law changes. Step 4: Therefore, the focal length f remains the same numerical value, even though the surrounding medium has changed. Step 5: Conclude that the focal length of the convex mirror in the liquid is still f. Verification / Alternative check: Contrast this with lenses: the focal length of a lens depends on the refractive index ratio between the lens material and its surroundings, as described by the lens maker formula. If a lens is immersed in a liquid whose refractive index is close to that of the lens, the focal length changes significantly. For mirrors, any ray diagram you draw in different media still uses the same reflection angles and curvature, leading to identical convergence or divergence behaviour. This thought experiment confirms that the focal length remains unchanged for a mirror. Why Other Options Are Wrong: (4/3)f: Suggests scaling of focal length with refractive index, which would apply to lenses, not mirrors. (3/4)f: Similarly implies dependence on the index in the opposite direction, which is incorrect for mirrors. (7/3)f: Has no standard basis in mirror optics and is clearly unrelated to the simple relation f = R / 2. Common Pitfalls: The main pitfall is confusing the behaviour of mirrors with that of lenses. Many students automatically assume that any optical element will have its focal length changed by immersion in a liquid. This is true for refracting elements like lenses but not for reflecting elements like mirrors. Another error is to think that the speed of light in the medium alone determines focal length; for mirrors, the path geometry and reflection law dominate. Always remember: mirrors depend on shape and reflection, lenses depend on shape and refractive indices. Final Answer: The focal length of the convex mirror in the liquid remains f.

Question 11

Which of the following statements about displacement in physics is false?

Accepted Answer

Introduction / Context: This question checks conceptual understanding of displacement, a key quantity in kinematics. Displacement is a vector quantity that describes the change in position of an object in a specified direction. Distinguishing displacement from distance, and understanding its sign and units, is crucial for correctly solving motion problems in one and two dimensions. Given Data / Assumptions: We are working in standard one dimensional or two dimensional motion. Displacement is defined as final position minus initial position, with direction. Distance is the total path length traveled and is always non negative. The question asks which statement about displacement is false. Concept / Approach: Displacement is a vector quantity with magnitude and direction, while distance is a scalar quantity with only magnitude. Because displacement can point in the positive or negative direction along a chosen axis, its value can be positive, negative, or zero. Displacement is always less than or equal to distance in magnitude because distance measures the full path, whereas displacement is a straight line between start and end. The SI unit of displacement is the metre, same as distance. Therefore, any statement claiming that displacement is always positive must be false. Step-by-Step Solution: Step 1: Examine the statement that displacement can be positive, negative or zero. This matches the vector nature of displacement and is true. Step 2: Check the statement that displacement is never greater than distance. For any motion, the straight line distance between two points is the shortest path, so this is true. Step 3: Confirm that the SI unit of displacement is metre, which is correct since it measures length with direction. Step 4: Evaluate the statement that displacement is always positive. This ignores the possibility of negative displacement (movement opposite to chosen positive direction) and zero displacement (returning to the starting point), so it is false. Step 5: Conclude that the false statement is that displacement is always positive. Verification / Alternative check: Consider a simple example of one dimensional motion along a straight line with right taken as positive. If a person walks 5 m to the right and then 3 m to the left, the displacement is +2 m, which is positive. If the person walks 3 m to the left from the origin, the displacement is -3 m, which is negative. If they walk 5 m right and then 5 m left, they return to their starting position and the displacement is zero. These examples clearly show that displacement can take positive, negative, or zero values, contradicting the claim that it is always positive. Why Other Options Are Wrong (but true statements about displacement): It can be positive, negative or zero: This is a correct description of displacement as a vector quantity. Displacement is never greater than Distance: Distance is the length of the actual path; displacement is the shortest path between initial and final positions, so its magnitude cannot exceed distance. Its SI unit is metre: Both displacement and distance are measured in metres in the SI system. Common Pitfalls: Students sometimes treat displacement like distance and forget about direction. This leads them to think that displacement must also always be positive. Another common mistake is to confuse the terms and use them interchangeably in answers. Remember that distance answers the question "How much ground was covered?" while displacement answers "How far and in which direction is the final position from the starting point?" Keeping this difference in mind helps to avoid conceptual errors. Final Answer: The false statement is: It is always positive.

Question 12

In the International System of Units (SI), what is the unit of force?

Accepted Answer

Introduction / Context: This general knowledge physics question asks about the correct SI unit of force. Force is a fundamental concept in mechanics, defined as any interaction that can change the motion of an object. Using the correct unit is essential when applying Newton’s laws of motion, calculating weight, and solving a wide range of engineering and physics problems. Given Data / Assumptions: Force is defined by Newton’s second law: F = m * a. Mass is measured in kilograms (kg) and acceleration in metres per second squared (m/s^2). The question provides several different physical unit names, only one of which is correct for force. Concept / Approach: By Newton’s second law, a force of 1 newton is defined as the force required to accelerate a mass of 1 kilogram at 1 m/s^2. This definition leads to the unit newton, symbol N, where 1 N = 1 kg * m/s^2. Pascal is the unit of pressure, watt is the unit of power, and Boyle refers to a scientist, not a unit. Therefore, among the given options, newton is the correct SI unit of force. Step-by-Step Solution: Step 1: Write Newton’s second law: F = m * a. Step 2: Recognise that in SI units, mass m is in kilograms (kg) and acceleration a is in metres per second squared (m/s^2). Step 3: Define the unit of force such that 1 N = 1 kg * 1 m/s^2. Step 4: Note that this unit is named newton in honour of Sir Isaac Newton. Step 5: Conclude that the SI unit of force is the newton. Verification / Alternative check: Physics textbooks, engineering handbooks, and scientific calculators all use the newton as the standard unit for force. Weight calculations such as W = m * g give weight in newtons when mass is in kilograms and g is in m/s^2. Pressure is defined as force per unit area, with SI unit pascal where 1 Pa = 1 N/m^2, again reinforcing that newton is the unit of force. Why Other Options Are Wrong: Pascal: SI unit of pressure, defined as 1 newton per square metre. Boyle: The name of a scientist associated with Boyle’s law in gas behaviour, not a physical unit. Watt: SI unit of power, defined as 1 joule per second, or 1 N * m/s. Common Pitfalls: Students sometimes confuse force and pressure, especially because both involve newtons in their definitions. However, force is a vector quantity measured in newtons, while pressure is scalar and measured in pascals. Others may confuse power with force, but power measures the rate of doing work in watts. Remember the basic relationships: F = m * a, P = W / t, and p = F / A, and associate newton with force, pascal with pressure, and watt with power. Final Answer: The SI unit of force is the newton.

Question 13

Which one of the following instruments is specifically used to measure soil water tension (the suction with which water is held in the soil)?

Accepted Answer

Introduction / Context: This general knowledge question relates to physics and agricultural science. Soil water tension, also called soil water suction or matric potential, indicates how strongly water is held by soil particles and how available it is to plant roots. Measuring this tension helps farmers and researchers make decisions about irrigation and crop management. Different instruments are used to measure light, temperature, humidity, and soil moisture related quantities. Given Data / Assumptions: We want to measure soil water tension, not just soil moisture content. The options include photometer, pyrometer, psychrometer, and tensiometer. Each instrument name refers to measurement of a different physical quantity. Concept / Approach: A tensiometer is a device used in soil physics to measure the tension or suction with which water is held in the soil. It typically consists of a porous ceramic cup attached to a water filled tube and a vacuum gauge. As water moves between the soil and the cup, the gauge reflects the tension. Photometers measure light intensity, pyrometers measure very high temperatures, and psychrometers measure humidity. Therefore, tensiometer is the only instrument among the options that is designed to measure soil water tension. Step-by-Step Solution: Step 1: Identify the target quantity: soil water tension, which is a measure of how tightly water is held in the soil. Step 2: Recall that tensiometers are commonly used in agriculture to monitor soil moisture status in terms of tension, guiding irrigation scheduling. Step 3: Recognise that photometers are used in optics to measure light intensity. Step 4: Remember that pyrometers are used in thermal physics to measure extremely high temperatures, such as in furnaces. Step 5: Know that psychrometers measure humidity in the air, not soil water tension. Therefore, tensiometer is the correct choice. Verification / Alternative check: Agricultural manuals and soil science textbooks describe tensiometers in detail as tools to measure soil matric potential in units like centibars or kilopascals. They are often installed at root zone depths and read periodically to decide when to irrigate. In contrast, photometers appear in optics labs, pyrometers in metallurgical or furnace monitoring, and psychrometers in meteorology and HVAC applications. This clear separation of applications confirms that tensiometer is uniquely associated with soil water tension. Why Other Options Are Wrong: Photometer: Used to measure light or illumination, not soil water properties. Pyrometer: Measures very high temperatures by radiation, not soil water tension. Psychrometer: Measures relative humidity and dew point using wet and dry bulb thermometers, not soil moisture suction. Common Pitfalls: Because the names sound similar, students may confuse psychrometer and tensiometer, both of which are used in environmental measurements. Another trap is thinking that pyrometer might be related to soil due to heat and sunlight, but its specific role is temperature measurement at high values. Carefully linking each instrument name to its main field of use helps avoid such confusion: photometer for light, pyrometer for high temperature, psychrometer for humidity, and tensiometer for soil water tension. Final Answer: The instrument used to measure soil water tension is the tensiometer.

Question 14

Which physical quantity has the SI unit siemens (symbol S)?

Accepted Answer

Introduction / Context: This question focuses on SI units in electricity and magnetism. In circuit theory, resistance and conductance are closely related but have different unit names. Conductance measures how easily electric current flows through a component. Recognising that siemens is the unit of conductance helps students correctly interpret circuit specifications and formulas involving reciprocal resistance. Given Data / Assumptions: We are working in SI units. Electric potential is measured in volts, resistance in ohms. Siemens is a named unit used for a specific electrical quantity. The options list electric potential, electrical conductance, magnetic flux, and refractive index. Concept / Approach: Electrical conductance, usually denoted by G, is defined as the reciprocal of resistance R: G = 1 / R. If resistance is measured in ohms, conductance is measured in siemens, symbol S, where 1 S = 1 / ohm. Electric potential is measured in volts, magnetic flux in webers, and refractive index is dimensionless. Since siemens is directly associated with conductance, that must be the correct answer. Step-by-Step Solution: Step 1: Recall that resistance R has SI unit ohm. Step 2: Recognise that conductance G is defined as G = 1 / R. Step 3: The unit of G must therefore be 1 / ohm, which has been given the special name siemens, symbol S. Step 4: Electric potential uses volt, magnetic flux uses weber, and refractive index has no unit (it is a ratio). Step 5: Conclude that siemens is the SI unit of electrical conductance. Verification / Alternative check: Electronic component datasheets for devices such as transconductance amplifiers or certain sensors sometimes specify conductance in siemens or microsiemens. Textbooks also describe the relation between ohm and siemens as reciprocal units. Practical applications such as water quality testing often quote electrical conductivity in siemens per metre, linking the basic unit siemens to conductance and conductivity, not to voltage or magnetic flux. Why Other Options Are Wrong: Electric potential: Measured in volts (V), not siemens. Magnetic flux: Measured in webers (Wb), not siemens. Refractive index: Dimensionless ratio, e.g., n = c / v, so it has no SI unit like siemens. Common Pitfalls: Students sometimes confuse conductivity and conductance. Conductance is for a specific component, with unit siemens, while conductivity is an intrinsic material property, often measured in siemens per metre. Another pitfall is to guess that siemens might relate to voltage or current because those are the most familiar electrical quantities. Remember the pair: ohm for resistance, siemens for conductance (its reciprocal), to keep these concepts straight. Final Answer: The SI unit siemens is used to measure electrical conductance.

Question 15

An electric current of 0.6 A flows through a circuit for 6 minutes. How much electric charge passes through the circuit in this time?

Accepted Answer

Introduction / Context: This numerical question is about the relationship between electric current, charge, and time. Electric current is defined as the rate of flow of electric charge. By knowing the current and the time for which it flows, we can calculate the total charge that has passed a given point in a circuit. This concept is fundamental in basic electricity and is widely used in designing and analysing electrical systems. Given Data / Assumptions: Current I = 0.6 A (ampere). Time t = 6 minutes. We assume constant current during this time interval. We are asked to find charge Q in coulombs. Concept / Approach: The definition of current is I = Q / t, where Q is charge in coulombs and t is time in seconds. Rearranging, we obtain Q = I * t. Before using this formula, we must convert time from minutes to seconds, because the ampere is defined in terms of coulombs per second. Once time is converted and multiplied by current, the result will give the total charge. Step-by-Step Solution: Step 1: Convert time from minutes to seconds: t = 6 minutes = 6 * 60 seconds = 360 seconds. Step 2: Use the relation Q = I * t. Step 3: Substitute the values: Q = 0.6 A * 360 s. Step 4: Calculate the product: 0.6 * 360 = 216. Step 5: Attach the correct unit: Q = 216 C (coulombs). Verification / Alternative check: We can check the calculation by breaking it into smaller steps: 0.6 * 100 = 60, 0.6 * 300 = 180, and 0.6 * 60 = 36. Adding 180 and 36 gives 216, matching the previous computation. Another way is to recognise that 0.6 A means 0.6 coulombs per second. In 360 seconds, the total charge is 360 times 0.6, again yielding 216 C. Both methods agree, confirming the answer. Why Other Options Are Wrong: 360 C: This would be the charge for a current of 1 A flowing for 360 seconds, not 0.6 A. 60 C: This corresponds to a much shorter time or smaller current, indicating a calculation error. 36 C: This is also far too small and likely comes from misplacing a decimal or incorrect multiplication. Common Pitfalls: Students often forget to convert minutes to seconds and mistakenly use 6 instead of 360 in the formula, giving Q = 0.6 * 6 = 3.6 C, which is incorrect. Others might miscalculate 0.6 * 360 due to handling decimals. To avoid errors, always convert time to seconds for current calculations and multiply carefully. Understanding that 1 A equals 1 C per second helps reinforce the need for consistent time units. Final Answer: The amount of electric charge flowing through the circuit is 216 C.

Question 16

Which of the following materials cannot be used for making optical lenses because it is not transparent?

Accepted Answer

Introduction / Context: This question belongs to basic optics and materials. Lenses are transparent optical elements that refract light to form images. To function as a lens, a material must allow light to pass through it with minimal scattering and absorption. Understanding which materials can or cannot be used for lenses reinforces the idea that transparency is essential for image formation. Given Data / Assumptions: The question asks about materials used for making lenses. Options given are plastic, soil, glass, and water. We assume normal visible light and everyday optical lenses. Concept / Approach: Lenses require materials that are transparent so that light can pass through and be refracted at the surfaces to focus or diverge beams. Common lens materials include glass and various plastics, and even water can act as a lens in simple demonstrations because it is transparent. Soil, however, is an opaque and heterogeneous mixture of particles and cannot transmit light clearly, so it cannot be shaped into a functional optical lens. Step-by-Step Solution: Step 1: Check which materials are transparent to visible light. Glass, many plastics, and clean water are transparent. Step 2: Recognise that soil, composed of sand, clay, organic matter, and stones, blocks light and is not transparent. Step 3: Note that lenses must transmit light; otherwise they cannot form images by refraction. Step 4: Identify that glass is widely used for high quality lenses in spectacles, cameras, and microscopes. Step 5: Realise that plastics are used for lightweight, shatter resistant lenses, and even water can form a temporary lens in demonstrations. Therefore, soil is the only material listed that cannot be used for making lenses. Verification / Alternative check: In practical life, we see glass lenses in cameras and spectacles, plastic lenses in inexpensive optics and sunglasses, and water drops acting as small lenses on surfaces. No optical devices are made from soil, because it does not permit the formation of clear images. Simple experiments with a spoonful of soil show that it blocks light instead of transmitting it, contrasting sharply with the behaviour of glass or clear plastic. Why Other Options Are Wrong: Plastic: Transparent plastics like acrylic and polycarbonate are commonly moulded into lenses. Glass: Traditional and high quality optical lenses are made from specially formulated glass. Water: Although not used in permanent instruments, water droplets and containers of water can act as lenses due to transparency and refraction. Common Pitfalls: Students sometimes focus on the solid or liquid state of a substance rather than its optical properties. They may assume only solids can be lenses, overlooking water lenses. The critical property is transparency and refractive index, not whether the material is a solid or liquid. Soil is a mixture of opaque particles and does not meet this requirement. Remember: if light cannot pass through a material, it cannot be used as a lens in the usual optical sense. Final Answer: Soil cannot be used for making lenses because it is not transparent.

Question 17

When a card placed over the mouth of a tumbler is flicked sharply with a finger, the coin resting on the card falls straight down into the tumbler. Which law of motion does this demonstrate?

Accepted Answer

Introduction / Context: This classic classroom demonstration with a card, coin, and tumbler illustrates a fundamental principle of mechanics. When the card is flicked away quickly, the coin appears to fall almost straight down into the tumbler instead of moving sideways with the card. This behaviour is a vivid example of inertia, as described by Newton's first law of motion. Given Data / Assumptions: A coin is placed on a card, which itself rests horizontally on the mouth of a tumbler. The card is flicked sharply in a horizontal direction. Air resistance and friction between coin and card are relatively small during the very short flick. The question asks which law of motion this setup demonstrates. Concept / Approach: Newton's first law of motion, also known as the law of inertia, states that a body at rest will remain at rest, and a body in uniform motion will continue in that state of motion, unless acted upon by a net external force. Initially, the coin is at rest. When the card is flicked quickly, only a small horizontal frictional force acts on the coin for a very short time, so the coin tends to remain in its state of rest. As the card moves out of the way, the coin no longer has support and falls vertically into the tumbler under gravity. This shows the tendency of matter to resist changes in its state of motion, that is, inertia. Step-by-Step Solution: Step 1: Observe that initially both the card and coin are at rest on top of the tumbler. Step 2: When the card is flicked horizontally, a large acceleration is given to the card, but the frictional force between card and coin is small and acts for a very short time. Step 3: Because of inertia, the coin tends to remain at rest and does not acquire the same horizontal speed as the card. Step 4: As soon as the card moves away, the coin loses its support and falls vertically downward under gravity, landing in the tumbler. Step 5: This behaviour directly illustrates Newton's first law or the law of inertia, not energy or momentum conservation laws. Verification / Alternative check: Similar demonstrations of inertia include pulling a tablecloth quickly from under dishes or yanking a sheet of paper from under a glass. In each case, objects resist change in their state of motion and tend to remain at rest if the applied force acts for only a short time. Textbooks explicitly describe the coin and card experiment as an example of Newton's first law, reinforcing that the correct law here is the law of inertia. Why Other Options Are Wrong: Law of conservation of energy: Total energy is indeed conserved, but this experiment is not primarily used to illustrate energy transformations. Newton's third law of motion: This law deals with action and reaction forces, which are not the main focus of the demonstration. Law of conservation of momentum: While momentum concepts are present in many interactions, this simple setup is not usually framed as a momentum conservation demonstration. Common Pitfalls: Students sometimes choose Newton's third law for any situation involving motion and forces, but it is important to match the key idea. Here the key is that the coin tends to remain at rest despite the motion of the card. Another error is to think that because the coin falls into the tumbler, energy or momentum laws are the main lesson. In reality, the experiment is used in most textbooks specifically to show inertia and Newton's first law. Final Answer: This demonstration illustrates the law of inertia, that is, Newton's first law of motion.

Question 18

A person does 1,000 J of mechanical work in 2 seconds. What is the power developed by the person?

Accepted Answer

Introduction / Context: This straightforward numerical question is about the concept of power in physics. Power measures how quickly work is done or energy is transferred. It is commonly used for rating electrical appliances, engines, and human physical performance. Correctly using the relation between work, time, and power is a fundamental skill in basic mechanics and energy calculations. Given Data / Assumptions: Work done, W = 1,000 J (joules). Time taken, t = 2 s (seconds). We assume the rate of doing work is constant over this time interval. We want to find power P in watts (W). Concept / Approach: Power is defined as the rate of doing work or the rate of energy transfer. Mathematically, P = W / t, where W is work done and t is time taken. In SI units, 1 watt is equal to 1 joule per second. By substituting the given values of work and time into this formula, we can directly calculate the power in watts. Step-by-Step Solution: Step 1: Write the formula for power: P = W / t. Step 2: Substitute the given values: W = 1,000 J and t = 2 s. Step 3: Calculate the quotient: P = 1,000 / 2. Step 4: Perform the division: 1,000 / 2 = 500. Step 5: Attach the correct unit: P = 500 W. Verification / Alternative check: Think in terms of rates: if a device uses 500 J of energy every second, then in 2 seconds it would use 1,000 J, corresponding to a power of 500 W. This matches the numerical calculation. Everyday devices like a 500 W mixer or a 500 W heater use energy at a similar rate, making the answer physically reasonable for the amount of work described over 2 seconds. Why Other Options Are Wrong: 1000 W: This would correspond to doing 1,000 J of work in 1 second, not in 2 seconds. 25 W and 50 W: These are much smaller power values; at 50 W, it would take 20 seconds to do 1,000 J of work, and at 25 W, even longer. Common Pitfalls: Students sometimes confuse power with work and pick 1,000 W simply because 1,000 J appears in the problem statement. Others may incorrectly multiply work and time instead of dividing. Some also forget that the SI unit of power is the watt and mix it up with joule. Remember the basic definition P = W / t in joules per second, and you will consistently obtain correct power values with the proper unit. Final Answer: The power developed by the person is 500 W.

Question 19

In materials science and basic physics, which mechanical property of metals allows them to be beaten into very thin metallic foils without breaking?

Accepted Answer

Introduction / Context: In basic materials science and school level physics, metals are often described using a set of characteristic mechanical properties such as malleability, ductility, hardness, and brittleness. These properties explain why metals can be made into wires, sheets, utensils, or structural members. The question focuses specifically on metallic foils, which are extremely thin sheets of metal used for packaging, cooking, electrical shielding, and many other applications. Understanding which property allows a block of metal to be beaten or rolled into a thin sheet without cracking is important for linking real life products to underlying physical concepts. Given Data / Assumptions: • We are dealing with metals being converted into very thin foils or sheets. • The options mention malleability, ductility, sonority, and conductivity. • Metallic foils are assumed to remain in one continuous piece without shattering. • Standard textbook definitions of these mechanical properties are used. Concept / Approach: Malleability is defined as the ability of a material, especially a metal, to be hammered, rolled, or pressed into thin sheets without breaking or cracking. Ductility is a related property but specifically refers to the ability to be drawn into wires. Sonority describes the ringing sound metals make when struck. Conductivity refers to the ability of metals to conduct heat and electricity. Because metallic foils are thin sheets, the property that directly explains their production is malleability. Therefore, the correct option must be the one that explicitly associates malleability with hammering or rolling into sheets. Step-by-Step Solution: Step 1: Recall the formal definition of malleability as the capacity of a metal to be shaped into thin sheets by hammering or rolling. Step 2: Recall the formal definition of ductility as the capacity of a metal to be drawn into long thin wires. Step 3: Note that metallic foils such as aluminium foil are extremely thin sheets, not wires. Step 4: Recognise that the manufacturing processes for foils use rolling mills and pressing operations which test malleability. Step 5: Observe that sonority relates only to sound and conductivity relates to heat and electricity, not shaping. Step 6: Conclude that malleability is the key property that enables the production of metallic foils. Verification / Alternative check: If you look at common examples, aluminium is highly malleable and is widely used to make kitchen foil and packaging. Gold is extremely malleable and can be beaten into gold leaf only a few atoms thick. Copper and silver are both malleable and ductile, allowing them to be used both as sheets and wires. In contrast, brittle materials such as glass or ceramic cannot be hammered into foils because they crack instead of deforming plastically. This real world evidence confirms that the foil forming ability is tied to malleability rather than to ductility or other properties. Why Other Options Are Wrong: Option B, ductility, is about drawing metal into wires, such as copper electrical wiring, and does not by itself explain sheet formation. Option C, sonority, only describes the ringing sound produced when metal is struck and has no relation to shaping or deformation. Option D, conductivity, explains why metals are good electrical and thermal conductors but not why they can be beaten into foils. None of these alternatives matches the sheet forming behaviour described in the question. Common Pitfalls: A common confusion among students is mixing up malleability and ductility because both describe plastic deformation. It is helpful to remember the association wires go with ductility and sheets or foils go with malleability. Another pitfall is to assume that any familiar property of metals, such as conductivity or sonority, must be the answer simply because it sounds characteristic. Always relate the property directly to the type of mechanical shaping mentioned. This simple check quickly guides you to the right term in exam questions. Final Answer: The correct choice is Malleability, the ability to be hammered into thin sheets, because malleability is the mechanical property that specifically allows metals to be beaten or rolled into very thin metallic foils without breaking.

Question 20

A person lifts a piece of luggage of mass 20 kg from the ground and places it on his head at a height of 2 m above the ground. Taking g = 10 m/s^2, what is the work done by him against gravity?

Accepted Answer

Introduction / Context: Work, energy and power are core topics in mechanics. In everyday life, lifting a suitcase or a bucket of water seems like a simple action, but in physics it directly illustrates the concept of work done against gravitational force. Whenever a person raises an object vertically, they increase its gravitational potential energy. This question asks you to calculate the work done by a person when lifting a 20 kg piece of luggage from the ground to a height of 2 m on his head, using a rounded value for the acceleration due to gravity. It is a standard example of work done in raising a body to a certain height. Given Data / Assumptions: • Mass of the luggage, m = 20 kg. • Vertical height raised, h = 2 m above the ground. • Acceleration due to gravity, g = 10 m/s^2 (approximate value). • The luggage is lifted slowly so that its kinetic energy change is negligible and motion is vertical. • Air resistance and other small losses are ignored, so work done equals gain in gravitational potential energy. Concept / Approach: In mechanics, the work W done against gravity in lifting a body of mass m through a vertical height h is equal to the increase in gravitational potential energy and is given by the formula W = m * g * h. This is because the gravitational force acting downward on the object has magnitude m * g, and moving the object upward through distance h against this force requires doing positive work. By substituting the given mass, gravitational acceleration, and height into this formula, we can directly compute the work done in joules. This approach is a straightforward application of the definition of work done by a constant force. Step-by-Step Solution: Step 1: Write the formula for work done against gravity when lifting an object vertically: W = m * g * h. Step 2: Substitute the given mass m = 20 kg into the formula. Step 3: Substitute the given acceleration due to gravity g = 10 m/s^2. Step 4: Substitute the given height h = 2 m. Step 5: Calculate W = 20 * 10 * 2 = 400 joules. Step 6: Conclude that the person does 400 J of work against gravity in lifting the luggage to his head. Verification / Alternative check: To check the calculation, you can first calculate the weight of the luggage. The weight is the gravitational force F = m * g = 20 * 10 = 200 newtons. Then, use the work formula W = F * s, where s is the displacement in the direction of the force. Here, the force that must be overcome is 200 newtons, and the displacement is 2 m upward. So W = 200 * 2 = 400 joules, which matches the earlier result. This confirms that 400 J is consistent whether you use the potential energy method or the direct force displacement approach. Why Other Options Are Wrong: Option A, 20 J, is far too small because it effectively uses only the mass without multiplying by g and height. Option C, 200 J, corresponds to using either half of the correct height or forgetting one factor of 2 when multiplying. Option D, 40 J, is also much too small and may come from confusing kilograms with newtons. None of these incorrect options satisfy the correct formula W = m * g * h with the given values of mass, gravity, and height. Common Pitfalls: A common mistake is to forget to convert mass into force by multiplying by g, leading students to use W = m * h instead of W = m * g * h. Another error is mixing units such as treating kilograms as newtons directly. Some learners also confuse work with power and attempt to include time, which is unnecessary unless the rate of doing work is being asked. Always remember that for vertical lifting in a uniform gravitational field, the work done against gravity depends only on the mass, gravitational acceleration, and vertical height raised. Final Answer: The correct choice is 400 J, because the work done against gravity in lifting the 20 kg luggage through a vertical height of 2 m with g = 10 m/s^2 is W = 20 * 10 * 2 = 400 joules.

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