Question 1

A household washing machine separates water from wet clothes during the spin or drying cycle. On which physical principle does this process mainly work?

Accepted Answer

Introduction / Context: This question links a common household appliance, the washing machine, with a basic physics principle. During the spin or drying cycle, the drum rotates rapidly and water is thrown out of the clothes. Understanding which physical process is responsible for this separation helps relate textbook physics to daily life and clarifies the meaning of terms like centrifugation, diffusion, dialysis and reverse osmosis. Given Data / Assumptions: We are analysing the spin or drying cycle of a washing machine. The drum rotates at high speed, and water moves out through holes in the drum. The options list four processes: dialysis, diffusion, reverse osmosis, and centrifugation. We assume the machine is functioning normally as a clothes washer and spinner. Concept / Approach: In the spin cycle, the washing machine drum rotates rapidly, and clothes along with water are forced outward against the drum wall. Due to centripetal motion, water experiences an outward effective force and passes through the perforations, while the clothes remain inside. This process of separating components of a mixture using rapid circular motion is called centrifugation. Dialysis separates solutes across a membrane, diffusion is random spreading of particles, and reverse osmosis uses pressure to force solvent through a semipermeable membrane. None of these describe the spinning separation in a washing machine as accurately as centrifugation. Step-by-Step Solution: Step 1: Observe that in a spin dryer, the drum rotates at high angular speed. Step 2: Clothes and water are pressed against the drum wall; water is forced through small holes due to the outward effect of circular motion. Step 3: This type of separation by rapid rotation is defined in physics and chemistry as centrifugation. Step 4: Dialysis involves selective diffusion of solutes through a membrane, not spinning of a drum. Step 5: Reverse osmosis uses high pressure across a semipermeable membrane, which does not match the washing machine spin action. Thus, centrifugation is the correct principle. Verification / Alternative check: Centrifuges in laboratories work on the same idea: they spin test tubes at high speed to separate heavier components from lighter ones. The washing machine spin dryer is often described in textbooks as a real life example of centrifugation. In both devices, rotation creates a large outward effective force on the liquid, helping it separate from solids. This close analogy confirms that the washing machine operates on the principle of centrifugation during the spin cycle. Why Other Options Are Wrong: Dialysis: Used in medical treatment and chemistry for separating small solutes from large molecules across a membrane, not for spinning clothes. Diffusion: A slow process driven by random motion of particles from high to low concentration, not responsible for rapid water removal in a spin cycle. Reverse osmosis: Involves forcing solvent through a semipermeable membrane using pressure; washing machines do not use such membranes to remove water in the spin stage. Common Pitfalls: Students sometimes mix centrifugation with diffusion or osmosis because all deal with separation. However, the key sign of centrifugation is rapid circular motion and outward separation due to rotation. Whenever you see spinning devices used to separate mixtures, such as cream separators, blood centrifuges or spin dryers, think of centrifugation rather than membrane based processes. Final Answer: The washing machine works mainly on the principle of centrifugation during the spin cycle.

Question 2

A ball is thrown vertically upward into the air. Neglecting air resistance, which of the following physical quantities does not change during its entire motion up and down?

Accepted Answer

Introduction / Context: This question tests understanding of motion under gravity. When a ball is thrown vertically upward, it rises, slows down, momentarily comes to rest at the top, and then falls back down. The behaviour of speed, potential energy, distance from the ground, and acceleration during this motion is a key topic in kinematics and helps students apply Newton laws of motion to real situations. Given Data / Assumptions: A ball is thrown vertically upward from near the Earth surface. Air resistance is neglected for simplicity. The only significant force during the motion is the gravitational force acting downward. We are asked which quantity does not change throughout the motion up and down. Concept / Approach: Near the Earth surface, and neglecting air resistance, the acceleration due to gravity has a constant magnitude g directed downwards. As the ball goes up, its speed decreases due to this downward acceleration. Its height above the ground increases, so its gravitational potential energy increases. On the way down, the reverse happens. However, at every instant during the flight, the gravitational acceleration is the same in magnitude and direction. Thus, acceleration remains constant, while speed, potential energy, and distance from the ground all change with time. Step-by-Step Solution: Step 1: Identify the force acting on the ball during flight. The only significant force is weight, mg, acting downward. Step 2: From Newton second law, this constant downward force produces a constant downward acceleration of magnitude g. Step 3: As the ball moves upward, velocity is upward but acceleration is downward, so the speed decreases until it becomes zero at the highest point. Step 4: As the ball falls back down, velocity is downward and acceleration remains downward, so speed increases. Step 5: The height from the ground and gravitational potential energy both change continuously, while acceleration remains unchanged in magnitude and direction, so acceleration is the quantity that does not change. Verification / Alternative check: The standard equations of motion for constant acceleration, such as v = u + a t and s = u t + (1/2) a t^2, are used to describe freely falling bodies and vertical throws. In these equations, acceleration a is constant and equal to minus g for upward motion when upward is taken as positive. Experimental measurements using motion sensors also show that the acceleration magnitude remains roughly constant near the Earth surface, confirming that acceleration is unchanged while other quantities vary. Why Other Options Are Wrong: Speed: Speed decreases on the way up, becomes zero at the top, and increases on the way down, so it clearly changes. Potential energy: As height changes, gravitational potential energy m g h also changes, increasing and then decreasing. Distance from the ground: The vertical distance from the ground is constantly changing, except at two instants when it may pass the same height. Common Pitfalls: Many learners instinctively think that at the highest point acceleration becomes zero because the speed is zero. This is incorrect. Acceleration is related to change of velocity, not current velocity value. At the top, the velocity is zero for an instant, but acceleration is still downward, causing the ball to start falling. Confusing velocity with acceleration leads to wrong answers. Remember that in ideal free fall problems near Earth, acceleration due to gravity is treated as constant in magnitude and direction. Final Answer: When a ball is thrown vertically upward, the quantity that does not change is the acceleration due to gravity.

Question 3

A body moves along a circular path and completes one full revolution in one second. What is the average velocity of the body over this complete cycle?

Accepted Answer

Introduction / Context: This question examines the difference between average speed and average velocity for circular motion. A body moving around a circle may have non zero speed at all times, yet its average velocity over a full revolution can be zero. Understanding this difference is fundamental in kinematics because speed is a scalar, while velocity is a vector quantity that depends on displacement between initial and final positions. Given Data / Assumptions: A body moves along a circular path. It completes exactly one full revolution in one second. The motion is along a closed path, starting and ending at the same point. We are asked for average velocity over the complete cycle, not instantaneous velocity. Concept / Approach: Average velocity is defined as total displacement divided by total time. Displacement is the straight line vector from initial position to final position. For one complete revolution in a circle, the starting point and ending point coincide, so the overall displacement is zero, even though the body has travelled a non zero distance. Since average velocity uses displacement, not path length, the average velocity over one complete cycle is zero. The time taken, one second, does not change this fact because zero displacement divided by any non zero time is zero. Step-by-Step Solution: Step 1: Recall the definition of average velocity: average velocity = total displacement / total time. Step 2: For one full revolution in a circle, the body returns to its starting point, so total displacement is zero. Step 3: The time taken for this full revolution is given as one second. Step 4: Substitute into the formula: average velocity = 0 / 1 second = 0. Step 5: Therefore, the average velocity over one complete cycle is zero, even though the body has a non zero speed throughout. Verification / Alternative check: If the body goes around the circle multiple times and always returns to its starting point, then after each complete number of revolutions its displacement is still zero. For any such interval, average velocity remains zero. However, the average speed is total distance travelled divided by time, which is circumference multiplied by the number of revolutions divided by time, and is non zero. This clearly shows that average speed and average velocity can be very different for circular motion. Why Other Options Are Wrong: Average velocity depends only on the time taken to complete one cycle: This is incorrect because displacement is the primary factor. Time alone cannot determine average velocity. One metre per second: This is an arbitrary value and does not follow from the given information about displacement. Average velocity is equal to average speed: False for circular motion over a full revolution because average speed is non zero while average velocity is zero. Common Pitfalls: Students often confuse distance and displacement. They know the body has moved, so they assume average velocity must be non zero. The key is to remember that displacement is a vector from start to end. If start and end coincide, displacement is zero, regardless of the path taken. Another error is to think that velocity must reflect the continuous motion along the path rather than the net change in position. Keeping the definitions clear for scalar speed and vector velocity helps avoid these misunderstandings. Final Answer: The average velocity of the body over one complete circular revolution is zero.

Question 4

According to Archimedes principle, the upward force acting on a body that is wholly or partly immersed in a fluid is called what?

Accepted Answer

Introduction / Context: This question is from fluid mechanics and tests knowledge of the basic term used for the upward force experienced by a body immersed in a liquid or gas. Archimedes principle explains why objects float or sink and is widely applied in ship design, hydrometers, and many practical situations. Knowing the correct name of this upward force is important for understanding floating bodies and related calculations. Given Data / Assumptions: A body is wholly or partly immersed in a fluid such as water or air. An upward force from the fluid acts on the body. We are asked for the standard term used to describe this upward force. Archimedes principle and basic fluid concepts are assumed. Concept / Approach: Archimedes principle states that a body immersed in a fluid experiences an upward force equal to the weight of the fluid displaced by it. This upward force is called buoyant force, and the effect is called buoyancy. This force acts opposite to the weight of the body and is responsible for floating and sinking behaviour. The term jerk refers to the rate of change of acceleration, anti gravity is a speculative or science fiction idea, and Archimedean force is not the standard textbook term, even though it refers to the same concept informally. The accepted term is buoyancy or buoyant force. Step-by-Step Solution: Step 1: Recall Archimedes principle: an immersed body experiences an upward force equal to the weight of displaced fluid. Step 2: Identify that this upward force is usually called the buoyant force in physics texts. Step 3: Recognise that buoyancy refers to the tendency of a fluid to exert this upward force on objects. Step 4: Check the options and note that buoyancy matches the standard term. Step 5: Conclude that the upward force on a floating body is called buoyancy or buoyant force. Verification / Alternative check: Examples include ships that float because the buoyant force balances their weight, and balloons that rise in air. Textbooks refer to the upward force as buoyant force and to the phenomenon as buoyancy. Hydrometers that measure relative density also rely on buoyant force. No text uses jerk or anti gravity as the name of this force. While the phrase Archimedean force might be used informally to honour Archimedes, it is not the standard term used in definitions or formulas. Why Other Options Are Wrong: Jerk: A kinematics term meaning rate of change of acceleration, unrelated to floating objects in fluids. Archimedean force: Not a standard textbook term, even though it hints at Archimedes principle. Anti gravity: A popular term in science fiction, not a recognised force in elementary physics. Common Pitfalls: Learners sometimes choose anti gravity because it sounds like a force opposite to gravity. However, in real physics, the upward force in fluids is explained entirely by pressure differences and is called buoyant force. Another confusion arises from mixing the principle name with the force name and selecting Archimedean force. Remember that the principle is called Archimedes principle, but the force is called buoyant force and the effect is buoyancy. Final Answer: The upward force on a floating or immersed body is called buoyancy or buoyant force.

Question 5

Women are often perceived to have a more shrill voice than men. From a physics point of view, this is mainly because their vocal sound waves generally have what characteristic?

Accepted Answer

Introduction / Context: This question connects basic wave physics with human speech. In everyday language, we say a voice is shrill, deep, or high pitched. Physics explains these perceptions using the concepts of frequency and wavelength of sound waves. Understanding how pitch relates to frequency helps link human voices to fundamental properties of sound. Given Data / Assumptions: Women are often described as having a higher pitched or shriller voice compared with many men. Pitch in physics is primarily related to the frequency of sound waves. The options list low frequency, high frequency, and some biological or vague descriptions. We assume normal speaking voices, not special training or individual exceptions. Concept / Approach: Pitch of a sound is directly related to the frequency of the sound wave. Higher frequency sounds are perceived as high pitched or shrill, while lower frequency sounds are perceived as deep or bass. On average, female vocal cords are shorter and thinner, which tend to produce sound waves with higher fundamental frequency. Hence, women often have higher frequency voices, which are heard as shriller. This is a general trend and not a strict rule for every individual, but it explains the typical perception. Step-by-Step Solution: Step 1: Recall that frequency is the number of sound wave cycles produced per second and is measured in hertz. Step 2: Understand that higher frequency corresponds to higher pitch and more shrill sound. Step 3: Recognise that average female speech has a higher fundamental frequency than average male speech because of differences in vocal cord size and tension. Step 4: Therefore, the shrill quality of many women voices is mainly due to higher frequency sound waves. Step 5: Compare this with options and identify high frequency as the correct physics explanation. Verification / Alternative check: Measured typical fundamental frequency ranges for adult male voices are often around 85 to 180 hertz, while adult female voices commonly range from about 165 to 255 hertz. These values show that female voices typically have higher frequency. Audio software that displays sound frequency also shows this difference. Speakers or instruments that produce high frequency tones are perceived as shrill regardless of the source, confirming that pitch is tied to frequency, not some separate loosly defined property. Why Other Options Are Wrong: Low frequency: Low frequency sounds are deep, not shrill, and usually associated with male voices or bass instruments. Shrill vocals as a separate property: Shrillness is not a separate physical quantity; it is simply our perception of higher frequency sounds. Strong epiglottis muscles: The epiglottis protects the airway but does not directly define pitch. The main physics factor is frequency of vibration of vocal cords. Common Pitfalls: Some students confuse loudness and pitch. Louder sounds have greater amplitude but may not be high frequency. A loud male voice can still be deep and not shrill. Others think anatomical terms such as epiglottis or diaphragm control pitch directly. While anatomy influences how vocal cords vibrate, the physics explanation of shrillness rests on frequency. Keeping amplitude and frequency as separate properties helps clarify the difference between loudness and pitch. Final Answer: Women commonly have shrill voices because their vocal sound waves generally have higher frequency.

Question 6

When a block of ice floats on water in a pond or glass, approximately what fraction of its total volume remains above the water surface?

Accepted Answer

Introduction / Context: This question is an application of Archimedes principle to the case of ice floating on water. Students often observe that ice cubes or icebergs have only a small portion visible above the water, while most of the volume remains submerged. Understanding the fraction of volume above water reinforces the relationship between density, buoyant force, and floating equilibrium. Given Data / Assumptions: A piece or block of ice floats at rest on the surface of liquid water. We assume pure ice and pure water at conditions where both phases can coexist. The density of ice is slightly less than that of water, often approximated as about 0.9 times water density. We are interested in what fraction of the ice volume remains above the water surface. Concept / Approach: For a floating body in equilibrium, the weight of the body equals the buoyant force, which equals the weight of displaced fluid. Let rho_ice be the density of ice and rho_water be the density of water. The fraction of ice volume submerged is approximately rho_ice / rho_water. With rho_ice roughly 0.9 times rho_water, about 0.9 of the volume is submerged and 0.1 of the volume remains above water. Thus, around one tenth of the ice block sticks out above the surface, which matches common observations of floating ice and icebergs. Step-by-Step Solution: Step 1: Use Archimedes principle: weight of floating ice = weight of displaced water. Step 2: Let V be the total volume of the ice block and V_sub the submerged volume. Then rho_ice * V * g = rho_water * V_sub * g. Step 3: Simplify to V_sub / V = rho_ice / rho_water. Step 4: Use typical densities: rho_ice is about 0.9 times rho_water, so V_sub / V is approximately 0.9. Step 5: Therefore, the fraction above water is 1 minus 0.9, which is about 0.1, or one tenth of the total volume. Verification / Alternative check: Practical observation supports this theoretical estimate. An ice cube in a glass of water appears mostly submerged, with only a small cap visible. Photographs of icebergs at sea also show relatively small tops compared to their total size, with most of their mass underwater. These real world observations consistently suggest that roughly one tenth of the volume is above the surface. While the exact fraction depends slightly on temperature and purity, 0.1 is a good approximation for school level questions. Why Other Options Are Wrong: 0.5 (half of the volume): If half the ice were above water, then the density ratio would be drastically different and ice would be much lighter than water, which is not true. 0.3 (about one third): This overestimates the above water portion; the actual fraction is closer to one tenth. 1 (the whole block remains outside): This would mean ice does not sink at all and would not be floating in the usual sense, which contradicts observation. Common Pitfalls: Students may guess that half the block is above water because they think of symmetry, not density. Others may confuse the idea of floating with fully staying outside the liquid. Remember that floating means the weight of the displaced fluid balances the weight of the object. For materials like ice with density slightly less than water, most of the volume must be submerged to displace enough water, leaving only a small fraction above. Using the density ratio is a reliable method to avoid guessing. Final Answer: When ice floats on water, roughly 0.1 or about one tenth of its volume remains above the water surface.

Question 7

In the International System of Units, one nanometre is equal to how many metres?

Accepted Answer

Introduction / Context: This question tests knowledge of metric prefixes and very small length scales. The nanometre is widely used in nanotechnology, semiconductor fabrication, and molecular physics to describe atomic and molecular dimensions. Being comfortable with powers of ten and unit conversions is essential in all branches of science and engineering. Given Data / Assumptions: We are working with standard SI prefixes for length based on the metre. The term nanometre uses the prefix nano. We need to convert one nanometre into metres using powers of ten. Options list various powers of ten in metres. Concept / Approach: In the SI system, prefixes represent powers of ten. The prefix nano stands for 10^-9. Thus, one nanometre is 10^-9 of a metre. For comparison, micro is 10^-6, milli is 10^-3, and pico is 10^-12. Remembering the order of these prefixes helps in converting between units. Since the question is specifically about nanometre, we apply the nano factor directly. Step-by-Step Solution: Step 1: Recall that the SI prefix nano means 10^-9. Step 2: One nanometre is therefore one unit times 10^-9, expressed relative to metre. Step 3: Write this as 1 nm = 10^-9 m. Step 4: Compare this with the given options 10^3, 10^-6, 10^-9, and 10^-12 metres. Step 5: The correct match is 10^-9 m. Verification / Alternative check: You can place nano in context with nearby prefixes. A micrometre is 10^-6 m, a nanometre is 10^-9 m, and a picometre is 10^-12 m. Thus, 1 nm is one thousand times smaller than 1 micrometre and one thousand times larger than 1 picometre. In atomic physics, typical atomic diameters are a fraction of a nanometre, which again fits the range implied by 10^-9 m. These cross checks confirm that nano corresponds to 10^-9. Why Other Options Are Wrong: 10^3 m: This equals 1000 metres and corresponds to the prefix kilo, used for kilometre, not nanometre. 10^-6 m: This is the micrometre scale and corresponds to the prefix micro, not nano. 10^-12 m: This is the picometre scale and corresponds to the prefix pico, smaller than a nanometre. Common Pitfalls: Learners may confuse the order of prefixes, especially micro and nano, because both refer to very small lengths. Another error is to misread the negative exponent and treat 10^-9 as 10^9. To avoid confusion, it is helpful to memorise a short list of key prefixes: milli 10^-3, micro 10^-6, nano 10^-9, pico 10^-12, and kilo 10^3. Writing them out in order on a number line of exponents can also reinforce the correct relationships. Final Answer: One nanometre is equal to 10^-9 metre.

Question 8

Which of the following physical quantities is a vector quantity rather than a scalar quantity?

Accepted Answer

Introduction / Context: This question tests the understanding of the difference between scalar and vector quantities in physics. Scalars have magnitude only, whereas vectors have both magnitude and direction. Being able to classify physical quantities correctly is essential for setting up equations of motion, torque, rotational dynamics, and energy calculations. Given Data / Assumptions: The options include heat, angular momentum, work, and time. We must identify which quantity is a vector. Standard school level definitions of these quantities are assumed. Direction and sign conventions follow usual physics usage. Concept / Approach: Heat, work, and time are scalar quantities. They have magnitude but no intrinsic direction in space. For example, time flows forward, but it is not treated as a vector in mechanics. Anglular momentum, on the other hand, is a vector quantity related to rotational motion. It has a magnitude and a direction given by the right hand rule, perpendicular to the plane of rotation. Thus, among the given options, angular momentum is the only vector quantity. Step-by-Step Solution: Step 1: Recall that a vector quantity has both magnitude and direction and obeys vector addition rules. Step 2: Heat is a scalar quantity representing energy transfer due to temperature difference; it has no fixed direction in space. Step 3: Work is also scalar, defined as force dot displacement and can be positive or negative, but it does not have a spatial direction vector. Step 4: Time is a scalar parameter, not a vector quantity, even though it increases in one sense. Step 5: Angular momentum is associated with rotational motion and has direction according to the right hand rule, so it is a vector. Hence, angular momentum is the correct answer. Verification / Alternative check: Mathematically, angular momentum L is defined as r cross p, where r is position vector and p is linear momentum vector. The cross product of two vectors is always a vector, confirming that angular momentum is a vector. In contrast, heat and work are measured just by magnitudes in joules. Laws of addition for heat and work are simple algebraic addition, not vector addition. Time is not represented by arrows in space diagrams, reinforcing that it is scalar. Why Other Options Are Wrong: Heat: Scalar quantity with units of joule, representing energy transfer, no direction in space. Work: Scalar quantity equal to force component along displacement times displacement, also in joules. Time: A scalar parameter that orders events but is not treated as a vector in mechanics problems. Common Pitfalls: Students sometimes think any quantity that can be positive or negative must be a vector, which is not correct. Scalars like work can be negative, but they still have no associated spatial direction. Another confusion is between momentum and angular momentum. Both are vectors, but this question includes only angular momentum. Recognising definitions and how quantities appear in vector equations is the safest way to classify them correctly. Final Answer: Angular momentum is the vector quantity among the given options.

Question 9

If the speed of an object moving along a straight line keeps changing with time, what is the nature of its motion called?

Accepted Answer

Introduction / Context: This question is about basic classification of motion in kinematics. When describing how an object moves along a path, we often distinguish between uniform and non uniform motion. Recognising these types helps in selecting the correct equations and graphs, such as distance time and velocity time relations. Given Data / Assumptions: An object moves along a straight line path. The speed of the object is not constant; it keeps changing with time. We are asked to name the type of motion based on this behaviour of speed. We assume one dimensional motion with no complications like oscillations. Concept / Approach: Uniform motion in a straight line means that the object covers equal distances in equal intervals of time and has constant speed. Non uniform motion means that the speed changes from one moment to another, so distances covered in equal time intervals are not the same. Periodic motion refers to motion that repeats after a fixed time, and circular motion refers to motion along a circular path. Since the object in this question is in a straight line and its speed keeps changing, the correct description is non uniform motion. Step-by-Step Solution: Step 1: Recall that uniform motion in a straight line implies constant speed. Step 2: The question states that the speed keeps changing, which directly rules out uniform motion. Step 3: Periodic motion requires repeating cycles, such as oscillations, which are not mentioned here. Step 4: Circular motion requires a circular path, whereas the problem clearly states a straight line path. Step 5: Since speed changes along a straight line, the motion is classified as non uniform motion. Verification / Alternative check: Consider a car driving in city traffic along a straight road. The driver speeds up, slows down, and sometimes stops, so the speed is not constant. Distances covered in each minute vary. This is a real life example of non uniform motion. By contrast, a vehicle on a cruise control system on a straight highway at steady speed is in uniform motion. Drawing distance time graphs also shows that uniform motion has a straight line graph, whereas non uniform motion produces a curve, further emphasising the difference. Why Other Options Are Wrong: Uniform motion: Requires constant speed, which contradicts the information that speed keeps changing. Periodic motion: Describes motion that repeats at regular time intervals, such as pendulum oscillation, not stated here. Circular motion: Involves a curved circular path, but the question specifies a straight line path. Common Pitfalls: Students sometimes confuse non uniform motion with random motion, but non uniform simply means speed or direction changes. Others might mistake periodic motion for any motion that changes, but periodic is more specific and requires regular repetition. Keeping the definitions precise helps classify motions correctly: constant speed means uniform, changing speed or direction means non uniform, repeated cycles mean periodic, and curved closed path suggests circular motion. Final Answer: If the speed along a straight line keeps changing, the motion is called non uniform motion.

Question 10

The substances that are applied between moving surfaces to reduce friction and allow smoother motion are called what?

Accepted Answer

Introduction / Context: This question addresses the concept of friction and the materials used to control it in machines and everyday devices. Friction between surfaces can cause wear, heating, and energy loss. Engineers and technicians use certain substances to reduce friction, making movement smoother and extending the life of components. Knowing the correct term for these substances is an important part of basic mechanics. Given Data / Assumptions: Two solid surfaces move or slide relative to each other. Contact between them produces friction that opposes motion. A substance can be applied between the surfaces to reduce this friction. We are asked what such friction reducing substances are called. Concept / Approach: Lubricants are substances, often oils, greases or special fluids, that are placed between moving surfaces to reduce direct contact between solid asperities and therefore decrease friction and wear. The lubricant creates a thin film or layer which allows surfaces to slide more easily. Irregularities are microscopic bumps on surfaces that actually cause friction, adhesives increase sticking, and viscosity describes the internal resistance to flow of fluids, not the friction between solid surfaces directly. Thus, the correct term for friction reducing substances is lubricants. Step-by-Step Solution: Step 1: Recall that friction arises from contact between microscopic irregularities on surfaces. Step 2: To reduce friction, engineers introduce a fluid or semi fluid layer between the surfaces so they no longer grind directly against each other. Step 3: Substances used for this purpose, such as engine oil, grease or graphite, are collectively called lubricants. Step 4: Recognise that irregularities are the cause of friction, not a remedy. Step 5: Adhesives increase sticking between surfaces, and viscous describes fluid behaviour, so lubricants is the only correct term for friction reducing agents. Verification / Alternative check: Everyday examples support this. Engines require engine oil, cycle chains use lubricating oil, and door hinges may be oiled or greased to stop squeaking and make movement easier. In each case, the substance is known as a lubricant. Textbooks on friction explicitly state that lubrication is a standard method to reduce friction, and the materials used are called lubricants. No reference calls them irregularities or adhesives in this context. Why Other Options Are Wrong: Irregularities: These are microscopic bumps on surfaces that cause friction, not substances that reduce it. Adhesives: Materials like glue or tape increase bonding or sticking between surfaces, increasing resistance to motion. Viscous substances: Viscosity refers to internal resistance to flow in fluids. Some lubricants are viscous, but the general term for friction reducing substances is lubricants, not viscous. Common Pitfalls: Students sometimes associate viscous liquids with lubrication and incorrectly think viscosity is the main label. While some lubricants are viscous oils, the defining feature is their role in reducing friction, not just their thickness. Another error is to overlook that irregularities and adhesives play opposite roles, increasing friction instead of reducing it. Always focus on the function: if the substance is used specifically to reduce friction between moving parts, it is a lubricant. Final Answer: The substances used to reduce friction between moving surfaces are called lubricants.

Question 11

In practical energy calculations, the calorific value of a fuel is usually expressed in which unit?

Accepted Answer

Introduction / Context: This question is about the unit used for calorific value of fuels. Calorific value measures how much energy is released when a unit mass of fuel is completely burned. It is widely used in evaluating coal, petrol, diesel, LPG, and other fuels for engines, stoves, and power plants. Using the correct unit is important in thermodynamics and energy engineering. Given Data / Assumptions: Calorific value refers to energy released per unit mass of fuel on complete combustion. We are working in the SI framework and also familiar with some practical units. Options list kN/kg, kJ/kg, kW/s, and kcal/s. We assume mass based calorific value, not per unit volume. Concept / Approach: In SI units, energy is measured in joules and mass in kilograms. Therefore, calorific value is naturally expressed as kilojoules per kilogram (kJ/kg) when dealing with large amounts of energy. This tells us how many kilojoules of energy are obtained from one kilogram of fuel. Occasionally, older systems use kilocalories per kilogram, but in strict SI, kJ/kg is preferred. Units involving newtons, watts or per second are not appropriate for calorific value, as they relate to force, power or rate rather than energy per mass. Step-by-Step Solution: Step 1: Recall that calorific value is energy per unit mass, so we need a unit of energy divided by kilogram. Step 2: In SI, energy is measured in joules, so kilojoules (kJ) are a convenient multiple for fuel energies. Step 3: Combining these, the natural unit is kJ per kg, written as kJ/kg. Step 4: Check the options and identify kJ/kg as matching energy per mass. Step 5: Conclude that kJ/kg is the correct unit for calorific value in SI based practice. Verification / Alternative check: Standard tables of fuel properties list calorific values such as 42000 kJ/kg for petrol or around 33000 kJ/kg for coal. These values confirm that kJ/kg is widely used. Engineering handbooks may also show values in kcal/kg, but when converted to SI they become kJ/kg. None of these sources use kN/kg or kW/s as units for calorific value, since newton and watt represent different physical quantities. Why Other Options Are Wrong: kN/kg: Newton per kilogram is related to acceleration (g) and is not an energy per mass unit. kW/s: Kilowatt per second would be a rate of change of power, not energy per unit mass. kcal/s: Kilocalorie per second would represent a power output, similar to watt, not energy released per kilogram of fuel. Common Pitfalls: A common mistake is to confuse power and energy. Power units involve time in the denominator, like watt (joule per second), while energy units do not. Calorific value is an energy per mass concept, so time should not appear in the unit. Another confusion occurs between kilojoules and kilocalories, as both are used in nutrition and fuel tables. Remember that one kilocalorie is approximately 4.2 kilojoules, and in pure SI usage, kJ/kg is preferred for calorific value. Final Answer: The calorific value of a fuel is usually expressed in kilojoules per kilogram, that is kJ/kg.

Question 12

If a force is applied on a moving object in a direction opposite to the direction of its motion, what happens to the speed of the object?

Accepted Answer

Introduction / Context: This question checks understanding of how force and motion are related through Newton second law. When a force acts in the same direction as motion, it speeds an object up, and when it acts opposite to the direction of motion, it tends to slow the object down. Recognising this basic relationship is essential for analysing braking, friction, and many everyday motion situations. Given Data / Assumptions: An object is already moving in some direction with a certain speed. A force is applied in the direction opposite to the direction of motion. No specific numerical values are given for mass, speed, or force. We assume that the applied force has a non zero magnitude and acts for some time. Concept / Approach: According to Newton second law, a net force on an object produces acceleration in the direction of the net force. If the net force is opposite to the direction of motion, the acceleration is negative relative to the velocity and is often called retardation or deceleration. This reduces the magnitude of the velocity vector, meaning the object slows down. If such a force acts long enough and is strong enough, the object can eventually come to rest, but the immediate effect is that speed decreases. Therefore, the correct qualitative answer is that speed decreases. Step-by-Step Solution: Step 1: Represent the motion with a velocity vector pointing in the direction of motion. Step 2: Apply a force vector in the opposite direction. Newton second law states that acceleration is in the direction of this net force. Step 3: Because acceleration is opposite to velocity, it reduces the magnitude of the velocity over time. Step 4: As a result, the speed (magnitude of velocity) decreases gradually while the force acts. Step 5: Thus, the correct description is that the speed of the object decreases. Verification / Alternative check: Consider a car moving forward and the driver applying brakes. The frictional force from the brakes acts in the direction opposite to motion, causing the car to slow down. Eventually, if braking continues, the car may stop, but the first effect is a reduction in speed. Similarly, a ball rolling on the ground slows down due to friction, which acts opposite to its motion. These real life examples match the theoretical conclusion from Newton second law. Why Other Options Are Wrong: The speed of the object increases: This would occur if the force were in the same direction as motion, not in the opposite direction. The object stops instantly in all cases: Stopping requires time and depends on the magnitude and duration of the force. It is not instant in the general case described. There is no effect on the speed of the object: Any non zero net force must change the state of motion, so no effect is not possible under Newton laws if a real opposite force is applied. Common Pitfalls: Some learners think only friction can slow objects down, but any net force opposite to motion has the same effect. Others interpret questions simplistically and think opposite force always means immediate stopping, which is only true if the force is extremely large or acts for a long enough time. Remember to distinguish between the immediate trend (slowing down) and the final outcome (possibly stopping) when a retarding force acts. Final Answer: When a force is applied opposite to the direction of motion, the speed of the object decreases.

Question 13

In the International System of Units, what is the SI unit of acceleration?

Accepted Answer

Introduction / Context: This question is about the correct SI unit for acceleration, a fundamental quantity in mechanics. Acceleration describes how quickly velocity changes with time and appears in Newton second law, equations of motion, and many real world calculations such as vehicle performance or free fall problems. Using the correct unit is essential when applying these formulas. Given Data / Assumptions: Acceleration is defined as rate of change of velocity with respect to time. Velocity itself has the SI unit metres per second. We work within the standard SI unit system. The options combine metres and seconds in different ways. Concept / Approach: Acceleration a is defined as change in velocity delta v divided by change in time delta t. If velocity is measured in metres per second (m/s) and time in seconds (s), then acceleration has units of (m/s) per second, which simplifies to m/s^2. This expresses how many metres per second the velocity changes in each second. Any unit that does not reflect this twofold division by time is incorrect for acceleration. Step-by-Step Solution: Step 1: Write the definition of average acceleration: a = delta v / delta t. Step 2: Use the SI unit for velocity: metres per second, written as m/s. Step 3: Use the SI unit for time: second, written as s. Step 4: Combine them to get the unit of acceleration: (m/s) per second = m/s^2. Step 5: Identify m/s^2, or metres per second squared, as the correct SI unit of acceleration. Verification / Alternative check: In common problems, gravitational acceleration near the Earth surface is approximately 9.8 m/s^2. This value and unit combination appears in almost every physics textbook and example involving free fall, confirming that metres per second squared is the standard unit. In dimensional analysis, acceleration has dimensions of length divided by time squared, consistent with m/s^2. These checks reinforce that no other combination given in the options is correct for acceleration. Why Other Options Are Wrong: Metres per second (m/s): This is the unit of velocity, not acceleration, as it involves only one division by time. Seconds per metre (s/m): This unit would correspond to the inverse of speed, sometimes used in wave contexts but not for acceleration. Seconds per metre squared (s/m^2): This has no standard interpretation as a basic kinematics unit. Common Pitfalls: Students may confuse velocity and acceleration because both involve metres and seconds. The key is to remember that acceleration is about how velocity changes, so it involves time twice. Writing acceleration as change in velocity over time and substituting units explicitly helps avoid mistakes. Keeping a small list of fundamental units in mind, such as m for distance, s for time, m/s for velocity and m/s^2 for acceleration, is a useful strategy. Final Answer: The SI unit of acceleration is metres per second squared, written as m/s^2.

Question 14

Which of the following is considered an inexhaustible natural resource, meaning it is not likely to be depleted by normal human use?

Accepted Answer

Introduction / Context: This question belongs to environmental physics and earth science. Natural resources are often classified as exhaustible (non renewable) or inexhaustible (practically unlimited). Understanding which resources can run out and which are continuously available helps guide sustainable development, conservation efforts, and energy planning. Given Data / Assumptions: The options list air, minerals, forests, and coal as natural resources. Inexhaustible resources are those that are continuously available and are not significantly depleted by normal human use over practical timescales. Exhaustible resources are limited in quantity and can be depleted by excessive use. We assume typical school level classifications of resources. Concept / Approach: Air is generally classified as an inexhaustible natural resource because the atmosphere is vast and continuously renewed by natural processes such as photosynthesis, respiration, and circulation. Although air quality can be degraded, the total supply is not easily depleted. In contrast, minerals, forests, and coal are considered exhaustible. Minerals and coal take geological timescales to form and can be used up. Forests, while renewable with careful management, can be depleted faster than they regrow and so are often treated as exhaustible resources in practice. Therefore, air is the correct answer. Step-by-Step Solution: Step 1: Recall that inexhaustible resources include sunlight, air, and often water in the global sense. Step 2: Consider air, which surrounds the planet and is constantly circulated and replenished; its total quantity is not easily reduced by human activity. Step 3: Minerals such as metals and ores exist in limited deposits in the Earth crust and can be mined until they are economically exhausted. Step 4: Forests can be cut down faster than they regenerate, so they are not inexhaustible even though they are biologically renewable. Step 5: Coal is a fossil fuel formed over millions of years and is clearly non renewable on human timescales, so it is exhaustible. Thus, air is the only inexhaustible resource among the options. Verification / Alternative check: Environmental science textbooks commonly list air and solar energy as classic examples of inexhaustible or practically inexhaustible resources. They list coal, petroleum, natural gas, and mineral ores as exhaustible natural resources. Forests are often placed in a separate category of renewable but exhaustible if overused. This classification aligns with the reasoning above and leaves air as the correct choice. Why Other Options Are Wrong: Minerals: Available in limited quantities in ore bodies; excessive mining can deplete economically viable deposits. Forests: Can be destroyed by deforestation faster than natural regrowth, leading to serious loss of forest cover. Coal: A fossil fuel that takes millions of years to form and is being used far more rapidly than it forms, making it exhaustible. Common Pitfalls: Students may confuse renewable with inexhaustible. A resource can be renewable but still exhaustible if misused, such as forests and freshwater in local regions. Inexhaustible refers to resources that are effectively unlimited on human timescales, such as air moving in the atmosphere and solar energy from the Sun. Recognising this distinction helps in correctly categorising different types of natural resources. Final Answer: Air is considered an inexhaustible natural resource.

Question 15

A heating element has a resistance of 900 ohms. If a current of 0.3 A flows through it for 10 seconds, how much heat energy is generated in the element in joules?

Accepted Answer

Introduction / Context: This numerical physics question applies Joule heating or Joule law of electric heating. When an electric current flows through a resistor, electrical energy is converted into heat energy. Calculating the heat generated is important for the design of heaters, fuses, and electrical appliances to ensure efficiency and safety. Given Data / Assumptions: Resistance of the heating element, R = 900 ohms. Current flowing through the element, I = 0.3 A. Time for which current flows, t = 10 seconds. We assume constant current and resistance during the interval. We want heat energy in joules. Concept / Approach: According to Joule law of heating, the heat energy H produced in a resistor when current flows is given by H = I^2 * R * t, where I is current, R is resistance, and t is time. In SI units, if I is in amperes, R in ohms, and t in seconds, then H is in joules. The calculation involves squaring the current, multiplying by resistance, and then by time. Careful arithmetic and attention to units yield the correct answer. Step-by-Step Solution: Step 1: Write Joule heating formula: H = I^2 * R * t. Step 2: Substitute the given values: I = 0.3 A, R = 900 ohms, t = 10 s. Step 3: First compute I^2: I^2 = (0.3)^2 = 0.09. Step 4: Multiply by resistance: 0.09 * 900 = 81. Step 5: Multiply by time: H = 81 * 10 = 810 joules. Verification / Alternative check: As a rough check, note that 0.3 A is less than 1 A, so heat generated will be less than 900 * 10 = 9000 J, since that would correspond to 1 A. More precisely, I^2 is 0.09, less than one tenth, so we expect around 0.09 times 9000 J, which is about 810 J. This estimate matches the exact computation. Also, unit analysis confirms that amperes squared times ohms times seconds yields joules, as required. Why Other Options Are Wrong: 2700 J: Could result from using I instead of I^2 or a different mistaken combination in the formula. 405 J: This is half of the correct value and may arise from using half the correct time or resistance by mistake. 1350 J: This value does not follow from the correct substitution and may come from calculating 0.3 * 900 * 5 instead of using I^2 and full time. Common Pitfalls: Students sometimes confuse the formulas H = I^2 * R * t, H = V * I * t, and H = V^2 * t / R. While all are equivalent, using the wrong one without proper substitution leads to errors. Another frequent mistake is forgetting to square the current in H = I^2 * R * t. Misplacing decimal points when squaring 0.3 or multiplying also causes incorrect answers. To avoid these, write each step clearly, check powers carefully, and perform a quick magnitude estimate to see if the result is reasonable. Final Answer: The heat generated in the element is 810 J.

Question 16

In basic electronics, diodes are most commonly used for which one of the following purposes?

Accepted Answer

Introduction / Context: This question comes from basic electronics and asks about the primary function of a diode. Diodes are two terminal semiconductor devices which allow current to flow mainly in one direction. Recognising their most common application helps learners understand how power supplies convert alternating current to direct current for use in electronic circuits. Given Data / Assumptions: The device in question is a diode. The possible functions listed are rectification, amplification, modulation, and filtration. We consider standard usage in introductory electronics. We assume simple semiconductor diode behaviour. Concept / Approach: A diode conducts current easily in one direction (forward biased) and blocks it in the opposite direction (reverse biased). This direction selective property is used in rectifier circuits to convert alternating current, which changes direction periodically, into direct current, which flows in one direction only. Amplification is typically done by transistors or operational amplifiers, modulation involves varying a carrier signal, and filtration is mainly performed by capacitors, inductors, and filters. Therefore, the key standard application of diodes is rectification. Step-by-Step Solution: Step 1: Recall that a diode behaves like a one way valve for electric current. Step 2: In rectifier circuits, diodes are arranged so that during both halves of an AC cycle, current through the load flows in the same direction. Step 3: This process of converting AC to DC is called rectification. Step 4: Amplification requires devices that can control a large output with a smaller input, such as transistors, not simple diodes. Step 5: Modulation and filtration are specialised functions usually handled by other circuit elements. Hence rectification is the main general use of diodes. Verification / Alternative check: Power supply circuits in radios, televisions, computers and chargers all contain diode bridges or rectifier diodes. These change mains AC into DC before further smoothing and regulation. Textbooks and circuit diagrams label such networks as half wave or full wave rectifiers. Although diodes are used in some specialised roles like clipping, clamping or as part of modulators, their first and most common application taught to beginners is rectification, confirming that this is the correct answer. Why Other Options Are Wrong: Amplification: Usually achieved using transistors or integrated circuit amplifiers, not single simple diodes. Modulation: Involves varying amplitude, frequency or phase of a carrier signal; diodes may be part of mixers, but rectification remains their primary textbook use. Filtration: Filtering unwanted frequencies is mainly done by RC or LC filters and active filters, not by diodes alone. Common Pitfalls: Beginners sometimes think diodes amplify because they appear in many circuits. However, a diode by itself cannot provide power gain. Others may see diodes in radio circuits and assume modulation, but the core property of a diode is unidirectional conduction. Whenever a question asks for the general or primary application of diodes, rectification is almost always the expected answer at school level. Final Answer: Diodes are generally used for rectification, converting AC to DC.

Question 17

An anemometer is a meteorological instrument used to measure which of the following physical quantities?

Accepted Answer

Introduction / Context: This question is about identifying the correct use of an anemometer, a common instrument in meteorology. Weather stations and climate studies require accurate measurements of wind speed to forecast weather, study storms, and design buildings that can withstand strong winds. Knowing which quantity an anemometer measures is basic environmental physics knowledge. Given Data / Assumptions: The instrument named is an anemometer. Options list speeds of light, wind, water current, and satellites. We assume standard meteorological equipment definitions. We are concerned with the primary, textbook use of this instrument. Concept / Approach: An anemometer is specifically designed to measure wind speed and sometimes wind direction. Classic cup anemometers have rotating cups that spin faster when winds are stronger, and the rotational speed is converted into a wind speed reading. Other types include vane anemometers with propeller blades. None of these instruments are used to measure speed of light, water currents or satellite motion in space. Thus, the correct answer is speed of wind. Step-by-Step Solution: Step 1: Recall that meteorologists use anemometers at weather stations to monitor wind conditions. Step 2: The mechanical or electronic output of an anemometer is calibrated in units such as metres per second or kilometres per hour, indicating wind speed. Step 3: Measuring speed of light requires specialised optical experiments, completely different from anemometers. Step 4: Speed of water current is measured with flow meters or current meters, not with anemometers designed for air. Step 5: Satellite speeds are determined through orbital mechanics and tracking systems, not by simple ground based wind measuring instruments. Therefore, an anemometer measures speed of wind. Verification / Alternative check: Illustrations in geography and physics textbooks show anemometers mounted on masts alongside wind vanes. Weather reports often quote wind speeds measured in kilometres per hour or knots, derived from anemometer data. No serious scientific source associates anemometers with measuring speed of light or orbital velocities. This consistent usage confirms that the proper association is with wind speed. Why Other Options Are Wrong: Speed of light: Determined by optical and electromagnetic experiments, not by rotating cups or vanes in air. Speed of water current: Measured in rivers or oceans using current meters or flow sensors, different devices from anemometers. Speed of satellites: Calculated using orbital parameters and space tracking systems, not simple surface wind instruments. Common Pitfalls: Because the term anemometer may not be familiar, students sometimes guess based on sound and may link it to any kind of speed. The key hint is its frequent appearance in weather station pictures. Remember that anemos is related to wind in Greek, which can help you associate anemometer with wind measurement. Connecting instrument names to their typical context is a good strategy for such questions. Final Answer: An anemometer is used to measure the speed of wind.

Question 18

Which one of the following components cannot normally be fabricated as part of a single integrated circuit chip?

Accepted Answer

Introduction / Context: This question comes from electronics and integrated circuit technology. Modern integrated circuits contain many electronic components manufactured together on a single semiconductor wafer. However, not all circuit elements are easily integrated. Understanding which components can and cannot be included on a chip is important for circuit design and for appreciating the limitations of integrated technology. Given Data / Assumptions: The context is an integrated circuit chip, typically made on silicon. Options include diode, triode, transformer and transistor. We consider mainstream monolithic integrated circuits, not hybrid modules with external parts. The question asks which component cannot normally be developed within an integrated circuit. Concept / Approach: Diodes and transistors are semiconductor devices and are fundamental building blocks of integrated circuits. They are easily fabricated on the same silicon substrate using standard processes. A triode in classical electronics refers to a vacuum tube with three electrodes and is not a semiconductor device; however, the term triode is sometimes used loosely for three terminal semiconductor devices. The most clearly non integrable component among the options is the transformer, which requires coils of wire wound on a magnetic core to function efficiently. Such components are bulky and depend on magnetic coupling and cannot be realised effectively on a small monolithic chip in the same way as discrete transformers. Step-by-Step Solution: Step 1: Identify which components are inherently semiconductor junction devices. Diodes and transistors are both semiconductor junction devices. Step 2: Recall that integrated circuits commonly contain hundreds, thousands, or millions of diodes and transistors fabricated directly on silicon wafers. Step 3: Recognise that a transformer relies on electromagnetic induction between primary and secondary coils wound around a magnetic core, typically using copper wire and iron or ferrite core. Step 4: Such coils and cores are large and three dimensional compared with the thin planar structures used on chips and cannot be implemented in the same compact way on a monolithic IC. Step 5: Therefore, the transformer is the component that cannot normally be developed within a single integrated circuit chip. Verification / Alternative check: Looking at typical integrated circuits in power supplies or amplifiers, the silicon chip is accompanied by discrete transformers, inductors and capacitors that handle energy storage and voltage conversion. Textbooks on microelectronics emphasise that resistors, diodes, transistors and some capacitors can be integrated, but inductors and especially transformers are difficult or impractical to integrate efficiently. Vacuum tube triodes historically were not part of silicon integrated circuits, but they are also rarely used in modern IC design. In contrast, transformers remain discrete components, confirming they are not normally fabricated on IC chips. Why Other Options Are Wrong: Diode: A basic semiconductor device routinely included in large numbers on integrated circuits. Triode: In the classical vacuum tube sense it is not a semiconductor device, but in the context of this typical question, the focus is on the clearly non integrable transformer. Modern semiconductor "triode" like devices are effectively implemented with transistors. Transistor: The fundamental active element of almost all integrated circuits, easily fabricated in huge numbers on a chip. Common Pitfalls: Some students may choose triode because they associate it with old vacuum tubes, which indeed are not integrated on silicon chips. However, standard competitive exam questions usually expect transformer as the answer, because transformers are physically large magnetic components and remain outside the chip even in modern systems. It is helpful to remember that ICs excel at integrating semiconductor devices but still rely on discrete magnetic components like transformers and large inductors. Final Answer: Among the given components, a transformer cannot normally be fabricated within a single integrated circuit chip.

Question 19

Why does pure diamond not conduct electricity even though it is made only of carbon atoms?

Accepted Answer

Introduction / Context: Diamond is a famous form of pure carbon that is extremely hard and widely used in jewellery and cutting tools. A common conceptual question in basic physics and chemistry is why this form of carbon does not conduct electricity, whereas other forms such as graphite can conduct. Understanding this contrast helps students connect ideas about bonding, electron movement, and electrical conduction in solids. Given Data / Assumptions: We are considering pure crystalline diamond at normal temperature and pressure. Diamond is made only of carbon atoms arranged in a three dimensional tetrahedral lattice. We are interested in the ability of diamond to conduct electric current when a potential difference is applied. Conduction in solids requires mobile charge carriers, usually free electrons or ions. No impurities or defects are assumed in the crystal to provide extra charge carriers. Concept / Approach: The basic idea is that electrical conduction in a solid requires electrons that are not tightly bound and can move freely through the material when an electric field is applied. In metals, outer electrons are delocalised and form an electron sea. In insulators, valence electrons are tightly bound in covalent bonds and there is a large energy gap to the empty conduction band. Diamond is a classic example of a covalent network solid where each carbon atom forms strong sigma bonds with four neighbours, leaving no free electrons. Step-by-Step Solution: Step 1: In diamond, each carbon atom uses its four valence electrons to form four strong covalent bonds in a tetrahedral arrangement.Step 2: Because all valence electrons are tied up in these sigma bonds, there are no loosely bound or delocalised electrons that can move under an electric field.Step 3: The energy gap between the filled valence band and the empty conduction band in diamond is very large, so thermal energy at ordinary temperatures cannot excite electrons into the conduction band.Step 4: With no mobile charge carriers, an applied voltage cannot produce a sustained electric current, so diamond behaves as an electrical insulator and does not conduct electricity. Verification / Alternative check: We can verify this explanation by comparing diamond with graphite, another allotrope of carbon. In graphite, each carbon atom forms only three sigma bonds and has one electron in a delocalised pi system that can move along the layers, so graphite conducts electricity. The difference in conduction is therefore not due to the presence of carbon itself, but due to the bonding arrangement and the presence or absence of free electrons. Experimental measurements of resistivity also show that diamond has extremely high resistance, confirming its insulating character. Why Other Options Are Wrong: Option a is incorrect because a compact structure alone does not prevent conduction; many metals are compact yet conduct well. Option b is wrong because being crystalline does not automatically imply insulating behaviour; metals are also crystalline. Option c is incorrect because the presence of only carbon atoms does not stop conduction, as shown by graphite. Option e is wrong because diamond does not contain ions; it is a neutral covalent network with no positive or negative ions that could cancel out. Common Pitfalls: Students sometimes think that any solid made of a single element should behave like a metal, which is not true. Others confuse the idea of hardness with electrical insulation and assume that a hard material cannot conduct. Another common misconception is to imagine that electrons are always free to move inside a solid, without considering how strongly they may be bound in covalent bonds. It is important to link conduction to the presence of free charge carriers and to band structure, not just to the element name or physical hardness. Remember that in diamond the key reason for non conduction is the lack of free electrons in its covalent lattice. Final Answer: No free electrons are present in its covalent crystal lattice.

Question 20

Why does an astronaut feel weightless while orbiting the Earth in a space ship despite the presence of gravity?

Accepted Answer

Introduction / Context: Astronauts in orbit around the Earth often appear to float freely inside their space craft, and this situation is commonly described as weightlessness. Many people wrongly believe this means there is no gravity in space. This conceptual physics question tests the correct explanation based on orbital motion and free fall rather than the mistaken idea of zero gravity. Given Data / Assumptions: The astronaut and the space ship are in a stable orbit around the Earth. Gravitational attraction from the Earth is still present at orbital altitude, only slightly weaker than at the surface. There is no physical contact with the ground or any support surface providing a normal reaction force. Air resistance is negligible inside the sealed cabin, and there is no significant thrust acting most of the time in stable orbit. We are interested in the apparent weight felt by the astronaut, not in the actual gravitational force magnitude. Concept / Approach: Weight is the force with which a body presses on a support, usually equal to the normal reaction exerted by that support. In orbit, both the astronaut and the space ship are in a continuous state of free fall around the Earth due to their orbital motion. They are constantly falling toward the Earth, but their tangential velocity keeps them moving around it. Because astronaut and ship accelerate together, there is no normal reaction on the astronaut feet or body, so the astronaut feels weightless. Gravity is still acting, but the apparent weight is zero. Step-by-Step Solution: Step 1: Recognise that in a circular orbit, gravity provides the centripetal force needed to keep the space ship and astronaut moving in a curved path.Step 2: Both astronaut and space craft have the same acceleration toward the centre of the Earth because both are acted upon by the same gravitational field.Step 3: Since there is no solid support resisting this common free fall, the normal reaction on the astronaut from the floor or walls is essentially zero.Step 4: With no normal reaction, the astronaut experiences no sensation of weight, so an apparent weightlessness is observed, even though gravity is still present. Verification / Alternative check: As a check, imagine briefly firing the rocket engine to produce an upward acceleration greater than gravity; the astronaut would then feel heavier as the normal reaction increases. In contrast, if the engine produced just the right upward acceleration to balance gravity in a stationary hover, the astronaut would feel normal weight. The orbital situation is different because there is no upward thrust balancing gravity; instead, both astronaut and ship are freely falling. Experiments on parabolic flights, which create short periods of free fall inside an airplane, show the same weightless sensation, confirming that free fall is the key reason. Why Other Options Are Wrong: Option b is wrong because gravity at typical orbital heights is still a large fraction of surface gravity. Option c is incorrect since inertia alone does not remove weight; it just resists acceleration. Option d is not accurate for a stable orbit, because the rocket engine is usually off and does not exactly cancel weight. Option e is irrelevant, since magnetic field has almost no effect on the weight sensation of an astronaut. Common Pitfalls: Many students confuse zero gravity with zero weight, assuming that if a person floats then gravity must be absent, which is not true. Another mistake is to think that being far from Earth automatically removes gravity, even though gravitational influence extends very far. It is essential to distinguish between gravitational force and apparent weight, which depends on the normal reaction from a support. Remember that in orbit the astronaut is in free fall, sharing the same acceleration as the space craft, and this is what produces weightlessness. Final Answer: Because the space ship is in continuous free fall due to its orbital motion around the Earth.

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