Chapter 2 — Work, energy, power and gravity
Begin with the physical quantity
In ordinary speech, a person may say they have “worked hard” while holding a heavy suitcase still. In mechanics, the work done on the suitcase by the supporting force is zero while the suitcase has no displacement. Mechanical work connects a force with displacement along the force's direction. For a constant force, write W = F s cos θ, where θ is the angle between the force and displacement. If force and displacement point together, cos θ = 1 and W = Fs. If they are perpendicular, cos θ = 0 and the work by that force is zero. The force can still matter; a supporting force may balance weight without transferring mechanical energy through displacement.
Work is measured in joules (J). One joule is one newton metre of work when a force of one newton acts along a displacement of one metre. The same combination of units can describe torque, but a torque is not simply called a joule because its physical meaning is rotational rather than energy transfer. A useful unit check is 1 N = 1 kg m/s², so 1 J = 1 kg m²/s². In a numerical MCQ, a result in newtons answers a force question; a result in joules answers a work or energy question.
Take a synthetic example: a cart is pulled 4 m horizontally with a constant 25 N horizontal force. Work by the pull is 25 × 4 = 100 J. If friction acts with 5 N opposite to motion, friction's work is −5 × 4 = −20 J. Net work is 80 J. The negative sign says friction removes kinetic energy from the cart; it does not mean the cart travelled a negative distance. If a question asks “work by the pulling force,” answer 100 J. If it asks “net work,” answer 80 J. Identify whose work is being requested before substituting numbers.
Separate the force from its useful component
When a force is slanted, only its component parallel to displacement performs work in the simple constant-force model. Suppose a person pulls a cart 6 m horizontally with a 20 N rope angled 60° above the horizontal. The parallel component is 20 cos 60°=10 N, so the rope's work is 10×6=60 J. Using 20×6=120 J would silently pretend the rope points along the journey. The vertical component may partly reduce the normal force and therefore affect friction, but that is a separate step; a question must provide enough information to calculate that effect. A perpendicular force can change direction of motion without increasing speed. In ideal uniform circular motion, the inward centripetal force is perpendicular to instantaneous displacement, so its instantaneous work is zero while velocity direction changes.
A quick sign table helps: force along displacement gives positive work; opposite force gives negative work; a perpendicular force gives zero work. The sign describes energy transfer to or from the selected body by that particular force. It does not describe whether the force itself is “good” or “bad.” During a downward fall, gravity does positive work because both force and displacement point downward. During an upward lift, gravity does negative work because they point in opposite directions. The same gravitational force changes the sign of its work when displacement reverses. Always mark the displacement arrow before deciding the sign.
Energy is a capacity, and forms can change
Kinetic energy is associated with motion: K = ½mv² for a mass m moving at speed v. Because speed is squared, doubling speed multiplies kinetic energy by four if mass is unchanged. Doubling mass only doubles kinetic energy if speed is unchanged. A 2 kg body at 3 m/s has kinetic energy ½ × 2 × 3² = 9 J. If its speed rises to 6 m/s, kinetic energy becomes 36 J. Do not use the arithmetic average of the two speeds to infer an energy ratio; square the specified speed in each state.
Near Earth's surface, gravitational potential energy relative to a chosen reference height is U = mgh, with g about 9.8 m/s² for common school-level calculations. If a 2 kg object is raised 5 m, the gain is approximately 98 J. If the question instructs you to take g = 10 m/s², the gain is 100 J instead. Use the stated approximation. Potential energy is relative: zero height may be the floor, a platform or another chosen level. A question about the change in potential energy depends on the height difference, so shifting the zero point does not alter the answer.
Mechanical energy is the sum of kinetic and potential energy for the system being modelled. When only conservative forces such as ideal gravity act, total mechanical energy remains constant. A falling body loses potential energy while gaining kinetic energy. Air resistance converts part of mechanical energy into thermal energy and sound, so kinetic energy at the bottom is then less than the initial gravitational potential energy. Total energy is conserved in the broader system, but mechanical energy alone need not be. This distinction prevents the common error of treating every real fall as a perfect conversion into motion.
The work–energy theorem says that net work on a body equals its change in kinetic energy. It connects a force calculation to a speed calculation. In the cart example above, 80 J of net work means an 80 J increase in kinetic energy if that is the entire net work over the displacement. If a cart already has 20 J of kinetic energy, it ends with 100 J, not 80 J. Write “final minus initial” beside a change question. If a force acts opposite to motion and the net work is negative, kinetic energy decreases.
Solve an energy change without time
The work–energy route is useful when a question gives force and distance but no time. A 4 kg trolley starts at 2 m/s and gains 30 J of kinetic energy. Its initial kinetic energy is ½×4×2²=8 J, so the final kinetic energy is 38 J. Hence ½×4×v²=38, or v²=19, giving v≈4.36 m/s. It would be wrong to set the final energy directly to 30 J because 30 J was the gain. If the trolley instead loses 30 J while starting with only 8 J, the stated physical situation is impossible without another energy input or a different initial state: kinetic energy cannot become negative. This plausibility check can expose an incorrectly copied number before an option is selected.
For an ideal drop from rest through height h, mgh=½mv² gives v=√(2gh). Mass cancels, so two objects dropped from the same height in the same ideal gravitational field reach the same speed when air effects are ignored. That does not mean a feather and a stone fall identically through ordinary air; drag can matter greatly. With h=5 m and g=10 m/s², v=√100=10 m/s. If a question asks for the energy gained instead, mass no longer cancels: a 2 kg object gains 100 J of kinetic energy while a 4 kg object gains 200 J over that same ideal drop. Equal final speeds do not imply equal energies.
When a rough surface dissipates energy, write an energy balance instead of forcing mechanical-energy conservation. If a 2 kg object starts from rest with 100 J of gravitational potential energy above a chosen zero level and friction removes 25 J during descent to that level, it arrives with 75 J of kinetic energy if no other energy transfers occur. Solving ½×2×v²=75 gives v=√75≈8.66 m/s. The missing 25 J has not disappeared; it has been transferred mostly into thermal energy of the contacting materials. The exam word “rough” is often a warning that the ideal mgh=½mv² shortcut cannot be used without a friction term.
Power measures a rate
Power is work or energy transfer per unit time: P = W/t for an average over the stated interval. Its SI unit is the watt (W), with 1 W = 1 J/s. Two machines can perform the same work but have different power if one completes it faster. If a hoist transfers 600 J in 10 s, its average power is 60 W. If it transfers the same 600 J in 5 s, average power is 120 W. A larger watt value is a faster energy-transfer rate, not automatically a larger total energy transfer; duration still matters.
A kilowatt-hour is a unit of energy, not power. One kilowatt used for one hour transfers 1 kW × 1 h = 1 kWh = 3.6 million joules. The factor follows from 1,000 watts and 3,600 seconds. A 100 W lamp operating for 10 hours consumes 1 kWh in the idealised calculation. Read whether an electricity question asks for instantaneous power, duration, or energy billed. If an efficiency is supplied, input energy is greater than useful output energy: efficiency = useful output/input × 100%. For 80% efficiency, 800 J useful output requires 1,000 J input. Do not invert that fraction.
Keep power, energy and efficiency in separate columns
Consider two pumps that each raise 200 kg of water through 5 m, using g=10 m/s². The useful energy delivered to the water is mgh=200×10×5=10,000 J for either pump. Pump A takes 20 s, giving useful power 500 W; pump B takes 50 s, giving useful power 200 W. The pumps do the same useful work, but A transfers it faster. If A is 50% efficient, its input energy for that run is 20,000 J and its average input power is 1,000 W. “Useful power” and “input power” are distinct values. Without an efficiency, do not assume they are equal in a real device.
To convert an electricity bill-style quantity, multiply power in kilowatts by time in hours. A 2 kW heater used for 30 minutes consumes 2×0.5=1 kWh. A 2 kW label is a rate; it does not say 2 kWh has already been consumed. If a question gives several appliances, compute each energy use and add them: a 0.1 kW lamp for 5 h uses 0.5 kWh, and a 1 kW motor for 1 h uses 1 kWh, totalling 1.5 kWh. Only multiply by a tariff if the tariff is explicitly supplied and the item asks for a cost. Actual household use may differ from nameplate values; the practice example assumes constant rated power throughout.
Machines, levers and mechanical advantage
A lever is a rigid bar that can turn about a fixed point called the fulcrum. The input force is effort; the force or weight to be moved is load. In an ideal lever at equilibrium, clockwise and anticlockwise moments about the fulcrum balance. Moment equals force multiplied by perpendicular distance from the pivot. If a 20 N effort acts 0.5 m from the fulcrum and a load acts 0.2 m away, equilibrium requires load × 0.2 = 20 × 0.5, so load = 50 N. The lever does not create energy. The greater load force is balanced by a smaller movement of the load compared with the effort point.
Mechanical advantage is load divided by effort. In the ideal example, it is 50/20 = 2.5. Real machines have friction and deformation, so useful output work is below input work. A pulley that changes force direction can make lifting more convenient without necessarily reducing the required ideal force. A system with multiple supporting rope segments can reduce ideal effort at the cost of pulling a greater rope length. Count the supporting segments only for the appropriate idealised arrangement; do not assume “any pulley halves the force.”
Velocity ratio describes how far the effort moves compared with the load: distance moved by effort divided by distance moved by load. In an ideal machine, work in equals work out, so effort×effort distance=load×load distance and mechanical advantage equals velocity ratio. In a real machine, efficiency is mechanical advantage divided by velocity ratio, expressed as a fraction or percentage. If effort moves 3 m while the load rises 1 m, the velocity ratio is 3. If load is 120 N and effort 50 N, mechanical advantage is 2.4, so efficiency is 2.4/3=0.8 or 80%. An answer above 100% efficiency for an ordinary passive machine is a signal that load and effort distances or the fraction have been reversed.
For a lever, classify the arrangement by what lies between the other two: a first-class lever has the fulcrum between effort and load, a second-class lever has the load between fulcrum and effort, and a third-class lever has the effort between fulcrum and load. Do not infer mechanical advantage from class alone without distances. A first-class lever can multiply force or speed depending on arm lengths; a third-class lever can trade force for a larger or faster movement of the load end. The moment equation gives the quantitative answer when arm lengths are supplied. The class name is only a structural clue.
Gravity, mass and weight
Gravitation is the mutual attraction between masses. For two point masses or two non-overlapping spherically symmetric bodies separated by distance r between their centres, Newton's law gives F = Gm₁m₂/r². This inverse-square dependence is the exam-relevant relationship: doubling one mass doubles the force; doubling both masses multiplies it by four; doubling centre-to-centre separation divides it by four. If both masses double and distance doubles, the mass factor four and distance factor four cancel. The formula uses centre-to-centre distance, not the gap between two surface edges.
Mass is measured in kilograms and does not change merely because an astronaut travels to the Moon. Weight is the gravitational force on that mass, W = mg in a local field. The same person's weight changes when g changes. For a 90 kg astronaut, mass remains 90 kg on the Moon. A school-level statement that lunar surface gravity is about one sixth of Earth's implies lunar weight is about one sixth of Earth weight, not one sixth of mass. In free fall, an object can feel weightless because it and its surroundings accelerate together; gravity has not vanished.
At Earth's centre, contributions from the surrounding Earth cancel by symmetry in the ideal spherical model, so gravitational acceleration is zero there. At the surface, the standard classroom approximation is about 9.8 m/s². It decreases with altitude above the surface, and in a simple uniform-density model it also decreases on moving below the surface toward the centre. These are model statements: Earth is neither perfectly spherical nor uniformly dense. Use the assumptions supplied in an exam item and avoid turning an approximation into an exact global fact.
Apply inverse-square reasoning carefully
The inverse-square law concerns separation between the centres of two masses. If two objects are separated by r and then moved to 3r with their masses fixed, their attraction becomes one ninth of its original value. If one mass simultaneously doubles, the new force is two ninths of the original. Writing a ratio is safer than inserting a value for G: Fnew/Folder=(m₁,new/m₁,old)(m₂,new/m₂,old)(rold/rnew)². This method works even when no individual mass or distance is supplied. If a planet has twice Earth's mass and twice Earth's radius, its ideal surface gravity relative to Earth's is (2)/(2²)=1/2, using g=GM/R². A larger planet can therefore have weaker surface gravity if its radius grows enough.
For an object above Earth's surface, distance in the inverse-square relation is measured from Earth's centre. At height h=R above a model Earth of radius R, the object is 2R from the centre, so the field strength is one quarter its surface value. Substituting h=R directly as the centre distance would incorrectly claim no change. Similarly, “twice as far from the surface” is ambiguous unless the original height is known. Draw a centre, radius and height before squaring the distance ratio. Many gravity distractors exploit this difference between altitude and centre-to-centre distance.
Orbits as a continuing fall
An orbiting satellite needs inward centripetal acceleration because its direction changes continually. Earth's gravity provides the inward force for an ideal Earth satellite. “There is no gravity in orbit” is false. The satellite falls around Earth while its sideways motion keeps it from hitting the surface under suitable conditions. A circular orbit can have constant speed but changing velocity, so the motion is accelerated even though the speedometer value stays fixed. No outward force needs to balance gravity in a simple inertial-frame description of the satellite's circular motion.
Kepler's first law describes a planet's orbit as an ellipse with the Sun at one focus in the ideal two-body model. The second law says a line from the Sun to a planet sweeps equal areas in equal times; the planet moves faster near perihelion, where it is closer to the Sun, and slower near aphelion. The third law relates orbital period T and semi-major axis a for planets orbiting the same central body: T² is proportional to a³. These laws describe orbits; Newtonian gravity explains them within the model. In a police exam, recognise the law and its relationship rather than memorising a derivation unless the operative syllabus asks for it.
Worked exam route and recall
Read this mixed item: a 5 kg box is raised vertically 2 m in 4 s at steady speed, taking g = 10 m/s². The gain in gravitational potential energy is mgh = 5 × 10 × 2 = 100 J. If losses are ignored, the lifting force does 100 J of work and the average useful power is 100/4 = 25 W. At steady speed, acceleration is zero, so net force is zero even though the lifting force and gravity both act. Their works over upward displacement are +100 J and −100 J respectively; net work is zero, consistent with no change in kinetic energy. The same example can therefore produce 100 J, 25 W or 0 J depending on which quantity and which force the stem asks about.
Before the practice set, answer these without options: What distinguishes mass from weight? When is mechanical work zero although a force acts? What happens to kinetic energy if speed doubles? Why is a kilowatt-hour energy rather than power? Which body supplies a satellite's inward orbital force? If two masses are each doubled, what happens to their gravitational attraction at fixed separation? Label every answer with its assumption and unit. That is a stronger recall check than recognising a single memorised formula.
Figure 2. Original chapter diagram.
Practice 36 questions
Choose one answer for each item. The answer key and explanations follow this chapter set.
1. In a lever, the fixed point about which the rigid rod turns is called the:
A. Effort B. Load C. Fulcrum D. Wedge
2. If the masses of both bodies are doubled while the distance between them stays the same, the gravitational force between them becomes:
A. Two times B. Half C. Four times D. Unchanged
3. Kepler's third law (law of periods) relates the orbital period T of a planet to the semi-major axis a of its orbit as:
A. T ∝ a B. T³ ∝ a² C. T² ∝ a³ D. T ∝ a²
4. According to Kepler's second law, a planet moves fastest in its orbit when it is at:
A. Aphelion (farthest from the Sun) B. Perihelion (closest to the Sun) C. Any point, since its speed is constant D. The point midway between perihelion and aphelion
5. An astronaut has a mass of 90 kg on Earth. What will be the astronaut's mass on the Moon?
A. 15 kg B. 540 kg C. 0 kg D. 90 kg
6. What is the value of the acceleration due to gravity at the centre of the Earth?
A. 9.8 m/s² B. Zero C. Maximum D. Half of its surface value
7. Which is the weakest of the four fundamental forces of nature?
A. Weak nuclear force B. Electromagnetic force C. Strong nuclear force D. Gravitational force
8. According to Kepler's first law, every planet moves around the Sun in:
A. An elliptical orbit with the Sun at one focus B. A circular orbit with the Sun at the centre C. An elliptical orbit with the Sun at the centre D. A parabolic path with the Sun at the vertex
9. Match List I (quantity) with List II (approximate value) and select the correct code.
List I:
A. Escape velocity from Earth
B. Escape velocity from the Moon
C. Standard acceleration due to gravity on Earth
D. Altitude of a geostationary orbit
List II:
1. 36,000 km
2. 2.4 km/s
3. 11.2 km/s
4. 9.8 m/s²
A. A-3, B-2, C-4, D-1 B. A-2, B-3, C-4, D-1 C. A-3, B-4, C-2, D-1 D. A-3, B-2, C-1, D-4
10. Ignoring air resistance and Earth's rotation, the escape speed from Earth's surface is approximately:
A. 11.2 km/s B. 7.9 km/s C. 2.4 km/s D. 9.8 km/s
11. What is the approximate value of the universal gravitational constant G?
A. 6.674 × 10⁻¹¹ N m²/kg² B. 6.674 × 10⁻⁹ N m²/kg² C. 9.8 N m²/kg² D. 6.022 × 10²³ N m²/kg²
12. The distance between two masses is tripled while the masses stay the same. The gravitational force between them becomes:
A. 1/3 of the original B. 3 times the original C. 9 times the original D. 1/9 of the original
13. What is the weight of a person of mass 60 kg on the Earth's surface? (Take g = 9.8 m/s²)
A. 60 N B. 98 N C. 588 N D. 6.1 N
14. Johannes Kepler formulated his laws of planetary motion using the observational data of:
A. Nicolaus Copernicus B. Galileo Galilei C. Claudius Ptolemy D. Tycho Brahe
15. What is the SI unit of the universal gravitational constant G?
A. N m/kg B. N m²/kg² C. N kg²/m² D. N/kg
16. Consider the following statements:
Statement I: Mass is a scalar quantity measured in kilograms.
Statement II: Weight is a vector quantity measured in newtons.
Which of the above statement(s) is/are correct?
A. Only I B. Only II C. Both I and II D. Neither I nor II
17. In the simple escape-speed model that ignores air resistance and planetary rotation, consider these statements.
Statement I: Escape speed from Earth depends on the mass of the object launched.
Statement II: Escape speed from a planet depends on the planet's mass and radius.
Which statement(s) is/are correct?
A. Only I B. Only II C. Both I and II D. Neither I nor II
18. Statement I: Kepler's first law states that every planet moves around the Sun in a circular orbit with the Sun at its centre.
Statement II: According to Kepler's law of areas, a planet moves faster at aphelion than at perihelion.
Which of the statements is/are correct?
A. Only I B. Only II C. Both I and II D. Neither I nor II
19. Which of the following is NOT one of Kepler's laws of planetary motion?
A. Law of orbits B. Law of areas C. Law of periods D. Law of inertia
20. Which of the following statements about an astronaut floating inside a spacecraft orbiting the Earth is NOT correct?
A. The astronaut and the spacecraft are in continuous free fall around the Earth B. Gravity does not act on the astronaut at that altitude C. The astronaut's mass is the same as on the ground D. No supporting surface pushes back on the astronaut
21. Which of the following statements about geostationary satellites is NOT correct?
A. They take about 24 hours to complete one orbit B. They appear fixed in the sky from the ground C. They orbit at an altitude of about 500–800 km D. They are mainly used for communication and broadcasting
22. A geostationary satellite always orbits directly above which of the following?
A. The equator B. The poles C. The Tropic of Cancer D. The Prime Meridian
23. At which of the following places on Earth's surface is the value of g (acceleration due to gravity) the least?
A. Equator B. North Pole C. South Pole D. Arctic Circle
24. According to ISRO's mission information, IRS-1A, India's first indigenous operational remote-sensing satellite, was placed in which type of orbit?
A. Geostationary orbit B. Polar (sun-synchronous) orbit C. Lunar orbit D. Geosynchronous transfer orbit
25. An electric motor converts:
A. Mechanical energy into electrical energy B. Chemical energy into electrical energy C. Heat energy into electrical energy D. Electrical energy into mechanical energy
26. The main benefit of using a single fixed pulley to draw water from a well is that it:
A. Halves the effort needed B. Reduces the total work done C. Reduces the weight of the bucket D. Changes the direction of the applied effort
27. If the speed of a moving body is tripled, its kinetic energy becomes:
A. 9 times B. 3 times C. 6 times D. 27 times
28. Match List I (source/situation) with List II (form of energy mainly involved) and select the correct code.
List I:
A. A stretched bow
B. A flowing river
C. Petrol
D. The Sun's core
List II:
1. Kinetic energy
2. Chemical energy
3. Potential energy
4. Nuclear energy (fusion)
A. A-1, B-3, C-2, D-4 B. A-3, B-1, C-4, D-2 C. A-3, B-1, C-2, D-4 D. A-2, B-1, C-3, D-4
29. One mechanical horsepower (hp) is approximately equal to:
A. 74.6 W B. 1000 W C. 7460 W D. 746 W
30. A boy of mass 50 kg climbs a staircase of vertical height 6 m in 10 s. Taking g = 10 m/s², his average power is:
A. 30 W B. 500 W C. 3,000 W D. 300 W
31. A force of 20 N moves a box through 5 m in the direction of the force. How much work is done?
A. 4 J B. 25 J C. 100 J D. 15 J
32. An electric heater of 2 kW is used for 3 hours. How many units (kWh) of electrical energy does it consume?
A. 1.5 units B. 6 units C. 5 units D. 0.6 units
33. What is the kinetic energy of a body of mass 4 kg moving with a speed of 5 m/s?
A. 100 J B. 50 J C. 20 J D. 10 J
34. A porter supports a suitcase on his head while walking on a level platform. Why is the mechanical work done by his upward support force on the suitcase zero?
A. The force he applies is perpendicular to the displacement B. The suitcase has no weight while being carried C. He moves with constant speed D. Friction cancels his effort
35. In which of the following situations is work done in the physics sense?
A. A boy pulls a toy cart along the floor through 5 m B. A porter carries a suitcase on his head while walking on level ground C. A person pushes with all his strength against a fixed wall D. A person holds a heavy bag steadily on his shoulder
36. A device receives 2.4 kJ of energy and delivers 1.8 kJ as useful output. What is its efficiency?
A. 25% B. 60% C. 75% D. 133.3%
Answers and explanations
1. C. A lever is a rigid rod that turns about a fixed point called the fulcrum. A see-saw, scissors and a crowbar are levers; the effort is the force applied and the load is the weight being moved.
2. C. F = G m₁m₂/r². Doubling both masses multiplies the product m₁m₂ by 2 × 2 = 4, with r unchanged, so the force becomes four times. Doubling only one mass would double it.
3. C. The law of periods (harmonic law): the square of the orbital period is proportional to the cube of the semi-major axis, T² ∝ a³. So planets farther from the Sun take disproportionately longer to go round once.
4. B. The law of areas says the Sun–planet line sweeps equal areas in equal times. To do so, the planet must move faster when closer to the Sun (perihelion) and slower when farther (aphelion).
5. D. Mass is the amount of matter and does not change with location, so it stays 90 kg on the Moon. Only weight changes: on the Moon it becomes about one-sixth of the Earth weight. 15 kg wrongly divides mass by 6.
6. B. At the centre of an ideal spherically symmetric Earth, gravitational contributions from opposite directions cancel, giving g = 0. A uniform-density model also predicts a linear decrease in g with depth; that depth trend is not exact for the real layered Earth. Above a spherical Earth, the field decreases with the square of the distance from its centre.
7. D. Gravitation is by far the weakest fundamental force, despite the "weak" name of the weak nuclear force. It still dominates at astronomical distances because it is always attractive and never cancels out like electric forces.
8. A. Kepler's first law (law of orbits): planetary orbits are ellipses with the Sun at one of the two foci, not at the centre. A circle is only a special case, and the 'Sun at centre' option is a common trap.
9. A. Earth's escape velocity is about 11.2 km/s, the Moon's about 2.4 km/s, standard g is 9.8 m/s², and geostationary satellites orbit at roughly 36,000 km. Code: A-3, B-2, C-4, D-1.
10. A. For an ideal launch from Earth's surface with air resistance and rotation ignored, escape speed is about 11.2 km/s. In this model it depends on Earth's mass and radius, not on the launched object's mass. About 7.9 km/s is the ideal circular-orbit speed close to Earth's surface, not escape speed.
11. A. G ≈ 6.674 × 10⁻¹¹ N m²/kg², a very small number, which is why gravitational pull between everyday objects is unnoticeable. 9.8 is the value of g in m/s², and 6.022 × 10²³ is the Avogadro number.
12. D. F = G m₁m₂/r², so F ∝ 1/r². Tripling r divides the force by 3² = 9. The force depends on the square of the distance; doubling the distance would reduce it to one-fourth.
13. C. Weight W = mg = 60 kg × 9.8 m/s² = 588 N. Mass (60 kg) is not weight; weight is a force measured in newtons. About 98 N would be the same person's weight on the Moon.
14. D. Kepler worked with Tycho Brahe in Prague and used Tycho's precise naked-eye observations, especially of Mars, to derive his laws (1609–1619). Copernicus proposed the heliocentric model; Galileo pioneered telescopic observation.
15. B. From F = G m₁m₂/r², G = F r²/(m₁m₂), so its unit is N × m² ÷ kg² = N m²/kg². N/kg is the unit of gravitational field strength (equal to m/s²).
16. C. Both are correct. Mass is the amount of matter, a scalar in kilograms that never changes with place. Weight is the gravitational force on that mass (W = mg), directed toward the planet's centre, so it is a vector measured in newtons.
17. B. In the stated ideal model, escape speed v = sqrt(2GM/R). The launched object's mass cancels, so I is false. The planet's mass M and radius R appear in the expression, so II is true. Real rocket launches also involve air, rotation and propulsion, which this school model omits.
18. D. Kepler's first law says orbits are ellipses with the Sun at one focus, not circles centred on the Sun. By the law of areas a planet sweeps equal areas in equal times, so it moves fastest at perihelion (nearest the Sun) and slowest at aphelion; both statements are false.
19. D. Kepler's three laws are the law of orbits (elliptical orbits), law of areas (equal areas in equal times) and law of periods (T² ∝ a³). The law of inertia is Newton's first law of motion.
20. B. Earth's gravity is still strong at orbital height (it supplies the centripetal force), so the claim that it does not act is false. The astronaut floats because the person and the spacecraft fall together around the Earth, so nothing pushes back on the body; mass is unchanged.
21. C. Geostationary satellites are at about 36,000 km, far higher than polar satellites (about 500–800 km). The other statements are true: a 24-hour period makes them appear fixed, which suits communication and broadcasting (INSAT).
22. A. A geostationary satellite moves in a circular orbit above the equator at about 36,000 km with a 24-hour period in the direction of Earth's rotation, so it appears fixed over one point. Polar satellites pass over the poles.
23. A. Earth bulges at the equator, so a point there is farthest from the centre, and the rotational (centrifugal) effect is greatest there. Both reduce g, making it least at the equator and greatest at the poles.
24. B. ISRO records IRS-1A in a polar sun-synchronous orbit. Such an orbit supports repeated Earth observation under broadly comparable local sunlight. This exact mission statement avoids implying that every Indian remote-sensing satellite has the same orbit.
25. D. A motor (in a fan, mixer or pump) turns electrical energy into mechanical energy of rotation. The reverse conversion, mechanical to electrical, is done by a generator; a cell converts chemical energy to electrical.
26. D. A single fixed pulley has mechanical advantage of about 1: it does not reduce the effort, but lets you pull down to lift up, which is more convenient. A system of multiple pulleys is needed for real mechanical advantage.
27. A. Kinetic energy = ½mv², so it varies as the square of speed. Tripling v multiplies KE by 3² = 9. Similarly, doubling speed makes KE four times.
28. C. A stretched bow stores potential energy of configuration; a flowing river has kinetic energy of motion; petrol stores chemical energy in its bonds; the Sun's core releases nuclear energy through fusion. Code: A-3, B-1, C-2, D-4.
29. D. One mechanical horsepower is about 745.7 W, conventionally rounded to 746 W. Metric horsepower is a different unit of about 735.5 W, so the qualifier 'mechanical' matters. The SI power unit is the watt (J/s).
30. D. Work done against gravity is W = mgh = 50 × 10 × 6 = 3,000 J. Power = work ÷ time = 3,000 ÷ 10 = 300 W. The figure 3,000 is the work in joules, not the power.
31. C. Work = force × displacement = 20 N × 5 m = 100 J. One joule is the work done when a force of one newton moves an object one metre in its own direction.
32. B. Energy = power × time = 2 kW × 3 h = 6 kWh = 6 units. The kilowatt-hour is the energy unit counted by household meters; dividing (2 ÷ 3) or adding (2 + 3) gives the wrong options.
33. B. KE = ½mv² = ½ × 4 × 5² = ½ × 4 × 25 = 50 J. Forgetting the ½ gives 100 J, and forgetting to square the speed gives 10 J.
34. A. The upward support force and the suitcase's horizontal displacement are perpendicular. Work by that force is Fd cos(90°) = 0. This does not mean the porter uses no metabolic energy; it identifies the work by the stated force on the stated object.
35. A. Work needs a force component along the displacement. Pushing a wall or holding a bag involves no displacement, and the porter's supporting force is perpendicular to his horizontal displacement, so those three do zero work. Only the pulled cart has force and displacement along the same line.
36. C. Efficiency = useful output / total input × 100 = 1.8 / 2.4 × 100 = 75%. The remaining 0.6 kJ is transferred in forms outside the useful output specified. Dividing input by output gives a value above 100% and reverses the definition.