Science & Technology · Chapter 07

72 solved questions

Chapter 7: Motion and Force

Every question in this chapter, answered and explained — step-by-step solutions drawn live from the chapter library across 8 question sections.

Very Short Answer Questions12 Short Answer Questions10 Give Reason Questions9 Differentiate Between2 Long Answer Questions9 Numerical Problems15 Diagram-Based / Conceptual Questions3 Application / Conceptual Questions12

Very Short Answer Questions

1SEE Boardqb-q0

Who propounded the Universal Law of Gravitation, and in which year?

Ans.Sir Isaac Newton propounded the Universal Law of Gravitation in 1687.

2SEE Boardqb-q1

What is the SI unit of the universal gravitational constant (G)?

Ans.The SI unit of G is N m²/kg².

3SEE Boardqb-q2

What is the value of the universal gravitational constant G?

Ans.The value of G is 6.67 × 10⁻¹¹ N m²/kg².

4SEE Boardqb-q3

Who first measured the value of G, and using what apparatus?

Ans.Henry Cavendish first measured the value of G in 1798 using the Cavendish balance.

5SEE Boardqb-q4

What is the SI unit of acceleration due to gravity?

Ans.The SI unit of acceleration due to gravity is metre per second squared (m/s²).

6SEE Boardqb-q5

What is the average value of acceleration due to gravity on Earth?

Ans.The average value of acceleration due to gravity on Earth is 9.81 m/s².

7SEE Boardqb-q6

Is mass a scalar or vector quantity? What is its SI unit?

Ans.Mass is a scalar quantity, and its SI unit is the kilogram (kg).

8SEE Boardqb-q7

Is weight a scalar or vector quantity? What is its SI unit?

Ans.Weight is a vector quantity, and its SI unit is the newton (N).

9SEE Boardqb-q8

What instrument is used to measure weight?

Ans.A spring balance is used to measure weight.

10SEE Boardqb-q9

What is the formula relating weight, mass, and acceleration due to gravity?

Ans.The formula is W = mg, where W is weight, m is mass, and g is acceleration due to gravity.

11SEE Boardqb-q10

What is the acceleration of an object in free fall equal to?

Ans.The acceleration of an object in free fall is equal to the acceleration due to gravity (g).

12SEE Boardqb-q11

Where is the value of 'g' greater — the equator or the poles?

Ans.The value of 'g' is greater at the poles than at the equator.

Short Answer Questions

1SEE Boardqb-q12

State Newton's universal law of gravitation.

Ans.The gravitational force produced between any two objects in the universe is directly proportional to the product of their masses and inversely proportional to the square of the distance between them: F = G m1m2/d².

2SEE Boardqb-q13

Define the universal gravitational constant (G).

Ans.The gravitational constant G is the magnitude of the gravitational force produced between two unit masses (1 kg each) separated by a unit distance (1 m). Its value is 6.67 × 10⁻¹¹ N m²/kg².

3SEE Boardqb-q14

Write the nature of gravitational force.

Ans.Gravitational force is always attractive, acts between every pair of objects in the universe (however small), acts along the line joining their centers, and does not require any medium to act — it can act through a vacuum.

4SEE Boardqb-q15

Write two effects (consequences) of gravitational force.

Ans.Gravitational force between the sun and planets keeps the planets revolving around the sun, and gravitational force between the earth and objects on its surface keeps objects stuck to the surface (and causes thrown objects to fall back).

5SEE Boardqb-q16

Define acceleration due to gravity.

Ans.Acceleration due to gravity is the acceleration produced in a freely falling object due to the force of gravity of a planet or satellite. It is denoted by 'g' and its average value on Earth's surface is 9.81 m/s².

6SEE Boardqb-q17

What is free fall? Give two examples of it.

Ans.Free fall is the motion of an object falling under the influence of gravity alone, without any obstruction such as air resistance. Examples: a stone dropped in a vacuum, and a crumpled paper ball dropped from a height (where air resistance is negligible).

7SEE Boardqb-q18

Under what conditions is an object said to be in free fall?

Ans.An object is in free fall only when the only force acting on it is gravity, with no air resistance or any other opposing force, so its acceleration exactly equals g.

8SEE Boardqb-q19

Write the conclusions of the feather and coin experiment.

Ans.In the presence of air, the coin falls faster than the feather because air resistance affects the feather more (due to its larger surface area). In a vacuum, with no air resistance, both the feather and coin fall together with the same acceleration, showing that acceleration due to gravity is independent of mass.

9SEE Boardqb-q20

What is weightlessness?

Ans.Weightlessness is the state in which the weight of an object in free fall becomes zero, because there is no supporting force to balance gravity. It is experienced by astronauts in orbiting satellites and space stations.

10SEE Boardqb-q21

Mention any four effects of gravitational force.

Ans.1) It keeps planets revolving around the sun. 2) The moon's gravity, being closer to Earth, causes tides more visibly than the sun's. 3) It keeps objects stuck to the Earth's surface. 4) It causes thrown objects to eventually fall back to the ground.

Give Reason Questions

1SEE Boardqb-q22

Acceleration due to gravity is not the same in all parts of the earth. Why?

Ans.The earth is not a perfect sphere — it is flattened at the poles and bulges at the equator, so its radius is smaller at the poles and larger at the equator. Since g is inversely proportional to the square of the radius, g is greater at the poles (9.83 m/s²) than at the equator (9.78 m/s²).

2SEE Boardqb-q23

Jumping from a significant height may cause more injury. Why?

Ans.As the height of the fall increases, the object (or person) falls for a longer time under constant acceleration due to gravity, gaining a greater final velocity (v² = u² + 2gh). This greater velocity means a greater impact force on landing, causing more injury.

3SEE Boardqb-q24

The mass of Jupiter is about 319 times the mass of the Earth, but its acceleration due to gravity is only about 2.6 times that of the Earth. Why?

Ans.Acceleration due to gravity depends on both mass and radius (g = GM/R²). Jupiter's radius is about 11 times Earth's radius, and since g is inversely proportional to the square of the radius, this greatly reduces the effect of Jupiter's larger mass: 319/11² = 319/121 ≈ 2.6.

4SEE Boardqb-q25

Among objects dropped from the same height in the polar region and the equatorial region of the earth, the object dropped in the polar region falls faster. Why?

Ans.The value of acceleration due to gravity (g) is greater in the polar region (9.83 m/s²) than in the equatorial region (9.78 m/s²) because Earth's radius is smaller at the poles. A greater g produces greater velocity in the same time, so the object falls faster in the polar region.

5SEE Boardqb-q26

Out of two paper sheets, one is folded to form a ball. If the paper ball and the flat sheet of paper are dropped simultaneously in the air, the folded paper will fall faster. Why?

Ans.The flat sheet has a much larger surface area exposed to air, so it experiences greater air resistance relative to its weight, slowing it down. The folded paper ball has a much smaller surface area, so air resistance has less effect on it, allowing it to fall faster and closer to true free-fall acceleration.

6SEE Boardqb-q27

When a marble and a feather are dropped simultaneously in a vacuum, they reach the ground together (at the same time). Why?

Ans.In a vacuum, there is no air resistance acting on either object. Since acceleration due to gravity does not depend on mass (g = GM/R² has no mass term for the falling object), both the marble and the feather experience exactly the same acceleration and therefore fall together, reaching the ground at the same time.

7SEE Boardqb-q28

As you climb Mount Everest, the weight of the goods that you carry decreases. Why?

Ans.As height above Earth's surface increases, the distance from Earth's center increases, so acceleration due to gravity (g) decreases (since g is inversely proportional to the square of the distance from the center). Since weight W = mg and mass stays the same, a smaller g means a smaller weight.

8SEE Boardqb-q29

It is difficult to lift a big stone on the surface of the earth, but it is easy to lift a smaller one. Why?

Ans.Weight is directly proportional to mass (W = mg, with g constant at a given place). A bigger stone has a larger mass and therefore a greater weight, requiring more force to lift it, while a smaller stone has less mass and less weight, making it easier to lift.

9SEE Boardqb-q30

Mass of an object remains constant but its weight varies from place to place. Why?

Ans.Mass is the amount of matter in an object and does not change with location. Weight (W = mg) depends on both mass and the local acceleration due to gravity (g), and since g varies from place to place (and from planet to planet), the weight of the same mass varies accordingly even though the mass itself stays constant.

Differentiate Between

1SEE Boardqb-q31

Gravitational constant G and Acceleration due to gravity g

Gravitational Constant (G)	Acceleration Due to Gravity (g)
Universal constant, same value throughout the universe.	Varies with planet, location, and height.
SI unit: N m²/kg².	SI unit: m/s².
Value: 6.67 × 10⁻¹¹ N m²/kg².	Average value on Earth: 9.81 m/s².

2SEE Boardqb-q32

Mass and Weight

Mass	Weight
Total quantity of matter in a body; scalar quantity.	Force of gravity on a body (W = mg); vector quantity.
SI unit: kilogram (kg). Constant everywhere.	SI unit: newton (N). Varies with location (value of g).
Measured using a beam balance.	Measured using a spring balance.

Long Answer Questions

1SEE Boardqb-q33

What is gravity? Write any two effects of gravitational force.

Ans.Earth and other planets and satellites pull nearby objects towards their centers; the force exerted by a planet or satellite on nearby objects is often called the force of gravity, also called the weight of the object. It decreases with increasing height and becomes negligible at a certain distance.

2SEE Boardqb-q34

Prove that acceleration due to the gravity of the Earth is inversely proportional to the square of its radius (g ∝ 1/R²).

Ans.Consider a body of mass 'm' on the surface of a planet of mass 'M' and radius 'R'. The gravitational force between them, from Newton's law, is:

F = GMm / R² ......... (i)

3SEE Boardqb-q35

Mention the factors that influence acceleration due to gravity.

Ans.Acceleration due to gravity (g = GM/R²) is influenced by:

4SEE Boardqb-q36

Mass of the Moon is about 1/81 times the mass of the Earth and its radius is about 37/100 times the radius of the Earth. If the earth is squeezed to the size of the moon, what will be the effect on its acceleration due to gravity? Explain with the help of mathematical calculation.

Ans.If Earth's mass stays the same (M) but its radius is squeezed to the Moon's radius, which is 37/100 of Earth's original radius (R), the new radius R' = 0.37R.

New acceleration due to gravity: g' = GM/(R')² = GM/(0.37R)² = GM/(0.1369 R²) = (1/0.1369) × (GM/R²) = 7.31 × g

5SEE Boardqb-q37

The acceleration due to gravity of an object of mass 1 kg in outer space is 2 m/s². What is the acceleration due to gravity of another object of mass 10 kg at the same point? Justify with arguments.

Ans.The acceleration due to gravity of another object of mass 10 kg at the same point will also be 2 m/s².

This is because acceleration due to gravity (g = GM/R², where M and R refer to the massive body creating the gravitational field, such as a planet) does NOT depend on the mass of the object experiencing the gravity. At any given point in space, all objects — regardless of their own mass — experience the same acceleration due to gravity, as confirmed by Galileo's experiment and the feather-coin experiment.

6SEE Boardqb-q38

A man first measures the mass and weight of an object in the mountain and then in the Terai. Compare the data that he obtains.

Ans.The mass of the object measured (using a beam balance) will be the same in both the mountain and the Terai, since mass does not depend on location.

However, the weight of the object (measured using a spring balance) will be slightly different: since the mountain is at a greater distance from the Earth's center than the Terai (lower altitude), the value of g is slightly smaller in the mountain, so the weight measured in the mountain will be slightly less than the weight measured in the Terai, even though the mass is identical in both places.

7SEE Boardqb-q39

A student suggests a trick for gaining profit in a business. He suggests buying oranges from the mountain and selling them in the Terai at the cost price. If a beam balance is used during this transaction, explain, based on scientific fact, whether his trick goes wrong or right.

Ans.The trick would NOT work, and the student would make no extra profit from this method.

A beam balance compares the mass of the oranges against a known mass, and mass does not change with location — a beam balance's reading is unaffected by changes in g, since both sides of the balance experience the same g at the same place, and the comparison is a ratio (mass), not an absolute force (weight). So the same mass of oranges will show the same reading on a beam balance whether weighed in the mountain or in the Terai.

8SEE Boardqb-q40

How is it possible to have a safe landing while jumping from a flying airplane using a parachute? Is it possible to have a safe landing on the moon in the same way? Explain with reasons.

Ans.While jumping from an airplane with a parachute, air resistance increases as the parachute's falling speed increases. Eventually, air resistance becomes equal to the weight of the person and parachute, making the net force (and thus acceleration) zero. From that point, the parachute falls at a constant (uniform) speed, allowing a safe landing on the ground.

On the Moon, this would NOT be possible, because the Moon has no atmosphere and therefore no air resistance. Without air resistance to balance the weight, a person with a parachute on the Moon would be in true free fall the entire way down, with speed increasing continuously until impact at very high speed — making a safe landing with a parachute impossible on the Moon.

9SEE Boardqb-q41

The acceleration of an object moving on the earth is inversely proportional to the mass of the object, but for an object falling towards the surface of the earth, the acceleration does not depend on the mass of the object. Why?

Ans.For an object being pushed or pulled by a fixed external force F along the ground, Newton's second law gives a = F/m — since F is fixed and does not depend on the object, a larger mass m results in smaller acceleration, so acceleration is inversely proportional to mass in that case.

However, for a freely falling object, the force acting on it (gravity) is not fixed — it is itself proportional to the object's mass: F(gravity) = GMm/R² = mg. When this is substituted into a = F/m, the mass m cancels out completely: a = (mg)/m = g. Because the gravitational force scales exactly with the object's own mass, the resulting acceleration due to gravity is the same for all masses — it depends only on the mass and radius of the planet, not on the falling object's mass.

Numerical Problems

1SEE Boardqb-q42

The masses of two objects A and B are 20 kg and 40 kg respectively. If the distance between their centers is 5 m, calculate the gravitational force produced between them.

Ans.Given: m1 = 20 kg, m2 = 40 kg, d = 5 m, G = 6.67 × 10⁻¹¹ N m²/kg²

F = G m1 m2 / d² = (6.67 × 10⁻¹¹ × 20 × 40) / 5² = (6.67 × 10⁻¹¹ × 800) / 25 = 2.134 × 10⁻⁹ N

2SEE Boardqb-q43

Mass of the Sun and Jupiter are 2 × 10³⁰ kg and 1.9 × 10²⁷ kg respectively. If the distance between the Sun and Jupiter is 1.8 × 10⁸ km, calculate the gravitational force between the Sun and Jupiter.

Ans.Given: m1 = 2 × 10³⁰ kg, m2 = 1.9 × 10²⁷ kg, d = 1.8 × 10⁸ km = 1.8 × 10¹¹ m

F = G m1 m2 / d² = (6.67 × 10⁻¹¹ × 2 × 10³⁰ × 1.9 × 10²⁷) / (1.8 × 10¹¹)² = 4.17 × 10²³ N

3SEE Boardqb-q44

Gravitational force produced between the Earth and Moon is 2.01 × 10²⁰ N. If the distance between these two masses is 3.84 × 10⁵ km and the mass of the earth is 5.972 × 10²⁴ kg, calculate the mass of the moon.

Ans.Given: F = 2.01 × 10²⁰ N, d = 3.84 × 10⁵ km = 3.84 × 10⁸ m, m1 (Earth) = 5.972 × 10²⁴ kg

From F = G m1 m2/d²: m2 = F d² / (G m1) = (2.01 × 10²⁰ × (3.84 × 10⁸)²) / (6.67 × 10⁻¹¹ × 5.972 × 10²⁴)

4SEE Boardqb-q45

Gravitational force produced between the Earth and the Sun is 3.54 × 10²² N. If the masses of the Earth and Sun are 5.972 × 10²⁴ kg and 2 × 10³⁰ kg respectively, what is the distance between them?

Ans.Given: F = 3.54 × 10²² N, m1 = 5.972 × 10²⁴ kg, m2 = 2 × 10³⁰ kg

From F = G m1 m2/d²: d² = G m1 m2 / F, so d = √(G m1 m2 / F)

5SEE Boardqb-q46

The mass of the moon is 7.342 × 10²² kg. If the average distance between the earth and the moon is 384400 km, calculate the gravitational force exerted by the moon on every kilogram of water on the surface of the earth.

Ans.Given: m1 (Moon) = 7.342 × 10²² kg, m2 (water) = 1 kg, d = 384400 km = 3.844 × 10⁸ m

F = G m1 m2 / d² = (6.67 × 10⁻¹¹ × 7.342 × 10²² × 1) / (3.844 × 10⁸)²

6SEE Boardqb-q47

If the mass of the moon is 7.342 × 10²² kg and its radius is 1737 km, calculate its acceleration due to gravity.

Ans.Given: M = 7.342 × 10²² kg, R = 1737 km = 1.737 × 10⁶ m

g = GM/R² = (6.67 × 10⁻¹¹ × 7.342 × 10²²) / (1.737 × 10⁶)²

7SEE Boardqb-q48

Mass of the Earth is 5.972 × 10²⁴ kg and the diameter of the moon is 3474 km. If the earth is compressed to the size of the moon, how many times will be the change in acceleration due to gravity of the earth so formed compared to that of the real Earth?

Ans.Radius of Moon = 3474/2 = 1737 km = 1.737 × 10⁶ m. New g' = GM(earth)/R(moon)²; original g = GM(earth)/R(earth)²

g'/g = R(earth)² / R(moon)² = (6371/1737)² ≈ 13.47

8SEE Boardqb-q49

If the mass of Mars is 6.4 × 10²³ kg and its radius is 3389 km, calculate its acceleration due to gravity. What is the weight of an object of mass 200 kg on the surface of Mars?

Ans.Given: M = 6.4 × 10²³ kg, R = 3389 km = 3.389 × 10⁶ m

g = GM/R² = (6.67 × 10⁻¹¹ × 6.4 × 10²³) / (3.389 × 10⁶)² = 3.75 m/s²

9SEE Boardqb-q50

The acceleration due to gravity of the earth is 9.8 m/s². If the mass of Jupiter is 319 times the mass of the Earth and its radius is 11 times the radius of the Earth, calculate the acceleration due to gravity of Jupiter. What is the weight of an object of mass 100 kg on Jupiter?

Ans.g(Jupiter) = g(Earth) × (M(Jupiter)/M(Earth)) / (R(Jupiter)/R(Earth))² = 9.8 × 319/121 = 25.83 m/s²

Weight = mg = 100 × 25.83 = 2583 N

10SEE Boardqb-q51

Earth's mass is 5.972 × 10²⁴ kg and its radius is 6371 km. Calculate the acceleration due to the gravity of the earth at the height of an artificial satellite orbiting at approximately 36000 km above the surface.

Ans.Given: M = 5.972 × 10²⁴ kg, R = 6371 km, h = 36000 km (approx, as per the figure), so (R+h) ≈ 42371 km = 4.2371 × 10⁷ m

g1 = GM/(R+h)² = (6.67 × 10⁻¹¹ × 5.972 × 10²⁴) / (4.2371 × 10⁷)²

11SEE Boardqb-q52

Mass of the earth is 5.972 × 10²⁴ kg and its radius is 6371 km. If the height of Mt. Everest is 8848.86 m from sea level, calculate the weight of an object of mass 10 kg at the peak of Mt. Everest.

Ans.(R+h) = 6371000 + 8848.86 = 6379848.86 m

g1 = GM/(R+h)² = (6.67 × 10⁻¹¹ × 5.972 × 10²⁴) / (6379848.86)² = 9.787 m/s² (approx)

12SEE Boardqb-q53

The acceleration due to gravity of Mars is 3.75 m/s². How much mass can a weight-lifter lift on Mars who can lift 100 kg mass on the Earth?

Ans.Given: M(earth) = 100 kg, g(earth) = 9.8 m/s², g(mars) = 3.75 m/s²

Weight liftable is equal in both places: m(mars) × g(mars) = M(earth) × g(earth)

13SEE Boardqb-q54

When a stone is dropped from a bridge over a river into the water, after 2.5 seconds the sound of the stone hitting the surface of the water is heard. Calculate the height of the bridge from the surface of the water. (g = 9.8 m/s²)

Ans.Given: u = 0, t = 2.5 s, g = 9.8 m/s²

h = ut + ½gt² = 0 + ½ × 9.8 × 2.5² = 30.625 m

14SEE Boardqb-q55

If a stone is dropped from a height of 15 m, how long will it take to reach the ground? Calculate the velocity of the stone when it hits the ground.

Ans.Given: h = 15 m, u = 0, g = 9.8 m/s²

From h = ut + ½gt²: 15 = ½ × 9.8 × t², so t² = 30/9.8 = 3.06, t = 1.75 s

15SEE Boardqb-q56

If a cricket ball is thrown vertically upwards into the sky with a velocity of 15 m/s, to what maximum height will the ball reach?

Ans.Given: u = 15 m/s, v = 0 (at maximum height), g = -9.8 m/s²

From v² = u² + 2gh: 0 = 15² + 2×(-9.8)×h, so h = 225/19.6 = 11.48 m

Diagram-Based / Conceptual Questions

1SEE Boardqb-q57

Draw a labelled diagram showing the gravitational force between two masses A and B.

Ans.Draw two spheres A and B with masses m1 and m2, separated by distance d, with two equal and opposite force arrows F pointing from each sphere toward the other, labelled along the line joining their centers.

2SEE Boardqb-q58

Mathematically present the difference in the gravitational force between two objects when the mass of each is made double and the distance between them is made one-fourth of their initial distance.

Ans.Let the initial force be F1 = G m1 m2 / d². If both masses are doubled (m1' = 2m1, m2' = 2m2) and distance becomes d/4:

F2 = G (2m1)(2m2) / (d/4)² = G × 4 m1 m2 / (d²/16) = 4 × 16 × G m1 m2/d² = 64 F1

3SEE Boardqb-q59

Draw and explain the diagram used to compare acceleration due to gravity with increasing distance from the center of the earth.

Ans.The diagram shows the Earth with markers at distances R, 2R, 3R, and 4R from the center. Since g ∝ 1/d², at d = 2R, g = g/4; at d = 3R, g = g/9; at d = 4R, g = g/16 (where g is the surface value at d = R).

Application / Conceptual Questions

1SEE Boardqb-q60

What is the relation between the distance between two objects (d) and the gravitational force (F) produced between them?

Ans.F ∝ 1/d² — the gravitational force is inversely proportional to the square of the distance between the two objects.

2SEE Boardqb-q61

What is the change in the gravitational force between two objects when their mass is doubled?

Ans.If the mass of both objects is doubled (distance kept constant), the gravitational force becomes four times the initial force.

3SEE Boardqb-q62

If the gravitational force between two objects on Earth is 60 N, what is the gravitational force between those two objects on the Moon (assuming the distance between them stays the same)?

Ans.The gravitational force between two objects (F = G m1m2/d²) depends only on their masses and the distance between them — not on which planet or moon they are on, since G is a universal constant. Therefore, the gravitational force between the same two objects at the same distance would still be 60 N on the Moon.

4SEE Boardqb-q63

Which one of the following statements is correct: (i) g increases going deeper into Earth, (ii) g decreases as height above the surface increases, (iii) g is less in the polar region than the equatorial region, (iv) g is highest at the highest place on Earth?

Ans.Statement (ii) is correct — the value of acceleration due to gravity decreases as the height above the surface of the earth increases, since g is inversely proportional to the square of the distance from Earth's center.

5SEE Boardqb-q64

At which of the following places do you weigh the most: peak of Mount Everest, peak of Api Himal, Kechanakalwal of Jhapa, or Chandragiri Hills?

Ans.You would weigh the most at Kechanakalwal of Jhapa, since it is at the lowest elevation (closest to sea level, i.e., closest to Earth's center) among the given places, giving it the highest value of g and therefore the greatest weight for the same mass.

6SEE Boardqb-q65

The radius of the Earth is 6371 km and the weight of an object on the earth's surface is 800 N. What is the weight of the object at a height of 6371 km (i.e., one Earth radius) from the surface of the earth?

Ans.At height h = R (one Earth radius above the surface), the distance from the center becomes (R+h) = 2R, so g1 = GM/(2R)² = (1/4) × GM/R² = g/4.

New weight = (1/4) × 800 N = 200 N.

7SEE Boardqb-q66

If the mass and the radius of a celestial body are two times the mass and the radius of the earth respectively, what is the value of acceleration due to the gravity of that body?

Ans.g(body) = G × (2M) / (2R)² = G × 2M / 4R² = (2/4) × GM/R² = 0.5 × g(earth) = 0.5 × 9.8 = 4.9 m/s².

8SEE Boardqb-q67

What will be the weight of a man on the moon, if his weight on earth is 750 N? (Acceleration due to gravity of the moon = 1.63 m/s²)

Ans.Mass of the man = W(earth)/g(earth) = 750/9.8 = 76.53 kg. Weight on Moon = mass × g(moon) = 76.53 × 1.63 ≈ 124.74 N.

9SEE Boardqb-q68

The mass of planet B is twice the mass of planet A but its radius is half of the radius of planet A. Similarly, the mass of planet C is half of the mass of planet A, but its radius is twice the radius of planet A. If the weight of an object on planets A, B, and C is W1, W2, and W3 respectively, which order is correct?

Ans.Let g(A) = GM/R². Then g(B) = G(2M)/(R/2)² = G(2M)/(R²/4) = 8 × GM/R² = 8g(A). And g(C) = G(M/2)/(2R)² = G(M/2)/(4R²) = (1/8) × GM/R² = g(A)/8.

So g(B) > g(A) > g(C), and since weight is proportional to g for the same mass object, W2 > W1 > W3.

10SEE Boardqb-q69

Which one of the following conclusions is correct while observing a freely falling object every second: distance covered increases uniformly, velocity increases uniformly, acceleration increases uniformly, or translation takes place uniformly?

Ans.Velocity increases uniformly is correct. For a freely falling object, acceleration (g) is constant, so velocity increases by the same amount (g) every second — but the distance covered each second is NOT the same (it increases each second, but not uniformly/linearly — it follows h = ut + ½gt², a quadratic relation), and acceleration itself remains constant (does not increase).

11SEE Boardqb-q70

Under what conditions is the value of gravitational force equal to the gravitational constant (F = G)?

Ans.The gravitational force F equals the gravitational constant G only when both masses are unit masses (m1 = m2 = 1 kg) and the distance between them is unit distance (d = 1 m), since F = G(1×1)/1² = G.

12SEE Boardqb-q71

One will have an eerie (unusual/floating) feeling when he/she moves down while playing a Rote Ping (a type of amusement ride/free-fall drop tower). Why?

Ans.A Rote Ping or similar free-fall style ride drops passengers rapidly, approaching a state close to free fall, where the ride's acceleration approaches g. In this condition, the seat and body fall together with little to no supporting force between them, so the person experiences a sensation close to weightlessness — an unusual, floating, 'eerie' feeling — similar to what astronauts feel in orbit.

Review the Notes Back to Chapter