Newton’s Laws of Motion Class 9 Question Answer Advanced Science Chapter 3

Students can use NCERT Class 9 Advanced Science Solutions Chapter 3 Newton’s Laws of Motion Question Answer to understand complex concepts with ease.

Newton’s Laws of Motion Class 9 Questions and Answers

Newton’s Laws of Motion Question Answer Class 9

Quick Check

Question 1.
In which type of reference frame are Newton’s laws valid?
Answer:
Newton’s laws of motion are valid only in inertial reference frames.

Question 2.
Define pseudo force and write its formula.
Answer:
Pseudo force: A pseudo force (or fictitious force) is an apparent force that acts on objects when they are observed from an accelerating or non-inertial frame of reference. It is not caused by any physical interaction.
Formula. F_pseudo = -m × a_pseudo

Question 3.
A lift accelerates upward at 4.5 m s^-2. Calculate the pseudo force experienced by a 60 kg person inside the lift.
Answer:
F_pseudo = -m × a_pseudo
Substituting, m = 60 kg and a_pseudo = 4.5 m/s2
|F_pseudo| = 60 kg × 4.5 m/s² = 270 N

Question 4.
Why does pseudo force disappear in an inertial frame?
Answer:
It disappears in inertial frame pseudo force disappears since it is not a “real” force and not caused by any physical interaction; it is merely a mathematical correction factor. In inertial frame, an observer can see exactly what is physically happening; so no correction factor is needed.

Question 5.
Where does the acceleration due to gravity reach its maximum value—on the surface, above, or below the Earth?
Answer:
The acceleration due to gravity is maximum at the surface of the Earth.

• Above the surface, gravity decreases with height.
• Below the surface, gravity decreases with depth. Therefore, it is greatest at the Earth’s surface.

Question 6.
What happens to g at the centre of the Earth?
Answer:
At the centre of the Earth, the acceleration due to gravity (g) becomes zero.
This is because the gravitational pull from all directions cancels out, resulting in no net force acting on an object.

Question 7.
Calculate g at a height of 400 km if R = 6400 km.
Answer:
g_h = g\(\left(\frac{R}{R+h}\right)^2\)
Given,
R = 6400 km,
h = 400 km,
g = 9.8 m/s²
g_h = 9.8\(\left(\frac{6400}{6400+400}\right)^2\) = 9.8\(\left(\frac{6400}{6800}\right)^2\)
g_h= 9.8\(\left(\frac{16}{17}\right)^2\) = 9.8 \(\frac{256}{289}\) ≈ 9.8 × 0.886
g_h = 8.7 m/s²

Question 8.
At what depth will g become half of its surface value?
Answer:
We know the formula for the variation of gravity with depth,
g_d = g(1 – \(\frac{d}{R}\))
For g_d
\(\frac{g}{2}\) = g(1 – \(\frac{d}{R}\))
\(\frac{1}{2}\) = 1 – \(\frac{d}{R}\)
\(\frac{d}{R}\) = \(\frac{1}{2}\)
d = \(\frac{R}{2}\) = \(\frac{6400}{2}\) km
d = 3200 km

Question 9.
Why does gravity decrease both above and below the surface of the Earth?
Answer:
Gravity decreases above the Earth’s surface because the distance from the centre of the Earth increases. According to the law of gravitation, gravitational force decreases with an increase in distance, so the value of g becomes smaller at higher altitudes.

Gravity also decreases below the Earth’s surface because the mass of the Earth that contributes to gravitational pull becomes less as we go deeper. At any depth, only the mass enclosed within that radius acts on the object, while the outer layers do not contribute. Hence, g decreases and becomes zero at the centre of the Earth.

Check Your Understanding

Question 1.
Why is it easier to open a door when you push at the handle rather than near the hinges?
Answer:
It is easier to open a door when we push at the handle because the handle is farther from the hinges (the pivot point). The turning effect of a force, called torque, depends on both the force applied and the distance from the pivot. When we push at the handle, the distance is greater, so even a small force produces a larger torque, making the door open easily. Near the hinges, the distance is very small, so more force is needed to produce the same turning effect.

Question 2.
A force is applied to a wrench at different angles. At which angle will the rotating force be maximum? What happens to the turning effect when the force is applied parallel to the wrench?
Answer:
The rotating force (torque) is maximum at 90° when the force is applied perpendicular to the wrench.
When the force is applied parallel (0° or 180°) to the wrench, the turning effect becomes zero, so no rotation occurs.

Question 3.
Two students apply the same force to open a gate. One pushes perpendicular to the gate at 20 cm from the hinge. The other pushes perpendicular to the gate at 80 cm. Who produces greater torque? Justify.
Answer:
Torque is the turning effect of a force and is calculated using the formula,
𝜏 = F × d
Where, F is the applied force and d is the perpendicular distance from the axis of rotation (the hinge).
Comparison:

1. Student 1: Pushes at a distance of 20 cm (0.2 m).
2. Student 2: Pushes at a distance of 80 cm (0.8 m).

As both students apply the same force (F), the torque depends entirely on the distance from the hinge. Since 80 cm is four times farther from the hinge than 20 cm, the second student generates four times as much torque with the same effort.

Question 4.
Is it possible for a force to act on a body and still produce zero turning about a given fixed point? Give a real-life example.
Answer:
Yes, it is possible for a force to act on a body and still produce zero turning effect (torque) about a fixed point. This happens when the force is applied along the line passing through the pivot (distance = 0) or when the angle is 0° or 180°.
Example:
Pushing a door exactly at the hinges does not make it rotate, even though a force is applied.

Question 5.
Two forces act on a rod pivoted at its centre:

I. 10 N downward at 0.5 m on the left
II. 10 N downward at 0.5 m on the right Will the rod rotate? Explain your reasoning.
Answer:
No, the rod will not rotate.
Reason: Both forces produce equal torque but in opposite directions about the pivot (centre).

• Left side: Torque = 10 N × 0.5 m (anticlockwise)
• Right side: Torque = 10 N × 0.5 m (clockwise)

As the magnitudes are equal and directions are opposite, they cancel each other, resulting in net torque = 0. Hence, the rod remains in equilibrium and does not rotate.

Question 6.
How can a mechanic loosen a tight bolt using a long spanner instead of applying a very large force? Explain using the torque formula.
Answer:
A mechanic can loosen a tight bolt more easily by using a long spanner because torque depends on the distance from the pivot.
Torque = Force × Distance from pivot
When a longer spanner is used, the distance (lever arm) increases. For the same applied force, a larger distance produces greater torque, which makes it easier to rotate and loosen the bolt. Thus, instead of applying a very large force, increasing the length of the spanner increases the turning effect.

Question 7.
A force of 20 N is applied to a door at 0.8 m from the hinge. Calculate the torque when the force is applied at (a)- 90°, (b) 60° (c) 30° to the door surface.
Answer:
Given, F = 20 N, d = 0.8 m
We know,
𝜏 = F × d × sinθ
(a) At 90°
i = 20 N × 0.8 m × sin 90° = 16 Nm × 1 = 16 Nm

(b) At 60°
𝜏 = 20 N × 0.8 m × sin 60° = 16 Nm × \(\frac{\sqrt{3}}{2}\) = 13.86 Nm

(c) At 30°
𝜏 = 20 N × 0.8 m × sin 30° = 16 Nm × 0.5 = 8 Nm

Newton’s Laws of Motion Class 9 Extra Questions and Answers

Short Answer Type Questions

Question 1.
A book is resting on the passenger seat of a car. When the driver slams on the brakes, the book slides forward onto the floorboards. Explain this event from the perspective of an observer standing on the sidewalk outside.
Answer:
To an observer on the sidewalk (an inertial frame), the car experiences a real backward braking force and slows down. The book, however, possesses inertia of motion. According to Newton’s First Law, the book attempts to maintain its constant forward velocity vector. Because there is insufficient friction between the book and the seat to slow the book down at the same rate as the car, it continues moving forward relative to the decelerating car until it falls.

Question 2.
The Sun pulls on the Earth with a massive gravitational force, yet the Earth maintains a stable orbit instead of falling straight into the Sun. Explain why.
Answer:
The Earth possesses a very high tangential velocity, meaning its inertia naturally drives it to move forward in a straight line. The Sun’s gravity acts perpendicularly to this forward path, pulling the Earth toward its centre as a centripetal force. The balance between the Earth’s forward momentum and this continuous inward gravitational pull results in a stable, curved orbital path.

Question 3.
A crate with a mass of 12 kg is sitting on the floor of a delivery truck. The truck suddenly accelerates forward at a uniform linear rate of +3.5 m/s². Calculate the absolute magnitude and state the precise direction of the pseudo-force acting on the crate from the perspective of a worker sitting inside the truck cabin.
Answer:
Mass of the crate (m) = 12 kg
Acceleration of the vehicle frame (a_frame) = +3.5 m/s² (Directed Forward)
Apply the foundational pseudo-force vector formula:
F_pseudo = -m.a_frame
Substituting the given parameters directly into the equation:
F_pseudo = -(12 kg) × ( + 3.5 m/s²)
F_pseudo = -42 N

Question 4.
An astronaut on the Moon (which mimics a perfect atmospheric vacuum environment) drops a heavy metal hammer from a scaffolding structure. If the hammer takes exactly 2.0 second to hit the lunar surface starting from rest, and the local acceleration due to gravity on the Moon is 1.6 m/s², calculate the total physical height of the scaffolding.
Answer:
Initial velocity (u) = 0 m/s (Dropped cleanly from rest)

• Total falling time interval (f) = 2.0 second
• Local lunar acceleration due to gravity (g_moon) = 1.6 m/s²

Utilise the second equation of motion configured for linear displacement/height (s):
S = ut + 1/2at²
Substitute the values into the polynomial expansion:
S = (0 × 2.0 s) + [ 1/2 × 1.6 m/s² × (2.0 s)² ]
S = 0 + [ 0.8 × 4.0 ]
S = 3.2 metre

Question 5.
A teacher drops a solid metal marble and a light feather inside an air-filled classroom, and the marble hits the floor first. If the teacher then repeats this experiment inside a high-vacuum chamber, both objects land simultaneously. Analyse this two-part scenario to explain:
(a) Why the marble falls faster in the presence of air despite having the same gravitational acceleration (g) as the feather.
(b) What physical change in the environment allows them to land together in the second test.
Answer:
(a) Real-World condition (In air): In an air- tilled classroom, both objects experience an identical downward gravitational acceleration (g = 9.8 m/s²). Flowever, as they move downward, they experience an upward fluid friction force called air resistance (drag). The feather has a much larger cross-sectional surface area relative to its light mass, causing the upward drag force to quickly counteract a significant portion of its weight. This dramatically reduces its net downward acceleration [a_net = (mg – F_drag)/m] compared to the streamlined, dense metal marble, allowing the marble to strike the floor first.

(b) Ideal Condition (In vacuum): When the air is completely evacuated to create a perfect vacuum, the upward opposing drag force (F_drag) drops to exactly zero. Without atmospheric molecules to push out of the way, gravity becomes the sole force acting on both bodies. According to Newton’s Second Law, an objects acceleration is its net force divided by its mass (a = mg/m). The mass variable cancels out completely from the numerator and denominator, forcing both the marble and the feather to accelerate at the exact same uniform rate (a = g) and hit the ground simultaneously.

Question 6.
A scientific research satellite monitoring atmospheric shifts climbs from an initial low Earth orbit to an altitude three times farther away from the centre of the Earth.
(a) State the fundamental law that governs how the Earth’s pulling force scales with height.
(b) Determine mathematically how much the satellites weight changes at this new, higher position.
Answer:
(a) Scaling Principle: The Earth’s gravitational pull and its resulting local acceleration due to gravity (g) are governed by an inverse-square law. This means that the strength of the local gravitational field is inversely proportional to the square of the distance (r) measured directly from the centre of the Earth [F ∝ 1/r²].

(b) Mathematical Derivation: Let the initial distance from the Earths centre be r₁ and the initial gravitational force (weight) be F₁. When the satellite climbs to a position three times farther away, its new distance parameter becomes r₂ = 3r₁. Substituting this new distance into the inverse-square relationship yields:
\(\frac{F_2}{F_1}=\frac{\left(\frac{1}{9 r_1^2}\right)}{\left(\frac{1}{r_1^2}\right)}\)
Comparing this output to the baseline state shows that F₂ = (1/9)F₁. Therefore, at this triple- distance altitude, the satellites weight decreases to exactly 1/9th of its original low-orbit value.

Long Answer Type Questions

Question 1.
A meteorological weather satellite is launched from the surface of the Earth to an operational altitude h that is exactly equal to the radius of the Earth (h = R). If the acceleration due to gravity on the Earth’s surface is g = 9.8 m/s²:
(a) Formulate the mathematical ratio relationship showing how the value of g alters with altitude.
(b) Calculate the exact numerical value of the new acceleration due to gravity (g) acting on the satellite at this altitude. □
Answer:
(a) The acceleration due to gravity on the surface of the Earth is given by Newton’s laws as:
g = \(\frac{G \cdot M}{R^2}\) ………….. (1)
Where G is the universal gravitational constant, M is the mass of the Earth, and R is the radius of the Earth.
When an object is lifted to a height h above the surface, its total straight-line distance from the centre of the Earth becomes (R + h). Consequently, the modified acceleration due to gravity (g’) is expressed as:
g’ = \(\frac{G \cdot M}{(R+h)^2}\) …………… (2)
Dividing Equation 2 by Equation 1 establishes the universal scaling ratio for altitude variations:
\(\frac{g^{\prime}}{g}=\frac{R^2}{(R+h)^2}\)
g’ = G . \(\left[\frac{R}{(R+h)}\right]^2\)
The problem mandates that the satellite is at an altitude equal to the Earths radius, meaning h = R. Substituting ‘R’ for ‘h’ in our relationship yields: 1
g’ = \(g\left[\frac{R}{(R+R)}\right]^2\)
g’ = g . \(\left[\frac{R}{2 R}\right]^2\)
Canceling out the radius variable ‘R’ inside the brackets simplifies the ratio expression:
g’ = g . \(\left[\frac{1}{2}\right]^2\)
g’ = \(\frac{g}{4}\)
Given that: g = 9.8 m/s²
g’ = \(\left(\frac{9.8}{4}\right)\) m/s²
g’ = 2.45 m/s²

Question 2.
(a) Explain the concept of vector cancellation that occurs at the centre of the Earth and distinguish how a pod’s ‘mass’ and ‘weight’ will alter.
(b) If an internal component inside the pod falls a tiny distance from a shelf while the pod rests exactly at the Earth’s centre, calculate its resulting acceleration and describe its immediate kinetic behaviour.
(c) Graphically or textually describe the continuous behaviour trend of the pod’s weight as it travels from a high-altitude space station down through the surface, and ultimately to the centre of the Earth.
Answer:
1. Vector Field Cancellation & Mass vs. Weight Dynamics:
Vector cancellation: Gravitational force is a vector quantity pointing toward the source mass.
When the pod is at the exact centre of the Earth, it is completely surrounded by the planet’s spherical mass. The earth matter pulling from the north is perfectly balanced by the matter pulling from the south, and the same occurs along every radial axis. Because of this perfect spatial symmetry, the directional pull vectors sum up to a net resultant force of zero.
Consequently, the local acceleration due to gravity drops to zero (g = 0).

Mass vs. Weight change: Mass is an intrinsic measure of the amount of matter (inertia) in a body and is inde¬pendent of location. Therefore, the pod’s mass remains exactly 2500 kg at the core. Weight, however, is a derived force dependent on local gra vity (W = mg). At the core, since g = 0, the pod’s weight completely collapses to 0 newton.

(b) Kinetic behaviour calculation of the falling component:
Mathematical Calculation: According to Newton’s Second Law of Motion, the acceleration of an object is defined by the net external force acting on it divided by its mass (a = F_net/m). At the centre of the Earth, the net gravitational force is zero, meaning:
F_net = 0 N
a = 0 N/m = 0 m/s²

Behaviour Description: Because the local acceleration parameter is exactly 0 m/s², the internal component will not accelerate downward or drop. Instead, it will enter a state of true environmental weightlessness and float perfectly stationary in mid-air at the exact coordinate position where it was released.

(c) The pod’s weight transitions through a distinct two- stage curve during its complete descent:
Zone 1 (Space station altitude to surface): As the pod moves from the high altitude station down toward the surface, the distance V to the Earths centre decreases. Because gravity obeys an inverse-square law (g ∝ 1/ r²), the acceleration due to gravity increases causing the pod’s weight to continuously increase, reaching its absolute peak value at the Earth’s surface.

Zone 2 (Surface down to the core centre): As the pod drills below the surface toward the core, the net inward gravitational field drops linearly with depth. Thus, the pod’s weight steadily decreases from its maximum surface value until it reaches exactly zero at the core centre.

Case-Based MCQs

I. Imagine a massive cosmic game of tetherball where the Sun stands at the centre, holding an invisible rope attached to the Earth. Both bodies possess substantial mass, meaning they exert an equal and opposite gravitational pull on each other according to Newton’s third law. However, because the Sun is overwhelmingly massive compared to the Earth, it undergoes negligible movement, while the Earth experiences a massive inward acceleration. This gravitational pull functions as a centripetal force, acting perpendicularly to the Earth’s motion.

The Earth manages to avoid crashing into the Sun because it possesses an immense tangential velocity. Due to its natural inertia, the Earth attempts to fly off forward in a straight line, while the Sun continuously pulls it inward. The resulting compromise of this dynamic “tug-of-war” is a stable, curved elliptical path known as an orbit. If a catastrophic cosmic event if suddenly forward inertia drops, it will be unable to maintain the curve, the gravitational balance would rupture, causing the Earth to spiral inward toward the centre of the solar system.

Question 1.
According to the text, why does the Earth experience a significantly larger acceleration than the Sun, despite the gravitational forces between them being completely equal in strength?
(A) The Earth is positioned farther away from the centre of the solar system.
(B) The Sun’s massive size and mass result in a much smaller acceleration for itself under the same force.
(C) The centripetal force only acts on the smaller object in a two-body system.
(D) Earth’s high tangential velocity amplifies the gravitational force acting on it.
Answer:
Option (B) is correct.
Explanation: According to Newton’s laws of motion, acceleration is inversely proportional to mass for a given force (a = F/m). Since the Sun possesses a tremendously massive size compared to the Earth, the same mutual gravitational force produces an incredibly small, negligible acceleration on the Sun, while causing a distinct, large orbital acceleration on the less massive Earth.

Question 2.
What serves as the “invisible tether” that prevents the Earth from breaking away from its path and drifting into deep, open space?
(A) The forward linear inertia of the Earth.
(B) The perpendicular tangential velocity of the planet.
(C) The Sun’s gravitational pull acting as a centripetal force.
(D) The shielding effect of the Earth’s atmospheric resis-tance.
Answer:
Option (C) is correct.
Explanation: The passage states that the constant gravitational pull from the Sun acts as a centripetal force, functioning exactly like an invisible tether. This inward- directed force continuously bends the Earth’s path toward the Sun, preventing its inertia from carrying it away into deep space.

Question 3.
If the Earth’s forward momentum and tangential velocity naturally drive it to travel in a straight line, why does it follow a stable, curved orbit instead?
(A) The Sun’s gravity acts perpendicularly to the Earth’s motion, pulling it continuously toward the centre.
(B) Gravitational forces double in strength every time the Earth tries to change its direction.
(C) Space coordinates are naturally curved by the magnetic field lines of the Sun.
(D) The Earth experiences a pseudo-force that pushes it sideways along its track.
Answer:
Option (A) is correct.
Explanation: The stable orbit is described as a “tug-of-war” balance. While the Earth’s inertia tries to drive it in a straight line, the Sun’s gravity acts perpendicularly to that forward velocity vector, constantly pulling the Earth inward and successfully bending its straight-line path into a continuous orbit.

Question 4.
What would be the immediate mechanical consequence if the Earth’s tangential speed were to suddenly decrease? (A) The Earth would immediately fly away from the Sun in a straight line.
(B) The Earth’s inertia would overcome the Sun’s gravity, making the orbit larger.
(C) The gravitational pull of the Sun would drop to zero due to the loss of speed.
(D) The Earth’s forward inertia would fail to resist gravity, causing it to spiral inward toward the Sun.
Answer:
Option (D) is correct.
Explanation: The text explains that orbital stability depends on a perfect balance between speed and gravity. If the tangential speed drops, the planet’s forward inertia is no longer strong enough to resist the inward pull; the balance breaks, and the Sun’s gravity pulls the Earth inward, causing it to spiral towards the centre.

Case-Based Subjective Questions

I. Imagine you are standing inside a city bus on your way to school. The bus is initially at rest at a traffic light, and you are balancing comfortably without holding onto any handrails. Suddenly, the light turns green, the driver steps on the accelerator, and the bus surges forward. Instantly, you feel an invisible force “push” your body backward toward the rear of the bus, causing you to lose your balance.

To a friend watching you from the sidewalk (a stationary observer on the road), your feet simply moved forward along with the floor of the accelerating bus, while your upper body tried to stay exactly where it was. EJowever, to you, trapped inside the moving vehicle, the bus feels stationary and it seems as though a mysterious, physical force actively threw you backward. This classic phenomenon highlights how our perception of motion depends entirely on our frame of reference, forcing physicists to introduce specific rules and “imaginary forces” to make sense of mechanical laws inside environments that speed up, slow down, or turn.

Question 1.
Identify the type of reference frame (inertial or non- inertial) an observer is in if they are standing completely still on a sidewalk, and state whether Newton’s laws of motion hold true in that frame without any modifications.
Answer:
The observer is in an inertial frame of reference, and Newton’s laws of motion work perfectly here without any modifications.

Question 2.
A delivery van accelerates forward along a straight road at a uniform rate of +2.5 m/s2. State the exact vector direction of the pseudo-force experienced by a cargo box resting inside the van.
Answer:
The pseudo-force acts in the backward direction (or directly opposite to the forward acceleration of the van).

Question 3.
Explain why a pseudo-force is scientifically classified as an “apparent” or “fictitious” force rather than a real physical interaction, and state how its formula accounts for its direction relative to the environment.
Answer:
A pseudo-force is considered fictitious because it does not arise from any actual physical interaction between two objects, such as a gravitational pull, contact friction, or a physical push. It only appears to exist because the observer is looking at the motion from inside an accelerating frame of reference.

In its mathematical formula, F_pseudo = -ma, the negative sign explicitly accounts for its direction, indicating that the pseudo-force always acts in the vector direction directly opposite to the acceleration of the reference frame.

Newton’s Laws of Motion Class 9 MCQ

Question 1.
The Sun exerts a powerful gravitational pull on the Earth. According to Newton’s laws of motion, which of the following statements accurately describes the gravitational force that the Earth simultaneously exerts back on the Sun?
(A) The Earth exerts a significantly weaker gravitational force due to its smaller mass.
(B) The Earth’s gravitational force fluctuates depending on its distance from deep space.
(C) The Earth exerts an equal and opposite gravitational force on the Sun.
(D) The Earth exerts no gravitational force on the Sun be-cause the Sun is the centre of the orbit.
Answer:
Option (C) is correct.
Explanation: According to Newton’s Third Law of Motion (action and reaction pairs) and Newton’s Law of Universal Gravitation, the gravitational attraction between any two interacting bodies is entirely mutual. Both exerts equal force on each other.

Question 2.
You whirl a stone tied to a thread in a steady horizontal circle. What physical quantity directly provides the inward centripetal force keeping the stone on its curved track?
(A) The magnetic attraction of the metal bob to your fingers.
(B) The gravitational force between your hand and the stone.
(C) The mechanical tension maintained within the tight thread.
(D) The upward buoyant force of the surrounding air molecules.
Answer:
Option (C) is correct.
Explanation: For an object to maintain a curved or circular track, an active inward force must pull it continuously toward the centre of rotation. The pulling force is entirely mechanical and transmitted directly through the string as tension. When the stone is whirled at a steady speed, its forward inertia tries to make it fly off in a straight line, while the taut string tension continuously bends its path into a perfect circle.

Question 3.
Astronauts living inside the International Space Station (ISS) experience a sensation of weightlessness. What is the true scientific explanation for this phenomenon?
(A) The Earth’s gravity at the altitude of the ISS has weak-ened to a true value of absolute zero.
(B) Astronauts spin the space station to cancel out plan-etary gravitational pulls mechanically.
(C) The gravity field has decreased with altitude, but weightlessness occurs because the station and astronauts are in a constant state of free fall around Earth.
(D) The thick metal hull of the space station blocks out all external gravitational fields.
Answer:
Option (C) is correct.
Explanation: A common misconception is that gravity in space is zero. In reality, Earth’s gravity at the altitude of the International Space Station (ISS) is only slightly weaker than it is on the ground. The true cause of weightlessness is that the ISS possesses a very high tangential velocity, causing it and the astronauts inside to be in a continuous state of free fall around the curve of the Earth. Because the station and the astronauts drop at the exact same rate, there is no normal reaction force from the floor, creating the physical sensation of floating.

Question 4.
Which mathematical expression demonstrates that in a perfect vacuum, the acceleration (ag) of any falling object is entirely independent of its individual mass (m)?
(A) a_g = m/g
(B) a_g = mg
(C) a_g = F/m = mg/m = g
(D) a_g = F × m = mg²
Answer:
Option (C) is correct.
Explanation: This expression combines Newton’s Second Law of Motion (a = \(\frac{F}{m}\)) with the formula for gravitational force or weight (F = mg). When an object falls in a perfect vacuum, air resistance is completely absent, meaning gravity is the only force acting on it. By substituting the weight formula into the acceleration equation, the mass variable (m) appears in both the numerator and the denominator, canceling out entirely. This proves algebraically that the net downward acceleration (a_g) equals the gravitational constant (g), meaning a heavy stone and a light feather will accelerate at identical rates and hit the ground simultaneously.

Question 5.
Match the statements in Column I with the correct physical principles or directions in Column II:

Column I (Mechanical Scenario / Property)	Column II (Physical Concept / Vector Direction)
P. A passenger bus deceler­ates suddenly to a halt	1. Pseudo-force acts in the forward direction
Q. An observer standing still on a stationary side­walk	2. Inertial frame of refer­ence
R. A passenger bus initiates a rapid forward launch	3. Pseudo-force acts in the backward direction
S. A vehicle enters a sharp, high-speed circular curve	4. Non-inertial frame of reference

Select the correct matching combination option below:
(A) P → 1, Q → 2, R → 3, S → 4
(B) P → 3, Q → 2, R → 1, S → 4
(C) P → 1, Q → 4, R → 3, S → 2
(D) P → 3, Q → 4, R → 1, S → 2
Answer:
Option (A) is correct
Explanation: The column matching is mapped out seamlessly based on directional rules: (P → 1) When a bus brakes or decelerates. A pseudo-force acts forward. When an observer standing still on a stationary side¬walk, then the frame is at rest i.e., inertial. (Q → 2) When a passenger bus initiates a rapid forward launch, a pseudo-force acts backward. (R → 3) A vehicle taking a turn alters its velocity vector continuously, making it an accelerating or non-inertial frame of reference. (S → 4)

Question 6.
A student conducts a precision physics experiment inside a vehicle moving at a constant speed of 80 km/h along a perfectly straight highway segment. Which of the following statements accurately characterises this setup?
(A) Newton’s laws of motion will fail to hold true because the vehicle is moving at a high speed velocity.
(B) The vehicle acts as an inertial frame of reference, so Newton’s laws apply perfectly without introducing pseu¬do-forces.
(C) A measurable pseudo-force will act on the experimental apparatus in a direction directly opposite to the vehi¬cle’s velocity vector.
(D) The internal mechanics can only be explained by introducing an imaginary force due exclusively to the inertia of rest.
Answer:
Option (B) is correct.
Explanation: Since the vehicle is moving at a uniform speed of 80 km/h along a perfectly straight line, its acceleration frame parameter is zero (a_frame = 0). Consequently, no pseudo-forces appear (F_pseud0 = -m × 0 = 0), and standard classical Newton laws work perfectly on their own without structural corrections.

Assertion-Reason Questions

Directions: In the following questions, a statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as:
(A) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.

Question 1.
Assertion (A): When a school bus standing at rest suddenly launches forward, a student standing inside feels an immediate backward push.
Reason (R): A pseudo-force always acts on an object in the direction diametrically opposite to the acceleration of the non-inertial reference frame.
Answer:
Option (A) is correct.
Explanation: When the bus launches forward, it becomes an accelerating, non-inertial frame of reference. From an observer inside the bus, any object with mass experiences an apparent pseudo-force acting opposite to the frame’s motion (F_pseudo = ~m × a_frame). Because the bus accelerates forward, the pseudo-force acts directly backward, pushing the student toward the rear. Therefore, both statements are true and the reason provides the perfect mechanical explanation.

Question 2.
Assertion (A): If the Earth’s high tangential velocity were to suddenly drop, it would begin to spiral inward toward the Sun.
Reason (R): A stable planetary orbit requires a constant, precise balance between the object’s forward inertia and the inward centripetal gravitational pull.
Answer:
Option (A) is correct.
Explanation: The assertion is true because the Earth’s high tangential velocity gives it the forward momentum (inertia) needed to resist falling directly into the Sun. If this speed drops, its forward inertia is no longer strong enough to maintain its curved track against the Sun’s continuous inward gravitational pull. The reason perfectly explains this mechanism by stating that orbital stability relies on this dynamic “tug-of-war” balance.

Question 3.
Assertion (A): As weather satellites launch to higher altitudes above the Earth’s surface, the local acceleration due to gravity (g) acting on them steadily decreases. Reason (R): Earth’s gravitational force is inversely proportional to the square of the distance from the Earth’s centre.
Answer:
Option (A) is correct.
Explanation: The assertion is true; the Earth’s gravitational pull is not a fixed constant everywhere and weakens as one travel 3 farther up. The reason is also true and provides the exact mathematical justification. Since gravity follows an inverse-square relationship with distance, increasing your altitude (distance from the Earth’s centre) directly forces the gravitational field strength and the resulting acceleration (g) to drop.