Class 9 · Advanced ScienceNCERT 2026 Revised Syllabus
Chapter 3: Newton’s Laws of Motion
Science Advanced
Newton’s laws in inertial frames, pseudo force, gravitation and variation of g with height and depth, and turning forces (torque) around a pivot.
Find complete NCERT solutions for Class 9 Advanced Science, Science Advanced Chapter 3. Every question has a clear, step-by-step answer.
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Activity 3.1 (Reflective Questions)(1 questions)
1
Why does a passenger feel pushed backward when a bus suddenly accelerates? Is there really a backward force?
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Ans.
Solution:- No real backward force acts on the passenger.
- Due to inertia, the passenger's body tends to remain at rest while the bus accelerates forward.
- This causes an apparent backward motion relative to the bus.Answer: No real backward force acts on the passenger; inertia causes an apparent backward motion relative to the accelerating bus.
Quick Check(4 questions)
1
In which type of reference frame are Newton’s laws valid?
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Solution:Newton's laws are strictly valid only in an inertial (non-accelerating) frame of reference.Answer: Newton's laws are strictly valid only in an inertial (non-accelerating) frame of reference.
2
Define pseudo force and write its formula.
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Solution:- A pseudo force is an apparent, imaginary force felt only in an accelerating (non-inertial) frame.
- Formula:
$F_{pseudo} = -m \times a_{frame}$, acting opposite to the frame's acceleration.Answer: $F_{pseudo} = -m \times a_{frame}$, acting opposite to the frame's acceleration.3
Hard
A lift accelerates upward at 4.5 m/s2. Calculate the pseudo force experienced by a 60 kg person inside the lift.
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Solution:
$F = m \times a = 60 \times 4.5 = 270 \text{ N}$Directed downward, opposite to the lift's upward acceleration.Answer: Pseudo force = 270 N, directed downward.4
Why does pseudo force disappear in an inertial frame?
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Ans.
Solution:- Pseudo force is only introduced to make Newton's laws appear valid within an accelerating frame.
- In an inertial frame, real forces alone fully explain the motion.Answer: In an inertial frame, real forces alone fully explain the motion, so pseudo force is not needed.
Gravitation (Let Us Think)(1 questions)
1
What would happen if Earth’s tangential speed decreases suddenly while the Sun’s gravitational pull stays the same?
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Ans.
Solution:- The required centripetal force would exceed the actual pull needed for a stable orbit.
- Gravity would pull Earth inward.
- Earth would drift closer to the Sun instead of maintaining its orbit.Answer: Gravity would pull Earth inward, causing it to drift closer to the Sun instead of maintaining its orbit.
Quick Check (Gravitation)(5 questions)
1
Where does the acceleration due to gravity reach its maximum value - on the surface, above, or below the Earth?
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Solution:Acceleration due to gravity is maximum at the Earth's surface; it decreases both above and below the surface.Answer: Maximum at the Earth's surface.
2
What happens to g at the centre of the Earth?
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Solution:The value of g becomes zero at the centre of the Earth.Answer: g = 0 at the centre of the Earth.
3
Hard
Calculate g at a height of 400 km if R = 6400 km.
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Solution:
$g(h) = g \times \frac{R^{2}}{(R + h)^{2}} = 9.8 \times \left(\frac{6400}{6800}\right)^{2} \approx 8.68 \text{ m/s}^{2}$Answer: $g \approx 8.68 \text{ m/s}^{2}$.4
Hard
At what depth will g become half of its surface value?
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Solution:
$g(d) = g\left(1 - \frac{d}{R}\right)$Setting $g(d) = g/2$ gives $1 - \frac{d}{R} = \frac{1}{2}$, so $d = \frac{R}{2} = 3200 \text{ km}$.Answer: g becomes half its surface value at a depth of 3200 km.5
Why does gravity decrease both above and below the surface of the Earth?
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Solution:- Above the surface, distance from Earth's centre increases, weakening the pull.
- Below the surface, only the mass enclosed within that radius contributes, reducing effective gravity.Answer: Above, distance from Earth's centre increases; below, only the enclosed mass contributes, reducing effective gravity.
Check Your Understanding (Turning Forces)(7 questions)
1
Why is it easier to open a door when you push at the handle rather than near the hinges?
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Ans.
Solution:- Torque depends on distance from the pivot.
- Pushing farther from the hinge (at the handle) produces greater turning effect for the same applied force.Answer: Pushing at the handle gives a larger distance from the hinge, producing greater torque for the same force.
2
A force is applied to a wrench at different angles. At which angle will the rotating force be maximum? What happens when force is applied parallel to the wrench?
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Ans.
Solution:- Torque is maximum when the force is applied at 90 degrees (perpendicular to the wrench).
- When force is applied parallel to the wrench, the turning effect (torque) becomes zero.Answer: Maximum at 90°; torque becomes zero when the force is parallel to the wrench.
3
Two students apply the same force to open a gate, one perpendicular at 20 cm from the hinge, the other at 80 cm. Who produces greater torque? Justify.
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Solution:- Torque = force × distance from the pivot.
- For the same force, a larger distance gives a larger torque.Answer: The student pushing at 80 cm produces greater torque.
4
Is it possible for a force to act on a body and still produce zero turning about a fixed point? Give a real-life example.
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Solution:- Yes, when the line of action of the force passes through the pivot point (angle 0° or 180°).
- Example: pushing a door directly toward its hinge line produces no rotation.Answer: Yes, when the line of action passes through the pivot; e.g. pushing a door directly toward its hinge line.
5
Two forces act on a rod pivoted at its centre: 10 N downward at 0.5 m on the left, and 10 N downward at 0.5 m on the right. Will the rod rotate? Explain.
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Ans.
Solution:- Left torque
$= 10 \times 0.5 = 5 \text{ N m}$ (clockwise).
- Right torque $= 10 \times 0.5 = 5 \text{ N m}$ (anticlockwise).
- Torques are equal in magnitude but opposite in direction, so they cancel.Answer: No, the rod will not rotate; the torques cancel and the rod remains balanced.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.
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Ans.
Solution:- Torque = force × distance from the pivot.
- Increasing the spanner's length increases the distance from the pivot.
- A larger torque is produced with a smaller applied force.Answer: A long spanner increases the distance from the pivot, producing a larger torque with a smaller force.
7
Hard
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 degrees (b) 60 degrees (c) 30 degrees.
- a.(a) 90 degrees
- b.(b) 60 degrees
- c.(c) 30 degrees
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Ans.
Solution:Torque
$= F \times d \times \sin\theta$(a) 90°: $20 \times 0.8 \times 1 = 16 \text{ N m}$(b) 60°: $20 \times 0.8 \times 0.866 = 13.86 \text{ N m}$(c) 30°: $20 \times 0.8 \times 0.5 = 8 \text{ N m}$Answer: (a) 16 Nm, (b) 13.86 Nm, (c) 8 Nm.Activities — How to Do(4 questions)
1
Activity 3.3: where should the boy push the door to open it easily — (a) near the handle, (b) near the hinges, (c) at the centre?
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Ans.
Push (a) near the handle. Turning effect (moment) = force × distance from hinges; the handle is farthest, so the same push produces the largest moment and the door opens easiest. Near the hinges the lever arm is nearly zero.
2
Why are door handles fixed far away from the hinges and not near them?
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Ans.
Handles sit far from hinges to give a long lever arm. Since moment = F × d, a large d means even a small push creates enough turning effect. Near the hinges you would need an impractically large force for the same opening.
3
The boy pushes in a straight direction, yet the door rotates. Why does this happen?
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The hinges act as a fixed pivot: they hold one edge while the handle is pushed sideways, and the push has a perpendicular distance from the hinge line. An off-centre force about a pivot always produces rotation (moment) instead of straight-line motion.
4
How is it possible to rotate such a heavy object by applying force at just one end?
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Ans.
Because the push acts at the maximum distance from the hinges, its moment is large even when the force is modest — enough to beat hinge friction and the door’s inertia. Heavy objects rotate easily when force is applied far from the pivot.