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

By @learncbse · July 16, 2026 · 16 min read

Chapter-wiseClass 9 Advanced Science Solutionsand NCERT Class 9 Advanced Science Chapter 3 Newton’s Laws of Motion Question Answer are useful for focused study. Newton’s Laws of Motion Class 9 Questions and AnswersNewton’s Laws of Motion Question Answer Class 9Quick CheckQuestion 1. In which type of reference frame are Newton’s laws valid?Answer:Newton’s laws of motion are valid only in an inertial frame of reference. Question 2. Define pseudo force and write its formula. Answer:Pseudo force (also called fictitious force or inertial force) is an imaginary force that appears to act on an object when observed from a non-inertial (accelerating) reference frame. FormulaFpseudo= – m × aframewhere m = mass of the object,aframe= acceleration of the non-inertial frame with respect to an inertial frame. The negative sign shows that the pseudo force acts opposite to the acceleration of the frame. Question 3. A lift accelerates upward at 4. 5 ms-2. Calculate the pseudo force experienced by a 60 kg person inside the lift. Answer:Given,Mass of person, m = 60 kgAcceleration of lift, a = 45 ms-2upwardThe lift is a non-inertial frame accelerating upward. Pseudo force on the person,Fpseudo= maSince the frame (lift) accelerates upward, the pseudo force acts downward (opposite direction). Force, Fpseudo= 60 × 4. 5 = 270 N (downward)Question 4. Why does pseudo force disappear in an inertial frame?Answer:Pseudo force disappears in an inertial frame because it is not a real force that only exists to make Newton’s law in an accelerating frame. It represents the effects of the frame’s acceleration itself rather than on actual interaction between objects. Inertial frames are stationary or move at a constant velocity, means they have zero acceleration(aframe= 0)The pseudo force from Fp= m . aframe∵ aframe= 0Fp= 0. Question 5. Where does the acceleration due to gravity (g) reach its maximum value?Answer:Acceleration due to gravity g is maximum at the surface of the earth. The acceleration due to gravity g decreases as we go below or above from the earth. Question 6. What happens to g at the centre of the earth?Answer:At the centre of the earth, g = 0 (gravity becomes zero). Question 7. Calculate g at a height of 400 km if R = 6400 km. Answer:g’= \(\frac{g R^2}{(R+h)^2}\)= \(\frac{9. 8 \times\left(6400 \times 10^3\right)^2}{(400+6400)^2 \times 10^6}\)= \(\frac{9. 8 \times(6400)^2}{(6800)^2}\)= 8. 6 m/s2Question 8. At what depth will g become half of its surface value?Answer:g’ = g (1 – \(\frac{d}{R}\))\(\frac{g}{2}\) = g (1 – \(\frac{d}{R}\))\(\frac{1}{2}\) = (1 – \(\frac{d}{R}\))\(\frac{d}{R}\) = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)d = \(\frac{R}{2}\)Question 9. Why does gravity decrease both above and below the surface of the earth?Answer:Gravity decreases as height above the earth, surface increases because gravitational acceleration is inversely proportional to the surface of the distance from the earth’s centreg = GM / (R + h)2. Gravity decreases with depth below the earth’s surface because the effective mass attracting an object toward the centre decreases faster than the distance decreases. Check Your UnderstandingQuestion 1. Why is it easier to open a door when push at the handle rather than near the hinges?Answer:Torque is given by, τ = Fd sin θTorque depends upon the force and distance from the pivot. Therefore, it is easier to open a door with its handle than pushing it near the hinges because the torque produced is greater when force is applied further from the hinges. 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 when the force is applied perpendicular (90°) to the wrench. Torque, τ = F × r × sin θ (r = length of wrench). sin θ is maximum (= 1) at θ = 90° → maximum torque. When the force is applied parallel to the wrench (θ = 0° or 180°), sin θ = 0 → torque = 0. No turning effect, the force only pushes or pulls along the wrench without rotating it. 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, τ = Fr sin θBoth apply forces perpendicular (θ = 90°, sin θ = 1),so torque simplifies to τ = Fr. Student 1:r = 20 cm = 0. 2 m → τ1= F × 0. 2Student 2:r = 80 cm = 0. 8 m → τ2= F × 0. 8Since F is same, student 2 produces greater torque (4 times more, because r is 4 times larger). Greater distance from hinge gives greater turning effect. 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-like example. Answer:Yes, it is possible for a force to act on a body and produce zero turning about a fixed point. Real life example:If you push a door with Significant force but your push is directly towards the hinge, the door will not open or close. Because the perpendicular distance (d) from the axis of rotation to the line of action of the force is zero (τ = F × d sin θ). 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 nr on the right. Will the rod rotate?Explain your reasoning. Answer:Both forces are equal (10 N), at equal distance (0. 5 m) from the pivot (centre), but on opposite sides. They produce equal torque in opposite directions (one clockwise, one anti-clockwise). Net torque = 0The rod will not rotate, it remains in equilibrium (like a balanced see-saw). The two torques cancel each other. 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 long spanner increases the length r of the momentum. Since, torque is directly proportional to the length of the spanner, using a longer spanner means more torque is generated for the same applied force this greater torque helps in loosing a tight ball with less effort. 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(i) 90°,(ii) 60° and(iii) 30° to the door surface. Answer:Here, r = 0. 8 m,F = 20 NTorque,(i) τ = Fr sin θ= 20 × 0. 8 × sin 90°= 16 Nm(ii) τ = Fr sin θ= 20 × 0. 8 × sin 60°= 13. 856 Nm(iii) τ = Fr sin θ= 20 × 0. 8 × sin 30°= 8 NmNewton’s Laws of Motion Class 9 MCQQuestion 1. A person standing in a bus feels a backward push when the bus suddenly starts moving forward. This is because(A) A real physical force is pulling the person back. (B) The person wants to remain at rest due to inertia. (C) The force of gravity increases suddenly. (D) Friction between the feet and the floor is zero. Answer:(B) The person wants to remain at rest due to inertia. When the bus accelerates forward suddenly, the lower part of the body (in contact with the floor) moves forward with the bus. However, the upper part of the body tends to remain in its state of rest due to inertia of rest. This makes the passenger feel as if they are being pushed backward. Question 2. Which of the following is the correct formula for pseudo force fpseudo(A) F = m a(B) F = – m. a(C) F = \(\frac{G m_1 m_2}{r^2}\)(D) F = m. gAnswer:(B) F = – m. aA pseudo force is a tool used to make Newton’s laws work in an accelerating (non-inertial) frame. Its magnitude is the product of the mass of the object (m) and the acceleration of the frame. The negative sign indicates that the direction of the pseudo force is always opposite to the direction of the frame’s acceleration. For example, if a lift accelerates upward, the pseudo force acts downward. Question 3. A pseudo force is also known as a “fictitious force” because(A) it is a very strong force like nuclear force. (B) it only exists in an inertial (non-accelerating) frame. (C) it does not arise due to any physical contact or interaction. (D) it is always directed towards the center of the earth. Answer:(C) it does not arise due to any physical contact or interaction. Real forces like gravity, friction, or tension involve interactions between two objects. A pseudo force (Fictitious Force) is purely an “apparent” force. It exists only because the observer is in an accelerating environment. There is no actual physical object pushing or pulling you, it is just your inertia being viewed from an accelerating frame. Question 4. According to the section on gravitation, why does the Earth move in a curved orbit instead of a straight line?(A) Because of the Earth’s magnetic field. (B) Because there is no friction in space. (C) Because the Sun’s gravitational pull provides the necessary centripetal force. (D) Because the Sun is moving faster than the earth. Answer:(C) Because the Sun’s gravitational pull provides the necessary centripetal force. The earth has a high tangential velocity and wants to move in a straight line due to inertia. However, the Sun’s gravitational pull acts as a centripetal force, constantly pulling the earth toward the center of the Sun. This balance between inertia and gravity results in a stable curved orbit. Assertion-Reason QuestionsDirections (Question Nos. 1 – 3): In each of the following questions, a statement of Assertion is given by the corresponding statement of Reason mark the correct answer as(A) Both Assertion and Reason are true and Reason is the correct explanation of Assertion. (B) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion. (C) Assertion is true but Reason is false. (D) Assertion is false but Reason is true. Question 1. Assertion (A): An astronaut in the International Space Station (ISS) feels weightless. Reason (R): The gravitational force of the earth becomes zero at the height, where the international space station orbits. Answer:(C) Assertion is true but Reason is false. The astronaut feels weightless because they are in a state of continuous free fall while orbiting the earth. We know that gravity decreases with height but it does not become zero. Gravity at that height is still about 90% of what it is on the surface. Question 2. Assertion (A): It is easier to open a heavy door by pushing it at the edge (near the handle) rather than near the hinges. Reason (R): Torque (moment of force) increases when the distance between the pivot and the point of application of force increases. Answer:(A) Both Assertion and Reason are true and Reason is the correct explanation of Assertion. The formula for torque is τ = Fd sin θ. By increasing the distance (d) from the hinges (pivot), arise more “turning effect” (torque) with the same amount of force, making the door easier to open. Question 3. Assertion (A): As a ball is thrown higher above the Earth’s surface, the acceleration due to gravity acting on it decreases. Reason (R): Gravitational force is inversely proportional to the square of the distance from the centre of the earth. Answer:(A) Both Assertion and Reason are true and Reason is the correct explanation of Assertion. We have gravitation force,F = G \(\frac{m_1 m_2}{r^2}\)F ∝ \(\frac{1}{r^2}\)Newton’s Laws of Motion Class 9 Extra Questions and AnswersCase/Source Based QuestionsDirections (Question No. 1): Answer the questions on the basis of your understanding of the following passage and related studied concepts. Question 1. During a science class, a student performed an experiment to understand circular motion. He tied a small ring or a bob to one end of a thread of length about 1 m and held the other end firmly with his finger. He swung the bob in a horizontal circle at a steady speed. The thread remained tight and the bob moved in a perfect circular path. When he slowly reduced the speed of rotation, the thread became loose and the bob started moving inward and finally fell down. Answer the following questionsi. What provides the centripetal force in this experiment? Explain briefly. ii. Why does the bob tend to move outward while rotating in a circle?iii. What happens when the speed of the bob is decreased? Why does this happen?iv. How is this experiment similar to the orbital motion of the earth around the Sun? Explain with the help of centripetal force and inertia. v. Why is the outward tendency in circular motion called centrifugal force? Is it a real force or a pseudo-force? Give a reason. Answer:i. The centripetal force is provided by the tension in the thread. The thread pulls the bob towards the centre. This inward pull keeps the bob moving in a circle. ii. The bob tends to move out word due to its inertia. When the bob is forced in circular motion, it resists this change and tries to move tangentially (straight outward). iii. When the speed decreases, the thread becomes loose and the bob moves inward or falls down. This happens because the inward pull (centripetal force) becomes weak. Without enough inward force, the bob cannot stay in the circle. iv. This experiment is similar to the earth moving around the Sun. In the experiment, the tension in the thread acts as centripetal force and pulls the bob inward. In the same way, the Sun’s gravity pulls the earth inward and acts as centripetal force. The bob tries to move outward due to inertia. The earth also tries to move in a straight line due to inertia. The balance between these two forces keeps the bob and the earth moving in a circular path. v. The outward tendency is called centrifugal force because it feels like a force is pushing the object outward. It is not a real force. It is a pseudo force. Reason:There is no actual push from any object. It only appears when we are in a rotating frame. In reality, it is just the effect of inertia. Very Short Answer Type QuestionsQuestion 1. Write the formula for the acceleration due to gravity at the height h in terms of G, M and R. Answer:g = G \(\frac{G M}{(R+h)^2}\)Question 2. Does the gravitational force remain the same everywhere? Explain. Answer:No, the gravitational force depends on the distance from the centre of the earth. As the distance from the centre of the earth increases and the gravitational force decreases. Question 3. What do you mean by variation of acceleration due to gravity?Answer:Variation of acceleration due to gravity means how the value of changes when we move away from the surface of the earth either upwards (at height) or downwards (at depth). Question 4. At what angle is the turning effect of a force considered maximum?Answer:τ = Fd sin θ. sin 90° = 1τmax= F . dHence, at 90° angle is the turning effect of a force considered maximum. Short Answer Type QuestionsQuestion 1. What are the centripetal and centrifugal forces?Answer:Centripetal and Centrifugal ForceCentripetal Force:Earlier, we have studied that acceleration of a body moving in a circle of radius with uniform speed v is \(\frac{v^2}{r}\) directed towards the centre. From second law of motion, the force Fcis given bycentripetal force, Fc= \(\frac{m v^2}{r}\)where, m is the mass of the body. This force directed forward the centre is called centripetal force. But \(\frac{v^2}{r}\) is centripetal acceleration or radial r acceleration. ∴ Centripetal force, Fc= mrω2[∵ v = ωr]Centrifugal force:There are certain situations in which we feel that a body is acted upon by a force, but actually there is no force on the body. Such an apparent force is called centrifugal force. It is not a real force. It is it a pseudo force that L appears only in non-inertial frames. Various force acting as centripetal force. Question 2. If two objects with different cross-sectional areas are dropped from the same height in the presence of air, which object will reach the ground first and why?Answer:The object with the smaller cross-sectional area will reach the ground first. This is because an object with a larger surface area experiences greater air resistance (drag force), which slows down its downward speed. Question 3. In a vacuum, does the acceleration of an object depend on its mass or shape? What is the result of this?Answer:No, there is no air resistance in a vacuum, so the acceleration depends. The result is that all objects, regardless of their mass or shape, will reach the ground at the exact same time, with acceleration a = g. Question 4. Why do not astronauts appears to float inside a spacecraft? Does it mean gravity disappears in space?Answer:No, gravity does not disappear in space. Astronauts appear to float because both the astronauts and the spacecraft are in continuous free fall towards the earth together. They are falling at the same rate, so they feel weightless. Gravity is still acting on them, but they do not feel it due to free fall. Long Answer Type QuestionsQuestion 1. Describe how the acceleration due to gravity changes as we move below the surface of the earth towards its centre. Answer:Now, consider a point A located at a depth d inside the earth. Assuming the Earth has uniform density,let us determine the acceleration due to gravity at that interior point. Let,Radius of earth = RDepth below surface = dDistance from centre = R – dDensity of earth = ρ (uniform)If density is uniform, then mass of the earth can be calculated byMass of earth, M = \(\frac{4}{3}\) πR3ρAt depth d, the object is at a distance (R-d) from the centre, only the mass inside radius (R-d) contributes to gravity. (This is by Newton’s shell theorem, the gravitational effect of a uniform spherical shell on a point inside it is exactly zero. Therefore, at depth d, only the sphere of radius (R-d) centred at earth’s core contributes to gravity-the outer shell of thickness d has no net effect)Md= \(\frac{4}{3}\) π(R – d)3ρFrom Newton’s law of gravitation,gd= \(\frac{G M_d}{(d)^2}\)Substitute Md,gd= \(\frac{G\left(\frac{4}{3} \pi(R-d)^3 \rho\right)}{(d)^2}\)gd= \(\frac{4}{3}\) πGρ (R – d)At earth’s surface,gd= \(\frac{4}{3}\) πGρRDividing the two equations\(\frac{g_d}{g}=\frac{R-d}{R}\)gd= g (1 – \(\frac{d}{R}\))Question 2. What will be value of g at the bottom of the sea 7 km deep?Diameter of the earth is 12800 km and g on the surface of the earth is 9. 8 m/s2. Answer:Depth of Sea = 7 kmg = 9. 8 m/s2Radius of the earth,R = \(\frac{D}{2}\)= \(\frac{12800}{2}\) km = 6400 kmValue of g at the bottom of Seagd= d (1 – \(\frac{d}{R}\))= 9. 8 (1 – \(\frac{7}{6400}\)) m/s2gd= \(\frac{9. 8 \times 6393}{6400}\)= 9. 789 m/s2

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

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