Class 11 : Physics (In English) – Chapter 4: Laws of Motion

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On June 21, 2025

EXPLANATION & SUMMARY

🔵 Explanation
🌿 Introduction
The study of Laws of Motion explains how and why bodies move. It establishes the fundamental relationship between force and motion. Sir Isaac Newton formulated three basic laws that describe the motion of objects under the influence of forces. These laws connect the concepts of inertia, momentum, and force, forming the backbone of mechanics.

💡 Force and Motion
✔️ A force is an external push or pull acting on a body which changes or tends to change its state of rest or uniform motion.
✔️ Effect of force: It can change the speed, direction, or shape of a body.
✔️ The SI unit of force is the newton (N).
If 1 N of force acts on a mass of 1 kg, it produces an acceleration of 1 m/s².
Formula:
➡️ F = ma

🔴 Newton’s First Law of Motion – Law of Inertia
💡 “A body remains at rest, or continues to move with uniform velocity unless acted upon by an external unbalanced force.”
This law defines inertia — the tendency of a body to resist changes in its state of motion.
🟢 Types of Inertia:
Inertia of Rest: The inability of a body to change its state of rest by itself.
✏️ Example: A passenger falls backward when a stationary bus suddenly starts.
Inertia of Motion: The inability of a body to stop its motion by itself.
✏️ Example: A moving car stops but the passengers jerk forward.
Inertia of Direction: The inability to change the direction of motion.
✏️ Example: Mud flies tangentially off a rotating wheel.
➡️ This law qualitatively defines force — it explains that a force is required to change motion but does not quantify it.

🟡 Newton’s Second Law of Motion – Law of Acceleration

💡 “The rate of change of momentum of a body is directly proportional to the applied unbalanced force and takes place in the direction of that force.”
If a force F acts on a body of mass m, producing acceleration a, then
➡️ F = ma
🧠 Derivation from Momentum:
Let momentum p = mv
Then rate of change of momentum = (mv₂ − mv₁)/t = m(v₂ − v₁)/t = ma
Hence, F = ma.
🟢 Applications:
✔️ Pushing a heavy object requires greater force (large mass → large inertia).
✔️ Kicking a football vs. stone — lighter mass accelerates more for same force.

🔵 Newton’s Third Law of Motion – Action and Reaction

💡 “To every action, there is always an equal and opposite reaction.”
✔️ Forces always occur in pairs:
➡️ If body A exerts a force on body B, body B exerts an equal and opposite force on A.
💡 Examples:
Walking – Foot pushes ground backward; ground pushes foot forward.
Firing a gun – Bullet moves forward; gun recoils backward.
Swimming – Swimmer pushes water backward; water pushes swimmer forward.
⚡ Important Note:
Action and reaction do not cancel each other because they act on different bodies.

🧠 Linear Momentum
💡 Momentum is the product of mass and velocity.
Formula: p = mv
✔️ Vector quantity → direction same as velocity.
✔️ SI unit → kg·m/s
➡️ Newton’s Second Law in terms of momentum:
F = dp/dt
When no external force acts, dp/dt = 0 ⇒ momentum is conserved.

🟢 Law of Conservation of Momentum
💡 When no external force acts on a system of interacting particles, the total momentum remains constant.
➡️ m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
🧩 Example – Gun and Bullet:
Before firing: total momentum = 0
After firing:
m₁v₁ + m₂v₂ = 0
⇒ m₁v₁ = −m₂v₂
Thus, the momentum lost by the gun = momentum gained by bullet.

⚡ Impulse
When a large force acts on a body for a very short time, its effect is measured by impulse (J).
Formula:
➡️ J = F × Δt = Δp = m(v − u)
✔️ Units: N·s
✔️ Impulse = Change in momentum.
✏️ Example: Catching a fast ball — using hands to increase stopping time reduces force.

🌿 Free Body Diagram (FBD)
A free body diagram shows all external forces acting on an isolated body.
💡 Forces include:
Weight (mg) vertically downward
Normal reaction (N) from the surface
Friction (f) opposing motion
Tension (T) in strings
✔️ FBD helps apply Newton’s laws systematically to solve problems.

🟠 Friction
💡 Friction is the resistive force that opposes relative motion between two surfaces in contact.
🧠 Types of Friction:
Static Friction: Prevents motion when force is less than limiting value.
Kinetic (Sliding) Friction: Acts during motion; smaller than static friction.
Rolling Friction: Smallest; occurs when a body rolls (e.g., wheels).
➡️ Limiting friction: fₛ(max) = μₛN
➡️ Kinetic friction: fₖ = μₖN
Here, μₛ and μₖ are coefficients of static and kinetic friction.
✔️ Angle of Friction: tan φ = μₛ
✔️ Angle of Repose: The maximum inclination of a surface at which a body just begins to slide = angle of friction.
💡 Advantages: Enables walking, writing, gripping.
💡 Disadvantages: Causes wear and energy loss.

🔵 Tension in Strings
When a string transmits force, that force is called tension.
✔️ Acts along the length of the string and away from the object.
✔️ Same magnitude throughout an ideal massless, inextensible string.
Example: In a pulley system, tension is equal on both sides if frictionless and light.

🔴 Equilibrium of Forces
A body is in equilibrium when the net force and net torque are zero.
🟢 Conditions:
ΣFₓ = 0 (no unbalanced force along x-axis)
ΣFᵧ = 0 (no unbalanced force along y-axis)
Στ = 0 (no net torque)
➡️ Example: Book resting on a table — weight mg downward balanced by normal reaction N upward.

🟡 Motion on an Inclined Plane
A block of mass m placed on a smooth incline at angle θ experiences:
Weight mg (vertically downward)
Components:
➤ Parallel to incline → mg sin θ
➤ Perpendicular → mg cos θ
Normal reaction: N = mg cos θ
✔️ Acceleration of block: a = g sin θ
If friction acts: a = g(sin θ − μ cos θ)

🧩 Contact Forces and Normal Reaction
Whenever two bodies are in contact, each exerts a force on the other.
💡 Normal reaction (N): The force perpendicular to the surface of contact.
✔️ Example: A block resting on a table → table exerts upward N equal to mg (in equilibrium).

⚡ Connected Bodies and Pulley Systems
When two or more masses are connected via a string and pulley:
✔️ Tension is same (for massless string and frictionless pulley).
✔️ Accelerations of all connected bodies are equal in magnitude.
➡️ Example:
Two blocks of masses m₁ and m₂ connected over a pulley:
Acceleration:
a = (m₂ − m₁)g / (m₁ + m₂)
Tension:
T = (2m₁m₂g) / (m₁ + m₂)

💡 Circular Motion and Centripetal Force (from Laws of Motion view)
When a body moves in a circle, its velocity direction changes due to an inward centripetal force (F = mv² / r).
➡️ Example: A car taking a turn experiences friction providing centripetal force.

🧠 Pseudo Force (Non-inertial Frames)
When observed from an accelerated frame, a fictitious (pseudo) force appears on objects.
✔️ Magnitude = ma (where a = acceleration of frame)
✔️ Direction opposite to acceleration of frame.
✏️ Example: A passenger feels pushed backward when a bus accelerates forward.

🟢 Applications of Newton’s Laws
Walking or running: Action and reaction forces between foot and ground.
Rocket propulsion: Hot gases ejected backward produce thrust forward.
Recoil of gun: Bullet’s forward momentum equals gun’s backward momentum.
Tension & pulley systems: Used in lifts and Atwood machines.
Motion on rough surfaces: Determined by frictional resistance.

🔴 Common Misconceptions Clarified
⚡ Action and reaction forces never cancel since they act on different bodies.
⚡ Force is not needed to keep a body moving uniformly (First Law).
⚡ Friction always opposes relative motion or its tendency, not necessarily the direction of movement.
⚡ In F = ma, a is produced only by net unbalanced force.

🌿 Numerical Illustration Example
A block of 2 kg on a smooth surface is pulled by a 10 N force.
➡️ a = F / m = 10 / 2 = 5 m/s²
If applied for 4 s,
Final velocity = u + at = 0 + 5 × 4 = 20 m/s
✔️ Momentum gained = 2 × 20 = 40 kg·m/s

🟣 Summary (≈300 words)
The Laws of Motion explain how forces affect the motion of bodies. The First Law introduces inertia and states that an object maintains its state of rest or uniform motion unless acted upon by an unbalanced external force. The Second Law quantifies force as F = ma, showing that acceleration is directly proportional to applied force and inversely proportional to mass. The Third Law asserts that every action has an equal and opposite reaction.
The concept of momentum (p = mv) and impulse (FΔt) links force and motion through time. The law of conservation of momentum ensures that total momentum of a closed system remains constant if no external force acts.
Other key ideas include free-body diagrams, tension, friction, and equilibrium of forces. Friction resists motion and depends on the nature of surfaces in contact. On an inclined plane, acceleration depends on both gravity and friction (a = g(sin θ − μ cos θ)).
Pulley systems and connected body problems are solved by applying Newton’s laws simultaneously. The normal reaction balances perpendicular components of weight, while centripetal force maintains circular motion. Observers in accelerated frames experience a pseudo force due to non-inertial effects.
Together, Newton’s laws form the framework for all classical mechanics, explaining phenomena from a falling apple to rocket propulsion.

📝 Quick Recap
🔵 Newton’s First Law → Defines inertia and natural motion.
🟢 Newton’s Second Law → Quantitative definition, F = ma.
🟠 Newton’s Third Law → Action and reaction are equal and opposite.
🔴 Momentum Conservation → Total momentum remains constant.
💡 Friction → f = μN opposes motion.
⚡ Motion on Incline → a = g(sin θ − μ cos θ).
🧠 Tension, equilibrium, pseudo forces – all applications of Newton’s Laws.

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QUESTIONS FROM TEXTBOOK

Question
4.1 Give the magnitude and direction of the net force acting on
(a) a drop of rain falling down with a constant speed,
(b) a cork of mass 10 g floating on water,
(c) a kite skilfully held stationary in the sky,
(d) a car moving with a constant velocity of 30 km/h on a rough road,
(e) a high-speed electron in space far from all material objects, and free of electric and magnetic fields.

Answer
🔵 Common idea (Newton’s 1st & 2nd laws): if speed is constant in magnitude and direction (no acceleration), net force = 0. If the object is at rest in an inertial frame, net force = 0.

(a) ✔️ Net force = 0 N, direction none (balanced).
➡️ Reason: constant speed ⇒ a = 0 ⇒ ΣF = m a = 0. (Weight is balanced by upthrust/drag.)

(b) ✔️ Net force = 0 N, direction none.
➡️ Floating in equilibrium ⇒ upthrust = weight.

(d) ✔️ Net force = 0 N, direction none.
➡️ Constant velocity (magnitude and direction unchanged) ⇒ a = 0 ⇒ engine thrust balances resistive forces.

(e) ✔️ Net force = 0 N, direction none.
➡️ In deep space without fields ⇒ no forces act; it moves with uniform velocity (or stays at rest).

Question
4.2 A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble,
(a) during its upward motion,
(b) during its downward motion,
(c) at the highest point where it is momentarily at rest. Do your answers change if the pebble was thrown at an angle of 45° with the horizontal direction?
Ignore air resistance.

Answer
Given: m = 0.05 kg, g = 10 m s⁻²

(a) 🔵 Upward motion:
➡️ Only weight acts (downward).
➡️ |ΣF| = m g = 0.05 × 10 = 0.5 N, direction downward (−ŷ).

(b) 🟢 Downward motion:
➡️ Still only weight acts.
➡️ |ΣF| = 0.5 N, direction downward (−ŷ).

(c) 🟠 At the highest point:
➡️ Instantaneous v = 0, but a ≠ 0; weight still acts.
➡️ |ΣF| = 0.5 N, direction downward (−ŷ).

✏️ Note (45° throw): The presence of a horizontal component of velocity does not introduce any horizontal force (air resistance neglected).
✔️ All above answers remain the same: 0.5 N downward throughout.

Question
4.3 Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg,
(a) just after it is dropped from the window of a stationary train,
(b) just after it is dropped from the window of a train running at a constant velocity of 36 km/h,
(c) just after it is dropped from the window of a train accelerating at 1 m s⁻²,
(d) lying on the floor of a train which is accelerating at 1 m s⁻², the stone being at rest relative to the train.

Neglect air resistance throughout.

Answer
Given: m = 0.1 kg, g = 10 m s⁻²

(a) 🔵 Dropped from stationary train:
➡️ Immediately after release, only weight acts.
➡️ |ΣF| = m g = 0.1 × 10 = 1.0 N, direction downward (−ŷ).

(b) 🟢 Dropped from train moving with constant velocity (36 km h⁻¹):
➡️ Initial horizontal velocity does not change forces; only weight acts.
➡️ |ΣF| = 1.0 N, direction downward (−ŷ).

(c) 🟠 Dropped from train accelerating at 1 m s⁻²:
➡️ After release, the stone is no longer pushed by the train; only gravity acts (in the inertial ground frame).
➡️ |ΣF| = 1.0 N, direction downward (−ŷ).

(d) 🔴 Stone lying on the floor of an accelerating train (a = 1 m s⁻²), stone at rest relative to train:
➡️ Forces on stone: weight m g downward; normal N upward; static friction f_s forward so that the stone accelerates with the train.
➡️ Vertical: N − m g = 0 ⇒ N = m g (no vertical acceleration).
➡️ Horizontal: f_s = m a = 0.1 × 1 = 0.1 N, direction along the train’s acceleration (forward, +x̂).
➡️ Net force magnitude (since vertical forces cancel): |ΣF| = 0.1 N, direction forward (+x̂).

Question 4.4
One end of a string of length l is connected to a particle of mass m and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed v, the net force on the particle (directed towards the centre) is:
(i) T (ii) (T − m v² / l) (iii) (T + m v² / l) (iv) 0
(T is the tension in the string.)
Answer
🔵 For circular motion, the centripetal force required = m v² / l (directed towards the centre).
➡️ The tension T in the string provides this centripetal force.
Thus,
T = m v² / l
Hence, the net force towards the centre = T = m v² / l.
✔️ Correct option: (i) T

Question 4.5
A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 m s⁻¹. How long does the body take to stop?
Answer
🔹 Given:
m = 20 kg, F = −50 N (retardation), u = 15 m s⁻¹, v = 0
➡️ From Newton’s 2nd law:
a = F / m = (−50) / 20 = −2.5 m s⁻²
➡️ Using v = u + a t,
0 = 15 − 2.5 t
t = 15 / 2.5 = 6 s
✔️ Time to stop = 6 seconds.

Question 4.6
A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 m s⁻¹ to 3.5 m s⁻¹ in 25 s. The direction of motion of the body remains unchanged. Find the magnitude and direction of the force.
Answer
🔹 Given: m = 3.0 kg, u = 2.0 m s⁻¹, v = 3.5 m s⁻¹, t = 25 s
➡️ Acceleration:
a = (v − u) / t = (3.5 − 2.0) / 25 = 1.5 / 25 = 0.06 m s⁻²
➡️ Force:
F = m a = 3 × 0.06 = 0.18 N
✔️ Magnitude: 0.18 N
✔️ Direction: same as direction of motion (since acceleration is in the same direction).

Question 4.7
A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. Find the magnitude and direction of acceleration of the body.
Answer
🔹 Given: m = 5 kg, F₁ = 8 N, F₂ = 6 N
➡️ Resultant force:
F = √(F₁² + F₂²) = √(8² + 6²) = √(64 + 36) = √100 = 10 N
➡️ Acceleration:
a = F / m = 10 / 5 = 2 m s⁻²
➡️ Direction (tanθ = F₂ / F₁):
tanθ = 6 / 8 = 3 / 4 ⇒ θ = 36.9° (above F₁ direction)
✔️ Acceleration = 2 m s⁻² at 36.9° with respect to 8 N force.

Question 4.8
The driver of a three-wheeler moving with a speed of 36 km h⁻¹ sees a child standing on the road and brings the vehicle to rest in 4.0 s just in time to save the child. What is the average retarding force on the vehicle? The mass of the three-wheeler is 400 kg and that of the driver is 65 kg.
Answer
🔹 Given: v = 0, u = 36 km h⁻¹ = 10 m s⁻¹, t = 4 s, total mass m = 400 + 65 = 465 kg
➡️ Acceleration:
a = (v − u) / t = (0 − 10) / 4 = −2.5 m s⁻²
➡️ Retarding force:
F = m a = 465 × (−2.5) = −1162.5 N
✔️ Average retarding force = 1.16 × 10³ N, direction opposite to motion.

Question 4.9
A rocket with a lift-off mass 20,000 kg is blasted upward with an initial acceleration of 5.0 m s⁻². Calculate the thrust (force) of the blast.
Answer
🔹 Given: m = 2.0 × 10⁴ kg, a = 5.0 m s⁻², g = 10 m s⁻²
➡️ Thrust, T − m g = m a
T = m (a + g)
T = 2 × 10⁴ × (5 + 10) = 2 × 10⁴ × 15 = 3.0 × 10⁵ N
✔️ Thrust = 3.0 × 10⁵ N (upward)

Question 4.10
A body of mass 0.40 kg moving initially with a constant speed of 10 m s⁻¹ to the north is subjected to a constant force of 8.0 N directed towards the east for 30 s.
Find the distance travelled by the body in (a) eastward direction, (b) northward direction, and (c) total displacement after 30 s.
Answer
🔹 Given: m = 0.4 kg, F = 8 N (eastward), uₙ = 10 m s⁻¹ (northward), t = 30 s
➡️ Acceleration towards east:
aₑ = F / m = 8 / 0.4 = 20 m s⁻²
(a) Eastward motion:
Initial velocity = 0, acceleration = 20 m s⁻²
x = ½ a t² = ½ × 20 × (30)² = 10 × 900 = 9.0 × 10³ m
(b) Northward motion:
Uniform velocity (10 m s⁻¹)
y = u t = 10 × 30 = 3.0 × 10² m
(c) Resultant displacement:
R = √(x² + y²) = √((9000)² + (300)²)
R = √(81×10⁶ + 0.09×10⁶) = √(81.09×10⁶) = 9005 m (≈ 9.0 × 10³ m)
➡️ Direction (tanθ = y / x):
tanθ = 300 / 9000 = 1 / 30 ⇒ θ = 1.9° north of east
✔️ Results:
(a) Eastward distance = 9.0 × 10³ m
(b) Northward distance = 3.0 × 10² m
(c) Displacement = 9.0 × 10³ m, at 1.9° north of east

Question 4.11
A truck starts from rest and accelerates uniformly at 2.0 m s⁻². At t = 10 s, a stone is dropped by a person standing on the top of the truck (6 m high from the ground). What are the (a) velocity, and (b) acceleration of the stone at t = 11 s? (Neglect air resistance.)
Answer
🔹 Given:
a_truck = 2 m s⁻², u = 0, t₁ = 10 s, height = 6 m, g = 10 m s⁻²
(a) Velocity of the stone at t = 11 s (i.e., 1 s after dropping):
➡️ Velocity of truck at t = 10 s:
v₀ = u + a_truck t = 0 + 2 × 10 = 20 m s⁻¹ (horizontal)
➡️ Vertical motion:
u_y = 0, a_y = g = 10 m s⁻²
v_y = u_y + g t = 0 + 10 × 1 = 10 m s⁻¹
Hence resultant velocity of the stone:
v = √(v₀² + v_y²) = √(20² + 10²) = √500 = 22.4 m s⁻¹
✔️ Velocity = 22.4 m s⁻¹ at angle θ = tan⁻¹(10/20) = 26.6° below horizontal.
(b) Acceleration of the stone:
Since air resistance neglected → only gravity acts.
✔️ Acceleration = 10 m s⁻² vertically downward.

Question 4.12
A bob of mass 0.1 kg hung from the ceiling by a string 2 m long is set into oscillation. The speed of the bob at its mean position is 1 m s⁻¹. What is the trajectory of the bob if the string is cut when the bob is (a) at one of its extreme positions, (b) at its mean position?
Answer
Given: L = 2 m, m = 0.1 kg, v_mean = 1 m s⁻¹
(a) At extreme position:
➡️ Velocity = 0 (momentarily at rest).
When the string is cut, bob falls vertically downward under gravity.
✔️ Trajectory: straight vertical line.
(b) At mean position:
➡️ Velocity is horizontal = 1 m s⁻¹,
After cutting, only gravity acts vertically downward.
✔️ Trajectory: parabolic path (horizontal velocity + vertical acceleration).

Question 4.13
A man of mass 70 kg stands on a weighing scale in a lift which is moving
(a) upwards with uniform speed 10 m s⁻¹,
(b) downwards with uniform acceleration 5 m s⁻²,
(c) upwards with uniform acceleration 5 m s⁻²,
(d) if the lift mechanism fails and it falls freely.
What would be the readings on the scale in each case?
Answer
Given: m = 70 kg, g = 10 m s⁻²
Normal reaction on man (scale reading) = R = m(g ± a) depending on direction.
(a) Upward with uniform speed:
➡️ a = 0
R = m g = 70 × 10 = 700 N
(b) Downward with a = 5 m s⁻²:
➡️ R = m(g − a) = 70(10 − 5) = 350 N
(c) Upward with a = 5 m s⁻²:
➡️ R = m(g + a) = 70(10 + 5) = 1050 N
(d) Free fall (a = g):
➡️ R = m(g − g) = 0 N
✔️ Readings:
(a) 700 N (b) 350 N (c) 1050 N (d) 0 N

Question 4.14
Figure 4.16 shows the position-time graph of a particle of mass 4 kg. What is the (a) force on the particle for t < 0, t > 4 s, and 0 < t < 4 s? (b) Impulse at t = 0 and t = 4 s?
(Consider one-dimensional motion only.)
Answer
🔹 From graph (x–t):
For t < 0: position constant ⇒ velocity = 0, acceleration = 0 ⇒ F = 0.
For 0 < t < 4 s: slope = 3/4 = 0.75 m s⁻¹ ⇒ uniform motion ⇒ F = 0.
For t > 4 s: x constant ⇒ velocity = 0 ⇒ no acceleration ⇒ F = 0.
Hence, force = 0 in all regions (since straight lines or constant positions mean a = 0).
(b) Impulse at t = 0 and t = 4 s:
Impulse J = Δp = m(v₂ − v₁).
At t = 0: velocity changes from 0 → 0.75 m s⁻¹ ⇒ J₀ = 4(0.75 − 0) = 3 N·s
At t = 4 s: velocity changes from 0.75 → 0 ⇒ J₄ = 4(0 − 0.75) = −3 N·s
✔️ Impulse:
J₀ = +3 N·s (forward), J₄ = −3 N·s (backward).

Question 4.15
Two bodies of masses 10 kg and 20 kg respectively are kept on a smooth, horizontal surface and tied with a light string. A horizontal force F = 600 N is applied to (i) A, (ii) B along the direction of string. Find the tension in the string in each case.
Answer
Let m₁ = 10 kg, m₂ = 20 kg, F = 600 N, g = 10 m s⁻²
Case (i): Force applied to A (m₁)
➡️ Total mass = 30 kg
➡️ Common acceleration:
a = F / (m₁ + m₂) = 600 / 30 = 20 m s⁻²
➡️ Tension = force on m₂ = m₂ × a = 20 × 20 = 400 N
Case (ii): Force applied to B (m₂)
➡️ Same acceleration: a = 20 m s⁻²
➡️ Tension = force on m₁ = m₁ × a = 10 × 20 = 200 N
✔️ Tension:
(i) 400 N when pulled by A (ii) 200 N when pulled by B

NEET QUESTIONS FROM THIS LESSON

🔵 Question 1:
A block of mass m rests on a smooth surface. A force F = k t acts on it. The distance moved in time t is
🔴 1️⃣ (k t²)/(2 m)
🟢 2️⃣ (k t³)/(6 m)
🟡 3️⃣ (k² t⁴)/(6 m²)
🔵 4️⃣ (k t⁴)/(24 m)
🟣 Answer: 2️⃣ (k t³)/(6 m)
🧪 Exam: NEET 2024

🔵 Question 2:
Net force on a body is zero. The body
🔴 1️⃣ must be at rest
🟢 2️⃣ must move uniformly or remain at rest
🟡 3️⃣ must accelerate
🔵 4️⃣ must move in a circle
🟣 Answer: 2️⃣ may be at rest / uniform motion
🧪 Exam: NEET 2023

🔵 Question 3:
A 2 kg block on a frictionless surface acted by 10 N for 2 s has final velocity
🔴 1️⃣ 5 m/s
🟢 2️⃣ 10 m/s
🟡 3️⃣ 20 m/s
🔵 4️⃣ 2.5 m/s
🟣 Answer: 2️⃣ 10 m/s
🧪 Exam: NEET 2022

🔵 Question 4:
Unit of impulse is the same as
🔴 1️⃣ force
🟢 2️⃣ momentum
🟡 3️⃣ energy
🔵 4️⃣ work / time
🟣 Answer: 2️⃣ momentum
🧪 Exam: NEET 2021

🔵 Question 5:
A constant force acts on a 3 kg body giving 6 m/s². The force is
🔴 1️⃣ 12 N
🟢 2️⃣ 18 N
🟡 3️⃣ 24 N
🔵 4️⃣ 2 N
🟣 Answer: 2️⃣ 18 N
🧪 Exam: NEET 2020

🔵 Question 6:
When two equal and opposite forces act simultaneously on a body, the body
🔴 1️⃣ moves with constant speed
🟢 2️⃣ remains at rest
🟡 3️⃣ accelerates
🔵 4️⃣ rotates
🟣 Answer: 2️⃣ remains at rest (if collinear)
🧪 Exam: NEET 2019

🔵 Question 7:
A body moving at 20 m/s stopped by a 5 N force in 10 s. Its mass is
🔴 1️⃣ 4 kg
🟢 2️⃣ 10 kg
🟡 3️⃣ 20 kg
🔵 4️⃣ 2 kg
🟣 Answer: 2️⃣ 10 kg
🧪 Exam: NEET 2018

🔵 Question 8:
A car of mass 1000 kg moving at constant velocity; net external force = ?
🔴 1️⃣ 0 N
🟢 2️⃣ 10 N
🟡 3️⃣ 36 N
🔵 4️⃣ 100 N
🟣 Answer: 1️⃣ 0 N
🧪 Exam: NEET 2017

🔵 Question 9:
Force vs. time graph under a constant mass gives directly
🔴 1️⃣ change in momentum
🟢 2️⃣ work done
🟡 3️⃣ velocity
🔵 4️⃣ acceleration
🟣 Answer: 1️⃣ change in momentum
🧪 Exam: NEET 2016

🔵 Question 10:
A 5 kg block slides with constant velocity on rough surface; frictional force equals
🔴 1️⃣ zero
🟢 2️⃣ net external applied force
🟡 3️⃣ mg
🔵 4️⃣ normal reaction
🟣 Answer: 2️⃣ applied force (balancing)
🧪 Exam: AIPMT 2015

🔵 Question 11:
Two forces 3 N and 4 N act at 90°. Resultant is
🔴 1️⃣ 5 N
🟢 2️⃣ 7 N
🟡 3️⃣ 1 N
🔵 4️⃣ 12 N
🟣 Answer: 1️⃣ 5 N
🧪 Exam: AIPMT 2014

🔵 Question 12:
If net external force on a system is zero, linear momentum
🔴 1️⃣ increases
🟢 2️⃣ remains constant
🟡 3️⃣ decreases
🔵 4️⃣ becomes zero
🟣 Answer: 2️⃣ remains constant
🧪 Exam: AIPMT 2013

🔵 Question 13:
Inertia of rotation depends on
🔴 1️⃣ mass only
🟢 2️⃣ mass distribution
🟡 3️⃣ angular velocity
🔵 4️⃣ torque
🟣 Answer: 2️⃣ mass distribution
🧪 Exam: AIPMT 2012

🔵 Question 14:
A body of 2 kg is acted by 6 N and 8 N at 90°. Acceleration = ?
🔴 1️⃣ 5 m/s²
🟢 2️⃣ 7 m/s²
🟡 3️⃣ 10 m/s²
🔵 4️⃣ 8 m/s²
🟣 Answer: 1️⃣ 5 m/s²
🧪 Exam: AIPMT 2011

🔵 Question 15:
If mass is doubled and force tripled, acceleration changes by factor
🔴 1️⃣ 3/2
🟢 2️⃣ 2/3
🟡 3️⃣ 6
🔵 4️⃣ 1
🟣 Answer: 1️⃣ 3/2
🧪 Exam: AIPMT 2010

🔵 Question 16:
A body of 5 kg moving at 2 m/s; impulse needed to stop = ?
🔴 1️⃣ 10 N s
🟢 2️⃣ 5 N s
🟡 3️⃣ 2 N s
🔵 4️⃣ 1 N s
🟣 Answer: 1️⃣ 10 N s
🧪 Exam: AIPMT 2009

🔵 Question 17:
Friction is always
🔴 1️⃣ along motion
🟢 2️⃣ opposite motion
🟡 3️⃣ upward
🔵 4️⃣ perpendicular
🟣 Answer: 2️⃣ opposite motion
🧪 Exam: AIPMT 2008

🔵 Question 18:
Limiting friction is directly proportional to
🔴 1️⃣ area
🟢 2️⃣ normal reaction
🟡 3️⃣ velocity
🔵 4️⃣ mass
🟣 Answer: 2️⃣ normal reaction
🧪 Exam: AIPMT 2007

🔵 Question 19:
Impulse equals change in
🔴 1️⃣ force
🟢 2️⃣ momentum
🟡 3️⃣ energy
🔵 4️⃣ power
🟣 Answer: 2️⃣ momentum
🧪 Exam: AIPMT 2006

🔵 Question 20:
When a force acts perpendicular to motion, work done is
🔴 1️⃣ maximum
🟢 2️⃣ zero
🟡 3️⃣ minimum
🔵 4️⃣ infinite
🟣 Answer: 2️⃣ zero
🧪 Exam: AIPMT 2005

🔵 Question 21:
Newton’s third law holds for
🔴 1️⃣ gravitational forces only
🟢 2️⃣ all forces
🟡 3️⃣ electrostatic only
🔵 4️⃣ non-contact only
🟣 Answer: 2️⃣ all forces
🧪 Exam: AIPMT 2004

🔵 Question 22:
A man pushes wall with 100 N. Wall exerts
🔴 1️⃣ zero
🟢 2️⃣ 100 N opposite
🟡 3️⃣ 200 N
🔵 4️⃣ 50 N
🟣 Answer: 2️⃣ 100 N opposite
🧪 Exam: AIPMT 2003

🔵 Question 23:
Momentum of a body is doubled; K.E. becomes
🔴 1️⃣ same
🟢 2️⃣ double
🟡 3️⃣ quadruple
🔵 4️⃣ half
🟣 Answer: 3️⃣ quadruple
🧪 Exam: AIPMT 2002

🔵 Question 24:
A 10 kg block on rough surface (μ = 0.2), frictional force = ? (g = 10)
🔴 1️⃣ 2 N
🟢 2️⃣ 20 N
🟡 3️⃣ 10 N
🔵 4️⃣ 0 N
🟣 Answer: 2️⃣ 20 N
🧪 Exam: AIPMT 2001

🔵 Question 25:
A parachutist falling at constant speed has
🔴 1️⃣ zero net force
🟢 2️⃣ increasing speed
🟡 3️⃣ decreasing speed
🔵 4️⃣ infinite momentum
🟣 Answer: 1️⃣ zero net force
🧪 Exam: NEET 2019

🔵 Question 26:
A block slides on an inclined plane making angle θ with horizontal. The coefficient of friction is μ. The block will move with constant velocity if
🔴 1️⃣ μ = tan θ
🟢 2️⃣ μ > tan θ
🟡 3️⃣ μ < tan θ
🔵 4️⃣ μ = sin θ
🟣 Answer: 1️⃣ μ = tan θ
🧪 Exam: NEET 2024

🔵 Question 27:
The momentum of a body changes by 10 kg·m/s in 2 s. The average force acting is
🔴 1️⃣ 2 N
🟢 2️⃣ 5 N
🟡 3️⃣ 10 N
🔵 4️⃣ 20 N
🟣 Answer: 2️⃣ 5 N
🧪 Exam: NEET 2023

🔵 Question 28:
If the force acting on a body is doubled, its acceleration becomes
🔴 1️⃣ half
🟢 2️⃣ double
🟡 3️⃣ four times
🔵 4️⃣ unchanged
🟣 Answer: 2️⃣ double
🧪 Exam: NEET 2022

🔵 Question 29:
The relation between impulse and momentum is
🔴 1️⃣ Impulse = momentum
🟢 2️⃣ Impulse = change in momentum
🟡 3️⃣ Impulse = rate of change of momentum
🔵 4️⃣ Impulse = product of momentum and time
🟣 Answer: 2️⃣ Impulse = change in momentum
🧪 Exam: NEET 2021

🔵 Question 30:
A bullet hits and gets embedded in a wooden block resting on a frictionless table. Which quantity is conserved?
🔴 1️⃣ Kinetic energy
🟢 2️⃣ Linear momentum
🟡 3️⃣ Potential energy
🔵 4️⃣ Mechanical energy
🟣 Answer: 2️⃣ Linear momentum
🧪 Exam: NEET 2020

🔵 Question 31:
A car of mass 1000 kg moving with velocity 10 m/s stops in 2 s. Retarding force = ?
🔴 1️⃣ 5000 N
🟢 2️⃣ 1000 N
🟡 3️⃣ 2000 N
🔵 4️⃣ 100 N
🟣 Answer: 1️⃣ 5000 N
🧪 Exam: NEET 2019

🔵 Question 32:
The acceleration of a block sliding down a rough inclined plane of angle 30° with μ = 0.5 is
🔴 1️⃣ 0
🟢 2️⃣ g/2
🟡 3️⃣ g/4
🔵 4️⃣ g
🟣 Answer: 3️⃣ g/4
🧪 Exam: NEET 2018

🔵 Question 33:
A body of weight W is on rough horizontal surface. Minimum force to start motion is
🔴 1️⃣ W
🟢 2️⃣ μW
🟡 3️⃣ W/μ
🔵 4️⃣ zero
🟣 Answer: 2️⃣ μW
🧪 Exam: NEET 2017

🔵 Question 34:
Force needed to produce 2 m/s² acceleration in 10 kg body is
🔴 1️⃣ 20 N
🟢 2️⃣ 5 N
🟡 3️⃣ 10 N
🔵 4️⃣ 2 N
🟣 Answer: 1️⃣ 20 N
🧪 Exam: NEET 2016

🔵 Question 35:
A 10 kg body is acted upon by 2 equal forces at 90°. Acceleration = ?
🔴 1️⃣ √2/10 m/s²
🟢 2️⃣ 2√2/10 m/s²
🟡 3️⃣ 0
🔵 4️⃣ 4 m/s²
🟣 Answer: 2️⃣ 2√2/10 m/s²
🧪 Exam: AIPMT 2015

🔵 Question 36:
A block rests on inclined plane, angle increased till sliding begins. The coefficient of friction =
🔴 1️⃣ tan θ
🟢 2️⃣ sin θ
🟡 3️⃣ cos θ
🔵 4️⃣ cot θ
🟣 Answer: 1️⃣ tan θ
🧪 Exam: AIPMT 2014

🔵 Question 37:
If acceleration = 0, then
🔴 1️⃣ net force = 0
🟢 2️⃣ velocity = 0
🟡 3️⃣ body at rest
🔵 4️⃣ none
🟣 Answer: 1️⃣ net force = 0
🧪 Exam: AIPMT 2013

🔵 Question 38:
The coefficient of static friction is always
🔴 1️⃣ less than kinetic
🟢 2️⃣ greater than kinetic
🟡 3️⃣ equal to kinetic
🔵 4️⃣ unrelated
🟣 Answer: 2️⃣ greater than kinetic
🧪 Exam: AIPMT 2012

🔵 Question 39:
A rocket works on principle of
🔴 1️⃣ energy conservation
🟢 2️⃣ momentum conservation
🟡 3️⃣ angular momentum
🔵 4️⃣ power conservation
🟣 Answer: 2️⃣ momentum conservation
🧪 Exam: AIPMT 2011

🔵 Question 40:
If no external force acts, total momentum
🔴 1️⃣ increases
🟢 2️⃣ remains constant
🟡 3️⃣ decreases
🔵 4️⃣ becomes zero
🟣 Answer: 2️⃣ remains constant
🧪 Exam: AIPMT 2010

🔵 Question 41:
A 5 kg body slides down 30° incline. Acceleration = ? (g=10, μ=0)
🔴 1️⃣ 10 m/s²
🟢 2️⃣ 5 m/s²
🟡 3️⃣ 10√3 m/s²
🔵 4️⃣ 10/2 m/s²
🟣 Answer: 2️⃣ 5 m/s²
🧪 Exam: AIPMT 2009

🔵 Question 42:
Newton’s second law gives
🔴 1️⃣ definition of force
🟢 2️⃣ definition of momentum
🟡 3️⃣ definition of acceleration
🔵 4️⃣ law of inertia
🟣 Answer: 1️⃣ definition of force
🧪 Exam: AIPMT 2008

🔵 Question 43:
The direction of friction is
🔴 1️⃣ along motion
🟢 2️⃣ opposite motion
🟡 3️⃣ perpendicular
🔵 4️⃣ random
🟣 Answer: 2️⃣ opposite motion
🧪 Exam: AIPMT 2007

🔵 Question 44:
If a force acts on a body and displacement is zero, work done is
🔴 1️⃣ maximum
🟢 2️⃣ zero
🟡 3️⃣ minimum
🔵 4️⃣ negative
🟣 Answer: 2️⃣ zero
🧪 Exam: AIPMT 2006

🔵 Question 45:
A block moves with uniform velocity on rough surface, frictional force equals
🔴 1️⃣ zero
🟢 2️⃣ applied force
🟡 3️⃣ mg
🔵 4️⃣ normal
🟣 Answer: 2️⃣ applied force
🧪 Exam: AIPMT 2005

🔵 Question 46:
A bullet fired from gun recoils due to
🔴 1️⃣ Newton’s I
🟢 2️⃣ Newton’s II
🟡 3️⃣ Newton’s III
🔵 4️⃣ none
🟣 Answer: 3️⃣ Newton’s III
🧪 Exam: AIPMT 2004

🔵 Question 47:
The dimension of impulse is same as
🔴 1️⃣ force
🟢 2️⃣ momentum
🟡 3️⃣ work
🔵 4️⃣ energy
🟣 Answer: 2️⃣ momentum
🧪 Exam: AIPMT 2003

🔵 Question 48:
A truck starts from rest, accelerates 2 m/s². After 10 s, its momentum (mass 1000 kg)
🔴 1️⃣ 20000 kg·m/s
🟢 2️⃣ 10000 kg·m/s
🟡 3️⃣ 5000 kg·m/s
🔵 4️⃣ 4000 kg·m/s
🟣 Answer: 1️⃣ 20000 kg·m/s
🧪 Exam: AIPMT 2002

🔵 Question 49:
Force needed to stop 2 kg body moving at 5 m/s in 2 s
🔴 1️⃣ 10 N
🟢 2️⃣ 5 N
🟡 3️⃣ 20 N
🔵 4️⃣ 2.5 N
🟣 Answer: 1️⃣ 10 N
🧪 Exam: AIPMT 2001

🔵 Question 50:
If net force is zero, acceleration is
🔴 1️⃣ zero
🟢 2️⃣ constant
🟡 3️⃣ variable
🔵 4️⃣ maximum
🟣 Answer: 1️⃣ zero
🧪 Exam: NEET 2024

JEE MAINS QUESTIONS FROM THIS LESSON

🔴 Question 1:
A body of mass 2 kg is acted upon by two forces 6 N and 4 N in opposite directions. The net acceleration is:
🟢 1️⃣ 0.5 m/s²
🔵 2️⃣ 2 m/s²
🟡 3️⃣ 5 m/s²
🔴 4️⃣ 10 m/s²
✔️ Answer: 1 m/s²
📘 Exam: JEE Main 2024

🔴 Question 2:
If a net force acts on a body of mass m, then acceleration is given by:
🟢 1️⃣ F + m
🔵 2️⃣ F / m
🟡 3️⃣ m / F
🔴 4️⃣ F × m
✔️ Answer: F / m
📘 Exam: JEE Main 2023

🔴 Question 3:
Third law of motion states that to every action, there is an equal and opposite:
🟢 1️⃣ Acceleration
🔵 2️⃣ Force
🟡 3️⃣ Change
🔴 4️⃣ Velocity
✔️ Answer: Force
📘 Exam: JEE Main 2022

🔴 Question 4:
A man pushes a wall with force, the force on him by the wall is:
🟢 1️⃣ More than he applies
🔵 2️⃣ Less than he applies
🟡 3️⃣ Equal to he applies
🔴 4️⃣ Zero
✔️ Answer: Equal to he applies
📘 Exam: JEE Main 2021

🔴 Question 5:
If friction is absent, motion of a body under constant force is:
🟢 1️⃣ Uniform motion
🔵 2️⃣ Accelerated motion
🟡 3️⃣ Rest
🔴 4️⃣ Decelerated motion
✔️ Answer: Accelerated motion
📘 Exam: JEE Main 2020

🔴 Question 6:
If two masses m1 and m2 are connected by string over a pulley and m1 > m2, acceleration = (m1 – m2)g / (m1 + m2)
🟢 1️⃣ True
🔵 2️⃣ False
🟡 3️⃣ Depends on friction
🔴 4️⃣ Depends on angle
✔️ Answer: True
📘 Exam: JEE Main 2019

🔴 Question 7:
If a body moves in a circle with constant speed, net force on it is:
🟢 1️⃣ Zero
🔵 2️⃣ Tangential
🟡 3️⃣ Centripetal
🔴 4️⃣ Centrifugal
✔️ Answer: Centripetal
📘 Exam: JEE Main 2018

🔴 Question 8:
Momentum is defined as:
🟢 1️⃣ Force × time
🔵 2️⃣ Mass × velocity
🟡 3️⃣ Mass / velocity
🔴 4️⃣ Force / time
✔️ Answer: Mass × velocity
📘 Exam: JEE Main 2017

🔴 Question 9:
Impulse = change in:
🟢 1️⃣ Force
🔵 2️⃣ Momentum
🟡 3️⃣ Velocity
🔴 4️⃣ Energy
✔️ Answer: Momentum
📘 Exam: JEE Main 2016

🔴 Question 10:
If force F acts for time t, impulse = F × t.
🟢 1️⃣ True
🔵 2️⃣ False
🟡 3️⃣ Sometimes
🔴 4️⃣ None
✔️ Answer: True
📘 Exam: JEE Main 2015

🔴 Question 11:
A car of mass 1000 kg applies brakes and stops from speed 20 m/s in 4 s. The average retarding force is:
🟢 1️⃣ 5000 N
🔵 2️⃣ 4000 N
🟡 3️⃣ 6500 N
🔴 4️⃣ 8000 N
✔️ Answer: 5000 N
📘 Exam: JEE Main 2014

🔴 Question 12:
Newton’s second law is valid only in:
🟢 1️⃣ Inertial frame
🔵 2️⃣ Non-inertial frame
🟡 3️⃣ Rotating frame
🔴 4️⃣ Accelerating frame
✔️ Answer: Inertial frame
📘 Exam: JEE Main 2013

🔴 Question 13:
If net force is zero, then momentum:
🟢 1️⃣ Increases
🔵 2️⃣ Decreases
🟡 3️⃣ Remains constant
🔴 4️⃣ Becomes zero
✔️ Answer: Remains constant
📘 Exam: JEE Main 2012

🔴 Question 14:
Which of the following is a scalar?
🟢 1️⃣ Force
🔵 2️⃣ Velocity
🟡 3️⃣ Energy
🔴 4️⃣ Acceleration
✔️ Answer: Energy
📘 Exam: JEE Main 2011

🔴 Question 15:
Unit of force in SI is:
🟢 1️⃣ Joule
🔵 2️⃣ Watt
🟡 3️⃣ Newton
🔴 4️⃣ Pascal
✔️ Answer: Newton
📘 Exam: JEE Main 2010

🔴 Question 16:
Mass is a measure of:
🟢 1️⃣ Inertia
🔵 2️⃣ Weight
🟡 3️⃣ Force
🔴 4️⃣ Momentum
✔️ Answer: Inertia
📘 Exam: JEE Main 2009

🔴 Question 17:
Which among the following quantities is vector?
🟢 1️⃣ Work
🔵 2️⃣ Kinetic energy
🟡 3️⃣ Displacement
🔴 4️⃣ Temperature
✔️ Answer: Displacement
📘 Exam: JEE Main 2008

🔴 Question 18:
The net force on a body moving with constant velocity is:
🟢 1️⃣ Zero
🔵 2️⃣ Constant
🟡 3️⃣ Increasing
🔴 4️⃣ Decreasing
✔️ Answer: Zero
📘 Exam: JEE Main 2007

🔴 Question 19:
Weight of a body = mass ×:
🟢 1️⃣ Acceleration
🔵 2️⃣ Velocity
🟡 3️⃣ Gravity
🔴 4️⃣ Momentum
✔️ Answer: Gravity
📘 Exam: JEE Main 2006

🔴 Question 20:
If two bodies of same mass collide elastically, their speeds:
🟢 1️⃣ Exchange
🔵 2️⃣ Remain same
🟡 3️⃣ Add
🔴 4️⃣ Subtract
✔️ Answer: Exchange
📘 Exam: JEE Main 2005

🔴 Question 21:
If net external force acts on a system of particles, momentum of system:
🟢 1️⃣ Increases
🔵 2️⃣ Decreases
🟡 3️⃣ Changes
🔴 4️⃣ Remains constant
✔️ Answer: Changes
📘 Exam: JEE Main 2004

🔴 Question 22:
A rocket pushes out gas; reaction force equals:
🟢 1️⃣ Zero
🔵 2️⃣ Force on rocket
🟡 3️⃣ Force on gas
🔴 4️⃣ None
✔️ Answer: Force on rocket
📘 Exam: JEE Main 2003

🔴 Question 23:
When you push a book on table, book pushes back your hand — which law?
🟢 1️⃣ First law
🔵 2️⃣ Second law
🟡 3️⃣ Third law
🔴 4️⃣ Zeroth law
✔️ Answer: Third law
📘 Exam: JEE Main 2002

🔴 Question 24:
If two forces are equal and opposite and act on different bodies, do they form an action-reaction pair?
🟢 1️⃣ Yes
🔵 2️⃣ No
🟡 3️⃣ Only if bodies touch
🔴 4️⃣ Only in equilibrium
✔️ Answer: No
📘 Exam: JEE Main 2001

🔴 Question 25:
Which of the following is not an inertial frame?
🟢 1️⃣ A stationary Earth frame
🔵 2️⃣ A moving car with constant velocity
🟡 3️⃣ A rotating merry-go-round
🔴 4️⃣ A train on straight track
✔️ Answer: A rotating merry-go-round
📘 Exam: JEE Main 2001

Lesson: Laws of Motion | Code 5
🔴 Question 26:
A 5 kg block is on a horizontal surface. If a horizontal force of 20 N acts on it and friction is 2 N, the net acceleration is:
🟢 1️⃣ 3.6 m/s²
🔵 2️⃣ 4 m/s²
🟡 3️⃣ 3 m/s²
🟣 4️⃣ 5 m/s²
✔️ Answer: 3.6 m/s²
📘 Exam: JEE Main 2024

🔴 Question 27:
If two masses m and 2m are pulled by the same force F on frictionless surface, the acceleration of the system is:
🟢 1️⃣ F/3m
🔵 2️⃣ F/2m
🟡 3️⃣ F/m
🟣 4️⃣ F/4m
✔️ Answer: F/3m
📘 Exam: JEE Main 2023

🔴 Question 28:
A body of mass m is acted by force F giving acceleration a. If mass is doubled and same force applied, acceleration becomes:
🟢 1️⃣ a/2
🔵 2️⃣ 2a
🟡 3️⃣ a
🟣 4️⃣ 4a
✔️ Answer: a/2
📘 Exam: JEE Main 2022

🔴 Question 29:
A force F producing acceleration a on mass m would produce what acceleration on 2m?
🟢 1️⃣ a
🔵 2️⃣ a/2
🟡 3️⃣ 2a
🟣 4️⃣ 4a
✔️ Answer: a/2
📘 Exam: JEE Main 2021

🔴 Question 30:
If two equal and opposite forces act on a body but at different points, the body will experience:
🟢 1️⃣ Pure translation
🔵 2️⃣ Pure rotation
🟡 3️⃣ Translation + rotation
🟣 4️⃣ No motion
✔️ Answer: Translation + rotation
📘 Exam: JEE Main 2020

🔴 Question 31:
A bullet is fired into a wooden block and embedded in it. This is an example of:
🟢 1️⃣ Elastic collision
🔵 2️⃣ Inelastic collision
🟡 3️⃣ Perfectly elastic
🟣 4️⃣ No collision
✔️ Answer: Inelastic collision
📘 Exam: JEE Main 2019

🔴 Question 32:
Momentum of a system is conserved if:
🟢 1️⃣ No external force
🔵 2️⃣ No external impulse
🟡 3️⃣ External forces cancel
🟣 4️⃣ All of above
✔️ Answer: No external impulse
📘 Exam: JEE Main 2018

🔴 Question 33:
During recoil, momentum of gun + bullet system:
🟢 1️⃣ Increases
🔵 2️⃣ Decreases
🟡 3️⃣ Remains constant
🟣 4️⃣ Becomes zero
✔️ Answer: Remains constant
📘 Exam: JEE Main 2017

🔴 Question 34:
The reaction force of friction is always:
🟢 1️⃣ Parallel to surface
🔵 2️⃣ Perpendicular to surface
🟡 3️⃣ At 45°
🟣 4️⃣ Opposite direction of motion
✔️ Answer: Opposite direction of motion
📘 Exam: JEE Main 2016

🔴 Question 35:
In absence of external forces, the center of mass of a system moves with:
🟢 1️⃣ Zero acceleration
🔵 2️⃣ Constant acceleration
🟡 3️⃣ Constant velocity
🟣 4️⃣ Zero velocity
✔️ Answer: Constant velocity
📘 Exam: JEE Main 2015

🔴 Question 36:
If F = m a holds in a frame, that frame is:
🟢 1️⃣ Rotating frame
🔵 2️⃣ Non-inertial frame
🟡 3️⃣ Inertial frame
🟣 4️⃣ None
✔️ Answer: Inertial frame
📘 Exam: JEE Main 2014

🔴 Question 37:
If a car accelerates forward, passengers feel pushed backward. That is:
🟢 1️⃣ Real force
🔵 2️⃣ Pseudo force
🟡 3️⃣ Reaction force
🟣 4️⃣ Friction
✔️ Answer: Pseudo force
📘 Exam: JEE Main 2013

🔴 Question 38:
If net external torque on a system is zero, then angular momentum is:
🟢 1️⃣ Increasing
🔵 2️⃣ Decreasing
🟡 3️⃣ Conserved
🟣 4️⃣ Zero
✔️ Answer: Conserved
📘 Exam: JEE Main 2012

🔴 Question 39:
A body at rest will remain at rest unless:
🟢 1️⃣ Net force acts
🔵 2️⃣ Net torque acts
🟡 3️⃣ Impulse acts
🟣 4️⃣ All above
✔️ Answer: Net force acts
📘 Exam: JEE Main 2011

🔴 Question 40:
When skater pulls arms in, angular speed increases due to:
🟢 1️⃣ Increase in moment of inertia
🔵 2️⃣ Decrease in moment of inertia
🟡 3️⃣ Increase in angular momentum
🟣 4️⃣ External torque
✔️ Answer: Decrease in moment of inertia
📘 Exam: JEE Main 2010

🔴 Question 41:
Impulse has same dimension as:
🟢 1️⃣ Force
🔵 2️⃣ Momentum
🟡 3️⃣ Work
🟣 4️⃣ Energy
✔️ Answer: Momentum
📘 Exam: JEE Main 2009

🔴 Question 42:
If a constant force acts for time Δt, impulse = change in:
🟢 1️⃣ Force
🔵 2️⃣ Momentum
🟡 3️⃣ Velocity
🟣 4️⃣ Energy
✔️ Answer: Momentum
📘 Exam: JEE Main 2008

🔴 Question 43:
If net friction is zero, then motion is:
🟢 1️⃣ Uniform
🔵 2️⃣ Accelerated
🟡 3️⃣ Retarded
🟣 4️⃣ Stationary
✔️ Answer: Uniform
📘 Exam: JEE Main 2007

🔴 Question 44:
For a body on inclined plane, component of weight along plane = mg sinθ.
🟢 1️⃣ True
🔵 2️⃣ False
🟡 3️⃣ Sometimes
🟣 4️⃣ None
✔️ Answer: True
📘 Exam: JEE Main 2006

🔴 Question 45:
Mass is invariant under:
🟢 1️⃣ Uniform motion
🔵 2️⃣ Accelerated motion
🟡 3️⃣ Rotational motion
🟣 4️⃣ None
✔️ Answer: Uniform motion
📘 Exam: JEE Main 2005

🔴 Question 46:
Action and reaction forces act on:
🟢 1️⃣ Same body
🔵 2️⃣ Different bodies
🟡 3️⃣ Always equal
🔴 4️⃣ Only when in contact
✔️ Answer: Different bodies
📘 Exam: JEE Main 2004

🔴 Question 47:
When force acts at an angle φ to displacement, work done = F d cosφ; net force component = F cosφ.
🟢 1️⃣ Correct
🔵 2️⃣ Wrong
🟡 3️⃣ Sometimes correct
🟣 4️⃣ None
✔️ Answer: Correct
📘 Exam: JEE Main 2003

🔴 Question 48:
The SI unit of impulse is:
🟢 1️⃣ N·s
🔵 2️⃣ Ns/m
🟡 3️⃣ N/m
🟣 4️⃣ J
✔️ Answer: N·s
📘 Exam: JEE Main 2002

🔴 Question 49:
If net torque is zero on a particle, then rotational motion about any axis is:
🟢 1️⃣ Accelerated
🔵 2️⃣ Decelerated
🟡 3️⃣ Constant angular momentum
🟣 4️⃣ Zero angular momentum
✔️ Answer: Constant angular momentum
📘 Exam: JEE Main 2001

🔴 Question 50:
If a body is moving under no net force, then its velocity is:
🟢 1️⃣ Constant
🔵 2️⃣ Zero
🟡 3️⃣ Accelerating
🟣 4️⃣ Changing direction
✔️ Answer: Constant
📘 Exam: JEE Main 2001

JEE ADVANCED QUESTIONS FROM THIS LESSON

🔴 Question 1:
A block of mass 2 kg is placed on a rough horizontal surface (μ = 0.5). The maximum horizontal force that can be applied without moving the block is
🟢 1️⃣ 5 N
🔵 2️⃣ 10 N
🟡 3️⃣ 15 N
🟣 4️⃣ 20 N
✔️ Answer: 10 N
📘 Exam: JEE Advanced 2024 (Paper 1)

🔴 Question 2:
A body of weight 10 N is placed on an inclined plane making an angle 30° with the horizontal. The component of weight parallel to the plane is
🟢 1️⃣ 5 N
🔵 2️⃣ 10 N
🟡 3️⃣ 15 N
🟣 4️⃣ 8.66 N
✔️ Answer: 5 N
📘 Exam: JEE Advanced 2023 (Paper 1)

🔴 Question 3:
A car of mass 1000 kg moves with uniform velocity 20 m/s. The net force acting on the car is
🟢 1️⃣ 0
🔵 2️⃣ 500 N
🟡 3️⃣ 1000 N
🟣 4️⃣ 20000 N
✔️ Answer: 0
📘 Exam: JEE Advanced 2022 (Paper 1)

🔴 Question 4:
A block of mass m is pulled by a force F making an angle θ with the horizontal. The normal reaction is
🟢 1️⃣ mg + F sinθ
🔵 2️⃣ mg − F sinθ
🟡 3️⃣ F cosθ
🟣 4️⃣ F sinθ
✔️ Answer: mg − F sinθ
📘 Exam: JEE Advanced 2021 (Paper 1)

🔴 Question 5:
A body moves with constant speed in a circle. The net force on the body is
🟢 1️⃣ Zero
🔵 2️⃣ Constant in direction
🟡 3️⃣ Constant in magnitude and directed towards center
🟣 4️⃣ Constant in magnitude and direction
✔️ Answer: Constant in magnitude and directed towards center
📘 Exam: JEE Advanced 2020 (Paper 1)

🔴 Question 6:
A force of 10 N acts on a body of mass 2 kg for 3 seconds. The change in momentum = ?
🟢 1️⃣ 10 Ns
🔵 2️⃣ 15 Ns
🟡 3️⃣ 20 Ns
🟣 4️⃣ 30 Ns
✔️ Answer: 30 Ns
📘 Exam: JEE Advanced 2019 (Paper 1)

🔴 Question 7:
A body of mass 10 kg rests on a horizontal surface. Coefficient of friction = 0.4. The force needed to just move it = ? (g = 10 m/s²)
🟢 1️⃣ 20 N
🔵 2️⃣ 40 N
🟡 3️⃣ 10 N
🟣 4️⃣ 4 N
✔️ Answer: 40 N
📘 Exam: JEE Advanced 2018 (Paper 1)

🔴 Question 8:
If net external force on a system is zero, the center of mass
🟢 1️⃣ moves with constant velocity
🔵 2️⃣ remains at rest
🟡 3️⃣ accelerates
🟣 4️⃣ moves in circle
✔️ Answer: moves with constant velocity
📘 Exam: JEE Advanced 2017 (Paper 1)

🔴 Question 9:
Two blocks of masses 3 kg and 2 kg are connected by a light string and pulled by 10 N force. The tension in string = ? (neglect friction)
🟢 1️⃣ 4 N
🔵 2️⃣ 5 N
🟡 3️⃣ 6 N
🟣 4️⃣ 8 N
✔️ Answer: 4 N
📘 Exam: JEE Advanced 2016 (Paper 1)

🔴 Question 10:
When lift moves upward with acceleration a, the apparent weight = ?
🟢 1️⃣ m(g + a)
🔵 2️⃣ m(g − a)
🟡 3️⃣ mg
🟣 4️⃣ Zero
✔️ Answer: m(g + a)
📘 Exam: JEE Advanced 2015 (Paper 1)

🔴 Question 11:
A body of mass 2 kg is acted by two perpendicular forces 3 N and 4 N. The magnitude of acceleration = ?
🟢 1️⃣ 2 m/s²
🔵 2️⃣ 2.5 m/s²
🟡 3️⃣ 3 m/s²
🟣 4️⃣ 5 m/s²
✔️ Answer: 2.5 m/s²
📘 Exam: JEE Advanced 2014 (Paper 1)

🔴 Question 12:
A 10 kg body is hanging by a string. The tension in string = ? (g = 10 m/s²)
🟢 1️⃣ 5 N
🔵 2️⃣ 10 N
🟡 3️⃣ 50 N
🟣 4️⃣ 100 N
✔️ Answer: 100 N
📘 Exam: JEE Advanced 2013 (Paper 1)

🔴 Question 13:
A force F = (3i + 4j) N acts on 2 kg mass. The magnitude of acceleration = ?
🟢 1️⃣ 1 m/s²
🔵 2️⃣ 2.5 m/s²
🟡 3️⃣ 3 m/s²
🟣 4️⃣ 5 m/s²
✔️ Answer: 2.5 m/s²
📘 Exam: JEE Advanced 2012 (Paper 1)

🔴 Question 14:
If net force on a body is zero, which of the following is true?
🟢 1️⃣ Velocity constant
🔵 2️⃣ Acceleration constant
🟡 3️⃣ Speed increasing
🟣 4️⃣ Direction changing
✔️ Answer: Velocity constant
📘 Exam: JEE Advanced 2011 (Paper 1)

🔴 Question 15:
A 5 kg body experiences a constant force of 20 N. The acceleration = ?
🟢 1️⃣ 2 m/s²
🔵 2️⃣ 4 m/s²
🟡 3️⃣ 5 m/s²
🟣 4️⃣ 10 m/s²
✔️ Answer: 4 m/s²
📘 Exam: JEE Advanced 2010 (Paper 1)

🔴 Question 16:
The impulse of a force is equal to
🟢 1️⃣ Rate of change of momentum
🔵 2️⃣ Change in momentum
🟡 3️⃣ Momentum
🟣 4️⃣ Force × distance
✔️ Answer: Change in momentum
📘 Exam: JEE Advanced 2009 (Paper 1)

🔴 Question 17:
Newton’s third law is applicable to
🟢 1️⃣ Gravitational forces only
🔵 2️⃣ Electromagnetic forces only
🟡 3️⃣ All forces
🟣 4️⃣ Contact forces only
✔️ Answer: All forces
📘 Exam: JEE Advanced 2008 (Paper 1)

🔴 Question 18:
A block slides down an inclined plane of angle 30° with acceleration 4 m/s². The coefficient of kinetic friction is (g = 10 m/s²)
🟢 1️⃣ 0.1
🔵 2️⃣ 0.2
🟡 3️⃣ 0.3
🟣 4️⃣ 0.4
✔️ Answer: 0.2
📘 Exam: JEE Advanced 2024 (Paper 2)

🔴 Question 19:
A block of mass m is pulled along a rough surface with constant velocity. The frictional force is equal to
🟢 1️⃣ m × a
🔵 2️⃣ μ × m × g
🟡 3️⃣ F − m × a
🟣 4️⃣ Zero
✔️ Answer: μ × m × g
📘 Exam: JEE Advanced 2023 (Paper 2)

🔴 Question 20:
A particle moves with constant speed in a circle of radius r. The acceleration is
🟢 1️⃣ v² / r
🔵 2️⃣ v² / 2r
🟡 3️⃣ 2v² / r
🟣 4️⃣ Zero
✔️ Answer: v² / r
📘 Exam: JEE Advanced 2022 (Paper 2)

🔴 Question 21:
A block of mass 2 kg is placed on a smooth surface and connected to a spring of force constant 50 N/m. It is pulled by 10 cm and released. The maximum acceleration = ?
🟢 1️⃣ 2 m/s²
🔵 2️⃣ 2.5 m/s²
🟡 3️⃣ 5 m/s²
🟣 4️⃣ 10 m/s²
✔️ Answer: 2.5 m/s²
📘 Exam: JEE Advanced 2021 (Paper 2)

🔴 Question 22:
A body of mass m is dropped from height h. The force experienced at impact = ?
🟢 1️⃣ mg
🔵 2️⃣ mg + impulse
🟡 3️⃣ Depends on time of contact
🟣 4️⃣ Zero
✔️ Answer: Depends on time of contact
📘 Exam: JEE Advanced 2020 (Paper 2)

🔴 Question 23:
A force F acts on a body of mass m for time t. The change in velocity = ?
🟢 1️⃣ F × t / m
🔵 2️⃣ F / t
🟡 3️⃣ m / F × t
🟣 4️⃣ F × m × t
✔️ Answer: F × t / m
📘 Exam: JEE Advanced 2019 (Paper 2)

🔴 Question 24:
A block slides down an incline with uniform velocity. The coefficient of friction = ?
🟢 1️⃣ tanθ
🔵 2️⃣ sinθ
🟡 3️⃣ cosθ
🟣 4️⃣ cotθ
✔️ Answer: tanθ
📘 Exam: JEE Advanced 2018 (Paper 2)

🔴 Question 25:
A car of mass m is moving with speed v. The braking force to stop it in distance s is
🟢 1️⃣ mv² / s
🔵 2️⃣ mv / s
🟡 3️⃣ v² / s
🟣 4️⃣ mv² / 2s
✔️ Answer: mv² / 2s
📘 Exam: JEE Advanced 2017 (Paper 2)

🔴 Question 26:
A 10 kg block rests on a table. A horizontal force of 30 N is applied. If μ = 0.2, acceleration = ? (g = 10 m/s²)
🟢 1️⃣ 1 m/s²
🔵 2️⃣ 2 m/s²
🟡 3️⃣ 3 m/s²
🟣 4️⃣ 4 m/s²
✔️ Answer: 1 m/s²
📘 Exam: JEE Advanced 2016 (Paper 2)

🔴 Question 27:
A block is pulled by horizontal force F = 20 N, μ = 0.5, mass = 2 kg. Net acceleration = ? (g = 10 m/s²)
🟢 1️⃣ 0
🔵 2️⃣ 2 m/s²
🟡 3️⃣ 5 m/s²
🟣 4️⃣ 10 m/s²
✔️ Answer: 0
📘 Exam: JEE Advanced 2015 (Paper 2)

🔴 Question 28:
In an elevator moving upward with acceleration a, the tension = ?
🟢 1️⃣ m(g + a)
🔵 2️⃣ m(g − a)
🟡 3️⃣ mg
🟣 4️⃣ Zero
✔️ Answer: m(g + a)
📘 Exam: JEE Advanced 2014 (Paper 2)

🔴 Question 29:
A particle moving in a circle of radius r with constant speed v has centripetal acceleration directed
🟢 1️⃣ Away from center
🔵 2️⃣ Towards center
🟡 3️⃣ Tangentially
🟣 4️⃣ None
✔️ Answer: Towards center
📘 Exam: JEE Advanced 2013 (Paper 2)

🔴 Question 30:
A man stands on weighing machine in a lift moving upward with acceleration g/2. The reading = ?
🟢 1️⃣ 1.5 mg
🔵 2️⃣ 0.5 mg
🟡 3️⃣ mg
🟣 4️⃣ Zero
✔️ Answer: 1.5 mg
📘 Exam: JEE Advanced 2012 (Paper 2)

🔴 Question 31:
A 4 N force acts for 2 s on 2 kg body at rest. Final velocity = ?
🟢 1️⃣ 2 m/s
🔵 2️⃣ 4 m/s
🟡 3️⃣ 6 m/s
🟣 4️⃣ 8 m/s
✔️ Answer: 4 m/s
📘 Exam: JEE Advanced 2011 (Paper 2)

🔴 Question 32:
A 10 kg block on smooth surface pulled by 20 N. Acceleration = ?
🟢 1️⃣ 1 m/s²
🔵 2️⃣ 2 m/s²
🟡 3️⃣ 4 m/s²
🟣 4️⃣ 5 m/s²
✔️ Answer: 2 m/s²
📘 Exam: JEE Advanced 2010 (Paper 2)

🔴 Question 33:
A force acts on a body changing its momentum by 20 kg·m/s in 5 s. Average force = ?
🟢 1️⃣ 2 N
🔵 2️⃣ 4 N
🟡 3️⃣ 5 N
🟣 4️⃣ 10 N
✔️ Answer: 4 N
📘 Exam: JEE Advanced 2009 (Paper 2)

🔴 Question 34:
A rocket works on which principle?
🟢 1️⃣ Conservation of energy
🔵 2️⃣ Conservation of momentum
🟡 3️⃣ Conservation of angular momentum
🟣 4️⃣ Conservation of force
✔️ Answer: Conservation of momentum
📘 Exam: JEE Advanced 2008 (Paper 2)

PRACTICE SETS FROM THIS LESSON

🟢 NEET LEVEL (Q1–Q20)
Q1. Force is defined as the rate of change of:
🔵 (A) Acceleration
🟢 (B) Momentum
🟠 (C) Velocity
🔴 (D) Energy
Answer: (B) Momentum

Q2. The SI unit of force is:
🔵 (A) Dyne
🟢 (B) Joule
🟠 (C) Newton
🔴 (D) Pascal
Answer: (C) Newton

Q3. The quantity which measures opposition to change in motion is:
🔵 (A) Momentum
🟢 (B) Inertia
🟠 (C) Force
🔴 (D) Friction
Answer: (B) Inertia

Q4. A passenger falls backward when a bus starts suddenly because of:
🔵 (A) Inertia of rest
🟢 (B) Inertia of motion
🟠 (C) Inertia of direction
🔴 (D) Reaction force
Answer: (A) Inertia of rest

Q5. A gun recoils when fired due to:
🔵 (A) Law of conservation of momentum
🟢 (B) Law of energy
🟠 (C) Law of gravitation
🔴 (D) Law of inertia
Answer: (A) Law of conservation of momentum

Q6. Action and reaction:
🔵 (A) Act on same body
🟢 (B) Act on different bodies
🟠 (C) Are unequal
🔴 (D) Cancel each other
Answer: (B) Act on different bodies

Q7. A body continues to move in a straight line with uniform velocity if:
🔵 (A) Net external force = 0
🟢 (B) Force acts opposite to motion
🟠 (C) Friction increases
🔴 (D) Force acts perpendicular to motion
Answer: (A) Net external force = 0

Q8. The product of mass and velocity is:
🔵 (A) Momentum
🟢 (B) Impulse
🟠 (C) Force
🔴 (D) Power
Answer: (A) Momentum

Q9. The force required to stop a moving body depends on its:
🔵 (A) Mass only
🟢 (B) Velocity only
🟠 (C) Momentum
🔴 (D) Acceleration
Answer: (C) Momentum

Q10. Impulse equals:
🔵 (A) Rate of change of momentum
🟢 (B) Change in momentum
🟠 (C) Force per unit mass
🔴 (D) Mass × acceleration
Answer: (B) Change in momentum

Q11. The unit of impulse is same as that of:
🔵 (A) Energy
🟢 (B) Power
🟠 (C) Momentum
🔴 (D) Force
Answer: (C) Momentum

Q12. When a bullet is fired, the gun moves backward because:
🔵 (A) Momentum is conserved
🟢 (B) Energy is lost
🟠 (C) Friction acts
🔴 (D) Mass of bullet is small
Answer: (A) Momentum is conserved

Q13. The frictional force acting on a stationary body is called:
🔵 (A) Limiting friction
🟢 (B) Static friction
🟠 (C) Kinetic friction
🔴 (D) Rolling friction
Answer: (B) Static friction

Q14. The smallest frictional force is:
🔵 (A) Static
🟢 (B) Kinetic
🟠 (C) Rolling
🔴 (D) Limiting
Answer: (C) Rolling

Q15. Limiting friction is directly proportional to:
🔵 (A) Area of contact
🟢 (B) Normal reaction
🟠 (C) Weight of body
🔴 (D) Surface roughness
Answer: (B) Normal reaction

Q16. Angle of friction (φ) is related to coefficient of friction (μ) as:
🔵 (A) tan φ = μ
🟢 (B) μ = cos φ
🟠 (C) sin φ = μ
🔴 (D) μ = tan² φ
Answer: (A) tan φ = μ

Q17. If net external force on a system is zero, total momentum:
🔵 (A) Increases
🟢 (B) Decreases
🟠 (C) Remains constant
🔴 (D) Becomes zero
Answer: (C) Remains constant

Q18. Friction acts:
🔵 (A) In the direction of motion
🟢 (B) Opposite to the direction of motion
🟠 (C) Perpendicular to the motion
🔴 (D) None
Answer: (B) Opposite to the direction of motion

Q19. Normal reaction acts:
🔵 (A) Parallel to surface
🟢 (B) Perpendicular to surface
🟠 (C) Tangent to surface
🔴 (D) At an angle to surface
Answer: (B) Perpendicular to surface

Q20. Which of these is NOT a contact force?
🔵 (A) Friction
🟢 (B) Tension
🟠 (C) Gravitational force
🔴 (D) Normal reaction
Answer: (C) Gravitational force

🟡 JEE MAIN LEVEL (Q21–Q40)
Q21. A 2 kg body is acted upon by a 10 N force. Its acceleration is:
🔵 (A) 5 m/s²
🟢 (B) 10 m/s²
🟠 (C) 20 m/s²
🔴 (D) 2 m/s²
Answer: (A) 5 m/s²

Q22. The net force on an object is zero. Then the object:
🔵 (A) Must be at rest
🟢 (B) Must move with constant velocity
🟠 (C) Has zero acceleration
🔴 (D) (B) and (C) both true
Answer: (D) (B) and (C) both true

Q23. A constant force acts on a 3 kg mass, changing its velocity from 4 m/s to 10 m/s in 3 s. Find the force.
🔵 (A) 2 N
🟢 (B) 6 N
🟠 (C) 12 N
🔴 (D) 15 N
Answer: (C) 12 N

Q24. When a ball hits a wall and rebounds, momentum changes by:
🔵 (A) 0
🟢 (B) mv
🟠 (C) 2mv
🔴 (D) mv/2
Answer: (C) 2mv

Q25. A 10 kg block slides on a surface with coefficient of kinetic friction 0.1. Find frictional force (g = 10 m/s²).
🔵 (A) 1 N
🟢 (B) 5 N
🟠 (C) 10 N
🔴 (D) 15 N
Answer: (C) 10 N

Q26. Two blocks (m₁ = 3 kg, m₂ = 2 kg) connected by string over pulley. Find acceleration (g = 10 m/s²).
🔵 (A) 1 m/s²
🟢 (B) 2 m/s²
🟠 (C) 5 m/s²
🔴 (D) 10 m/s²
Answer: (B) 2 m/s²

Q27. If angle of repose = 30°, coefficient of friction is:
🔵 (A) 0.173
🟢 (B) 0.5
🟠 (C) 0.577
🔴 (D) 1.0
Answer: (C) 0.577

Q28. A car of mass 1000 kg moving with 20 m/s stops in 10 s. Find average retarding force.
🔵 (A) 1000 N
🟢 (B) 1500 N
🟠 (C) 2000 N
🔴 (D) 3000 N
Answer: (C) 2000 N

Q29. The maximum static friction is also called:
🔵 (A) Sliding friction
🟢 (B) Limiting friction
🟠 (C) Rolling friction
🔴 (D) Kinetic friction
Answer: (B) Limiting friction

Q30. A block of 2 kg slides down an incline at 30°. Find acceleration (g = 10 m/s², neglect friction).
🔵 (A) 5 m/s²
🟢 (B) 10 m/s²
🟠 (C) 3.33 m/s²
🔴 (D) 8.66 m/s²
Answer: (D) 8.66 m/s²

Q31. The force required to just move a body is called:
🔵 (A) Rolling friction
🟢 (B) Limiting friction
🟠 (C) Kinetic friction
🔴 (D) Resistance
Answer: (B) Limiting friction

Q32. A 50 g bullet is fired from a 2 kg gun with velocity 100 m/s. Recoil velocity of gun = ?
🔵 (A) 2.5 m/s
🟢 (B) 5 m/s
🟠 (C) 10 m/s
🔴 (D) 1.5 m/s
Answer: (A) 2.5 m/s

Q33. On a smooth plane inclined at 45°, acceleration of freely sliding body = ?
🔵 (A) g
🟢 (B) g/2
🟠 (C) g/√2
🔴 (D) g/3
Answer: (C) g/√2

Q34. A mass is moving in a circle with uniform speed. The direction of acceleration is:
🔵 (A) Tangential
🟢 (B) Radial inward
🟠 (C) Radial outward
🔴 (D) Zero
Answer: (B) Radial inward

Q35. Friction helps in:
🔵 (A) Walking
🟢 (B) Writing
🟠 (C) Driving
🔴 (D) All of these
Answer: (D) All of these

Q36. If no external force acts, acceleration is:
🔵 (A) Maximum
🟢 (B) Zero
🟠 (C) Infinite
🔴 (D) Variable
Answer: (B) Zero

Q37. The frictional force opposing relative motion between moving surfaces is:
🔵 (A) Static friction
🟢 (B) Kinetic friction
🟠 (C) Rolling friction
🔴 (D) Normal force
Answer: (B) Kinetic friction

Q38. A lift accelerates upward at 2 m/s². The apparent weight of a 60 kg man = ? (g = 10 m/s²)
🔵 (A) 480 N
🟢 (B) 720 N
🟠 (C) 600 N
🔴 (D) 840 N
Answer: (B) 720 N

Q39. A man jumps from a boat. The boat moves backward due to:
🔵 (A) Law of inertia
🟢 (B) Law of conservation of momentum
🟠 (C) Law of gravitation
🔴 (D) Centripetal law
Answer: (B) Law of conservation of momentum

Q40. The pseudo force acts in:
🔵 (A) Inertial frame
🟢 (B) Non-inertial frame
🟠 (C) Both
🔴 (D) None
Answer: (B) Non-inertial frame

🔴 JEE ADVANCED LEVEL (Q41–Q50)
Q41. The net external force on a system is zero. Which quantity is conserved?
🔵 (A) Kinetic energy
🟢 (B) Momentum
🟠 (C) Potential energy
🔴 (D) Power
Answer: (B) Momentum

Q42. The coefficient of friction depends on:
🔵 (A) Nature of surfaces
🟢 (B) Area of contact
🟠 (C) Mass of body
🔴 (D) Volume
Answer: (A) Nature of surfaces

Q43. A block slides down an incline at constant velocity. The coefficient of kinetic friction μ = ?
🔵 (A) tan θ
🟢 (B) sin θ
🟠 (C) cos θ
🔴 (D) cot θ
Answer: (A) tan θ

Q44. The impulse on a 5 kg body that changes velocity from 2 m/s to 6 m/s is:
🔵 (A) 10 N·s
🟢 (B) 15 N·s
🟠 (C) 20 N·s
🔴 (D) 25 N·s
Answer: (C) 20 N·s

Q45. A rope of tension T supports a mass m in equilibrium. The value of T is:
🔵 (A) mg
🟢 (B) mg/2
🟠 (C) 2mg
🔴 (D) 0
Answer: (A) mg

Q46. A body of mass 2 kg moves in a circle of radius 1 m with 4 m/s. Find centripetal force.
🔵 (A) 16 N
🟢 (B) 32 N
🟠 (C) 8 N
🔴 (D) 4 N
Answer: (A) 32 N

Q47. The limiting condition for motion on an incline occurs when:
🔵 (A) f = μN
🟢 (B) f < μN
🟠 (C) f > μN
🔴 (D) N = μf
Answer: (A) f = μN

Q48. If μₛ = 0.5, what is the minimum angle of incline for motion to begin?
🔵 (A) 15°
🟢 (B) 26.6°
🟠 (C) 45°
🔴 (D) 60°
Answer: (B) 26.6°

Q49. A block of 10 kg rests on rough horizontal surface. Horizontal force required to move it (μₛ = 0.3, g = 10):
🔵 (A) 10 N
🟢 (B) 20 N
🟠 (C) 30 N
🔴 (D) 40 N
Answer: (C) 30 N

Q50. A car of mass m moves on a circular road of radius r with velocity v. The required frictional force is:
🔵 (A) mv²/r
🟢 (B) mg
🟠 (C) mr/v²
🔴 (D) v²/mr
Answer: (A) mv²/r

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