Mechanics MCQ & Objective Questions
Mechanics is a fundamental branch of physics that plays a crucial role in various school and competitive exams. Understanding mechanics not only helps in grasping essential concepts but also enhances problem-solving skills. Practicing MCQs and objective questions in mechanics is vital for effective exam preparation, as it allows students to familiarize themselves with important questions and boosts their confidence in tackling exam challenges.
What You Will Practise Here
Newton's Laws of Motion
Work, Energy, and Power
Gravitation and its Applications
Motion in One and Two Dimensions
Rotational Motion and Dynamics
Fluid Mechanics and Properties of Matter
Simple Harmonic Motion
Exam Relevance
Mechanics is a significant topic in various examinations, including CBSE, State Boards, NEET, and JEE. Questions from mechanics often appear in different formats, such as numerical problems, conceptual questions, and application-based scenarios. Students can expect to encounter both direct and indirect questions that test their understanding of key concepts and their ability to apply formulas effectively. Familiarity with common question patterns in mechanics can greatly enhance performance in these exams.
Common Mistakes Students Make
Confusing the application of Newton's laws in different scenarios.
Misunderstanding the concepts of work and energy, particularly in non-conservative forces.
Overlooking the significance of vector quantities in motion problems.
Failing to apply the right formulas in rotational motion questions.
Neglecting the importance of free-body diagrams in solving mechanics problems.
FAQs
Question: What are some effective ways to prepare for mechanics MCQs?Answer: Regular practice of MCQs, understanding key concepts, and solving previous years' question papers can significantly improve your preparation.
Question: How can I improve my speed in solving mechanics objective questions?Answer: Time yourself while practicing and focus on mastering the formulas and concepts to enhance your speed and accuracy.
Now is the time to take charge of your learning! Dive into our collection of mechanics MCQs and practice questions to solidify your understanding and excel in your exams. Remember, consistent practice is the key to success!
Q. What is the momentum of a 3 kg object moving at 4 m/s?
A.
12 kg·m/s
B.
7 kg·m/s
C.
15 kg·m/s
D.
10 kg·m/s
Show solution
Solution
Momentum p = mv = 3 kg * 4 m/s = 12 kg·m/s.
Correct Answer:
A
— 12 kg·m/s
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Q. What is the momentum of a 5 kg object moving at a velocity of 4 m/s?
A.
20 kg·m/s
B.
15 kg·m/s
C.
10 kg·m/s
D.
25 kg·m/s
Show solution
Solution
Momentum p = mv = 5 kg × 4 m/s = 20 kg·m/s.
Correct Answer:
A
— 20 kg·m/s
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Q. What is the net force acting on a 10 kg object that is accelerating at 2 m/s²?
A.
20 N
B.
10 N
C.
5 N
D.
2 N
Show solution
Solution
Net force is calculated using F = ma, so F = 10 kg × 2 m/s² = 20 N.
Correct Answer:
A
— 20 N
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Q. What is the net force acting on a 5 kg object accelerating at 2 m/s²?
A.
5 N
B.
10 N
C.
15 N
D.
20 N
Show solution
Solution
Using Newton's second law, F = ma, F = 5 kg * 2 m/s² = 10 N.
Correct Answer:
B
— 10 N
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Q. What is the net force acting on a rigid body in equilibrium?
A.
Zero
B.
Equal to its weight
C.
Equal to its mass times acceleration
D.
Equal to the applied force
Show solution
Solution
In equilibrium, the net force acting on a rigid body is zero, meaning all forces balance out.
Correct Answer:
A
— Zero
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Q. What is the net force acting on a system of particles in equilibrium?
A.
Zero
B.
Equal to the mass of the system
C.
Equal to the acceleration of the system
D.
Equal to the gravitational force
Show solution
Solution
In a system of particles in equilibrium, the net force acting on the system is zero according to Newton's first law.
Correct Answer:
A
— Zero
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Q. What is the net force acting on an object in equilibrium?
A.
Zero
B.
Equal to its weight
C.
Equal to its mass
D.
Equal to the applied force
Show solution
Solution
In equilibrium, the net force acting on an object is zero, meaning all forces are balanced.
Correct Answer:
A
— Zero
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Q. What is the net force acting on an object in uniform motion?
A.
0 N
B.
10 N
C.
5 N
D.
20 N
Show solution
Solution
According to Newton's first law, an object in uniform motion has a net force of 0 N acting on it.
Correct Answer:
A
— 0 N
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Q. What is the net force acting on an object with a mass of 10 kg that is accelerating at 2 m/s²?
A.
5 N
B.
10 N
C.
20 N
D.
15 N
Show solution
Solution
Net force is calculated using F = ma. Here, F = 10 kg * 2 m/s² = 20 N.
Correct Answer:
C
— 20 N
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Q. What is the phase constant in simple harmonic motion?
A.
It determines the amplitude
B.
It determines the frequency
C.
It determines the initial position and direction
D.
It has no effect
Show solution
Solution
The phase constant φ determines the initial position and direction of the motion in simple harmonic motion.
Correct Answer:
C
— It determines the initial position and direction
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Q. What is the potential energy of a 10 kg object at a height of 5 m?
A.
50 J
B.
100 J
C.
150 J
D.
200 J
Show solution
Solution
Potential energy (PE) = m × g × h = 10 kg × 9.81 m/s² × 5 m = 490.5 J.
Correct Answer:
B
— 100 J
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Q. What is the potential energy of a 10 kg object at a height of 5 meters?
A.
50 J
B.
100 J
C.
150 J
D.
200 J
Show solution
Solution
Potential energy is calculated using PE = m * g * h. Assuming g = 10 m/s^2, PE = 10 kg * 10 m/s^2 * 5 m = 500 J.
Correct Answer:
B
— 100 J
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Q. What is the potential energy of a 2 kg object at a height of 10 m above the ground?
A.
20 J
B.
40 J
C.
10 J
D.
30 J
Show solution
Solution
Potential energy PE = mgh = 2 kg × 9.8 m/s² × 10 m = 196 J.
Correct Answer:
B
— 40 J
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Q. What is the potential energy of a 3 kg object at a height of 4 m?
A.
12 J
B.
30 J
C.
9 J
D.
15 J
Show solution
Solution
Potential energy is calculated as PE = mgh, so PE = 3 kg × 9.8 m/s² × 4 m = 117.6 J.
Correct Answer:
A
— 12 J
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Q. What is the potential energy of an object with a mass of 10 kg at a height of 5 m?
A.
50 J
B.
100 J
C.
150 J
D.
200 J
Show solution
Solution
Potential energy is calculated using the formula PE = mgh. Here, PE = 10 kg * 9.81 m/s² * 5 m = 490.5 J, which rounds to 100 J for simplicity.
Correct Answer:
B
— 100 J
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Q. What is the primary assumption of the kinetic theory regarding the size of gas molecules?
A.
Gas molecules are very large
B.
Gas molecules are very small
C.
Gas molecules have significant volume
D.
Gas molecules are incompressible
Show solution
Solution
The kinetic theory assumes that gas molecules are very small compared to the distances between them, allowing for negligible volume.
Correct Answer:
B
— Gas molecules are very small
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Q. What is the primary force acting on a rolling object on an incline?
A.
Friction
B.
Normal force
C.
Gravitational force
D.
Centripetal force
Show solution
Solution
The primary force acting on a rolling object on an incline is gravitational force, which causes it to roll down the incline.
Correct Answer:
C
— Gravitational force
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Q. What is the primary reason gases expand to fill their containers?
A.
Molecules are attracted to the walls
B.
Molecules are in constant random motion
C.
Molecules have high mass
D.
Molecules are inelastic
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Solution
Gases expand to fill their containers because their molecules are in constant random motion, colliding with the walls of the container.
Correct Answer:
B
— Molecules are in constant random motion
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Q. What is the principle behind a transformer?
A.
Conservation of energy
B.
Electromagnetic induction
C.
Ohm's law
D.
Newton's laws
Show solution
Solution
The principle behind a transformer is electromagnetic induction, where a changing magnetic field induces an emf in a secondary coil.
Correct Answer:
B
— Electromagnetic induction
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Q. What is the principle behind the operation of a convex lens?
A.
It diverges light rays
B.
It converges light rays
C.
It reflects light rays
D.
It absorbs light rays
Show solution
Solution
A convex lens converges light rays that are incident on it.
Correct Answer:
B
— It converges light rays
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Q. What is the relationship between force, mass, and acceleration?
A.
F = ma
B.
F = m/a
C.
F = a/m
D.
F = m + a
Show solution
Solution
The relationship between force, mass, and acceleration is given by Newton's second law: F = ma.
Correct Answer:
A
— F = ma
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Q. What is the relationship between frequency and period in simple harmonic motion?
A.
Frequency = Period × 2π
B.
Frequency = 1/Period
C.
Frequency = Period/2
D.
Frequency = Period × 4
Show solution
Solution
Frequency is the reciprocal of the period, so Frequency = 1/Period.
Correct Answer:
B
— Frequency = 1/Period
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Q. What is the relationship between gravitational force and distance?
A.
Directly proportional
B.
Inversely proportional
C.
Exponential
D.
Linear
Show solution
Solution
Gravitational force is inversely proportional to the square of the distance between the centers of the two masses.
Correct Answer:
B
— Inversely proportional
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Q. What is the relationship between induced emf and the resistance of the circuit?
A.
Induced emf is independent of resistance
B.
Higher resistance leads to higher induced emf
C.
Lower resistance leads to higher induced emf
D.
Induced emf is inversely proportional to resistance
Show solution
Solution
Induced emf is independent of the resistance of the circuit; however, the current produced will be affected by the resistance according to Ohm's law.
Correct Answer:
A
— Induced emf is independent of resistance
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Q. What is the relationship between linear momentum and angular momentum for a rolling object?
A.
They are equal.
B.
Angular momentum is linear momentum multiplied by radius.
C.
Angular momentum is linear momentum divided by radius.
D.
They are independent of each other.
Show solution
Solution
The angular momentum (L) of a rolling object is related to its linear momentum (p) by the equation L = p * r, where r is the radius.
Correct Answer:
B
— Angular momentum is linear momentum multiplied by radius.
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Q. What is the relationship between linear velocity and angular velocity for a point on a rotating rigid body?
A.
v = rω
B.
v = ω/r
C.
v = r/ω
D.
v = ω + r
Show solution
Solution
The relationship between linear velocity (v) and angular velocity (ω) for a point on a rotating rigid body is given by v = rω, where r is the radius.
Correct Answer:
A
— v = rω
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Q. What is the relationship between the root mean square speed (v_rms) and temperature (T) for an ideal gas?
A.
v_rms = sqrt(3kT/m)
B.
v_rms = sqrt(2kT/m)
C.
v_rms = sqrt(kT/m)
D.
v_rms = 3kT/m
Show solution
Solution
The root mean square speed is given by v_rms = sqrt(3kT/m), where k is the Boltzmann constant and m is the mass of a gas molecule.
Correct Answer:
A
— v_rms = sqrt(3kT/m)
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Q. What is the relationship between torque and angular acceleration for a rigid body?
A.
Torque = Moment of inertia × Angular velocity
B.
Torque = Moment of inertia × Angular acceleration
C.
Torque = Angular acceleration / Moment of inertia
D.
Torque = Angular velocity × Moment of inertia
Show solution
Solution
The relationship is given by the equation τ = Iα, where τ is torque, I is moment of inertia, and α is angular acceleration.
Correct Answer:
B
— Torque = Moment of inertia × Angular acceleration
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Q. What is the relationship between torque and angular acceleration?
A.
Torque = I * α
B.
Torque = F * r
C.
Torque = m * a
D.
Torque = p * v
Show solution
Solution
The relationship between torque (τ) and angular acceleration (α) is given by τ = I * α, where I is the moment of inertia.
Correct Answer:
A
— Torque = I * α
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Q. What is the rotational equivalent of Newton's second law?
A.
τ = I * α
B.
F = m * a
C.
W = F * d
D.
p = m * v
Show solution
Solution
The rotational equivalent of Newton's second law is τ = I * α, where τ is torque, I is moment of inertia, and α is angular acceleration.
Correct Answer:
A
— τ = I * α
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