Back to LAWS OF MOTION

Your Progress

0 / 34 Answered

Correct

Incorrect

Accuracy

JEE • PHYSICS • LAWS OF MOTION

Laws of motion practice · Easy

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

Take a chapter quiz →

Showing 31-34 of 34 questions

Preparing practice and saved progress…

31 Easy

31 of 34

Understanding +4 | -1

Assertion: To conserve linear momentum of a system, no force should act on the system.

Reason: If net force on a system is zero, its linear momentum should remain constant.

View answer and explanation

Correct option: D

If Assertion is false but the Reason is true

Option D is correct because the reason accurately describes the condition for linear momentum conservation, while the assertion is too restrictive.

Solution approach

Correct Answer + Core Reason: If Assertion is false but the Reason is true. Net force being zero implies constant linear momentum.

Step-by-Step Solution:

Step 1: Analyze the assertion: To conserve linear momentum of a system, no force should act on the system.

Step 2: The assertion is false because the *net* force on the system must be zero for linear momentum to be conserved. Individual forces can act, but they must cancel each other out.

Step 3: Analyze the reason: If net force on a system is zero, its linear momentum should remain constant.

Step 4: The reason is true according to Newton's first law (law of inertia) and the definition of linear momentum conservation.

Why other options are wrong:
- Option A: The assertion is false.
- Option B: The assertion is false.
- Option C: The reason is true.

Key Concepts:
- Conservation of linear momentum.
- Newton's first law of motion.

Memory Aids:
- Net force = 0 => Constant linear momentum.

Quick Tips:
- Remember that it's the *net* force that determines linear momentum conservation.

32 Easy

32 of 34

Application +4 | -1

A force is applied for a duration of 10 seconds on a body of mass 5 kg that is at rest. As a result the body acquires a velocity of

2\,m{s^{ - 1}}

. Find the magnitude of the force applied

View answer and explanation

Correct option: A

1 N

Option A (1 N) is correct because the force is calculated using Newton's second law and the given values of mass, velocity, and time.

Solution approach

Correct Answer + Core Reason: 1 N. The force is calculated using $F = ma = \frac{{mv}}{t}$ .

Step-by-Step Solution:

Step 1: Identify the given values: mass $m = 5$ kg, final velocity $v = 2$ m/s, time $t = 10$ s.

Step 2: Apply the formula for force: $F = ma$ . Since the initial velocity is 0, we can use $a = \frac{{v - u}}{t} = \frac{v}{t}$ . Thus, $F = m \frac{v}{t}$ .

Step 3: Substitute the given values: $F = 5 \times \frac{2}{10}$ .

Step 4: Calculate the force: $F = 1$ N.

Why other options are wrong:
- Option B (2 N): This would be the answer if the time was 5 seconds instead of 10.
- Option C (4 N): This is incorrect and doesn't relate to the given values in a straightforward manner.
- Option D (5 N): This would be the answer if the velocity was 10 m/s and time was 10 seconds, or if the mass was incorrectly used as the force.

Key Concepts:
- Newton's Second Law of Motion: $F = ma$
- Definition of acceleration: $a = \frac{{\Delta v}}{{\Delta t}}$

Memory Aids:
- F = ma (Force equals mass times acceleration)

Quick Tips:
- Always check the units of the given values before substituting them into the formula.
- Remember that if the initial velocity is zero, the change in velocity is simply the final velocity.

33 Easy

33 of 34

Application +4 | -1

An object is placed on the surface of a smooth inclined plane of inclination

\theta

. It takes time

t

to reach the bottom of the inclined plane. If the same object is allowed to slide down rough inclined plane of same inclination

\theta

, it takes time nt to reach the bottom where

\mathrm{n}

is a number greater than 1 . The coefficient of friction

\mu

is given by

View answer and explanation

Correct option: A

\mu=\tan \theta\left(1-\frac{1}{\mathrm{n}^2}\right)

Option A (

\mu=\tan \theta\left(1-\frac{1}{\mathrm{n}^2}\right)

) is correct because it is derived from comparing the distances traveled on smooth and rough inclined planes using kinematic equations.

Solution approach

Correct Answer + Core Reason: $\mu=\tan \theta\left(1-\frac{1}{\mathrm{n}^2}\right)$ . The coefficient of friction is found by comparing the motion on smooth and rough inclines.

Step-by-Step Solution:

Step 1: Define the distance traveled along the inclined plane as $s$ .

Step 2: For the smooth inclined plane, the acceleration is $a_1 = g \sin \theta$ . Using the kinematic equation $s = \frac{1}{2} a_1 t^2$ , we have $s = \frac{1}{2} g \sin \theta t^2$ .

Step 3: For the rough inclined plane, the acceleration is $a_2 = g \sin \theta - \mu g \cos \theta$ . Using the kinematic equation $s = \frac{1}{2} a_2 (nt)^2$ , we have $s = \frac{1}{2} (g \sin \theta - \mu g \cos \theta) n^2 t^2$ .

Step 4: Equate the distances traveled: $\frac{1}{2} g \sin \theta t^2 = \frac{1}{2} (g \sin \theta - \mu g \cos \theta) n^2 t^2$ .

Step 5: Simplify and solve for $\mu$ :
$\sin \theta = (\sin \theta - \mu \cos \theta) n^2 \Rightarrow \frac{\sin \theta}{n^2} = \sin \theta - \mu \cos \theta \Rightarrow \mu \cos \theta = \sin \theta - \frac{\sin \theta}{n^2} \Rightarrow \mu = \frac{\sin \theta}{\cos \theta} - \frac{\sin \theta}{n^2 \cos \theta} \Rightarrow \mu = \tan \theta \left(1 - \frac{1}{n^2}\right)$ .

Why other options are wrong:
- **Option B ( $\mu=\cot \theta\left(1-\frac{1}{n^2}\right)$ ):** This is incorrect because it uses $\cot \theta$ instead of $\tan \theta$ .
- **Option C ( $\mu=\tan \theta\left(1-\frac{1}{n^2}\right)^{1 / 2}$ ): This is incorrect because the expression is not derived correctly from the equations of motion.
- Option D ( $\mu=\cot \theta\left(1-\frac{1}{\mathrm{n}^2}\right)^{1 / 2}$ ):** This is incorrect because it uses $\cot \theta$ instead of $\tan \theta$ and the expression is not derived correctly.

Key Concepts:
- Kinematics on an inclined plane
- Newton's second law
- Friction

Memory Aids:
- Remember the kinematic equations: $s = ut + \frac{1}{2}at^2$ .
- Friction opposes motion.

Quick Tips:
- Draw free-body diagrams to identify forces.
- Use consistent coordinate systems.

34 Easy

34 of 34

Understanding +4 | -1

A rider on horse falls backward when horse starts running all of a sudden because of

View answer and explanation

Correct option: C

Inertia of rest

Option C (Inertia of rest) is correct because the rider's upper body tends to remain at rest while the horse moves forward.

Solution approach

Correct Answer + Core Reason: Inertia of rest. The rider's upper body tends to remain at rest due to inertia.

Step-by-Step Solution:

Step 1: Define inertia.

Step 2: Inertia is the tendency of an object to resist changes in its state of motion.

Step 3: Inertia of rest is the tendency of an object to remain at rest.

Step 4: When the horse starts running suddenly, the rider's lower body moves forward with the horse, but the upper body tends to remain at rest due to inertia of rest.

Step 5: This causes the rider to fall backward.

Why other options are wrong:
- Option A: Inertia of motion is incorrect because it applies to objects already in motion.
- Option B: Inertia of direction is incorrect because it applies to changes in direction, not changes in speed from rest.
- Option D: Both (1) and (2) is incorrect because only inertia of rest is relevant here.

Key Concepts:
- Inertia
- Inertia of rest
- Newton's first law of motion

Memory Aids:
- Inertia resists changes in motion.

Quick Tips:
- Remember the different types of inertia and their applications.

Revision notes and key takeaways

Review LAWS OF MOTION in your textbook or class notes, then use this set to check your understanding.

Showing 31-34 of 34 questions

Your Progress

Laws of motion practice · Easy

Continue Your Practice

Your practice