JEE

# JEE Important Chapter - Mechanics

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## Mechanics for JEE

Mechanics is a branch of physics that deals with energy and forces and their effect on bodies. As far as JEE Advanced exam is concerned, this chapter covers a vast amount of topics and is thus a very important chapter. This chapter requires conceptual understanding as well as numerical skills to score good marks in JEE Advanced.

In Mechanics, we learn about one dimensional and two dimensional and practical applications of newton’s laws of motion and force system in mechanics. Concepts related to kinetic energy and potential energy are also discussed in detail along with the work-energy theorem.

Problems related to rotational dynamics and gravitation are also important in Mechanics. Concepts related to the properties of solids and liquids as well as thermodynamics are included in this chapter mechanics.

Now, let us see what is mechanics, mechanics definitions, laws of mechanics, important concepts and formulae related to JEE advanced exams along with a few solved examples.

### Important Topics of Mechanics

• Motion along two dimension

• Laws of motion

• Mass Pulley system

• Work done by a constant force and variable force

• Kepler’s Law of planetary motion

• Stress-strain curve

• Rotational motion

• Bernoulli’s theorem

• Laws of thermodynamics

### Solved Examples

1. A bomb of mass30 kg at rest explodes into two pieces of masses 18 kg and 12 kg. The velcoity of 18 kg mass is 6 m/s. The kinetic energy of the other mass is ________.

Sol:

The total momentum of the bomb before and after exploision is constant.

$0=m_1v_1+m_2v_2$

$0=18\times 6+12\times v_2$

$v_2=-9~ m/s$

The kinetic energy of the mass having 12 kg is given by,

$K=\dfrac{1}{2}mv^2$

$K=\dfrac{1}{2}\times 12\times 9^2$

$K=486~J$

Therefore, kinetic energy of the other mass is 486 J

Key Point: The total momentum is conserved in the case of explosion of a bomb.

2. The instantaneous angular position of a point on a rotating wheel is given by the equation, $\theta(t)=2t^3-6t^2$ . The torque on the wheel becomes zero at ______.

Sol:

We can find the equation of angular velocity from the angular position as given below,

$\omega=\dfrac{d\theta}{dt}$

$\omega=\dfrac{d}{dt}(2t^3-6t^2)$

$\omega=6t^2-12t$

The equation for angular acceleration of the point is obtained by,

$\alpha=\dfrac{d\omega}{dt}$

$\alpha=\dfrac{d}{dt}(6t^2-12t)$

$\alpha=12t-12$

The torque becomes zero when angular acceleration becomes zero

$\alpha=12t-12$

$0=12t-12$

$t=1~s$

Therefore, torque becomes zero at t =1 s.

Key Point: Differentiating angular velocity with respect to time gives angular acceleration.

### Previous Year Questions from JEE Advanced Paper

1. A wire which passes through the hole in a small bead, is bent in the form of quarter of a circle. The wire is fixed vertically on the ground as shown in the figure. The ball is released from near  the top of the wire and it slides along the wire without any friction. As the bead moves from A to B, the force it applies on the wire is  (JEE Advanced 2010)

Sol:

Applying newton’s second law of motion equation,

$mg\cos\theta-\dfrac{mv^2}{R}=N$...(1)

Applying conservation of energy,

$\dfrac{1}{2}mv^2=mgR(1-\cos\theta)$...(2)

Using equation (1) and (2),

$N=mg(3\cos\theta-2)$

So, N is positive when

$0\lt \theta\lt \cos^{-1}\left(\dfrac{2}3{}\right)$

N is negative when

$\theta\gt \cos^{-1}\left(\dfrac{2}3{}\right)$

Therefore, force exerted  on the wire by the bead is radially inwards initially and radially outwards later. So, correct answer is option (D)

Key Point: For a body to move in a circular path, there must be a centripetal force acting on it.

2. Two identical uniform discs roll without slipping on two different surfaces,  AB and CD starting at A and C with linear speeds v1 and v2 respectively, and always remain in contact with surfaces. If they reach B and D with same linear speed v1= 3 m/s and v2= ______.(JEE Advanced 2015)

Sol:

Both discs reach B and D with same speed. Hence, final kinetic energy is same for both disc.

Applying conservation of energy,

For first disc,

$\dfrac{1}{2}I\omega_1^2+mg\times30=K$...(1)

For second disc,

$\dfrac{1}{2}I\omega_2^2+mg\times27=K$.....(2)

Solving equation (1) and equation (2), we get,

$\dfrac{1}{2}\times\left(\dfrac{3}{2}mR^2\right)\times\left(\dfrac{3}{R}\right)^2+mg\times30=\dfrac{1}{2}\times\left(\dfrac{3}{2}mR^2\right)\times\left(\dfrac{v_2}{R}\right)^2+mg\times27$

$v_2=7~m/s$

Key Point: The total mechanical energy is constant if the work is done only by the conservative force.

### Practice Questions

1. A constant torque of 1000 Nm turns a wheel of moment of inertia 200 kgm2 about an axis through its centre. Its angular velocity after 3 seconds is  (Ans: 15 rad/s)

2. For a satellite moving in an orbit around the earth, the ratio of kinetic energy to the potential energy is (Ans: 1:2)

### Conclusion

In this article we have provided important information regarding the chapter mechanics such as important concepts, formulae, etc.. Students should work on more solved examples along with  previous year question papers for scoring good grades in the JEE advanced exams.

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JEE Advanced 2022 exam date and revised schedule have been announced by the NTA. JEE Advanced 2022 will now be conducted on 28-August-2022, and the exam registration closes on 11-August-2022. You can check the complete schedule on our site. Furthermore, you can check JEE Advanced 2022 dates for application, admit card, exam, answer key, result, counselling, etc along with other relevant information.
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IIT Bombay has announced the JEE Advanced 2022 application form release date on the official website https://jeeadv.ac.in/. JEE Advanced 2022 Application Form is available on the official website for online registration. Besides JEE Advanced 2022 application form release date, learn about the application process, steps to fill the form, how to submit, exam date sheet etc online. Check our website for more details.
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It is crucial for the the engineering aspirants to know and download the JEE Advanced 2022 syllabus PDF for Maths, Physics and Chemistry. Check JEE Advanced 2022 syllabus here along with the best books and strategies to prepare for the entrance exam. Download the JEE Advanced 2022 syllabus consolidated as per the latest NTA guidelines from Vedantu for free.
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## JEE Advanced 2022 Study Material

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JEE Advanced 2022 Study Materials: Strengthen your fundamentals with exhaustive JEE Advanced Study Materials. It covers the entire JEE Advanced syllabus, DPP, PYP with ample objective and subjective solved problems. Free download of JEE Advanced study material for Physics, Chemistry and Maths are available on our website so that students can gear up their preparation for JEE Advanced exam 2022 with Vedantu right on time.
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IIT Bombay is responsible for the release of the JEE Advanced 2022 cut off score. The qualifying percentile score might remain the same for different categories. According to the latest trends, the expected cut off mark for JEE Advanced 2022 is 50% for general category candidates, 45% for physically challenged candidates, and 40% for candidates from reserved categories. For the general category, JEE Advanced qualifying marks for 2021 ranged from 17.50%, while for OBC/SC/ST categories, they ranged from 15.75% for OBC, 8.75% for SC and 8.75% for ST category.
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IIT Bombay is responsible for the release of the JEE Advanced 2022 cut off score. The qualifying percentile score might remain the same for different categories. According to the latest trends, the expected cut off mark for JEE Advanced 2022 is 50% for general category candidates, 45% for physically challenged candidates, and 40% for candidates from reserved categories. For the general category, JEE Advanced qualifying marks for 2021 ranged from 17.50%, while for OBC/SC/ST categories, they ranged from 15.75% for OBC, 8.75% for SC and 8.75% for ST category.
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The JEE Advanced 2022 result will be published by IIT Bombay on https://jeeadv.ac.in/ in the form of a scorecard. The scorecard will include the roll number, application number, candidate's personal details, and the percentile, marks, and rank of the candidate. Only those candidates who achieve the JEE Advanced cut-off will be considered qualified for the exam.
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Want to know which Engineering colleges in India accept the JEE Advanced 2022 scores for admission to Engineering? Find the list of Engineering colleges accepting JEE Advanced scores in India, compiled by Vedantu. There are 1622 Colleges that are accepting JEE Advanced. Also find more details on Fees, Ranking, Admission, and Placement.
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