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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.

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

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.

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)

Radially outwards

Always radially inwards

Radially outwards initially and radially inwards later.

Radially inwards initially and radially outwards later.

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.

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)

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

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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Download JEE Advanced Question Papers & Answer Keys of 2022, 2021, 2020, 2019, 2018 and 2017 PDFs. JEE Advanced Question Paper are provided language-wise along with their answer keys. We also offer JEE Advanced Sample Question Papers with Answer Keys for Physics, Chemistry and Maths solved by our expert teachers on Vedantu. Downloading the JEE Advanced Sample Question Papers with solutions will help the engineering aspirants to score high marks in the JEE Advanced examinations.

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In order to prepare for JEE Advanced 2023, candidates should know the list of important books i.e. RD Sharma Solutions, NCERT Solutions, HC Verma books and RS Aggarwal Solutions. They will find the high quality readymade solutions of these books on Vedantu. These books will help them in order to prepare well for the JEE Advanced 2023 exam so that they can grab the top rank in the all India entrance exam.

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IIT Bombay is responsible for the release of the JEE Advanced 2023 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 2023 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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FAQ

1. What is the weightage of mechanics in JEE Advanced?

Nearly 25% questions are asked in the JEE Advanced exam from this chapter. Mechanics is one of the most important topics and fundamental topic in JEE Advanced.

2. What is the difficulty level of Mechanics in JEE Advanced exam?

Theory of Mechanics is a vast topic and one of the fundamentals of physics required for JEE advanced. The mixed concepts from different types of mechanics topics can also asked by relating to other topics of JEE Asvanced. Students who are good in mechanics can score good marks in JEE advanced. Overall, mechanics is a moderate difficult which becomes easier with proper practise and effort. It is better if yo keep mechanics of structure pdf readily available for reference.

3. Are previous year's questions Important for JEE Advanced?

Yes, previous year's JEE Advanced questions are important. Once the student is finished learning concepts and solving problems, they must go through the previous year's question papers the last 20 years to understand the test pattern and frequent questions asked in the exam.

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