The national testing agency (NTA) has scheduled the second session of JEE Main 2019 from 7^{th} April to 20^{th} April. In January session, 9,41,117 candidates appeared in 10 shifts. The number of appearing students will definitely increase as it is the last and final chance for the engineering aspirants to fulfill their dream of getting best engineering colleges like IITs, IIITs, NITs, CFTIs. The Physics section was the most difficult in all shifts according to the students who attempted January exam. Physics is an application based subject. Students cannot solve the JEE level questions without clearing the basic concepts. But if students prepare well, then they can easily score good marks in Physics section. After doing a detailed analysis of JEE Main Previous years’ papers, Physics experts at jagranjosh.com have collected 100 solved questions which are very important from the examination point of view. The elaborated solutions for all the questions are also provided so that students can easily understand the concept used. Appearing students must solve these questions before writing JEE Main 2019 April exam.

**JEE Main Question Papers 2019 with Official Answer Keys (January exam)**

**JEE Main 2019: How to improve your NTA Score in April exam?**

1. Focus only on the important topics before a few days of the examination. To know the important topics, students need to do the thorough analysis of JEE Main Previous Years’ Papers.

2. Clear all your basic concepts before attempting any numerical problem.

3. Practice more and more numerical problems which will help you to master any concept or topic.

4. Clear all your basic concepts of calculus i.e., basic rules of integration and differentiation.

5. Never ignore units in CGS and MKS systems while solving numerical problems.

A circular disc rolls down an inclined plane. The ratio of the total kinetic energy to the rotational kinetic energy is

(a) 1 : 3

(b) 3 : 1

(c) 2 : 3

(d) 3 : 2

**Most Important 100 Chemistry Solved Questions for JEE Main 2019**

A satellite is moving with a constant speed 'V’ in a circular orbit about the earth. An object of mass ‘m’ is ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time of its ejection, the kinetic energy of the object is

What is the minimum energy required to launch a satellite of mass *m* from the surface of a planet of mass *M* and radius *R* in a circular orbit at an altitude of 2*R*?

In circular orbit of a satellite, potential energy = −2 × Kinetic Energy = −*mV*^{2}. So, just to escape from the gravitational pull, its total mechanical energy should be zero. Therefore its kinetic energy should be *mV*^{2}.

The radiation corresponding to 3 → 2 transition of hydrogen atom falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field of 3 × 10^{−4} T. If the radius of the largest circular path followed by these electrons is 10.0 mm, the work function of the metal is close to:

(a) 0.8 eV

(b) 1.6 eV

(c) 1.8 eV

(d) 1.1 eV

When an electron moves in a circular path,

**Most Important 100 Maths Solved Questions for JEE Main 2019**

If a semiconductor has an intrinsic carrier concentration of 1.41 × 10^{16}/m^{3}, when doped with 10^{21}/m^{3} phosphorous atoms, then the concentration of holes/m^{3} at room temperature will be

(a) 2 × 10^{21 }

(b) 2 × 10^{11 }

(c) 1.41 × 10^{10 }

(d) 1.41 × 10^{16}

Doping will increase the number of electrons onl*y *and not the holes. So, number of holes will be equal to number of intrinsic carrier concentration i.e., 1.41 × 10^{16}/m^{3}.

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**Most Important 100 Physics Solved Questions for JEE Main 2019**