After doing a detailed analysis of JEE Main Previous years’ papers, Physics experts have collected some 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. A total of 90 questions are asked from Physics, Chemistry and Mathematics in JEE Main exam. Therefore, every subject is important for students who are going to appear for JEE Main 2019. Students cannot clear JEE Main 2019 without studying all three subjects. Physics is an application based subject. Students cannot solve the JEE level questions without clearing the basic concepts. If students prepare well, then they can definitely score very good marks in Physics section. The level of difficulty of Physics is not different than the other two subjects.
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
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 2R?
In circular orbit of a satellite, potential energy = −2 × Kinetic Energy = −mV2. So, just to escape from the gravitational pull, its total mechanical energy should be zero. Therefore its kinetic energy should be mV2.
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,
If a semiconductor has an intrinsic carrier concentration of 1.41 × 1016/m3, when doped with 1021/m3 phosphorous atoms, then the concentration of holes/m3 at room temperature will be
(a) 2 × 1021
(b) 2 × 1011
(c) 1.41 × 1010
(d) 1.41 × 1016
Doping will increase the number of electrons only and not the holes. So, number of holes will be equal to number of intrinsic carrier concentration i.e., 1.41 × 1016/m3.
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