In this article, engineering section of Jagranjosh is providing you JEE Main Solved Physics Practice Paper 2018. This paper consists of 30 multiple choice questions with only one correct option.
The questions have been taken from different topics like Units and Measurements, Motion in a Straight Line, Laws of Motion, Rotational Motion, Gravitation, Thermodynamics, Kinetic Theory of gases, Waves, Electric Charges and Fields, Electrostatic Potential and Capacitance, Current Electricity, Moving Charges and Magnetism, Magnetism and Matter, Electromagnetic Induction & Alternating Current, Electromagnetic Induction, Alternating Current, Electromagnetic Waves, Dual Nature of Radiation and Matter, Atoms & Nuclei, Semiconductor Electronics, Communication Systems.
The solution of each question is explained by Subject Experts of Physics in a very detailed manner.
Few sample questions from the Solved Physics Practice Paper are given below:
A hemisphere is uniformly charged positively. The electric field at a point on a diameter away from the centre is directed
(a) Perpendicular to the diameter
(c) Parallel to the diameter
(c) At an angle tilted towards the diameter
(c) At an angle tilted away from the diameter
When the point is situated at a point on diameter away from the centre of hemisphere charged uniformly positively, the electric field is perpendicular to the diameter. The component of electric intensity parallel to the diameter cancels out.
Consider a beam of electrons (each electron with energy E0) incident on a metal surface kept in an evacuated chamber. Then,
(a) No electrons will be emitted as only photons can emit electrons
(b) Electrons can be emitted but all with an energy, E0
(c) Electrons can be emitted with any energy, with a maximum of E0 − f (f is the work function)
(d) Electrons can be emitted with any energy, with a maximum of E0
When a beam of electrons of energy E0 is incident on a metal surface kept in an evacuated chamber electrons can be emitted with maximum energy E0 (due to elastic collision) and with any energy less than E0, when part of incident energy of electron is used in liberating the electrons from the surface of metal.
Two eelis of emfs approximately 5V and 10 V are to be accurately compared using a potentiometer of length 400 cm.
(a)The battery that runs the potentiometer should have võitage of 8V
(b)The battery of potentiometer can have a votage of 15 V and R adjusted so that the potential drop across the wire slightly exceeds 10 V
(c)The first portion of 50 cm of wire itself should have a potential drop of 10 V
(d) Potentiometer is usually used for comparing resistances and not voltages
In a potentiometer experiment, the emf of a cells can be measured, if the potential drop along the potentiometer wire is more than the emf of the cells to be determined. Here, values of emfs of two cells are given as 5V and 10V, therefore, the potential drop along the potentiometer wire must be more than 10V.
A male voice after modulation-transmission sounds like that of a female to the receiver. The problem is due to
(a) Poor selection of modulation index (selected 0 < m < 1)
(b) Poor bandwidth selection of amplifiers
(c) Poor selection of carrier frequency
(d) Loss of energy in transmission.
Here, in this question, the frequency of modulated signal received becomes more, which is possible with the poor bandwidth selection of amplifiers. This happens because bandwidth in amplitude modulation is equal to twice the frequency of modulating signal. But, the frequency of male voice is less than that of a female.
The conductivity of a semiconductor increases with increase in temperature, because
(a) Number density of free current carries increases
(b) Relaxation time increases
(c) Both number density of carries and relaxation time increase
(d) Number density of carries increases, relaxation time decreases but effect of decrease in relaxation time is much less than increase in number density
The conductivity of a semiconductor increases with increase in temperature, because the number density of current carries increases, relaxation time decreases but effect of decrease in relaxation is much less than increases in number density.
You are given four sources of light each one providing a light of a single colour - red, blue, green and yellow. Suppose the angle of refraction for a beam of yellow Light corresponding to a particular angle of incidence at the interface of two media is 90°. Which of the following statements is correct if the source of yellow light is replaced with that of other lights without changing the angle of incidence?
(a) The beam of red light would undergo total internal reflection
(b) The beam of red light would bend towards normal while it gets refracted through the second medium
(c) The beam of blue light would undergo total internal reflection
(d) The beam of green light would bend away from the normal as it gets refracted through the second medium
According to VIBGYOR, among all given sources of light, the blue light have smallest wavelength. According to Cauchy relationship, smaller the wavelength higher the refractive index and consequently smaller the critical angle.
So, corresponding to blue colour, the critical angle is least which facilitates total internal reflection for the beam of blue light. The beam of green light would also undergo total internal reflection.
This solved practice paper will help the students to understand the pattern and difficulty level of the upcoming JEE Main 2018. It will also help students in their final level of preparation. Detailed solution of each question has been provided to save precious time of students.