# JEE Main Solved Physics Practice Paper 2017 - 2018 Set - VI

Find JEE Main Solved Physics Practice Paper based on the latest syllabus in this article. It will help you to track your progress of coming JEE Main exam.

Mar 19, 2018 17:46 IST
JEE Main Solved Physics Practice Paper 2017 - 2018 Set - VI

In this article, JEE aspirants will find fully solved practice paper based on the latest pattern and syllabus for JEE Main 2018.

It will also help students to get familiar with the difficulty level of upcoming JEE Main 2018 exam. After attempting this practice paper students can through with the detailed solutions of all the questions provided here and revise their concepts.

1. There are 30 multiple choice questions out of which only one correct option is correct.

2. Questions have been taken from different chapters 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.

3. This paper is prepared by experienced Subject Experts of Physics.

Few sample questions from the solved practice paper are given below:

Question:

A box of mass 8 kg is placed on a rough inclined plane of inclination. Its downward motion can be prevented by applying an upward pull F and it can be made to slide upwards by applying a force 2 F. The coefficient of friction between the box and the inclined plane is

Question:

Two short bar magnets of length 1 cm each have magnetic moments 1.20 Am2 and 1.00 Am2 respectively. They are placed on a horizontal table parallel to each other with their N poles pointing towards the south. They have a common magnetic equator and are separated by a distance of 20.0 cm. The value of the resultant horizontal magnetic induction at the midpoint O of the line joining their centres is close to (Horizontal component of earth's magnetic induction is 3.6 × 10−5 Wb/m2)

(a) 3.6 × 10−5 Wb/m2

(b) 2.56 × 10−4 Wb/m2

(c) 3.50 × 10−4 Wb/m2

(d) 5.80 × 10−4 Wb/m2

Sol. (b)

Question:

A green light is incident from the water to the air-water interface at the critical angle (q). Select the correct statement.

(a) The spectrum of visible light whose frequency is less than that of green light will come out to the air medium.

(b) The spectrum of visible light whose frequency is more than that of green light will come out to the air medium.

(c) The entire spectrum of visible light will come out of the water at various angles to the normal.

(d) The entire spectrum of visible light will come out of the water at an angle of 90° to the normal.

Sol. (c)

For total internal reflection angle of incidence must be greater than critical angle. If frequency is less then wavelength will be larger and hence the refractive index will be less and therefore critical angle increases.

So, they do not suffer reflection and come out at an angle less than.

Question:

The electric field at a point is

(a) Always continuous

(b) Continuous if there is no charge at that point

(c) Discontinuous only if there is a negative charge at that point

(d) Continuous if there is a charge at that point

Sol. (b)

The electric field due to a charge Q at a point in space may be defined as the force that a unit positive charge would experience if placed at that point. Thus, electric field due to the charge Q will be continuous, if there is no charge at that point. It will be discontinuous if there is a charge at that point.

Question: An astronomical refractive telescope has an objective of focal length 20m and an eyepiece of focal length 2 cm

(a) The length of the telescope tube is 20 m

(b) The magnification is 100

(c) The image formed is inverted

(d) An objective of a larger aperture will increase the brightness and reduce chromatic aberration of the image

Sol. (c)

The length of the telescope tube is fo + fe = 20 + (0.02) = 20.02 m

Also, m = 20/0.02 = 1000

Also, the image formed is inverted.

Conclusion:

After doing a lot of research on previous year papers of JEE Main, the subject experts of Physics have designed this solved practice paper. It will help you to practice the new set of questions. It will also help students to manage the speed and accuracy for upcoming JEE Main 2018.