Engineering aspirants have already completed the JEE Main Syllabus 2019 of Physics, Chemistry and Mathematics. Now, it time to solve JEE Main Previous years’ papers to get an idea about the paper format and difficulty level of the questions. Students must try solve the question papers of last 5 years so that they aware with the most important topics from which the questions are being asked in the exams. Here, students will get JEE Main Physics Question Paper 2018 along with the solution done by experts. Appearing students can also get the complete chapter wise analysis of the Physics paper in this article. Students can download the compete paper in PDF format with the help of the link available at the end of this article.
JEE Main Physics Paper 2018: Chapter Wise Analysis
Chapters/Topics |
Number of Questions |
Error and Significant Figure |
1 |
1 dimensional Motion |
1 |
Work Power and Energy |
1 |
Momentum Collision |
2 |
Centre of Mass and Moment of Inertia |
1 |
Rotational Motion |
1 |
Gravitation |
1 |
Simple Harmonic motion |
1 |
Elasticity |
1 |
Heat |
2 |
Wave and Sound |
4 |
Current Electricity |
2 |
Electrostatics |
2 |
Capacitor |
2 |
Electromagnetism |
2 |
Geometrical Optics |
1 |
Wave Optics |
1 |
Modern Physics |
4 |
Total |
30 |
Few sample questions are given below:
Question:
The density of a material in the shape of a cube is determined by measuring three sides of the cube and its mass. If the relative errors in measuring the mass and length are respectively 1.5% and 1%, the maximum error in determining the density is:-
(a) 3.5 %
(b) 4.5 %
(c) 6 %
(d) 2.5 %
Solution:
JEE Main 2019: Most Important Physics Solved Question
Question:
A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of 10^{12}/sec. What is the force constant of the bonds connecting one atom with the other ? (Mole wt. of silver =108 and Avagadro number = 6.02 x 10^{23} gm mole^{-1})
(a) 7.1 N/m
(b) 2.2 N/m
(c) 5.5 N/m
(d) 6.4 N/m
Solution:
Question:
A telephonic communication service is working at carrier frequency of 10 GHz. Only 10% of it is utilized for transmission. How many telephonic channels can be transmitted simultaneously if each channel requires a bandwidth of 5 kHz ?
(a) 2 × 10^{4}
(b) 2 × 10^{5}
(c) 2 × 10^{6}
(d) 2 × 10^{3}
Solution:
We are given that the carrier frequency is distributed as band width frequency, so
∴ 10% of 10 GHz = n × 5 kHz , where n = no of channels
⇒ n = 2 × 10^{5} telephonic channels
Hence, the correct option is (b).
Question:
Unpolarized light of intensity I passes through an ideal polarizer A. Another identical polarizer B is placed behind A. The intensity of light 1 beyond B is found to be 1/2. Now another identical polarizer C is placed between A and B. The intensity beyond B is now found to be I/8. The angle between polarizer A and C is :
(a) 30°
(b) 45°
(c) 60°
(d) 0°
Solution:
JEE Main 2017: Question Paper and Detailed Solution
Question:
If the series limit frequency of the Lyman series is v_{L}, then the series limit frequency of the P fund series is :
(a) 16 v_{L }
(b) v_{L }/16
(c) v_{L} / 25
(d) 25 v_{L }
Solution:
To download the compete paper in PDF format, students can click on the below link:
JEE Main Physics Solved Paper 2018
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