In this article, students will get JEE Main Physics Solved Question Paper 2018. There are 30 objective type questions with only one correct option. For each correct response one mark is awarded, while 1/4 mark will be deducted for each incorrect answer. Also, no marks will be awarded for unattempted questions.

The best way to get the idea about the examination pattern is to study the previous year papers. This paper will be helpful for those aspirants who are going to appear for the coming engineering entrance exams. Generally questions in JEE Main are not repeated but the concept on which questions are being asked keeps on repeating year after year. The students must go through the complete paper in order to understand the examination pattern and the level of questions asked in the examination.

**JEE Advanced 2018 mock tests 1,2,3,4, 5 and 6 are available now**

**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:**

**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:**

**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:**

- JoSAA Counselling 2018: Institute wise seat allotment and Opening & Closing Rank
- JoSAA Seat Allotment 2018: Round 4 Result released on josaa.nic.in
- JoSAA Counselling 2018 restarted, check revised counselling schedule
- UPSEE Seat Allotment 2018: Round 2 Result declared on upsee.nic.in
- Important topics in Mathematics for JEE Main Examination 2018