JEE has been divided into two exams: JEE Main and JEE Advanced. JEE Main 2018 is the preliminary stage of JEE, which is scheduled to be held on April 8, 2018 in offline mode and JEE Main 2018 computer based test will be conducted on April 15 and 16, 2018. Those who will qualify JEE Main 2018 will be eligible to appear for JEE Advanced. Students who qualify both exams will be short listed for admissions to engineering courses at IITs, NITs, and CFTIs.
Above is the very brief information about the JEE Main and JEE advanced. Once the candidates have submitted the application forms, it becomes very challenging to make a right study plan to crack these exams.
Most of the aspirants preparing for JEE Main and JEE Advanced are fresh candidates which means they are going to appear in the exam right after their board exams.
The other category is of repeaters who are going to take the JEE Main test for the second time to improve their rank. Such candidates would have been preparing for this entrance exam by taking various coaching classes or simply doing self-studies.
Let us discuss first about the JEE Main, the preliminary test which is scheduled to be held on April 8, 2018. The syllabus of the JEE Main exam is based on the class 11th and class 12th NCERT books which is a favorable factor for the fresher students. But repeaters can also take a lead by revising their whole year’s learning in a proper way.
All the aspirants need to keep a track of their preparedness by practicing more and more questions based on each important topic for JEE Main test.
Here in this article we are presenting the 5 most important questions from the topic Crystal Lattices and Unit Cells. About 1-2 questions have always been asked from this topic in the examination.
Every day you will get the entirely FRESH 5 Questions for JEE Main.
9th Feb, 2018: 5 Important Questions of Crystal Lattices and Unit Cells for JEE Main 2018
Q 1. Silver crystallises in FCC lattice. If edge length of the cell is 4.07 × 10−8 cm and density is 10.5 g cm−3, then the atomic mass of silver is
Options:
(a) 110.15
(b) 107.15
(c) 117.15
(d) 120.15
Ans. Option (b) is correct
Explanation:
a = 4.07 × 10−8 cm
d = 10.5 g cm−3
Lattice is fcc type, z = 4
NA = 6.022 × 1023 mol−1
Hence, atomic mass of silver = 107.15 u
Q 3. A cubic solid is made of two elements P and Q. Atoms of Q are at the corners of the cube and P at the body-centre. What is the formula of the compound?
Options:
(a) PQ
(b) P2Q3
(c) PQ2
(d) P2Q
Ans. Option (b) is correct
Explanation:
Atoms of Q are present at the corners of the cube.
Q 4. Gold (atomic radius = 0.144 nm) crystallises in a face-centred unit cell. What is the length of a side of the cell?
Options:
(a) 0.107 nm
(b) 0.207 nm
(c) 0.307 nm
(d) 0.407 nm
Ans. Option (d) is correct
Explanation:
Q 5. The total number of tetrahedral voids in the face centred unit cell is ………..
Options:
(a) 6
(b) 8
(c) 10
(d) 12
Ans. Option (b) is correct
Explanation:
Fcc unit cell contains 8 tetrahedral voids at centre of each 8 smaller cube of an unit cell as shown below:
Number of atoms in fcc unit cell = 4
Number of octahedral voids = 4
Number of tetrahedral voids = 8
Above questions, answers and solutions are 100% unique and easy to understand.
Keep visiting this article daily to get the other 5 QUESTIONS on the next topic.
Students can click on the link given below to download the PDF of the important questions from Feb 2 to Feb 8, 2018.