Open App

JEE Exam > JEE Notes > Chemistry for JEE Main & Advanced > JEE Advanced Previous Year Questions (2021-2023): Solid State

JEE Advanced Previous Year Questions (2021-2023): Solid State | Chemistry for JEE Main & Advanced PDF Download

Download, print and study this document offline

Please wait while the PDF view is loading

 Page 1


JEE Advanced Previous Year 
Questions (2021-2023): Solid State 
2023 
Multiple Correct Type 
JEE Advanced 2023 Paper 2 Online 
Q3. Atoms of metals ?? , ?? , and ?? form face-centred cubic ( ?????? ) unit cell of edge 
length ?? ?? , body-centred cubic (bcc) unit cell of edge length ?? ?? , and simple cubic 
unit cell of edge length ?? ?? , respectively. If ?? ?? =
v?? ?? ?? ?? ; ?? ?? =
?? v?? ?? ?? ; ?? ?? =
?? ?? ?? ?? and 
?? ?? = ?? ?? ?? , then the correct statement(s) is(are): 
[Given: ?? ?? , ?? ?? , and ?? ?? are molar masses of metals ?? , ?? , and ?? , respectively. ?? ?? , ?? ?? , 
and ?? ?? are atomic radii of metals ?? , ?? , and ?? , respectively.] 
 
A. Packing efficiency of unit cell of ?? > Packing efficiency of unit cell of ?? > 
Packing efficiency of unit cell of ?? 
 
B. ?? ?? > ?? ?? 
C. ?? ?? > ?? ?? 
D. Density of ?? > Density of ?? 
Ans: (A,B,D) 
Atomic radii relations : 
Given, 
?? ?? =
8
v3
?? ?? . … (1)
?? ?? =
v3
2
?? ?? . … (2)
 
Substituting Eq. (1) into Eq. (2), we get : 
?? ?? =
v3
2
×
8
v3
?? ?? = 4?? ?? … . (3) 
Edge lengths of the unit cells: 
The edge lengths (?? ) of the unit cells for ?? , ?? , and ?? are given by : 
?? ?? = 2v2?? ?? … (4) 
(for FCC) 
Page 2


JEE Advanced Previous Year 
Questions (2021-2023): Solid State 
2023 
Multiple Correct Type 
JEE Advanced 2023 Paper 2 Online 
Q3. Atoms of metals ?? , ?? , and ?? form face-centred cubic ( ?????? ) unit cell of edge 
length ?? ?? , body-centred cubic (bcc) unit cell of edge length ?? ?? , and simple cubic 
unit cell of edge length ?? ?? , respectively. If ?? ?? =
v?? ?? ?? ?? ; ?? ?? =
?? v?? ?? ?? ; ?? ?? =
?? ?? ?? ?? and 
?? ?? = ?? ?? ?? , then the correct statement(s) is(are): 
[Given: ?? ?? , ?? ?? , and ?? ?? are molar masses of metals ?? , ?? , and ?? , respectively. ?? ?? , ?? ?? , 
and ?? ?? are atomic radii of metals ?? , ?? , and ?? , respectively.] 
 
A. Packing efficiency of unit cell of ?? > Packing efficiency of unit cell of ?? > 
Packing efficiency of unit cell of ?? 
 
B. ?? ?? > ?? ?? 
C. ?? ?? > ?? ?? 
D. Density of ?? > Density of ?? 
Ans: (A,B,D) 
Atomic radii relations : 
Given, 
?? ?? =
8
v3
?? ?? . … (1)
?? ?? =
v3
2
?? ?? . … (2)
 
Substituting Eq. (1) into Eq. (2), we get : 
?? ?? =
v3
2
×
8
v3
?? ?? = 4?? ?? … . (3) 
Edge lengths of the unit cells: 
The edge lengths (?? ) of the unit cells for ?? , ?? , and ?? are given by : 
?? ?? = 2v2?? ?? … (4) 
(for FCC) 
?? ?? = 4?? ?? =
32
v3
?? ?? … . (5) (for BCC, using Eq. (1)) 
?? ?? = 2?? ?? = 8?? ?? … . (6)  (for simple cubic, using Eq. (3)) 
 
Comparing the edge lengths : 
From Eq. (4), (5) and (6), we find : 
?? ?? > ?? ?? > ?? ?? 
This means Option B: ?? ?? > ?? ?? is correct. 
 
Packing Efficiencies: 
The packing efficiencies for FCC, BCC, and simple cubic are calculated as follows: 
? FCC : 
?? ?? ?????? =
4 ×
4
3
?? ?? ?? 3
(?? ?? )
3
=
?? v2
 
? BCC: 
?? ?? ?????? =
2 ×
4
3
?? ?? ?? 3
(?? ?? )
3
=
3?? v3
32
 
? Simple Cubic : 
?? ?? ????
=
1 ×
4
3
?? ?? ?? 3
(?? ?? )
3
=
?? 6
 
Comparing these packing efficiencies, we find: 
?? ?? ?????? > ?? ?? ?????? > ?? ?? ????
 
This means Option A : Packing efficiency of unit cell of ?? > ?? > ?? is correct. 
5 Calculating Densities : 
The densities of x and y are given by the formula ?? =
?? ·?? ?? ?? ·?? 3
, where ?? is the number of 
atoms in the unit cell, ?? is the molar mass of the metal, ?? ?? is Avogadro's number, and ?? 
is the edge length of the unit cell. 
? For ?? (FCC), ?? = 4 : 
?? ?? =
4?? ?? ?? ?? (?? ?? )
3
 
? For y (BCC), ?? = 2 : 
Page 3


JEE Advanced Previous Year 
Questions (2021-2023): Solid State 
2023 
Multiple Correct Type 
JEE Advanced 2023 Paper 2 Online 
Q3. Atoms of metals ?? , ?? , and ?? form face-centred cubic ( ?????? ) unit cell of edge 
length ?? ?? , body-centred cubic (bcc) unit cell of edge length ?? ?? , and simple cubic 
unit cell of edge length ?? ?? , respectively. If ?? ?? =
v?? ?? ?? ?? ; ?? ?? =
?? v?? ?? ?? ; ?? ?? =
?? ?? ?? ?? and 
?? ?? = ?? ?? ?? , then the correct statement(s) is(are): 
[Given: ?? ?? , ?? ?? , and ?? ?? are molar masses of metals ?? , ?? , and ?? , respectively. ?? ?? , ?? ?? , 
and ?? ?? are atomic radii of metals ?? , ?? , and ?? , respectively.] 
 
A. Packing efficiency of unit cell of ?? > Packing efficiency of unit cell of ?? > 
Packing efficiency of unit cell of ?? 
 
B. ?? ?? > ?? ?? 
C. ?? ?? > ?? ?? 
D. Density of ?? > Density of ?? 
Ans: (A,B,D) 
Atomic radii relations : 
Given, 
?? ?? =
8
v3
?? ?? . … (1)
?? ?? =
v3
2
?? ?? . … (2)
 
Substituting Eq. (1) into Eq. (2), we get : 
?? ?? =
v3
2
×
8
v3
?? ?? = 4?? ?? … . (3) 
Edge lengths of the unit cells: 
The edge lengths (?? ) of the unit cells for ?? , ?? , and ?? are given by : 
?? ?? = 2v2?? ?? … (4) 
(for FCC) 
?? ?? = 4?? ?? =
32
v3
?? ?? … . (5) (for BCC, using Eq. (1)) 
?? ?? = 2?? ?? = 8?? ?? … . (6)  (for simple cubic, using Eq. (3)) 
 
Comparing the edge lengths : 
From Eq. (4), (5) and (6), we find : 
?? ?? > ?? ?? > ?? ?? 
This means Option B: ?? ?? > ?? ?? is correct. 
 
Packing Efficiencies: 
The packing efficiencies for FCC, BCC, and simple cubic are calculated as follows: 
? FCC : 
?? ?? ?????? =
4 ×
4
3
?? ?? ?? 3
(?? ?? )
3
=
?? v2
 
? BCC: 
?? ?? ?????? =
2 ×
4
3
?? ?? ?? 3
(?? ?? )
3
=
3?? v3
32
 
? Simple Cubic : 
?? ?? ????
=
1 ×
4
3
?? ?? ?? 3
(?? ?? )
3
=
?? 6
 
Comparing these packing efficiencies, we find: 
?? ?? ?????? > ?? ?? ?????? > ?? ?? ????
 
This means Option A : Packing efficiency of unit cell of ?? > ?? > ?? is correct. 
5 Calculating Densities : 
The densities of x and y are given by the formula ?? =
?? ·?? ?? ?? ·?? 3
, where ?? is the number of 
atoms in the unit cell, ?? is the molar mass of the metal, ?? ?? is Avogadro's number, and ?? 
is the edge length of the unit cell. 
? For ?? (FCC), ?? = 4 : 
?? ?? =
4?? ?? ?? ?? (?? ?? )
3
 
? For y (BCC), ?? = 2 : 
?? ?? =
2?? ?? ?? ?? (?? ?? )
3
 
?? ?? =
3
2
?? ?? and ?? ?? = 3?? ?? . This implies that ?? ?? =
?? ?? 2
. Thus, 2?? ?? = ?? ?? , we can write : 
?? ?? ?? ?? =
2?? ?? ?? ?? × (
?? ?? ?? ?? )
3
 
Substituting the given relation 2?? ?? = ?? ?? into Eq. (7), we get : 
?? ?? ?? ?? = 2 × 1 × (
32/3
2v2
)
3
. 
After simplifying Eq. (8), we find ?? ?? > ?? ?? . 
This means Option D : Density of ?? > ?? is correct. 
So, the correct answers are Options A, B, and D. 
 
 
 
 
 
 
 
 
 
 
Page 4


JEE Advanced Previous Year 
Questions (2021-2023): Solid State 
2023 
Multiple Correct Type 
JEE Advanced 2023 Paper 2 Online 
Q3. Atoms of metals ?? , ?? , and ?? form face-centred cubic ( ?????? ) unit cell of edge 
length ?? ?? , body-centred cubic (bcc) unit cell of edge length ?? ?? , and simple cubic 
unit cell of edge length ?? ?? , respectively. If ?? ?? =
v?? ?? ?? ?? ; ?? ?? =
?? v?? ?? ?? ; ?? ?? =
?? ?? ?? ?? and 
?? ?? = ?? ?? ?? , then the correct statement(s) is(are): 
[Given: ?? ?? , ?? ?? , and ?? ?? are molar masses of metals ?? , ?? , and ?? , respectively. ?? ?? , ?? ?? , 
and ?? ?? are atomic radii of metals ?? , ?? , and ?? , respectively.] 
 
A. Packing efficiency of unit cell of ?? > Packing efficiency of unit cell of ?? > 
Packing efficiency of unit cell of ?? 
 
B. ?? ?? > ?? ?? 
C. ?? ?? > ?? ?? 
D. Density of ?? > Density of ?? 
Ans: (A,B,D) 
Atomic radii relations : 
Given, 
?? ?? =
8
v3
?? ?? . … (1)
?? ?? =
v3
2
?? ?? . … (2)
 
Substituting Eq. (1) into Eq. (2), we get : 
?? ?? =
v3
2
×
8
v3
?? ?? = 4?? ?? … . (3) 
Edge lengths of the unit cells: 
The edge lengths (?? ) of the unit cells for ?? , ?? , and ?? are given by : 
?? ?? = 2v2?? ?? … (4) 
(for FCC) 
?? ?? = 4?? ?? =
32
v3
?? ?? … . (5) (for BCC, using Eq. (1)) 
?? ?? = 2?? ?? = 8?? ?? … . (6)  (for simple cubic, using Eq. (3)) 
 
Comparing the edge lengths : 
From Eq. (4), (5) and (6), we find : 
?? ?? > ?? ?? > ?? ?? 
This means Option B: ?? ?? > ?? ?? is correct. 
 
Packing Efficiencies: 
The packing efficiencies for FCC, BCC, and simple cubic are calculated as follows: 
? FCC : 
?? ?? ?????? =
4 ×
4
3
?? ?? ?? 3
(?? ?? )
3
=
?? v2
 
? BCC: 
?? ?? ?????? =
2 ×
4
3
?? ?? ?? 3
(?? ?? )
3
=
3?? v3
32
 
? Simple Cubic : 
?? ?? ????
=
1 ×
4
3
?? ?? ?? 3
(?? ?? )
3
=
?? 6
 
Comparing these packing efficiencies, we find: 
?? ?? ?????? > ?? ?? ?????? > ?? ?? ????
 
This means Option A : Packing efficiency of unit cell of ?? > ?? > ?? is correct. 
5 Calculating Densities : 
The densities of x and y are given by the formula ?? =
?? ·?? ?? ?? ·?? 3
, where ?? is the number of 
atoms in the unit cell, ?? is the molar mass of the metal, ?? ?? is Avogadro's number, and ?? 
is the edge length of the unit cell. 
? For ?? (FCC), ?? = 4 : 
?? ?? =
4?? ?? ?? ?? (?? ?? )
3
 
? For y (BCC), ?? = 2 : 
?? ?? =
2?? ?? ?? ?? (?? ?? )
3
 
?? ?? =
3
2
?? ?? and ?? ?? = 3?? ?? . This implies that ?? ?? =
?? ?? 2
. Thus, 2?? ?? = ?? ?? , we can write : 
?? ?? ?? ?? =
2?? ?? ?? ?? × (
?? ?? ?? ?? )
3
 
Substituting the given relation 2?? ?? = ?? ?? into Eq. (7), we get : 
?? ?? ?? ?? = 2 × 1 × (
32/3
2v2
)
3
. 
After simplifying Eq. (8), we find ?? ?? > ?? ?? . 
This means Option D : Density of ?? > ?? is correct. 
So, the correct answers are Options A, B, and D. 
 
 
 
 
 
 
 
 
 
 
 
2022 
Single Correct Type 
JEE Advanced 2022 Paper 2 Online 
Q1. Atom X occupies the fcc lattice sites as well as alternate tetrahedral voids of 
the same lattice. The packing efficiency (in %) of the resultant solid is closest to 
A. 25 
B. 35 
C. 55 
D. 75 
Ans: (B) 
Atom ?? occupies FCC lattice sites as well as alternate tetrahedral voids of FCC. 
In FCC, tetrahedral voids are 8 (in a unit cell) 
Hence 
Atom ?? in a unit cell (FCC lattice sites) = 8 ×
1
8
+ 6 ×
1
2
= 4 
Atom ?? in a unit cell (in T.V.) =
1
2
× 8 = 4 
Total atom X in one unit cell = 8 
For relation between a and ?? , since T.V. forms at 
1
4
 th of body diagonal, 
?? v3
4
= 2?? 
? ?? =
8?? v3
 
Packing efficiency 
=
8 ×
4
3
?? ?? 3
?? 3
× 100 =
8 ×
4
3
?? ?? 3
(
8?? v3
)
3
× 100 ? 35% 
 
 
 
Page 5


JEE Advanced Previous Year 
Questions (2021-2023): Solid State 
2023 
Multiple Correct Type 
JEE Advanced 2023 Paper 2 Online 
Q3. Atoms of metals ?? , ?? , and ?? form face-centred cubic ( ?????? ) unit cell of edge 
length ?? ?? , body-centred cubic (bcc) unit cell of edge length ?? ?? , and simple cubic 
unit cell of edge length ?? ?? , respectively. If ?? ?? =
v?? ?? ?? ?? ; ?? ?? =
?? v?? ?? ?? ; ?? ?? =
?? ?? ?? ?? and 
?? ?? = ?? ?? ?? , then the correct statement(s) is(are): 
[Given: ?? ?? , ?? ?? , and ?? ?? are molar masses of metals ?? , ?? , and ?? , respectively. ?? ?? , ?? ?? , 
and ?? ?? are atomic radii of metals ?? , ?? , and ?? , respectively.] 
 
A. Packing efficiency of unit cell of ?? > Packing efficiency of unit cell of ?? > 
Packing efficiency of unit cell of ?? 
 
B. ?? ?? > ?? ?? 
C. ?? ?? > ?? ?? 
D. Density of ?? > Density of ?? 
Ans: (A,B,D) 
Atomic radii relations : 
Given, 
?? ?? =
8
v3
?? ?? . … (1)
?? ?? =
v3
2
?? ?? . … (2)
 
Substituting Eq. (1) into Eq. (2), we get : 
?? ?? =
v3
2
×
8
v3
?? ?? = 4?? ?? … . (3) 
Edge lengths of the unit cells: 
The edge lengths (?? ) of the unit cells for ?? , ?? , and ?? are given by : 
?? ?? = 2v2?? ?? … (4) 
(for FCC) 
?? ?? = 4?? ?? =
32
v3
?? ?? … . (5) (for BCC, using Eq. (1)) 
?? ?? = 2?? ?? = 8?? ?? … . (6)  (for simple cubic, using Eq. (3)) 
 
Comparing the edge lengths : 
From Eq. (4), (5) and (6), we find : 
?? ?? > ?? ?? > ?? ?? 
This means Option B: ?? ?? > ?? ?? is correct. 
 
Packing Efficiencies: 
The packing efficiencies for FCC, BCC, and simple cubic are calculated as follows: 
? FCC : 
?? ?? ?????? =
4 ×
4
3
?? ?? ?? 3
(?? ?? )
3
=
?? v2
 
? BCC: 
?? ?? ?????? =
2 ×
4
3
?? ?? ?? 3
(?? ?? )
3
=
3?? v3
32
 
? Simple Cubic : 
?? ?? ????
=
1 ×
4
3
?? ?? ?? 3
(?? ?? )
3
=
?? 6
 
Comparing these packing efficiencies, we find: 
?? ?? ?????? > ?? ?? ?????? > ?? ?? ????
 
This means Option A : Packing efficiency of unit cell of ?? > ?? > ?? is correct. 
5 Calculating Densities : 
The densities of x and y are given by the formula ?? =
?? ·?? ?? ?? ·?? 3
, where ?? is the number of 
atoms in the unit cell, ?? is the molar mass of the metal, ?? ?? is Avogadro's number, and ?? 
is the edge length of the unit cell. 
? For ?? (FCC), ?? = 4 : 
?? ?? =
4?? ?? ?? ?? (?? ?? )
3
 
? For y (BCC), ?? = 2 : 
?? ?? =
2?? ?? ?? ?? (?? ?? )
3
 
?? ?? =
3
2
?? ?? and ?? ?? = 3?? ?? . This implies that ?? ?? =
?? ?? 2
. Thus, 2?? ?? = ?? ?? , we can write : 
?? ?? ?? ?? =
2?? ?? ?? ?? × (
?? ?? ?? ?? )
3
 
Substituting the given relation 2?? ?? = ?? ?? into Eq. (7), we get : 
?? ?? ?? ?? = 2 × 1 × (
32/3
2v2
)
3
. 
After simplifying Eq. (8), we find ?? ?? > ?? ?? . 
This means Option D : Density of ?? > ?? is correct. 
So, the correct answers are Options A, B, and D. 
 
 
 
 
 
 
 
 
 
 
 
2022 
Single Correct Type 
JEE Advanced 2022 Paper 2 Online 
Q1. Atom X occupies the fcc lattice sites as well as alternate tetrahedral voids of 
the same lattice. The packing efficiency (in %) of the resultant solid is closest to 
A. 25 
B. 35 
C. 55 
D. 75 
Ans: (B) 
Atom ?? occupies FCC lattice sites as well as alternate tetrahedral voids of FCC. 
In FCC, tetrahedral voids are 8 (in a unit cell) 
Hence 
Atom ?? in a unit cell (FCC lattice sites) = 8 ×
1
8
+ 6 ×
1
2
= 4 
Atom ?? in a unit cell (in T.V.) =
1
2
× 8 = 4 
Total atom X in one unit cell = 8 
For relation between a and ?? , since T.V. forms at 
1
4
 th of body diagonal, 
?? v3
4
= 2?? 
? ?? =
8?? v3
 
Packing efficiency 
=
8 ×
4
3
?? ?? 3
?? 3
× 100 =
8 ×
4
3
?? ?? 3
(
8?? v3
)
3
× 100 ? 35% 
 
 
 
 
2021 
JEE Advanced 2021 Paper 1 Online 
Q2. For the given close packed structure of a salt made of cation ?? and anion ?? 
shown below (ions of only one face are shown for clarity), the packing fraction is 
approximately (packing fraction = 
 packing efficiency 
??????
 ) 
 
A. 0.74 
B. 0.63 
C. 0.52 
D. 0.48 
Ans: (B) 
 
Radius of cation X = r
+
 
Number of cation ?? present = 6 ×
1
2
= 3 
Radius of anion ?? = ?? -

	Chemistry for JEE Main & Advanced 352 videos\|596 docs\|309 tests

Chemistry for JEE Main & Advanced

352 videos|596 docs|309 tests

Join Course for Free

Top Courses for JEE

View all

Related Exams

JEE

About this Document

Oct 24, 2024 Last updated

Document Description: JEE Advanced Previous Year Questions (2021-2023): Solid State for JEE 2024 is part of Chemistry for JEE Main & Advanced preparation. The notes and questions for JEE Advanced Previous Year Questions (2021-2023): Solid State have been prepared according to the JEE exam syllabus. Information about JEE Advanced Previous Year Questions (2021-2023): Solid State covers topics like and JEE Advanced Previous Year Questions (2021-2023): Solid State Example, for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for JEE Advanced Previous Year Questions (2021-2023): Solid State.

Introduction of JEE Advanced Previous Year Questions (2021-2023): Solid State in English is available as part of our Chemistry for JEE Main & Advanced for JEE & JEE Advanced Previous Year Questions (2021-2023): Solid State in Hindi for Chemistry for JEE Main & Advanced course. Download more important topics related with notes, lectures and mock test series for JEE Exam by signing up for free. JEE: JEE Advanced Previous Year Questions (2021-2023): Solid State | Chemistry for JEE Main & Advanced

Description

Full syllabus notes, lecture & questions for JEE Advanced Previous Year Questions (2021-2023): Solid State | Chemistry for JEE Main & Advanced - JEE | Plus excerises question with solution to help you revise complete syllabus for Chemistry for JEE Main & Advanced | Best notes, free PDF download

Information about JEE Advanced Previous Year Questions (2021-2023): Solid State

In this doc you can find the meaning of JEE Advanced Previous Year Questions (2021-2023): Solid State defined & explained in the simplest way possible. Besides explaining types of JEE Advanced Previous Year Questions (2021-2023): Solid State theory, EduRev gives you an ample number of questions to practice JEE Advanced Previous Year Questions (2021-2023): Solid State tests, examples and also practice JEE tests

	Chemistry for JEE Main & Advanced 352 videos\|596 docs\|309 tests

Chemistry for JEE Main & Advanced

352 videos|596 docs|309 tests

Join Course for Free

Download as PDF

Explore Courses for JEE exam

Top Courses for JEE

Explore Courses

Signup for Free!

Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.

Start learning for Free

10M+ students study on EduRev

Sample Paper

study material

Extra Questions

Viva Questions

Important questions

Summary

Exam

Previous Year Questions with Solutions

video lectures

shortcuts and tricks

ppt

practice quizzes

pdf

JEE Advanced Previous Year Questions (2021-2023): Solid State | Chemistry for JEE Main & Advanced

Semester Notes

JEE Advanced Previous Year Questions (2021-2023): Solid State | Chemistry for JEE Main & Advanced

past year papers

JEE Advanced Previous Year Questions (2021-2023): Solid State | Chemistry for JEE Main & Advanced

Objective type Questions

Free

MCQs

mock tests for examination

;

Additional Information about JEE Advanced Previous Year Questions (2021-2023): Solid State for JEE Preparation

JEE Advanced Previous Year Questions (2021-2023): Solid State Free PDF Download

The JEE Advanced Previous Year Questions (2021-2023): Solid State is an invaluable resource that delves deep into the core of the JEE exam. These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective. With the help of these notes, you can grasp complex subjects quickly, revise important points easily, and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner, allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material, or toppers' notes, this PDF has got you covered. Download the JEE Advanced Previous Year Questions (2021-2023): Solid State now and kickstart your journey towards success in the JEE exam.

Importance of JEE Advanced Previous Year Questions (2021-2023): Solid State

The importance of JEE Advanced Previous Year Questions (2021-2023): Solid State cannot be overstated, especially for JEE aspirants. This document holds the key to success in the JEE exam. It offers a detailed understanding of the concept, providing invaluable insights into the topic. By knowing the concepts well in advance, students can plan their preparation effectively. Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.

JEE Advanced Previous Year Questions (2021-2023): Solid State Notes

JEE Advanced Previous Year Questions (2021-2023): Solid State Notes offer in-depth insights into the specific topic to help you master it with ease. This comprehensive document covers all aspects related to JEE Advanced Previous Year Questions (2021-2023): Solid State. It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation. Practice papers and question papers enable you to assess your progress effectively. Additionally, the paper analysis provides valuable tips for tackling the exam strategically. Access to Toppers' notes gives you an edge in understanding complex concepts. Whether you're a beginner or aiming for advanced proficiency, JEE Advanced Previous Year Questions (2021-2023): Solid State Notes on EduRev are your ultimate resource for success.

JEE Advanced Previous Year Questions (2021-2023): Solid State JEE

The "JEE Advanced Previous Year Questions (2021-2023): Solid State JEE Questions" guide is a valuable resource for all aspiring students preparing for the JEE exam. It focuses on providing a wide range of practice questions to help students gauge their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation. The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level. Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.

Study JEE Advanced Previous Year Questions (2021-2023): Solid State on the App

Students of JEE can study JEE Advanced Previous Year Questions (2021-2023): Solid State alongwith tests & analysis from the EduRev app, which will help them while preparing for their exam. Apart from the JEE Advanced Previous Year Questions (2021-2023): Solid State, students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc. The EduRev App will make your learning easier as you can access it from anywhere you want. The content of JEE Advanced Previous Year Questions (2021-2023): Solid State is prepared as per the latest JEE syllabus.

Education Revolution