jee advanced 2020 exam date
JEE Advanced 2020 Exam Date, Syllabus, Information Brochure (Released)

JEE Advanced 2020 will be organised by IIT Delhi this year. The Indian Institute of Technology, (IIT) Delhi has released the information brochure for JEE Advanced 2020 on its official website. JEE Advanced 2020 Exam Date has been announced. The Exam Will be held in 2 Sessions, Paper 1 and Paper 2.

JEE Advanced 2020 Exam Date: 17 May 2020 (Announced)

  • Session 1: Paper 1 – 9 am to 12 pm
  • Session 2: Paper 2 – 2 pm to 5 pm

As per the Released Information on Website, The Online Registration for JEE Advanced Exam will held from May 1 to May 6, 2020. JEE Advanced is organised by one of seven Zonal Coordinating IIT’s guided by the Joint Admission Board (JAB). It is a National Level Engineering Entrance Exam for admission in the Bachelors, Integrated Masters and Dual Degree Programs. The exam will be conducted in online mode. Candidate needs to clear the JEE Main Exam to be eligible for the JEE Advanced Examination.

IIT will release the admit card for the JEE Advanced 2020 on May 12, 2020, on its official website. It will be available to download until May 17, 2020.

 

Eligibility Criteria

Candidates must fulfill one of the following two criteria in order to be eligible for admission at IIT:

  1. Candidates must have secured at least 75% aggregate marks in the Class XII (or an equivalent) Board examination. The aggregate marks for SC, ST and PwD candidates should be at least 65%. Physics, Chemistry, and Mathematics are required as compulsory subjects in Class XII (or equivalent) Board examination in 2019 or 2020.
  2. Candidates must be within the category-wise top 20 percentile of successful candidates in their respective Class XII (or equivalent) board examination in 2019 or 2020 with Physics, Chemistry, and Mathematics as compulsory subjects.

 

Online Crash Course for JEE Advanced 2020 Preparation : CLICK HERE FOR DETAILS

Information Bulletin (Brochure) for JEE Advanced 2020

JEE Advanced 2020 Information Bulletin available now : Download PDF [Online Registration, Examination and Admission Details].

JEE Advanced Exam Pattern

The details regarding the JEE Advanced Exam pattern is given below:

  • Online Mode: From 2020, the exam is conducted via online mode only.
  • Number of Papers: JEE Advanced consists of two compulsory papers (Paper 1 and Paper 2).
  • Duration of Exam: 3 hours/each paper. Some extra time will be given to the PwD candidates.
  • Type of Questions: Objective type (MCQs).
  • Language of Question Paper: English and Hindi.
  • Subjects: Physics, Chemistry, and Mathematics subjects will be asked in the exam.
  • Negative Marking: Yes, there is a provision of negative marking and different for paper 1 & 2.

Previous Year Question Papers are available here.

 

JEE Advanced 2020 Syllabus

See theComplete & Detailed syllabus of JEE Advanced 2020 here : JEE Advanced Complete Syllabus

JEE Advanced Previous Year papers are available here: IIT JEE Papers

 

All The Best 🙂

JEE Advanced Previous Year Questions
JEE Advanced Previous Year Question Papers – Download PDF

Aspirants who are willing to clear Joint Entrance Examination JEE 2020 have to solve Previous Year question Papers of Both JEE Main & Advanced. Here you will get JEE Advanced Previous Year Question Papers PDF’s. This is probably the final and important step of completing the overall JEE Exam preparation. Considering the numerous benefits of it owns, eSaral offers a wide range of topic wise Previous Year questions for JEE Main and Advanced.

The exam conducting authority NTA will conduct JEE main twice a year in computer-based mode. If you are aiming to successfully crack JEE Advanced Exams, eSaral offers JEE Advanced previous year question papers from Year (2007-2019).

Year

PAPER 1

PAPER 2

2007 Download Paper 1 Download Paper 2
2008 Download Paper 1 Download Paper 2
2009 Download Paper 1 Download Paper 2
2010 Download Paper 1 Download Paper 2
2011 Download Paper 1 Download Paper 2
2012 Download Paper 1 Download Paper 2
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2015 Download Paper 1 Download Paper 2
2016 Download Paper 1 Download Paper 2  
2017 Download Paper 1 Download Paper 2
2018 Download Paper 1 Download Paper 2
2019 Download Paper 1 Download Paper 2

 

Previous Year Questions For JEE will help the aspirants in:

  1. Analyzing the Real Exam Scenario. the types of questions asked in Exam.
  2. A fair Idea of On-going Exam Preparation
  3. Getting an idea of essential topics from the examination point of view.
  4. Improve time management and also the speed and Accuracy level.
  5. Follow a methodical study schedule
  6. Take practice assignments Develop key analytical skills
  7. Strengthen application-based learning
  8. Evaluate their strengths and weaknesses.
  9. Understand question patterns and marking patterns.
  10. And finally to Score higher marks.

 

👉Physics Revision Series

👉Chemistry Revision Series

👉Mathematics Revision Series

 

JEE Question Papers with Solutions PDF Free Download Benefits

  • Gives you a fair idea on your on-going preparation.
  • They pretty much cover everything because questions are asked from the entire syllabus.
  • Improves both your speed and accuracy level.
  • They are real papers that have appeared in previous exams which means you will be solving a paper like this in future.
  • You can closely anlyze the pattern and then make your own strategy to score high marks.
  • Downloading the solutions makes you aware of each step of solving a question. Your analytical abilities will improve.
  • You become friendly with the actual exam which boosts your confidence

Wave Optics – JEE Advanced Previous Year Questions with Solutions

JEE Advanced Previous Year Questions of Physics with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Physics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

Get detailed Class 11th & 12th Physics Notes to prepare for Boards as well as competitive exams like IIT JEE, NEET etc.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects.

 

 

Click Here for JEE main Previous Year Topic Wise Questions of Physics with Solutions 

 

Download eSaral app  for free study material and video tutorials.

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Previous Years JEE Advanced Questions

Q. Column I shows four situations of standard Young’s double slit arrangement with the screen placed far away from the slits $S_{1}$ and $S_{2}$. In each of these cases $S_{1} P_{0}$ = $S_{2} P_{0}$, $S_{1} P_{1}$$S_{2} P_{1}$ = $\lambda / 4$ and $S_{1} P_{2}$$S_{2} P_{2}=\lambda / 3$, where $\lambda$ is the wavelength of the light used. In the cases B, C and D, a transparent sheet of refractive index $\mu$ and thickness t is pasted on slit S2. The thicknesses of the sheets are different in different cases. The phase difference between the light waves reaching a point P on the screen from the two slits is denoted by $\delta$ (P) and the intensity by I(P). Match each situation given in Column I with the statement(s) in Column II valid for that situation.

[IIT-JEE-2009]

Sol. ((A) $p, s ;(B) q ;(C) t ;(D) r, s, t$)

(A) $\Delta \mathrm{x}=\mathrm{S}_{2} \mathrm{P}-\mathrm{S}_{1} \mathrm{P}=0$

$\delta\left(\mathrm{P}_{0}\right)=\frac{2 \pi}{\lambda} \Delta \mathrm{x}=0$

$\Delta \mathrm{x}=\mathrm{S}_{1} \mathrm{P}_{1}-\mathrm{S}_{2} \mathrm{P}_{1}=\frac{\lambda}{4}$

$\delta\left(\mathrm{P}_{1}\right)=\frac{2 \pi}{\lambda} \times \frac{\lambda}{4}=\frac{\pi}{2}$

$\mathrm{I}=\mathrm{I}_{\max } \cos ^{2}\left(\frac{\Delta \phi}{2}\right)$

$\mathrm{I}\left(\mathrm{P}_{1}\right)=\mathrm{I}_{1}=\mathrm{I}_{\max } \cos ^{2} \frac{\delta}{2}=\frac{\mathrm{I}_{\max }}{2}$

$\delta\left(\mathrm{P}_{2}\right)=\frac{2 \pi}{\lambda} \times \frac{\lambda}{3}=\frac{2 \pi}{3}$

$\mathrm{I}\left(\mathrm{P}_{2}\right)=\mathrm{I}_{2}=\mathrm{I}_{\max } \cos ^{2} \frac{\pi}{3}=\frac{\mathrm{I}_{\max }}{4}$

$\mathrm{I}\left(\mathrm{P}_{0}\right)>\mathrm{I}\left(\mathrm{P}_{1}\right)$

$(\mathrm{B}) \Delta \mathrm{x}=\mathrm{S}_{1} \mathrm{P}-\left[\mathrm{S}_{2} \mathrm{P}+(\mu-1) \mathrm{t}\right]$

$\Delta \mathrm{x}_{1}=\mathrm{S}_{1} \mathrm{P}_{1}-\mathrm{S}_{2} \mathrm{P}_{1}-(\mu-1) \mathrm{t}$

$\Delta \mathrm{x}_{1}=\frac{\lambda}{4}-\frac{\lambda}{4}=0$

$8\left(\mathrm{P}_{1}\right)=0 ; \mathrm{I}\left(\mathrm{P}_{1}\right)=\mathrm{I}_{\max }$

$8\left(\mathrm{P}_{0}\right)=\frac{\pi}{2} \delta\left(\mathrm{P}_{0}\right) \neq 0$

$\mathrm{I}\left(\mathrm{P}_{0}\right)=\mathrm{I}_{\max } / 2$

$\Delta \mathrm{x}=\mathrm{S}_{1} \mathrm{P}_{2}-\mathrm{S}_{1} \mathrm{P}_{2}-(\mu-1) \mathrm{t}$

$=\frac{\lambda}{3}-\frac{\lambda}{4}=\frac{\lambda}{12}$

$8\left(\mathrm{P}_{2}\right)=\frac{2 \pi}{\lambda} \times \frac{\lambda}{12}=\frac{\pi}{6}$

$\mathrm{I}\left(\mathrm{P}_{2}\right)=\mathrm{I}_{\max } \cos ^{2}\left(\frac{\pi}{12}\right)$


Q. Young’s double slit experiment is carried out by using green, red and blue light, one color at a time. The fringe widths recorded are $\beta_{G}, \beta_{R}$ and $\beta_{B},$ respectively. Then

(A) $\beta_{G}>\beta_{B}>\beta_{R}$

(B) $\beta_{B}>\beta_{G}>\beta_{R}$

(C) $\beta_{R}>\beta_{B}>\beta_{G}$

(D) $\beta_{R}>\beta_{G}>\beta_{B}$

[IIT-JEE-2012]

Sol. (D)

$\beta=\frac{\mathrm{D} \lambda}{\mathrm{d}}$

$\lambda_{\mathrm{R}}>\lambda_{\mathrm{a}}>\lambda_{\mathrm{B}}$


Q. In the Young’s double slit experiment using a monochromatic light of wavelength $\lambda$, the path difference (in terms of an integer n) corresponding to any point having half the peak intensity is :-

(A) $(2 n+1) \frac{\lambda}{2}$

(B) $(2 n+1) \frac{\lambda}{4}$

(C) $(2 n+1) \frac{\lambda}{8}$

$(D)(2 n+1) \frac{\lambda}{16}$

[JEE Advanced 2013]

Sol. (B)

$\frac{\mathrm{I}_{\max }}{2}=\mathrm{I}_{\max } \cos ^{2}\left(\frac{\pi}{\lambda} \Delta \mathrm{x}\right)$

$\cos ^{2}\left(\frac{\pi}{\lambda} \Delta \mathrm{x}\right)=\frac{1}{2}$

$\cos \left(\frac{\pi}{\lambda} \Delta \mathrm{x}\right)=\pm \frac{1}{\sqrt{2}}$

$\frac{\pi}{\lambda} \Delta \mathrm{x}=\mathrm{n} \pi \pm \frac{\pi}{4}$

$\Delta \mathrm{x}=\left(\mathrm{n} \pm \frac{1}{4}\right) \lambda$


Q. A light source, which emits two wavelengths $\lambda_{1}=400 \mathrm{nm}$ and $\lambda_{2}=600 \mathrm{nm},$ is used in a Young’s double slit experiment. If recorded fringe widths for $\lambda_{1}$ and $\lambda_{2}$ are $\beta_{1}$ and $\beta_{2}$ and the number of fringes for them within a distance y on one side of the central maximum are $\mathrm{m}_{1}$ and $\mathrm{m}_{2},$ respectively, then :-

(A) $\beta_{2}>\beta_{1}$

(B) $\mathrm{m}_{1}>\mathrm{m}_{2}$

(C) From the central maximum, $3^{\mathrm{rd}}$ maximum of $\lambda_{2}$ overlaps with $5^{\text {th }}$ minimum of $\lambda_{1}$

(D) The angular separation of fringes of $\lambda_{1}$ is greater than $\lambda_{2}$

[JEE Advanced 2014]

Sol. (A,B,C)

$\beta=\frac{\mathrm{D} \lambda}{\mathrm{d}}$

$\mathrm{B}_{2}>\beta_{1}$

$\mathrm{y}=\mathrm{m}_{1} \frac{\mathrm{D} \lambda_{1}}{\mathrm{d}}=\mathrm{m}_{2} \frac{\mathrm{D} \lambda_{2}}{\mathrm{d}}$

$\frac{\mathrm{nD} \times \lambda_{2}}{\mathrm{d}}=\left(\mathrm{n}^{\prime}+\frac{1}{2}\right) \frac{\mathrm{D} \lambda_{1}}{\mathrm{d}} \Rightarrow 600 \mathrm{n}=\left(\mathrm{n}^{\prime}+\frac{1}{2}\right) \times 4$


Q. A Young’s double slit interference arrangement with slits $S_{1}$ and $S_{2}$ is immersed in water (refractive index $=4 / 3$ ) as shown in the figure. The positions of maxima on the surface of water are given by $x^{2}=p^{2} m^{2} \lambda^{2}-d^{2},$ where $\lambda$ is the wavelength of light in air (refractive index $=1$, $2 d$ is the separation between the slits and $m$ is an integer. The value of p is.

[JEE Advanced 2015]

Sol. 3


Q. While conducting the Young’s double slit experiment, a student replaced the two slits with a large opaque plate in the x-y plane containing two small holes that act as two coherent point sources $\left(\mathrm{S}_{1}, \mathrm{S}_{2}\right)$ emitting light of wavelength 600 nm. The student mistakenly placed the screen parallel to the x-z plane (for z > 0) at a distance D = 3m from the mid-point of $\mathrm{S}_{1} \mathrm{S}_{2}$, as shown schematically in the figure. The distance between the sources d = 0.6003 mm. The origin O is at the intersection of the screen and the line joining $\mathrm{S}_{1} \mathrm{S}_{2}$. Which of the following is (are) true of the intensity pattern on the screen ?

(A) Hyperbolic bright and dark bands with foci symmetrically placed about O in the x-direction

(B) Semi circular bright and dark bands centered at point O

(C) The region very close to the point O will be dark

(D) Straight bright and dark bands parallel to the x-axis

[JEE-Mains 2016]

Sol. (B,C)

Path difference at point O = d = .6003 mm = 600300 nm

$=\frac{2001}{2}(600 \mathrm{nm})=1000 \lambda+\frac{\lambda}{2}$

$\Rightarrow$ minima form at point $\mathrm{O}$

Line $S_{1} S_{2}$ and screen are $\perp$ to each other so fringe pattern is circular (semi-circular because only half of screen is available)


Q. Two coherent monochromatic  point sources $\mathrm{S}_{1}$ and $\mathrm{S}_{2}$ of wavelength $\lambda$ = 600 nm are placed symmetrically on either side of the center of the circle as shown. The sources are separated by a distance d = 1.8mm. This arrangement produces interference fringes visible as alternate bright and dark spots on the circumference of the circle. The angular separation between two consecutive bright spots is $\Delta \theta$. Which of the following options is/are correct ?

(A) A dark spot will be formed at the point $\mathrm{P}_{2}$

(B) The angular separation between two consecutive bright spots decreases as we move from $\mathrm{P}_{1}$ to $\mathrm{P}_{2}$ along the first quadrant

(C) At $\mathrm{P}_{2}$ the order of the fringe will be maximum

(D) The total number of fringes produced between $P_{1}$ and $\mathrm{P}_{2}$ in the first quadrant is close to 3000

[JEE Advanced 2017]

Sol. (C,D)


Caboxylic Acid – JEE Advanced Previous Year Questions with Solutions

JEE Advanced Previous Year Questions of Chemistry with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Chemistry will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

Simulator

Previous Years JEE Advance Questions

Q. Match each of the compound in Column I with its characteristic reaction(s) in Column II.

[IIT 2009]

Sol.


Q. Match each of the compounds given in Column I with the reaction(s), that they can undergo, given in Column II.

[IIT 2009]

Sol. ($(A) \rightarrow P, Q, S, T:(B) \rightarrow P, S, T ;(C) \rightarrow P:(D) \rightarrow R$)


Q. The major product of the following reaction is

[IIT 2011]

Sol. ($(\mathrm{A}) \rightarrow \mathrm{P}, \mathrm{Q}, \mathrm{T} ;(\mathrm{B}) \rightarrow \mathrm{P}, \mathrm{Q}, \mathrm{S}, \mathrm{T},(\mathrm{C}) \rightarrow \mathrm{R}, \mathrm{S}, ;(\mathrm{D}) \rightarrow \mathrm{P}$)


Q. With reference the scheme given, which of the given statement(s) about T, U, V & W is (are) correct ?

(A) ‘T’ is soluble in hot aq NaOH

(B) ‘U’ is optically active

(C) mol formula of $\mathrm{W}$ is $\mathrm{C}_{10} \mathrm{H}_{18} \mathrm{O}_{4}$

(D) V gives effervescence with aq NaHCO $_{3}$

[IIT 2012]

Sol. (A)


Q. Identify the binary mixtures (s) that can be separated into the individual compounds, by differential extraction, as shown in the given scheme –

[IIT 2012]

Sol. (A,C,D)


Q. The total number of carboxylic acid groups in the product P is

[IIT 2013]

Sol. (B,D)


Q. In the reaction shown below, the major product(s) formed is / are :

[IIT 2014]

Sol. 2


Q. Different possible thermal decomposition pathways for peroxyesters are shown below. Match each pathway from List-I with an appropriate structure from List-II and select the correct answer using the code given below the lists.

[IIT 2014]

Sol. (A)


Q. The major product of the reaction is :

[IIT 2015]

Sol. (A)


Q. Aniline reacts with mixed acid (conc. HNO, and conc. $\mathrm{H}_{2} \mathrm{SO}_{4}$ ) at 288 K to give P (51%), Q (47%) and R (2%). The major product(s) the following reaction sequence is (are) :-

[JEE Adv. 2018]

Sol. (C)


Q. In the following reaction sequence, the correct structure(s) of X is (are)

[JEE Adv. 2018]

Sol. (D)


Treatment of benzene with CO/HCl in the presence of anhydrous $\mathrm{AlCl}_{3} / \mathrm{CuCl}$ followed by reaction with Ac2O/NaOAc gives compound X as the major product. Compound X upon reaction with

$\mathrm{Br}_{2} / \mathrm{Na}_{2} \mathrm{CO}_{3}$, followed by heating at 473 K with moist KOH furnishes Y as the major product. Reaction of X with $\mathrm{H}_{2} / \mathrm{Pd}-\mathrm{C}$, followed by $\mathrm{H}_{3} \mathrm{PO}_{4}$ treatment gives Z as the major product.

(There are two questions based on PARAGRAPH “X”, the question given below is one of them)

Q. The compound Y is :-

[JEE Adv. 2018]

Sol. (B)


Q. The compound Z is :-

[JEE Adv. 2018]

Sol. (C)


An organic acid P $\left(\mathrm{C}_{1} \mathrm{H}_{12} \mathrm{O}_{2}\right)$ can easily be oxidized to a dibasic acid which reacts with ethyleneglycol to produce a polymer dacron. Upon ozonolysis, P gives an aliphatic ketone as one of the products. P undergoes the following reaction sequences to furnish R via Q. The compound P also undergoes another set of reactions to produce S.

(There are two questions based on PARAGRAPH “A”, the question given below is one of them)

Q.

[JEE Adv. 2018]

Sol. (A)


Q. The compound R is

(A)

(B)

(C)

(D)

[JEE Adv. 2018]

Sol. (A)


Q. The compound S is

[JEE Adv. 2018]

Sol. (B)


 

Radioactivity – JEE Advanced Previous Year Questions with Solutions

JEE Advanced Previous Year Questions of Chemistry with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Chemistry will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

Simulator

Previous Years JEE Advance Questions

Q. The total number of $\alpha$ and $\beta$particles emitted in the nuclear reaction $_{92}^{238} \mathrm{U} \rightarrow_{82}^{214} \mathrm{Pb}$ is

[JEE 2009]

Sol. 8


Q. The number of neutrons emitted when $_{92}^{235} \mathrm{U}$ undergoes controlled nuclear fission to $_{54}^{142} \mathrm{Xe}$ and is –

[JEE 2010]

Sol. 4


Q. Bombardment of aluminium by $\alpha$ -particle leads to its artificial disintegration in two ways,

(i) and (ii) as shown. Products X, Y and Z respectively are :

(A) proton, neutron, positron

(B) neutron, positron, proton

(C) proton, positron, neutron

(D) positron, proton, neutron

[JEE 2011]

Sol. (A)


Q. The periodic table consists of 18 groups. An isotope of copper, on bombardment with protons, undergoes a nuclear reaction yielding element X as shown below. To which group , element X belongs in the periodic table ?

[JEE 2012]

Sol. 8


Q. In the nuclear transmutation

(X, Y) is(are)

$(\mathrm{A})(\gamma, \mathrm{n})$

(B) (p, D)

(C) (n, D)

$(\mathrm{D})(\gamma, \mathrm{D})$

[JEE 2013]

Sol. (AB)


Q. A closed vessel with rigid walls contains 1 mol of 238

92 $\mathrm{U}$ U and 1 mol of air at 298 K. Considering complete decay of $^{238}_{92} \mathrm{U}$ the ratio of the final pressure to the initial pressure of the system at 298 K is –

[JEE 2015]

Sol. 9


Metallurgy – JEE Advanced Previous Year Questions with Solutions

JEE Advanced Previous Year Questions of Chemistry with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Chemistry will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

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Previous Years JEE Advance Questions

Paragraph for questions 1 to 3

Copper is the most nobel of the first row transition metals and occurs in small deposits in several

countries. Ores of copper include chalcanthite $\left(\mathrm{CuSO}_{4}, 5 \mathrm{H}_{2} \mathrm{O}\right),$ atacamite $\left(\mathrm{Cu}_{2} \mathrm{Cl}(\mathrm{OH})_{3}\right),$ cuprite $\left(\mathrm{Cu}_{2} \mathrm{O}\right),$ copper glance (Cu.S) and malachite $\left(\mathrm{Cu}_{2}(\mathrm{OH})_{2} \mathrm{CO}_{3}\right) .$ However, $80 \%$ of the world copper production comes from the ore chalcopyrite (CuFeS_{2} ) . \text { The extraction of copper from chalcopyrite involves partial roasting, removal of iron and self-reduction.

Q. Partial roasting of chalcopyrite produces :-

(A) $\mathrm{Cu}_{2} \mathrm{S}$ and $\mathrm{FeO}$

(B) $\mathrm{Cu}_{2} \mathrm{O}$ and FeO

(C) $\mathrm{CuS}$ and $\mathrm{Fe}_{2} \mathrm{O}_{3}$

(D) $\mathrm{Cu}_{2} \mathrm{O}$ and $\mathrm{Fe}_{2} \mathrm{O}_{3}$

[JEE-2010]

Sol. (A)


Q. Iron is removed from chalcopyrite as :-

(A) FeO

(B) FeS

(C) $\mathrm{Fe}_{2} \mathrm{O}_{3}$

(D) FeSiO $_{3}$

[JEE-2010]

Sol. (D)


Q. In self-reduction, the reducing species is :-

(A) S

(B) $\mathrm{O}^{2-}$

(C) $\mathrm{S}^{2-}$

(D) $\mathrm{SO}_{2}$

[JEE-2010]

Sol. (C)


Q. Extraction of metal from the ore cassiterite involves –

(A) carbon reduction of an oxide ore

(B) self-reduction of a sulphide ore

(C) removal of copper impurity

(D) removal of iron impurity

[JEE-2011]

Sol. (A,C,D)

$\mathrm{SnO}_{2}+2 \mathrm{C} \rightarrow \mathrm{Sn}+2 \mathrm{CO}$

Cassetrite contains impurity of Wolframite (FeWO,), Which is a ferro magnatic.


Q. Oxidation states of the metal in the minerals haematite and magnetite, respectively, are –

(A) II, III in haematite and III in magnetite

(B) II, III in haematite and II in magnetite

(C) II in haematite and II, III in magnetite

(D) III in haematite and II, III in magnetite

[JEE-2011]

Sol. (D)

Haematite Ore $\Rightarrow \mathrm{Fe}_{2} \mathrm{O}_{3} ;$ Oxidation state $(+3)$ magnetite ore $\Rightarrow \mathrm{Fe}_{3} \mathrm{O}_{4}\left(\mathrm{FeO}+\mathrm{Fe}_{2} \mathrm{O}_{3}\right)=$ Oxidation

state of $(+2 \&+3)$


Q. In the cyanide extraction process of silver from argentite ore, the oxidizing and reducing agents used are :

(A) $\mathrm{O}_{2}$ and CO respectively.

(B) $\mathrm{O}_{2}$ and $\mathrm{Zn}$ dust respectively.

(C) $\mathrm{HNO}_{3}$ and $\mathrm{Zn}$ dust respectively.

(D) $\mathrm{HNO}_{3}$ and CO respectively.

[JEE-2012]

Sol. (B)

$2 \mathrm{Ag}+4 \mathrm{NaCN}+\mathrm{H}_{2} \mathrm{O}+\frac{1}{2} \mathrm{O}_{2}(\text { air }) \longrightarrow 2 \mathrm{Na}\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]+2 \mathrm{NaOH},$ Here $\mathrm{O}_{2}$ is oxidising agent $\& \mathrm{Zn}^{-}$

dust act as reducing agent.


Q. Sulfide ores are common for the metals –

(A) Ag, Cu and Pb

(B) Ag, Cu and Sn

(B) Ag, Cu and Sn

(D) Al, Cu and Pb

[JEE-2013]

Sol. (A)


Q. The carbon-based reduction method is NOT used for the extraction of

(A) $\operatorname{tin}$ from $\mathrm{SnO}_{2}$

(B) Iron from $\mathrm{Fe}_{2} \mathrm{O}_{3}$

(C) aluminium from $\mathrm{Al}_{2} \mathrm{O}_{3}$

(D) magnesium from $\mathrm{MgCO}_{3} . \mathrm{CaCO}_{3}$

[JEE-2013]

Sol. (C,D)

Being a more reactive metal “Al” and “Mg” can be reduced by electrolytic reduction.


Q. Upon heating with $\mathrm{Cu}_{2} \mathrm{S},$ the reagent(s) that give copper metal is/are

(A) CuFeS $_{2}$

(B) CuO

(C) $\mathrm{Cu}_{2} \mathrm{O}$

(D) $\mathrm{CuSO}_{4}$

[JEE Adv. 2015]

Sol. (A,C,D)


Q. Copper is purified by electrolytic refining of blister copper. The correct statement(s) about this process is (are) –

(A) Impure Cu strip is used as cathode

(B) Acidified aqueuous $\mathrm{CuSO}_{4}$ is used as electrolyte

(C) Pure Cu deposits at cathode

(D) Impurities settle as anode-mud

[JEE Adv. 2015]

Sol. (B,C,D)

Impure “Cu” act as a Anode . While ” Pure thin strips of Cu act as cathode and aqueous solution of $\mathrm{CuSO}_{4}$ act as electrolyte.


Q. Match the anionic species given in Column-I that are present in the ore(s) given in

Column-II –

[JEE Adv. 2015]

Sol. $(A \rightarrow P, Q, S, B \rightarrow T, C \rightarrow Q, R, D \rightarrow R)$

Name of ores & their metals.


Q. Galena (an ore) is partially oxidized by passing air through it at high temperature. After some time, the passage of air is stopped, but the heating is continued in a closed furnance such that the contents undergo self-reduction. The weight (in kg) of Pb produced per kg of $\mathrm{O}_{2}$ consumed is ______ .

(Atomic weights in $\left.\mathrm{g} \text { mol }^{-1}: \mathrm{O}=16, \mathrm{S}=32, \mathrm{Pb}=207\right)$

[JEE Adv. 2018]

Sol. 6.47


Liquid Solution – JEE Advanced Previous Year Questions with Solutions

JEE Advanced Previous Year Questions of Chemistry with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Chemistry will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

Simulator

Previous Years JEE Advance Questions

Q. The Henry’s law constant for the solubility of $\mathrm{N}_{2}$ gas in water at 298 K is 1.0 × $10^{5}$ atm. The mole fraction of N2 in air is 0.8. The number of moles of $\mathrm{N}_{2}$ from air dissolved in 10 moles of waterat 298 K and 5 atm pressure is-

(A) $4.0 \times 10^{-4}$

(B) $4.0 \times 10^{-5}$

(C) $5.0 \times 10^{-4}$

(D) $4.0 \times 10^{-5}$

[JEE 2009]

Sol. (A)

$\mathrm{P}_{\mathrm{N}_{2}}=\mathrm{K}_{\mathrm{H}} \mathrm{X}_{\mathrm{N}_{2}}$

$\mathrm{Y}_{\mathrm{N}_{2}} \cdot \mathrm{P}_{\mathrm{T}}=\mathrm{K}_{\mathrm{H}} \times \mathrm{N}_{2}$

$0.8 \times 5=1 \times 10^{5} \times \frac{\mathrm{n}}{\mathrm{n}+10}$

$4=10^{5} \times \frac{\mathrm{n}}{10}$

$\mathrm{n}=4 \times 10^{-4}$


Q. For a dilute solution containing 2.5 g of a non-volatile non-electrolyte solute in 100 g of water, the elevation in boiling point at 1 atm pressure is $2^{\circ} \mathrm{C}$. Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure (mm of Hg) of the solution is-(take $\left.\mathrm{K}_{\mathrm{b}}=0.76 \mathrm{K} \mathrm{kg} \mathrm{mol}^{-1}\right)$

(A) 724             (B) 740            (C) 736             (D) 718

[JEE 2011]

Sol. (A)


Q. The freezing point (in °C) of a solution containing 0.1 g of $\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ (Mol. Wt. 329) in

100 g of water $\left(\mathrm{K}_{\mathrm{f}}=1.86 \mathrm{K} \mathrm{kg} \mathrm{mol}^{-1}\right)$ is –

(A) $-2.3 \times 10^{-2}$

(B) $-5.7 \times 10^{-2}$

(C) $-5.7 \times 10^{-3}$

(D)-1.2 \times 10^{-2}

[JEE 2011]

Sol. (A)

$\mathrm{T}_{\mathrm{f}}^{\prime}=\mathrm{T}_{\mathrm{f}}-\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{T}_{\mathrm{f}}-\mathrm{i} \mathrm{K}_{\mathrm{f}} \cdot \mathrm{m}$

$=0^{\circ} \mathrm{C}-4 \times 1.86 \times \frac{0.1 / 329}{100 / 1000}$

$=-0.023^{\circ} \mathrm{C}=-2.3 \times 10^{2} \mathrm{C}$


Q. For a dilute solution containing 2.5 g of a non-volatile non-electrolyte solute in 100 g of water, the elevation in boiling point at 1 atm pressure is 2^{\circ} \mathrm{C}. Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure (mm of Hg) of the solution is \text { (take }\left.\mathrm{K}_{\mathrm{b}}=0.76 \mathrm{K} \mathrm{kg} \mathrm{mol}^{-1}\right)

(A) 724 (B) 740 (C) 736 (D) 718

[JEE 2012]

Sol. (A)

$\Delta \mathrm{T}_{\mathrm{b}}=\mathrm{K}_{\mathrm{b}} \cdot \mathrm{m}$

$2=0.76 \times \frac{\mathrm{n}}{100 / 1000}$

$\mathrm{n}=0.263 \mathrm{mol}$

$\mathrm{P}_{\mathrm{S}}=\mathrm{P}^{\circ} \mathrm{x}_{\text {solvent }}=760 \times \frac{\frac{100}{18}}{\frac{100}{18}+0.263}=760 \times \frac{5.55}{5.82}=724.7$ torr


Q. Benzene and naphthalene form an ideal solution at room temperature. For this process, the true statement(s) is(are)

(A) $\Delta \mathrm{G}$ is positive

(B) $\Delta S_{\text {system }}$ is positive

(C) $\Delta \mathrm{S}_{\text {surroundings }}=0$

(D) $\Delta \mathrm{H}=0$

[J-Adv. 2013]

Sol. (B,C,D)


Q. A compound $\mathrm{H}_{2} \mathrm{X}$ with molar weight of 80 g is dissolved in a solvent having density of

$0.4 \mathrm{g} \mathrm{mL}^{-1}$, Assuming no change in volume upon dissolution, the molality of a 3.2 molar solution is

[JEE-Adv. 2014]

Sol. 8

$\mathrm{m}=\frac{3.2 \mathrm{mol}}{0.4 \mathrm{Kg}}=8 \mathrm{mol} / \mathrm{kg}$


Q. If the freezing point of a 0.01 molal aqueous solution of a cobalt (III) chloride-ammonia complex (which behaves as a strong eletrolyte) is $-0.0558^{\circ} \mathrm{C}$ , the number of chloride (s) in the coordination sphere of the complex is- $\left[\mathrm{K}_{\mathrm{f}} \text { of water }=1.86 \mathrm{K} \mathrm{kg} \mathrm{mol}^{-1}\right]$

[JEE-Adv. 2015]

Sol. 1

0.0558 = i × 1.86 × 0.01

i = 3

$\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2}$


Q. Mixture(s) showing positive deviation from Raoult’s law at $35^{\circ}$C is (are)

(A) carbon tetrachloride + methanol

(B) carbon disulphide + acetone

(C) benzene + toluene

(D) phenol + aniline

[JEE – Adv. 2016]

Sol. (A,C)

(A) H-bonding of methanol breaks when $\mathrm{CCl}_{4}$ is added so bonds become weaker, resulting positive

deviation.

(B) Mixing of polar and non-polar liquids will produce a solution of weaker interaction, resulting positive deviation

(C) Ideal solution

(D) –ve deviation because stronger H-bond is formed.


Q. For a solution formed by mixing liquids L and M, the vapour pressure of L plotted against the mole fraction of M in solution is shown in the following figure, Here $\mathrm{x}_{\mathrm{L}}$ and $\mathbf{X}_{\mathrm{M}}$ represent mole fractions of L and M, respectively, in the solution. the correct statement(s) applicable to this system is(are) –

(A) Attractive intramolecular interactions between L–L in pure liquid L and M–M in pure liquid M are stronger than those between L–M when mixed in solution

(B) The point $Z$ represents vapour pressure of pure liquid $\mathrm{M}$ and Raoult’s law is obeyed when $\quad \mathrm{x}_{\mathrm{L}} \rightarrow 0$

(C) The point Z represents vapour pressure of pure liquid L and Raoult’s law is obeyed when $\mathrm{x}_{\mathrm{L}} \rightarrow 1$

(D) The point Z represents vapour pressure of pure liquid M and Raoult’s law is obeyed from $\mathrm{x}_{\mathrm{L}}=0$ to $\mathrm{x}_{\mathrm{L}}=1$

[JEE – Adv. 2017]

Sol. (A,C)


Q. Pure water freezes at 273 K and 1 bar. The addition of 34.5 g of ethanol to 500 g of water changes the freezing point of the solution. Use the freezing point depression constant of water as 2 K kg $\mathrm{mol}^{-1}$ . The figures shown below represents plots of vapour pressure (V.P.) versus temperature (T). [Molecular weight of ethanol is 46 g $\mathrm{mol}^{-1}$]

Among the following, the option representing change in the freezing point is –

[JEE – Adv. 2017]

Sol. (D)

Ethanol should be considered non volatile as per given option

$\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{K}_{\mathrm{f}} \times \mathrm{m}$

$\Delta \mathrm{T}_{\mathrm{f}}=2 \times \frac{34.5}{46 \times 0.5}=3 \mathrm{K}$


Q. Liquids A and B form ideal solution over the entire range of composition. At temperature T, equimolar binary solution of liquids A and B has vapour pressure 45 Torr. At the same temperature, a new solution of A and B having mole fractions $\mathbf{X}_{\mathrm{A}}$ and $\mathrm{x}_{\mathrm{B}}$, respectively, has vapour pressure of 22.5 Torr. The value of $\mathrm{x}_{\mathrm{A}} / \mathrm{x}_{\mathrm{B}}$ in the new solution is____.

(given that the vapour pressure of pure liquid A is 20 Torr at temperature T)

[JEE – Adv. 2018]

Sol. 19

$45=\mathrm{P}_{\mathrm{A}}^{\mathrm{o}} \times \frac{1}{2}+\mathrm{P}_{\mathrm{B}}^{\mathrm{o}} \times \frac{1}{2}$$\mathrm{P}_{\mathrm{A}}^{\mathrm{o}}+\mathrm{P}_{\mathrm{B}}^{\circ}=90 \ldots \ldots(1)$

given $\mathrm{P}_{\mathrm{A}}^{\circ}=20$ torr

$\mathrm{P}_{\mathrm{B}}^{\circ}=70 \mathrm{torr}$

$\Rightarrow 22.5$ torr $=20 \mathrm{x}_{\mathrm{A}}+70\left(1-\mathrm{x}_{\mathrm{A}}\right)$

$=70-50 \mathrm{x}_{\mathrm{A}}$

$\mathrm{x}_{\mathrm{A}}=\left(\frac{70-22.5}{50}\right)=0.95$

$\mathrm{x}_{\mathrm{B}}=0.05$

So $\frac{\mathrm{x}_{\mathrm{A}}}{\mathrm{x}_{\mathrm{B}}}=\frac{0.95}{0.05}=19$


Q. The plot given below shows P–T curves (where P is the pressure and T is the temperature) for two solvents X and Y and isomolal solutions of NaCl in these solvents. NaCl completely dissociates in both the solvents.

On addition of equal number of moles a non-volatile solute S in equal amount (in kg) of these solvents, the elevation of boiling point of solvent X is three times that of solvent Y. Solute S is known to undergo dimerization in these solvents. If the degree of dimerization is 0.7 in solvent Y, the degree of dimerization in solvent X is ___.

[JEE – Adv. 2018]

Sol. 0.05

From graph


Isomerism – JEE Advanced Previous Year Questions with Solutions

JEE Advanced Previous Year Questions of Chemistry with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Chemistry will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

Simulator

Previous Years JEE Advance Questions

Q. The correct statement(s) about the compound $\mathrm{H}_{3} \mathrm{C}$ (HO)HC – CH = CH – CH(OH) $\mathrm{CH}_{3}$ (X) is (are) :

(A) The total number of stereoisomers possible for X is 6

(B) The total number of diastereomers possible for X is 3

(C) If the stereochemistry about the double bond in X is trans, the number of enantiomers possible for X is 4

(D) If the stereochemistry about the double bond in X is cis, the number of enantiomers possible for X is 2

[iit-2009]

Sol. (A,D)


Q. In the Newman projection for 2,2–dimethylbutane –

X and Y can respectively be –

(A) H and H

(B) $\mathrm{H}$ and $\mathrm{C}_{2} \mathrm{H}_{5}$

(C) $\mathrm{C}_{2} \mathrm{H}_{5}$ and $\mathrm{H}$

(D) $\mathrm{CH}_{3}$ and $\mathrm{CH}_{3}$

[iit-2010]

Sol. (B,D)


Q. Amongst the given option, the compound(s) in which all the atoms are in one plane in all the possible conformations (if any), is (are) –

[iit-2011]

Sol. (B,C)


Q. Which of the given statement(s) about N,O,P and Q with respect to M is (are) correct ?

(A) M and N are non-mirror image stereoisomers

(B) M and O are identical

(C) M and P are enantiomers

(D) M and Q are identical

[JEE-2012]

Sol. (A,B,C)


Q. The total number(s) of stable conformers with non-zero dipole moment for the following compound is (are) :

[JEE-2014]

Sol. 3


ElectroChemistry – JEE Advanced Previous Year Questions with Solutions

JEE Advanced Previous Year Questions of Chemistry with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Chemistry will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

Simulator

Previous Years JEE Advance Questions

Q. For the reaction of $\mathrm{NO}_{3}^{-}$ ion in an aqueous solution, $\mathrm{E}^{\circ}$ is +0.96 V. Values of $\mathrm{E}^{\circ}$ for some metal ions are given below

The pair(s) of metal that is(are) oxidised by $\mathrm{NO}_{3}^{-}$ in aqueous solution is(are)

(A) V and Hg (B) Hg and Fe (C) Fe and Au (D) Fe and V

[JEE 2009]

Sol. (A,B,D)

(A,B,D) as $\mathrm{E}^{\circ}$ will be positive


Paragraph for Questions 2 to 3

The concentration of potassium ions inside a biological cell is at least twenty times higher than the outside. The resulting potential difference across the cell is important in several processes such as transmission of nerve impulses and maintaining the ion balance. A simple model for such a concentration cell involving a metal M is :

$\mathrm{M}(\mathrm{s}) | \mathrm{M}^{+}\left(\mathrm{aq} ; 0.05 \text { molar) } \| \mathrm{M}^{+}(\mathrm{aq} ; 1 \mathrm{molar}) | \mathrm{M}(\mathrm{s})\right.$

For the above electrolytic cell the magnitude of the cell potential $\left|\mathrm{E}_{\mathrm{cell}}\right|$ = 70 mV.

Q. For the above cell :-

(A) $\mathrm{E}_{\text {cell }}<0 ; \Delta \mathrm{G}>0$

(B) $\mathrm{E}_{\text {cell }}>0 ; \Delta \mathrm{G}<0$

(C) $\mathrm{E}_{\text {cell }}<0 ; \Delta \mathrm{G}^{0}>0$

(D) $\mathrm{E}_{\text {cell }}>0 ; \Delta \mathrm{G}^{\text {o }}<0$

[JEE 2010]

Sol. (C)

$\mathrm{E}_{1}=-\frac{059}{1} \log \frac{05}{1}=(+) \mathrm{ve} \Rightarrow \mathrm{so}$


Q. If the 0.05 molar solution of $\mathrm{M}^{+}$ is replaced by a 0.0025 molar $\mathrm{M}^{+}$ solution, then the magnitude of the cell potential would be :-

(A) 35 mV              (B) 70 mV             (C) 140 mV              (D) 700 mV

[JEE 2010]

Sol. (B)


Q. Consider the following cell reaction :

$2 \mathrm{Fe}_{(\mathrm{s})}+\mathrm{O}_{2(\mathrm{g})}+4 \mathrm{H}_{(\mathrm{aq})}^{+} \rightarrow 2 \mathrm{Fe}_{(\mathrm{aq})}^{2+}+2 \mathrm{H}_{2} \mathrm{O}(\ell) \quad \mathrm{E}^{\circ}=1.67 \mathrm{V}$

$\mathrm{At}\left[\mathrm{Fe}^{2+}\right]=10^{-3} \mathrm{M}, \mathrm{P}\left(\mathrm{O}_{2}\right)=0.1$ atm and $\mathrm{pH}=3,$ the cell potential at $25^{\circ} \mathrm{C}$ is –

(A) 1.47 V (B) 1.77 V (C) 1.87 V (D) 1.57 V

[JEE 2011]

Sol. (D)


Q. $\mathrm{AgNO}_{3}$ (aq.) was added to an aqueous KCl solution gradually and the conductivity of the solution was measured. the plot of conductance () versus the volume of $\mathrm{AgNO}_{3}$ is –

(A) (P) (B) (Q) (C) (R) (D) (S)

[JEE 2011]

Sol. (D)


Paragraph for Question 6 and 7

The electrochemical cell shown below is a concentration cell.

$\mathrm{M} | \mathrm{M}^{2+}$ (saturated solution of a sparingly soluble salt,

$\mathrm{MX}_{2}$) | | $\mathbf{M}^{2+}$ (0.001 mol $\mathrm{dm}^{-3}$) | M

The emf of the cell depends on the difference in concentrations of M2+ ions at the two electrodes. The emf of the cell at 298 K is 0.059V.

Q. The value of G $\left(\mathrm{kJ} \mathrm{mol}^{-1}\right)$ for the given cell is (take If = 96500 C $\mathrm{mol}^{-1}$)

(A) –5.7

(B) 5.7

(C) 11.4

(D) –11.4.

[JEE 2012]

Sol. (B)


Q. The solubility product $\left(\mathrm{K}_{\mathrm{sp}} ; \mathrm{mol}^{3} \mathrm{dm}^{-9}\right)$ of $\mathrm{MX}_{2}$ at 298 K based on the information available for the given concentration cell is (take 2.303 × R × 298/F = 0.059 V)

(A) $1 \times 10^{-15}$

(B) $4 \times 10^{-15}$

(C) $1 \times 10^{-12}$

(D) $1 \times 10^{-12}$

Sol. (D)

$\Delta \mathrm{G}=\frac{-2 \times 96500 \times .059}{1000}=11.4$


Q. The standard reduction potential data at $25^{\circ} \mathrm{C}$ is given below

Match $\mathrm{E}^{\mathrm{o}}$ of the redox pair in List-I with the values given in List-II and select the correct answer using the code given below the lists :

[JEE-Adv. 2013]

Sol. (D)


Q. An aqueous solution of X is added slowly to an aqueous solution of Y as shown in List-I. The variation in conductivity of these reactions is given in List-II. Match List-I with List-II and select the correct answer using the code given below the lists :

[JEE-Adv. 2013]

Sol. (A)

$\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{3} \mathrm{N}+\mathrm{CH}_{3} \mathrm{COOH} \Rightarrow$ Weak acid and weak base so conductivity increases and then does not change much so option 3 hence and (a)


Q. In a galvanic cell , the salt bridge –

(A) Does not participate chemically in the cell reaction

(B) Stops the diffusion of ions from one electrode to another

(C) Is necessary for the occurence of the cell reaction

(D) Ensures mixing of the two electrolytic solutions

[JEE-Adv. 2014]

Sol. (A,B)

Fact


Q. The molar conductivity of a solution of a weak acid HX (0.01 M) is 10 times smaller than the molar conductivity of a solution of a weak acid HY (0.1 M). If $\lambda_{\mathrm{X}^{-}}^{0} \approx \lambda_{\mathrm{Y}^{-}}^{0}$ the difference in their $\mathrm{pK}_{\mathrm{a}}$ values $, \mathrm{pK}_{\mathrm{a}}(\mathrm{HX})-\mathrm{p} \mathrm{K}_{\mathrm{a}}(\mathrm{HY}),$ is (consider degree of ionization of both acids to be <<1).

[JEE-Adv. 2015]

Sol. 3


Q. All the energy released from the reaction X $\rightarrow \mathrm{Y}, \Delta_{\mathrm{r}} \mathrm{G}^{\circ}=-193 \mathrm{kJ} \mathrm{mol}^{-1}$ is used for the oxidizing $\mathrm{M}^{+}$ and $\mathrm{M}^{+} \rightarrow \mathrm{M}^{3+}+2 \mathrm{e}^{-}, \mathrm{E}^{\circ}=-0.25 \mathrm{V}$

Under standard conditions, the number of moles of M+ oxidized when one mole of X is converted to Y is $\left[\mathrm{F}=96500 \mathrm{C} \mathrm{mol}^{-1}\right]$

[JEE-Adv. 2015]

Sol. 4


Q. For the electrochemical cell,

$\mathrm{Mg}(\mathrm{s})\left|\mathrm{Mg}^{2+}(\mathrm{aq}, 1 \mathrm{M}) \| \mathrm{Cu}^{2+}(\mathrm{aq}, 1 \mathrm{M})\right| \mathrm{Cu}(\mathrm{s})$

the standard emf of the cell is 2.70 V at 300 K. When the concentration of $\mathrm{Mg}^{2+}$ is changed to x M, the cell potential changes to 2.67 V at 300 K. The value of x is____.

(given, $\frac{\mathrm{F}}{\mathrm{R}}=11500 \mathrm{KV}^{-1}$ where F is the Faraday constant and R is the gas constant, ln(10) = 2.30)

[JEE-Adv. 2018]

Sol. 10


Q. Consider an electrochemical cell: $\mathrm{A}(\mathrm{s})\left|\mathrm{A}^{\mathrm{n}+}(\mathrm{aq}, 2 \mathrm{M}) \| \mathrm{B}^{2 \mathrm{n}+}(\mathrm{aq}, 1 \mathrm{M})\right| \mathrm{B}(\mathrm{s})$. The value of $\Delta \mathrm{H}^{\theta}$ for the cell reaction is twice that of $\Delta \mathrm{G}^{\theta}$ at 300 K. If the emf of the cell is zero, the $\Delta \mathrm{S}^{\theta}$ $\left(\text { in } \mathrm{JK}^{-1} \mathrm{mol}^{-1}\right)$ of the cell reaction per mole of B formed at 300 K is___.
(Given : ln (2) = 0.7, R (universal gas constant) = $8.3 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$. H, S and G are enthalpy, entropy and Gibbs energy, respectively.)

[JEE-Adv. 2018]

Sol. (-11.62)


Chemical Kinetics – JEE Advanced Previous Year Questions with Solutions

JEE Advanced Previous Year Questions of Chemistry with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Chemistry will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

Simulator

Previous Years JEE Advance Questions

Q. Plots showing the variation of the rate constant (k) with temperature (T) are given

below. The plot that follows Arrhenius equation is –

[JEE 2010]

Sol. (A)

Arrhenius equation : $\mathrm{k}=\mathrm{Ae}^{-\mathrm{E}_{\mathrm{a}} / \mathrm{RT}}$

(taking A and $\mathrm{E}_{\mathrm{a}}$ to be constant, differentiating w.r.t. T)

$\frac{\mathrm{d} \mathrm{k}}{\mathrm{d} \mathrm{T}}=\left[\frac{\mathrm{AE}_{\mathrm{a}}}{\mathrm{RT}^{2}}\right] \mathrm{e}^{-\mathrm{E}_{\mathrm{a}} / \mathrm{RT}}$

assuming A and $\mathrm{E}_{\mathrm{a}}$ to be constant theoretically, the plot should be

ButNo such option is given. Now experimentally, A and $\mathrm{E}_{\mathrm{a}}$ both vary with temperature and k increases as T increases and become very large at infinite temperature. Hence option (A) is correct.


Q. The concentration of R in the reaction $\mathrm{R} \rightarrow \mathrm{P}$ was measured as a function of time and the following data is obtained :

[JEE 2010]

Sol. 0

Zero order reaction because rate is constant with time.

Alternative solution : By hit and trial method, assuming the reaction is of zero order, putting given data in integrated expression for zero order.

the value of k is same from different data so reaction is zero order reaction.


Q. For the first order reaction

$2 \mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g}) \longrightarrow 4 \mathrm{NO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})$

(A) the concentration of the reactant decreases exponentially with time

(B) the half-life of the reaction decreases with increasing temperature.

(C) the half-life of the reaction depends on the initial concentration of the reactant.

(D) the reaction proceeds to 99.6% completion in eight half-life duration.

[JEE 2011]

Sol. (A,B,D)

$\mathrm{A}_{\mathrm{t}}=\mathrm{A}_{0} \mathrm{e}^{-\mathrm{kt}}$

for option (D)

$\frac{1}{\mathrm{t}_{1 / 2}} \ln \frac{100}{50}=\frac{1}{\mathrm{t}_{99.6 \%}} \ln \frac{100}{0.4}$


Q. An organic compound undergoes first-order decomposition . The time taken for its decomposition to 1/8 and 1/10 of its initial concentration are $$

\mathrm{t}_{1 / 8} \text { and } \mathrm{t}_{1 / 10}

$$ What is the value of $$

\frac{\left[\mathrm{t}_{1 / 8}\right]}{\mathrm{t}_{1 / 10}} \times 10 ?\left(\text { take } \log _{10} 2=0.3\right)

$$

[JEE 2012]

Sol. 9


Q. In the reaction :

$$

\mathrm{P}+\mathrm{Q} \longrightarrow \mathrm{R}+\mathrm{S}

$$

the time taken for 75% reaction of P is twice the time taken for 50% reaction of P. The

concentration of Q varies with reaction time as shown in the figure. The overall order of the

reaction is –

(A) 2 (B) 3 (C) 0 (D) 1

[JEE 2013]

Sol. 9


Q. For the elementary reaction $$

\mathrm{M} \rightarrow \mathrm{N}$$, the rate of disappearance of M increases by a factor of 8 upon doubling the concentration of M. The order of the reaction with respect to M is

(A) 4 (B) 3(C) 2(D) 1

[JEE 2014]

Sol. (D)


Q. In dilute aqueous $$

\mathrm{H}_{2} \mathrm{SO}_{4}$$ , the complex diaquodioxalatoferrate(II) is oxidized by $\mathrm{MnO}_{4}^{-}$.For this reaction, the ratio of the rate of change of $$

\left[\mathrm{H}^{+}\right]$$ to the rate of change of [MnO4–] is.

[JEE 2015]

Sol. (B)


Q. The % yield of ammonia as a function of time in the reaction

If this reaction is conducted at $\left(P, T_{2}\right)$, with $\mathrm{T}_{2}>\mathrm{T}_{1}$, the % yield of ammonia as a function of time is represented by –

[JEE 2015]

Sol. (B)

$\therefore$ % yield will increase in initial stages due to increase in net speed

As time proceeds $\Rightarrow \mathrm{r}_{\mathrm{net}}=\mathrm{k}_{\mathrm{f}}\left[\mathrm{N}_{2}\right]\left[\mathrm{H}_{2}\right]^{3}-\mathrm{k}_{\mathrm{b}}\left[\mathrm{NH}_{3}\right]^{2}$

On increasing temp., $\mathrm{k}_{\mathrm{f}} \& \mathrm{k}_{\mathrm{b}}$ increase

but increase of $\mathrm{k}_{\mathrm{b}}$ is more

so % yield will decrease

% yield will increase in initial stage due to enhance speed but as time proceeds , final yield is governed by thermodynamics due to which yield decrease since reaction is exothermic


Q. For a first order reaction A(g)  2B(g) + C(g) at constant volume and 300 K, the

total pressure at the beginning (t = 0) and at time t are P0 and Pt, respectively. Initially, only A

is present with concentration [A]0, and $t_{1 / 3}$ is the time required for the partial pressure

of A to reach $1 / 3^{\mathrm{rd}}$ of its initial value. The correct option(s) is (are) :-

(Assume that all these gases behave as ideal gases)

[JEE Advance 2018]

Sol. (A,D)


Q. Consider the following reversible reaction,

$\mathrm{A}(\mathrm{g})+\mathrm{B}(\mathrm{g}) \square \mathrm{AB}(\mathrm{g})$

The activition energy of the backward reaction exceeds that of the forward reaction by 2RT $\left(\text { in } \mathrm{J} \mathrm{mol}^{-1}\right)$. If the pre-exponential factor of the forward reaction is 4 times that of the reverse reaction, the absolute value of $\Delta \mathrm{G}^{\theta}$(in J mol–1) for the reaction at 300 K is____.

(Given ; ln (2) = 0.7, RT = 2500 J $\mathrm{mol}^{-1}$ at 300 K and G is the Gibbs energy)

[JEE Advance 2018]

Sol. 8500


Coordination Compounds – JEE Advanced Previous Year Questions with Solutions

JEE Advanced Previous Year Questions of Chemistry with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Chemistry will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

Simulator

Previous Years JEE Advance Questions

Q. The spin only magnetic moment value (in Bohr magneton units) of \mathrm{Cr}(\mathrm{CO})_{6} is

(A) 0

(B) 2.84

(C) 4.90

(D) 5.92

[JEE 2009]

Sol. (A)


Q. The compound(s) that exhibit(s) geometrical isomerism is (are) :

(A) $\left[\mathrm{Pt}(\mathrm{en}) \mathrm{Cl}_{2}\right]$

(B) $\left[\mathrm{Pt}(\mathrm{en})_{2}\right] \mathrm{Cl}_{2}$

(C) $\left[\mathrm{Pt}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right] \mathrm{Cl}_{2}$

(D) $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\right]$

[JEE 2009]

Sol. (C,D)


Q. The number of water molecule(s) directly bonded to the metal centre in $\mathrm{CuSO}_{4}$. $5 \mathrm{H}_{2} \mathrm{O}$ is.

[JEE 2009]

Sol. 4


Q. The ionization isomer of $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4} \mathrm{Cl}\left(\mathrm{NO}_{2}\right)\right] \mathrm{Cl}$ is –

(A) $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\left(\mathrm{O}_{2} \mathrm{N}\right)\right] \mathrm{Cl}_{2}$

(B) $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4} \mathrm{Cl}_{2}\right]\left(\mathrm{NO}_{2}\right)$

(C) $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4} \mathrm{Cl}(\mathrm{ONO})\right] \mathrm{Cl}$

(D) $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4} \mathrm{Cl}_{2}\left(\mathrm{NO}_{2}\right)\right] \cdot \mathrm{H}_{2} \mathrm{O}$

Sol. (B)

Ionisation isomers differ in ions in solution thus, ionisation isomer of $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4} \mathrm{Cl}\left(\mathrm{NO}_{2}\right)\right] \mathrm{Cl}$ is $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4} \mathrm{Cl}_{2}\right]\left(\mathrm{NO}_{2}\right)$. Because

$\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4} \mathrm{Cl}\left(\mathrm{NO}_{2}\right)\right] \mathrm{Cl} \longrightarrow\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4} \operatorname{Cl}\left(\mathrm{NO}_{2}\right)\right]^{+}+\mathrm{Cl}^{-}$ (Given compound)

$\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4} \mathrm{Cl}_{2}\right]\left(\mathrm{NO}_{2}\right) \longrightarrow\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4} \mathrm{Cl}_{2}\right]^{+}+\mathrm{NO}_{2}^{-}$ Ionisation isomer of given compound.


Q. Total number of geometrical isomers for the complex $\left[\mathrm{RhCl}(\mathrm{CO})\left(\mathrm{PPh}_{3}\right)\left(\mathrm{NH}_{3}\right)\right]$ is.

[JEE 2010]

Sol. 3

$\left[\mathrm{RhCl}(\mathrm{Co})\left(\mathrm{PPh}_{3}\right)\left(\mathrm{NH}_{3}\right)\right]$

$\mathrm{dsp}^{2},$ square planar, total 3 geometrical isomer.


Q. The correct structure of ethylenediaminetetraacetic acid (EDTA) is –

[JEE 2010]

Sol. (C)

The correct structure of ethylenediaminetetra acetic acid (EDTA) is


Q. Geometrical shapes of the complexes formed by the reaction of $\mathrm{Ni}^{2+}$ with $\mathrm{Cl}^{-}, \mathrm{CN}$ and $\mathrm{H}_{2} \mathrm{O}$ respectively, are –

(A) octahedral, tetrahedral and square planar

(B) tetrahedral, square planar and octahedral

(C) square planar, tetrahedral and octahedral

(D) octahedral, square planar and octahedral

[JEE 2011]

Sol. (B)


Q. Among the following complexes (K–P)

$\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right](\mathbf{K}),\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}(\mathrm{L}), \mathrm{Na}_{3}\left[\mathrm{Co}\left(\text { oxalate) }_{3}\right](\mathrm{M}),\left[\mathrm{Ni}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}\right] \mathrm{Cl}_{2}(\mathrm{N})\right.$$\mathrm{K}_{2}\left[\mathrm{Pt}(\mathrm{CN}) {4}\right](\mathbf{O})$ and $\left[\mathrm{Zn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]\left(\mathrm{NO}_{3}\right)_{2}(\mathbf{P})$

The diamagnetic complex are –

(A) K, L, M, N               (B) K, M, O, P               (C) L, M, O, P                (D) L, M, N, O

[JEE 2011]

Sol. (C)


Q. The volume (in mL) of 0.1M AgNO3 required for complete precipitation of chloride ions present in 30 mL of 0.01M solution of $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6} \mathrm{Cl}\right] \mathrm{Cl}_{2}$, as silver chloride is close to.

[JEE 2011]

Sol. 6


Q. As per IUPAC nomenclature, the name of the complex $\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\left(\mathrm{NH}_{3}\right)_{2}\right] \mathrm{Cl}_{3}$ is :

(A) Tetraaquadiaminecobalt(III) chloride

(B) Tetraaquadiamminecobalt(III) chloride

(C) Diaminetetraaquacobalt(III) chloride

(D) Diamminetetraaquacobalt(III) chloride

[JEE 2012]

Sol. (D)

$\left[\mathrm{C}_{\alpha}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{3}\right.$

Diamminetetraaquacobalt(III) chloride

$\frac{V I B G Y O R}{\lambda-v^{-} E^{-}}$


Q. The colour of light absorbed by an aqueous solution of $\mathrm{CuSO}_{4}$ is –

(A) orange-red (B) blue-green (C) yellow (D) violet

[JEE 2012]

Sol. (A)


Q. $\mathrm{NiCl}_{2}\left\{\mathrm{P}\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2}\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)\right\}_{2}$ exhibits temperature dependent magnetic behavior (paramagnetic/diamagnetic). The coordination geometries of Ni2+ in the paramagnetic and diamagnetic states are respectively :

(A) tetrahedral and tetrahedral

(B) square planar and square planar

(C) tetrahedral and square planar

(D) square planar and tetrahedral

[JEE 2012]

Sol. (C)


Q. Consider the following complex ions P, Q and R ,

$\mathbf{P}=\left[\mathrm{FeF}_{6}\right]^{3-}, \mathbf{Q}=\left[\mathrm{V}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$ and $\mathbf{R}=\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$

The correct order of the complex ions, according to their spin-only magnetic moment values (in B.M.) is –

(A) R < Q < P (B ) Q < R < P (C) R < P < Q (D) Q < P < R

[JEE 2013]

Sol. (B)


Q. EDTA $^{4}$ is ethylenediaminetetraacetate ion. The total number of $\mathrm{N}-\mathrm{Co}-\mathrm{O}$ bond angles in $[\mathrm{Co}(\mathrm{EDTA})]^{-1}$ complex ion is –

[JEE 2013]

Sol. 8


Q. The pair(s) of coordination complex/ion exhibiting the same kind of isomerism is(are) –

(A) $\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2}$ and $\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{Cl}$

(B) $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right]^{+}$ and $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2}\left(\mathrm{H}_{2} \mathrm{O}\right) \mathrm{Cl}\right]^{+}$

(C) $\left[\mathrm{CoBr}_{2} \mathrm{Cl}_{2}\right]^{2-}$ and $\left[\mathrm{PtBr}_{2} \mathrm{Cl}_{2}\right]^{2-}$

(D)$\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{3}\left(\mathrm{NO}_{3}\right)\right] \mathrm{Cl}$ and $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{3} \mathrm{Cl}\right] \mathrm{Br}$

[JEE 2013]

Sol. (B,D)


Q. Match each coordination compound in List-I with an appropriate pair of characteristics from List-II and select the correct answer using the code given below the lists.

[JEE Adv. 2014]

Sol. (B)

(P) $\left[\mathrm{Cr}^{\mathrm{III}}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{Cl}:$

(1) Complex given in (P) is Paramagnetic & show two geometrical

(3 unpaired electrons) isomerism (cis and trans) (does not show ionization isomer)

(Q) $\left[\mathrm{Ti}^{\mathrm{III}}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{Cl}\right]\left(\mathrm{NO}_{3}\right)_{2}$

(2) Complex given in (Q) is paramagnetic show ionization(1 unpaired electrons) isomerism

(R) $\left[\mathrm{Pt}^{\mathrm{Il}}(\mathrm{en})\left(\mathrm{NH}_{3}\right) \mathrm{Cl}\right] \mathrm{NO}_{3}$

(3) Complex given in (R) is diamagnetic and show ionization(1 unpaired electrons) isomerism

(S)$\left[\mathrm{Co}^{\mathrm{III}}\left(\mathrm{NH}_{3}\right)_{4}\left(\mathrm{NO}_{3}\right)_{2}\right] \mathrm{NO}_{3}$

(4) Complex given in (S) is diamagnetic does not show ionization (0 unpaired electrons) isomerism show geometrical isomerism


Q. A list of species having the formula $\mathrm{XZ}_{4}$ is given below :

$\mathrm{XeF}_{4}, \mathrm{SF}_{4}, \mathrm{SF}_{4}, \mathrm{BF}_{4}^{-}, \mathrm{BrF}_{4}^{-},\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+},\left[\mathrm{FeCl}_{4}\right]^{2-},\left[\mathrm{CoCl}_{4}\right]^{2-}$ and $\left[\mathrm{PtCl}_{4}\right]^{2-}$.

Defining shape on the basis of the location of X and Z atoms, the total number of species having a square planar shape is

[JEE Adv. 2014]

Sol.


Q. The geometries of the ammonia complexes of $\mathrm{Ni}^{2+}, \mathrm{Pt}^{2+}$ and $\mathrm{Zn}^{2+}$ , respectively , are :

(A) octahedral, square planar and tetrahederal

(B) square planar, octahederal and tetrahederal

(C) tetrahederal, square planar and octahederal

(D) octahederal , tetrahederal and square planar

[JEE – Adv. 2016]

Sol. (A)


Q. Among $\left[\mathrm{Ni}(\mathrm{CO}), \mathrm{I},\left[\mathrm{NiCl}_{4}\right]^{2},\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right), \mathrm{Cl}_{2}\right] \mathrm{Cl}, \mathrm{Na}_{3}\left[\mathrm{CoF}_{6}\right], \mathrm{NaO}_{2} \mathrm{and} \mathrm{Co}_{2}\right.$, the total number of paramagnetic compounds is –

(A) 2             (B) 3              (C) 4                (D) 5

[JEE – Adv. 2016]

Sol. (B)


Q. The number of geometric isomers possible for the complex$\left[\mathrm{CoL}_{2} \mathrm{Cl}_{2}\right]^{-}\left(\mathrm{L}=\mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{CH}_{2} \mathrm{O}^{-}\right)$ is

[JEE – Adv. 2016]

Sol. 5


Q. Addition of excess aqueous ammonia to a pink coloured aqueous solution of $\mathrm{MCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}$ (X) and $\mathrm{NH}_{4} \mathrm{Cl}$ gives an octahedral complex Y in the presence of air. In aqueous solution, complex Y behaves as 1 : 3 electrolyte. The reaction of X with excess HCl at room temperature results in the formation of a blue coloured complex Z. The calculated spin only magnetic moment of X and Z is 3.87 B.M., whereas it is zero for complex Y.

Among the following options, which statements is(are) correct ?

(A) The hybridization of the central metal ion in Y is d2sp3

(B) Z is tetrahedral complex

(C) Addition of silver nitrate to Y gives only two equivalents of silver chloride

(D) When X and Z are in equilibrium at 0°C, the colour of the solution is pink

[JEE – Adv. 2017]

Sol. (A,B,D)

(A) Hybridisation of $(\mathrm{Y})$ is $\mathrm{d}^{2} \mathrm{sp}^{3}$ as $\mathrm{NH}_{3}$ is strong field ligand

(B) $\left[\mathrm{CoCl}_{4}\right]^{2-}$ have $\mathrm{sp}^{3}$ hybridisation as $\mathrm{Cl}^{-}$ is weak field ligand

When ice is added to the solution the equilibrium shifts right hence pink colour will remain predominant

So, correct answer is (A,B& D)


A

Q. The correct statement(s) regarding the binary transition metal carbonyl compounds is (are)

(Atomic numbers : Fe = 26, Ni = 28)

(A) Total number of valence shell electrons at metal centre in $\mathrm{Fe}(\mathrm{CO})_{5}$ or $\mathrm{Ni}(\mathrm{CO})_{4}$ is 16

(B) These are predominantly low spin in nature

(C) Metal – carbon bond strengthens when the oxidation state of the metal is lowered

(D) The carbonyl C–O bond weakens when the oxidation state of the metal is increased

[JEE – Adv. 2018]

Sol. (B,C)

(A) $\left[\mathrm{Fe}\left(\mathrm{CO}_{5}\right)\right] \&\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]$ complexes have 18-electrons in their valence shell.

(B) Carbonyl complexes are predominantly low spin complexes due to strong ligand field.

(C) As electron density increases on metals (with lowering oxidation state on metals), the extent of synergic bonding increases. Hence M–C bond strength increases

(D) While positive charge on metals increases and the extent of synergic bond decreases and hence C–O bond becomes stronger.


Q. Among the species given below, the total number of diamagnetic species is____.

H atom, $\mathrm{NO}_{2}$ monomer, $\mathrm{O}_{2}^{-}$ (superoxide), dimeric sulphur in vapour phase,

$\mathrm{Mn}_{3} \mathrm{O}_{4},\left(\mathrm{NH}_{4}\right)_{2}\left[\mathrm{FeCl}_{4}\right],\left(\mathrm{NH}_{4}\right)_{2}\left[\mathrm{NiCl}_{4}\right], \mathrm{K}_{2} \mathrm{MnO}_{4}, \mathrm{K}_{2} \mathrm{CrO}_{4}$

[JEE – Adv. 2018]

Sol. (1)


Q. The ammonia prepared by treating ammonium sulphate with calcium hydroxide is completely used by $\mathrm{NiCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}$ to form a stable coordination compound. Assume that both the reactions are 100% complete. If 1584 g of ammonium sulphate and 952g of NiCl2.6H2O are used in the preparation, the combined weight (in grams) of gypsum and the nickel-ammonia coordination compound thus produced is___.

(Atomic weights in g $\mathrm{mol}^{-1}$: H = 1, N = 14, O = 16, S = 32, Cl = 35.5, Ca = 40, Ni = 59)

(A) It has two geometrical isomers

(B) It will have three geometrical isomers if bidentate ‘en’ is replaced by two cyanide ligands

(C) It is paramagnetic

(D) It absorbs light at longer wavelength as compared to $\left[\mathrm{Co}(\mathrm{en})\left(\mathrm{NH}_{3}\right)_{4}\right]^{3+}$

[JEE – Adv. 2018]

Sol. 2992

Total mass = 12 × 172 + 4 × 232 = 2992 g


Q. The correct option(s) regarding the complex $\left[\mathrm{Co}(\mathrm{en})\left(\mathrm{NH}_{3}\right)_{3}\left(\mathrm{H}_{2} \mathrm{O}\right)\right]^{3+}:-$

$\left(\mathrm{en}=\mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\right)$ is (are)

[JEE – Adv. 2018]

Sol. (A,B,D)


Q. Match each set of hybrid orbitals from LIST-I with complex (es) given in LIST-II.

[JEE – Adv. 2018]

Sol. (C)


Solid State – JEE Advanced Previous Year Questions with Solutions

JEE Advanced Previous Year Questions of Chemistry with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Chemistry will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

Simulator

Previous Years JEE Advance Questions

Q. The correct statement(s) regarding defects in solid is (are)

(A) Frenkel defect is usually favoured by a very small difference in the sizes of cation and

anion.

(B) Frenkel defect is a dislocation defect

(C) Trapping of an electron in the lattice leads to the formation of F‑center.

(D) Schottky defects have no effect on the physical properties of solids.

[JEE 2009]

Sol. (B,C)


Q. The packing effeciency of the two-dimensional square unit cell shown below is

(A) 39.27%         (B) 68.02%            (C) 74.05%             (D) 78.54%

[JEE-2010]

Sol. (D)


Q. The number of hexagonal faces that present in a truncated octahedron is.

[JEE-2011]

Sol. 8


Q. A compound $\mathrm{M}_{\mathrm{p}} \mathrm{X}_{\mathrm{q}}$ has cubic close packing (ccp) arrangement of X. Its unit cell structure is shown below. The empirical formula of the compound is :

(A) MX

(B) $\mathrm{MX}_{2}$

(C) $\mathrm{M}_{2} \mathrm{X}$

(D) $\mathrm{M}_{5} \mathrm{X}_{14}$

[JEE-2012]

Sol. (B)


Q. The arrangement of $\mathrm{X}^{-}$ ions around $\mathrm{A}^{+}$ ion in solid AX is given in the figure (not drawn to scale). If the radius of $\mathrm{X}^{-}$ is 250 pm, the radius of $\mathrm{A}^{+}$ is –

(A) 104 pm         (B) 125 pm          (C) 183 pm         (D) 57 pm

[JEE-2013]

Sol. (A)


Q. The correct statement(s) for cubic close packed (ccp) three dimensional

structure is (are)

(A) The number of the nearest neighbours of an atom present in the topmost layer is 12

(B) The efficiency of atom packing is 74%

(C) The number of octahedral and tetrahedral voids per atom are 1 and 2, respectively

(D) The unit cell edge length is $2 \sqrt{2}$ times the radius of the atom

[JEE – Adv. 2016]

Sol. (B,C,D)

CCP is ABC ABC ….. type packing

(A) In topmost layer, each atom is in contact with 6 atoms in same layer and 3 atoms below

this layer.


Q. A crystalline solid of a pure substance has a face-centred cubic structure with a cell edge of

400 pm. If the density of the substance in the crystal is 8g $\mathrm{cm}^{-3}$, then the number of atoms present in 256g of the crystal is $\mathrm{N} \times 10^{24}$. The value of N is

[JEE – Adv. 2017]

Sol. 2


Q. Consider an ionic solid MX with NaCl structure. Construct a new structure (Z) whose unit cell is constructed from the unit cell of MX following the sequential instructions given below. Neglect the charge balance.

(i) Remove all the anions (X) except the central one

(ii) Replace all the face centered cations (M) by anions (X)

(iii) Remove all the corner cations (M)

(iv) Replace the central anion (X) with cation (M)

[JEE – Adv. 2017]

Sol. 3


Surface Chemistry – JEE Advanced Previous Year Questions with Solutions

JEE Advanced Previous Year Questions of Chemistry with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Chemistry will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

Simulator

Previous Years JEE Advance Questions

Q. Among the electrolytes $\mathrm{Na}_{2} \mathrm{SO}_{4}, \mathrm{CaCl}_{2}, \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}$ and $\mathrm{NH}_{4} \mathrm{Cl}$, the most effective coagulation agent for $\mathrm{Sb}_{2} \mathrm{S}_{3}$ sol is

(A) $\mathrm{Na}_{2} \mathrm{SO}_{4}$

(B) $\mathrm{CaCl}_{2}$

(C) $\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}$

(D) $\mathrm{NH}_{4} \mathrm{Cl}$

JEE 2009

Sol. (C)

$\mathrm{Sb}_{2} \mathrm{S}_{3}$ forms negatively charged sol. Hence cation with

greatest charge density is most effective in bringing about coagulation C.F. Hardy sulz rule.


Q. The correct statement(s) pertaining to the adsorption of a gas on a solid surface is

(are) –

(A) Adsorption is always exothermic

(B) Physisorption may transform into chemisorption at high temperature

(C) Physisorption increases with increasing temperature but chemisorption decreases with

increasing temperature

(D) Chemisorption is more exothermic than physisorption, however it is very slow due to higher energy of activation

JEE 2011

Sol. (A,B,D)

general properties.


Q. Choose the correct reason(s) for the stability of the lyophobic colloidal particle.

(A) Preferential adsorption of ions on their surface from the solution

(B) Preferential adsorption of solvent on their surface from the solution

(C) Attraction between different particles having opposite charges on their surface

(D) Potential difference between the fixed layer and the diffused layer of opposite charges around the colloidal particles

JEE 2012

Sol. (A,D)

Theory Based


Q. The given graphs / data I, II, III and IV represent general trends observed for different physisorption and chemisorption processes under mild conditions of temperature and pressure. Which of the following choice(s) about I, II, III and IV is (are) correct ?

(A) I is physisorption and II is chemisorption

(B) I is physisorption and III is chemisorption

(C) IV is chemisorption and II is chemisorption

(D) IV is chemisorption and III is chemisorption

JEE 2012

Sol. (A,C)

For physical adsorption $\rightarrow$ favourable conditions is decrease in temp.

for chemical adsorption $\rightarrow$ chemical bond

Occurs $\rightarrow$ PE $\downarrow$ with bonding and increase in temp..


Q. Methylene blue, from its aqueous solution, is adsorbed on activated charcoal at $25^{\circ} \mathrm{C}$. For this process, the correct statement is

(A) The adsorption requires activation at $25^{\circ} \mathrm{C}$

(B) The adsorption is accompanied by a decrease in enthalpy

(C) The adsorption increases with increase of temperature

(D) The adsorption is irreversible

JEE Adv. 2013

Sol. (B)

Explanation for physical adsorption

• Activation energy is very low

• Physical adsorption is an exothermic process

• Physical adsorption decreases with increase in temperature

• Physical adsorption is reversible


Q. The qualitative sketches I , II and III given below show the variation of surface tension with molar concentration of three different aqueous solutions of KCl, $\mathrm{CH}_{3} \mathrm{OH}$ and $\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{11} \mathrm{OSO}_{3}^{-} \mathrm{Na}^{+}$ at room temperature. The correct assignment of the sketches is –

JEE Adv. 2016

Sol. (D)

Water has large surface tension due to very strong interaction. Generally adding organic

derivatives to water decreases its surface tension due to hydrophobic interaction.

In case III, hydrophobic interaction is stronger than case I causing surface tension to decrease more rapidly.


Q. The correct statement(s) about surface properties is (are)

(A) Cloud is an emulsion type of colloid in which liquid is dispersed phase and gas is dispersion

medium

(B) Adsorption is accompanied by decrease in enthalpy and decrease in entropy of the system.

(C) Brownian motion of colloidal particles does not depend on the size of the particles but

depends on viscosity of the solution.

(D) The critical temperatures of ethane and nitrogen and 563 K and 126 K, respectively. The adsorption of ethane will be more than that of nitrogen on same amount of activated charcoal at a given temperature.

JEE Adv. 2017

Sol. (B,D)

(A) Emulsion is liquid in liquid type colloid.

(B) For adsorption, $\Delta \mathrm{H}<0 \& \Delta \mathrm{S}<0$

(C) Smaller the size and less viscous the dispersion medium, more will be the brownian motion.

(D) Higher the $\mathrm{T}_{\mathrm{C}}$ , greater will be the extent of adsorption.


Monotonicity – JEE Advanced Previous Year Questions with Solutions

JEE Advanced Previous Year Questions of Math with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of mathematics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral also provides complete chapter-wise notes of Class 11th and 12th both for all subjects.

Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more.

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Mathematics

Maxima and Minima – JEE Advanced Previous Year Questions with Solutions

JEE Advanced Previous Year Questions of Math with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of mathematics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral also provides complete chapter-wise notes of Class 11th and 12th both for all subjects.

Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more.

Download eSaral app for free study material and video tutorials.

Mathematics

3D Geometry- JEE Advanced Previous Year Questions with Solutions

JEE Advanced Previous Year Questions of Math with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of mathematics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral also provides complete chapter-wise notes of Class 11th and 12th both for all subjects.

Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more.

Download eSaral app for free study material and video tutorials.

Q. (A) Let $\mathrm{P}(3,2,6)$ be a point in space and $\mathrm{Q}$ be a point on the line $\overrightarrow{\mathrm{r}}=(\hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}})+\mu(-3 \hat{\mathrm{i}}+\hat{\mathrm{j}}+5 \hat{\mathrm{k}}) .$ Then the value of $\mu$ for which the vector $\overline{\mathrm{PQ}}$ is parallel to the plane $x-4 y+3 z=1$ is –

(A) $\frac{1}{4}$

(B) $-\frac{1}{4}$

(C) $\frac{1}{8}$

(D) $-\frac{1}{8}$

(B) A line with positive direction cosines passes through the point P (2, –1, 2) and makes equal angles with the coordinate axes. The line meets the plane 2x + y + z = 9 at point Q. The length of the line segment PQ equals –

(A) 1

(B) $\sqrt{2}$

(C) $\sqrt{3}$

(D) 2

(C) Let $(\mathrm{x}, \mathrm{y}, \mathrm{z})$ be points with integer coordinates satisfying the system of homogeneous equations $: 3 x-y-z=0 ;-3 x+z=0 ;-3 x+2 y+z=0 .$ Then the number of such points for which $\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2} \leq 100$ is

[JEE 2009, 3+3+4]

Sol. ( (a) $A ;(b) C ;(c) 7$ )


Q. (A) Equation of the plane containing the straight line $\frac{\mathrm{x}}{2}=\frac{\mathrm{y}}{3}=\frac{\mathrm{z}}{4}$ and perpendicular to the plane containing the straight lines $\frac{x}{3}=\frac{y}{4}=\frac{z}{2}$ and $\frac{x}{4}=\frac{y}{2}=\frac{z}{3}$

(A) x + 2y – 2z = 0

(B) 3x + 2y – 2z = 0

(C) x – 2y + z = 0

(D) 5x + 2y – 4z = 0

(B) If the distance of the point $\mathrm{P}(1,-2,1)$ from the plane $\mathrm{x}+2 \mathrm{y}-2 \mathrm{z}=\alpha$ where $\alpha>0,$ is $5,$ then the foot of the perpendicular from $P$ to the plane is-

(A) $\left(\frac{8}{3}, \frac{4}{3},-\frac{7}{3}\right)$

(B) $\left(\frac{4}{3},-\frac{4}{3}, \frac{1}{3}\right)$

(C) $\left(\frac{1}{3}, \frac{2}{3}, \frac{10}{3}\right)$

(D) $\left(\frac{2}{3},-\frac{1}{3}, \frac{5}{2}\right)$

(C) If the distance between the plane Ax – 2y + z = d and the plane containing the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}$ is $\sqrt{6},$ then $|d|$ is

(D) Match the statements in Column-I with the values in Column-II.

[JEE 2010, 3+5+3+(2+2+2+2)]

Sol. ((A) $\mathrm{C} ;(\mathrm{B}) \mathrm{A} ;(\mathrm{C}) 6 ;(\mathrm{D})(\mathrm{A}) \mathrm{t}(\mathrm{B}) \mathrm{p}, \mathrm{r}(\mathrm{C}) \mathrm{q}(\mathrm{D}) \mathrm{r}$ )

(a) Normal vector to the plane containing the

lines $\frac{x}{3}=\frac{y}{4}=\frac{z}{2}$ and $\frac{x}{4}=\frac{y}{2}=\frac{z}{3}$ is

$\hat{n}=\left|\begin{array}{lll}{\hat{i}} & {\hat{j}} & {\hat{k}} \\ {3} & {4} & {2} \\ {4} & {2} & {3}\end{array}\right|=8 \hat{i}-\hat{j}-10 \hat{k}$

Let direction ratios of required plane be a, b, c.

Now 8a – b – 10c = 0

and $2 \mathrm{a}+3 \mathrm{b}+4 \mathrm{c}=0\left(\because \text { plane contains the line } \frac{\mathrm{x}}{2}=\frac{\mathrm{y}}{3}=\frac{\mathrm{z}}{4}\right)$

$\Rightarrow \frac{a}{1}=\frac{b}{-2}=\frac{c}{1}$

$\cdot$ plane contains the line, which passes through origin, hence origin lies on a plane.

$\Rightarrow$ equation of required plane is $x-2 y+z=0$

(b) $\quad \because \quad\left|\frac{1-4-2-\alpha}{3}\right|=5$

$\Rightarrow \alpha=10,-20$

$\Rightarrow \alpha=10 \because \alpha>0$


Q. (A) The point P is the intersection of the straight line joining the points Q(2,3,5) and R(1,–1,4) with the plane 5x – 4y – z = 1. If S is the foot of the perpendicular drawn from the point T(2,1,4) to QR, then the length of the line segment PS is –

(A) $\frac{1}{\sqrt{2}}$

(B) $\sqrt{2}$

(C) 2

(D) $2 \sqrt{2}$

(B) The equation of a plane passing through the line of intersection of the planes x + 2y $+3 \mathrm{z}=2$ and $\mathrm{x}-\mathrm{y}+\mathrm{z}=3$ and at a distance $\frac{2}{\sqrt{3}}$ from the point $(3,1,-1)$ is

(A) $5 x-11 y+z=17$

(B) $\sqrt{2} x+y=3 \sqrt{2}-1$

(C) $x+y+z=\sqrt{3}$

(D) $x-\sqrt{2} y=1-\sqrt{2}$

(C) If the straight lines $\frac{x-1}{2}=\frac{y+1}{k}=\frac{z}{2}$ and $\frac{x+1}{5}=\frac{y+1}{2}=\frac{z}{k}$ are coplanar, then the plane(s) containing these two lines is(are)

(A) y + 2z = –1

(B) y + z = –1

(C) y – z = –1

(D) y – 2z = –1

[JEE 2012, 3+3+4]

Sol. ((a) $A ;(b) A ;(c) B, C$)

(a) Line QR :

$\frac{x-2}{1}=\frac{y-3}{4}=\frac{z-5}{1}=\lambda$

Any point on line QR :

$(\lambda+2,4 \lambda+3, \lambda+5)$

$\therefore$ Point of intersection with plane :

$5 \lambda+10-16 \lambda-12-\lambda-5=1$

$\Rightarrow \lambda=-\frac{2}{3}$

$\therefore \mathrm{P}\left(\frac{4}{3}, \frac{1}{3}, \frac{13}{3}\right)$


Q. Perpendiculars are drawn from points on the line $\frac{x+2}{2}=\frac{y+1}{-1}=\frac{z}{3}$ to the plane $x$ $+\mathrm{y}+\mathrm{z}=3 .$ The feet of perpendiculars lie on the line

(A) $\frac{x}{5}=\frac{y-1}{8}=\frac{z-2}{-13}$

(B) $\frac{x}{2}=\frac{y-1}{3}=\frac{z-2}{-5}$

(C) $\frac{x}{4}=\frac{y-1}{3}=\frac{z-2}{-7}$

(D) $\frac{x}{2}=\frac{y-1}{-7}=\frac{z-2}{5}$

[JEE-Advanced 2013, 2]

Sol. (D)


Q. A line $\ell$ passing through the origin is perpendicular to the lines

$\ell_{1}:(3+\mathrm{t}) \hat{\mathrm{i}}+(-1+2 \mathrm{t}) \hat{\mathrm{j}}+(4+2 \mathrm{t}) \hat{\mathrm{k}},-\infty<\mathrm{t}<\infty$

$\ell_{2}:(3+2 s) \hat{\mathrm{i}}+(3+2 \mathrm{s}) \hat{\mathrm{j}}+(2+\mathrm{s}) \hat{\mathrm{k}},-\infty<\mathrm{s}<\infty$ Then, the coordinate(s) of the point(s) on $\ell_{2}$

at a distance of $\sqrt{17}$ from the point of intersection of $\ell$ and $\ell_{1}$ is (are) –

(A) $\left(\frac{7}{3}, \frac{7}{3}, \frac{5}{3}\right)$

(B) (–1,–1,0)

(C) (1,1,1)

(D) $\left(\frac{7}{9}, \frac{7}{9}, \frac{8}{9}\right)$

[JEE-Advanced 2013, 4, (–1)]

Sol. (B,D)


Q. Two lines $L_{1}: x=5, \frac{y}{3-\alpha}=\frac{z}{-2}$ and $L_{2}: x=\alpha, \frac{y}{-1}=\frac{z}{2-\alpha}$ are coplanar. Then $\alpha$ can take value(s)

(A) 1              (B) 2             (C) 3                  (D) 4

[JEE-Advanced 2013, 3, (–1)]

Sol. (A,D)

$\mathrm{L}_{1}: \frac{\mathrm{x}-5}{0}=\frac{\mathrm{y}}{3-\alpha}=\frac{\mathrm{z}}{-2}$

$\mathrm{L}_{2}: \frac{\mathrm{x}-\alpha}{0}=\frac{\mathrm{y}}{-1}=\frac{\mathrm{z}}{2-\alpha}$

for lines to be coplanar

$\left|\begin{array}{ccc}{5-\alpha} & {0} & {0} \\ {0} & {3-\alpha} & {-2} \\ {0} & {-1} & {2-\alpha}\end{array}\right|=0$

$\Rightarrow \quad(5-\alpha)((3-\alpha)(2-\alpha)-2)=0$

$\Rightarrow \quad(5-\alpha)\left(\alpha^{2}-5 \alpha+4\right)=0$

$\Rightarrow \quad \alpha=1,4,5$


Q. Consider the lines $\mathrm{L}_{1}: \frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}}{-1}=\frac{\mathrm{z}+3}{1}, \mathrm{L}_{2}: \frac{\mathrm{x}-4}{1}=\frac{\mathrm{y}+3}{1}=\frac{\mathrm{z}+3}{2}$ and the planes $\mathrm{P}_{1}: 7 \mathrm{x}+\mathrm{y}+2 \mathrm{z}=3, \mathrm{P}_{2}: 3 \mathrm{x}+5 \mathrm{y}-6 \mathrm{z}=4 .$ Let $\mathrm{ax}+\mathrm{by}+\mathrm{cz}=\mathrm{d}$ be the

equation of the plane passing through the point of intersection of lines $L_{1}$ and $\mathrm{L}_{2}$ and perpendicular to planes $\mathrm{P}_{1}$ and $\mathrm{P}_{2} .$ Match List-I with List-II and select the correct answer using the code given below the lists.

[JEE-Advanced 2013, 3, (–1)]

Sol. (A)

For point of intersection of $L_{1}$ and $L_{2}$

$\left\{\begin{array}{l}{2 \lambda+1=\mu+4} \\ {-\lambda=\mu-3} \\ {\lambda-3=2 \mu-3}\end{array}\right.$

$\Rightarrow \mu=1$

$\Rightarrow$ point of intersction is $(5,-2,-1)$

Now, vector normal to the plane is $\overrightarrow{\mathrm{n}}_{1} \times \overrightarrow{\mathrm{n}}_{2}=\left|\begin{array}{ccc}{\hat{\mathrm{i}}} & {\hat{\mathrm{j}}} & {\hat{\mathrm{k}}} \\ {7} & {1} & {2} \\ {3} & {5} & {-6}\end{array}\right|$

$=-16(\hat{i}-3 \hat{j}-2 \hat{k})$

Let equation of required plane be $x-3 y-2 z=\alpha$

$\because$ it passes through $(5,-2,-1)$

$\therefore \alpha=13$

$\Rightarrow$ equation of plane is $x-3 y-2 z=13$


Q. From a point $\mathrm{P}(\lambda, \lambda, \lambda),$ perpendiculars $\mathrm{PQ}$ and $\mathrm{PR}$ are drawn respectively on the lines $\mathrm{y}=$ $\mathrm{x}, \mathrm{z}=1$ and $\mathrm{y}=-\mathrm{x}, \mathrm{z}=-1 .$ If $\mathrm{P}$ is such that $\angle \mathrm{QPR}$ is a right angle, then the possible value(s) of $\lambda$ is (are)

(A) $\sqrt{2}$              (B) 1            (C) –1              (D) $-\sqrt{2}$

[JEE(Advanced)-2014, 3]

Sol. (C)

Line $\mathrm{L}_{1}$ given by $\mathrm{y}=\mathrm{x} ; \mathrm{z}=1$ can be expressed as

$\mathrm{L}_{1}: \frac{\mathrm{x}}{1}=\frac{\mathrm{y}}{1}=\frac{\mathrm{z}-1}{0}$

Similarly $\mathrm{L}_{2}(\mathrm{y}=-\mathrm{x} ; \mathrm{z}=-1)$ can be expressed as

$\mathrm{L}_{2}: \frac{\mathrm{x}}{1}=\frac{\mathrm{y}}{-1}=\frac{\mathrm{z}+1}{0}$

Let any point $\mathrm{Q}(\alpha, \alpha, 1)$ on $\mathrm{L}_{1}$ and $\mathrm{R}(\beta,-\beta,-1)$ on $\mathrm{L}_{2}$

Given that $\mathrm{PQ}$ is perpendicular to $\mathrm{L}_{1}$

$\Rightarrow(\lambda-\alpha) .1+(\lambda-\alpha) \cdot 1+(\lambda-1) \cdot 0=0 \Rightarrow \lambda=\alpha$

$\therefore \mathrm{Q}(\lambda, \lambda, 1)$

Similarly PR is perpendicular to L $_{2}$ $(\lambda-\beta) \cdot 1+(\lambda+\beta)(-1)+(\lambda+1) \cdot 0=0 \Rightarrow \beta=0$

$\therefore \mathrm{R}(0,0,-1)$

Now as given

$\Rightarrow \overrightarrow{\mathrm{PR}} \cdot \overrightarrow{\mathrm{PQ}}=0$

$0 . \lambda+0 . \lambda+(\lambda-1)(\lambda+1)=0$

$\lambda \neq 1$ as $\mathrm{P} \& \mathrm{Q}$ are different points

$\Rightarrow \lambda=-1$


Q. In $\mathbb{D}^{3},$ consider the planes $P_{1}: y=0$ and $P_{2}: x+z=1 .$ Let $P_{3}$ be a plane, different from $\mathrm{P}_{1}$ and $\mathrm{P}_{2},$ which passes through the intersection of $\mathrm{P}_{1}$ and $\mathrm{P}_{2} .$ If the distance of the point $(0,1,0)$ from $P_{3}$ is 1 and the distance of a point $(\alpha, \beta, \gamma)$ from $P_{3}$ is $2,$ then which of the following relations is (are) true?

(A) $2 \alpha+\beta+2 \gamma+2=0$

(B) $2 \alpha-\beta+2 \gamma+4=0$

(C) $2 \alpha+\beta-2 \gamma-10=0$

(D) $2 \alpha-\beta+2 \gamma-8=0$

[JEE 2015, 4M, –2M]

Sol. (B,D)


Q. In $\square^{3},$ let $L$ be a straight line passing through the origin. Suppose that all the points on $L$ are at a constant distance from the two planes $P_{1}: x+2 y-z+1=0$ and $P_{2}: 2 x-y+$ $\mathrm{z}-1=0 .$ Let $\mathrm{M}$ be the locus of the feet of the perpendiculars drawn from the points on L to the plane $P_{1} .$ Which of the following points lie(s) on M?

[JEE 2015, 4M, –2M]

Sol. (A,B)

Straight line ‘L’ is parallel to line of intersection of plane $\mathrm{P}_{1} \&$ plane $\mathrm{P}_{2}$

$\therefore$ Equation of line $^{\prime} \mathbf{L}^{\prime}$ is

$\frac{x}{1}=\frac{y}{-3}=\frac{z}{-5}=\lambda$

$\frac{\alpha-\lambda}{1}=\frac{\beta+3 \lambda}{2}=\frac{\gamma+5 \lambda}{-1}=\mathrm{k}$

$\left.\begin{array}{l}{\alpha=\mathrm{k}+\lambda} \\ {\beta=2 \mathrm{k}-3 \lambda} \\ {\mathrm{y}=-\mathrm{k}-5 \lambda}\end{array}\right\}$ …(1)

satisfying in plane $\mathrm{P}_{1}$

$\mathrm{k}+\lambda+4 \mathrm{k}-6 \lambda+\mathrm{k}+5 \lambda+1=0$

$6 k=-1$

putting in ( 1) required locus is

$\mathrm{x}=-\frac{1}{6}+\lambda$

$y=-\frac{1}{3}-3 \lambda$

$z=\frac{1}{6}-5 \lambda$

Now check the options.


Q. Consider a pyramid OPQRS located in the first octant $(\mathrm{x} \geq 0, \mathrm{y} \geq 0, \mathrm{z} \geq 0)$ with $\mathrm{O}$ as origin, and OP and OR along the x-axis and the y-axis, respectively. The base OPQR of the pyramid is a square with $\mathrm{OP}=3 .$ The point $\mathrm{S}$ is directly above the mid-point $\mathrm{T}$ of diagonal OQ such that TS $=3 .$ Then-

(A) the acute angle between $\mathrm{OQ}$ and $\mathrm{OS}$ is $\frac{\mathrm{K}}{3}$

(B) the equaiton of the plane containing the triangle $\mathrm{OQS}$ is $\mathrm{x}-\mathrm{y}=0$

(C) the length of the perpendicular from $P$ to the plane containing the triangle OQS is $\frac{3}{\sqrt{2}}$

(D) the perpendicular distance from $\mathrm{O}$ to the straight line containing RS is $\sqrt{\frac{15}{2}}$

[JEE(Advanced) 2016]

Sol. (B,C,D)


Q. Let $P$ be the image of the point $(3,1,7)$ with respect to the plane $x-y+z=3 .$ Then the equation of the plane passing through $\mathrm{P}$ and containing the straight line $\frac{\mathrm{x}}{1}=\frac{\mathrm{y}}{2}=\frac{\mathrm{z}}{1}$ is

(A) x + y – 3z = 0

(B) 3x + z = 0

(C) x – 4y + 7z = 0

(D) 2x – y = 0

[JEE(Advanced) 2016]

Sol. (C)

$\therefore x-4 y+7 z=0$


Q. The equation of the plane passing through the point (1,1,1) and perpendicular to the planes 2x + y – 2z = 5 and 3x – 6y – 2z = 7, is-

(A) 14x + 2y + 15z = 31

(B) 14x + 2y – 15z = 1

(C) –14x + 2y + 15z = 3

(D) 14x – 2y + 15z = 27

[JEE(Advanced) 2017]

Sol. (A)


Q. Let $P_{1}: 2 x+y-z=3$ and $P_{2}: x+2 y+z=2$ be two planes. Then, which of the following statement(s) is (are) TRUE ?

(A) The line of intersection of $\mathrm{P}_{1}$ and $\mathrm{P}_{2}$ has direction ratios $1,2,-1$

(B) The line $\frac{3 x-4}{9}=\frac{1-3 y}{9}=\frac{z}{3}$ is perpendicular to the line of intersection of $P_{1}$ and $P_{2}$

(C) The acute angle between $\mathrm{P}_{1}$ and $\mathrm{P}_{2}$ is $60^{\circ}$

(D) If $P_{3}$ is the plane passing through the point $(4,2,-2)$ and perpendicular to the line of intersection of $\mathrm{P}_{1}$ and $\mathrm{P}_{2},$ then the distance of the point $(2,1,1)$ from the plane $\mathrm{P}_{2}$ is $\frac{2}{\sqrt{3}}$

[JEE(Advanced) 2018]

Sol. (C,D)

D.C. of line of intersection $(a, b, c)$

$\begin{aligned} \Rightarrow \quad & 2 \mathrm{a}+\mathrm{b}-\mathrm{c}=0 \\ & \mathrm{a}+2 \mathrm{b}+\mathrm{c}=0 \end{aligned}$

$\frac{a}{1+2}=\frac{b}{-1-2}=\frac{c}{4-1}$

$\therefore \quad \mathrm{D} . \mathrm{C} .$ is $(1,-1,1)$

B) $\frac{3 x-4}{9}=\frac{1-3 y}{9}=\frac{z}{3}$

$\Rightarrow \quad \frac{x-4 / 3}{3}=\frac{y-1 / 3}{-3}=\frac{z}{3}$

$\Rightarrow \quad$ lines are parallel.

(C) Acute angle between $\mathrm{P}_{1}$ and $\mathrm{P}_{2}=\cos ^{-1}\left(\frac{2 \times 1+1 \times 2-1 \times 1}{\sqrt{6} \sqrt{6}}\right)$ \[ =\cos ^{-1}\left(\frac{3}{6}\right)=\cos ^{-1}\left(\frac{1}{2}\right)=60^{\circ} \]

(D) Plane is given by $(x-4)-(y-2)+(z+2)=0$ \[ \Rightarrow \quad x-y+z=0 \]

Distance of $(2,1,1)$ from plane $=\frac{2-1+1}{\sqrt{3}}=\frac{2}{\sqrt{3}}$


Q. Consider the cube in the first octant with sides $\mathrm{OP}, \mathrm{OQ}$ and $\mathrm{OR}$ of length $1,$ along the x-axis,

y-axis and z-axis, respectively, where $\mathrm{O}(0,0,0)$ is the origin. Let $\mathrm{S}\left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right)$ be the centre

of the cube and $T$ be the vertex of the cube opposite to the origin $O$ such that $S$ lies on the diagonal OT.If $\overrightarrow{\mathrm{p}}=\overrightarrow{\mathrm{SP}}, \overrightarrow{\mathrm{q}}=\overrightarrow{\mathrm{SQ}}, \overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{SR}}$ and $\overrightarrow{\mathrm{t}}=\overrightarrow{\mathrm{ST}},$ then the value of $|(\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}) \times(\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{t}})|$ is

[JEE(Advanced) 2018]

Sol. 8


Q. Let P be a point in the first octant, whose image Q in the plane x + y = 3 (that is, the line segment PQ is perpendicular to the plane x + y = 3 and the mid-point of PQ lies in the plane x + y = 3) lies on the z-axis. Let the distance of P from the x-axis be 5. If R is the image of P in the xy-plane, then the length of PR is

Sol. 8


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Q. Let ƒ be a non-negative function defined on the interval $[0,1] .$ If $\int_{0}^{x} \sqrt{1-\left(f^{\prime}(t)\right)^{2}} d t=\int_{0}^{x} f(t) d t$ $0 \leq \mathrm{x} \leq 1,$ and $f(0)=0,$ then $-$

(A) $f\left(\frac{1}{2}\right)<\frac{1}{2}$ and $f\left(\frac{1}{3}\right)>\frac{1}{3}$

(B) $f\left(\frac{1}{2}\right)>\frac{1}{2}$ and $f\left(\frac{1}{3}\right)>\frac{1}{3}$

(C) $f\left(\frac{1}{2}\right)<\frac{1}{2}$ and $f\left(\frac{1}{3}\right)<\frac{1}{3}$

(D) $f\left(\frac{1}{2}\right)>\frac{1}{2}$ and $f\left(\frac{1}{3}\right)<\frac{1}{3}$

[JEE 2009, 3]

Sol. (C)

$\int_{0}^{x} \sqrt{1-\left(f^{\prime}(t)\right)^{2}} \mathrm{d} t=\int_{0}^{x} f(t) d t, 0 \leq x \leq 1$

differentiating both the sides & squreing

$\Rightarrow 1-\left(f^{\prime}(\mathrm{x})\right)^{2}=f^{2}(\mathrm{x})$

$\Rightarrow \frac{f^{\prime}(x)}{\sqrt{1-f^{2}(x)}}=1$

$\Rightarrow \sin ^{-1} f(\mathrm{x})=\mathrm{x}+\mathrm{c}$

$f(0)=0$

$\Rightarrow f(\mathrm{x})=\sin \mathrm{x}$

$\Rightarrow \because \sin \mathrm{x} \leq \mathrm{x}$ for $\mathrm{x} \in[0,1]$

$\Rightarrow f\left(\frac{1}{2}\right)<\frac{1}{2}$ and $f\left(\frac{1}{3}\right)<\frac{1}{3}$


Q. If $\mathrm{I}_{\mathrm{n}}=\int_{-\pi}^{\pi} \frac{\sin \mathrm{nx}}{\left(1+\pi^{\mathrm{x}}\right) \sin \mathrm{x}} \mathrm{d} \mathrm{x}, \mathrm{n}=0,1,2, \ldots, \mathrm{then}-$

(A) $\mathrm{I}_{\mathrm{n}}=\mathrm{I}_{\mathrm{n}+2}$

(B) $\sum_{\mathrm{m}=1}^{10} \mathrm{I}_{2 \mathrm{m}+1}=10 \pi$

(C) $\sum_{\mathrm{m}=1}^{10} \mathrm{I}_{2 \mathrm{m}}=0$

(D) $\mathrm{I}_{\mathrm{n}}=\mathrm{I}_{\mathrm{n}+1}$

[JEE 2009, 4]

Sol. (A,B,D)

$\mathrm{I}_{\mathrm{n}}=\int_{-\pi}^{\pi} \frac{\sin \mathrm{nx}}{\left(1+\pi^{\mathrm{x}}\right) \sin \mathrm{x}} \mathrm{dx}$

$\mathrm{I}_{\mathrm{n}}=\int_{-\pi}^{\pi} \frac{\pi^{\mathrm{x}} \sin \mathrm{nx}}{\left(1+\pi^{\mathrm{x}}\right) \sin \mathrm{x}} \mathrm{dx}$

$2 \mathrm{I}_{\mathrm{n}}=\int_{-\pi}^{\pi} \frac{\sin \mathrm{nx}}{\sin \mathrm{x}} \mathrm{dx}$ ..(i)

$2 \mathrm{I}_{\mathrm{n}+2}=\int_{-\pi}^{\pi} \frac{\sin (\mathrm{n}+2) \mathrm{x}}{\sin \mathrm{x}} \mathrm{dx} \quad \ldots(\mathrm{i})$

(ii) – (i)

$\Rightarrow 2\left(\ln _{+2}-\mathrm{I}_{\mathrm{n}}\right)=\int_{-\pi}^{\pi} \cos (\mathrm{n}+1) \mathrm{x}=0$

$\Rightarrow \quad \mathrm{I}_{\mathrm{n}+2}=\mathrm{I}_{\mathrm{n}}$

$\sum_{m=1}^{10} \mathrm{I}_{2 \mathrm{m}}=10 \sum_{\mathrm{m}=1}^{10} \mathrm{I}_{2}=\frac{10}{2} \int_{-\pi}^{\pi} \frac{\sin 2 \mathrm{x}}{\sin \mathrm{x}} \mathrm{d} \mathrm{x}=0$

Put n = 1 in equation (i)

$2 \mathrm{I}_{1}=\int_{-\pi}^{\pi} \frac{\sin \mathrm{x} \mathrm{d} \mathrm{x}}{\sin \mathrm{x}}=2 \pi$

$\mathrm{I}_{1}=\pi$

$\sum_{m=1}^{10} I_{2 m+1}=10 \pi$


Q. Let $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ be a continuous function which satisfies $\mathrm{f}(\mathrm{x})=\int_{0}^{\mathrm{x}} \mathrm{f}(\mathrm{t}) \mathrm{d} \mathrm{t}$ Then the value of f(ln 5) is……..

[JEE 2009, 4]

Sol. 0

$\mathrm{f}(\mathrm{x})=\int_{0}^{\mathrm{x}} \mathrm{f}(\mathrm{t}) \mathrm{dt}$

$\mathrm{f}^{\mathrm{l}}(\mathrm{x})=\mathrm{f}(\mathrm{x})$

$\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{y}$

$\Rightarrow \int \frac{d y}{y}=\int d x$

$\Rightarrow \ln y=x+c$

$\Rightarrow y=e^{x+c}$

$\Rightarrow y=0$

$\left(\begin{array}{c}{\text { at } x=0, y=0} \\ {c \rightarrow-\infty}\end{array}\right)$

$f(x)=0$

$f(\ell n 5)=0$


Q. The value of $\lim _{x \rightarrow 0} \frac{1}{x^{3}} \int_{0}^{x} \frac{t \ell n(1+t)}{t^{4}+4} d t$ is

(A) 0

(B) $\frac{1}{12}$

(C) $\frac{1}{24}$

(D) $\frac{1}{64}$

[JEE 2010, 3 (–1)]

Sol. (B)

Applying L-Hospital rule,


Q. The value(s) of $\int_{0}^{1} \frac{\mathrm{x}^{4}(1-\mathrm{x})^{4}}{1+\mathrm{x}^{2}} \mathrm{dx}$ is (are)

(A) $\frac{22}{7}-\pi$

(B) $\frac{2}{105}$

(C) 0

(D) $\frac{71}{15}-\frac{3 \pi}{2}$

[JEE 2010, 3]

Sol. (A)


Q. Let $f$ be a real-valued function defined on the interval $(-1,1)$ such that

$e^{-x} f(x)=2+\int_{0}^{x} \sqrt{t^{4}+1} d t,$ for all $x \in(-1,1),$ and let $f^{-1}$ be the inverse function of $f$ Then $\left(f^{-1}\right)^{\prime}(2)$ is equal to-

(A) 1

(B) $\frac{1}{3}$

(C) $\frac{1}{2}$

(D) $\frac{1}{\mathrm{e}}$

[JEE2010, 5 (–2)]

Sol. (B)

from $(2), f^{-1}(2)=\frac{1}{3}$


Q. For any real number x, let [x] denote the largest integer less than or equal to x. Let f be a real valued function defined on the interval [–10, 10] by $\mathrm{f}(\mathrm{x})=\left\{\begin{aligned} \mathrm{x}-[\mathrm{x}] & \text { if }[\mathrm{x}] \text { is odd } \\ 1+[\mathrm{x}]-\mathrm{x} & \text { if }[\mathrm{x}] \text { is even } \end{aligned}\right.$ Then the value of $\frac{\pi^{2}}{10} \int_{-10}^{10} \mathrm{f}(\mathrm{x}) \cos \pi \mathrm{x} \mathrm{d} \mathrm{x}$ is

[JEE 2010, 3]

Sol. 4


Q. The value of $\int_{\sqrt{\mathrm{in} 2}}^{\sqrt{\mathrm{n} 3}} \frac{\mathrm{x} \sin \mathrm{x}^{2}}{\sin \mathrm{x}^{2}+\sin \left(\mathrm{ln} 6-\mathrm{x}^{2}\right)} \mathrm{dx}$ is

(A) $\frac{1}{4} \ln \frac{3}{2}$

(B) $\frac{1}{2} \ln \frac{3}{2}$

(C) $\ln \frac{3}{2}$

(D) $\frac{1}{6} \ln \frac{3}{2}$

[JEE 2011, 3 (–1)]

Sol. (A)


Q. Let $S$ be the area of the region enclosed by $y=e^{-x^{2}}, y=0, x=0,$ and $x=1 .$ Then

(A) $\mathrm{S} \geq \frac{1}{\mathrm{e}}$

(B) $\mathrm{S} \geq 1-\frac{1}{\mathrm{e}}$

(C) $S \leq \frac{1}{4}\left(1+\frac{1}{\sqrt{\mathrm{e}}}\right)$

(D) $S \leq \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{\mathrm{e}}}\left(1-\frac{1}{\sqrt{2}}\right)$

[JEE 2012, 4M]

Sol. (A,B,D)


Q. The value of the integral $\int_{-\pi / 2}^{\pi / 2}\left(\mathrm{x}^{2}+\ln \frac{\pi+\mathrm{x}}{\pi-\mathrm{x}}\right) \cos \mathrm{xd} \mathrm{x}$ is

(A) 0

(B) $\frac{\pi^{2}}{2}-4$

(C) $\frac{\pi^{2}}{2}+4$

(D) $\frac{\pi^{2}}{2}$c

[JEE 2012, 3M, –1M]

Sol. (B)


Q. For a $\in \mathrm{R}$ (the set of all real numbers), a\neq-1.

$\lim _{n \rightarrow \infty} \frac{\left(1^{a}+2^{a}+\ldots \ldots+n^{a}\right)}{(n+1)^{a-1}[(n a+1)+(n a+2)+\ldots \ldots+(n a+n)]}=\frac{1}{60}$ Then $a=$

(A) 5 (B) 7 (C) $\frac{-15}{2}$ (D) $\frac{-17}{2}$

[JEE(Advanced) 2013, 3, (–1)]

Sol. (B)


Q. Let $f:[\mathrm{a}, \mathrm{b}] \rightarrow[1, \infty)$ be a continuous function and let $\mathrm{g}: \square \rightarrow \square$ be defined as

Then

(A) g(x) is continuous but not differentiable at a

(B) g(x) is differentiable on 

(C) g(x) is continuous but not differentiable at b

(D) g(x) is continuous and differentiable at either a or b but not both.

[JEE(Advanced)-2014, 3]

Sol. (A,C)


Q. The value of $\int_{0}^{1} 4 x^{3}\left\{\frac{d^{2}}{d x^{2}}\left(1-x^{2}\right)^{5}\right\} d x$ is

[JEE(Advanced)-2014, 3]

Sol. 2


Q. The following integral $\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}(2 \csc x)^{17} d x$ is equal to

(A) $\int_{0}^{\log (1+\sqrt{2})} 2\left(e^{\mathfrak{u}}+e^{-\mathfrak{u}}\right)^{16} \mathrm{d} \mathfrak{u}$

(B) $\int_{0}^{\log (1+\sqrt{2})}\left(\mathrm{e}^{\mathrm{u}}+\mathrm{e}^{-\mathrm{u}}\right)^{17} \mathrm{du}$

(C) $\int_{0}^{\log (1+\sqrt{2})}\left(e^{\mathfrak{u}}-e^{-\mathfrak{u}}\right)^{17} \mathrm{d} \mathfrak{u}$

(D) $\int_{0}^{\log (1+\sqrt{2})} 2\left(e^{\mathfrak{u}}-e^{-\mathfrak{u}}\right)^{16} d \mathfrak{u}$

[JEE(Advanced)-2014, 3(–1)]

Sol. (A)


Q. Let $f:[0,2] \rightarrow \square$ be a function which is continuous on $[0,2]$ and is differentiable on $(0,2)$ with $f(0)=1 .$ Let $F(x)=\int_{0}^{x^{2}} f(\sqrt{t}) d t$ for $x \in[0,2] .$ If $F^{\prime}(x)=f^{\prime}(x)$ for all $x \in(0,2)$ then $F(2)$ equals $-$

(A) $\mathrm{e}^{2}-1$

(B) $\mathrm{e}^{4}-1$

(C) e – 1

(D) e $^{4}$

[JEE(Advanced)-2014, 3(–1)]

Sol. (B)


Given that for each a $\in(0,1)$, $\lim _{\mathrm{h} \rightarrow 0^{+}} \int_{\mathrm{h}}^{1-\mathrm{h}} \mathrm{t}^{-\mathrm{a}}(1-\mathrm{t})^{\mathrm{a}-1}$ $\mathrm{dt}$ exists. Let this limit be g(a). In addition,

it is given that the function g(a) is differentiable on (0,1).

Q. The value of $\mathrm{g}\left(\frac{1}{2}\right)$ is –

(A) $\pi$

(B) $2 \pi$

(C) $\frac{\pi}{2}$

(D) $\frac{\pi}{4}$

[JEE(Advanced)-2014, 3(–1)]

Sol. (A)


Q. The value of $\mathrm{g}^{\prime}\left(\frac{1}{2}\right)$ is-

(A) $\frac{\pi}{2}$

(B) $\pi$

(C) $-\frac{\pi}{2}$

(D) 0

[JEE(Advanced)-2014, 3(–1)]

Sol. (D)


Q.

[JEE(Advanced)-2014, 3(–1)]

Sol. (C)


Q. Let $f: \square \rightarrow \square$ be a function defined by $f(x)$ $=\left\{\begin{array}{ccc}{[\mathrm{x}]} & {,} & {\mathrm{x} \leq 2} \\ {0} & {,} & {\mathrm{x}>2}\end{array}\right.$ where [x] is the greatest integer less than or equal to x. If $\mathrm{I}=\int_{-1}^{2} \frac{\mathrm{x} f\left(\mathrm{x}^{2}\right)}{2+f(\mathrm{x}+1)} \mathrm{dx}$ , then the value of (4I – 1) is

[JEE 2015, 4M, –0M]

Sol. (A)


Q. If $\alpha=\int_{0}^{1}\left(\mathrm{e}^{9 \mathrm{x}+3 \tan ^{-1} \mathrm{x}}\right)\left(\frac{12+9 \mathrm{x}^{2}}{1+\mathrm{x}^{2}}\right) \mathrm{d} \mathrm{x}$ where $\tan ^{-1} \mathrm{x}$ takes only principal values, then the value of $\left(\log _{\mathrm{e}}|1+\alpha|-\frac{3 \pi}{4}\right)$ is

[JEE 2015, 4M, –0M]

Sol. 0


Q. Let $f: \mathbb{U} \rightarrow \square$ be a continuous odd function, which vanishes exactly at one point and $f(1)=\frac{1}{2}$ Suppose that $\mathrm{F}(\mathrm{x})=\int_{-1}^{\mathrm{x}} f(\mathrm{t}) \mathrm{dt}$ for all $\mathrm{x} \in[-1,2]$ and $\mathrm{G}(\mathrm{x})$ $=\int_{-1}^{x} \mathfrak{t}|f(f(\mathfrak{t}))| d \mathfrak{t}$ for all $x \in[-1,2]$. If $\lim _{x \rightarrow 1} \frac{F(x)}{G(x)}=\frac{1}{14},$ then the value of $f\left(\frac{1}{2}\right)$ is

[JEE 2015, 4M, –0M]

Sol. 9


Q. The option(s) with the values of a and L that satisfy the following equation is(are)

(A) $a=2, L=\frac{e^{4 \pi}-1}{e^{\pi}-1}$

(B) $a=2, L=\frac{e^{4 \pi}+1}{e^{\pi}+1}$c

(C) $a=4, L=\frac{e^{4 \pi}-1}{e^{\pi}-1}$

(D) $a=4, L=\frac{e^{4 \pi}+1}{e^{\pi}+1}$

[JEE 2015, 4M, –0M]

Sol. 7


Q. Let $f(\mathrm{x})=7 \tan ^{8} \mathrm{x}+7 \tan ^{6} \mathrm{x}-3 \tan ^{4} \mathrm{x}-3 \tan ^{2} \mathrm{x}$ for all $\mathrm{x} \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) .$ Then the correct expression(s)is(are)

(A) $\int_{0}^{\pi / 4} \mathrm{x} f(\mathrm{x}) \mathrm{d} \mathrm{x}=\frac{1}{12}$

(B) $\int_{0}^{\pi / 4} f(\mathrm{x}) \mathrm{d} \mathrm{x}=0$

(C) $\int_{0}^{\pi / 4} \mathrm{x} f(\mathrm{x}) \mathrm{d} \mathrm{x}=\frac{1}{6}$

(D) $\int_{0}^{\pi / 4} f(\mathrm{x}) \mathrm{d} \mathrm{x}=1$

[JEE 2015, 4M, –0M]

Sol. (A,C)


Q. Let $f^{\prime}(x)=\frac{192 x^{3}}{2+\sin ^{4} \pi x}$ for all $\mathrm{x} \in \square$ with $f$ $\left(\frac{1}{2}\right)$ $=0 .$ If $\mathrm{m} \leq \int_{1 / 2}^{1} f(\mathrm{x}) \mathrm{d} \mathrm{x} \leq \mathrm{M}$ then the possible values of m and M are

(A) m = 13, M = 24

(B) $\quad \mathrm{m}=\frac{1}{4}, \mathrm{M}=\frac{1}{2}$

(C) m = –11, M = 0

(D) m = 1, M = 12

[JEE 2015, 4M, –0M]

Sol. (A,B)


Let $\mathrm{F}: \mathbb{U} \rightarrow \square$ be a thrice differentiable function. Suppose that $\mathrm{F}(1)=0, \mathrm{F}(3)=-4 \mathrm{F}^{\prime}(\mathrm{x})<$ 0 for all $\mathrm{x} \in(1 / 2,3) .$ Let $f(\mathrm{x})=\mathrm{xF}(\mathrm{x})$ for all $\mathrm{x} \in \mathbb{D}$.

Q. The correct statement(s) is(are)

(A) $f^{\prime}(1)<0$

(B) $f(2)<0$

(C) $f^{\prime}(\mathrm{x}) \neq 0$ for any $\mathrm{x} \in(1,3)$

(D) $f^{\prime}(x)=0$ for some $x \in(1,3)$

[JEE 2015, 4M, –0M]

Sol. (D)


Q. If $\int_{1}^{3} \mathrm{x}^{2} \mathrm{F}^{\prime}(\mathrm{x}) \mathrm{d} \mathrm{x}=-12$ and $\int_{1}^{3} \mathrm{x}^{3} \mathrm{F}^{\prime \prime}(\mathrm{x}) \mathrm{d} \mathrm{x}=40,$ then the correct expression(s) is (are)

(A) 9ƒ'(3) + ƒ'(1) – 32 = 0

(B) $\int_{1}^{3} f(\mathrm{x}) \mathrm{d} \mathrm{x}=12$

(C) 9ƒ'(3) – ƒ'(1) + 32 = 0

(D) $\left.\int_{1}^{3} f(x) d x=-12\right]$

[JEE 2015, 4M, –0M]

Sol. (A,B,C)


Q. The value of $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x^{2} \cos x}{1+e^{x}} d x$ is equal to

(A) $\frac{\pi^{2}}{4}-2$

(B) $\frac{\pi^{2}}{4}+2$

(C) $\pi^{2}-\mathrm{e}^{\frac{\pi}{2}}$

(D) $\pi^{2}+\mathrm{e}^{\frac{\pi}{2}}$

[JEE(Advanced)2016]

Sol. (C,D)


Q. Let $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ be a differentiable function such that $\mathrm{f}(0)=0, \mathrm{f}\left(\frac{\pi}{2}\right)=3$ and $\mathrm{f}^{\prime}(0)=1$ If $\mathrm{g}(\mathrm{x})=\int_{\mathrm{x}}^{\frac{\pi}{2}}\left[\mathrm{f}^{\prime}(\mathrm{t}) \csc \mathrm{t}-\cot t \csc \mathrm{t} \mathrm{f}(\mathrm{t})\right] \mathrm{d} \mathrm{t}$ for $\mathrm{x} \in\left(0, \frac{\pi}{2}\right],$ then $\lim _{\mathrm{x} \rightarrow 0} \mathrm{g}(\mathrm{x})=$

[JEE(Advanced)-2017]

Sol. 2


Q. If $\mathrm{I}=\sum_{\mathrm{k}=1}^{98} \int_{\mathrm{k}}^{\mathrm{k}+1} \frac{\mathrm{k}+1}{\mathrm{x}(\mathrm{x}+1)} \mathrm{d} \mathrm{x},$ then

(A) $\mathrm{I}<\frac{49}{50}$

(B) $\mathrm{I}<\log _{\mathrm{e}} 99$

(C) $\mathrm{I}>\frac{49}{50}$

(D) $\mathrm{I}>\log _{\mathrm{e}} 99$

[JEE(Advanced)-2017]

Sol. (B,C)


Q. If $\mathrm{g}(\mathrm{x})=\int_{\sin \mathrm{x}}^{\sin (2 \mathrm{x})} \sin ^{-1}(\mathrm{t}) \mathrm{dt},$ then

(A) $\mathrm{g}^{\prime}\left(\frac{\pi}{2}\right)=-2 \pi$

(B) $\mathrm{g}^{\prime}\left(-\frac{\pi}{2}\right)=2 \pi$

(C) $\mathrm{g}^{\prime}\left(\frac{\pi}{2}\right)=2 \pi$

(D) $\mathrm{g}^{\prime}\left(-\frac{\pi}{2}\right)=-2 \pi$

[JEE(Advanced)-2017]

Sol. (Bonus)


Q. The value of the integral $\int_{0}^{\frac{1}{2}} \frac{1+\sqrt{3}}{\left((x+1)^{2}(1-x)^{6}\right)^{\frac{1}{4}}} d x$ is

[JEE(Advanced)-2018]

Sol. 2


Kinematics 1D- JEE Advanced Previous Year Questions with Solutions

JEE Advanced Previous Year Questions of Physics with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Physics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

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Previous Years JEE Advanced Questions

Q. A train is moving along a straight line with a constant acceleration ‘a’. A boy standing in the train throws a ball forward with a speed of 10 m/s, at an angle of $60^{\circ}$ to the horizontal. The boy has to move forward by 1.15 m inside the train to catch the ball back at the initial height. The acceleration of the train, in $\mathrm{m} / \mathrm{s}^{2}$, is

[IIT-JEE 2011]

Sol. 5


Q. A rocket is moving in a gravity free space with a constant acceleration of 2 $\mathrm{ms}^{-2}$ along + x direction (see figure). The length of a chamber inside the rocket is 4m. A ball is thrown from the left end of the chamber in + x direction with a speed of 0.3 $\mathrm{ms}^{-1}$ relative to the rocket. At the same time, another ball is thrown in –x direction with a speed of 0.2 $\mathrm{ms}^{-1}$ from its right end relative to the rocket. The time in seconds when the two balls hit each other is

[JEE Advanced 2014]

Sol. 8 or 2

Assuming open chamber

$\mathrm{V}_{\text {relative }}=0.5 \mathrm{m} / \mathrm{s}$

$\mathrm{S}_{\text {relative }}=4 \mathrm{m}$

time $=\frac{4}{0.5}=8 \mathrm{m} / \mathrm{s}$

Alternate

Assuming closed chamber

In the frame of chamber :

Maximum displacement of ball A from its left end is $\frac{\mathrm{u}_{\mathrm{A}}^{2}}{2 \mathrm{a}}=\frac{(0.3)^{2}}{2(2)}=0.0225 \mathrm{m}$

This is negligible with respect to the length of chamber i.e. 4m. So, the collision will be verym close to the left end.

Hence, time taken by ball B to reach left end will be given by

$\mathrm{S}=\mathrm{u}_{\mathrm{B}} \mathrm{t}+\frac{1}{2} \mathrm{at}^{2}$

$4=(0.2)(\mathrm{t})+\frac{1}{2}(2)(\mathrm{t})^{2}$

Solving this, we get

$\mathrm{t} \approx 2 \mathrm{s}$


Q. Airplanes A and B are flying with constant velocity in the same vertical plane at angles $30^{\circ}$ and $60^{\circ}$ with respect to the horizontal respectively as shown in figure. The speed of A is $\mathrm{ms}^{-1}$. At time t = 0 s, an observer in A finds B at a distance of 500 m. This observer sees B moving with a constant velocity perpendicular to the line of motion of A. If at t = $\mathrm{t}_{0}$, A just escapes being hit by B, $\mathrm{t}_{0}$ in seconds is

[JEE Advanced-2014]

Sol. 5

As observed from A, B moves perpendicular to line of motion of A. It means velocity of B along A is equal to velocity of A

$\mathrm{V}_{\mathrm{B}} \cos 30=100 \sqrt{3}$

$\mathrm{V}_{\mathrm{B}}=200$

If A is observer A remains stationary therefore

$\mathrm{t}=\frac{500}{\mathrm{V}_{\mathrm{B}} \sin 30}=\frac{500}{100}=5$


Q. Consider an expanding sphere of instantaneous radius R whose total mass remains constant. The expansion is that the instantaneous density $\rho$ remains uniform throughout the volume.

the rate of fractional change in density is $\left(\frac{1}{\rho} \frac{d \rho}{d t}\right)$ constant. the velocity v of any point on the surface of the expanding sphere is proportional to

(A) $R^{3}$

(B) $\frac{1}{R}$

(C) $\mathrm{R}$

(D) $R^{2 / 3}$

[JEE Advanced-2017]

Sol. (C)


Vector- JEE Advanced Previous Year Questions with Solutions

JEE Advanced Previous Year Questions of Physics with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Physics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

Get detailed Class 11th & 12th Physics Notes to prepare for Boards as well as competitive exams like IIT JEE, NEET etc.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects.

 

 

Click Here for JEE main Previous Year Topic Wise Questions of Physics with Solutions 

 

Download eSaral app  for free study material and video tutorials.

 

 

Simulator

 

Previous Years JEE Advanced Questions

Q. Three vectors $\overrightarrow{\mathrm{P}}, \overrightarrow{\mathrm{Q}}$ and $\overrightarrow{\mathrm{R}}$ are shown in the figure. Let S be any point on the vector $\overrightarrow{\mathrm{R}}$. The distance between the points P and S is b $|\overrightarrow{\mathrm{R}}|$. The general relation among vectors $\overrightarrow{\mathrm{P}}, \overrightarrow{\mathrm{Q}}$ and $\overrightarrow{\mathrm{S}}$ is :

$(\mathrm{A}) \overrightarrow{\mathrm{S}}=(1-\mathrm{b}) \overrightarrow{\mathrm{P}}+\mathrm{b}^{2} \overrightarrow{\mathrm{Q}}$

(B) $\overrightarrow{\mathrm{S}}=(b-1) \overrightarrow{\mathrm{P}}+b \overrightarrow{\mathrm{Q}}$

(C) $\overrightarrow{\mathrm{S}}=(1-\mathrm{b}) \overrightarrow{\mathrm{P}}+\mathrm{b} \overrightarrow{\mathrm{Q}}$

$(\mathrm{D}) \overrightarrow{\mathrm{S}}=\left(1-\mathrm{b}^{2}\right) \overrightarrow{\mathrm{P}}+\mathrm{b} \overrightarrow{\mathrm{Q}}$

[JEE Advanced – 2017]

Sol. (C)

Let vector from point P to point S be $\overrightarrow{\mathrm{c}}$

$\Rightarrow \overrightarrow{\mathrm{c}}=\mathrm{b}|\overrightarrow{\mathrm{R}}| \hat{\mathrm{R}}=\mathrm{b}|\overrightarrow{\mathrm{R}}|\left(\frac{\overrightarrow{\mathrm{R}}}{|\overrightarrow{\mathrm{R}}|}\right)=\mathrm{b} \overrightarrow{\mathrm{R}}=\mathrm{b}(\overrightarrow{\mathrm{Q}}-\overrightarrow{\mathrm{P}})$

from triangle rule of vector addition

$\overrightarrow{\mathrm{P}}+\overrightarrow{\mathrm{C}}=\overrightarrow{\mathrm{S}}$

$\overrightarrow{\mathrm{P}}+\mathrm{b}(\overrightarrow{\mathrm{Q}}-\overrightarrow{\mathrm{P}})=\overrightarrow{\mathrm{S}}$

$\Rightarrow \overrightarrow{\mathrm{S}}=(1-\mathrm{b}) \overrightarrow{\mathrm{P}}+\mathrm{b} \overrightarrow{\mathrm{Q}}$


Q. Consider an expanding sphere of instantaneous radius R whose total mass remains constant. The expansion is such that the instantaneous density  remains uniform throughout the volume. The rate of fractional change in density $\left(\frac{1}{\rho} \frac{\mathrm{d} \rho}{\mathrm{dt}}\right)$ is constant. The velocity v of any point on the surface of the expanding sphere is proportional to :

(A) $\mathrm{R}^{3}$

(B) $\frac{1}{\mathrm{R}}$

(C) R

(D) $\mathrm{R}^{2 / 3}$

[JEE Advanced – 2017]

Sol. (C)

Density of sphere is $\rho=\frac{\mathrm{m}}{\mathrm{v}}=\frac{3 \mathrm{m}}{4 \pi \mathrm{R}^{3}}$

$\Rightarrow \frac{1}{\rho} \frac{\mathrm{d} \rho}{\mathrm{dt}}=-\frac{3}{\mathrm{R}} \frac{\mathrm{d} \mathrm{R}}{\mathrm{dt}}$

since $\Rightarrow \frac{1}{\rho} \frac{\mathrm{d} \rho}{\mathrm{dt}}$ is constant

$\therefore \frac{\mathrm{d} \mathrm{R}}{\mathrm{dt}} \propto \mathrm{R}$

Velocity of any point on the circumfrence V is equal to $\frac{\mathrm{d} \mathrm{R}}{\mathrm{dt}}$ (rate of change of radius of outer layer)


Ray Optics – JEE Advanced Previous Year Questions with Solutions

JEE Advanced Previous Year Questions of Physics with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Physics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.

Get detailed Class 11th & 12th Physics Notes to prepare for Boards as well as competitive exams like IIT JEE, NEET etc.

eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects.

 

 

Click Here for JEE main Previous Year Topic Wise Questions of Physics with Solutions 

 

Download eSaral app  for free study material and video tutorials.

Simulator

Previous Years JEE Advanced Questions

Q. A ball is dropped from a height of 20 m above the surface of water in a lake. The refractive index of water is 4/3. A fish inside the lake, in the line of fall of the ball, is looking at the ball. At an instant, when the ball is 12.8 m above the water surface, the fish sees the speed of ball as $\left[\mathrm{g}=10 \mathrm{m} / \mathrm{s}^{2} .\right]$

(A) 9 m/s (B) 12 m/s (C) 16 m/s (D) 21.33 m/s

[IIT-JEE 2009]

Sol. (C)


Q. A student performed the experiment of determination of focal length of a concave mirror by u-v method using an optical bench of length 1.5 m. The focal length of the mirror used is 24 cm. The maximum error in the location of the image can be 0.2 cm. The 5 sets of (u, v) values recorded by the student (in cm) are : (42, 56), (48, 48), (60, 40), (66, 33), (78, 39). The data set(s) that cannot come from experiment and is (are) incorrectly recorded, is (are) :

(A) (42, 56) (B) (48, 48) (C) (66, 33) (D) (78, 39)

[IIT-JEE 2009]

Sol. (C,D)

$\mathrm{V}=\frac{\mathrm{uf}}{\mathrm{u}-\mathrm{f}} \quad$ by sustituting the value of u and $\mathrm{f}$

$\mathrm{V}_{42} \Rightarrow \frac{24 \times 42}{18} \Rightarrow 56 ; \quad \mathrm{V}_{60}=\frac{24 \times 60}{36} \Rightarrow 40$

$\mathrm{v}_{48} \Rightarrow \frac{48 \times 24}{24} \Rightarrow 48 ; \mathrm{V}_{66}=\frac{66 \times 24}{42} \Rightarrow 37.71$

$\mathrm{V}_{78} \Rightarrow \frac{78 \times 24}{54} \Rightarrow 34.3$

(66, 33) ; (78, 39)


Q. A biconvex lens of focal length 15 cm is in front of a plane mirror. The distance between the lens and the mirror is 10 cm. A small object is kept at a distance of 30 cm from the lens. The final image is –

(A) virtual and at a distance of 16 cm from the mirror

(B) real and at a distance of 16 cm from the mirror

(C) virtual and at a distance of 20 cm from the mirror

(D) real and at a distance of 20 cm from the mirror

[IIT-JEE 2010]

Sol. (B)

$\frac{1}{15}=\frac{1}{v}-\frac{1}{10}$


Q. A ray OP of monochromatic light is incident on the face AB of prism ABCD near vertex B at an incident angle of $60^{\circ}$ (see figure). If the refractive index of the material of the prism is $\sqrt{3}$, which of the following is (are) correct?

(A) The ray gets totally internally reflected at face CD

(B) The ray comes out through face AD

(C) The angle between the incident ray and the emergent ray is $90^{\circ}$

(D) The angle between the incident ray and the emergent ray is $120^{\circ}$

[IIT-JEE 2010]

Sol. (A,B,C)


Q. Two transparent media of refractive indices $\mu_{1}$ and $\mu_{3}$ have a solid lens shaped transparent material of refractive index $\mu_{2}$ between them as shown in figures in Column II. A ray traversing these media is also shown in the figures. In Column I different relationships between $\mu_{1}, \mu_{2}$ and $\mu_{3}$ are given. Match them to the ray diagrams shown in Column II.

[IIT-JEE 2010]

Sol. $(\mathrm{A})-\mathrm{pr},(\mathrm{B})-\mathrm{qst},(\mathrm{C})-\mathrm{prt},(\mathrm{D})-\mathrm{qs}$


Q. The focal length of a thin biconvex lens is 20 cm. When an object is moved from a distance of 25 cm in front of it to 50 cm, the magnification of its image changes from $\mathrm{m}_{25}$ to $\mathrm{m}_{50} .$ The ratio $\frac{m_{25}}{m_{50}}$ is –

[IIT-JEE 2010]

Sol. 6

$\mathrm{m}=\frac{\mathrm{f}}{\mathrm{f}+\mathrm{u}}$


Q. Image of an object approaching a convex mirror of radius of curvature 20 m along its optical axis is observed to move from $\frac{25}{3}$ m to $\frac{50}{7}$ m in 30 seconds. What is the speed of the object in km per hour ?

[IIT-JEE 2010]

Sol. 3

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v} \Rightarrow \frac{1}{10}=\frac{1}{u}+\frac{3}{25} \Rightarrow u=-50 \mathrm{m}$

$\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{u}}+\frac{1}{\mathrm{v}} \Rightarrow \frac{1}{10}=\frac{1}{\mathrm{u}}+\frac{7}{50} \Rightarrow \mathrm{u}=-25 \mathrm{m}$

Speed $=\frac{25}{30} \times \frac{18}{5}=3$


Q. A large glass slab $\left(\mu=\frac{5}{3}\right)$ of thickness 8 cm is placed over a point source of light on a plane surface. It is seen that light emerges out of the top surface of the slab from a circular area of radius R cm. What is the value of R?

[IIT-JEE 2010]

Sol. 6

$r=\frac{h}{\sqrt{\mu^{2}-1}}=\frac{8}{\sqrt{\left(\frac{5}{3}\right)^{2}-1}}=6 \mathrm{cm}$


Q. A light ray traveling in glass medium is incident on glass-air interface at an angle of incidence $\theta$. The reflected (R) and transmitted (T) intensities, both as function of $\theta$, are plotted. The correct sketch is –

[IIT-JEE 2011]

Sol. (C)

When $\theta=0^{\circ},$ maximum light is transmitted. At $\theta>\theta_{\mathrm{C}}$ (critical angle), no further light is transmitted


Q. Water (with refractive index = $\frac{4}{3}$) in a tank is 18 cm deep. Oil of refractive index $\frac{7}{4}$ lies on water making a convex surface of radius of curvature ‘R = 6 cm’ as shown. Consider oil to act as a thin lens. An object ‘S’ is placed 24 cm above water surface. The location of its image is at ‘x’ cm above the bottom of the tank. Then ‘x’ is

[IIT-JEE 2011]

Sol. 2

First refraction [ Lens-air interface]

$\frac{\mu_{2}}{v}-\frac{\mu_{1}}{u}=\frac{\mu_{2}-\mu_{1}}{R}$ where $\mu_{1}=1, \mathrm{u}=-24, \mu_{2}=\frac{7}{4}, \mathrm{R}=+6$

After solving v = 21

Now for second refraction [Lens-water interface]

$\frac{4 / 3}{v_{2}}-\frac{7 / 4}{21}=0 \Rightarrow v_{2}=h=16 \mathrm{cm}$

So, from bottom $18-16=2 \Rightarrow \mathrm{x}=2$


Q. A biconvex lens is formed with two thin plano-convex lenses as shown in the figure, Refractive index n of the first lens is 1.5 and that of the second lens is 1.2. Both the curved surfaces are of the same radius of curvature R = 14 cm. For this biconvex lens, for an object distance of 40 cm, the image distance will be :-

(A) –280.0 cm             (B) 40.0 cm            (C) 21.5 cm             (D) 13.3 cm

[IIT-JEE 2012]

Sol. (B)

$\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{f}_{1}}+\frac{1}{\mathrm{f}_{2}}$

$\frac{1}{\mathrm{f}}=\left(\frac{\mu_{1}-1}{\mathrm{R}_{1}}\right)+\left(\frac{\mu_{2}-1}{\mathrm{R}_{2}}\right) \Rightarrow \frac{1}{\mathrm{f}}=\frac{5}{14}+\frac{2}{14}$

f = 20 c.m.

$\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}} \Rightarrow \frac{1}{\mathrm{v}}-\frac{1}{-40}=\frac{1}{20}=40 \mathrm{c.m}$


Paragraph for Questions 12 and 13

Most materials have the refractive index, n>1. So, when a light ray from air enters a naturally occurring material, then by Snell’s law, $\frac{\sin \theta_{1}}{\sin \theta_{2}}=\frac{n_{2}}{n_{1}}$ , it is understood that the refracted ray bends towards the normal. But it never emerges on the same side of the normal as the incident ray. According to electromagnetism, the refractive index of the medium is given by the relation, $n=\left(\frac{c}{v}\right)=\pm \sqrt{\varepsilon_{r} \mu_{r}}$ , where c is the speed of electromagnetic waves in vacuum, v its speed in the medium, $\varepsilon_{\mathrm{r}}$ and $\mu_{\mathrm{r}}$ are the relative permittivity and permeability of the medium respectively.

In normal materials, both $\varepsilon_{\mathrm{r}}$ and $\mu_{\mathrm{r}}$ are positive, implying positive n for the medium. When both $\varepsilon_{\mathrm{r}}$ and $\mu_{\mathrm{r}}$ are negative, one must choose the negative root of n. Such negative refractive index materials can now be artificially prepared and are called meta-materials. They exhibit significantly different optical behaviour, without violating any physical laws. Since n is negative, it results in a change in the direction of propagation of the refracted light. However, similar to normal materials, the frequency of light remains unchanged upon refraction even in meta materials.

Q. For light incident from air on a meta-material, the appropriate ray diagram is –

[IIT-JEE 2012]

Sol. (C)

$_{1} \mu_{2}=\frac{\mu_{2}}{\mu_{1}}=\frac{\sin \mathrm{i}}{\sin \mathrm{r}}$

$\frac{(-)}{1}=\frac{\sin \mathrm{i}}{\sin \mathrm{r}}$

sini = sin(–r)


Q. Choose the correct statement.

(A) The speed of light in the meta-material is $\mathrm{v}=\mathrm{c}|\mathrm{n}|$

(B) The speed of light in the meta-material is v = $\frac{c}{|n|}$

(C) The speed of light in the meta-material is v = c.

(D) The wavelength of the light in the meta-material $\left(\lambda_{\mathrm{m}}\right)$ is given by $\lambda_{\mathrm{m}}=\lambda_{\mathrm{air}}|\mathrm{n}|,$ where $\lambda_{\mathrm{air}}$ is the wavelength of the light in air.

[IIT-JEE 2012]

Sol. (B)

$\mu=\frac{\mathrm{c}}{\mathrm{v}} \Rightarrow \mathrm{v}=\frac{\mathrm{c}}{\mu}=\frac{\mathrm{c}}{\mathrm{n}}$


Q. A ray of light travelling in the direction $\frac{1}{2}(\hat{i}+\sqrt{3} \hat{j})$ is incident on a plane mirror. After reflection, it travels along the direction $\frac{1}{2}(\hat{i}-\sqrt{3} \hat{j})$. The angle of incidence is :-

(A) $30^{\circ}$

(B) $45^{\circ}$

(C) $60^{\circ}$

(D) $75^{\circ}$

[JEE-Advance-2013]

Sol. (A)

Here normal is along $\hat{j}$

Angle between incident ray and normal $\cos \theta=\frac{\frac{1}{2}(\hat{i}+\sqrt{3} \hat{j}) \cdot \hat{j}}{(1)(1)}=\frac{\sqrt{3}}{2} \Rightarrow \theta=30^{\circ}$


Q. The image of an object, formed by a plano-convex lens at a distance of 8 m behind the lens, is real and is one-third the size of the object. The wavelength of light inside the lens is $\frac{2}{3}$ times the wavelength in free space. The radius of the curved surface of the lens is :

(A) 1 m            (B) 2 m              (C) 3 m               (D) 6 m

[JEE-Advance-2013]

Sol. (C)

$\mathrm{m}=-\frac{1}{3}=\frac{\mathrm{v}}{\mathrm{u}}$

v = 8m

u = – 24 m

$\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}} \rightarrow \frac{1}{8}-\frac{1}{-24}=\frac{1}{\mathrm{f}} \Rightarrow \mathrm{f}=6 \mathrm{m}$

$\mathrm{f}=\frac{\mathrm{R}}{(\mu-1)}$

$\mu=\frac{\lambda_{\text {vaceum }}}{\lambda_{\text {medium }}}=\frac{3}{2}$

$6 \mathrm{m}=\left(\frac{\mathrm{R}}{3 / 2-1}\right) \Rightarrow \mathrm{R}=3 \mathrm{m}$


Q. A right angled prism of refractive index $\mu_{1}$ is placed in a rectangular block of refractive index $\mu_{2}$, which is surrounded by a medium of refractive index $\mu_{3}$, as shown in the figure. A ray of light ‘e’ enters the rectangular block at normal incidence. Depending upon the relationships between $\mu_{1}, \mu_{2},$ and $\mu_{3}$, it takes one of the four possible paths ‘ef’, ‘eg’, ‘eh’ or ‘ei’.

Match the paths in List I with conditions of refractive indices in List II and select the correct answer using the codes given below the lists :

[JEE-Advance-2013]

Sol. (D)

(P) at prism surface ray moving towards normal so $\left(\mu_{2}>\mu_{1}\right)$at block surface ray moving away from normal so $\left(\mu_{3}<\mu_{2}\right)$

(Q) No deflection of ray on both surface; so $\mu_{1}=\mu_{2}=\mu_{3}$

(R) At prism surface ray moving away from normal so $\mu_{2}<\mu_{1}$. At block surface ray movign away from normal so µ3 < µ2 but since on total internal reflection not takes place on prism surface

$\mu_{1} \sin 45^{\circ}<\mu_{2} \sin 90^{\circ} \Rightarrow \mu_{1}<\sqrt{2} \mu_{2}$

(S) Total internal reflection takes place so $\mu_{1} \sin 45^{\circ}>\mu_{2} \sin 90^{\circ} \Rightarrow \mu_{1}>\sqrt{2} \mu_{2}$


Q. A transparent thin film of uniform thickness and refractive index $\mathrm{n}_{1}=1.4$ is coated on the convex spherical surface of radius R at one end of a long solid glass cylinder of refractive index $\mathrm{n}_{2}=1.5$, as shown in the figure. Rays of light parallel to the axis of the cylinder traversing through the film from air to glass get focused at distance $\mathrm{f}_{1}$ from the film, while rays of light traversing from glass to air get focused at distance $\mathrm{f}_{2}$ from the film. Then

$(\mathrm{A})\left|\mathrm{f}_{1}\right|=3 \mathrm{R}$

(B) $\left|\mathrm{f}_{1}\right|=2.8 \mathrm{R}$

(C) $\left|\mathrm{f}_{2}\right|=2 \mathrm{R}$b

$(\mathrm{D})\left|\mathrm{f}_{2}\right|=1.4 \mathrm{R}$

[JEE-Advance-2014]

Sol. (A,C)

When rays are moving from air to glass,

$\frac{1.5}{\mathrm{f}_{1}}=\frac{(1.4-1)}{+\mathrm{R}}+\frac{(1.5-1.4)}{+\mathrm{R}}$

$\frac{1.5}{\mathrm{f}_{1}}=\frac{0.4}{\mathrm{R}}+\frac{0.1}{\mathrm{R}}=\frac{0.5}{\mathrm{R}}$

$\left|\mathrm{f}_{1}\right|=3 \mathrm{R}$

When rays are moving from glass to air,

$\frac{1}{\mathrm{F}_{2}}=\frac{(1-1.4)}{-\mathrm{R}}+\frac{(1.4-1.5)}{-\mathrm{R}}=\frac{0.5}{\mathrm{R}}$

$\left|\mathrm{f}_{2}\right|=2 \mathrm{R}$


Q. A point source S is placed at the bottom of a transparent block of height 10 mm and refractive index 2.72. It is immersed in a lower refractive index liquid as shown in the figure. It is found that the light emerging from the block to the liquid forms a circular bright spot of diameter 11.54 mm on the top of the block. The refractive index of the liquid is :-

(A) 1.21            (B) 1.30              (C) 1.36               (D) 1.42

[JEE-Advance-2014]

Sol. (C)

From the given situation, at critical angle,

$\tan \theta=\frac{(\mathrm{d} / 2)}{\mathrm{h}}=\frac{5.77}{10}$

$\therefore \sin \theta_{\mathrm{C}} \approx \frac{1}{2}$

$\mu_{\text {denser }} \sin \theta_{\mathrm{C}}=\mu_{\text {raver }} \sin (\pi / 2)$

$\Rightarrow 2.72 \times \frac{1}{2}=\mu_{\mathrm{r}} \Rightarrow \mu_{\mathrm{r}}=1.36$


Q. Four combinations of two thin lenses are given in List I. The radius of curvature of all curved surfaces is r and the refractive index of all the lenses is 1.5. Match lens combinations in List I with their focal length in List II and select the correct answer using the code given below the lists.

[JEE-Advance-2014]

Sol. (B)


Q. Consider a concave mirror and a convex lens (refractive index = 1.5) of focal length 10 cm each, separated by a distance of 50 cm in air (refractive index = 1) as shown in the figure. An object is placed at a distance of 15 cm from the mirror. Its erect image formed by this combination has magnification $\mathrm{M}_{1}$. When the set-up is kept in a medium of refractive index 7/6 the magnification becomes $\mathrm{M}_{2}$. The magnitude $\left|\frac{M_{2}}{M_{1}}\right|$ is.

[JEE-Advance-2015]

Sol. 7

For reflectionfrom concave mirror,

$\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}} \Rightarrow \frac{1}{\mathrm{v}}-\frac{1}{15}=\frac{-1}{10}$

$\frac{1}{\mathrm{v}}=\frac{1}{15}-\frac{1}{10}=\frac{-1}{30}$

$\therefore \mathrm{v}=-30$

magnification $\left(\mathrm{m}_{1}\right)=-\frac{\mathrm{v}}{\mathrm{u}}=-2$

Now for refraction from lens,

$\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}} \Rightarrow \frac{1}{\mathrm{v}}=\frac{1}{10}-\frac{1}{20}=\frac{1}{20}$

$\therefore$ magnification $\left(\mathrm{m}_{2}\right)=\frac{\mathrm{v}}{\mathrm{u}}=-1$

$\therefore \mathrm{M}_{1}=\mathrm{m}_{1} \mathrm{m}_{2}=2$

Now when the set-up is immersed in liquid, no effect for the image formed by mirror.

we have $\left(\mu_{\mathrm{L}}-1\right)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)=\frac{1}{10}$

$\Rightarrow\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)=\frac{1}{5}$

when lens is immersed in liquid,

$\frac{1}{\mathrm{f}_{\mathrm{lens}}}=\left(\frac{\mu_{\mathrm{L}}}{\mu_{\mathrm{S}}}-1\right)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)=\frac{2}{7} \times \frac{1}{5}=\frac{2}{35}$

$\therefore \frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}_{\mathrm{Liquid}}}$

$\Rightarrow \frac{1}{\mathrm{v}}=\frac{2}{35}-\frac{1}{20}=\frac{8-7}{140}=\frac{1}{140}$

$\therefore$ magnification $=-\frac{140}{20}=-7$

$\therefore \mathrm{M}_{2}=2 \times 7=14$

$\therefore\left|\frac{\mathrm{M}_{2}}{\mathrm{M}_{1}}\right|=7$


Q. Two identical glass rods $\mathrm{S}_{1}$ and $\mathrm{S}_{2}$ (refractive index = 1.5) have one convex end of radius of curvature 10 cm. They are placed with the curved surfaces at a distance d as shown in the figure, with their axes (shown by the dashded line) aligned. When a point source of light P is placed inside rod $\mathrm{S}_{1}$ on its axis at a distance of 50 cm from the curved face, the light rays emanating from it are found to be parallel to the axis inside $\mathrm{S}_{2}$. The distance d is :

(A) 60 cm         (B) 70 cm           (C) 80 cm           (D) 90 cm

[JEE-Advance-2015]

Sol. (B)

For first surface

$\frac{1}{V}-\frac{1.5}{-50}=\frac{1-1.5}{-10}$

V = 50 cm

for second surface

$\frac{1.5}{\infty}-\frac{1}{-(\mathrm{d}-50)}=\frac{1.5-1}{10}$

d = 70 cm

$\therefore(\mathrm{B})$


Q. A monochromatic beam of light is incident at $60^{\circ}$ on one face of an equilateral prism of refractive index n and emerges from the opposite face making an angle $\theta$(n) with the normal (see the figure). For n = $\sqrt{3}$ the value of $\theta$ is $60^{\circ}$ and $\frac{\mathrm{d} \theta}{\mathrm{dn}}=\mathrm{m}$. The value of m is.

[JEE-Advance-2015]

Sol. 2

By snell’s law

$1 \sin 60=n \sin r_{1} \Rightarrow \sin r_{1}=\frac{1}{2} r_{1}=30^{\circ} \quad \ldots \ldots(i)$

By differentiating ‘w.r.t’ n

$\mathrm{O}=\sin \mathrm{r}_{1}+\mathrm{n} \cos \mathrm{r}_{1}\left(\frac{\mathrm{dr}_{1}}{\mathrm{dn}}\right)$

$=\frac{1}{2}+\sqrt{3}(\sqrt{\frac{3}{2}}) \frac{\mathrm{dr}_{1}}{\mathrm{dn}}$

$\frac{\mathrm{d} \mathrm{r}_{1}}{\mathrm{dn}}=\frac{1}{3}$ ….(ii)

By applying snell’s law

$n \sin r_{2}=1 \sin \theta$

$n \sin \left(60-r_{1}\right)=1 \sin \theta\left[\therefore A=r_{1}+r_{2}\right]$

By diffrentiating ‘w.r.t’ n

$\sin \left(60-r_{1}\right)-n \cos \left(60-r_{1}\right) \frac{d r_{1}}{d n}=\cos \theta \frac{d \theta}{d n}$$\sin \left(60-r_{1}\right)-n \cos \left(60-r_{1}\right) \frac{d r_{1}}{d n}=\cos \theta \frac{d \theta}{d n}$

By substituting value of $\mathrm{r}_{1}^{\prime}$ and $\frac{\mathrm{dr}_{1}}{\mathrm{dn}}$ from ( 1) and ( 2)

$\frac{\mathrm{d} \theta}{\mathrm{dn}}=2$


Paragraph for Question No. 23 and 24

Light guidance in an optical fiber can be understood by considering a structure comprising of thin solid glass cylinder of refractive index $\mathbf{n}_{1}$ surrounded by a medium of lower refractive index $\mathrm{n}_{2}$. The light guidance in the structure takes place due to successive total internal reflections at the interface of the media $\mathbf{n}_{1}$ and $\mathrm{n}_{2}$ as shown in the figure. All rays with the angle of incidence i less than a particular value of $\dot{l}_{\mathrm{m}}$ are confined in the medium of refractive index $\mathrm{n}_{1}$. The numerical aperture (NA) of the structure is defined as sin $i_{\mathrm{m}}$.

Q. For two structures namely $\mathrm{S}_{1}$ with $\mathrm{n}_{1}=\sqrt{45} / 4$ and $\mathrm{n}_{2}=3 / 2,$ and $\mathrm{S}_{2}$ with $\mathrm{n}_{1}=8 / 5$ and $\mathrm{n}_{2}=7 / 5$ and

taking the refractive index of water to be $4 / 3$ and that of air to be $1,$ the correct option(s) is (are)

(A) NA of $\mathrm{S}_{1}$ immersed in water is the same as that of $\mathrm{S}_{2}$ immersed in liquid of refractive index $\frac{16}{3 \sqrt{15}}$

(B) NA of $\mathrm{S}_{1}$ immersed in liquid of refractive index $\frac{6}{\sqrt{15}}$ is the same as that of $\mathrm{S}_{2}$ immersed in water.

(C) NA of $\mathrm{S}_{1}$ placed in air is the same as that of $\mathrm{S}_{2}$ immersed in liquid of refractive index .

(D) NA of $\mathrm{S}_{1}$ placed in air is the same as that of $\mathrm{S}_{2}$ placed in water.

[JEE-Advance-2015]

Sol. (A,C)

Let the whole structure is placed in a medium of refractive index n’, then

$n^{\prime} \sin i=n_{1} \sin (90-\theta)$

$\mathrm{n}^{\prime} \sin \mathrm{i}=\mathrm{n}_{1} \cos \theta \quad \ldots(\mathrm{i})$

Here for $\mathrm{i}_{\mathrm{m}} ; \quad \theta=\mathrm{C}$ and $\sin \mathrm{C}=\frac{\mathrm{n}_{2}}{\mathrm{n}_{1}}$

from eq. (i), $\mathrm{n}^{\prime} \sin \mathrm{i}_{\mathrm{m}}=\mathrm{n}_{1} \sqrt{\frac{1-\mathrm{n}_{2}^{2}}{\mathrm{n}_{1}^{2}}}=\sqrt{\mathrm{n}_{1}^{2}-\mathrm{n}_{2}^{2}}$

$\Rightarrow \sin i_{m}=\frac{\sqrt{n_{1}^{2}-n_{2}^{2}}}{n^{\prime}}$

Now, for $(\mathrm{A})(\mathrm{NA})_{\mathrm{s} 1}=\frac{3}{4} \sqrt{\frac{45}{16}-\frac{9}{4}}=\frac{3}{4} \times \frac{3}{4}=\frac{9}{16}$

$(\mathrm{NA})_{\mathrm{s} 2}=\frac{3 \sqrt{15}}{16} \sqrt{\frac{64}{25}-\frac{49}{25}}=\frac{3 \sqrt{15}}{16} \frac{1}{5} \sqrt{15}=\frac{9}{16}$

For (B) $\quad(\mathrm{NA})_{\mathrm{s} 1}=\frac{\sqrt{15}}{6} \times \frac{3}{4}=\frac{\sqrt{15}}{8}$

$(\mathrm{NA})_{\mathrm{s} 2}=\frac{3}{4}=\frac{\sqrt{15}}{5}$ Not equal

For $(\mathrm{C})(\mathrm{NA})_{\mathrm{s} 1}=1 \times \frac{3}{4}=\frac{3}{4}$

$(\mathrm{NA})_{\mathrm{s} 2}=\frac{\sqrt{15}}{4} \times \frac{\sqrt{15}}{5}=\frac{15}{4 \times 5}=\frac{3}{4}$

For (D) $(\mathrm{NA})_{\mathrm{s} 1}=\frac{3}{4}$

$(\mathrm{NA})_{\mathrm{s} 2}=\frac{3}{4} \frac{\sqrt{15}}{5}$ Not equal


Q. If two structures of same cross-sectional area, but different numerical apertures $\mathrm{NA}_{1}$ and $\mathrm{NA}_{2}$ $\left(\mathrm{NA}_{2}<\mathrm{NA}_{1}\right)$ are joined longitudinally, the numerical aperture of the combined structure is

(A) $\frac{\mathrm{NA}_{1} \mathrm{NA}_{2}}{\mathrm{NA}_{1}+\mathrm{NA}_{2}}$

$(\mathrm{B}) \mathrm{NA}_{1}+\mathrm{NA}_{2}$

$(\mathrm{C}) \mathrm{NA}_{1}$

$(\mathrm{D}) \mathrm{NA}_{2}$

[JEE-Advance-2015]

Sol. (D)

$\begin{array}{ll}{\text { It is given that }} & {\mathrm{NA}_{2}<\mathrm{NA}_{1}} \\ {\Rightarrow \mathrm{i}_{\mathrm{m} 2}<\mathrm{i}_{\mathrm{m} 1}}\end{array}$

Hence if the combination can be placed both ways i.e. 1st structure & then 2nd structure and then reversed also, then the condition of TIR is satisfied for lower $\dot{\mathbf{1}}_{\mathrm{m}}$ then it can be satisfied for all other less angler as well.

Hence $\mathrm{NA}_{2}$ will be the numerical aperture of the combined structure.


Q. A parallel beam of light is incident from air at an angle  on the side PQ of a right angled triangular prism of refractive index $\mathrm{n}=\sqrt{2}$. Light undergoes total internal reflection in the prism at the face PR when  has a minimum value of $45^{\circ}$. The angle $\theta$ of the prism is :

(A) $15^{\circ}$

(B) $22.5^{\circ}$

(C) $30^{\circ}$b

(D) $45^{\circ}$

[JEE-Advance-2016]

Sol. (A)

$1 \sin 45^{\circ}=\sqrt{2} \sin \mathrm{r}_{1}$

$\mathrm{r}_{2}-\mathrm{r}_{1}=\theta$

$\theta=45^{\circ}-30^{\circ}$

$\Rightarrow \theta=15^{\circ}$


Q. A transparent slab of thickness d has a refractive index n(z) that increases with z. Here z is the vertical distance inside the slab, measured from the top. The slab is placed between two media with uniform refractive indices $\mathrm{n}_{1}$ and $\mathrm{n}_{2}\left(>\mathrm{n}_{1}\right)$, as shown in the figure. A ray of light is incident with angle $\theta_{\mathrm{i}}$ from medium 1 and emerges in medium 2 with refraction angle $\theta_{\mathrm{f}}$ with a lateral displacement . Which of the following statement(s) is(are) true ?

(A) $\ell$ is independent of $\mathrm{n}_{2}$

(B) $\ell$ is dependent on $\mathrm{n}(\mathrm{z})$

(C) $\mathrm{n}_{1} \sin \theta_{\mathrm{i}}=\left(\mathrm{n}_{2}-\mathrm{n}_{1}\right) \sin \theta_{\mathrm{f}}$

(D) $\mathrm{n}_{1} \sin \theta_{\mathrm{i}}=\mathrm{n}_{2} \sin \theta_{\mathrm{f}}$

[JEE-Advance-2016]

Sol. (A,B,D)


Q. A plano-convex lens is made of a material of refractive index n. When a small object is placed 30 cm away in front of the curved surface of the lens, an image of double the size of the object is produced. Due to reflection from the convex surface of the lens, another faint image is observed at a distance of 10 cm away from the lens. Which of the following statement(s) is(are) true?

(A) The refractive index of the lens is 2.5

(B) The radius of curvature of the convex surface is 45 cm

(C) The faint image is erect and real

(D) The focal length of the lens is 20 cm.

[JEE-Advance-2016]

Sol. (A,D)


Q. A small object is placed 50 cm to the left of thin convex lens of focal length 30 cm. A convex spherical mirror of radius of curvature 100 cm is placed to the right of the lens at a distance of 50 cm. The mirror is tilted such that the axis of the mirror is at an angle  = $30^{\circ}$ to the axis of the lens, as shown in the figure. If the origin of the coordinate system is taken to be at the centre of the lens, the coordinates (in cm) of the point (x, y) at which the image is formed are :

(A) $(25,25 \sqrt{3})$

$(\mathrm{B})\left(\frac{125}{3}, \frac{25}{\sqrt{3}}\right)$

(C) $(50-25 \sqrt{3}, 25)$

(D) (0, 0)

[JEE-Advance-2016]

Sol. (A)

For lens $\mathrm{V}=\frac{(-50)(30)}{-50+30}=75$

For mirror $\mathrm{V}=\frac{\left(\frac{25 \sqrt{3}}{2}\right)(50)}{\frac{25 \sqrt{3}}{2}-50}=\frac{-50 \sqrt{3}}{4-\sqrt{3}}$

$\mathrm{m}=-\frac{\mathrm{v}}{\mathrm{u}}=\frac{\mathrm{h}_{2}}{\mathrm{h}_{1}} \Rightarrow \mathrm{h}_{2}=-\left(\frac{\frac{-50 \sqrt{3}}{4-\sqrt{3}}}{\frac{25 \sqrt{3}}{2}}\right) \cdot \frac{25}{2}$

$\mathrm{h}_{2}=\frac{+50}{4-\sqrt{3}}$

The x coordinate of the images $=50-v \cos 30+h_{2} \cos 60 \approx 25$

The y coordinate of the images $=v \sin 30+h_{2} \sin 60 \approx 25 \sqrt{3}$


Q. For an isosceles prism of angle A and refractive index µ, it is found that the angle of minimum deviation $\delta_{\mathrm{m}}$ = A. Which of the following options is/are correct ?

(A) At minimum deviation, the incident angle $\dot{\mathbf{1}}_{1}$ and the refracting angle $\mathrm{r}_{1}$ at the first refracting surface are related by $\mathbf{r}_{1}=\left(\mathbf{i}_{1} / 2\right)$

(B) For this prism, the refractive index µ and the angle of prism A are related as $\mathrm{A}=\frac{1}{2} \cos ^{-1}\left(\frac{\mu}{2}\right)$

(C) For this prism, the emergent ray at the second surface will be tangential to the surface when the angle of incidence at the first surface is $\mathrm{i}_{1}=\sin ^{-1}[\sin \mathrm{A} \sqrt{4 \cos ^{2} \frac{\mathrm{A}}{2}-1}-\cos \mathrm{A}]$

(D) For the angle of incidence $\mathbf{i}_{1}$ = A, the ray inside the prism is parallel to the base of the prism.

[JEE-Advance-2017]

Sol. (A,C,D)

i = e (for minimum deviation)

$\mathrm{r}_{1}+\mathrm{r}_{2}=\mathrm{A}, \mathrm{r}_{1}=\mathrm{r}_{2}$

(A) $\delta_{\mathrm{m}}=2 \mathrm{i}-\mathrm{A}=\mathrm{A}$ (given)

$\Rightarrow \mathrm{i}=\mathrm{A}$

$\Rightarrow \mathrm{r}_{1}=\frac{\mathrm{A}}{2}=\frac{\mathrm{i}}{2}$

(B) $\mu=\frac{\sin (\mathrm{A})}{\sin (\mathrm{A} / 2)}=2 \cos \frac{\mathrm{A}}{2} \Rightarrow \mathrm{A}=2 \cos ^{-1}\left(\frac{\mu}{2}\right)$

(C) $\mu \sin \left(\mathrm{r}_{2}\right)=1$

$\sin \left(\mathrm{r}_{2}\right)=\frac{1}{\mu}$

$\mathrm{r}_{1}+\mathrm{r}_{2}=\mathrm{A}$

$\mathrm{r}_{1}=\mathrm{A}-\mathrm{r}_{2}$

$=\mathrm{A}-\sin ^{-1}\left[\frac{1}{\mu}\right]$

$\sin (\mathrm{i})=\mu \sin \left(\mathrm{r}_{1}\right)$

$\mathrm{i}=\sin ^{-1}\left[\mu \sin \left[\mathrm{A}-\sin ^{-1}\left[\frac{1}{\mu}\right]\right]\right.$

$\mathrm{i}_{\mathrm{g}}=\sin ^{-1}[\sqrt{\mu^{2}-1} \sin \mathrm{A}-\cos \mathrm{A}]=\sin ^{-1}\left[\mu \sin \left(\mathrm{A}-\theta_{\mathrm{C}}\right)\right]$

(Here $\left.\mu=2 \cos \frac{\mathrm{A}}{2}\right)$

(D) Condition of min. deviation $\mathrm{i}=\mathrm{e} \& \mathrm{r}_{1}=\mathrm{r}_{2}=\frac{\mathrm{A}}{2}$

Rays will be parallel to base.


Q. A monochromatic light is travelling in a medium of refractive index n = 1.6. It enters a stack of glass layers from the bottom side at an angle $\theta=30^{\circ}$. The interfaces of the glass layers are parallel to each other. The refractive indices of different glass layers are monotonically decreasing as $\mathrm{n}_{\mathrm{m}}=\mathrm{n}-\mathrm{m} \Delta \mathrm{n}$, where $\mathrm{n}_{\mathrm{m}}$ is the refractive index of the mth slab and $\Delta \mathrm{n}=0.1$ (see the figure). The ray is refracted out parallel to the interface between the (m – 1)th and mth slabs from the right side of the stack. What is the value of m ?

[JEE-Advance-2017]

Sol. 8

Applying snell’s law between entry & exit surfaces,

n $\sin \theta=\mu \sin \left(\frac{\pi}{2}\right)$


Q. Sunlight of intensity $1.3 \mathrm{kW} \mathrm{m}^{-2}$ is incident normally on a thin convex lens of focal length 20 cm. Ignore the energy loss of light due to the lens and assume that the lens aperture size is much smaller than its focal length. The average intensity of light, in kW m–2, at a distance 22 cm from the lens on the other side is

[JEE-Advance-2018]

Sol. 130

$\frac{\mathrm{r}}{\mathrm{R}}=\frac{2}{20}=\frac{1}{10}$

$\therefore$ Ratio of area $=\frac{1}{100}$

Let energy incident on lens be E.

$\therefore$ Given $\frac{\mathrm{E}}{\mathrm{A}}=1.3$

So final, $\frac{\mathrm{E}}{\mathrm{a}}=? ?$

E = A × 1.30

Also $\frac{\mathrm{a}}{\mathrm{A}}=\frac{1}{100}$

$\therefore$ Average intensity of light at $22 \mathrm{cm}=\frac{\mathrm{E}}{\mathrm{a}}=\frac{\mathrm{A} \times 1.3}{\mathrm{a}}=100 \times 1.3=130 \mathrm{kW} / \mathrm{m}^{2}$


Q. A wire is bent in the shape of a right angled triangle and is placed in front of a concave mirror of focal length f, as shown in the figure. Which of the figures shown in the four options qualitatively represent(s) the shape of the image of the bent wire ? (These figures are not to scale.) ?

[JEE-Advance-2018]

Sol. (D)

Distance of point A is f/2

Let A’ is the image of A from mirror, for this image

$\frac{1}{\mathrm{v}}+\frac{1}{-\mathrm{f} / 2}=\frac{1}{-\mathrm{f}}$

$\frac{1}{\mathrm{v}}=\frac{2}{\mathrm{f}}-\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{f}}$

image of line AB should be perpendicular to the principle axis & image of F will form at infinity, therefor correct image diagram is