- Some paramagnetic substances are $A \ell, N a,$ Sb, Pt, $\operatorname{CuC} \ell_{2}, M n,$ Cr, liquid oxygen etc.
Paramagnetic Substances – Magnetism & Matters | Class 12 Physics Notes
Paramagnetic substances
The substances which when placed in a magnetic field are feebly magnetised in the direction of magnetising field are called paramagnetic substances.
(1) The property of paramagnetism is found to exist in substances whose atoms or molecules have an excess of electron spinning in same direction.
(2) Atoms of paramagnetic substances possess a permanent magnetic dipole moment and thus behave like small bar magnets called atomic magnets.
(3) In absence of external magnetic field paramagnetic substances to not show any magnetism because atomic magnets are randomly oriented so net magnetic dipole moment is zero.
(4) In presence of external field each atomic magnet experiences a torque which tries to rotate and align them parallel to direction of magnetic field. The substance acquires net dipole moment and thus gets magnetised in direction of field.
(5) The property of paramagnetism is temperature dependent. The thermal agitation on increase of temperature spoils the alignment of atomic magnets which reduces net magnetic dipole moment.
JEE Advanced 2020 Admit Card (Postponed) | Exam on 23rd August ’20
IIT Delhi has postponed the JEE Advanced 2020 Admit Card release along with other events. Candidates who will complete the registration process before the last date, will be able to access the JEE Advanced admit card 2020 in online mode by providing their registration number, date of birth, mobile number and email ID. Information regarding the examination such as the venue address, timings, date and more will be made known through JEE Advanced admit card 2020. Without the admit card of JEE Advanced 2020, the candidates will not be allowed to attempt the examination. Since it is an important document, the candidates are advised to keep the JEE Advanced 2020 admit card safely till the end of the admission procedure.
How to Download JEE Advanced 2020 Admit Card?
To download JEE Advanced hall ticket 2020, candidates need to follow the steps given below:- Visit the official website of JEE Advanced 2020
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JEE Advanced Admit Card Important Dates:
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Dates |
| Release of JEE Advanced 2020 admit card/hall ticket | To be announced |
| Last date to download JEE Advanced 2020 admit card | To be announced |
| JEE Advanced 2020 exam | 23-Aug-2020 |
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Documents to Carry to JEE Advanced Test Centre
Along with JEE Advanced 2020 admit card, candidates need to carry one of the following original photo identity proofs:- Aadhaar Card
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JEE Advanced Previous Year Question Papers are available here.
JEE Advanced 2020 Syllabus
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JEE Advanced 2020 Exam Date Announced, Check Syllabus – Important Information
Joint Entrance Examination, JEE Advanced or IIT JEE Exam dates 2020 have been announced. Union HRD Minister Ramesh Pokhriyal today shared that the JEE Advanced 2020 would be conducted on August 23, 2020. The dates for JEE Main 2020 were announced on May 5 by the HRD Minister during a webinar with the students.
JEE Main 2020 would be conducted from July 18 to July 23. Given the date of the JEE Advanced 2020, it can be suggested that the results of JEE Main 2020 are expected to be announced by August 10. The online application process for JEE Advanced 2020 would begin only after the final results of JEE Main 2020 are announced along with the JEE Main 2020 Ranks
All The Best 🙂
JEE Advanced 2020 Exam Date: 23rd August ’20
JEE Main 2020 (Revised dates by MHRD): July 18th to July 23rd ’20
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.JEE Advanced Result 2020
Results for JEE Advanced 2020 will tentatively be released by the IIT Delhi in the second week of August. The result will comprise the exam scores, All India Rank (AIR) of candidates, subject-wise marks, etc.
Students may please note that the timeline provided above is only based on the usual time frame followed by the IITs in regards the examination. The actual schedule would be announced by IIT Delhi in a few days time. IIT JEE or JEE Advanced 2020 is conducted by the IITs while JEE Main, the qualifier is conducted by National Testing Agency. Both JEE Main and JEE Advanced 2020 would be online or computer based tests.
Eligibility Criteria
Candidates must fulfill one of the following two criteria in order to be eligible for admission at IIT:
- 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.
- 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.
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.
JEE Advanced Previous Year Question Papers are available here.
JEE Advanced 2020 Syllabus
Click Here for JEE Advanced Complete Syllabus
All The Best 🙂
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.
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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.
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Previous Years JEE Advanced Questions
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Sol. ((A) $p, s ;(B) q ;(C) t ;(D) r, s, t$) (A) $\Delta \mathrm{x}=\mathrm{S}_{2} \mathrm{P}-\mathrm{S}_{1} \mathrm{P}=0$ $\delta\left(\mathrm{P}_{0}\right)=\frac{2 \pi}{\lambda} \Delta \mathrm{x}=0$ $\Delta \mathrm{x}=\mathrm{S}_{1} \mathrm{P}_{1}-\mathrm{S}_{2} \mathrm{P}_{1}=\frac{\lambda}{4}$ $\delta\left(\mathrm{P}_{1}\right)=\frac{2 \pi}{\lambda} \times \frac{\lambda}{4}=\frac{\pi}{2}$ $\mathrm{I}=\mathrm{I}_{\max } \cos ^{2}\left(\frac{\Delta \phi}{2}\right)$ $\mathrm{I}\left(\mathrm{P}_{1}\right)=\mathrm{I}_{1}=\mathrm{I}_{\max } \cos ^{2} \frac{\delta}{2}=\frac{\mathrm{I}_{\max }}{2}$ $\delta\left(\mathrm{P}_{2}\right)=\frac{2 \pi}{\lambda} \times \frac{\lambda}{3}=\frac{2 \pi}{3}$ $\mathrm{I}\left(\mathrm{P}_{2}\right)=\mathrm{I}_{2}=\mathrm{I}_{\max } \cos ^{2} \frac{\pi}{3}=\frac{\mathrm{I}_{\max }}{4}$ $\mathrm{I}\left(\mathrm{P}_{0}\right)>\mathrm{I}\left(\mathrm{P}_{1}\right)$ $(\mathrm{B}) \Delta \mathrm{x}=\mathrm{S}_{1} \mathrm{P}-\left[\mathrm{S}_{2} \mathrm{P}+(\mu-1) \mathrm{t}\right]$ $\Delta \mathrm{x}_{1}=\mathrm{S}_{1} \mathrm{P}_{1}-\mathrm{S}_{2} \mathrm{P}_{1}-(\mu-1) \mathrm{t}$ $\Delta \mathrm{x}_{1}=\frac{\lambda}{4}-\frac{\lambda}{4}=0$ $8\left(\mathrm{P}_{1}\right)=0 ; \mathrm{I}\left(\mathrm{P}_{1}\right)=\mathrm{I}_{\max }$ $8\left(\mathrm{P}_{0}\right)=\frac{\pi}{2} \delta\left(\mathrm{P}_{0}\right) \neq 0$ $\mathrm{I}\left(\mathrm{P}_{0}\right)=\mathrm{I}_{\max } / 2$ $\Delta \mathrm{x}=\mathrm{S}_{1} \mathrm{P}_{2}-\mathrm{S}_{1} \mathrm{P}_{2}-(\mu-1) \mathrm{t}$ $=\frac{\lambda}{3}-\frac{\lambda}{4}=\frac{\lambda}{12}$ $8\left(\mathrm{P}_{2}\right)=\frac{2 \pi}{\lambda} \times \frac{\lambda}{12}=\frac{\pi}{6}$ $\mathrm{I}\left(\mathrm{P}_{2}\right)=\mathrm{I}_{\max } \cos ^{2}\left(\frac{\pi}{12}\right)$
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Sol. (D) $\beta=\frac{\mathrm{D} \lambda}{\mathrm{d}}$ $\lambda_{\mathrm{R}}>\lambda_{\mathrm{a}}>\lambda_{\mathrm{B}}$
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Sol. (B) $\frac{\mathrm{I}_{\max }}{2}=\mathrm{I}_{\max } \cos ^{2}\left(\frac{\pi}{\lambda} \Delta \mathrm{x}\right)$ $\cos ^{2}\left(\frac{\pi}{\lambda} \Delta \mathrm{x}\right)=\frac{1}{2}$ $\cos \left(\frac{\pi}{\lambda} \Delta \mathrm{x}\right)=\pm \frac{1}{\sqrt{2}}$ $\frac{\pi}{\lambda} \Delta \mathrm{x}=\mathrm{n} \pi \pm \frac{\pi}{4}$ $\Delta \mathrm{x}=\left(\mathrm{n} \pm \frac{1}{4}\right) \lambda$
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Sol. (A,B,C) $\beta=\frac{\mathrm{D} \lambda}{\mathrm{d}}$ $\mathrm{B}_{2}>\beta_{1}$ $\mathrm{y}=\mathrm{m}_{1} \frac{\mathrm{D} \lambda_{1}}{\mathrm{d}}=\mathrm{m}_{2} \frac{\mathrm{D} \lambda_{2}}{\mathrm{d}}$ $\frac{\mathrm{nD} \times \lambda_{2}}{\mathrm{d}}=\left(\mathrm{n}^{\prime}+\frac{1}{2}\right) \frac{\mathrm{D} \lambda_{1}}{\mathrm{d}} \Rightarrow 600 \mathrm{n}=\left(\mathrm{n}^{\prime}+\frac{1}{2}\right) \times 4$
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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)
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Sol. (C,D)
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]
[IIT-JEE-2009]
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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]
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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]
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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]
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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]
[JEE Advanced 2015]
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]
[JEE-Mains 2016]
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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]
(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]
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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.
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Previous Years JEE Advance Questions
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Sol. ($(A) \rightarrow P, Q, S, T:(B) \rightarrow P, S, T ;(C) \rightarrow P:(D) \rightarrow R$)
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Sol. ($(\mathrm{A}) \rightarrow \mathrm{P}, \mathrm{Q}, \mathrm{T} ;(\mathrm{B}) \rightarrow \mathrm{P}, \mathrm{Q}, \mathrm{S}, \mathrm{T},(\mathrm{C}) \rightarrow \mathrm{R}, \mathrm{S}, ;(\mathrm{D}) \rightarrow \mathrm{P}$)
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Sol. (A,C,D)
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Sol. (B,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)
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. Match each of the compound in Column I with its characteristic reaction(s) in Column II.
[IIT 2009]
[IIT 2009]
Q. Match each of the compounds given in Column I with the reaction(s), that they can undergo, given in Column II.
[IIT 2009]
[IIT 2009]
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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]
[IIT 2011]
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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]
(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]
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]
[IIT 2012]
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Sol. (A,C,D)
Q. The total number of carboxylic acid groups in the product P is
[IIT 2013]
[IIT 2013]
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Sol. (B,D)
Q. In the reaction shown below, the major product(s) formed is / are :
[IIT 2014]
[IIT 2014]
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]
[IIT 2014]
Q. The major product of the reaction is :
[IIT 2015]
[IIT 2015]
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]
[JEE Adv. 2018]
Q. In the following reaction sequence, the correct structure(s) of X is (are)
[JEE Adv. 2018]
[JEE Adv. 2018]
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]
[JEE Adv. 2018]
Q. The compound Z is :-
[JEE Adv. 2018]
[JEE Adv. 2018]
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]
Q. The compound R is
(A)
(B)
(C)
(D)
[JEE Adv. 2018]
(B) 
(C) 
(D) 
[JEE Adv. 2018]
Q. The compound S is
[JEE Adv. 2018]
[JEE Adv. 2018]
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.
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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]
Q. The number of neutrons emitted when $_{92}^{235} \mathrm{U}$ undergoes controlled nuclear fission to $_{54}^{142} \mathrm{Xe}$ and is –
[JEE 2010]
is –
[JEE 2010]
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]
(A) proton, neutron, positron
(B) neutron, positron, proton
(C) proton, positron, neutron
(D) positron, proton, neutron
[JEE 2011]
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]
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]
(X, Y) is(are)
$(\mathrm{A})(\gamma, \mathrm{n})$
(B) (p, D)
(C) (n, D)
$(\mathrm{D})(\gamma, \mathrm{D})$
[JEE 2013]
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]
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.
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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.
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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 $(+2 \&+3)$
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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.
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Sol. (C,D) Being a more reactive metal “Al” and “Mg” can be reduced by electrolytic reduction.
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Sol. (A,C,D)
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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.
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Sol. $(A \rightarrow P, Q, S, B \rightarrow T, C \rightarrow Q, R, D \rightarrow R)$ Name of ores & their metals.
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]
Q. Iron is removed from chalcopyrite as :-
(A) FeO
(B) FeS
(C) $\mathrm{Fe}_{2} \mathrm{O}_{3}$
(D) FeSiO $_{3}$
[JEE-2010]
Q. In self-reduction, the reducing species is :-
(A) S
(B) $\mathrm{O}^{2-}$
(C) $\mathrm{S}^{2-}$
(D) $\mathrm{SO}_{2}$
[JEE-2010]
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]
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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]
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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]
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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]
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]
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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]
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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]
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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]
[JEE Adv. 2015]
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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]
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.
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Previous Years JEE Advance Questions
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Sol. (A) $\mathrm{P}_{\mathrm{N}_{2}}=\mathrm{K}_{\mathrm{H}} \mathrm{X}_{\mathrm{N}_{2}}$ $\mathrm{Y}_{\mathrm{N}_{2}} \cdot \mathrm{P}_{\mathrm{T}}=\mathrm{K}_{\mathrm{H}} \times \mathrm{N}_{2}$ $0.8 \times 5=1 \times 10^{5} \times \frac{\mathrm{n}}{\mathrm{n}+10}$ $4=10^{5} \times \frac{\mathrm{n}}{10}$ $\mathrm{n}=4 \times 10^{-4}$
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Sol. (A) $\mathrm{T}_{\mathrm{f}}^{\prime}=\mathrm{T}_{\mathrm{f}}-\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{T}_{\mathrm{f}}-\mathrm{i} \mathrm{K}_{\mathrm{f}} \cdot \mathrm{m}$ $=0^{\circ} \mathrm{C}-4 \times 1.86 \times \frac{0.1 / 329}{100 / 1000}$ $=-0.023^{\circ} \mathrm{C}=-2.3 \times 10^{2} \mathrm{C}$
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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
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Sol. (B,C,D)
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Sol. 8 $\mathrm{m}=\frac{3.2 \mathrm{mol}}{0.4 \mathrm{Kg}}=8 \mathrm{mol} / \mathrm{kg}$
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Sol. 1 0.0558 = i × 1.86 × 0.01 i = 3 $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2}$
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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.
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Sol. (A,C)
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Sol. (D) Ethanol should be considered non volatile as per given option $\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{K}_{\mathrm{f}} \times \mathrm{m}$ $\Delta \mathrm{T}_{\mathrm{f}}=2 \times \frac{34.5}{46 \times 0.5}=3 \mathrm{K}$
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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$
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Sol. 0.05 From graph
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]
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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]
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]
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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]
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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]
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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]
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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]
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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]
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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]
(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]
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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]
[JEE – Adv. 2017]
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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]
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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]
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]
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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.
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Previous Years JEE Advance Questions
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Sol. (A,D)
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Sol. (B,D)
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Sol. (B,C)
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Sol. (A,B,C)
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]
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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]
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]
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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]
[iit-2011]
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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]
(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]
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Sol. (A,B,C)
Q. The total number(s) of stable conformers with non-zero dipole moment for the following compound is (are) :
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[JEE-2014]
ElectroChemistry – JEE Advanced Previous Year Questions with Solutions
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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.
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Sol. (C) $\mathrm{E}_{1}=-\frac{059}{1} \log \frac{05}{1}=(+) \mathrm{ve} \Rightarrow \mathrm{so}$
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.
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Sol. (D) $\Delta \mathrm{G}=\frac{-2 \times 96500 \times .059}{1000}=11.4$
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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)
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Sol. (A,B) Fact
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Sol. (-11.62)
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]
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
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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) $\mathrm{E}_{\text {cell }}>0 ; \Delta \mathrm{G}^{\text {o }}<0$
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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]
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]
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]
(A) (P) (B) (Q) (C) (R) (D) (S)
[JEE 2011]
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]
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) $1 \times 10^{-12}$
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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]
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]
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]
[JEE-Adv. 2013]
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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]
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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]
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]
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]
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.)
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Sol. (-11.62)
Chemical Kinetics – JEE Advanced Previous Year Questions with Solutions
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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
ButNo 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.
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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.
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Sol. (A,B,D) $\mathrm{A}_{\mathrm{t}}=\mathrm{A}_{0} \mathrm{e}^{-\mathrm{kt}}$ for option (D) $\frac{1}{\mathrm{t}_{1 / 2}} \ln \frac{100}{50}=\frac{1}{\mathrm{t}_{99.6 \%}} \ln \frac{100}{0.4}$
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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
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Sol. (A,D)
Q. Plots showing the variation of the rate constant (k) with temperature (T) are given
below. The plot that follows Arrhenius equation is –
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[JEE 2010]
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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 :
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[JEE 2010]
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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.
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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]
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]
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]
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]
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]
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]
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]
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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]
[JEE Advance 2018]
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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]
Coordination Compounds – JEE Advanced Previous Year Questions with Solutions
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Sol. (C,D)
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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.
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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.
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Sol. (C) The correct structure of ethylenediaminetetra acetic acid (EDTA) is
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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 $\frac{V I B G Y O R}{\lambda-v^{-} E^{-}}$
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Sol. (B,D)
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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
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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
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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.
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Sol. 2992
Total mass = 12 × 172 + 4 × 232 = 2992 g
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Sol. (A,B,D)
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]
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]
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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]
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) $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4} \mathrm{Cl}_{2}\left(\mathrm{NO}_{2}\right)\right] \cdot \mathrm{H}_{2} \mathrm{O}$
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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]
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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]
[JEE 2010]
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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]
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]
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]
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]
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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]
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]
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]
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]
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]
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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]
[JEE Adv. 2014]
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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]
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]
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]
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]
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]
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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]
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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]
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]
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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]
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Sol. (A,B,D)
Q. Match each set of hybrid orbitals from LIST-I with complex (es) given in LIST-II.
[JEE – Adv. 2018]
[JEE – Adv. 2018]
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.
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Previous Years JEE Advance Questions
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Sol. (B,C)
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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. 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]
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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]
(A) 39.27% (B) 68.02% (C) 74.05% (D) 78.54%
[JEE-2010]
Q. The number of hexagonal faces that present in a truncated octahedron is.
[JEE-2011]
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]
(A) MX
(B) $\mathrm{MX}_{2}$
(C) $\mathrm{M}_{2} \mathrm{X}$
(D) $\mathrm{M}_{5} \mathrm{X}_{14}$
[JEE-2012]
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]
(A) 104 pm (B) 125 pm (C) 183 pm (D) 57 pm
[JEE-2013]
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]
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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]
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]
[JEE – Adv. 2017]
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.
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Previous Years JEE Advance Questions
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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.
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Sol. (A,B,D) general properties.
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Sol. (A,D) Theory Based
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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..
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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
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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.
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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.
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
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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
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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
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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
(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
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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
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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
JEE Adv. 2016
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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
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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.
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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.
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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.
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Sol. ( (a) $A ;(b) C ;(c) 7$ )
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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) $\quad \because \quad\left|\frac{1-4-2-\alpha}{3}\right|=5$ $\Rightarrow \alpha=10,-20$ $\Rightarrow \alpha=10 \because \alpha>0$
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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 : $5 \lambda+10-16 \lambda-12-\lambda-5=1$ $\Rightarrow \lambda=-\frac{2}{3}$ $\therefore \mathrm{P}\left(\frac{4}{3}, \frac{1}{3}, \frac{13}{3}\right)$
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Sol. (B,D)
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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 $\left|\begin{array}{ccc}{5-\alpha} & {0} & {0} \\ {0} & {3-\alpha} & {-2} \\ {0} & {-1} & {2-\alpha}\end{array}\right|=0$ $\Rightarrow \quad(5-\alpha)((3-\alpha)(2-\alpha)-2)=0$ $\Rightarrow \quad(5-\alpha)\left(\alpha^{2}-5 \alpha+4\right)=0$ $\Rightarrow \quad \alpha=1,4,5$
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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 $x-3 y-2 z=13$
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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 $\Rightarrow \lambda=-1$
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Sol. (B,D)
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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.
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Sol. (B,C,D)
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Sol. (C)
$\therefore x-4 y+7 z=0$
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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. (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]
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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)]
[JEE 2010, 3+5+3+(2+2+2+2)]
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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]
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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]
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)]
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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)]
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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)]
[JEE-Advanced 2013, 3, (–1)]
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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]
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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]
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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]
[JEE 2015, 4M, –2M]
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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]
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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]
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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]
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]
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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]
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
Definite integration – JEE Advanced Previous Year Questions with Solutions
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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 $f\left(\frac{1}{3}\right)<\frac{1}{3}$
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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) $2 \mathrm{I}_{1}=\int_{-\pi}^{\pi} \frac{\sin \mathrm{x} \mathrm{d} \mathrm{x}}{\sin \mathrm{x}}=2 \pi$ $\mathrm{I}_{1}=\pi$ $\sum_{m=1}^{10} I_{2 m+1}=10 \pi$
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Sol. 0 $\mathrm{f}(\mathrm{x})=\int_{0}^{\mathrm{x}} \mathrm{f}(\mathrm{t}) \mathrm{dt}$ $\mathrm{f}^{\mathrm{l}}(\mathrm{x})=\mathrm{f}(\mathrm{x})$ $\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{y}$ $\Rightarrow \int \frac{d y}{y}=\int d x$ $\Rightarrow \ln y=x+c$ $\Rightarrow y=e^{x+c}$ $\Rightarrow y=0$ $\left(\begin{array}{c}{\text { at } x=0, y=0} \\ {c \rightarrow-\infty}\end{array}\right)$ $f(x)=0$ $f(\ell n 5)=0$
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Sol. (B) Applying L-Hospital rule,
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Sol. (B)
from $(2), f^{-1}(2)=\frac{1}{3}$
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Sol. (A,B,D)
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Sol. (A,C)
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).
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Sol. (A,C)
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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 $\mathrm{x} \in \mathbb{D}$.
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Sol. (A,B,C)
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Sol. (C,D)
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Sol. (B,C)
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Sol. (Bonus)
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]
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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]
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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]
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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)]
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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]
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)]
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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]
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)]
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]
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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]
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)]
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]
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]
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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]
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)]
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)]
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)]
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)]
Q.
[JEE(Advanced)-2014, 3(–1)]
[JEE(Advanced)-2014, 3(–1)]
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]
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]
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]
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]
(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]
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]
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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]
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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]
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]
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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]
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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]
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]
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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]
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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]
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Previous Years JEE Advanced Questions
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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
$\mathrm{t} \approx 2 \mathrm{s}$
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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. 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]
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]
[JEE Advanced 2014]
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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]
[JEE Advanced-2014]
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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]
