datavorous/entrance-exam-dataset huggingface dataset:
Iron (III) catalyses the reaction between iodide and persulphate ions, in which
A. \(\mathrm{Fe}^{3+}\) oxidises the iodide ion
B. \(\mathrm{Fe}^{3+}\) oxidises the persulphate ion
C. \(\mathrm{Fe}^{2+}\) reduces the iodide ion
D. \(\mathrm{Fe}^{2+}\) reduces the persulphate ion
Choose the most appropriate answer from the options given below:

Options:
A B only
B A only
C B and C only
D A and D only

The correct answer is:
A and D only

Solution:
$\begin{aligned} & 2 \mathrm{Fe}^{3+}+2 \mathrm{I}^{-} \longrightarrow 2 \mathrm{Fe}^{2+}+\mathrm{I}_2 \\ & 2 \mathrm{Fe}^{2+}+\mathrm{S}_2 \mathrm{O}_8^{2-} \longrightarrow 2 \mathrm{Fe}^{3+}+2 \mathrm{SO}_4^{2-}\end{aligned}$
$\mathrm{Fe}^{+3}$ oxidises $\mathrm{I}^{-}$to $\mathrm{I}_2$ and convert itself into $\mathrm{Fe}^{+2}$. This $\mathrm{Fe}^{+2}$ reduces $\mathrm{S}_2 \mathrm{O}_8{ }^{2-}$ to $\mathrm{SO}_4{ }^{2-}$ and converts itself into $\mathrm{Fe}^{+3}$.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2024 (08 Apr Shift 1)
Irreversible precipitation of proteins is called

Options:
A Denaturation
B Hydrolysis
C Rearrangement
D Electrophoresis

The correct answer is:
Denaturation

Solution:

Topic: Chemistry, Biomolecules, JEE Main
____is a discrete packet of energy related to electromagnetic radiation (light), in which energy is E which is proportional to frequency of radiation $v$.

Options:
A Photon
B Thermion
C Neutron
D More than one of the above

The correct answer is:
Photon

Solution:
- Definition: A photon is a fundamental particle of light and other forms of electromagnetic radiation. It behaves both as a particle and as a wave, displaying characteristics of both.
- Energy and Frequency Relationship: The energy of a photon (E) is directly proportional to the frequency of the radiation \((v)\). This relationship is described by the equation \(\mathrm{E}=\) hv, where h is Planck's constant.
- Quantization of Energy: The concept of a photon arises from the quantization of energy in electromagnetic radiation. Energy is not continuous but exists in discrete packets called photons.
- Particle-like Behavior: Photons exhibit particle-like behavior, carrying a fixed amount of energy. They can interact with matter, transferring their energy and momentum in discrete units.

Topic: Physics, Atomic Physics, JEE Main
is a repeating unit for :

Options:
A Novolac
B Buna-N
C Acrilan
D Neoprene

The correct answer is:
Novolac

Solution:
Phenol+HCHO→H+Novolac (Given structure)→PolymerisationBakelite(Cross linked polymer)

Topic: Chemistry, Polymers, JEE Main, JEE Main 2021 (25 Jul Shift 1)
is called

Options:
A Etard reaction
B Perkin's reaction
C Claisen condensation
D Claisen Schmidt reaction

The correct answer is:
Etard reaction

Solution:
Etard's reaction

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
Is $g=\{(1,1),(2,3),(3,5),(4,7)\}$ a function ? If this is described by the formula $g(x)=\alpha x+\beta$, then what values should be assigned to $\alpha$ and $\beta$ ?

Options:

None

Solution:
As $\alpha=2, \beta=-1$
Domain $\mathrm{A}=\{1,2,3,4\}$
Range $\mathrm{B}=\{1,3,5,7\}$
Every element of domain has a unique image in $\mathrm{B}$ and hence $g$ is a function. Now $g(x)=\alpha x+\beta$ but $g(2)=3, g(3)=5$ $\therefore \quad 2=2 \alpha+\beta$ and $5=3 \alpha+\beta$ $\alpha=2, \beta=-1$

Topic: Mathematics, Relations and Functions (Class 11)
Is it possible that productivity and diversity of a natural community remains constant over a time period of, say one hundred years?

Options:

None

Solution:
Productivity and diversity of a natural community cannot remain constant over a certain time period.
This is due to the following reasons
(i) In real, the natural habitat is never maintained.
(ii) Resources available are never abundant, they are always in short supply or just enough.
(iii) Environmental conditions for survival and reproduction are changing continuously.

Topic: Biology, Biodiversity and its Conservation
_____ is observed when the magnetic moments of the domains in the substance are aligned in parallel and antiparallel directions in unequal numbers.

Options:
A Ferromagnetism
B Ferrimagnetism
C Antiferromagnetism
D Paramagnetism

The correct answer is:
Ferrimagnetism

Solution:
Correct Option is : 2
Ferrimagnetism

Topic: Chemistry, Solid State, NEET
_________ is present in cysteine and methionine and is the main constituent of several coenzymes.

Options:
A Magnesium
B Sulphur.
C Potassium
D Phosphorus

The correct answer is:
Sulphur.

Solution:
Sulphur is a macromolecule and component of cysteine, methionine, vitamins, ferredoxin and several coenzymes.

Topic: Biology, Mineral Nutrition, NEET
$\Delta \lambda$ is the difference between the wavelength of $k_\alpha$ line and the minimum wavelength of the continuous X-ray spectrum when the X-ray tube is operated at a voltage $V$. If the operating voltage is changed to $V / 3$, then the above difference is $\Delta \lambda^{\prime}$. Then

Options:
A $\Delta \lambda^{\prime}=5 \Delta \lambda$
B $\Delta \lambda^{\prime}=4 \Delta \lambda$
C $\Delta \lambda^{\prime}=3 \Delta \lambda$
D $\Delta \lambda^{\prime} < 3 \Delta \lambda$

The correct answer is:
$\Delta \lambda^{\prime}=3 \Delta \lambda$

Solution:
$\begin{aligned} & \text { We know } \mathrm{eV}=\frac{h c}{\lambda} \Rightarrow \frac{V}{V^{\prime}}=\frac{\Delta \lambda^{\prime}}{\Delta \lambda} \\ & \Rightarrow \quad \frac{V}{V / 3}=\frac{\Delta \lambda^{\prime}}{\Delta \lambda} \\ & \Rightarrow \quad \Delta \lambda^{\prime}=3(\Delta \lambda)\end{aligned}$

Topic: Physics, Electromagnetic Waves, JEE Main
Is the given relation a function? Give reason for your answer.
(i) $h=\{4,6),(3,9),(-11,6),(3,11)\}$
(ii) $f=\{(x, x) \mid x$ is a real number $\}$
(iii) $g=\left\{\left(x, \frac{1}{x}\right) x\right.$ is a positive integer $\}$
(iv) $s=\left\{\left(x, x^2\right) \mid x\right.$ is a positive integer $\}$
(v) $t=\{(x, 3) \mid x$ is a real number $\}$

Options:

None

Solution:
(i) Wehave, $h=\{(4,6),(3,9),(-11,6)(3,11)\}$.
Since, 3 has two images 9 and 11 . So, it is not a function.
(ii) We have, $f=\{(x, x) \mid x$ is a real number.
We observe that, every element in the domain has unique image. So, it is a function.
(iii) Given $g=\left\{\left(x, \frac{1}{x}\right) x\right.$ is a positive integer $\}$
For every $x$, it is a positive integer and $\frac{1}{x}$ is unique and distinct. Therefore, every element in the domain has unique image. So it is a function.
(iv) Given, $s=\left\{\left(x, x^2\right) \mid x\right.$ is a positive integer $\}$.
Since, the square of any positive integer is unique. So, every element in the domain has unique image, Hence, it is a function.
(v) Given. $t=\{(x, 3) \mid x$ is a real number $\}$.
Since, every element in the domain has the image 3 . So, it is a constant function.

Topic: Mathematics, Relations and Functions (Class 11)
Is there a species specific or region specific type of plastids? How does one distinguish one from the other?

Options:

None

Solution:
Plastids are specific to different species and are found in all plant cells and in euglenoids. They bear certain pigments that impart specific colours to the part of the plant possesing them. Plastids ar classified into three main types, based on the type of pigments- leucoplasts, chromoplast and chloroplast.
Leucoplasts are colourless plastids which store food material. They are of three types based on their storage products.
(a) Amyloplasts store starch, e.g., tuber of potato, grain of rice, grain of wheat.
(b) Elaioplasts store fats, e.g., rose
(c) Aleuroplasts are protein storing plastids, e.g., castor endosperm.
Chromoplast are non photosynthetic coloured plastids which synthesise and store carotenoid pigments. They appear orange, red or yellow. These mostly occur in ripe fruits (tomato and chilies) carrot roots, etc.
Chloroplasts are green color plastids which help in synthesising food material by photosyntheis. They contain chrophyll and carotenoid pigments which trap light energy. Each chloroplast is oval or spherical, double membrane bound cell organelle.
The space present inside inner membrane is called stroma. A number rof organised flattened membranous sacs called thylakoids are present in the stroma. Thylakoids are arranged in stacks called grana.
The thylakoids of different grana are connected by membranous tubules called the stroma lamellae. The stroma of the lamellae contain the enzymes that are required for the synthesis of carbohydrates and proteins.

Topic: Biology, Cell - The Unit of Life
Is there any change in the hybridisation of \(B\) and \(N\) atoms as a result of the following reaction?
\(\mathrm{BF}_3+\mathrm{NH}_3 \longrightarrow \mathrm{F}_3 \mathbf{B . N H}{ }_3\)

Options:

None

Solution:
In \(\mathrm{BF}_3, \mathrm{~B}\) is \(s p^2\) hybridized and in \(\mathrm{NH}_3, \mathrm{~N}\) is \(s p^3\) hybridized. After the reaction, hybridization of B changes to \(s p^3\) but that of \(\mathrm{N}\) remains unchanged.

Topic: Chemistry, Chemical Bonding and Molecular Structure
is used as:

Options:
A Antacid
B Insecticide
C Antihistamine
D Analgesic

The correct answer is:
Analgesic

Solution:
The given structure is the acetylsalicylic acid, which is known as Aspirin.

Aspirin, also known as acetylsalicylic acid, is a medication used to reduce pain, fever, or inflammation.

This is an example of an analgesic drug.

An analgesic or painkiller is any member of the group of drugs used to achieve analgesia, relief from pain. They are distinct from anesthetics, which temporarily affect, and in some instances eliminate, sensation.

Hence, Acetyl salicylic acid is analgesic.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main, JEE Main 2015 (10 Apr Online)
Isobutyraldehyde on reaction with formaldehyde and K2CO3 gives compound 'A'. Compound 'A' reacts with KCN and yields compound 'B', which on hydrolysis gives a stable compound 'C'. The compound 'C' is

Options:
A
B
C
D

The correct answer is:

Solution:
In the first step formaldehyde undergoes cross aldol reaction with isobutyraldehyde to form (A)and then the product(A) forms cyanohydrin (B)with KCN, which on hydrolysis forms a compound containing carboxylic acid group and finally this molecule undergoes intramolecular esterification to form (C) a stable compound.

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2022 (28 Jun Shift 2)
Isoelectric point is a

Options:
A Specific temperature
B Suitable concentration of amino acid
C Hydrogen ion concentration that does not allow migration of amino acid under electric field
D Melting point of an amino acid under the influence of electric field

The correct answer is:
Hydrogen ion concentration that does not allow migration of amino acid under electric field

Solution:
Isoelectric point is a pH at which zwitter ions do not migrate towards any of the electrode.

Topic: Chemistry, Biomolecules, JEE Main
Isopropyl alcohol on oxidation forms

Options:
A Acetone
B Ether
C Ethylene
D Acetaldehyde

The correct answer is:
Acetone

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Isopropyl amine with excess of acetyl chloride will give

Options:
A $\left(\mathrm{CH}_{3} \mathrm{CO}\right)_{2} \mathrm{~N}-\mathrm{CH}-\left(\mathrm{CH}_{3}\right)_{2}$
B < smiles>CCCNC(=O)OC < /smiles>
C $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHN}\left(\mathrm{COCH}_{3}\right)_{2}$
D none of these

The correct answer is:
$\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHN}\left(\mathrm{COCH}_{3}\right)_{2}$

Solution:
$\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH} \cdot \mathrm{NH}_{2}+2 \mathrm{CH}_{3} \mathrm{COCl}$
iso-propyl amine (pri amine)

Topic: Chemistry, Amines, JEE Main
Isotope(s) of hydrogen which emits low energy β- particles with t1/2 value >12 years is/are

Options:
A Protium
B Tritium
C Deuterium
D Deuterium and Tritium

The correct answer is:
Tritium

Solution:
H11 and H12 are stable while ${ }_{1}^{3} \mathrm{H}$ is radioactive.

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2021 (22 Jul Shift 1)
It has been found that for a chemical reaction with rise in temperature by 9 K the rate constant gets doubled. Assuming a reaction to be occurring at 300 K, the value of activation energy is found to be kJmol-1. [nearest integer] (Given ln10=2.3,R=8.3 J K-1 mol-1,log2=0.30)

Options:

The correct answer is:
59

Solution:
Arrhenius equation for the rate constant of a reaction in terms of energy of activation at two different temperatures is given by

logK2K1=Ea2.3R1T1-1T2

Rise in temperature =9 K

Initial temperature =300 K

logK309K300=Ea2.3R1300-1309

log2=Ea2.3R9300×309

Ea=2.3×0.30×300×3099

Ea=58.988×103 J=58.988kJ=59kJ

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2022 (27 Jun Shift 2)
It is a common practice to plant trees and shurbs near the boundary walls of buildings. What purpose do they serve?

Options:

None

Solution:
There is a common practice of growing and maintaining trees and shrubs near the boundary walls of residential, and official buildings to act as a barrier for sound and keep noise pollution under control. This green belt of trees and shrub also act as an effective measure to check primary air pollutants like dust, flyash, etc.

Topic: Biology, Environmental Issues
It is a disease which mainly affects mucous membrane of urinogenital tract. In males, burning feeling on passing urine, after a yellow discharge occurs, that is accompanied by fever, headache and feeling of illness. Its name is

Options:
A phenylketonuria
B gonorrhoea
C AIDS
D none of these.

The correct answer is:
gonorrhoea

Solution:
Gonorrhoea is caused by a bacterium, Neisseria gonorrhoeae. It spreads through sexual contact, common toilets and underclothes. Gonorrhoea affects both male and female genitals. The bacterium lives in genital tubes, produces pus-containing discharge, pain around genitalia and burning sensation during urination. It damages the eyes of babies born to infected mothers.

Topic: Biology, Excretory Products and their Elimination, AIIMS, AIIMS 2018 (27 May)
It is best to carry out reactions with sugars in neutral or acid medium and not in alkaline medium. This is because in alkaline medium sugars undergo one of the following changes

Options:
A Racemisation
B Decomposition
C Inversion
D Rearrangement

The correct answer is:
Rearrangement

Solution:
As sugars contain carbonyl groups which can be enolize by the $-\mathrm{OH}$ of alkaline medium. i.e it there may enolization can occur Hence rearrangement may possibly be acid-medium-and-not-in-alkaline-medium.

Topic: Chemistry, Biomolecules, JEE Main
It is desired to photograph the image of an object placed at a distance of $3 \mathrm{~m}$ from the plane mirror. The camera which is at a distance of $4.5 \mathrm{~m}$ from the mirror should be focussed for a distance of

Options:
A $3 m$
B $4.5 m$
C $6 m$
D $7.5 \mathrm{~m}$

The correct answer is:
$7.5 \mathrm{~m}$

Solution:
The distance between the object and the mirror is $3 \mathrm{~m}$, i.e., $\mathrm{d} 1=3 \mathrm{~m}$,
Hence image distance also equals $\mathrm{d} 1=3 \mathrm{~m}$
Also, the distance between the mirror and the camera is $4.5 \mathrm{~m}$, i.e., $\mathrm{d} 2=4.5 \mathrm{~m}$.
Hence the focus of the camera,
$\mathrm{f}=\mathrm{d} 1+\mathrm{d} 2=3+4.5=7.5 \mathrm{~m}$

Topic: Physics, Ray Optics, JEE Main
It is estimated that mainly ________ of the earthâs terrestrial surface is covered by lichens.

Options:
A 18%
B 12%
C 8%
D 4%

The correct answer is:
8%

Solution:
Ecologically the lichens are very important because they are pioneer colonizers and helps in soil formation. It is estimated that mainly 8% of the earthâs terrestrial surface is covered by lichens.

Topic: Biology, Biological Classification, NEET
It is found that if a neutron suffers an elastic collinear collision with a deuterium at rest, the fractional loss of its energy is Pd, while for its similar collision with a carbon nucleus at rest, the fractional loss of energy is Pc. The values of Pd and Pc are respectively

Options:
A 0, 1
B 0.89, 0.28
C 0.28, 0.89
D 0, 0

The correct answer is:
0.89, 0.28

Solution:
Since, the linear momentum is conserved, u=V1+2V2.
e=1=V2-V1u (where e⇒the coefficient of restitution), u=V2-V1
V2=2u3 ;V1=-u3. Initial energy, =12u2. Final energy, =12×1×u29=u218. Fractional change =u22-u218u22=0.88
u=V1+12V2⇒1=V2-V1u⇒u=V2-V1
⇒V2=2u13⇒V1=-11u13.
Change in energy,=u2-11u132×100=0.294

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2018 (08 Apr)
It is given that the events $A$ and $B$ are such that $P(A)=\frac{1}{4}, P\left(\frac{A}{B}\right)=\frac{1}{2}$ and $P\left(\frac{B}{A}\right)=\frac{2}{3}$. Then $P(B)$ is

Options:
A
$\frac{1}{6}$

B
$\frac{1}{3}$

C
$\frac{2}{3}$

D
$\frac{1}{2}$

The correct answer is:
$\frac{1}{3}$

Solution:
$$
\begin{aligned}
& \frac{P(A \cap B)}{P(B)}=\frac{1}{2}, \frac{P(A \cap B)}{P(A)}=\frac{2}{3} \\
& \text { Hence } \frac{P(A)}{P(B)}=\frac{3}{4} . \quad(\text { But } P(A)=1 / 4) \\
& \Rightarrow P(B)=\frac{1}{3} .
\end{aligned}
$$

Topic: Mathematics, Probability, JEE Main, JEE Main 2008
It is observed that characteristic X-ray spectra of elements show regularity. When frequency to the power 'n' i.e. vn of X-rays emitted is plotted against atomic number Z, the following graph is obtained.

The value of 'n' is

Options:
A 1
B 2
C 12
D 3

The correct answer is:
12

Solution:
The statement of moseley's law is: “The square root of the frequency of the x-ray emitted by an atom is proportional to its atomic number”.

v ∝ Z-b So n=12

Moseley is the scientist most directly responsible for realizing that the structure of the atom is key to understanding the elements themselves.

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2023 (24 Jan Shift 1)
It is observed that H2 and He gases always show positive deviation from ideal behaviour i.e., Z > 1. This is because

Options:
A the value of 'a' is very large due to high attractive forces
B the weak intermolecular forces of attraction due to which 'a' is very small and aV2 is negligible
C the value of 'b' is very large due to large size of the molecules
D both 'a' and 'b' are very small and negligible

The correct answer is:
the weak intermolecular forces of attraction due to which 'a' is very small and aV2 is negligible

Solution:
Due to weak intermolecular forces of attraction, H2 and He gases show the value of Z > 1 and always show positive deviation at any pressure.

Topic: Chemistry, States of Matter, JEE Main
It is possible to obtain oxygen from air by fractional distillation because

Options:
A Oxygen is in a different group of the periodic table from nitrogen
B Oxygen is more reactive than nitrogen
C Oxygen has higher b.p. than nitrogen
D Oxygen has a lower density than nitrogen

The correct answer is:
Oxygen has higher b.p. than nitrogen

Solution:
It is possible to obtain oxygen from air by fractional distillation because Oxygen has higher b.p than Nitrogen.
Hence, Option "3" is the correct answer.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
It is preferable to measure the e.m.f. of a cell by potentiometer than by a voltmeter because of the following possible reasons.
(i) In case of potentiometer, no current flows through the cell.
(ii) The length of the potentiometer allows greater precision.
(iii) Measurement by the potentiometer is quicker.
(iv) The sensitivity of the galvanometer, when using a potentiometer is not relevant.
Which of these reasons are correct?

Options:
A
(i),(iii),(iv)

B
(i),(iii),(iv)

C
(i),(ii)

D
(i),(ii),(iii),(iv)

The correct answer is:
(i),(ii)

Solution:
To measure the emf of a cell we prefer potentiometer rather than voltmeter because
(i) the length of potentiometer which allows greater precision.
(ii) in case of potentiometer, no current flows through the cell.
(iii) of high sensitivity.

Topic: Physics, Current Electricity, JEE Main, JEE Main 2012 (12 May Online)
It is recommended that ammonia bottle should be opened after cooling in ice for some time because

Options:
A it has high vapour pressure
B it comes out with brisk effervescence
C it is a corrosive fluid
D it vaporises at room temperature

The correct answer is:
it has high vapour pressure

Solution:
The vapour pressure of ammonia at room temperature is very high and hence the ammonia will evaporate unless the vapour pressure is decreased.Cooling decreases the vapour pressure so that the liquid remains in the same state. Hence, the bottle is cooled before opening.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
It is required to seat 5 men and 4 women in a row so that the men occupy odd places. Then the number of arrangements that are possible is

Options:
A $144$
B $362880$
C $2880$
D $1140$

The correct answer is:
$2880$

Solution:
$\frac{M}{1} \frac{W}{2} \frac{M}{3} \frac{W}{4} \frac{M}{5} \frac{W}{6} \frac{M}{7} \frac{W}{8} \frac{M}{9}$
5 men can be seated in 5 odd places in $5_{\mathrm{P}_5}=120$ ways and 4 women can be seated in remaining 4 places in $4_{p_4}=24$ ways $\Rightarrow$ Total possible arrangements $=120 \times 24=2880$

Topic: Mathematics, Permutation Combination, MHT CET, MHT CET 2022 (05 Aug Shift 1)
It is true that:

Options:
A A second order reaction is always a multistep reaction
B A zero order reaction is a multi-step reaction
C A first order reaction is always a single step reaction
D A zero order reaction is a single step reaction

The correct answer is:
A zero order reaction is a multi-step reaction

Solution:
For elementary reaction, order is equal to stoichiometric coefficient. Since reaction having zero stoichiometric coefficient is meaningless therefore, zero order reaction is complex reaction or multi-step reaction.

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2020 (03 Sep Shift 1)
It $x$ is real, then the maximum and minimum values of expression $\frac{x^2+14 x+9}{x^2+2 x+3}$ will be

Options:
A $4,-5$
B $5,=-4$
C $-4,5$
D $-4,-5$

The correct answer is:
$4,-5$

Solution:
Let $y=\frac{x^2+14 x+9}{x^2+2 x+3}$
$\Rightarrow y\left(x^2+2 x+3\right)-x^2-14 x-9=0$
$\Rightarrow(y-1) x^2+(2 y-14) x+3 y-9=0$
For real $x$, its discriminant $\geq 0$
i.e. $4(y-7)^2-4(y-1) 3(y-3) \geq 0$
$\Rightarrow y^2+y-20 \leq$ o or $(y-4)(y+5) \leq 0$
Now, the product of two factors is negative if these are of opposite signs. So following two cases arise:
Case I: $y-4 \geq 0$ or $y \geq 4$ and $y+5 \leq 0$ or $y \leq-5$ This is not possible.
Case II : $y-4 \leq 0$ or $y \leq 4$ and $y+5 \geq 0$ or $y \geq-5$ Both of these are satisfied if $-5 \leq y \leq 4$
Hence maximum value of $y$ is 4 and minimum value is $-5$

Topic: Mathematics, Quadratic Equation, JEE Main
IUPAC name of CCl 3 CHO is

Options:
A Chloral
B Trichloro acetaldehyde
C 1, 1, 1-trichloroethanal
D 2, 2, 2-trichloroethanal

The correct answer is:
2, 2, 2-trichloroethanal

Solution:
Acccording to functional group priorities for Nomenclature;
The carbon attached with the oxygen atom is given priority and numbered 1 and the halo-attached carbon is numbered 2.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
IUPAC name of $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_3\left(\mathrm{H}_2 \mathrm{O}\right)_2 \mathrm{Cl}\right] \mathrm{Cl}_2$ is

Options:
A Diaquachlorodiammine cobalt (III) chloride
B Triamminediaquachloro cobalt (III) chloride
C Chlorodiamminediaqua cobalt (III) chloride
D Diamminediaquachloro cobalt (II) chloride

The correct answer is:
Triamminediaquachloro cobalt (III) chloride

Solution:
The complex $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_3\left(\mathrm{H}_2 \mathrm{O}\right)_2 \mathrm{Cl}\right] \mathrm{Cl}_2$ contains cobalt in the +3 oxidation state. The ligands are three ammonia $\left(\mathrm{NH}_3\right)$, two water $\left(\mathrm{H}_2 \mathrm{O}\right)$, and one chloride ($\mathrm{Cl}$). According to IUPAC rules, the ligands are named alphabetically and the oxidation state of cobalt is indicated. Therefore, the correct name is Triamminediaquachloro cobalt (III) chloride.

Topic: Chemistry, Coordination Compounds, JEE Main
IUPAC name of complex ion CrCl2ox23- is

Options:
A dichlorodioxalatochromium (III)
B dioxalatodichlorochromate(III)
C dichlorodioxalatochromate(III)
D bisoxalaeodichlorochromate(III)

The correct answer is:
dichlorodioxalatochromate(III)

Solution:
IUPAC name is dichlorodioxalatochromate (III)

Topic: Chemistry, Coordination Compounds, JEE Main