datavorous/entrance-exam-dataset huggingface dataset:
Chemosensitive areas are situated close to respiratory rhythm centers and are found in CNS. These are mainly sensitive to

Options:
A CO2 but not H+ changes
B H+ but not CO2 changes
C CO2 and H+ changes both
D CO2 and O2 but not H+ changes

The correct answer is:
CO2 and H+ changes both

Solution:
A specialised centre present in the medulla region of the brain, called respiratory rhythm centre, is primarily responsible for this regulation. Another centre present in the pons region of the brain, called pneumotaxic centre, can moderate the functions of the respiratory rhythm centre. The neural signal from this centre can reduce the duration of inspiration, and thereby, alter the respiratory rate. A chemosensitive area is situated adjacent to the rhythm centre which is highly sensitive to CO2 and hydrogen ions. Increase in these substances can activate this centre, which in turn can signal the rhythm centre to make necessary adjustments in the respiratory process, by which these substances can be eliminated.

Topic: Biology, Breathing and Exchange of Gases, NEET
Chile saltpetre is an ore of

Options:
A Iodine
B Sodium
C Bromine
D Magnesium

The correct answer is:
Sodium

Solution:
Chile salt petre $\rightarrow \mathrm{NaNO}_3$

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
Chipko movement was launched for the protection of -

Options:
A Wet lands
B Grasslands
C Forests
D Livestock

The correct answer is:
Forests

Solution:
No Solution Available

Topic: Biology, Biodiversity and its Conservation, NEET, NEET 2009 (Mains)
Chloramphenicol is an

Options:
A antifertility drug
B antihistamin
C antiseptic and disinfectant
D antibiotic broad spectrum

The correct answer is:
antibiotic broad spectrum

Solution:
(a) Antifertility drugs are used to control pregnancy. These drugs prevent conception or fertilization e.g., Mifepristone, narethindrone, mestranol etc.
(b) Antihistamines are used for the relief of allergies. e.g., diphenlhydramine, chlorpheniramine, promethazine etc.
(c) Antiseptic are the used to reduce the number and growth of microorganism. e.g., dettol soap etc, whereas disinfectants kill bacteria and are used for sterilization of inanimate objects like instruments, utensils, cloths, floors etc. e.g., phenol, $\mathrm{H}_2 \mathrm{O}_2, \mathrm{SO}_2$, dettol, iodol etc.
(d) Broad spectrum antibiotics are such antibiotics that acts against a wide range of disease causing bacteria. They act against both Gram-positive and Gram-negative bacteria. e.g., ampicillin chloramphenicol etc.

Topic: Chemistry, Chemistry in Everyday Life, NEET, NEET 2012 (Mains)
Chloride of which of the following element is coloured?

Options:
A Ag
B Hg
C Zn
D Co

The correct answer is:
Co

Solution:
Atomic number of Ag, Hg, Zn and Co is 47, 80, 30 and 27 respectively.
Write the electronic configuration of each and see if there are any unpaired electrons.
In CoCl3-Co+3-[Ar]3d64s0
4 unpaired electrons. So, it will be coloured.

Topic: Chemistry, d and f Block Elements, JEE Main
Chlorides of which metal are soluble in organic solvents:

Options:
A Ca
B Mg
C K
D Be

The correct answer is:
Be

Solution:
As per Fajan's rule, Be cations has a higher charge and smaller size, as a result, have a higher charge density and covalent nature.

Ca, Mg, K are ionic compounds.

Hence, chlorides of Be will be more soluble in organic solvent.

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2023 (30 Jan Shift 2)
Chlorination of toluene in presence of light and heat followed by treatment with aqueous NaOH gives

Options:
A o-Cresol
B p-Cresol
C 2, 4-Dihydroxytoluene
D Benzoic acid

The correct answer is:
Benzoic acid

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
Chlorine atom, in its third excited state, reacts with fluorine to form a compound $X$. The formula and shape of $X$ are :

Options:
A $\mathrm{ClF}_5$, pentagonal
B $\mathrm{ClF}_4$, tetrahedral
C $\mathrm{ClF}_4$ pentagonal bipyramidal
D $\mathrm{ClF}_7$, pentagonal bipyramidal

The correct answer is:
$\mathrm{ClF}_7$, pentagonal bipyramidal

Solution:
${ }_{17} \mathrm{Cl}=1 s^2, 2 s^2, 2 p^6, 3 s^2, 3 p_x^2, 3 p_y^2, 3 p_z^1$
$\mathrm{Cl}=1 s^2, 2 s^2, 2 p^6, 3 s^1, 3 p_x^1, 3 p_y^1, 3 p_z^1, 3 d^1$
(3rd excited state)
$3 d^1 3 d^1$
Chlorine atom, in its third excited state, reacts with fluorine to form $\mathrm{ClF}_7$. Its shape is pentagonal bipyramidal.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Chlorine can be manufactured from

Options:
A Electrolysis of $\mathrm{NaCl}$
B Electrolysis of brine
C Electrolysis of bleaching powder
D All of these

The correct answer is:
Electrolysis of brine

Solution:
$2 \mathrm{NaCl}+2 \mathrm{H}_2 \mathrm{O} \xrightarrow{\text { Electrolysis }} \underset{\text { (aq) }}{2 \mathrm{NaOH}}+\underset{\text { (g) }}{\mathrm{Cl}_2}+\underset{(\mathrm{g})}{\mathrm{H}_2}$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Chlorine gas is prepared by reaction of H2SO4 with MnO2 and NaCl. What volume of Cl2 will be produced at STP if 50 g of NaCl is taken in the reaction?

Options:
A 19.14 L
B 22.4 L
C 11.2 L
D 9.57 L

The correct answer is:
9.57 L

Solution:
2NaCl2 moles2×58.5=117g+MnO2+3H2SO4→2NaHSO4+MnSO4+Cl21 mole22.4 LSTP+2H2O
117 g of NaCl ≡22.4 L of Cl2
50 g of NaCl≡22.4117×50=9.57 L of Cl2 at STP.

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
Chlorine on reaction with hot and concentrated sodium hydroxide gives.

Options:
A Cl- and ClO2-
B Cl- and ClO3-
C Cl- and ClO-
D ClO3- and ClO2-

The correct answer is:
Cl- and ClO3-

Solution:
Chlorine on reaction with hot and concentrated sodium hydroxide gives Cl-, ClO3-

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2019 (12 Jan Shift 2)
Chlorine reacts with hot and concentrated NaOH and produces compounds (P) and (Q). Compound (P) gives white precipitate with silver nitrate solution. The average bond order between Cl and O atoms in (Q) is Report your answer up to two decimal places.

Options:

The correct answer is:
1.67

Solution:
3Cl2+6NaOH→5NaCl+NaClO3+3H2O
NaCl +  AgNO 3 → AgCl White ppt  +  NaNO 3
Q is NaClO3→ClO3-
Bond order =53=1.67

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Chlorine undergoes disproportionation in alkaline medium as shown below :
aCl2( g)+bOH-(aq)→cClO-(aq)+dCl-(aq)+eH2O(l)

The values of a, b, c and d in a balanced redox reaction are respectively :

Options:
A 1, 2, 1 and 1
B 2, 2, 1 and 3
C 3,4, 4 and 2
D 2, 4, 1 and 3

The correct answer is:
1, 2, 1 and 1

Solution:
Chlorine undergoes disproportionation reaction in alkaline medium. It is the simultaneous oxidation-reduction. Chlorine is simultaneously reduced to chloride ion Cl− and is oxidised to ClO− ion. Halogens (e.g. Cl) have a strong tendency to accept electrons, so they act as strong oxidising agents.

The reaction is as follows,

\(\Rightarrow \mathrm{Cl}_2 - 2 \mathrm{O} \mathrm{H}^- \longrightarrow \mathrm{Cl}^{-}+\mathrm{ClO}^{-}+\mathrm{H}_2 \mathrm{O}\)

Hence the values of a,b,c,d are 1,2,1,1.

Topic: Chemistry, Redox Reactions, JEE Main, JEE Main 2024 (29 Jan Shift 1)
Chlorine water on standing loses its color and forms:

Options:
A HOCl and HOCl2
B HCl only
C HCl and HClO2
D HCl and HOCl

The correct answer is:
HCl and HOCl

Solution:
Chlorine water loses it's yellow color on standing in sunlight because it forms mixture of hypochlorous acid (HOCl) and hydrochloric acid HCl.

Cl2(g) Is soluble in H2O and form chlorine water, in presence of sunlight as
Cl2g+H2O→HCl+HOCl

This is a disproportionation reaction., where chlorine changes it's oxidation state from zero in Cl2 to -1 in HCl and +1 in HOCl. Hypochlorous acid formed in this reaction gives nascent oxygen which is responsible for bleaching action

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2015 (11 Apr Online)
Chlorobenzene is

Options:
A Less reactive than benzyl chloride
B More reactive than ethyl bromide
C Nearly as reactive as methyl chloride
D More reactive than isopropyl chloride

The correct answer is:
Less reactive than benzyl chloride

Solution:
Chlorobenzene is less reactive than benzyl chloride
In chlorobenzene the lone pairs present on $\mathrm{Cl}$ atom get involved in resonance with $\pi$ electrons of benzene due to which $C-C l$ bond acquires double bond character Hence, reactivity decreases.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Chlorobenzne reacts with trichloro acetaldehyde in the presence of $\mathrm{H}_2 \mathrm{SO}_4$.
The major product formed is:

Options:
A

B

C

D

The correct answer is:

Solution:
Chloral on reaction with chlorobenzene in the presence of a catalytic amount of sulphuric acid forms DDT (dichlorodiphenyl Trichloro ethane).

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2014 (11 Apr Online)
Chloroform can be obtained from

Options:
A Methanol
B Methanal
C Propanol-1
D Propanol-2

The correct answer is:
Propanol-2

Solution:
$\mathrm{CaOCl}_2+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{Ca}(\mathrm{OH})_2+\mathrm{Cl}_2$

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Chloroform for anesthetic purposes is tested for its purity with the reagent

Options:
A Silver nitrate
B Lead nitrate
C Ammoniacal $\mathrm{Cu}_2 \mathrm{Cl}_2$
D Lead nitrate

The correct answer is:
Silver nitrate

Solution:
If $\mathrm{CHCl}_3$ sample contains phosgene $\left(\mathrm{COCl}_2\right)$ then it will give a white ppt. When treated with cold $\mathrm{AgNO}_3$.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Chloromethane on treatment with excess of ammonia yields mainly

Options:
A
N, N-dimethylmethanamine

B
$\mathrm{N}$-methylmethanamine $\left(\mathrm{CH}_3-\mathrm{NH}-\mathrm{CH}_3\right)$

C
methanamine $\left(\mathrm{CH}_3 \mathrm{NH}_2\right)$

D
mixture containing all these in equal proportion

The correct answer is:
methanamine $\left(\mathrm{CH}_3 \mathrm{NH}_2\right)$

Solution:
$\mathrm{CH}_3 \mathrm{Cl}+\mathrm{NH}_3 \rightarrow \mathrm{CH}_3 \mathrm{NH}_2+\mathrm{HCl}$
Excess Methanamine
However, if the two reactants are present in the same amount, then the mixture of amines(i.e., primary, secondary and tertiary) are obtained.

Topic: Chemistry, Haloalkanes and Haloarenes
Chlorophyceae appears grass green due to the pigments

Options:
A Chlorophyll b, a
B Chlorophyll d, c
C Chlorophyll e
D Chlorophyll a, c

The correct answer is:
Chlorophyll b, a

Solution:
Chlorophyceae are commonly called as green algae, their main pigment is chlorophyll a and chlorophyll b and the accessory pigment is beta carotene and xanthophyll.

Topic: Biology, Plant Kingdom, NEET
Chlorophyceae are grass green due to the dominance of pigment

Options:
A Chlorophyll a, c
B Chlorophyll b, a
C Chlorophyll d, c
D Chlorophyll a, d

The correct answer is:
Chlorophyll b, a

Solution:
Chlorophyceae are commonly called as green algae, their main pigment is chlorophyll a and chlorophyll b and the accessory pigment is beta carotene and xanthophyll.

Topic: Biology, Plant Kingdom, NEET
Chlorophyll extracted from the crushed green leaves was dissolved in water to make 2 L solution of Mg of concentration $48 \mathrm{ppm}$. The number of atoms of Mg in this solution is x×1020 atoms. The value of x is (Nearest Integer) (Given : Atomic mass of Mg is 24 g mol-1, NA=6.02×1023 mol-1)

Options:

The correct answer is:
24

Solution:
Chlorophyll is the green pigment in plants, algae, and cyanobacteria that is essential for photosynthesis. Its central structure is an aromatic porphyrin ring system with a sequestered magnesium atom.

ppm=WMgVsol×106=48

⇒WMg=48×2×1000106

=48×2×10-3 g

nMg=WMg24=48×2×10-324

=4×10-3

Number of Mg atoms =4×10-3×6.02×1023=4×6.02×1020

=4×6.02×1020

=24.08×1020

∴ x=24.08

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2022 (26 Jul Shift 1)
Chloroprene is

Options:
A 2-chloro-1, 3-butadiene
B 3-chloro-2, 3-butadiene
C 2, 3-dichlorobutadiene
D None of these

The correct answer is:
2-chloro-1, 3-butadiene

Solution:
2-chloro-1, 3-butadiene
Chloroprene is the common name for 2-chlorobuta-1,3-diene (IUPAC name) with the chemical formula $\mathrm{CH}_2=\mathrm{CCl}-\mathrm{CH}+\mathrm{CH}_2$. Chloroprene is a colorless volatile liquid, almost exclusively used as a monomer for the production of the polymer polychloroprene, a type of synthetic rubber.

Topic: Chemistry, Hydrocarbons, JEE Main
Choke coil works on the principle of

Options:
A Transient current
B Self induction
C Mutual induction
D Wattless current

The correct answer is:
Self induction

Solution:
The choke coil works because it acts as an inductor. When the current passing through changes, as $\mathrm{AC}$ currents do, it typically creates a magnetic field in the coil that works against that current. This property, known as inductance.

Topic: Physics, Alternating Current, JEE Main
Choose from the indicated protons, the one that is most acidic

Options:
A 1
B 2
C 3
D 4

The correct answer is:
4

Solution:
Deprotonation of hydrogen labelled 4 produces a conjugate base which has more stable resonating structure and a resonance structure in which negative charge is present on the electronegative oxygen atom. Therefore, hydrogen labelled 4 is most acidic.

Topic: Chemistry, General Organic Chemistry, JEE Main
Choose the biodegradable pollutant out of the following.

Options:
A DDT
B Polyglycolic acid
C Alkyl benzene sulphonate
D Mercury

The correct answer is:
Polyglycolic acid

Solution:
Polyglycolic acid is a biodegradable pollutant. Rest are non biodegradable.

Topic: Chemistry, Polymers, JEE Main
Choose the connotations of the theory of special creation.

(i) All living organisms that we see today were created as such.
(ii) First form of life arose slowly through evolutionary forces from non-living molecules.
(iii) Earth is about 4000 years old.
(iv) Diversity has always been the same since creation and will remain thus in future.

Options:
A (i) and (ii)
B (i), (iii) and (iv)
C (i), (ii) and (iii)
D (i), (ii), (iii) and (iv)

The correct answer is:
(i), (iii) and (iv)

Solution:
Conventional religious literature tells us about the theory of special creation. This theory has three connotations. One, that all living organisms (species or types) that we see, today were created as such. Two, that the diversity was always the same since creation and will be the same in future also. Three, that earth is about 4000 years old,

Topic: Biology, Evolution, NEET
Choose the correct answer- A thermodynamic state function is a quantity:

Options:
A used to determine heat changes.
B whose value is independent of path.
C used to detemine pressure-volume work.
D whose value depends on temperature only.

The correct answer is:
whose value is independent of path.

Solution:
A thermodynamic state function is a quantity, whose value is independent of path. Its value depends only upon the state of the system.

Topic: Chemistry, Thermodynamics (C), JEE Main
Choose the correct answer from the options given below.

Options:
A (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
B (a)-(iii), (b)-(ii), (c)-(i), (d)-(iv)
C (a)-(i), (b)-(iv), (c)-(iii), (d)-(ii)
D (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)

The correct answer is:
(a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)

Solution:
a

This reaction is Gattermann- Koch reaction (ii) in which benzaldehye is formed.

b

This reaction is haloform reaction (iii) which is given by methyl ketones and produce haloform as one of the product.

c

This reaction is esterification (iv) reaction which involves the formation of ester by reaction of aldehyde and ketone.
d

This reaction is HVZ reaction (i), which involves alpha halogenation of carboxylic acid.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Choose the correct arrangement, where the symbols have their usual meanings

Options:
A $\bar{u} \gt u_p\gtu_{r m s}$
B $u_{r m s} \gt \bar{u} \gt u_p$
C $u_p\gt\bar{u} \gt u_{r m s}$
D $u_p\gtu_{r m s} \gt \bar{u}$

The correct answer is:
$u_{r m s} \gt \bar{u} \gt u_p$

Solution:
$u_{r m s} \gt \bar{u} \gt u_p$

Topic: Chemistry, States of Matter, JEE Main
Choose the correct length (L) versus square of time period (T2) ) graph for a simple pendulum executing simple harmonic motion.

Options:
A
B
C
D

The correct answer is:

Solution:
A simple pendulum consists of a point mass attached to an inextensible string and suspended from a support.

The formula to calculate the time period T of a simple pendulum is given by

T=2πlg.......................(1)

where, l is the length of the pendulum and g is the acceleration due to gravity.

Simplify equation (1) to obtain the relation between the time period and the length.

T2=4π2g×𝓁T2∝l............................(2)

Equation (2) suggests that the graph between the square of the time period versus the length of the pendulum is a straight line passing through the origin.

Topic: Physics, Oscillations, JEE Main, JEE Main 2023 (01 Feb Shift 2)
Choose the correct name for compound given below :

Options:
A (2E)-2-Bromo-hex-4-yn-2-ene
B (4E)-5-Bromo-hex-2-en-4-yne
C (4E)-5-Bromo-hex-4-en-2-yne
D (2E)-2-Bromohex-2-en-4-yne

The correct answer is:
(2E)-2-Bromohex-2-en-4-yne

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2021 (31 Aug Shift 1)
Choose the correct number of ways in which 15 different books can be divided into five heaps of equal number of books

Options:
A $\frac{15!}{5!(3!)^5}$
B $\frac{15!}{(3!)^5}$
C ${ }^{15} C_5$
D ${ }^{15} P_5$

The correct answer is:
$\frac{15!}{5!(3!)^5}$

Solution:
There are five heaps of an equal number of books.
Thus each heap contains $\frac{15}{5}=3$ books.
So the required number of ways
$={ }^{15} C_3 \times{ }^{12} C_3 \times{ }^9 C_3 \times{ }^6 C_3 \times{ }^3 C_3 \times$ $\times \frac{1}{5!}$
$=\frac{15!}{12!\times 3!} \times \frac{12!}{9!\times 3!} \times \frac{9!}{6!\times 3!} \times \frac{6!}{3!\times 3!} \times \frac{3!}{0!\times 3!}$ $\times \frac{1}{5!}$
$=\frac{5!}{\left[5!(3!)^5\right]}$

Topic: Mathematics, Permutation Combination, JEE Main
Choose the correct option about the matrices given below
\(\begin{aligned} & A=\left[\begin{array}{ccc}\cos \frac{\pi}{4} & \sin \frac{\pi}{4} & 0 \\ -\sin \frac{\pi}{4} & \cos \frac{\pi}{4} & 0 \\ 0 & 0 & 1\end{array}\right] \\ & B=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & \cos \frac{\pi}{3} & \sin \frac{\pi}{3} \\ 0 & -\sin \frac{\pi}{3} & \cos \frac{\pi}{3}\end{array}\right] \\ & C=\left[\begin{array}{ccc}\cos \frac{\pi}{6} & 0 & \sin \frac{\pi}{6} \\ 0 & 1 & 0 \\ -\sin \frac{\pi}{6} & \cos \frac{\pi}{6} & 0\end{array}\right]\end{aligned}\)
\(D=\left[\begin{array}{ccc}
\cos \frac{\pi}{2} & \sin \frac{\pi}{2} & 0 \\
-\sin \frac{\pi}{2} & \cos \frac{\pi}{2} & 0 \\
0 & 0 & 1
\end{array}\right]\)

Options:
A \(A^{2020}=1\)
B \(B^{2020}=1\)
C \(D^{2019}=l\)
D \(B^{2022}=1\)

The correct answer is:
\(B^{2022}=1\)

Solution:
Given matrix
\(\begin{aligned}
& A=\left[\begin{array}{ccc}
\cos \frac{\pi}{4} & \sin \frac{\pi}{4} & 0 \\
-\sin \frac{\pi}{4} & \cos \frac{\pi}{4} & 0 \\
0 & 0 & 1
\end{array}\right] \\
& \Rightarrow \quad A^2=\left[\begin{array}{ccc}
\cos \frac{\pi}{2} & \sin \frac{\pi}{2} & 0 \\
-\sin \frac{\pi}{2} & \cos \frac{\pi}{2} & 0 \\
0 & 0 & 1
\end{array}\right] \\
\end{aligned}\)
\(\begin{array}{ll}
\therefore & A^2=D=\left[\begin{array}{ccc}
\cos \frac{\pi}{2} & \sin \frac{\pi}{2} & 0 \\
-\sin \frac{\pi}{2} & \cos \frac{\pi}{2} & 0 \\
0 & 0 & 1
\end{array}\right]=\left[\begin{array}{ccc}
0 & 1 & 0 \\
-1 & 0 & 0 \\
0 & 0 & 1
\end{array}\right] \\
\therefore & A^4=\left[\begin{array}{ccc}
-1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & 1
\end{array}\right]=D^2 \Rightarrow A^8=\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right] \\
\therefore & A^{2020}=\left(A^{505}\right)^4=\left(\left(A^8\right)^{63} \cdot A\right)^4=\left(I^{63} \cdot A\right)^4=A^4 \neq I
\end{array}\)
Similarly, \(D^4=I\), so \(D^{2019} \neq I\)
Now, \(B=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & \cos \frac{\pi}{3} & \sin \frac{\pi}{3} \\ 0 & -\sin \frac{\pi}{3} & \cos \frac{\pi}{3}\end{array}\right]\)
\(\Rightarrow \quad B^2=\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & \cos \frac{2 \pi}{3} & \sin \frac{2 \pi}{3} \\
0 & -\sin \frac{2 \pi}{3} & \cos \frac{2 \pi}{3}
\end{array}\right]\)
\(\begin{aligned}
& \Rightarrow B^4=\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & \cos \frac{4 \pi}{3} & \sin \frac{4 \pi}{3} \\
0 & -\sin \frac{4 \pi}{3} & \cos \frac{4 \pi}{3}
\end{array}\right] \\
& \Rightarrow B^6=\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & \cos \frac{6 \pi}{3} & \sin \frac{6 \pi}{3} \\
0 & -\sin \frac{6 \pi}{3} & \cos \frac{6 \pi}{3}
\end{array}\right]=I \\
& \therefore B^{2020} \neq I \text { but } B^{2022}=\left(B^6\right)^{337}=I^{337}=I
\end{aligned}\)

Topic: Mathematics, Matrices, JEE Main
Choose the correct option for free expansion of an ideal gas under adiabatic condition from the following :

Options:
A q=0,ΔT≠0,w=0
B q=0,ΔT<0,w≠0
C q≠0,ΔT<0,w=0
D q=0,ΔT=0,w=0

The correct answer is:
q=0,ΔT=0,w=0

Solution:
For an adiabatic free expansion of an ideal gas, the gas is kept in an insulated container and then permitted to enlarge in a vacuum. Since there is no external pressure for the gas to enlarge against, the work done by or on the system is zero. Since this process does not involve any exchange of heat transfer or work, the first law of thermodynamics states that the net internal energy change of the system is zero.

For a free expansion, the work done will be zero.
w=0

For an adiabatic process, the heat will be zero.
q=0
And from the first law of thermodynamics,

∆T=0 or ∆S=0

Hence option D is the answer.

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2024 (01 Feb Shift 1)
Choose the correct option from the following options given below :

Options:
A

In the ground state of Rutherford's model electrons are in stable equilibrium. While in Thomson's model electrons always experience a net-force.

B

An atom has a nearly continuous mass distribution in a Rutherford's model but has a highly non-uniform mass distribution in Thomson's model

C A classical atom based on Rutherford's model is doomed to collapse.
D The positively charged part of the atom possesses most of the mass in Rutherford's model but not in Thomson's model.

The correct answer is:
A classical atom based on Rutherford's model is doomed to collapse.

Solution:
According to Rutherford's model, most of the mass of the atom is concentrated at the centre (nucleus) and the electron revolves around it.

As the electron is moving in a circular motion, it is accelerated. According to Maxwell accelerated charged particle loses its energy in the form of electro-magnetic radiation.

Therefore, a classical atom based on Rutherford's model is doomed to collapse.

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2022 (24 Jun Shift 1)
Choose the correct option having all the elements with d10 electronic configuration from the following:

Options:
A Co27,Ni28,Fe26,Cr24
B Cu29,Zn30,Cd48,Ag47
C Pd46,Ni28,Fe26,Cr24
D Ni28,Cr24,Fe26,Cu29

The correct answer is:
Cu29,Zn30,Cd48,Ag47

Solution:
Electronic Configuration is the distribution of electrons in the various available orbitals of an atom of an element.

The mentioned elements are present in d-block.

[Cr]=[Ar]4 s13 d5

[Cd]=[Kr]5 s24 d10

[Cu]=[Ar]4 s13 d10

[Ag]=[Kr]5 s14 d10

[Zn]=[Ar]4 s23 d10

Hence, the answer is B.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2024 (27 Jan Shift 2)
Choose the correct option relating wavelengths of different parts of electromagnetic wave spectrum:

Options:
A λvisible<λmicro waves<λradio waves<λx-rays
B λradio waves>λmicro waves>λvisible>λx-rays
C λx-rays<λmicro waves<λradio waves<λvisible
D λvisible>λx-rays>λradio waves>λmicro waves

The correct answer is:
λradio waves>λmicro waves>λvisible>λx-rays

Solution:
First, recall the spectrum of electromagnetic waves, and we know that in the case of electromagnetic waves the frequency is inversely proportional to wavelength.

Now from the above spectrum, we can say that, the relations of wavelength,

λradio waves>λmicro waves>λvisible>λx-rays.

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2020 (04 Sep Shift 1)