datavorous/entrance-exam-dataset huggingface dataset:
Which of the following is correct about the given figure?

Options:
A The length of the thick and thin myofilaments has changed.
B Length of both anisotropic and isotropic band has changed.
C The myosin cross-bridges move on the surface of actin and the thin and thick myofilaments slide past each other.
D Length of the sarcomere remains same.

The correct answer is:
The myosin cross-bridges move on the surface of actin and the thin and thick myofilaments slide past each other.

Solution:
No Solution Available

Topic: Biology, Locomotion and Movement, AIIMS, AIIMS 2010
Which of the following is correct explanation of chord of contact of ellipse

Options:
A The chord of contact of an ellipse is a line segment that joins two points on the ellipse.
B The chord of contact of an ellipse is a line segment that joins three points on the ellipse.
C The chord of contact is the only line segment in ellipse which passes from it's centre
D None of the above

The correct answer is:
The chord of contact of an ellipse is a line segment that joins two points on the ellipse.

Solution:
The chord of contact of an ellipse is a line segment that joins two points on the ellipse.

Topic: Mathematics, Ellipse, JEE Main
Which of the following is correct for critical temperature

Options:
A It is the highest temperature at which liquid and vapour can coexist
B Beyond the critical temperature, there is no distinction between the two phases and a gas cannot be liquefied compression
C At critical temperature $\left(T_c\right)$ the surface tension of the system is zero
D At critical temperature the gas and the liquid phases have different critical densities

The correct answers are:
It is the highest temperature at which liquid and vapour can coexist, Beyond the critical temperature, there is no distinction between the two phases and a gas cannot be liquefied compression

Solution:
It is the highest temperature at which liquid and vapour can coexist. Beyond the critical temperature, there is no distinction between the two phases and a gas cannot be liquefied compression These 2 correct for critical temperature

Topic: Chemistry, States of Matter, JEE Main
Which of the following is CORRECT match.

Column I	Column II

A	ADA - deficiency	i	Î±-1 antitrypsin

B	Emphysema	ii	Bone marrow transplanation

C	Insulin	iii	Diabetes mellitus

D	Insect resistance	iv	Ti - plasmid

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

The correct answer is:
(A)-(ii), (B)-(i), (C)-(iii), (D)-(iv)

Solution:
The first clinical gene therapy was given for treating Adenosine deaminase deficiency. It was conducted at Maryland in 1990 on a four year old girl .She was suffering from ADA deficiency. This enzyme is important for formation of immune system in the body. Hence biotechnology principles were applied to treat this condition. A functional ADA gene was introduced ex vivo in the marrow cells through a vector which had functional ADA gene. The treated cells were reintroduced into the patient's marrow.
Emphysema is a condition that affects the respiratory system and the risk increases by deficiency of alpha -1- antitrypsin.
Insulin is a hormone that is responsible to maintain the glucose levels and lack of it will cause diabetes mellitus.
Ti plasmid is used as a vector to induce insect resistance in plants.

Topic: Biology, Biotechnology and Its Applications, NEET
Which of the following is correct order of ligand field strength?

Options:
A CO<en<NH3<C2O42-<S2-
B S2-<C2O42-<NH3<en<CO
C NH3<en<CO<S2-<C2O42-
D S2-<NH3<en<CO<C2O42-

The correct answer is:
S2-<C2O42-<NH3<en<CO

Solution:
According to spectrochemical series, ligands can be arranged in a series in the order of increasing field strength as given below:

I – < Br– < SCN – < Cl – < S2– < F – < OH – < C2O4 2– < H2O < NCS – < edta4– < NH3 < en < CN – < CO

As per the above series, correct answer is S2-<C2O42-<NH3<en<CO

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2023 (30 Jan Shift 1)
Which of the following is correct order of their increasing boiling points?

Options:
A $10^{-4} \mathrm{M} \mathrm{NaCl}>10^{-3} \mathrm{M} \mathrm{MgCl}_{2}>10^{-2} \mathrm{M} \mathrm{NaCl}$ $>10^{-4} \mathrm{M}$ urea.
B $10^{-2} \mathrm{M} \mathrm{NaCl}>10^{-3} \mathrm{M} \mathrm{MgCl}_{2}>10^{-4} \mathrm{M} \mathrm{NaCl}$ $>10^{-4} \mathrm{M}$ urea.
C $10^{-4} \mathrm{M}$ urea $>10^{-4} \mathrm{M} \mathrm{NaCl}>10^{-3} \mathrm{M} \mathrm{MgCl}_{2}$ $>10^{-2} \mathrm{M} \mathrm{NaCl}$.
D $10^{-2} \mathrm{M} \mathrm{NaCl}>10^{-3} \mathrm{M} \mathrm{MgCl}_{2}>10^{-4} \mathrm{M} \mathrm{NaCl}$ $>10^{-4} \mathrm{M}$ urea.

The correct answer is:
$10^{-2} \mathrm{M} \mathrm{NaCl}>10^{-3} \mathrm{M} \mathrm{MgCl}_{2}>10^{-4} \mathrm{M} \mathrm{NaCl}$ $>10^{-4} \mathrm{M}$ urea.

Solution:
More the value of $i$, more will be the elevation in boiling point. Also, more dilute solution has low boiling point. Hence, the increasing order of boiling points is
$$
\begin{aligned}
&10^{-2} \mathrm{M} \mathrm{NaCl}>10^{-3} \mathrm{M} \mathrm{MgCl}_{2}>10^{-4} \mathrm{M} \mathrm{NaCl} \\
&(i=2) \quad(i=3) \\
&>10^{-4} \mathrm{M} \text { urea. } \\
&\quad(i=1)
\end{aligned}
$$

Topic: Chemistry, Solutions, JEE Main
Which of the following is correct regarding the internal energy of a substance.

Options:
A It is a state function
B It increases with increase in temperature
C It can be calculated by experiment
D It remains unaffected with change in temperature

The correct answer is:
It is a state function

Solution:
Internal energy of a system is a state function and extensive property and is independent of the path by which it is obtained.

Topic: Chemistry, Thermodynamics (C), JEE Main
Which of the following is correct regarding the largest part of the human brain?

Options:
A It maintains balance and coordination
B It controls involuntary reflexes and actions
C It contains nerve fibres connecting different regions of brain
D It controls all motor and sensory activities

The correct answer is:
It controls all motor and sensory activities

Solution:
The largest part of the human brain is the cerebrum. It is the part that has areas for controlling all motor and sensory activities and intersensory associations.

Topic: Biology, Neural Control and Coordination, NEET
Which of the following is correct regarding the organism given below?

Options:
A This organism has a cell wall made up of a protein called a pellicle.
B It is a flagellated protozoan.
C It is both phototrophic and heterotrophic in nature.
D Toxins released by them may even kill other marine animals such as fish.

The correct answer is:
Toxins released by them may even kill other marine animals such as fish.

Solution:
These organisms are mostly marine and photosynthetic. They appear yellow, green, brown, blue or red depending on the main pigments present in their cells. The cell wall has stiff cellulose plates on the outer surface. Most of them have two flagella, one lies longitudinally and the other transversely in a furrow between the wall plates. Very often, red dinoflagellates (example: Gonyaulax) undergo such rapid multiplication that they make the sea appear red (red tides). Toxins released by such large numbers may even kill other marine animals such as fishes.

Topic: Biology, Biological Classification, NEET
Which of the following is correct regarding tricuspid valve

Options:
A It has three muscular flaps
B It guards the opening between right atrium and right ventricle
C It is an uni-directional valve.
D All of the above

The correct answer is:
All of the above

Solution:
It prevents blood from flowing back into the right atrium during contraction of the ventricle.
The tricuspid valve is one of the two main valves on the right side of your heart. Normally, the tricuspid valve has three flaps (leaflets) that open and close, allowing blood to flow from the right atrium to the right ventricle in your heart and preventing blood from flowing backward.

Topic: Biology, Body Fluids and Circulation, NEET
Which of the following is correct structure of tyrosine?

Options:
A
B
C
D

The correct answer is:

Solution:
The structure of Tyrosine amino acid is

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2021 (17 Mar Shift 1)
Which of the following is correctly ordered according to the ascending order of band gap energy?

Options:
A Silicon, Graphite, Diamond
B Graphite, Silicon, Diamond
C Silicon, Diamond, Graphite
D Diamond, Graphite, Silicon

The correct answer is:
Graphite, Silicon, Diamond

Solution:
- In semiconductors, there exists a finite but small band gap between the conduction band and valence band ( $\mathbf{E}_9 \lt 3$ eV). Because of the small bandgap, at room temperature, some electrons from the valence band can acquire enough energy to cross the energy gap and enter the conduction band.
- Therefore, the forbidden energy bandgap in conductors, semiconductors, and insulators are in the relation insulator $\gt$ semiconductor $\gt$ conductor.

Topic: Physics, Semiconductors, JEE Main
Which of the following is deliquescent?

Options:
A ZnCl2
B Hg2Cl2
C HgCl2
D CdCl2

The correct answer is:
ZnCl2

Solution:
ZnCl2 is deliquescent as it absorbs moisture from atmosphere and gets dissolved in it.

Topic: Chemistry, d and f Block Elements, JEE Main
Which of the following is diamagnetic ?

Options:
A
$\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$

B
$\left[\mathrm{Co}(\mathrm{ox})_3\right]^{3-}$

C
$\left[\mathrm{FeF}_6\right]^{3-}$

D
$\left[\mathrm{CoF}_6\right]^{3-}$

The correct answer is:
$\left[\mathrm{Co}(\mathrm{ox})_3\right]^{3-}$

Solution:
$\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$ has magnetic moment of a single unpaired electron whereas $\left[\mathrm{FeF}_6\right]^{3-}$ has a magnetic moment of five unpaired electrons. $\left[\mathrm{CoF}_6\right]^{3-}$ is paramagnetic with four unpaired electrons while $\left[\mathrm{Co}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]^3$ is diamagnetic. This anomaly is explained by valence bond theory in terms of formation of inner and outer orbital coordination entities. $\left[\mathrm{Co}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]^{3-}$ is an inner orbital complexes having $d^2 s p^3$ hybridization.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2013 (22 Apr Online)
Which of the following is disproportionation reaction?

Options:
A \( \mathrm{CH}_{4}+2 \mathrm{O}_{2} \rightarrow \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O} \)
B \( \mathrm{CH}_{4}+4 \mathrm{Cl}_{2} \rightarrow \mathrm{CCl}_{4}+4 \mathrm{HCl} \)
C \( 2 \mathrm{~F}_{2}+2 \mathrm{OH}^{-} \rightarrow 2 \mathrm{~F}^{-}+\mathrm{OF}_{2}+\mathrm{H}_{2} \mathrm{O} \)
D \( 2 \mathrm{NO}_{2}+2 \mathrm{OH}^{-} \rightarrow \mathrm{NO}_{2}^{-}+\mathrm{NO}_{3}^{-}+\mathrm{H}_{2} \mathrm{O} \)

The correct answer is:
\( 2 \mathrm{NO}_{2}+2 \mathrm{OH}^{-} \rightarrow \mathrm{NO}_{2}^{-}+\mathrm{NO}_{3}^{-}+\mathrm{H}_{2} \mathrm{O} \)

Solution:
Reactions in which the same substance is oxidized as well as reduced are called disproportionation reaction. Writing the oxidation number of each element above its symbol in the given reactions.
C+4H4+1+2O20âC+4O2-2+2H2+1O-2
C-4H4+1+4Cl20âC+4Cl4+ 4 H +1 C −1 l
2F20+2O-2H-+1â2F-1+O+2F2-1+H2-1O-2
2N+4O2-2+2O-2H-+1âN+3O2--2+N+5O3--2+H2+1O-2
Thus, in the reaction of NO2 and OH-, N is both oxidized as well as reduced. Hence, the oxidation number increases from +4 in NO2 to +5 in NO3- and decreases from +4 in NO2 to +3 in NO2-.

Topic: Chemistry, Redox Reactions, JEE Main
Which of the following is equivalent to the Boolean expression p∧~q ?

Options:
A ~p→~q
B ~(q→p)
C ~(p→q)
D ~(p→~q)

The correct answer is:
~(p→q)

Solution:
Which of the following is equivalent to the Boolean expression p∧~q ?

For option 1

~p→~q≡p∨~q

≢p∧~q

For Option 2

~q→p≡~~q∨p≡q∧~p

≢p∧~q

For option 3

~(p→q)

≡~(~p∨q)

≡(p∧~q)

For option 4

~p→~q≡~~p∨~q≡p∧q

≢p∧~q

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2021 (01 Sep Shift 2)
Which of the following is FALSE regarding the lock and key concept of enzyme catalysis?

Options:
A An enzyme molecule has one or more active sites at which a specific substrate (reactant) molecule fits.
B The active site acts like a lock.
C The product acts as a key.
D The active site returns to its original state after the products are released.

The correct answer is:
The product acts as a key.

Solution:
The specific action of an enzyme with a single substrate can be explained using a lock and key analogy. In this analogy, the lock is the enzyme and the key is the substrate.

Topic: Chemistry, Surface Chemistry, JEE Main
Which of the following is fully fluorinated polymer?

Options:
A
Neoprene

B
Teflon

C
Thiokol

D
PFC

The correct answer is:
Teflon

Solution:
No Solution Available

Topic: Chemistry, Polymers, JEE Main, JEE Main 2005
Which of the following is general form equation of chord of contact of ellipse

Options:
A $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
B $\frac{x_1 x}{a^2}+\frac{y_1 y}{b^2}=1$
C $\frac{h x}{a^2}+\frac{k y}{b^2}=0$
D none of the above

The correct answer is:
$\frac{x_1 x}{a^2}+\frac{y_1 y}{b^2}=1$

Solution:
$\frac{x_1 x}{a^2}+\frac{y_1 y}{b^2}=1$

Topic: Mathematics, Ellipse, JEE Main
Which of the following is highly basic?

Options:
A Diphenylamine
B Benzylamine
C Aniline
D Triphenylamine

The correct answer is:
Benzylamine

Solution:
whereas lone pair of $\mathrm{N}$ in aniline, diphenylamine and triphenyl amine are delocalised over the benzene ring. hence are not available for donation, making them weaker bases.

Topic: Chemistry, Amines, JEE Main
Which of the following is incorrect?

Options:
A ~pâ~qâ¡pâq
B p⇔~q≡p⇒q
C ~pâqâ¡~p⇒q
D ~p⇒qâ¡pâ§~q

The correct answer is:
p⇔~q≡p⇒q

Solution:
Option C & D are theorems

Hence, C & D are correct
For option A,

~p⇔~q≡p⇔~~q
≡p⇔q
Hence, A is correct
For option B,

When p is true and q is false then p⇔~q is true and p⇒q is false

Hence, option B is incorrect

Topic: Mathematics, Mathematical Reasoning, JEE Main
Which of the following is incorrect ?

Options:
A IE1 of Li < IE1 of Be
B IE1 of Be < IE1 of B
C IE1Â ofÂ LiÂ >Â IE1Â ofÂ Na
D IE1Â ofÂ HeÂ >Â IE1Â ofÂ Ne

The correct answer is:
IE1 of Be < IE1 of B

Solution:
IE1 of Be > IE1 of B ; In Be removal occurs from 2s ; In B it occurs from 2p . More closer are subshell to nucleus more is IE .

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
Which of the following is incorrect?

Options:
A Regulators- show homeostasis
B Conformers - cannot maintain homeostasis
C Partial regulators - have an ability to regulate upto limited range
D Conformers - show homeostasis

The correct answer is:
Conformers - show homeostasis

Solution:
Homeostasis is the state of steady internal, physical and chemical conditions maintained by living systems. This is the condition of optimal functioning for the organism and includes many variables, such as body temperature and fluid balance, being kept within certain pre-set limits. Any organism whose internal environment is highly influenced by external factors is conformer. Many marine invertebrates are conformers: they have no need to control their internal environment since the external environment is fairly constant in terms of temperature, oxygen tension and nutrients. Conformers cannot maintain their constant internal environment.

Topic: Biology, Organisms and Populations, NEET
Which of the following is incorrect about a bacterial cell?

Options:
A The 70S ribosome is mainly associated with plasma membrane and many a times shows polysome assembly.
B The filament is the longest portion of the flagella and extends from the cell surface to the outside.
C The plasma membrane extensions containing chromatophores basically helps in cellular respiration.
D Gram staining is basically based on the thickness of peptidoglycan layer.

The correct answer is:
The plasma membrane extensions containing chromatophores basically helps in cellular respiration.

Solution:
In some prokaryotes like cyanobacteria, there are other membranous extensions into the cytoplasm called chromatophores which contain pigments. These pigments facilitate photosynthesis.

Topic: Biology, Cell - The Unit of Life, NEET
Which of the following is incorrect about the experiment performed by T. W. Engelmann?

Options:
A Cladophora (a green algae) is used to detect the sites of oxygen evolution.
B Suspension of aerobic bacteria is used.
C Bacteria accumulated mainly in the region of orange and green part of the light spectrum.
D Prism was used to split light into its spectral component.

The correct answer is:
Bacteria accumulated mainly in the region of orange and green part of the light spectrum.

Solution:
Engelmann, using a prism he split light into spectral components and then illuminated a green algae Cladophora, placed in a suspension of aerobic bacteria. The bacteria were used to detect the site of oxygen evolution. Bacteria accumulated mainly in the region of blue and red light of the split spectrum, this proved that maximum absorption spectrum is observed in blue and red region.

Topic: Biology, Photosynthesis in Higher Plants, NEET
Which of the following is incorrect about the experiment performed by T. W. Engelmann?

Options:
A The experimental organism used was Cladophora (green algae).
B For the experimental purpose, the suspension of aerobic bacteria was utilized.
C The bacteria accumulated mainly in the region of the orange and green parts of the light spectrum.
D For splitting of light into its spectral component prism was used.

The correct answer is:
The bacteria accumulated mainly in the region of the orange and green parts of the light spectrum.

Solution:
T.W. Engelmann experimented on Cladophora. He illuminated Cladophora with white light after passing through a prism. He used aerobic bacteria to detect the oxygen evolution site and observed that maximum bacteria accumulated in the blue and red region of the spectrum, indicating that maximum photosynthesis occurs in this region.

Topic: Biology, Photosynthesis in Higher Plants, NEET
Which of the following is incorrect regarding the first law of thermodynamics?

Options:
A
It is applicable to any cyclic process

B
It is a restatement of the principle of conservation of energy

C
It introduces the concept of the internal energy

D
It introduces the concept of the entropy

The correct answer is:
It introduces the concept of the entropy

Solution:
The first law of thermodynamics states the principle of conservation of energy. It introduces the concept of internal energy. And it is applicable for any cycle.

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2005
WHich of the following is incorrect?
The intensity of a progressing plane wave in loss-free medium is

Options:
A Directly proportional to the square of amplitude of the wave
B Directly proportional to the velocity of the wave
C Directly proportional to the square of frequency of the wave
D Inversely proportional to the density of the medium

The correct answer is:
Inversely proportional to the density of the medium

Solution:
$I=2 \pi n^2 a^2 \rho v \Rightarrow I \alpha n^2 a^2 v$

Topic: Physics, Waves and Sound, JEE Main
Which of the following is incorrect?

Options:
A FeCl3 is used in detection of phenol
B Fehling solution is used in detection of glucose
C Tollen's reagent is used in detection of unsaturation
D NaHSO3 is used in detection of carbonyl compound

The correct answer is:
Tollen's reagent is used in detection of unsaturation

Solution:
Tollen's reagent is a solutionof ammoniacal silver nitrate and used for the detection of -CHO group. Aldehydes reduce Tollen's reagent and itself get oxidised to give Ag+ ions to Ag powder which forms the silver coloured mirror in the test tube. So this test is also known as silver mirror test.
R−CHOPowder+AgNH32+→R−COO−+Ag

Topic: Chemistry, Hydrocarbons, JEE Main
Which of the following is isoelectronic?

Options:
A $\mathrm{CO}_2, \mathrm{NO}_2$
B $\mathrm{NO}_2^{-}, \mathrm{CO}_2$
C $\mathrm{CN}^{-}, \mathrm{CO}$
D $\mathrm{SO}_2, \mathrm{CO}_2$

The correct answer is:
$\mathrm{CN}^{-}, \mathrm{CO}$

Solution:
In $\mathrm{CN}^{-}$the number of electrons $=6$ $+7+1=14$
In $\mathrm{CO}$ the number of electrons $=6+$ $8=14$

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Which of the following is least basic?

Options:
A CH3CON¨HC2H5
B C2H53N¨
C CH3CO2N¨H
D C2H52N¨H

The correct answer is:
CH3CO2N¨H

Solution:
For the given compounds:

(1) ; L.P. on Nitrogen is delocalised.

(2) ; L.P. on Nitrogen is localised.

(3) ;

L.P. on Nitrogen is delocalised due to conjugation with both

(Hence least basic)

(4) CH3-CH2-N¨H-CH2-CH3; L.P. on Nitrogen is localised.

Topic: Chemistry, Amines, JEE Main, JEE Main 2021 (16 Mar Shift 2)
Which of the following is least basic?

Options:
A NF3
B NCl3
C NBr3
D NI3

The correct answer is:
NF3

Solution:
As the central atom in all the given molecule is same therefore the basicity depends on the surrounding atom. As fluorine is the most electronegative element thus it has the least tendency to donate its lone pair to central atom N hence, its basicity will be least.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Which of the following is least ionic?

Options:
A BaCl2
B AgCl
C KCl
D CoCl2

The correct answer is:
AgCl

Solution:
In inorganic chemistry, Fajans' rules, formulated by Kazimierz Fajans in 1923, are used to predict whether a chemical bond will be covalent or ionic, and depend on the charge on the cation and the relative sizes of the cation and anion. They can be summarised in the following table:

$\begin{array}{|l|l|} \hline \text{Ionic } & \text{Covalent } \\ \hline \text{Low positive charge on cation} & \text{High positive charge on cation} \\ \hline \text{Large cation} & \text{Small cation} \\ \hline \text{Small anion} & \text{Large anion} \\ \hline \end{array}$

In the given case all the anions are same in all the given compounds. So we need to think about cations.

The ionic radius of Ag+ is less than that of Cl− but their charges are equal. Hence, AgCl is a least ionic compound among the given options.

AgCl<CoCl2<BaCl2<KCl (ionic character)

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2024 (31 Jan Shift 2)
Which of the following is least likely to be the component of the cell wall in higher plants?

Options:
A Polymer of glucose
B Polymer of lipid
C Hemicellulose
D Polymer of amino acids

The correct answer is:
Polymer of lipid

Solution:
Lipid are monomers and they are mainly present in cell membrane not in cell wall. Polymers of glucose like cellulose, hemicelluloses, pectin are main components of cell wall.

Topic: Biology, Cell - The Unit of Life, NEET
Which of the following is least reactive towards SN1 ?

Options:
A
B
C
D

The correct answer is:

Solution:
Here is least reactive towards SN1 .
Rate of SN1∝ stability of carbocation
A⇒ Anti aromatic
B⇒ Non-aromatic (no resonance)
C⇒ Non-aromatic (resonance)
D⇒ Aromatic
Stability order
Aromatic > non-aromatic > anti-aromatic
So D > C > B > A.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Which of the following is less than zero during adsorption?

Options:
A ΔG
B ΔS
C ΔH
D ÎH and ÎS

The correct answer is:
ΔH and ΔS

Solution:
Adsorption is process where randomness decreases and energy is released. Hence, ΔS will be negative (as randomness is decreasing) and ΔH will also be negative ( ΔH is negative for exothermic processes).
Depending on the values of Entropy and enthalpy, the Gibbs energy can either be negative or positive.

Topic: Chemistry, Surface Chemistry, JEE Main
Which of the following is Lindlar catalyst?

Options:
A Zinc chloride and HCl
B Cold dilute solution of KMnO4
C Sodium and Liquid NH3
D Partially deactivated palladised charcoal

The correct answer is:
Partially deactivated palladised charcoal

Solution:
Partially deactivated palladised charcoal H2/Pd/CaCO3 or BaSO4 is Lindlar catalyst.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2021 (16 Mar Shift 1)
Which of the following is logically equivalent to ~(~p⇒q) ?

Options:
A p∧q
B p∧~q
C ~p∧q
D ~p ∧~q

The correct answer is:
~p ∧~q

Solution:
Since, ~(p⇒q)≡p ∧~q ~(~p⇒q)≡~p∧~q

Topic: Mathematics, Mathematical Reasoning, JEE Main