datavorous/entrance-exam-dataset huggingface dataset:
Choose the correct option.

Options:
A True dip is not mathematically related to apparent dip.
B True dip is less than apparent dip.
C True dip is always greater than the apparent dip.
D True dip is always equal to apparent dip.

The correct answer is:
True dip is less than apparent dip.

Solution:
If the apparent dip circle is at an angle α with true dip circle then,

tanϕ'>tanϕ ⇒ϕ'>ϕ

Therefore, apparent dip angle is greater than true dip angle.

Hence, true dip (ϕ) is less than apparent dip ϕ'

Topic: Physics, Magnetic Properties of Matter, JEE Main, JEE Main 2021 (22 Jul Shift 1)
Choose the correct order of density of the alkali metals

Options:
A Li< K<Na<Rb<Cs
B Li< Na<K <Rb<Cs
C Li<Na<K<Cs<Rb
D Cs<Rb<K<Na<Li

The correct answer is:
Li< K<Na<Rb<Cs

Solution:
As we move down the group from Li to Cs, both the atomic size and atomic mass increase. But the increase in atomic mass more than compensates the bigger atomic size. As a result, the densities of alkali metals increase from Li to Cs. Potassium is, however, lighter than sodium.

Hence, the increasing order of density is as given below:

Li< K<Na<Rb<Cs

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2022 (25 Jul Shift 1)
Choose the correct order of stability of carbocation using the concept of hyperconjugation.

Options:
A \( \mathrm{I} < \mathrm{II} < \mathrm{III} < \mathrm{IV} \)
B \( \mathrm{IV} < \mathrm{III} < \mathrm{II} < \mathrm{I} \)
C \( \mathrm{III} < \mathrm{IV} < \mathrm{II} < \mathrm{I} \)
D None of the above

The correct answer is:
\( \mathrm{IV} < \mathrm{III} < \mathrm{II} < \mathrm{I} \)

Solution:
Greater the number of alkyl groups attached to a positively charged carbon atom, the greater is the hyper conjugation interaction and stabilisation of the cation. Thus, we have the following relative stability of carbocation:

(CH3)3C+ > (CH3)2CH+ > CH3CH2+ > CH3+ 9 hyperconjugative structure 6 hyperconjugative structure 3 hyperconjugative structure

Topic: Chemistry, General Organic Chemistry, JEE Main
Choose the correct reagents used in the conversion

Options:
A pqrBr2alc.KOHNaOHÂ Â Â Â sAl2O3
B pqrHBralc.KOHCaC2Â Â Â Â sKMnO4
C pqrHBralc.KOHNaNH2Â Â Â Â sredÂ hotÂ ironÂ tube
D pqrBr2alc.KOHNaNH2 sred hot iron tube

The correct answer is:
pqrBr2alc.KOHNaNH2 sred hot iron tube

Solution:
Here NaNH2 is used for second time elimination as it is stronger reagent and suitable for this step.

Topic: Chemistry, Hydrocarbons, JEE Main
Choose the correct relationship between Poisson ratio σ, bulk modulus (K) and modulus of rigidity η of a given solid object:

Options:
A σ=3K-2η6K+2η
B σ=6K+2η3K-2η
C σ=3K+2η6K+2η
D σ=6K-2η3K-2η

The correct answer is:
σ=3K-2η6K+2η

Solution:
The relations of Young's modulus, Y, bulk modulus, K, the modulus of rigidity, η and Poisson's ratio, σ are given as

Y=3K(1-2σ) …1 and

Y=2η1+σ …2

From equation 1 and 2, we have

⇒2η1+σ=3K1-2σ

⇒σ=3K-2η6K+2η

Topic: Physics, Mechanical Properties of Solids, JEE Main
Choose the correct relationship between Poisson ratio σ, bulk modulus (K) and modulus of rigidity η of a given solid object:

Options:
A σ=3K-2η6K+2η
B σ=6K+2η3K-2η
C σ=3K+2η6K+2η
D σ=6K-2η3K-2η

The correct answer is:
σ=3K-2η6K+2η

Solution:
The relations of Young's modulus, Y, bulk modulus, K, the modulus of rigidity, η and Poisson's ratio, σ are given as

Y=3K(1-2σ) …1 and

Y=2η1+σ …2

From equation 1 and 2, we have

⇒2η1+σ=3K1-2σ

⇒σ=3K-2η6K+2η

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2023 (30 Jan Shift 1)
Choose the correct representation of conductometric titration of benzoic acid vs sodium hydroxide.

Options:
A
B
C
D

The correct answer is:

Solution:
C6H5CWAOOH+NaOSBH→C6H5CSaltOONa+H2O

Initially the conductance is low due to the feeble ionisation of acetic acid. On the addition of base, there is decrease in conductance(A→B) not only due to the replacement of H+ by Na+ but also suppresses the dissociation of benzoic acid due to common ion benzoate. But very soon, the conductance increases(B→C) on adding NaOH as NaOH neutralises the un-dissociated C6H5COOH to C6H5COONa which is the strong electrolyte. This increase in conductance continues raise up to the equivalence point. The graph near the equivalence point is curved due the hydrolysis of salt C6H5COONa. Beyond the equivalence point, conductance increases(C→D) more rapidly with the addition of NaOH due to the highly conducting OH− ions

Topic: Chemistry, Redox Reactions, JEE Main, JEE Main 2023 (24 Jan Shift 2)
Choose the correct sequence for the geometry of the given molecules Borazon, Borazole, B3O63-,Fe2Cl6 , trimer of FCN.
['P' stands for planar and 'NP' stands for non-planar]

Options:
A NP, NP, NP, P, P
B P, P, NP, NP, P
C NP, NP, NP, P, NP
D NP, P, P, NP, P

The correct answer is:
NP, P, P, NP, P

Solution:
Borazon-A crystalline form of boron nitride which has diamond like structure. (Non-planar)
Borazole-Inorganic benzene- B3N3H6 (Planar)

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Choose the correct sequence of events, with respect to the fungal sexual reproductive cycle.

Options:
A Karyogamy, Plasmogamy and Meiosis
B Plasmogamy, Karyogamy and Meiosis
C Meiosis, Plasmogamy and Karyogamy
D Meiosis, Karyogamy and Plasmogamy

The correct answer is:
Plasmogamy, Karyogamy and Meiosis

Solution:
In fungi vegetative, asexual, and sexual all three types of reproduction take place. The sexual reproduction takes place with the help of some spores like oospores, ascospores and basidiospores. Their sexual cycle is conducted mainly in three steps - a. Plasmogamay: Fusion of protoplasms between two motile or non-motile gametes b. Karyogamy: Fusion of two nuclei c. Meiosis: Zygote undergoes meiosis and forms haploid spores.

Topic: Biology, Biological Classification, NEET
Choose the correct set of reagents for the following conversion
transPh-CH=CH-CH3→ cis Ph-CH=CH-CH3

Options:
A Br2, alc KOH,NaNH2,NaLiqNH3
B

Br2, alc KOH,NaNH2,H2

Lindlar Catalyst

C Br2,aqKOH,NaNH2,H2 Lindlar Catalyst
D Br2, aq KOH,NaNH2,NaLiqNH3

The correct answer is:

Br2, alc KOH,NaNH2,H2

Lindlar Catalyst

Solution:
A trans alkene can be converted to a vicinal dihalide through a halogenation reaction. The vicinal dihalide can undergo double elimination to afford an alkyne. The alkyne can undergo partial reduction to give a cis alkene.

The reagents needed to convert trans(Ph - CH = CH - CH3) to cis-(Ph - CH = CH - CH3) are shown in the reaction scheme below.

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2023 (31 Jan Shift 1)
Choose the correct sketch of the magnetic field lines of a circular current loop shown by the dot
$$
\text { and the cross } \otimes \text {. }
$$

Options:
A

B

C

D

The correct answer is:

Solution:
If magnetic field is perpendicular and into the plane of the paper, it is represented by cross $\otimes$ and if the direction of the magnetic field is perpendicular out of the plane of the paper it is represented by dot $\odot$.

Topic: Physics, Magnetic Properties of Matter, JEE Main, JEE Main 2013 (22 Apr Online)
Choose the correct stability order of group 13 elements in their +1 oxidation state.

Options:
A Al<Ga<In<Tl
B Tl<In<Ga<Al
C In<Tl<Ga<Al
D Ga<In<Al<Tl

The correct answer is:
Al<Ga<In<Tl

Solution:
On moving down the group, the stability of +3 oxidation state decreases while +1 oxidation state increases due to inert pair effect.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2022 (26 Jun Shift 1)
Choose the correct statement.

Options:
A All proteins have 18 amino acids
B Both ends of a protein are similar
C All proteins are soluble
D Proteins are formed by amide C-N bonds

The correct answer is:
Proteins are formed by amide C-N bonds

Solution:
- Twenty-two amino acids are naturally incorporated into polypeptides and are called proteinogenic or natural amino acids. Of these, 20 are encoded by the universal genetic code. The remaining 2, selenocysteine and pyrrolysine, are incorporated into proteins by unique synthetic mechanisms.
- There is an N-terminal and C-terminal present in all the proteins.
- The solubility of a protein in water depends on the 3D shape of it. Usually, globular proteins are soluble, while fibrous ones are not.
- A peptide bond is a type of covalent bond or amide C-N bond formation between the carboxyl group of one amino acid and the amino group of another amino acid.

Topic: Chemistry, Biomolecules, NEET
Choose the correct statement:

Options:
A In frequency modulation the amplitude of the high frequency carrier wave is made to vary in proportion to the amplitude of the audio signal.
B In frequency modulation the amplitude of the high frequency carrier wave is made to vary in proportion to the frequency of the audio signal.
C In amplitude modulation the amplitude of the high frequency carrier wave is made to vary in proportion to the amplitude of the audio signal.
D In amplitude modulation the frequency of the high frequency carrier wave is made to vary in proportion to the amplitude of the audio signal.

The correct answer is:
In amplitude modulation the amplitude of the high frequency carrier wave is made to vary in proportion to the amplitude of the audio signal.

Solution:
As per properties of A.M. in amplitude modulation the amplitude of high frequency carrier wave is made to vary in proportion to the amplitude of the audio signal.

Topic: Physics, Communication System, JEE Main, JEE Main 2016 (03 Apr)
Choose the correct statement

Options:
A Ionization energy and electron affinity increases across a period
B Ionization energy increases but electron affinity decreases along a period
C Ionization energy decreases but electron affinity increases
D Both decreases along a period

The correct answer is:
Ionization energy and electron affinity increases across a period

Solution:
Ionization energy increase across a period due to decrease in size of atom and electron affinity for incoming electron increase across a period.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
Choose the correct statement about the protist given in the diagram below.

Options:
A These microscopic organisms float passively in water currents and are found in both marine and freshwater environments.
B These protists are generally biflagellates, of which one lies longitudinally and the other transversely in a furrow between the wall plates.
C These biflagellate protists are autotrophic in the sunlight and heterotrophic in its absence.
D Their cell wall has stiff cellulose plates on the outer surface and deposits of such is known as diatomaceous earth.

The correct answer is:
These protists are generally biflagellates, of which one lies longitudinally and the other transversely in a furrow between the wall plates.

Solution:
The given diagram is that of a dinoflagellate.
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
Choose the correct statement about two circles whose equations are given below:

x2+y2-10x-10y+41=0

x2+y2-22x-10y+137=0

Options:
A circles have same centre
B circles have no meeting point
C circles have only one meeting point
D circles have two meeting points

The correct answer is:
circles have only one meeting point

Solution:
x2+y2-10x-10y+41=0

A(5,5),R1=3

x2+y2-22x-10y+137=0

B(11,5),R2=3

AB=6=R1+R2

Touch each other externally

⇒ circles have only one meeting point.

Topic: Mathematics, Circle, JEE Main, JEE Main 2021 (18 Mar Shift 1)
Choose the correct statement about Zener diode:

Options:
A It works as a voltage regulator in reverse bias and behaves like simple p-n junction diode in forward bias
B It works as a voltage regulator in both forward and reverse bias
C It works a voltage regulator only in forward bias
D It works as a voltage regulator in forward bias and behaves like simple p-n junction diode in reverse bias

The correct answer is:
It works as a voltage regulator in reverse bias and behaves like simple p-n junction diode in forward bias

Solution:
Zener diode is a special type of p-n junction diode.

In a p-n junction diode, the majority carriers in p-side are the holes and that in n-side are the electrons. When a p-n junction diode is connected in such a way that the p-side is connected to the positive terminal and the n-side is connected to the negative terminal of a battery, there is a flow of current through the circuit. This condition is known as the forward bias condition of the junction.

If the connections are made in the opposite order, there is very less current that can pass through the p-n junction diode. This is called the reverse bias condition.

A Zener diode works in a reverse bias condition. It works in breakdown region of the applied voltage as a voltage regulator. Hence, it can act as simple p-n junction in forward bias

Topic: Physics, Semiconductors, JEE Main, JEE Main 2023 (01 Feb Shift 2)
Choose the correct statement applicable for the reaction: $2 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{~F}_{2} \rightarrow 4 \mathrm{HF}+\mathrm{O}_{2}$

Options:
A water is oxidised to $\mathrm{O}_{2}$.
B $\mathrm{F}_{2}$ is oxidised to HF.
C water is reduced to $\mathrm{HF}$.
D $\mathrm{F}_{2}$ acts as reducing agent in the reaction.

The correct answer is:
water is oxidised to $\mathrm{O}_{2}$.

Solution:
Water is oxidised to $\mathrm{O}_{2}$ and $\mathrm{F}_{2}$ is reduced to $\mathrm{HF}$ as follows.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Choose the correct statement for amplitude modulation

Options:
A Amplitude of modulating signal is varied in accordance with the information signal.
B Amplitude of modulated signal is varied in accordance with the information signal.
C Amplitude of carrier signal is varied in accordance with the information signal.
D Amplitude of modulated signal is varied in accordance with the modulating signal.

The correct answer is:
Amplitude of carrier signal is varied in accordance with the information signal.

Solution:
Amplitude modulation (AM) is a modulation technique used in electronic communication, most commonly for transmitting messages with a radio wave. In amplitude modulation, the amplitude (signal strength) of the carrier wave is varied in proportion to that of the information signal.

Topic: Physics, Communication System, JEE Main, JEE Main 2022 (26 Jun Shift 1)
Choose the correct statement for processes A & B shown in figure.

Options:
A PVγ=k for process B and PV=k for process A.
B PV=k for process B and A.
C Pγ-1Tγ=k for process B and T=k for process A.
D TγPγ-1=k for process A and PV=k for process B.

The correct answer is:
PVγ=k for process B and PV=k for process A.

Solution:
Steeper curve (B) is adiabatic
For adiabatic ⇒PVγ= constant

and curve (A) is isothermal. Therefore,
T= constant
⇒PV= constant

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2024 (30 Jan Shift 2)
Choose the correct statement for the work done in the expansion and heat absorbed or released when 5 litres of an ideal gas at 10 atmospheric pressure isothermally expands into vacuum until volume is 15 litres:

Options:
A Both the heat and work done will be greater than zero
B Heat absorbed will be|less than zero and work done will be positive
C Work done will be zero and heat will also be zero
D Work done will be greater than zero and heat will remain zero

The correct answer is:
Work done will be zero and heat will also be zero

Solution:
Since it is isothermal, $\Delta \mathrm{T}=0$
$\Delta \mathrm{U}=\mathrm{nC} \mathrm{C}_{\mathrm{v}} \Delta \mathrm{T}=0$
Since expansion is taking place against vacuum
$P_{\text {ext }}=0$
$W=-P$ ex $\Delta V=0$
From first law of thermodynamics,
$\Delta U=q+W$
$0=q+0$
$q=0$

Topic: Chemistry, Thermodynamics (C), NEET, NEET 2024 (Re-NEET)
Choose the CORRECT statement from the following.

Options:
A Stomata opens when guard cells become turgid
B Stomata opens when guard cells become flaccid
C Stomata opens when guard cells become turgid or flaccid
D Stomata closes when guard cells become turgid

The correct answer is:
Stomata opens when guard cells become turgid

Solution:
Stomata is a pore present in the epidermis section of the leaf surrounded by two cells called guard cells. These pores generally are present on the undersurface of the leaf but also found on upper surface in monocot plant leaves.
The activity of guard cells results in opening and closing of the stomata. The guard cells have a thickened cell wall on the inner of the pore. When the cells are at rest or flaccid, the pore remains closed that is stomata is closed and there is no movement of water and gases. When the guard cells absorb water via osmosis making the cell turgid. The outer cell wall bulges out and the inner thick cell wall takes a crescent shape opening the pore of stomata.

Topic: Biology, Transport in Plants, NEET
Choose the correct statement from the following :
Weightlessness of an astronaut moving in a satellite is a situation of

Options:
A Zero $g$
B No gravity
C Zero mass
D Free fall

The correct answer is:
Free fall

Solution:
Usually Weightlessness is defined as the feeling of zero gravity.
However, gravity is not zero anywhere in the space.
The astronaut in a satellite feels weightless because of the freefall above the earth.

Topic: Physics, Gravitation, JEE Main
Choose the correct statement regarding the formation of carbocations A and B given :-

Options:
A Carbocation B is more stable and formed relatively at faster rate
B Carbocation A is more stable and formed relatively at slow rate
C Carbocation B is more stable and formed relatively at slow rate
D Carbocation A is more stable and formed relatively at faster rate

The correct answer is:
Carbocation B is more stable and formed relatively at faster rate

Solution:
This is more stable due to secondary cation formation and formed with faster rate due to low activation energy.

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2021 (17 Mar Shift 2)
Choose the correct statement which describe the position of the point $(-6,2)$ relative to straight lines $2 x+3 y-4=0$ and $6 x+9 y+8=0$

Options:
A Below both the lines
B Above both the lines
C In between the lines
D None of these

The correct answer is:
Below both the lines

Solution:
$\begin{aligned}
& L \equiv 2 x+3 y-4=0, \quad L_{(-6,2)}=-12+6-4 \lt 0 \\
& L^{\prime}=6 x+9 y+8=0, \quad L_{(-6,2)}^{\prime}=-36+18+8 \lt 0
\end{aligned}$
Hence the point is below both the lines.

Topic: Mathematics, Straight Lines, JEE Main
Choose the correct statement with respect to the vapour pressure of a liquid among the following :

Options:
A Increases non-linearly with increasing temperature
B Decreases non-linearly with increasing temperature
C Decreases linearly with increasing temperature
D Increases linearly with increasing temperature

The correct answer is:
Increases non-linearly with increasing temperature

Solution:
The pressure exerted by the vapors over the liquid surface when liquid and vapour are in equilibrium is called vapour pressure. Vapour pressure increases with an increase with temperature.
Average kinetic energy of the gas molecules is directly proportional to the absolute temperature. At higher temperatures, more molecules from the liquid have enough KE to escape from the surface of the liquid. Which will increase vapor pressure.

For phase equilibrium for liquid & vapour phases
dP dT = Δ S Δ V = Δ H vap T . Δ V vap
dP dT = Δ H vap T RT P = Δ H vap × P RT 2 for 1 mole
dP P = Δ H vap RT 2 dT ⇒
The relation between temperature and vapor pressure is given by the Clausius-Clapeyron equation.
ln P 2 P 1 = Δ H vap R T 2 - T 1 T 1 T 2
undefined
ie. vapour increases exponentially with rise in temp ie.
Non linear increase.

Topic: Chemistry, Solutions, JEE Main
Choose the correct statement with respect to the vapour pressure of a liquid among the following :

Options:
A Increases non-linearly with increasing temperature
B Decreases non-linearly with increasing temperature
C Decreases linearly with increasing temperature
D Increases linearly with increasing temperature

The correct answer is:
Increases non-linearly with increasing temperature

Solution:
The pressure exerted by the vapors over the liquid surface when liquid and vapour are in equilibrium is called vapour pressure. Vapour pressure increases with an increase with temperature.
Average kinetic energy of the gas molecules is directly proportional to the absolute temperature. At higher temperatures, more molecules from the liquid have enough KE to escape from the surface of the liquid. Which will increase vapor pressure.

For phase equilibrium for liquid & vapour phases
dP dT = Δ S Δ V = Δ H vap T . Δ V vap
dP dT = Δ H vap T RT P = Δ H vap × P RT 2 for 1 mole
dP P = Δ H vap RT 2 dT ⇒
The relation between temperature and vapor pressure is given by the Clausius-Clapeyron equation.
ln P 2 P 1 = Δ H vap R T 2 - T 1 T 1 T 2
undefined
ie. vapour increases exponentially with rise in temp ie.
Non linear increase.

Topic: Chemistry, Solutions, JEE Main, JEE Main 2014 (19 Apr Online)
Choose the correct statements about the hydrides of group 15 elements.
A. The stability of the hydrides decreases in the order NH3>PH3>AsH3>SbH3>BiH3
B. The reducing ability of the hydrides increases in the order NH3<PH3<AsH3<SbH3<BiH3
C. Among the hydrides, NH3 is strong reducing agent while BiH3 is mild reducing agent.
D. The basicity of the hydrides increases in the order NH3<PH3<AsH3<SbH3<BiH3

Choose the most appropriate from the option given below:

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

The correct answer is:
A and B only

Solution:
On moving down the group from N to Bi, the atomic size increases. Consequently, the electron density on the central atom decreases and the basic strength decreases.

With an increase in size, the M-H bond becomes weaker, hence, the thermal stability decreases down the group.

The highest bond length will have less bond energy, indicating that it is a strong reducing agent.

On moving down the group, bond strength of M–H bond decreases, which reduces the thermal stability but increases reducing nature of hydrides,

Hence A and B are correct statements.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2024 (30 Jan Shift 2)
Choose the correct statements from the following

A. All group 16 elements form oxides of general formula EO2 and EO3 where E=S, Se, Te and Po. Both the types of oxides are acidic in nature.

B. TeO2 is an oxidising agent while SO2 is reducing in nature.

C. The reducing property decreases from H2S to H2Te down the group.

D. The ozone molecule contains five lone pairs of electrons.

Choose the correct answer from the options given below:

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

The correct answer is:
A and B only

Solution:
(A) All group 16 elements form oxides of the EO2 and EO3 type where E=S, Se, Te or Po. And these oxides produces acids on hydrolysis.

(B) SO2 is reducing while TeO2 is an oxidising agent. An oxidising agent is a substance that causes oxidation by accepting electrons; therefore, it gets reduced. A reducing agent is a substance that causes reduction by losing electrons; therefore, it gets oxidised.

(C) The reducing property increases from H2S to H2Te down the group. This is because the bond energy E-H(E=S,Se,Te) bond decreases down the group.

(D) have six lone pairs

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2024 (31 Jan Shift 2)