datavorous/entrance-exam-dataset huggingface dataset:
The ascending order of acidity of –OH group in the following compounds is :

(A) Bu – OH

Choose the correct answer from the options given below :

Options:
A (A) < (D) < (C) < (B) < (E)
B (C) < (A) < (D) < (B) < (E)
C (C) < (D) < (B) < (A) < (E)
D (A) < (C) < (D) < (B) < (E)

The correct answer is:
(A) < (C) < (D) < (B) < (E)

Solution:
The decrease in electron density of the O-H bond of o,p-dinitrophenol, the polarity of O-H bond decreases. The electron withdrawing group -NO2), withdraws electrons and disperses the negative charge. Therefore, -NO2 group stabilizes the phenoxide ion. Hence, o,p-dinitrophenol is more acidic than p-nitrophenol and phenol.

OMe is an electron donating group, hence it increases electron density, there why making it less acidic.

Bu-OH is an aliphatic alcohol. In phenoxide ion the negative charge is delocalized within the ring structure which makes it more stable than the aliphatic alcohols. In aliphatic alcohols there is no such delocalization stabilization. Therefore, phenols are more acidic and the resulting phenoxide ion is stabilized because of resonance.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2024 (27 Jan Shift 1)
The ascent of xylem sap in plants is mainly accomplished by the :

Options:
A size of the stomatal aperture
B distribution of stomata on the upper and lower epidermis
C cohesion and adhesion between water molecules
D root pressure

The correct answer is:
cohesion and adhesion between water molecules

Solution:
Ascent of sap in xylem is mainly due to transpiration pull which is supported by physical properties of water such as cohesion and adhesion between water molecules. Cohesion is also called Tensile strength of water molecule.

Topic: Biology, Anatomy of Flowering Plants, NEET, NEET 2022 (Phase 2)
The association areas of the brain are responsible for

Options:
A intersensory associations, memory and communication.
B the coordination for sensory and motor signaling.
C regulation of respiration and circulation.
D body temperature, urge for eating and drinking.

The correct answer is:
intersensory associations, memory and communication.

Solution:
An area of cerebral cortex that lies away from the main areas that are concerned with the reception of sensory impulses and the start of motor impulses but is linked to them by many neurons known as association fibres. The areas of association are thought to be responsible for the elaboration of the information received by the primary sensory areas and its correlation with the information fed in from memory and from other brain areas. They are thus responsible for the maintenance of many higher mental activities.

Topic: Biology, Neural Control and Coordination, NEET
The association of histone H1 with a nucleosome indicates:

Options:
A Transcription is occurring
B DNA replication is occurring
C The DNA is condensed into a chromatin fibre
D The DNA double helix is exposed

The correct answer is:
The DNA is condensed into a chromatin fibre

Solution:
The association of H1 protein indicates the complete formation of nucleosome which requires DNA condensation. The DNA is therefore in condensed form. Histone H1 is one of the five main histone protein families (H1, H2A, H2B, H3 and H4) which are components of chromatin in eukaryotic cells. H1 histones bind to the linker DNA exiting from the nucleosome core particle, thus making the nucleosome complete.

Topic: Biology, Molecular Basis of Inheritance, NEET, NEET 2017
The at wood machine shown is suspended from a spring balance. The mass on one hanger is \( M \), that on the other is \( (M+m) \). Suppose the heavier side (right side) hanger is fastened to the top of the pulley by a thread. The scale reads \( (2 M+m) g \). The thread is burned and the system accelerates. The reading of spring balance now will be

Options:
A the same as before.
B more than before.
C less than before.
D can't say.

The correct answer is:
less than before.

Solution:
Suppose once the block starts moving tension in the string is T, acceleration of the blocks is a and tension in the spring balance is TS. When the thread is burned, blocks start moving.

From FBD of the block of mass M+m kg,

M+mg-T=M+ma …i

From FBD of the block of mass M kg,

T-Mg=Ma …ii

From FBD of the pulley,

Since the pulley is in equilibrium.

∴ TS=2T …iii

From the equation i, ii and iii,

⇒TS=4MM+mg2M+m<2M+mg.

Hence, the reading of spring balance will be less than the initial reading.

Topic: Physics, Laws of Motion, JEE Main
The atmosphere is held to the earth by

Options:
A Winds
B Gravity
C Clouds
D None of the above

The correct answer is:
Gravity

Solution:
Gravity:
1. As gravity hugs a large blanket of air to the Earth's surface, what physicists call a density gradient is set up in the air
2. The air near the ground is pulled on by gravity and compressed by the air higher in the sky.
3. This causes the air near the ground to be denser and at greater pressure than air at higher elevations.
4. Earth's gravity is very strong to hold onto its atmosphere and keep it from drifting into space.
5. Various atom of gases present in the earth's atmosphere has masses and a gravitational force acts between these atoms and the earth.
6. Also, this is the force that holds the various gases near the surface and the atmosphere is present.
7. Due to gravity, the earth pulls all objects towarcls it. The air that surrounds our planet is held in place by gravity, and this envelope of air is referred to as the atmosphere
Thus, The atmosphere is held to the earth by Gravity.

Topic: Physics, Gravitation, JEE Main
The atmospheric pressure at the surface of a lake of depth h is P0. If the pressure at half the depth of the lake is equal to 23 of the pressure at the bottom of the lake, then the depth (h) of the lake is

Options:
A 10 m

B 20 m

C 60 m

D 30 m

The correct answer is:
20 m

Solution:
The pressure at half the depth = P 0 + h 2 dg

The pressure at the bottom = P 0 +hρg

According to given condition P 0 + h 2 dg= 2 3 ( P 0 +hdg)

⇒ 3 P 0 + 3h 2 dg=2 P 0 +2hdg

⇒h=2P0dg=2×105103×10=20 m

Topic: Physics, Mechanical Properties of Fluids, JEE Main
The atomic hydrogen emits a line spectrum consisting of various series. Which series of hydrogen atomic spectra is lying in the visible region?

Options:
A Brackett series
B Paschen series
C Lyman series
D Balmer series

The correct answer is:
Balmer series

Solution:
The transition from higher states to n=2 lead to the emission of radiation with wavelengths 656.3 nm and 365.0 nm. These wavelengths fall in the visible region and constitute the Balmer series.

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2021 (17 Mar Shift 2)
The atomic mass of C126 is 12.000000 u and that of C136 is 13.003354 u. The required energy to remove a neutron from C136, if mass of neutron is 1.008665 u, will be:

Options:
A 62.5MeV
B 6.25MeV
C 4.95MeV
D 49.5MeV

The correct answer is:
4.95MeV

Solution:
Given the nuclear reaction is

C136+Energy→C126+n10

The mass defect for the above reaction can be calculated as follows:

Δm=(12.000000+1.008665)-13.003354 u=-0.00531 u

Hence, the energy required is given by

E=0.00531×931.5 MeV=4.95 MeV

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2024 (27 Jan Shift 2)
The atomic mass unit is

Options:
A $1 / 12$th the mass of 1 mole of carbon atoms
B $1 / 12$th the mass of 1 mole of oxygen atoms
C $1 / 12$th the mass of an oxygen atom
D $1 / 12$th the mass of a carbon atom

The correct answer is:
$1 / 12$th the mass of a carbon atom

Solution:
The atomic mass unit is equal to $1 / 12$ th the mass of a carbon atom. It is denoted by the unit " $\mathrm{U}^{\text {" }}$ and is equal to $1.66 \times 10^{-27} \mathrm{~kg}$.

Topic: Physics, Nuclear Physics, JEE Main
The atomic masses of 12H and 24H are 2.0141 u and 4.0026 u respectively. Taking the velocity of light in vacuum to be 2.998×108 m s-1, calculate the amount of energy (in J) liberated when two moles of 12H undergo a fusion to form one mole of 24H. [u=1.66057×10-27 kg]

Options:
A 2.3×1012J
B 3.3×1012J
C 5.3×1012J
D 2.9×1012J

The correct answer is:
2.3×1012J

Solution:
2 12H → 24H+Energy
Mass defect =2×2.0141-4.0026= 0.0256 amu
∴ ∆E= ∆m×c2
=0.0256×1.66057×10-27×6.02×1023×2.998×1082
=2.3×1012J

Topic: Physics, Nuclear Physics, JEE Main
The atomic masses of He and Ne are 4 and 20 a.m.u, respectively. The value of the de Broglie wavelength of He gas at − 73oC is "M" times that of the de Broglie wavelength of Ne at 727oC . M is

Options:

The correct answer is:
5

Solution:
λ∝ 1 M.T.
Î»HeÎ»Ne=MNe.TNeMHe.THe=20Ã10004Ã200=5

Topic: Chemistry, Structure of Atom, JEE Main
The atomic masses of Li and K are 7 and 39, respectively. According to law of triads the atomic mass of Na will be

Options:
A 23
B 32
C 46
D 64

The correct answer is:
23

Solution:
Atomic mass of middle element of triad is equal to average of masses of terminal elements.
Li Na K
7ââ 23ââ 39ââ 7+392=23

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
The atomic number and the mass number of an atom remains unchanged when it emits

Options:
A a photon
B a neutron
C β-particle
D An α- particle

The correct answer is:
a photon

Solution:
When a γ-ray photon is emitted then atomic number and mass number remains unchanged.

Topic: Physics, Atomic Physics, NEET
The atomic number of elements A, B, C and D are Zâ1, âZ, âZâ+â1, âZâ+â2 respectively. If B is a noble gas, choose the correct option
(i) (A) has higher electron affinity
(ii) C exists in + 2 oxidation state
(iii) D is an alkaline earth metal

Options:
A i and ii
B ii and iii
C i and iii
D i, ii and ii

The correct answer is:
i and iii

Solution:
Atomic number of B=Z and is a noble gas
Atomic number of A=Z-1 (i.e., is a Halogen)
Atomic number C,=Z+1 (i.e., is a alkali metal)
At number of D=Z+2 (i.e., alkaline earth metal)
Hence, element A must be a Halogen, so has highest electron affinity among the given species. C. is an alkali metal and D is an alkaline earth metal

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
The atomic number of the element unnilennium is :

Options:
A 109
B 102
C 108
D 119

The correct answer is:
109

Solution:
un=1

nil=0

enn=9

So Atomic number =109

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2020 (03 Sep Shift 1)
The atomic number of the element with highest ionisation energy among the following is

Options:
A $Z=16$
B $Z=14$
C $Z=13$
D $Z=15$

The correct answer is:
$Z=15$

Solution:
The electronic configuration is
(a) $Z=16: 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{4}$
(b) $Z=14: 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{2}$
(c) $Z=13: 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{1}$
(d) $Z=15: 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{3}$
$\because$ For $Z=15, p$-orbital has stable half-filled configuration.
$\therefore$ It will have highest ionisation energy.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
The atomic number of Unnilumium is ______.

Options:

The correct answer is:
101

Solution:
According to IUPAC convention for naming of elements with atomic number more than 100, different digits are written in order and at the end ium is added. For digits following naming is used.

0-nil

1-un

2-bi

3-tri

and so on...

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2020 (06 Sep Shift 2)
The atomic number of Unnilunium is _____.

Options:

The correct answer is:
101

Solution:
According to IUPAC convention for naming of elements with atomic number more than 100, different digits are written in order and at the end ium is added. For digits following naming is used.

0-nil

1-un

2-bi

3-tri

and so on...

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
The atomic number of vanadium $(V)$, chromium $(\mathrm{Cr})$, manganese $(M n)$ and iron $(F e)$ are respectively $23,24,25$ and 26 which one of these may be expected to have the highest second ionization enthalpy

Options:
A $V$
B $\mathrm{Cr}$
C $M n$
D $\mathrm{Fe}$

The correct answer is:
$\mathrm{Cr}$

Solution:
\(\begin{aligned} & V=1 s^2, 2 s^2, 2 p^6, 3 s^2, 3 p^6, 3 d^3, 4 s^2 \\ & C r=1 s^2, 2 s^2, 2 p^6, 3 s^2, 3 p^6, 3 d^3, 4 s^1 \\ & V=1 s^2, 2 s^2, 2 p^6, 3 s^2, 3 p^6, 3 d^5, 4 s^2 \\ & F e=1 s^2, 2 s^2, 2 p^6, 3 s^2, 3 p^6, 3 d^6, 4 s^2 \end{aligned}\)
In second ionization enthalpy \(\mathrm{Cr}^{+}\) has exact half filled d-sub shell.

Topic: Chemistry, d and f Block Elements, JEE Main
The atomic numbers of elements $A, B, C$ and $D$ are $Z-1, Z, Z+1$ and $Z+2$, respectively. If ' $B$ ' is a noble gas, choose the correct answers from the following statements
(1) ' $A$ ' has higher electron affinity
(2) ' $C$ ' exists in +2 oxidation state
(3) ' $D$ ' is an alkaline earth metal

Options:
A (1) and (2)
B (2) and (3)
C (1) and (3)
D $(1),(2)$ and (3)

The correct answer is:
(1) and (3)

Solution:
Given,
Atomic number of element $B=Z$
( $\because$ noble gas $\quad \therefore$ belong to zero group)
Atomic number of element $A=Z-1$
(ie, halogens)
Atomic number of element $C=Z+1$
(ie, group IA)
Atomic number of element $D=Z+2$
(ie, group IIA)
$\because$ Element $B$ is a noble gas
$\therefore$ Element $A$ must be a halogen ie, have highest electron affinity.
and element $C$ must be an alkali metal and exist in +1 oxidation state.
and element $D$ must be an alkaline earth metal with +2 oxidation state.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
The atomic numbers of vanadium V , chromium Cr , manganese Mn and iron (Fe) are respectively 23, 24, 25 and 26. Which one of these may be expected to have the highest second ionization enthalpy?

Options:
A Mn
B V
C Cr
D Fe

The correct answer is:
Cr

Solution:
Consider Electronic configuration of M +
23 V + â 18 [Ar]â 4s 1 â 3d 3
24 Cr + â 18 [Ar]â 4s 0 â 3d 5
25 Mn + â 18 [Ar]â 4s 1 â 3d 5
26 Fe + ⇒ 18 [Ar]  3d 6   4s 1
Removal of second electron from stable configuration of 18 [Ar]â 4s 0 â 3d 5 will require very high energy input i.e. (Cr) as it will disturb the stable electronic configuration of 3d 5 (half filled subshell).

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
The atomic numbers of vanadium $(\mathrm{V})$, Chromium $(\mathrm{Cr})$, manganese $(\mathrm{Mn})$ and iron $(\mathrm{Fe})$ are respectively 23 , 24,25 and 26 . Which one of these may be expected to have the highest second ionization enthalpy?

Options:
A
$\mathrm{Cr}$

B
$\mathrm{Mn}$

C
$\mathrm{Fe}$

D
$\mathrm{V}$

The correct answer is:
$\mathrm{Cr}$

Solution:
Electronic configuration of $\mathrm{Cr}$ is
So due to half filled orbital I.P. is high of Cr.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2003
The atomic orbital is____

Options:
A The circular path of the electron
B Elliptical shaped Orbit
C Three-dimensional field around nucleus
D The region in which there is a maximum probability of finding an electron

The correct answer is:
The region in which there is a maximum probability of finding an electron

Solution:
Atomic orbitals are regions of space around the nucleus of an atom where an electron is likely to be found. Thus Atomic Orbitals are region around the nucleus where the probability of finding the electron is maximum.

Topic: Physics, Atomic Physics, JEE Main
The atomic radius of Ag is closest to

Options:
A Au
B Ni
C Cu
D Hg

The correct answer is:
Au

Solution:
Due to lanthanoid contraction, atomic radius of second transition series is almost same as third transition series.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2020 (07 Jan Shift 1)
The atomic radius of gallium is less than that of aluminium. This is due to

Options:
A Greater shielding power of s-electrons of gallium atom
B poor shielding power of s-electrons of gallium
C Poor shielding power of d-electrons of gallium
D Greater shielding power of d-electrons of gallium

The correct answer is:
Poor shielding power of d-electrons of gallium

Solution:
Due to poor shielding effect of d-electrons of gailium, it leads to increase in effective nuclear charge which inturn leads to decrease in the radius of gallium as compared to aluminium.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
The atomic ratio between the uranium isotopes 238U and 234U in a mineral sample is found to be 1.8×104 . The half life of 234U is T12234=2.5×105 years. The half-life of 238U is-

Options:
A 4.5×109 years
B 5.4×109 years
C 4.5 years
D 5.4 years

The correct answer is:
4.5×109 years

Solution:
In radioactive equilibrium
λ1N1=λ2N2
N1T1=N2T2
T2=N2N1×T1
T2=1.8×104×2.5×105
T2=4.5×109 years

Topic: Physics, Nuclear Physics, JEE Main
The atomic structure of He+ arises due to transition from n2 to n1 level. If n1+n2 is 3 and n2-n1 is 1. Find the λ in nm of transition for this series in He+ in nm.

Options:

The correct answer is:
30

Solution:
n1+n2=3 …(i)
n2-n1=1 …(ii)
2n2=4
So n2=2
Putting the value of n2 in (i) n1=3-n2=3-2=1
1λ=Rz21n12-1n22=R.22112-122
λ=13R=13×109678cm-1
=3×10-6cm=3×10-6×107=30 nm.

Topic: Chemistry, Structure of Atom, JEE Main
The atomic weights of two elements $A$ and $B$ are 40 and 80 respectively. If $x g$ of $A$ contains $y$ atoms, how many atoms are present in $2 \times g$ of $B$

Options:
A $\frac{y}{2}$
B $\frac{y}{4}$
C y
D $2 y$

The correct answer is:
y

Solution:
Number of moles of \(A=\frac{x}{40}\)
Number of atoms of \(A=\frac{x}{40} \times\) Avogadro no. \(=y\) (say)
Or \(x=\frac{40 y}{\text { Avogadrono. }}\)
Number of moles of \(B\) \(=\frac{2 x}{80}\)
Number of atoms of \(B\)
\(=\frac{2 x}{80} \times \text { Av. no. }=\frac{2}{80} \times \frac{40 y}{\text { Av. no. }} \times \text { Av. no. }=y\)

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
The attenuation in optical fibre is mainly due to

Options:
A absorption
B scattering
C neither absorption nor scattering
D Both (a) and (b)

The correct answer is:
Both (a) and (b)

Solution:
From an optical fibre due to absorption or light leaving the fibre area resulting scattering of light sideways by impurities in the glass fibre. And due to this reason a very small part of light energy is lost.

Topic: Physics, Waves and Sound, VITEEE, VITEEE 2010
The auxiliary equation of circle of hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, is

Options:
A $x^2+y^2=a^2$
B $x^2+y^2=b^2$
C $x^2+y^2=a^2+b^2$
D $x^2+y^2=a^2-b^2$

The correct answer is:
$x^2+y^2=a^2$

Solution:
The equation is $(x-0)^2+(y-0)^2=a^2$. As center (0,0) as according to hyperbola equation and radius $=a$ for auxilary condition.

Topic: Mathematics, Hyperbola, JEE Main
The Avena curvature is used for bioassay of:

Options:
A ABA
B GA3
C IAA
D Ethylene

The correct answer is:
IAA

Solution:
Among the plant growth regulators, Auxins are one of the major classes which is considered as growth promoters. Auxins are of mainly two types, natural and artificial. Avena curvature bioassay, first demonstrated by Went in 1928, is done to test the function of IAA, which is a natural auxin. The Avena geo-curvature test is a bioassay for auxin-type growth regulators.

Triple pea test is bioassay for Ethylene. Dwarf pea test is the bioassay for GA3.

Topic: Biology, Plant - Growth and Development, NEET, NEET 2016 (Phase 1)
The average concentration of SO2 in the atmosphere over a city on a certain day is 10 ppm, when the average temperature is 298 K. Given that average solubility of SO2 at 298 K is 1.3653 M atm-1 and pKa of H2SO3 is 1.92, estimate the pH of rain on that day.

Options:

The correct answer is:
4.86

Solution:
Partial pressure of SO2 in air = 10- 5 atm
[SO2 ] = 1.3653 x 10- 5 M
[H2SO3 ] = 1.3653 x 10- 5 M
H 2 SO 3 â H + + HSO 3 -
H2SO3 at low concentration is fully ionised.
[H+ ] = 1.3653 x 10- 5
pH = - log 1.3653 x 10- 5
= 5 - log 1.36
= 4.86

Topic: Chemistry, Ionic Equilibrium, JEE Main
The average daily income of workers of a factory including that of the owner is $â‚¹ 110$. However, if the income of the owner is excluded, the average daily income of the remaining 9 workers is $â‚¹ 76$. What is the daily income of the owner?

Options:
A $â‚¹ 300$
B $â‚¹ 316$
C $â‚¹ 322$
D $â‚¹ 416$

The correct answer is:
$â‚¹ 416$

Solution:
Daily income of owner $=10 \times 110-9 \times 76=1100-684=₹ 416$

Topic: Mathematics, Statistics, NDA, NDA 2011 (Phase 1)
The average depth of Indian ocean is about $3000 \mathrm{~m}$. The value of fractional compression $\left(\frac{\Delta V}{V}\right)$ of water at the bottom of the ocean is (given that the bulk modulus of water is $2.2 \times 10^9 \mathrm{Nm}^{-2}, g=9.8, \rho_{\mathrm{H}_2 \mathrm{O}}=1000 \mathrm{~kg} . \mathrm{m}^{-3}$ )

Options:
A $3.4 \times 10^{-2}$
B $1.34 \times 10^{-2}$
C $4.13 \times 10^{-2}$
D $13.4 \times 10^{-2}$

The correct answer is:
$1.34 \times 10^{-2}$

Solution:
We have,
$\begin{aligned}
& \frac{\Delta V}{V}=\frac{\mathrm{h} \rho \mathrm{g}}{\mathrm{B}}\left[\because \mathrm{B}=\frac{\Delta \mathrm{P}}{\Delta \mathrm{V} / \mathrm{V}} \Rightarrow \frac{\Delta \mathrm{V}}{\mathrm{V}}=\frac{\Delta \mathrm{P}}{\mathrm{B}}=\frac{\mathrm{h} \rho \mathrm{g}}{\mathrm{B}}\right] \\
& \Rightarrow \quad \frac{\Delta V}{V}=\frac{3 \times 10^3 \times 10^3 \times 9.8}{2.2 \times 10^9}=1.34 \times 10^{-2}
\end{aligned}$

Topic: Physics, Mechanical Properties of Solids, JEE Main
The average energy of molecules in a sample of oxygen gas at $300 \mathrm{~K}$ is $6.21 \times 10^{-21} \mathrm{~J}$. The corresponding values at $600 \mathrm{~K}$ are

Options:
A $12.12 \times 10^{-21} \mathrm{~J}$
B $8.78 \times 10^{-21} \mathrm{~J}$
C $6.21 \times 10^{-21} \mathrm{~J}$
D $12.42 \times 10^{-21} \mathrm{~J}$

The correct answer is:
$12.42 \times 10^{-21} \mathrm{~J}$

Solution:
At temperature, $T_{1}=300 \mathrm{~K}$
Average kinetic energy, $E_{1}=6.21 \times 10^{-21} \mathrm{~J}$
$$
\therefore \quad T_{2}=600 \mathrm{~K}, E_{2}=\text { ? }
$$
We know that, average kinetic energy of gas molecule,
$$
\begin{aligned}
\Rightarrow \quad \frac{E_{2}}{E_{1}} &=\frac{T_{2}}{T_{1}} \Rightarrow E_{2}=\frac{T_{2}}{T_{1}} \times E_{1} \\
&=\frac{600}{300} \times 6.21 \times 10^{-21} \\
&=2 \times 6.21 \times 10^{-21} \\
&=12.42 \times 10^{-21} \mathrm{~J}
\end{aligned}
$$

Topic: Physics, Kinetic Theory of Gases, JEE Main
The average force applied on the walls of a closed container depends on $\mathrm{T}^{\mathrm{x}}$ where $\mathrm{T}$ is the temperature of an ideal gas. The value of ' $x$ ' is

Options:
A $4$
B $3$
C $2$
D $1$

The correct answer is:
$1$

Solution:
$\begin{aligned}
& P=\frac{F}{A} \\
\therefore \quad P & \propto F .... (i)
\end{aligned}$
From Ideal Gas Equation,
$\mathrm{P}=\frac{\mathrm{nRT}}{\mathrm{V}}$
When $\mathrm{V}$ remains constant,
$\mathrm{P} \propto \mathrm{T}.... (ii)$
Comparing (i) and (ii)
$\mathrm{F} \propto \mathrm{T}^{\mathrm{x}}$
$\therefore \quad x=1$

Topic: Physics, Kinetic Theory of Gases, MHT CET, MHT CET 2023 (10 May Shift 1)
The average kinetic energy of a molecule of the gas is

Options:
A proportional to absolute temperature
B proportional to volume
C proportional to pressure
D dependent on the nature of the gas

The correct answer is:
proportional to absolute temperature

Solution:
The average kinetic energy of a molecule of the gas =32RT.

Average kinetic energy of one molecule of gas =32kT ( k=RNA)

Where k is Boltzmann constant and T is absolute temperature.

So, from above relation we can say average kinetic energy of molecule is directly proportional to absolute temperature.

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2023 (01 Feb Shift 1)
The average kinetic energy of a monatomic molecule is 0.414 eV at temperature:

(Use KB=1.38×10-23 J mol-1 K-1)

Options:
A 3000 K
B 3200 K
C 1600 K
D 1500 K

The correct answer is:
3200 K

Solution:
For monatomic molecule, degree of freedom =3.

The formula for the average kinetic energy is given by

Kavg=12f KBT ...1

From equation (1), it follows that

0.414 eV×1.6×10-19 J1 eV=32×1.38×10-23 J mol-1 K-1×T⇒T=0.414 eV×1.6×10-19 J1 eV×23×1.38×10-23 J mol-1 K-1=3200 K

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2024 (27 Jan Shift 1)