datavorous/entrance-exam-dataset huggingface dataset:
$\operatorname{adj}\left[\begin{array}{ccc}1 & 0 & 2 \\ -1 & 1 & -2 \\ 0 & 2 & 1\end{array}\right]=\left[\begin{array}{ccc}5 & a & -2 \\ 1 & 1 & 0 \\ -2 & -2 & b\end{array}\right], \quad$ then $\left[\begin{array}{ll}a & b\end{array}\right]$ is equal to

Options:
A [âˆ’ 4 1]
B [âˆ’ 4 âˆ’1]
C [4 1]
D [4 âˆ’1]

The correct answer is:
[4 1]

Solution:
Given that
$\begin{aligned}
& \text { Cofactor of }\left[\begin{array}{ccc}
1 & 0 & 2 \\
-1 & 1 & -2 \\
0 & 2 & 1
\end{array}\right] \text { are } \\
& C_{11}=5, C_{12}=1, C_{13}=-2 \\
& C_{21}=4, C_{22}=1, C_{23}=-2 \\
& C_{31}=-2, C_{32}=0, C_{33}=1 \\
& \Rightarrow\left[\begin{array}{ccc}
5 & 4 & -2 \\
1 & 1 & 0 \\
-2 & -2 & 1
\end{array}\right]=\left[\begin{array}{ccc}
5 & a & -2 \\
1 & 1 & 0 \\
-2 & -2 & b
\end{array}\right]
\end{aligned}$
On comparing the corresponding elements, we get
$\begin{array}{rlrl}
& a & =4, b=1 \\
\therefore & {\left[\begin{array}{ll}
a & b
\end{array}\right]} & =\left[\begin{array}{ll}
4 & 1
\end{array}\right]
\end{array}$

Topic: Mathematics, Matrices, JEE Main
Adjoint of the matrix
$N=\left[\begin{array}{ccc}-4 & -3 & -3 \\ 1 & 0 & 1 \\ 4 & 4 & 3\end{array}\right]$ is

Options:
A $N$
B $2 \mathrm{~N}$
C $-N$
D None of these

The correct answer is:
$N$

Solution:
The cofactors of $N=\left[\begin{array}{ccc}-4 & -3 & -3 \\ 1 & 0 & 1 \\ 4 & 4 & 3\end{array}\right]$ are
$c_{11}=-4, c_{12}=1, c_{13}=4 ; \quad c_{21}=-3, c_{22}=0, c_{23}=4$
$c_{31}=-3, c_{32}=1, c_{33}=3$
$\therefore \operatorname{adj} N=\left[\begin{array}{ccc}-4 & -3 & -3 \\ 1 & 0 & 1 \\ 4 & 4 & 3\end{array}\right]=N$

Topic: Mathematics, Matrices, JEE Main
ADP and ATP differ in the number of

Options:
A phosphate units
B ribose units
C adenine base
D nitrogen atom

The correct answer is:
phosphate units

Solution:
$\begin{array}{llll}\text {} & \text { ADP } & \text { is } & \text { adenosine } & \text { diphosphate }\end{array}$
(2-phosphate groups) and ATP is adenosine triphosphate ( 3 phosphate groups) Thus, they differ in number of phosphate units.

Topic: Chemistry, Biomolecules, JEE Main
Adsorption of a gas follows Freudlich adsorption isotherm. If x is the mass of the gas adsorbed on mass m of the adsorbent, the correct plot of xm versus p is:

Options:
A
B
C
D

The correct answer is:

Solution:
From Freundlich adsorption isotherm
xm∝P (At low pressure)
xm∝P° (At high pressure)
→ On increasing temperature physical adsorption decreases.

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2020 (05 Sep Shift 2)
Adsorption of a gas follows Freundlich adsorption isotherm. In the given plot, x is the mass of the gas adsorbed on mass m of the adsorbent at pressure P . xm is proportional to:

Options:
A p14
B p
C p2
D p12

The correct answer is:
p12

Solution:
According to the Freundlich adsorption isotherm, xm=Kp1n
Taking log on both sides.
logxm=logK+1nlogâ¡P
Slope from graph =y2-y1x2-x1=24=12 or, 1n=12
â´ xm=Kp12
or xmâp12

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2019 (09 Jan Shift 1)
Adsorption of a gas follows Freundlich adsorption isotherm. In the given plot, x is the mass of the gas adsorbed on mass m of the adsorbent at pressure (P). if xm∝P(1/y) find the magnitude of ‘y’ is ------.

Options:

The correct answer is:
2

Solution:
According to Freundlich adsorption isotherm,
xm=K×P1/n
logxm=logK+1nlogP
m=1n=24=12⇒n=2
So, xm=K×P1/2
Therefore, value of 'y' is 2.

Topic: Chemistry, Surface Chemistry, JEE Main
Adsorption of a gas follows Freundlich adsorption isotherm. x is the mass of the gas adsorbed on mass m of the adsorbent. The plot of logxm vs logp is shown in the given graph. xm is proportional to,

Options:
A p3
B p2
C p2/3
D p3/2

The correct answer is:
p2/3

Solution:
According to Freundlich adsorption Isotherm,
xm=kP1n
log⁡xm=log⁡k+1nlog⁡p
y=mx+c; m=1n;
from the graph, m=23
xm=kP23
xm∝P23.

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2019 (08 Apr Shift 1)
Adsorption of gas on a surface follows Freundlich adsorption isotherm. The plot of log⁡xm versus logP gives a straight line with slope equal to 0.5, then:
(xm is the mass of the gas adsorbed per gram of adsorbent)

Options:
A adsorption is proportional to the square root of pressure.
B adsorption is proportional to the square of pressure.
C adsorption is proportional to the pressure.
D adsorption is independent of pressure.

The correct answer is:
adsorption is proportional to the square root of pressure.

Solution:
log(xm)=12log(P)+log(K);
xm=KP12

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2017 (09 Apr Online)
Adsorption of gases on solid surface is generally exothermic because

Options:
A enthalpy is positive
B entropy decreases
C entropy increases
D free energy increases

The correct answer is:
entropy decreases

Solution:
Thermodynamics of adsorption:
i. ΔS<0 because gaseous phase decrease
ii. ΔG<0 because process is spontaneous
so ΔG=ΔH-TΔS
-ve=ΔH-+ve-ve
ΔH=-ve
iii. ΔH<0 so adsorption is an exothermic process.

Topic: Chemistry, Surface Chemistry, JEE Main
‘Adsorption’ principle is used for which of the following purification method?

Options:
A Extraction
B Chromatography
C Distillation
D Sublimation

The correct answer is:
Chromatography

Solution:
Adsorption chromatography is the oldest type of chromatography technique. It makes use of a mobile phase which is either in liquid or gaseous form. The mobile phase is adsorbed onto the surface of a stationary solid phase.

Adsorption Chromatography involves the analytical separation of a chemical mixture based on the interaction of the adsorbate with the adsorbent. The mixture of gas or liquid gets separated when it passes over the adsorbent bed that adsorbs different compounds at different rates.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2024 (31 Jan Shift 1)
Aerobic atmosphere is maintained by ______

Options:
A prokaryotes
B protists
C plants
D fungi

The correct answer is:
plants

Solution:
Photosynthesis is important due to two reasons: it is the primary source of all food on earth. It is also responsible for the release of oxygen into the atmosphere by green plants which helps in the maintenance of the aerobic atmosphere on earth.

Topic: Biology, Photosynthesis in Higher Plants, NEET
Aerobic respiration is called

Options:
A Fermentation
B Chemosynthesis
C Bio–oxidation
D Photorespiration

The correct answer is:
Bio–oxidation

Solution:
Respiration uses oxygen and completely oxidizes organic food mainly carbohydrates (Sugars) to carbon dioxide and water. It, therefore, releases the entire energy available in glucose.
It is divided into two phases: Glycolysis and aerobic oxidation of pyruvic acid. Because it takes place in living beings (biological matter) it is called a bio-oxidation process.

Topic: Biology, Respiration in Plants, NEET
Aerobic respiratory pathway is appropriately termed :-

Options:
A Anabolic
B Catabolic
C Parabolic
D Amphibolic

The correct answer is:
Amphibolic

Solution:
No Solution Available

Topic: Biology, Respiration in Plants, NEET, NEET 2009 (Mains)
Aestivation of petals in the flower of cotton is correctly shown in

Options:
A

B

C

D

The correct answer is:

Solution:
Petals in the flowers of cotton (Gossypium sp), China rose (Hibiscus rosa sinensis), etc, show twisted aestivation. In this type of aestivation, a regular overlapping of petals occurs where one margin of each petal overlaps with the next one petal.

Topic: Biology, Morphology of Flowering Plants, NEET, NEET 2010 (Mains)
After 280 days , the activity of a radioactive sample is 6000 dps. The activity reduces to 3000dps after another 140days . The initial activity of the sample(in dps) is

Options:
A 6000
B 9000
C 3000
D 24000

The correct answer is:
24000

Solution:
Activity reduces from 6000 dps to 3000 dps in 140 days. It implies that half-life of the radioactive sample is 140 days. In 280 days (or two half-lives)activity will remain 14 th of the initial activity . Hence the initial activity of the sample is
4Ã6000 dps=24000 dps

Topic: Physics, Nuclear Physics, JEE Main
After a perfectly inelastic collision between two identical particles moving with the same speed in different directions, the speed of the particles becomes half the initial speed. The angle between the velocities of the particles before the collision is

Options:
A 60°
B 45°
C 120°
D 30°

The correct answer is:
120°

Solution:
In a perfectly inelastic collision between two particles, linear momentum is conserved. Let θ be the angle between the velocities of the two particles before the collision. Then
p2=p12+p22+2p1p2cosθ
or 2mv⁡22=mv⁡2+mv2+2mvmvcosθ
or 1 = 1 + 1 + 2 cos θ
or cos θ = - 1 2
or θ=120°

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
After adding non-volatile solute freezing point of water decreases to \( -0.186^{\circ} \mathrm{C} . \) Calculate
\( \Delta T_{b} \) if \( K_{\mathrm{f}}=1.86 K \mathrm{~kg} \mathrm{~mol}^{-1} \) and \( K_{b}=0.521 K \mathrm{~kg} \mathrm{~mol}^{-1} \)

Options:
A \( 0.521 \mathrm{~K} \)
B \( 0.0521 \mathrm{~K} \)
C \( 1.86 \mathrm{~K} \)
D \( 0.0186 \mathrm{~K} \)

The correct answer is:
\( 0.0521 \mathrm{~K} \)

Solution:
Depression of freezing point for the water is,
\[
\begin{array}{l}
\Delta T_{f}=T_{f}^{\circ}-T_{f}=0-(-0.186)=0.186^{\circ} \mathrm{C} \\
\Delta T_{f}=K_{f} \times m
\end{array}
\]
Now, the molality of the solution,
\[
m=\frac{\Delta T_{f}}{K_{f}}=\frac{0.186}{1.86}=0.1 \mathrm{~m}
\]
Elevation of boiling point for water is,
\[
\begin{array}{l}
\Delta T_{b}=K_{b} \times m \\
=0.521 \times 0.1 \\
\Delta T_{b}=0.0521 K
\end{array}
\]

Topic: Chemistry, Solutions, JEE Main
After equilibrium is restored in a sealed container with increased volume, what happens to the rates of evaporation and condensation?

Options:
A The rates of evaporation and condensation become equal again.
B The rate of evaporation remains higher than the rate of condensation.
C The rate of condensation exceeds the rate of evaporation.
D The rates of evaporation and condensation continue to change.

The correct answer is:
The rates of evaporation and condensation become equal again.

Solution:
Correct Option is : (A)
The rates of evaporation and condensation become equal again.

Topic: Chemistry, Ionic Equilibrium, NEET
After how many seconds will the concentration of the reactant in a first order reaction be halved if the rate constant is $1.155 \times 10^{-3} \mathrm{~s}^{-1}$ ?

Options:
A 600
B 100
C 60
D 10

The correct answer is:
600

Solution:
Rate constant $k=1.155 \times 10^{-3} \mathrm{~s}^{-1}$
$k=\frac{2.303}{t} \log \frac{a}{(a-x)} \quad \because a=a,(a-x)=\frac{a}{2}$
$\begin{aligned} t_{1 / 2} &=\frac{2.303}{k} \log \frac{a}{a / 2} \\ &=\frac{2.303}{1.155 \times 10^{-3}} \log 2 \\ &=\frac{2.303}{1.155 \times 10^{-3}} \times 0.3010 \end{aligned}$
$\begin{aligned} &=\frac{0.693 \times 10^{3}}{1.155} \\ \text { or } t_{1 / 2} &=\frac{0.693}{k}=\frac{0.693}{1.155 \times 10^{-3}} \\ &=600 \mathrm{~s} \end{aligned}$

Topic: Chemistry, Chemical Kinetics, JEE Main
After inserting $n$ A.M.'s between 2 and 38 , the sum of the resulting progression is 200 . The value of $n$ is

Options:
A $10$
B $8$
C $9$
D None of these

The correct answer is:
$8$

Solution:
The resulting progression will have $n+2$ terms with $2$ as the first term and 38 as the last term. Therefore the sum of the progression
$=\frac{n+2}{2}(2+38)=20(n+2)$
By hypothesis, $20(n+2)=200 \Rightarrow n=8$

Topic: Mathematics, Sequences and Series, JEE Main
After passing through a polariser a linearly polarised light of intensity $\mathrm{I}$ is incident on an analyser making an angle of $30^{\circ}$ with that of the polariser. The intensity of light emitted from the analyser will be :

Options:
A $\frac{\mathrm{I}}{2}$
B $\frac{\mathrm{I}}{3}$
C $\frac{3 I}{4}$
D $\frac{2 I}{4}$

The correct answer is:
$\frac{3 I}{4}$

Solution:
As per Malus law, $\mathrm{I}=\mathrm{I}_0 \cos ^2 \phi$, where symbols have their usual meanings.
$\begin{aligned}
& I^{\prime}=I \cos ^2 30^{\circ} \\
& I^{\prime}=I\left(\frac{\sqrt{3}}{2}\right)^2 \\
& I^{\prime}=\frac{3 I}{4}
\end{aligned}$

Topic: Physics, Wave Optics, JEE Main
After prolonged treatment of A by D2O/DO-, the maximum possible difference in molecular weight of compound A and B is

Options:
A 2
B 3
C 4
D 8

The correct answer is:
4

Solution:
In four positions, Hydrogen is replaced by deuterium. So, maximum difference in molecular weights is 4.

Topic: Chemistry, General Organic Chemistry, JEE Main
After understanding the assertion and reason, choose the correct option.
Assertion: In the bonding molecular orbital (MO) of H 2 , electron density is increased between the nuclei.
Reason: The bonding MO is ψA+ψB , which shows destructive interference of the combining electron waves.

Options:
A Assertion and Reason are correct, but Reason is not the correct explanation for the Assertion.
B Assertion and Reason are correct and Reason is the correct explanation for the Assertion.
C Assertion is incorrect, Reason is correct.
D Assertion is correct, Reason is incorrect.

The correct answer is:
Assertion is correct, Reason is incorrect.

Solution:
Electron density between nuclei increased during formation of BMO in H2 .
BMO is ψA+ψB (Linear combination of Atomtextc orbitals) provides constructive interference.
While by distructive interference anti BMO ψ A − ψ B formed.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2015 (10 Apr Online)
After understanding the assertion and reason, choose the correct option.
Assertion: In the bonding molecular orbital (MO) of H 2 , electron density is increased between the nuclei.
Reason: The bonding MO is ψA+ψB , which shows destructive interference of the combining electron waves.

Options:
A Assertion and Reason are correct, but Reason is not the correct explanation for the Assertion.
B Assertion and Reason are correct and Reason is the correct explanation for the Assertion.
C Assertion is incorrect, Reason is correct.
D Assertion is correct, Reason is incorrect.

The correct answer is:
Assertion is correct, Reason is incorrect.

Solution:
Electron density between nuclei increased during formation of BMO in H2 .
BMO is ψA+ψB (Linear combination of Atomtextc orbitals) provides constructive interference.
While by distructive interference anti BMO ψ A − ψ B formed.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
After understanding the assertion and reason, choose the correct option.

Assertion: In the bonding molecular orbital (MO) of H 2 , electron density is increased between the nuclei.
Reason: The bonding MO is ψA+ψB , which shows destructive interference of the combining electron waves.

Options:
A Assertion and Reason are correct, but Reason is not the correct explanation for the Assertion.
B Assertion and Reason are correct and Reason is the correct explanation for the Assertion.
C Assertion is incorrect, Reason is correct.
D Assertion is correct, Reason is incorrect.

The correct answer is:
Assertion is correct, Reason is incorrect.

Solution:
Electron density between nuclei increased during formation of BMO in H2 .
BMO is ψA+ψB (Linear combination of Atomtextc orbitals) provides constructive interference.
While by distructive interference anti BMO ψ A − ψ B formed.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
After vaccination the body builds up:

Options:
A Toxins
B Lymph
C Antibodies
D Plasma

The correct answer is:
Antibodies

Solution:
An antibody, also known as an immunoglobulin (Ig), is a large, Y-shaped protein produced mainly by plasma cells that are used by the immune system to neutralize pathogens such as pathogenic bacteria and viruses.

Topic: Biology, Human Health and Diseases, NEET
Ag 2 S + NaCN → A A + Zn → B

(B) is a metal. Hence, (A) and (B) are

Options:
A Na 2 Zn (CN) 4 ,Zn
B Na 2 Ag (CN) 2 ,Ag
C Na 2 Ag (CN) 4 ,Ag
D Na 3 Ag (CN) 4 ,Ag

The correct answer is:
Na 2 Ag (CN) 2 ,Ag

Solution:
Ag 2 S + 4 NaCN → 2 Na Ag CN 2 + Na 2 S A
2 Na Ag CN 2 + Zn → 2 Ag + Na 2 Zn CN 4 B

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
$\mathrm{Ag}^{+}+\mathrm{NH}_3 \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)\right]^{+} ; k_1=3.5 \times 10^{-3}$ $\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)\right]^{+}+\mathrm{NH}_3 \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+} ; k_2=1.7 \times 10^{-3}$ then the formation constant of $\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+}$is

Options:
A
$6.08 \times 10^{-6}$

B
$6.08 \times 10^6$

C
$6.08 \times 10^{-9}$

D
None of these

The correct answer is:
$6.08 \times 10^{-6}$

Solution:
(i) $\mathrm{Ag}^{+}+\mathrm{NH}_3 \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)\right]^{+} ; k_1=3.5 \times 10^{-3}$
(ii) $\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)\right]^{+}+\mathrm{NH}_3 \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+}, k_2=1.7 \times 10^{-3}$
on the basis of above reaction
$$
\begin{aligned}
& k_1=\frac{\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)\right]^{+}}{\left[\mathrm{Ag}^{+}\right]\left[\mathrm{NH}_3\right]} \\
& k_2=\frac{\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+}}{\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)\right]^{+}\left[\mathrm{NH}_3\right]}
\end{aligned}
$$
For the formation of $\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+}$
$$
\mathrm{Ag}^{+}+2 \mathrm{NH}_3 \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+}
$$
Formation constant $(K)=\frac{\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+}}{\left[\mathrm{Ag}^{+}\right]\left[\mathrm{NH}_3\right]^2}$
From eq. (i) and (ii)
$$
K=k_1 \times k_2=3.5 \times 10^{-3} \times 1.7 \times 10^{-3}=5.95 \times 10^{-6} \approx 6.08 \times 10^{-6}
$$

Topic: Chemistry, Chemical Equilibrium, JEE Main
Ag s + Fe 3 + aq â¶ Ag + aq + Fe 2 + aq
Given standard electrode potentials- E Fe 3 + / Fe 2 + o = + 0 . 7 7 V and E Ag + / Ag s o = + 0 . 8 0 V
If the reaction is feasible, enter 1.00 as answer elsewise enter 0.00.

Options:

The correct answer is:
0.00

Solution:
Ag s + Fe 3 + aq â¶ Ag + aq + Fe 2 + aq
E cell o = E Fe 3 + / Fe 2 + o - E Ag + / Ag o = 0 . 7 7 - 0 . 8 0 = - 0 . 0 3 V
The reaction is not feasible.
Fe 2 + + Ag + â¶ Ag s + Fe 3 + aq

Topic: Chemistry, Electrochemistry, JEE Main
$\mathrm{AgCl}$ dissolves in a solution of $\mathrm{NH}_{3}$ but not in water because

Options:
A $\mathrm{NH}_{3}$ is a better solvent than $\mathrm{H}_{2} \mathrm{O}$
B $\mathrm{Ag}^{+}$forms a complex ion with $\mathrm{NH}_{3}$
C $\mathrm{NH}_{3}$ is a stronger base than $\mathrm{H}_{2} \mathrm{O}$
D the dipole moment of water is higher than $\mathrm{NH}_{3}$

The correct answer is:
$\mathrm{Ag}^{+}$forms a complex ion with $\mathrm{NH}_{3}$

Solution:
$\mathrm{AgCl}$ dissolves in a solution of $\mathrm{NH}_{3}$ but not in water because $\mathrm{Ag}^{+}$forms soluble complex ion with $\mathrm{NH}_{3}$.

Topic: Chemistry, Coordination Compounds, JEE Main
AgNO3aq was added to an aqueous KCl solution gradually and the conductivity of the solution was measured. The plot of conductivity Î versus the volume of AgNO3 is

Options:
A (P)
B (Q)
C (R)
D (S)

The correct answer is:
(S)

Solution:
AgNO3+KClaq→AgCls+KNO3aq
Initially there is aq. KCl solution now as solution of AgNO3 is added, AgCls is formed. Hence conductivity of solution is almost compensated (or slightly increase) by the formation of KNO3 . After end point conductivity increases more rapidly because addition of excess AgNO3 solution.

Topic: Chemistry, Electrochemistry, JEE Main
Air (density ρ ) is being blown on a soap film (surface tension T ) by a pipe of radius R with its opening right next to the film. The film is deformed and a bubble detaches from the film when the shape of the deformed surface is a hemisphere. Given that the dynamic pressure on the film due to the air blown at speed v is 12ρv2, the speed at which the bubble formed is

Options:
A TρR
B 2TρR
C 4TρR
D 8TρR

The correct answer is:
8TρR

Solution:
Bubble detached from the look at ends of circumference, when force of dynamic pressure exceeds force of surface tension.

⇒pdynamic× Area ≥ Force of surface tension

⇒12ρv2×πR2≥22πRT

Note- that factor 2 on right hand side appears as there are two sides of film surface.

⇒v≥8TρR

⇒vmin=8TρR

Topic: Physics, Mechanical Properties of Fluids, JEE Main
Air inside a closed container is saturated with water vapour. The air pressure is \(\mathrm{p}\) and the saturated vapour pressure of water is \(\bar{p}\). If the mixture is compressed to one half of its volume by maintaining temperature constant, the pressure becomes

Options:
A \(2(\mathrm{p}+\overline{\mathrm{p}})\)
B \(2 \mathrm{p}+\overline{\mathrm{p}}\)
C \((\mathrm{p}+\overline{\mathrm{p}}) / 2\)
D \(\mathrm{p}+2 \overline{\mathrm{p}}\)

The correct answer is:
\(2 \mathrm{p}+\overline{\mathrm{p}}\)

Solution:
Hints: \(P_f=2 P+\bar{P}\)
Saturated vapour pressure will not change if temperature remains constant

Topic: Physics, Kinetic Theory of Gases, JEE Main
Air inside a closed container is saturated with water vapour. The air pressure is \(\mathrm{p}\) and the saturated vapour pressure of water is \(\bar{p}\). If the mixture is compressed to one half of its volume by maintaining temperature constant, the pressure becomes

Options:
A \(2(\mathrm{p}+\overline{\mathrm{p}})\)
B \(2 \mathrm{p}+\overline{\mathrm{p}}\)
C \((\mathrm{p}+\overline{\mathrm{p}}) / 2\)
D \(\mathrm{p}+2 \overline{\mathrm{p}}\)

The correct answer is:
\(2 \mathrm{p}+\overline{\mathrm{p}}\)

Solution:
Hints: \(P_f=2 P+\bar{P}\)
Saturated vapour pressure will not change if temperature remains constant

Topic: Physics, Kinetic Theory of Gases, WBJEE, WBJEE 2011
Air is an example of $\qquad$

Options:
A Paramagnetic material
B Diamagnetic material
C Ferromagnetic material
D Antiferromagnetic material

The correct answer is:
Paramagnetic material

Solution:
Air is an example of paramagnetic material since under an external magnetic field it forms an electric dipole.

Topic: Physics, Magnetic Properties of Matter, JEE Main
Air is bad conductor of heat or partly conducts heat, still vacuum is to be placed between the walls of the thermos flask because

Options:
A It is difficult to fill the air between the walls of thermos flask
B Due to more pressure of air, the thermos can get crack
C By convection, heat can flow through air
D On filling the air, there is no advantage

The correct answer is:
By convection, heat can flow through air

Solution:
No flow of heat by convection in vacuum.

Topic: Physics, Thermal Properties of Matter, JEE Main
Air is filled at $60^{\circ} \mathrm{C}$ in a vessel of open mouth. The vessel is heated to a temperature $t^{\circ} \mathrm{C}$ so that $\frac{1}{4}$ th of the air is escaped from the vessel. Assuming air as ideal gas and the volume of the vessel remaining constant, then the value of ' $t$ ' is

Options:
A $80^{\circ} \mathrm{C}$
B $171^{\circ} \mathrm{C}$
C $333^{\circ} \mathrm{C}$
D $444^{\circ} \mathrm{C}$

The correct answer is:
$171^{\circ} \mathrm{C}$

Solution:
Initial temperature, $T_1=60^{\circ} \mathrm{C}=60+273=333 \mathrm{~K}$
Mass of the air left in the vessel, $M_2=M-\frac{M}{4}$
$$
=\frac{3 \mathrm{M}}{4}
$$
Here pressure and volume remain constant from ideal gas equation
$$
\begin{aligned}
& P V=n R T \\
& P V=\left(\frac{M}{m}\right) R T
\end{aligned}
$$
$$
\begin{aligned}
& \mathrm{MT}=\text { constant } \\
& \mathrm{M}_1 \mathrm{~T}_1-\mathrm{M}_2 \mathrm{~T}_2 \\
& \mathrm{~T}_2=\frac{\mathrm{M}_1 \mathrm{~T}_1}{\mathrm{M}_2}=\frac{\mathrm{M} \times 333}{3 \mathrm{M}} \times 4 \\
& =444 \mathrm{~K} \\
& \Rightarrow \mathrm{t}=\mathrm{T}_2=171^{\circ} \mathrm{C}
\end{aligned}
$$

Topic: Physics, Kinetic Theory of Gases, TS EAMCET, TS EAMCET 2023 (13 May Shift 1)
Air is filled in a bottle at atmospheric pressure and it is corked at 35 Â°C. If the cork can come out at a pressure 3 times the atmospheric pressure, then up to what temperature should the bottle be heated in order to remove the cork.

Options:
A 325.5 °C
B 851 °C
C 651 °C
D None of these

The correct answer is:
651 °C

Solution:
We know that, p1V=nRT1
⇒1×V=nR35+273
=nR×308 ....(i)
and p2V=nRT2
⇒ 3×V=nRT2 ....(ii)
On dividing Equation (ii) by Equation (i), we get
3=T2/308
⇒ T2=924 K
T2=924-273
=651 °C

Topic: Physics, Thermodynamics, JEE Main