datavorous/entrance-exam-dataset huggingface dataset:
There rotten apples are mixed accidently with seven good apples and four apples are drawn one by one without replacement. Let the random variable X denote the number of rotten apples. If μ and σ2 represent mean and variance of X, respectively, then 10μ2+σ2 is equal to

Options:
A 20
B 250
C 25
D 30

The correct answer is:
20

Solution:
There rotten apples are mixed accidently with seven good apples,

And four apples are drawn one by one without replacement.

Now the random variable X denote the number of rotten apples.

And μ and σ2 represent mean and variance of X, respectively,

So, now plotting the table of given data we get,

x	P(x)	xP(X)	x2P(X)

0	16	0	0

1	12	12	12

2	310	610	1210

3	130	110	930

Now mean is given by, ∑xP(x)=62=μ

And variance is given by σ2=∑x2P(x)-μ2

So, by putting the value from above table we get, σ2+μ2=0+12+1210+930=2

Hence, 10σ2+μ2=20

Topic: Mathematics, Probability, JEE Main, JEE Main 2023 (29 Jan Shift 1)
Thermal decomposition of a Mn compound X at 513 K results in compound Y, MnO2 and a gaseous product. MnO2 reacts with NaCl and concentrated H2SO4 to give a pungent gasZ . X, Y and Z, respectively, are:

Options:
A K3MnO4,K3MnO4 and Cl2
B K2MnO4, KMnO4 and Cl2
C K2MnO4,KMnO4 and SO2
D KMnO4, K2MnO4 and Cl2

The correct answer is:
KMnO4, K2MnO4 and Cl2

Solution:
Potassium manganate is water soluble and so is leached out as green filtrate. MnO2 is insoluble in water and gives black residue. The oxygen gas is released in this reaction.

In the first step potassium permanganate (KMnO4) the compound (X) decomposes on heating to give potassium manganate (K2MnO4) which is the compound (Y), MnO2 and the gaseous product is oxygen.

In the second step MnO2 further reacts with NaCl and concentrated sulfuric acid to give a pungent-smelling gas that is Chlorine gas (Cl2 ), MnSO4 , Na2SO4 and H2O.

2KMnO4X→Δ513 KΔ K2Mn(Y)O4+MnO2+O2g

MnO2+4NaCl+4H2SO4→MnCl2+4NaHSO4+2H2O+Cl2gZ Pungent gas

The compound X= KMnO4 , Y=K2MnO4, Z=Cl2.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2019 (12 Apr Shift 2)
Thermal power plants can lead to :

Options:
A Acid rain
B Blue baby syndrome
C Ozone layer depletion
D Eutrophication

The correct answer is:
Acid rain

Solution:
Burning of fossil fuels (which contain sulphur and nitrogenous matter) such as coal and oil in power stations and furnaces produce sulphur dioxide and nitrogen oxides which causes acid rain.

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2020 (03 Sep Shift 1)
Thermite is a mixture of $X$ parts of ferric oxide and $Y$ parts of aluminium powder. $X, Y$ respectively are

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

The correct answer is:
$3,1$

Solution:
No Solution Available

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
Thermococcus, Methanococcus and Methanobacterium examplify

Options:
A archabacteria that contain protein homologous to eukaryotic core histones
B archaebacteria that lack any histones resembling those found in eukaryotes but whose DNA is negatively supercoiled
C bacteria whose DNA is relaxed or positively supercoiled but which have a cytoskeleton as well as mitochondria
D bacteria that contain a cytoskeleton and ribosomes

The correct answer is:
archabacteria that contain protein homologous to eukaryotic core histones

Solution:
Inspection of domain Archaea shows that two sub-divisions exist the Euryarchaeota and the Crenarchaeota. The Archaea are chemotrophic although Halobacterium can use light to make ATP. Many others are chemolithotrophs with hydrogen gas being a widely used energy source. The Euryarchaeota includes Methanobacterium, Methanocaccus, Thermocaccus. The differential features among bacteria, Archaea and Eukarya are:

Topic: Biology, Biological Classification, NEET, NEET 2008 (Screening)
Thermococcus, Methanococcus and Methanobacterium exemplify

Options:
A Bacteria that contain a cytoskeleton and ribosomes
B Archaebacteria that contain protein homologous to eukaryotic core histones
C Archaebacteria that lack any histones resembling those found in eukaryoties but whose DNA is negatively supercoiled
D Bacteria whose DNA is relaxed or positively supercoiled but which have a cytoskeleton as well as mitochondria

The correct answer is:
Archaebacteria that lack any histones resembling those found in eukaryoties but whose DNA is negatively supercoiled

Solution:
Archaebacteria have -ve supercoiled DNA but histones are chemically different in composition from eukaryotes.

Topic: Biology, Biological Classification, NEET, NEET 2008 (Mains)
Thermodynamic process is shown below on a P-V diagram for one mole of an ideal gas. If V2=2V1, then the ratio of temperature T2T1 is :

Options:
A 2
B 12
C 12
D 2

The correct answer is:
2

Solution:
PV1/2=c

nRTVV1/2=c

T=c1 V1/2

T2 T1=V2 V11/2=2 V1 V11/2

T2 T1=2

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2021 (25 Feb Shift 2)
Thermosetting polymers:
(a) are formed by condensation reaction.
(b) have cross-linked structure.
(c) are hard, rigid and brittle.
Correct statements are.

Options:
A Both (a) and (b)
B Both (a) and (c)
C (a), (b) and (c)
D Both (b) and (c)

The correct answer is:
(a), (b) and (c)

Solution:
Thermosetting polymer, also known as a thermoset or thermosetting plastic, is a polymer consisting of cross-linked structure or heavily branched molecules.

These polymers are in the soft solid or viscous state and on heating undergo extensive cross-linking in moulds and become irreversibly hard as well as insoluble products.

Properties of thermosetting polymers

One of the main properties of thermosetting plastics or polymers is that they harden during the moulding process and after solidifying they cannot be softened.

When the polymers are moulded and given a shape, they acquire a three-dimensional cross-linked structure along with strong covalent bonds that help them retain their strength as well as structure when the temperatures are set high.

However, thermoset plastics are brittle and tend to char and burn when heat is applied for a prolonged period. Normally, thermosets decompose before melting.

Thermosetting plastics are synthesised by condensation polymerization.

Thermosetting polymers are formed by condensation polymerization, whereas thermoplastic are plastic polymers formed by the addition polymerization process.

Examples of thermosetting polymers are: vulcanised rubber, novolac, polyurethane etc.

Topic: Chemistry, Polymers, JEE Main
Ther.m.s. velocity of oxygen molecule at $16^{\circ} \mathrm{C}$ is $474 \mathrm{~m} / \mathrm{sec}$. The r.m.s. velocity in $\mathrm{m} / \mathrm{s}$ of hydrogen molecule at $127^{\circ} \mathrm{C}$ is

Options:
A 1603
B 1896
C 2230.59
D 2730

The correct answer is:
2230.59

Solution:
$v_{\operatorname{oxg} .}=\sqrt{\frac{3 R \times 289}{32}}\left(v_{\mathrm{rms}}=\sqrt{\frac{3 R T}{\mathrm{M}}}\right)$

$\mathrm{v}_{\mathrm{H}}=\sqrt{\frac{3 \mathrm{R} \times 400}{2}}$ so $\mathrm{v}_{\mathrm{H}}=2230.59 \mathrm{~m} / \mathrm{sec}$

Topic: Physics, Kinetic Theory of Gases, JEE Main
These are physical properties of an element
(A) Sublimation enthalpy
(B) Ionisation enthalpy
(C) Hydration enthalpy
(D) Electron gain enthalpy
The total number of above properties that affect the reduction potential is (Integer answer)

Options:

The correct answer is:
3

Solution:
So reduction potential depends on sublimation, ionisation and Hydration energy.

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2021 (26 Aug Shift 1)
Thiamine and pyridoxine are also known respectively as :

Options:
A
Vitamin $B_2$ and Vitamin $E$
B
Vitamin $E$ and Vitamin $B_2$
C
Vitamin $B_6$ and Vitamin $B_2$
D
Vitamin $B_1$ and Vitamin $B_6$

The correct answer is:
Vitamin $B_1$ and Vitamin $B_6$

Solution:
Vitamin $B_1$ - Thiamin Vitamin
$\mathrm{B}_6$ - Pyridoxine

Topic: Chemistry, Biomolecules, JEE Main
Thiamine and pyridoxine are also known respectively as:

Options:
A Vitamin B2 and Vitamin E
B Vitamin E and Vitamin B2
C Vitamin B6 and Vitamin B2
D Vitamin B1 and Vitamin B6

The correct answer is:
Vitamin B1 and Vitamin B6

Solution:
Vitamine- B1 is also known as thiamine while vitamin B-6 is known as pyridoxine

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2021 (22 Jul Shift 1)
Thin layer chromatography of a mixture shows the following observation:

The correct order of elution in the silica gel column chromatography is

Options:
A B, A, C
B B, C, A
C A, C, B
D C, A, B

The correct answer is:
A, C, B

Solution:
The process of extraction of different compounds absorbed on the column by means of a suitable solvent (eluent) is called 'elution'. Silica gel and solvent methylene chloride both are polar. The polar solvent and silica gel bind the polar compound, so it moves very slowly, whereas the non-polar compound moves faster on the silica gel.

Correct order of elution → A > C > B

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2023 (11 Apr Shift 1)
Thiol group is present in:

Options:
A Cysteine
B Methionine
C Cytosine
D Cystine

The correct answer is:
Cysteine

Solution:
Thiol group is -SH
(i) Cysteine â
(ii) Methionine
(iii) Cytosine
(iv) Cystine

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2016 (03 Apr)
Thiosulphate reacts differently with iodine and bromine in the reactions given below:
\(\begin{aligned}
& 2 \mathrm{~S}_2 \mathrm{O}_3^{2-}+\mathrm{I}_2 \rightarrow \mathrm{S}_4 \mathrm{O}_6^{2-}+2 \mathrm{I}^{-} \\
& \mathrm{S}_2 \mathrm{O}_3^{2-}+5 \mathrm{Br}_2+5 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{SO}_4^{2-}+4 \mathrm{Br}^{-}+10 \mathrm{H}^{+}
\end{aligned}\)
Which of the following statement justifies the above dual behaviour of thiosulphate?

Options:
A Bromine is a stronger oxidant than iodine
B Thiosulphate undergoes oxidation by bromine and reduction by iodine in these reaction
C Bromine is a weaker oxidant than iodine
D Bromine undergoes oxidation and iodine undergoes reduction in these reactions

The correct answer is:
Bromine is a stronger oxidant than iodine

Solution:
In the reaction of $\mathrm{S}_2 \mathrm{O}_3{ }^{2-}$ with $\mathrm{I}_2$, oxidation state of sulphur changes to +2 to +2.5
In the reaction of $\mathrm{S}_2 \mathrm{O}_3{ }^{2-}$ with $\mathrm{Br}_2$, oxidation state of sulphur changes from +2 to +6 .
$\therefore$ Both $\mathrm{I}_2$ and $\mathrm{Br}_2$ are oxidant (oxidising agent) and $\mathrm{Br}_2$ is stronger oxidant than $\mathrm{I}_2$.

Topic: Chemistry, Redox Reactions, JEE Main, JEE Main 2024 (08 Apr Shift 1)
Thirty two persons $X_{1}, X_{2}, \ldots, X_{32}$ are randomly seated around a circular table at equal intervals. Two persons $X_{i}$ and $X_{j}$ are said to be within earshot of each other if there are at most three persons between them on the minor arc joining $X_{i}$ and $X_{j}$. The probability that $X_{1}$ and $\mathrm{X}_{2}$ are within earshot of each other is, $\left(\right.$ Here $\left.\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right)=\frac{\mathrm{n} !}{(\mathrm{n}-\mathrm{r}) ! \mathrm{r} !}\right)$

Options:
A $\frac{\left(\begin{array}{c}32 \\ 2\end{array}\right) 30 !}{8(32 !)}$
B $\frac{\left(\begin{array}{c}32 \\ 2\end{array}\right) 30 !}{4(32 !)}$
C $\frac{8}{31}$
D $\frac{4}{31}$

The correct answer is:
$\frac{8}{31}$

Solution:
Case - 1
No person between $\mathrm{X}_{1} \& \mathrm{X}_{2}$
$\frac{30 ! \times 2 !}{31 !}=\frac{2}{31}$
Case-2
One person between $X_{1} \& X_{2}$
$\frac{{ }^{30} \mathrm{C}_{1}(29) ! \times 2 !}{31 !}=\frac{2}{31}$
Case-3
When 2 person between $\mathrm{X}_{1} \& \mathrm{X}_{2}$
$\frac{{ }^{30} \mathrm{C}_{2} \times 28 ! \times 2 ! \times 2 !}{31 !}=\frac{2}{31}$
Case-4
When 3 person between $X_{1} \& X_{2}$ $\frac{{ }^{30} \mathrm{C}_{3} \times 27 ! \times 2 ! \times 3 !}{31 !}=\frac{2}{31}$
Total $=\frac{8}{31}$

Topic: Mathematics, Permutation Combination, JEE Main
This compound given below on heating gives P (Major product)

is

Options:
A
B
C
D No reaction

The correct answer is:

Solution:
This dehydration reaction occurs as follows

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
This enzyme used for the degradation of chitin, which forms the main component of the cell wall of fungi is the function of ___.

Options:
A Origin of replication
B Restriction enzyme
C Bioreactors
D Chitinase

The correct answer is:
Chitinase

Solution:
This enzyme used for the degradation of chitin, which forms the main component of the cell wall of fungi is the function of Chitinase

Topic: Biology, Biotechnology - Principles and Processes, NEET
This question contains Statement $-1$ and Statement-2. Of the four choices given after the statements, choose the one that best describes the two statements.
Statement - I:
For a mass $M$ kept at the centre of a cube of side ' $a$ ', the flux of gravitational field passing through its sides is $4 \pi$ GM.
and
Statement - II
If the direction of a field due to a point source is radial and its dependence on the distance ' $r$ ' for the source is given as $1 / r^2$, its flux through a closed surface depends only on the strength of the source enclosed by the surface and not on the size or shape of the surface

Options:
A
Statement $-1$ is false, Statement $-2$ is true.

B
Statement $-1$ is true, Statement $-2$ is true; Statement $-2$ is correct explanation for Statement-1.

C
Statement $-1$ is true, Statement $-2$ is true; Statement $-2$ is not a correct explanation for Statement-1.

D
Statement $-1$ is true, Statement $-2$ is False.

The correct answer is:
Statement $-1$ is true, Statement $-2$ is true; Statement $-2$ is correct explanation for Statement-1.

Solution:
$$
\mathrm{g}=\mathrm{GM} / \mathrm{r}^2
$$

Topic: Physics, Gravitation, JEE Main, JEE Main 2008
This question contains Statement $-1$ and Statement-2. Of the four choices given after the statements, choose the one that best describes the two statements.
Statement - I:
Energy is released when heavy nuclei undergo fission or light nuclei undergo fusion. and
Statement - II
For heavy nuclei, binding energy per nucleon increases with increasing $Z$ while for light nuclei it decrease with increasing $Z$.

Options:
A
Statement $-1$ is false, Statement $-2$ is true.

B
Statement - 1is true, Statement $-2$ is true; Statement $-2$ is correct explanation for Statement-1.

C
Statement $-1$ is true, Statement $-2$ is true; Statement $-2$ is not a correct explanation for Statement-1.

D
Statement $-1$ is true, Statement $-2$ is False.

The correct answer is:
Statement $-1$ is true, Statement $-2$ is False.

Solution:
No Solution Available

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2008
This question contains Statement-1 and Statement-2. Of the four choices given after the statements, choose the one that best describes the two statements.
Statement-1: The temperature dependence of resistance is usually given as $R=R_0(1+\alpha \Delta t)$. The resistance of a wire changes from $100 \Omega$ to $150 \Omega$ when its temperature is increased from $27^{\circ} \mathrm{C}$ to $227^{\circ} \mathrm{C}$. This implies that $\alpha=2.5 \times 10^{-3} /{ }^{\circ} \mathrm{C}$.
Statement 2: $R=R_i(1+\alpha \Delta T)$ is valid only when the change in the temperature $\Delta T$ is small and $\Delta R=$ $\left(R-R_0\right) \ll R_0$.

Options:
A
Statement-1 is true, Statement-2 is false

B
Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1.

C
Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation of Statement-1.

D
Statement-1 is false, Statement-2 is true

The correct answer is:
Statement-1 is true, Statement-2 is false

Solution:
$$
\text { Directions: Question numbers } 12 \text { and } 13 \text { are based on the following paragraph. }
$$
A current loop $A B C D$ is held fixed on the plane of the paper as shown in the figure. The arcs $B C$ (radius $=b$ ) and $D A$ (radius $=a)$ of the loop are joined by two straight wires $A B$ and $C D$. A steady current $I$ is flowing in the loop. Angle made by $A B$ and $C D$ at the origin $O$ is $30^{\circ}$. Another straight thin wire with steady current $\mathrm{l}_1$ flowing out of the plane of the paper is kept at the origin.

Topic: Physics, Current Electricity, JEE Main, JEE Main 2009
This question contains Statement-1 and Statement-2. Of the four choices given after the statements, choose the one that best describes the two statements.
Statement - 1: For a charged particle moving from point $P$ to point $Q$, the net work done by an electrostatic field on the particle is independent of the path connecting point $P$ to point $Q$.
Statement-2: The net work done by a conservative force on an object moving along a closed loop is zero

Options:
A
Statement-1 is true, Statement-2 is false

B
Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1.

C
Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation of Statement-1.

D
Statement-1 is false, Statement-2 is true

The correct answer is:
Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1.

Solution:
Work done by conservative force does not depend on the path. Electrostatic force is a conservative force.

Topic: Physics, Electrostatics, JEE Main
This question contains Statement-1 and Statement-2. Of the four choices given after the statements, choose the one that best describes the two statements.
Statement - 1: For a charged particle moving from point $P$ to point $Q$, the net work done by an electrostatic field on the particle is independent of the path connecting point $P$ to point $Q$.
Statement-2: The net work done by a conservative force on an object moving along a closed loop is zero

Options:
A
Statement-1 is true, Statement-2 is false

B
Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1.

C
Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation of Statement-1.

D
Statement-1 is false, Statement-2 is true

The correct answer is:
Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1.

Solution:
Work done by conservative force does not depend on the path. Electrostatic force is a conservative force.

Topic: Physics, Electrostatics, JEE Main, JEE Main 2009
This question has statement $1$ and statement $2$ . Of the four choices given after the statements, choose the one that best describes the two statements.
An insulating solid sphere of radius $\mathrm{R}$ has a uniformly positive charge density $\rho$. As a result of this uniform charge distribution there is a finite value of electric potential at the centre of the sphere, at the surface of the sphere and also at a point out side the sphere. The electric potential at infinity is zero.
Statement $1$: When a charge $q$ is taken from the centre to the surface of the sphere, its potential energy changes by $\frac{\mathrm{qp}}{3 \varepsilon_0}$
Statement $2$: The electric field at a distance $r(r < R)$ from the centre of the sphere is $\frac{\rho r}{3 \varepsilon_0}$

Options:
A
Statement $1$ is true, Statement $2$ is true, Statement $2$ is not the correct explanation for statement $1$.

B
Statement $1$ is true, Statement $2$ is false

C
Statement $1$ is false, Statement $2$ is true

D
Statement $1$ is true, Statement $2$ is the correct explanation for statement $1$

The correct answer is:
Statement $1$ is false, Statement $2$ is true

Solution:
$\oint \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{d}} \mathrm{A}=\frac{1}{\varepsilon_0}\left(\rho \times \frac{4}{3} \pi \mathrm{r}^3\right)$
$\mathrm{E}=\frac{\rho \mathrm{r}}{3 \varepsilon_0}$
Statement $2$ is correct
$\Delta \mathrm{PE}=\left(\mathrm{V}_{\text {sur }}-\mathrm{V}_{\text {cent }}\right) \mathrm{q}=-\frac{\mathrm{q}}{6 \varepsilon_0} \rho \mathrm{R}^2$
Statement $1$ is incorrect

Topic: Physics, Electrostatics, JEE Main
This question has statement $1$ and statement $2$ . Of the four choices given after the statements, choose the one that best describes the two statements.
An insulating solid sphere of radius $\mathrm{R}$ has a uniformly positive charge density $\rho$. As a result of this uniform charge distribution there is a finite value of electric potential at the centre of the sphere, at the surface of the sphere and also at a point out side the sphere. The electric potential at infinity is zero.
Statement $1$: When a charge $q$ is taken from the centre to the surface of the sphere, its potential energy changes by $\frac{\mathrm{qp}}{3 \varepsilon_0}$
Statement $2$: The electric field at a distance $r(r < R)$ from the centre of the sphere is $\frac{\rho r}{3 \varepsilon_0}$

Options:
A
Statement $1$ is true, Statement $2$ is true, Statement $2$ is not the correct explanation for statement $1$.

B
Statement $1$ is true, Statement $2$ is false

C
Statement $1$ is false, Statement $2$ is true

D
Statement $1$ is true, Statement $2$ is the correct explanation for statement $1$

The correct answer is:
Statement $1$ is false, Statement $2$ is true

Solution:
$\oint \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{d}} \mathrm{A}=\frac{1}{\varepsilon_0}\left(\rho \times \frac{4}{3} \pi \mathrm{r}^3\right)$
$\mathrm{E}=\frac{\rho \mathrm{r}}{3 \varepsilon_0}$
Statement $2$ is correct
$\Delta \mathrm{PE}=\left(\mathrm{V}_{\text {sur }}-\mathrm{V}_{\text {cent }}\right) \mathrm{q}=-\frac{\mathrm{q}}{6 \varepsilon_0} \rho \mathrm{R}^2$
Statement $1$ is incorrect

Topic: Physics, Electrostatics, JEE Main, JEE Main 2012 (Offline)
This question has statement $1$ and statement $2$. Of the four choices given after the statements, choose the one that best describes the two statements.
If two springs $\mathrm{S_1}$ and $\mathrm{S_2}$ of force constants $k_1$ and $k_2$, respectively, are stretched by the same force, it is found that more work is done on spring $\mathrm{S_1}$ than on spring $\mathrm{S_2}$.
Statement $1$ : If stretched by the same amount, work done on $\mathrm{S}_1$, will be more than that on $\mathrm{S}_2$
Statement $2: k_1 < k_2$

Options:
A
Statement $1$ is false, Statement $2$ is true

B
Statement $1$ is true, Statement $2$ is false

C
Statement $1$ is true, Statement $2$ is the correct explanation for statement $1$

D
Statement $1$ is true, Statement $2$ is true, Statement $2$ is not the correct explanation for statement $1$.

The correct answer is:
Statement $1$ is false, Statement $2$ is true

Solution:
$\mathrm{F}=\mathrm{K}_1 \mathrm{S}_1=\mathrm{K}_2 \mathrm{S}_2$
$\mathrm{W}_1=\mathrm{FS}_1, \mathrm{~W}_2=\mathrm{FS}_2$
$\mathrm{K}_1 \mathrm{S}_1{ }^2>\mathrm{K}_2 \mathrm{S}_2^2$
$\mathrm{S}_1>\mathrm{S}_2$
$\mathrm{K}_1 < \mathrm{K}_2$
$\mathrm{W} \propto \mathrm{K}$
$\mathrm{W}_1 < \mathrm{W}_2$

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2012 (Offline)
This question has Statement 1 and Statement 2. Of the four choices given after the Statements, choose the one that best describes the two Statements.
Statement 1: In an adiabatic process, change in internal energy of a gas is equal to work done on/ by the gas in the process.
Statement 2: The temperature of a gas remains constant in an adiabatic process.

Options:
A
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation of Statement 1.

B
Statement 1 is true, Statement 2 is false.

C
Statement 1 is false, Statement 2 is true.

D
Statement 1 is false, Statement 2 is true, Statement 2 is not a correct explanation of Statement 1.

The correct answer is:
Statement 1 is true, Statement 2 is false.

Solution:
In an adiabatic process, $\delta \mathrm{H}=0$ And according to first law of thermodynamics
$$
\begin{aligned}
& \delta H=\delta U+W \\
& \therefore W=-\delta U
\end{aligned}
$$

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2012 (07 May Online)