datavorous/entrance-exam-dataset huggingface dataset:
What was the main challenge faced by India after independence?

Options:
A Lack of food for the increasing population
B Lack of cultivable land
C Lack of high yielding crops
D None of the above

The correct answer is:
Lack of food for the increasing population

Solution:
One of the main challenges faced by India after independence was to produce enough food for the increasing population.

Topic: Biology, Strategies for Enhancement in Food Production, NEET
What was the main conclusion of the experiments performed by Meselson and Stahl and Taylor and colleagues?

Options:
A DNA replicates conservatively
B DNA replicates randomly
C DNA replicates semiconservatively
D DNA replicates differently in different organisms

The correct answer is:
DNA replicates semiconservatively

Solution:
The experiments performed by Meselson and Stahl on E. coli and by Taylor and colleagues on Vicia faba (faba beans) both proved that DNA replicates semiconservatively.

Topic: Biology, Molecular Basis of Inheritance, NEET
What was the main contribution of Julius von Sachs to the field of photosynthesis?

Options:
A Discovery of chlorophyll as the green substance in plants
B Evidence for production of oxygen in plants
C Evidence for production of glucose in plants
D Splitting light into spectral components

The correct answer is:
Evidence for production of glucose in plants

Solution:
Julius von Sachs provided evidence for production of glucose in plants when they grow. He also showed that the green parts in plants is where glucose is made, and that the glucose is usually stored as starch.

Topic: Biology, Photosynthesis in Higher Plants, NEET
What was the main drawback of Mendeleev's periodic table?

Options:
A Many elements have several isotopes.
B Noble gases do not form compounds.
C Some groups stand divided into two sub groups A and B.
D Atomic weights of elements are not always whole numbers.

The correct answer is:
Many elements have several isotopes.

Solution:
Mendeleev's arranged all the known elements in increasing order of their atomic masses and gave the periodic law.

According to Mendeleev's Periodic law, The physical and chemical properties of the elements are the periodic function of their atomic masses.

Mendeleev designed a periodic table and arranged all the elements in increasing order of their atomic masses but not included the isotpes. As isotopes also have the different atomic masses. It is the major limitation of the Mendeleev's periodic table.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET
What was the main factor responsible for the success of Green Revolution in India?

Options:
A The implementation of better management practices
B Increase in acreage
C Plant breeding techniques
D Increased exports of food

The correct answer is:
Plant breeding techniques

Solution:
Green Revolution in India was dependent to a large extent on plant breeding techniques for the development of high-yielding and disease-resistant varieties of crops such as wheat, rice, and maize. Plant breeding techniques were the main factor responsible for the success of the Green Revolution in India.

Topic: Biology, Strategies for Enhancement in Food Production, NEET
What was the main finding of Julius von Sachs regarding the storage of glucose in plants?

Options:
A Glucose is stored in the roots of the plant
B Glucose is stored in the leaves of the plant
C Glucose is stored in special bodies within plant cells
D Glucose is stored in the air around the plant

The correct answer is:
Glucose is stored in special bodies within plant cells

Solution:
Julius von Sachs found that glucose is usually stored as starch in special bodies (chloroplasts) within plant cells.

Topic: Biology, Photosynthesis in Higher Plants, NEET
What was the main focus of Ernst Mayr's research?

Options:
A Ornithology
B Taxonomy
C The origin of species diversity
D All of the above

The correct answer is:
All of the above

Solution:
Ernst Mayr's research spanned a wide range of fields, including ornithology, taxonomy, zoogeography, evolution, systematics, and the history and philosophy of biology. He made the origin of species diversity the central question of evolutionary biology.

Topic: Biology, The Living World, NEET
What was the main focus of the Earth Summit held in Rio de Janeiro in 1992?

Options:
A Biodiversity conservation
B Sustainable utilization of biodiversity benefits
C Both A) and B)
D Reduction in the current rate of biodiversity loss

The correct answer is:
Both A) and B)

Solution:
The main focus of the Earth Summit held in Rio de Janeiro in 1992 was on both conservation of biodiversity and sustainable utilization of its benefits.

Topic: Biology, Biodiversity and its Conservation, NEET
What was the main goal of Green Revolution?

Options:
A To increase crop yields in India
B To export food from India
C To improve the quality of food crops
D All of the above

The correct answer is:
All of the above

Solution:
Green Revolution was responsible for increasing food production in India to the point that it could meet national requirements and even export fooD) The revolution was dependent on plant breeding techniques to develop high-yielding and disease-resistant varieties of crops such as wheat, rice, and maize.

Topic: Biology, Strategies for Enhancement in Food Production, NEET
What was the main goal of Hershey and Chase's experiments?

Options:
A To determine whether protein or DNA entered the bacteria from viruses
B To prove that RNA is the genetic material
C To isolate the transforming principle in bacteriophages
D To determine the chemical composition of bacteriophages

The correct answer is:
To determine whether protein or DNA entered the bacteria from viruses

Solution:
Hershey and Chase's experiments aimed to discover whether it was the DNA or protein from the viruses that entered the bacteria.

Topic: Biology, Molecular Basis of Inheritance, NEET
What was the main idea that influenced Darwin's theory of natural selection?

Options:
A The work of Thomas Malthus on populations
B The diversity of creatures in the Galapagos Islands
C The observation of convergent evolution in Australia
D The observation of adaptive radiation in Darwin's finches

The correct answer is:
The work of Thomas Malthus on populations

Solution:
Darwin's theory of natural selection was influenced by the observations made by Thomas Malthus on populations, which stated that population size will grow exponentially if everyone reproduced maximally, but population sizes in reality are limited

Topic: Biology, Evolution, NEET
What was the main limitation of Dobereiner's Law of Triads?

Options:
A It was too simple to be useful.
B It didn't apply to any known elements.
C It worked for only a few elements and was considered coincidental.
D It couldn't explain the atomic structure of elements.

The correct answer is:
It worked for only a few elements and was considered coincidental.

Solution:
Dobereiner's Law of Triads had a limitation in that it seemed to work for only a few elements and was considered coincidental.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET
What was the main limitation of Thomson's atomic model?

Options:
A It could not explain the overall neutrality of the atom.
B It did not account for the existence of electrons.
C It could not explain the behavior of gases.
D It did not consider the existence of a nucleus.

The correct answer is:
It could not explain the overall neutrality of the atom.

Solution:
The main limitation of Thomson's atomic model was that it could not explain the overall neutrality of the atom.

Topic: Chemistry, Structure of Atom, NEET
What was the main problem in producing insulin using rDNA techniques?

Options:
A Insulin was not assembled into a mature form.
B Insulin was not extracted from pancreas of slaughtered cattle and pigs.
C Insulin was not synthesised as a pro-hormone.
D Insulin was not introduced in plasmids of E. coli.

The correct answer is:
Insulin was not assembled into a mature form.

Solution:
The main challenge for production of insulin using rDNA techniques was getting insulin assembled into a mature form.

Topic: Biology, Biotechnology and Its Applications, NEET
What was the main problem with switching over to CNG in Delhi?

Options:
A Increase in number of vehicles
B Difficulty of laying down pipelines
C Uninterrupted supply
D Increase in cost

The correct answer is:
Difficulty of laying down pipelines

Solution:
The main problem with switching over to CNG was the difficulty of laying down pipelines to deliver CNG through distribution points/ pumps and ensuring uninterrupted supply.

Topic: Biology, Environmental Issues, NEET
What was the main task of the family planning program initiated in 1951 in India?

Options:
A To create awareness among people about various reproduction related aspects
B To provide facilities and support for building up a reproductively healthy society
C To assess the progress of the family planning program over the past decades
D To initiate action plans and programs to attain total reproductive health as a social goal

The correct answer is:
To create awareness among people about various reproduction related aspects

Solution:
The main task of the family planning program initiated in 1951 in India was to create awareness among people about various reproduction related aspects.

Topic: Biology, Reproductive Health, NEET
What was the objective of the Reproductive Child Health (RCH) programme in India?

Options:
A To bring down the population growth rate
B To increase the population growth rate
C To remain unchanged
D To increase the maternal mortality rate

The correct answer is:
To bring down the population growth rate

Solution:
Through our Reproductive Child Health (RCH) programme, though we could bring down the population growth rate, it was only marginal.

Topic: Biology, Reproductive Health, NEET
What was the outcome of T.W Engelmann's experiments on photosynthesis?

Options:
A He discovered the role of water in photosynthesis
B He described the first action spectrum of photosynthesis
C He discovered the role of oxygen in photosynthesis
D He discovered the role of chlorophyll in photosynthesis

The correct answer is:
He described the first action spectrum of photosynthesis

Solution:
T.W Engelmann observed that the bacteria accumulated mainly in the region of blue and red light of the split spectrum, leading to the description of the first action spectrum of photosynthesis.

Topic: Biology, Photosynthesis in Higher Plants, NEET
What was the primary source of airborne lead emissions in Indian cities, and how has it been addressed?

Options:
A Leaded petrol was the source, and it remains a major issue in Indian cities.
B Unleaded petrol was the source, and it is still a problem in Indian cities.
C Lead emissions are not a concern in Indian cities.
D Leaded petrol was the source, and it has been overcome by using unleaded petrol.

The correct answer is:
Leaded petrol was the source, and it has been overcome by using unleaded petrol.

Solution:
Lead used to be a major air pollutant emitted by vehicles in Indian cities, but this problem has been addressed by switching to unleaded petrol.

Topic: Chemistry, Environmental Chemistry, NEET
What was the purpose of the experiments conducted by T.W Engelmann?

Options:
A To discover the location of chlorophyll in plant cells
B To determine the main component of light that drives photosynthesis
C To determine the role of aerobic bacteria in photosynthesis
D To determine the role of water in photosynthesis

The correct answer is:
To determine the main component of light that drives photosynthesis

Solution:
T.W Engelmann used a prism to split light into its spectral components and then illuminated a green alga to determine the main component of light that drives photosynthesis. He observed that the bacteria accumulated mainly in the region of blue and red light of the split spectrum, leading to the description of the first action spectrum of photosynthesis.

Topic: Biology, Photosynthesis in Higher Plants, NEET
What weight of glucose must be dissolved in 100 g of water to lower the vapour pressure by 0.20 mm Hg?
(Assume dilute solution is being formed)
Given: Vapour pressure of pure water is 54.2 mm Hg at room temperature.Molar mass of glucose is 180 g mol-1

Options:
A 3.59 g
B 3.69 g
C 4.69 g
D 2.59 g

The correct answer is:
3.69 g

Solution:
Sol. Using the formula
Po-PsPo=Moles of soluteMoles of solventfor dilute solution

Po=Vapour pressure of pure water

Ps=Vapour pressure of solution
54.2-5454.2=w/18010018w=0.2×100054.2=20054.2=3.69

Topic: Chemistry, Solutions, JEE Main, JEE Main 2023 (11 Apr Shift 2)
What weight of H2C2O4.2H2O (molecular weight = 126 g/mol) should be dissolved to prepare 250 ml of centinormal solution to be used as reducing agent?

Options:
A 0.635 g
B 0.1575 g
C 0.1263 g
D 0.835 g

The correct answer is:
0.1575 g

Solution:
For mole of given acid C2O42- present = 1 mole
So C2O42-→CO2+2e-
So n-factor for H2C2O4.2H2O as a reducing agent = 2
Equivalent weight (E) =M2=1262=63
250 ml of centinormal solution (NV) =250100×10-3 equivalent
So Weight of H2C2O4.2H2O required (W) =NVE=250100×10-3×63=0.1575 g

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
What weight of $\mathrm{SO}_2$ can be made by burning sulphur in 5.0 moles of oxygen

Options:
A 640 grams
B 160 grams
C 80 grams
D 320 grams

The correct answer is:
320 grams

Solution:
$5 \mathrm{~S}+5 \mathrm{O}_2 \rightarrow 5 \mathrm{SO}_2 ; 5 \mathrm{O}_2 \equiv 5 \mathrm{SO}_2 ; 5 \times 64=320 \mathrm{gm}$.

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
What will be equation of that chord of hyperbola $25 x^2-16 y^2=400$, whose mid point is $(5,3)$

Options:
A $115 x-117 y=17$
B $125 x-48 y=481$
C $127 x+33 y=341$
D $15 x+121 y=105$

The correct answer is:
$125 x-48 y=481$

Solution:
According to question, $S \equiv 25 x^2-16 y^2-400=0$
Equation of required chord is $S_1=T$ ....(i)
Here,
$\begin{gathered}
S_1=25(5)^2-16(3)^2-400 \\
=625-144-400=81
\end{gathered}$
and $T=25 x x_1-16 y y_1-400$, where $x_1=5, y_1=3$
$=25(x)(5)-16(y)(3)-400=125 x-48 y-400$
So from (i), required chord is
$125 x-48 y-400=81 \text { or } 125 x-48 y=481 \text {. }$

Topic: Mathematics, Hyperbola, JEE Main
What will be product of the following given reaction?

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
What will be standard cell potential of galvanic cell with the following reaction?

2Cr s  +  3Cd aq 2+ → 2Cr aq 3+  +  3Cd s

[Given: ECr3+/Cro=-0.74V and ECd2+/Cdo=-0.40V ]

Options:
A 0.74 V
B 1.14 V
C 0.34 V
D -0.34 V

The correct answer is:
0.34 V

Solution:
Ecello=Ecathodeo-Eanodeo
E cell o  =  E Cd 2+ /Cd o − E Cr 3+ /Cr o
=−0.40− 0.74 =+0.34 V

Topic: Chemistry, Electrochemistry, JEE Main
What will be the additional volume of air that a person can expire by a forcible expiration?

Options:
A 500 mL
B 1100 mL
C 2300 mL
D 3000 mL

The correct answer is:
1100 mL

Solution:
Expiratory Reserve Volume : (ERV = 1000 mL to 1100 mL) - It is an additional volume of air that a person can expire by a forcible expiration.
Tidal Volume (TV = 500 mL) - It is the volume of air inspired or expired during normal respiration.
Inspiratory Reserve Volume (IRV = 2500 mL to 3000 mL) It is an additional volume of air that a person can inspire by a forcible inspiration.
Residual Volume (RV = 1100 mL to 1200 mL) - It is the volume of air remaining in the lungs even after a forcible expiration.
Functional Residual Capacity (FRC = ERV + RV) - It refers to the volume of air that remains in the lungs after a normal expiration.
Vital Capacity (VC = ERV + TV + IRV) - It is the maximum volume of air that a person can inspire after forced expiration.
Total Lung Capacity (TLC = ERV+TV+IRV+RV = VC+RV) - It refers to the total volume of air that can be accommodated in the lungs at the end of forced inspiration.

Topic: Biology, Breathing and Exchange of Gases, NEET
What will be the average value of energy along one degree of freedom for an ideal gas in thermal equilibrium at a temperature T ? (kB is Boltzmann constant)

Options:
A 12kBT
B 23kBT
C 32kBT
D kBT

The correct answer is:
12kBT

Solution:
Energy associated with each degree of freedom per molecule =12kBT.

Hence, the average value of energy along one degree of freedom for an ideal gas in thermal equilibrium at a temperature T =12kBT.

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2021 (18 Mar Shift 1)
What will be the compound A which can be used to carry out this synthesis of glycol?
Find the compound A which can be used to carry out the synthesis of glycol

Options:
A RCOR'
B $\mathrm{RCO}_2 \mathrm{OH}$
C $\mathrm{RCHO}$
D $\mathrm{RCOOH}$

The correct answer is:
$\mathrm{RCO}_2 \mathrm{OH}$

Solution:
Explanation: A peroxy acid can be used to carry out this synthesis of glycol as shown in below reaction.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
What will be the correct decreasing order of acid strength of the hydroxybenzoic acids? (Symbols and notations carry their usual meanings)

Options:
A p-Hydroxybenzoic acid > benzoic acid > m-hydroxybenzoic acid > o-hydroxybenzoic acid
B o-Hydroxybenzoic acid > m-hydroxybenzoic acid > benzoic acid > p-hydroxybenzoic acid
C o-Hydroxybenzoic acid > benzoic acid > m-hydroxybenzoic acid > p-hydroxybenzoic acid
D m-Hydroxybenzoic acid > benzoic acid > o-hydroxybenzoic acid > p-hydroxybenzoic acid

The correct answer is:
o-Hydroxybenzoic acid > m-hydroxybenzoic acid > benzoic acid > p-hydroxybenzoic acid

Solution:
In case of hydroxybenzoic acids, they display both kind of effect and there is a decrease in electron density at all positions due to inductive effect of -OH group, but increase in electron density at o- and p-positions due to resonance effect by -OH group. o-hydroxy benzoic acid is strongest among the given acids due to intramolecular hydrogen bonding in its conjugate base

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
What will be the decreasing order of basic strength of the following conjugate bases? \({ }^{-} \mathrm{OH}, \mathrm{R} \overline{\mathrm{O}}, \mathrm{CH}_3 \mathrm{CO} \overline{\mathrm{O}}, \mathrm{C} \overline{\mathrm{I}}\)

Options:
A \(\mathrm{R} \overline{\mathrm{O}}>{ }^{-} \mathrm{OH}>\mathrm{CH}_3 \mathrm{CO} \overline{\mathrm{O}}>\mathrm{C} \overline{\mathrm{I}}\)
B \(\mathrm{C} \overline{\mathrm{l}}>\mathrm{R} \overline{\mathrm{O}}>{ }^{-} \mathrm{OH}>\mathrm{CH}_3 \mathrm{CO} \overline{\mathrm{O}}\)
C \({ }^{-} \mathrm{OH}>\mathrm{R} \overline{\mathrm{O}}>\mathrm{CH}_3 \mathrm{CO} \overline{\mathrm{O}}>\mathrm{C} \overline{1}\)
D \(\mathrm{C} \overline{\mathrm{I}}>{ }^{-} \mathrm{OH}>\mathrm{R} \overline{\mathrm{O}}>\mathrm{CH}_3 \mathrm{CO} \overline{\mathrm{O}}\)

The correct answer is:
\(\mathrm{R} \overline{\mathrm{O}}>{ }^{-} \mathrm{OH}>\mathrm{CH}_3 \mathrm{CO} \overline{\mathrm{O}}>\mathrm{C} \overline{\mathrm{I}}\)

Solution:
Strong acid have weak conjugate base Acidic strength :
$\mathrm{H}-\mathrm{Cl}>\mathrm{CH}_3 \mathrm{COOH}>\mathrm{H}_2 \mathrm{O}>\mathrm{R}-\mathrm{OH}$
Conjugate base strength :
$\mathrm{Cl}^{-} < \mathrm{CH}_3 \mathrm{COO}^{-} < \overline{\mathrm{O}} \mathrm{H} < \mathrm{RO}^{-}$

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2024 (04 Apr Shift 1)
What will be the displacement equation of the simple harmonic motion obtained by combining the motions?
x1=2sinâ¡Ït, x2=4sinâ¡Ït+Ï6 and x3=6sinâ¡Ït+Ï3

Options:
A x=10.25sin⁡ωt+ϕ
B x=10.25sin⁡ωt-ϕ
C x=11.25sin⁡ωt+ϕ
D x=11.25sin⁡ωt-ϕ

The correct answer is:
x=11.25sin⁡ωt+ϕ

Solution:
The resultant equation is
x=Asin⁡ωt+ϕ
∑Ax=2+4cos⁡30°+6cos⁡60°=8.46
And ∑Ay=4sin⁡30°+6cos⁡30°=7.2
∴ A=∑Ax2+∑Ay2
= 8.462+7.22=11.25
And tan⁡ϕ=∑Ay∑Ax=7.28.46=0.85
⇒ ϕ=tan-1⁡0.85=40.4°
Thus, the displacement equation of combined motion is
x=11.25sin⁡ωt+ϕ
Where, ϕ=40.4°

Topic: Physics, Oscillations, JEE Main
What will be the DNA content and number of chromosomes in a cell of after S-phase compared to the gamete of the same organism?

Options:
A Same DNA content but double chromosome number
B Four times DNA content but double chromosome number
C Same DNA content but half chromosome number
D Half DNA content but double chromosome number

The correct answer is:
Four times DNA content but double chromosome number

Solution:
The DNA content will be four times and the chromosomes will be double in number in a cell after S - phase as compared to the gamete of the same organism.
Gametes are haploid(n) cells that contain one set of chromosomes and half amount of DNA content in it.
During S- phase of cell division, only the DNA content of cell gets doubled from 2 n to 4 n while number of chromosome remain same in the dividing cell.

Topic: Biology, Cell Cycle and Cell Division, JIPMER, JIPMER 2019
What will be the effect on the root mean square velocity of oxygen molecules if the temperature is doubled and oxygen molecule dissociates into atomic oxygen?

Options:
A The velocity of atomic oxygen remains same
B The velocity of atomic oxygen doubles
C The velocity of atomic oxygen becomes half
D The velocity of atomic oxygen becomes four times

The correct answer is:
The velocity of atomic oxygen doubles

Solution:
The RMS velocity of a gas is given by, vRMS=3RTM

When O2 dissociates into O its molecular mass becomes M2. Therefore,

Ratio of RMS speed, v1v2=T1T2×M2M1⇒v1v2=T2T×M/2M⇒v2v1=2

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2022 (28 Jun Shift 2)
What will be the emf for the given cell $P \mathrm{t}\left|\mathrm{H}_2\left(\mathrm{P}_1\right)\right| \mathrm{H}^{+}(\mathrm{aq})|| \mathrm{H}_2\left(\mathrm{P}_2\right) \mid \mathrm{Pt}$

Options:
A
$\frac{R T}{f} \log \frac{P_1}{P_2}$

B
$\frac{R T}{2 f} \log \frac{P_1}{P_2}$

C
$\frac{R T}{f} \log \frac{P_2}{P_1}$

D
none of these

The correct answer is:
none of these

Solution:
No Solution Available

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2002
What will be the emf of the given cell Pt | H 2 (P 1 ) | H + (aq) | H 2 (P 2 ) | Pt ?

Options:
A RT F âlnâ P 1 P 2
B RT 2F âlnâ P 1 P 2
C RT F  ln  P 2 P 1
D None of these

The correct answer is:
RT 2F  ln  P 1 P 2

Solution:
Pt| H 2 | H + | H 2 |Pt
Both the electrodes share the same [ H + ].
A: H 2 →2 H + +2 e −
Applying Nernst equation
E cell =0− RT 2F ln P H 2 c P H 2 a
= −RT 2F ln P 2 P 1 = RT 2F ln P 1 P 2

Topic: Chemistry, Electrochemistry, JEE Main
What will be the energy released in joule, in the process of fission by $1 \mathrm{mg}$ of ${ }_{92}^{240} \mathrm{U}$. Assume energy release per fission is $200 \mathrm{MeV}$.
[useAvogadro's number as $6 \times 10^{23}$ and $1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}$ ]

Options:
A $6.2 \times 10^7 \mathrm{~J}$
B $7.0 \times 10^7 \mathrm{~J}$
C $8.0 \times 10^7 \mathrm{~J}$
D $8.2 \times 10^7 \mathrm{~J}$

The correct answer is:
$8.0 \times 10^7 \mathrm{~J}$

Solution:
Let $\mathrm{n}=$ number of moles
Then, $\mathrm{n}=\frac{\mathrm{m}}{\mathrm{M}}=\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{A}}}$
$\begin{aligned}
& \Rightarrow \mathrm{N}=\frac{\mathrm{m}}{\mathrm{M}} \times \mathrm{N}_{\mathrm{A}} \\
& \Rightarrow \mathrm{N}=\frac{1 \times 10^{-3}}{240} \times\left(6 \times 10^{23}\right) \Rightarrow \mathrm{N}=2.5 \times 10^{18}
\end{aligned}$
So, energy released $=\mathrm{N} \times 200 \mathrm{MeV}$
$=2.5 \times 10^{18} \times 200 \times 1.6 \times 10^{-13}=8 \times 10^7 \mathrm{~J}$

Topic: Physics, Nuclear Physics, JEE Main
What will be the entropy change of the system when expansion of 1 mole of a gas takes place from 3 L to 6 L under isothermal conditions? Consider, R=2 cal K-1mol-1 and log⁡2=0.301.

Options:
A 2.84 cal K-1
B 1.386 cal K-1
C 0.37 cal K-1
D 5.26 cal K-1

The correct answer is:
1.386 cal K-1

Solution:
n=1,Vi=3L,Vf=6L
ΔS=2.303nRlog10⁡VfVi
ΔS=2.303×1×2log⁡63=2.303×2×0.3010
=1.386 cal K-1

Topic: Chemistry, Thermodynamics (C), JEE Main