datavorous/entrance-exam-dataset huggingface dataset:
The minimum force required to move a body up an inclined plane is three times the minimum force required to prevent it from sliding down the plane. If the coefficient of friction between the body and the inclined plane is 123, then the angle of the inclined plane is

Options:
A 60°
B 45°
C 30°
D 15°

The correct answer is:
30°

Solution:
Minimum force required to move a body up a rough inclined plane,
F1=mg(sin⁡θ+μcos⁡θ)
Minimum force required to prevent the body from sliding down the rough inclined plane,
F2=mgsin⁡θ-μcos⁡θ
According to the question,
F1=3F2
mgsin⁡θ+μcos⁡θ=3mgsin⁡θ-μmgcos⁡θ
4μcos⁡θ=2sin⁡θ
sin⁡θ=2μcos⁡θ
tan⁡θ=2μ=13
θ=30°

Topic: Physics, Laws of Motion, NEET
The minimum NOR gates are required to make one NAND gate are

Options:
A 1
B 2
C 3
D 4

The correct answer is:
4

Solution:
Three NOR gates form an AND gate, another NOR gate which behaves as NOT gate, And the combination of the two sets will form NAND gate.

Topic: Physics, Semiconductors, NEET
The minimum number of 8μF and 250 V capacitors which are used to make a combination of capacitance 16μF and voltage 1000 V is

Options:
A 4
B 32
C 8
D 3

The correct answer is:
32

Solution:
Let m rows of n series capacitor be taken then minimum number of capacitors required is
Also effective voltage is
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â V′′=1000=nÃ250
âÂ Â Â Â Â Â Â Â Â Â nÂ Â =1000250=4
Also these four capacitor are connected in series then effective capacitance is
Â Â Â Â Â Â Â Â Â Â Â Â 1C′′=18+18+18+18=48
âÂ Â Â Â Â Â Â C′′=2Î¼F
â´Â Â Â Â Â Â Â Â C′′′′=Â 16=2Ãm
âÂ Â Â Â Â Â Â Â m=162=8
Hence N=mÃn=8Ã4=32

Topic: Physics, Electrostatics, NEET
The minimum number of chiasmata in a bivalent is/are

Options:
A One
B Two
C Three
D Four

The correct answer is:
One

Solution:
Chiasmata is the point of contact or the X shaped structure formed between two non-sister chromatids of the homologous chromosome during the process of crossing over of genetic material. A bivalent is a pair of the homologous chromosome that has exchanged DNA in at least one site called the chiasmata. The minimum chiasmata found in a bivalent is one and can go up to three in rare cases.

Topic: Biology, Cell Cycle and Cell Division, NEET
The minimum number of meiotic divisions to obtain 1000 pollen of wheat are

Options:
A 250
B 500
C 1000
D 1250

The correct answer is:
250

Solution:
The meiotic division is also called reductional division. The diploid cells after meiosis become haploid. Microsporogenesis starts in the sporogenous tissue of anther. In sporogenous tissue, pollen mother cells are present that undergo meiosis and form microspore tetrad. Out of which one will get mature and form pollen grains that will be haploid in nature. 1 PMC will produce 4 pollen grains. Thus for 1000 pollens, meiotic division required will be 10004=250

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
The minimum pressure required to compress 600 dm3 of a gas at 1 bar to 150 dm3 at 40°C is

Options:
A 4.0 bar
B 0.2 bar
C 1.0 bar
D 2.5 bar

The correct answer is:
4.0 bar

Solution:
By Boyle's law
P1V1=P2V2
1 bar×600dm3=P2×150dm3
P2=4

Topic: Chemistry, States of Matter, NEET, NEET 2020 (Phase 2)
The minimum time taken by a spring block system (having time period T) to travel a distance equal to amplitude of motion is equal to

Options:
A T4
B T6
C T8
D T10

The correct answer is:
T6

Solution:
Velocity-time graph of a body executing S.H.M. is
So to cover a given distance in minimum time velocity should be greater. So it must be around mean position and minimum time taken to cover a distance a2 is T12
∴TRequired=T6

Topic: Physics, Oscillations, NEET
The minimum value of effective capacitance that can be obtained by combining 3 capacitors of capacitances 1 pF, 2 pF and 4 pF is

Options:
A 47 pF
B 1 pF
C 2Â pF
D 74 pF

The correct answer is:
47 pF

Solution:
The minimum value of effective capacitance is get in series combination.
Now, the equivalent capacitance,
1Ceq=11+12+14
1Ceq=4+2+14
1Ceq=74
Ceq=47 pF

Topic: Physics, Electrostatics, NEET
The minimum wavelength of X-rays produced by an electron accelerated through a potential difference of V volts is proportional to:

Options:
A V2
B V
C 1V
D 1V

The correct answer is:
1V

Solution:
Minimum wavelength means maximum energy as E=hcλ.

Therefore,

eV=hcλmin⇒λmin=hceV⇒λmin∝1V

Topic: Physics, Dual Nature of Matter, NEET, NEET 2023 (All India)
The Ministry of Environment and Forest has initiated which plan to save major rivers of our country?

Options:
A Ganga action plan
B Yamuna action plan
C Fermentation plan
D Both (a) and (b)

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

Solution:
Due to increasing urbanization sewage is being produced in large quantity and the untreated sewage is discharged directly into the rivers leading to their pollution and an increase in water-borne diseases. Ministry of Environment and forest has initiated Ganga action and Yamuna action plans to save these rivers from pollution.

Topic: Biology, Microbes in Human Welfare, NEET
The mirror in the below case is

Options:
A Convex
B Concave
C Plane
D None of these

The correct answer is:
Plane

Solution:
Parallel ray before incidance is parallel after reflection, Hence mirror is plane

Topic: Physics, Ray Optics, NEET
The mismatched pair among the following is:

Options:
A Monadelphous - Hibiscus
B Diadelphous - Cucurbita
C Polyadelphous - Citrus
D Variation in the length of stamen filaments - Salvia

The correct answer is:
Diadelphous - Cucurbita

Solution:
Adelphy is a condition in which the stamens are joined together only by their filaments while anthers are free. If there is only one bundle of stamens it is called monadelphy. This is commonly seen in Hibiscus. If there are two bundles of stamens it is called diadelphy. This is commonly seen in gram and pea. If there are more than two bundles of stamens it is called polyadelphy. This is commonly seen in Citrus and castor. In Cucurbita, anthers as well as filaments of stamens are united through their whole length. This is called synandry. Salvia has lever mechanism with variable length of filaments which helps in pollination by insects.

Topic: Biology, Morphology of Flowering Plants, NEET
The missing structures A and B in the reaction sequence:
R-CH2-CH2OH→350oCAl2O3→iO3iiZn,H3O⊕RCHO+A;RCHO→ReduceB;
are –

Options:
A CH3OH,RCOOH
B Methanal,RCH2OH
C Ethanal,RCOOH
D Methanal,RCHOHR

The correct answer is:
Methanal,RCH2OH

Solution:
R–CH2–CH2OH→Al2O3 350oCR–CH=CH2→O3/ZnRCHO+HCHO;
RCHO→ReduceRCH2OH

Topic: Chemistry, Aldehydes and Ketones, NEET
The mitochondria

Options:
A divide by budding
B divide by fission
C divide by meiosis
D do not divide

The correct answer is:
divide by fission

Solution:
Mitochondria are power houses of the cells. These divide by binary fission. Fission takes place with DNA replication. Proteins are needed for this process. These are semiautonomous organelles as they have their own DNA and thus can coordinate its division too.

Topic: Biology, Cell - The Unit of Life, NEET
The mitral valve is present between the

Options:
A right atria and right ventricle
B left atria and left ventricle
C right atria and left ventricle
D left atria and right ventricle

The correct answer is:
left atria and left ventricle

Solution:
Atrio-ventricular valves :
1. Tricuspid valve or TV (made up of 3 muscular flaps or cusps) - between the right atrium and the right ventricle.
2. Bicuspid valve or Mitral valve or MV (made up of 2 muscular flaps or cusps) - between the left atrium and the left ventricle.

Topic: Biology, Body Fluids and Circulation, NEET
The mixture of chloroxylenol and terpineol acts as

Options:
A analgestic.
B antiseptic.
C antipyretic.
D antibiotic.

The correct answer is:
antiseptic.

Solution:
Mixture of chloroxylenol and terpineol acts as antiseptic.

Topic: Chemistry, Chemistry in Everyday Life, NEET, NEET 2017
The mixture of compounds A and B is passed through a column of Al2O3by using alcohol as eluent. Compound A is eluted in preference to compound B. Which of the compounds A or B, is more readily adsorbed on the column?

Options:
A Since compound ‘A’ comes out before compound ‘B’, the compound ‘B’ is more readily adsorbed on column.
B Since compound ‘A’ comes out before compound ‘B’, the compound ‘A’ is more readily adsorbed on column.
C Compound ‘A’ and Compound ‘B’ are adsorbed on the column at the same rate
D Neither of Compound ‘A’ and Compound ‘B’ are adsorbed on the column

The correct answer is:
Since compound ‘A’ comes out before compound ‘B’, the compound ‘B’ is more readily adsorbed on column.

Solution:
Correct Option is : (A)
Since compound ‘A’ comes out before compound ‘B’, the compound ‘B’ is more readily adsorbed on column.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, NEET
The mixture of two liquids boils without changing its composition is called

Options:
A Zeotropic liquid mixture
B Binary liquid mixture
C Azeotropic liquid mixture
D Stable structure complex

The correct answer is:
Azeotropic liquid mixture

Solution:
Azeotropic mixture is a constant boiling mixture of two liquids, which boils at fixed temperature like a pure liquid and distils over in the same composition.

Topic: Chemistry, Solutions, NEET
The mixture that forms maximum boiling azeotrope is

Options:
A Water and Nitric Acid
B Ethanol and Water
C Acetone and Carbon disulphide
D Heptane and Octane

The correct answer is:
Water and Nitric Acid

Solution:
The solutions that show large negative deviation from Raoult’s law form maximum boiling azeotrope at a specific composition. Nitric acid and water is an example of this class of azeotrope. This azeotrope has the approximate composition, 68% nitric acid and 32% water by mass, with a boiling point of 393.5K.

Topic: Chemistry, Solutions, NEET, NEET 2019 (All India)
The mixture which shows positive deviation from Raoult’s law is

Options:
A Benzene + Toluene
B Acetone + Chloroform
C Chloroethane + Bromoethane
D Ethanol + Acetone

The correct answer is:
Ethanol + Acetone

Solution:
In solutions showing positive deviation partial pressure of each component A and B is higher than that calculated from Raoult's
law, and hence total pressure over the solution is also higher than if the solution were ideal.

The intermolecular interaction is weaker as interaction found in any one of the pure components.

Hydrogen bond of ethanol gets weakened by addition of acetone.

Topic: Chemistry, Solutions, NEET, NEET 2020 (Phase 1)
The Mn3+ion is unstable in solution and undergoes disproportionation to give Mn2+, MnO2, and H+ion. Write a balanced ionic equation for the reaction. What are the products formed when Mn3+undergoes disproportionation?

Options:
A Mn2+, MnO2, and H+
B Mn2+and MnO2
C Mn2+and H+
D MnO2and H+

The correct answer is:
Mn2+, MnO2, and H+

Solution:
Correct Option is : (A)
Mn2+, MnO2, and H+

Topic: Chemistry, Redox Reactions, NEET
The mode of action of an enzyme is:

Options:
A Decreasing potential energy
B Increasing activation energy
C Increasing kinetic energy
D Decreasing activation energy

The correct answer is:
Decreasing activation energy

Solution:
The activation energy is the energy required to start a reaction. Enzymes are proteins that bind to a molecule, or substrate, to modify it and lower the energy required to make it react. The rate of reaction increases if the activation energy decreases.

Topic: Chemistry, Biomolecules, NEET
The mode of arrangement of sepals or petals in floral buds with respect to the other members of the same whorl is known as ___.

Options:
A Polysepalous
B Aestivation
C Androecium
D Gynoecium

The correct answer is:
Aestivation

Solution:
The mode of arrangement of sepals or petals in floral buds with respect to the other members of the same whorl is known as Aestivation.

Topic: Biology, Morphology of Flowering Plants, NEET
The mode of excretion in which animal excrete urea as major excretory waste is called

Options:
A Ureotelism
B Uricotelism
C Ammonotelism
D Aminotelism

The correct answer is:
Ureotelism

Solution:
Urea is generally excreted outside the body through nephridia or kidney. Animals, which produce urea are known as ureotelic and mode of excretion as ureotelism, e.g., mammals, adults amphibian, etc.

Topic: Biology, Excretory Products and their Elimination, NEET
The modern day washing powders have some biological content. Which of the following will be mostly added to the powder as biological content?

Options:
A DNA ligase
B lactase
C pectinase
D lipase

The correct answer is:
lipase

Solution:
Biological washing powder are the mixed with lipase enzymes. These have the ability to break the fats and help in removing stains easily. Lipases are the group of enzymes which break down lipids to its constituents. Pectinase is useful in the juice industry to breakdown cell wall of fruit cells.

Topic: Chemistry, Biomolecules, NEET
The modified stem of Opuntia is

Options:
A Phyllode
B Phylloclade
C Cladode
D Staminode

The correct answer is:
Phylloclade

Solution:
In Opuntia, the stem becomes green and photosynthetic and performs the function of a leaf. Leaves are reduced to spines to reduce transpiration. Phyllode is petiole modification in which petiole becomes flat leaf-like and carries out photosynthesis.

Topic: Biology, Morphology of Flowering Plants, NEET
The molal lowering of vapour pressure for water at 100oC , is

Options:
A 760 mm
B 750 mm
C 13.43 mm
D 0.760 mm

The correct answer is:
13.43 mm

Solution:
P0-PSP0=n2n1+n2
P0-PS=11+100018×760=13.43mm

Topic: Chemistry, Solutions, NEET
The molality of 15% by wt solution of H2SO4 is

Options:
A 18
B 2.6
C 1.2
D 1.8

The correct answer is:
1.8

Solution:
15% w/w solution weight of solute = 15 g
Weight of solvent = 85 g
m=wA×1000MA×wB
=15×100098×85=1.8
A→Solute,B→Solvent

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
The molality of solution containing 15.20 g of urea, (molar mass = 60) dissolved in 150 g of water is

Options:
A 1.689 mol kg-1
B Â 0.1689Â molÂ kg-1
C 0.5922Â molÂ kg-1
D Â 0.2533Â molÂ kg-1

The correct answer is:
1.689 mol kg-1

Solution:
m=w2m2Ã1000w1
=15.2060Ã100050
=1.689 mol kg-1

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
The molar conductance of NaCl, HCl and CH3COONa at infinite dilution are 126.45, 426.16 and 91.0 Scm2 mol-1 respectively. The molar conductance of CH3COOH at infinite dilution is. Choose the right option for your answer.

Options:
A 201.28Scm2 mol-1
B 390.71 S cm2 mol-1
C 698.28Scm2 mol-1
D 540.48Scm2 mol-1

The correct answer is:
390.71 S cm2 mol-1

Solution:
Λm∞CH3COOH=Λm∞CH3COONa+Λm∞(HCl)-Λm∞(NaCl)

=91+426.16-126.45

=390.71 S cm2 mol-1

Topic: Chemistry, Electrochemistry, NEET, NEET 2021
The molar conductivity of 0.007 M acetic acid is 20 S cm2 mol-1. What is the dissociation constant of acetic acid? Choose the correct option.

λH+∘=350 S cm2 mol-1λCH3COO-∘=50 S cm2 mol-1

Options:
A 1.75×10-4 mol L-1
B 2.50×10-4 mol L-1
C 1.75×10-5 mol L-1
D 2.50×10-5 mol L-1

The correct answer is:
1.75×10-5 mol L-1

Solution:
λCH3COOH∘=λCH3COO-∘+λH+∘

λCH3COO-∘=350+50=400 S cm2 mol-1

λCH3COOH=20Scm2 mol-1 given

α=λCH3COOHλCH3COOH∘=20400=5×10-2

Ka=cα2=0.007×5×10-22=1.75×10-5 molL-1

Topic: Chemistry, Electrochemistry, NEET, NEET 2021
The molar conductivity of 0.025 mol L-1methanoic acid is 46.1 S cm2mol-1. Calculate its degree of dissociation and dissociation constant. Given λo(H+) = 349.6 S cm2mol-1and λo(HCOO-) = 54.6 S cm2mol-1

Options:
A Degree of dissociation = 0.013, Dissociation constant = 1.8 x 10-3
B Degree of dissociation = 0.114, Dissociation constant = 3.7 x 10-4
C Degree of dissociation = 0.024, Dissociation constant = 3.2 x 10-3
D Degree of dissociation = 0.024, Dissociation constant = 1.1 x 10-3

The correct answer is:
Degree of dissociation = 0.114, Dissociation constant = 3.7 x 10-4

Solution:
Correct Option is : (B)
Degree of dissociation = 0.114, Dissociation constant = 3.7 x 10-4

Topic: Chemistry, Electrochemistry, NEET
The molar conductivity of 0.00241 M acetic acid is 7.896 × 10-5S cm-1. What is its molar conductivity? If 0 Λ m for acetic acid is 390.5 S cm.2mol-1, what is its dissociation constant?

Options:
A Molar conductivity = 33 S cm2mol-1, Dissociation constant = 1.86 × 10-5
B Molar conductivity = 7.896 × 10-5S cm2mol-1, Dissociation constant = 390.5
C Molar conductivity = 390.5 S cm2mol-1, Dissociation constant = 7.896 × 10-5
D Molar conductivity = 1.86 × 10-5S cm2mol-1, Dissociation constant = 33

The correct answer is:
Molar conductivity = 7.896 × 10-5S cm2mol-1, Dissociation constant = 390.5

Solution:
Correct Option is : (B)
Molar conductivity = 7.896 × 10-5S cm2mol-1, Dissociation constant = 390.5

Topic: Chemistry, Electrochemistry, NEET
The molar conductivity of a 0.5 mol dm-3 solution of AgNO3 with electrolytic conductivity of 5.76×10-3 S cm-1 at 298 K

Options:
A 2.88 S cm2 mol-1
B 11.52 S cm2 mol-1
C 0.086 S cm2 mol-1
D 28.8 S cm2 mol-1

The correct answer is:
11.52 S cm2 mol-1

Solution:
∧m=k×1000M=5.76×10-3×10000.5
=11.52 S cm2 mol-1.

Topic: Chemistry, Electrochemistry, NEET, NEET 2016 (Phase 2)
The molar enthalpy of vapourisation of acetone is less than that of water. Why?

Options:
A Because of weak hydrogen bonding in water, its enthalpy of vapourisation is less.
B Because of strong hydrogen bonding in water, its enthalpy of vapourisation is more.
C Because of weak hydrogen bonding in water, its enthalpy of vapourisation is more.
D Because of strong hydrogen bonding in water, its enthalpy of vapourisation is less.

The correct answer is:
Because of strong hydrogen bonding in water, its enthalpy of vapourisation is more.

Solution:
Correct Option is : (B)
Because of strong hydrogen bonding in water, its enthalpy of vapourisation is more.

Topic: Chemistry, Thermodynamics (C), NEET
The molar heat capacity of oxygen gas at STP is nearly 2.5R. As the temperature is increased, it gradually increases and approaches 3.5R. The most appropriate reason for this behaviour is that at high temperature

Options:
A oxygen does not behave as an ideal gas
B oxygen molecules dissociate into atoms
C the molecules collide more frequently
D molecular vibrations gradually become effective

The correct answer is:
molecular vibrations gradually become effective

Solution:
The molar heat capacity depends upon the degree of freedom. At low temperature, only translatory degree of freedom is present. At high temperature all types of degrees of freedom (translatory, rotatory and vibratory) are present.

Topic: Physics, Thermodynamics, NEET
The molar heat capacity of water at constant, pressure, $\mathrm{C}$, is $75 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$. When 1.0 $\mathrm{kJ}$ of heat is supplied to $100 \mathrm{~g}$ of water which is free to expand, the increase in temperature of water is:

Options:
A $1.2 \mathrm{~K}$
B $2.4 \mathrm{~K}$
C $4.8 \mathrm{~K}$
D $6.6 \mathrm{~K}$

The correct answer is:
$2.4 \mathrm{~K}$

Solution:
Molar heat capacity $=75 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$
$\begin{aligned}
& 18 \mathrm{~g} \text { of water }=1 \text { mole }=75 \mathrm{~J} \mathrm{~K}^{-1} \\
& \mathrm{~mol}^{-1} \\
& 1 \mathrm{~g} \text { of water }=\frac{75}{18} \mathrm{JK}^{-1} \\
& \mathrm{Q}=\text { m.c. } \Delta t
\end{aligned}$
$\begin{aligned}
1000 & =100 \times \frac{75}{18} \times \Delta t \\
\Rightarrow \Delta t & =\frac{10 \times 18}{75}=2.4 \mathrm{~K}
\end{aligned}$

Topic: Chemistry, Thermodynamics (C), NEET, NEET 2003
The molar heat capacity of water at constant pressure P, is 75 JK-1mol-1 . When 1.0 kJ of heat is supplied to 100 g of water which is free to expand, the increase in temperature of water is

Options:
A 1.2 K
B 2.4 K
C 4.8 K
D 6.6 K

The correct answer is:
2.4 K

Solution:
q=n.CpΔT
1000=10018×75×ΔT
So ΔT=2.4 K.

Topic: Chemistry, Thermodynamics (C), NEET
The molar heat capacity of water in equilibrium with ice at constant pressure is

Options:
A negative
B zero
C infinity
D 40.45 kJ K-1mol-1

The correct answer is:
infinity

Solution:
When water is at equilibrium with ice, the temperature of system is of melting point of water at that pressure.
Here, ΔT=0 , while Heat exchange, ΔQ≠0
So molar heat capacity is, ΔQ/ΔT→∞ .
It can be also represented as follows
Cp,mLtT→0∂H∂Tp=∞

Topic: Chemistry, Thermodynamics (C), NEET
The molar ratio of Fe2+ to Fe3+ in a mixture of FeSO4 and Fe2(SO4)3 having equal number of sulphate ions in both ferrous and ferric sulphates is

Options:
A 1:2
B 3:2
C 2:3
D none of these

The correct answer is:
3:2

Solution:
3FeSO4→3Fe2++3SO42-
Fe2(SO4)3→2Fe3++3SO42-
Suppose x mole SO42- ions are furnished by both FeSO4 and Fe2(SO4)3 .
Number of moles of Fe2+=x
Number of moles of Fe3+=23x
Fe2+:Fe3+=x:23x
=3:2 .

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
The molar ratio of Fe++ to Fe+++ in a mixture of FeSO4 and Fe2(SO4)3 having equal number of sulphate ion in both ferrous and ferric sulphate is

Options:
A 1:2
B 3:2
C 2:3
D Can’t be determined

The correct answer is:
3:2

Solution:
Let’s take number of sulphate ion produced from both salt seperatly be X
Then number of Fe2+ions=X
And number Fe3+ions=2X/3
i.e. Fe2+:Fe3+=3:2

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
The molar solubility [in mol L-1] of a sparingly soluble salt MX4 is âsâ. The corresponding solubility product is Ksp.s is given in terms of Ksp by the relation

Options:
A s=Ksp2561/5
B s=128Ksp1/4
C s=Ksp1281/6
D s=256Ksp1/5

The correct answer is:
s=Ksp2561/5

Solution:
For the solute AxBy â xA+yB
Ksp=xxyysx+y
MX 4 ⇌ M 4+ +4 X −
x = 1, y = 4
Ksp=4411s5=256s5
s=Ksp2561/5

Topic: Chemistry, Ionic Equilibrium, NEET
The molar solubility of $\mathrm{CaF}_2$ $\left(\mathrm{K}_{\mathrm{sp}}=5.3 \times 10^{-11}\right)$ in $0.1 \mathrm{M}$ solution of $\mathrm{NaF}$ will be

Options:
A $5.3 \times 10^{11} \mathrm{~mol} \mathrm{~L}^{-1}$
B $5.3 \times 10^{-8} \mathrm{~mol} \mathrm{~L}^{-1}$
C $5.3 \times 10^{-9} \mathrm{~mol} \mathrm{~L}^{-1}$
D $5.3 \times 10^{-10} \mathrm{~mol} \mathrm{~L}^{-1}$

The correct answer is:
$5.3 \times 10^{-9} \mathrm{~mol} \mathrm{~L}^{-1}$

Solution:
Let the solubility of $\mathrm{CaF}_2$ in $0.1 \mathrm{M} \mathrm{NaF}$ is ' $\mathrm{S}$ ' $\mathrm{mol} \mathrm{L}^{-1}$

Topic: Chemistry, Ionic Equilibrium, NEET, NEET 2019 (Odisha)
The molar specific heat at constant pressure of an ideal gas is (7/2) $R$. The ratio of specific heat at constant pressure to that at constant volume is:

Options:
A $9 / 7$
B $7 / 5$
C $8 / 7$
D $5 / 7$

The correct answer is:
$7 / 5$

Solution:
Molar specific heat at constant pressure
$$
C_P=\frac{7}{2} R
$$
We know that $C_P-C_V=R$
$$
\Rightarrow \quad \begin{aligned}
C_V & =C_P-R \\
& =\frac{7}{2} R-R=\frac{5}{2} R
\end{aligned}
$$
from (i) and (ii),
$$
\therefore \quad \frac{C_P}{C_V}=\frac{7}{5} .
$$

Topic: Physics, Kinetic Theory of Gases, NEET, NEET 2006
The molar specific heat of oxygen at constant pressure is Cp=7.03 cal mol-1 K-1 and R=8.31 J mol-1ââK-1. The amount of heat taken by 5 moles of oxygen, when heated at constant volume from 10 Â°C to 20 Â°C, will be approximately

Options:
A 25Â cal
B 50Â cal
C 253 cal
D 500Â cal

The correct answer is:
253 cal

Solution:
Given, Cp=7.03 cal mol-1 K-1 ,R=8.315 J mol-1 K-1
n=5 moles ,T2=20 oC, T1=10 oC,
âQ= ?
â´ Heat, âQ=nCVâT
=nCp-RâT âµCp-Cv=R
=57.03-8.314.2Ã20-10
â253 cal

Topic: Physics, Thermal Properties of Matter, NEET
The molar specific heats of an ideal gas at constant pressure and volume are denoted by $C_p$ and $C_V$ respectively. If $\gamma=\frac{C_p}{C_V}$ and $R$ is the universal gas constant, then $C_V$ is equal to

Options:
A $\frac{1+\gamma}{1-\gamma}$
B $\frac{R}{(\gamma-1)}$
C $\frac{(\gamma-1)}{R}$
D $\gamma R$

The correct answer is:
$\frac{R}{(\gamma-1)}$

Solution:
The value of $C_V=\frac{R}{\gamma-1}$

Topic: Physics, Kinetic Theory of Gases, NEET, NEET 2013 (All India)
The molarity of HNO3 in a sample which has density 1.4 g/mL and mass percentage of 63% is
(Molecular Weight of HNO3=63 )

Options:
A 7 M
B 14 M
C 2.1 M
D 28 M

The correct answer is:
14 M

Solution:
d=1.4g/ml
63% by mass means 63 g of solute taken in 100 g of solution.
w=63 g
W=100 g
V=Wd=1001.4 ml
M=wmol.wt×V×1000
Molarity=63×1.463×100×1000 mole/litre
=14 mole/L

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET