datavorous/entrance-exam-dataset huggingface dataset:
Emf of a cell is

Options:
A the maximum potential difference between the terminals of a cell when no current is drawn from the cell
B the force required to push the electrons in the circuit.
C the potential difference between the positive and negative terminal of a cell in a closed circuit.
D less than terminal potential difference of the cell.

The correct answer is:
the maximum potential difference between the terminals of a cell when no current is drawn from the cell

Solution:
Electromotive force or emf is defined as the maximum potential difference between the terminals of a cell when no current is drawn from the cell. When current will be drawn from cell, due to internal resistance, voltage across the terminal will change.

Topic: Physics, Current Electricity, JEE Main
EMF of cell $\mathrm{Ni}\left|\mathrm{Ni}^{2+}(1.0 \mathrm{M}) \| A u^{3+}(1.0 \mathrm{M})\right| \mathrm{Au}$ (Where $E^{\circ}$ for $\mathrm{Ni}^{2+} \mid \mathrm{Ni}$ is $-0.25 \mathrm{~V} ; E^{\circ}$ for $A u^{+3} \mid A u$ is $\left.1.50 \mathrm{~V}\right)$ is

Options:
A $+1.25 \mathrm{~V}$
B $-1.75 V$
C $+1.75 V$
D $+4.0 \mathrm{~V}$

The correct answer is:
$+1.75 V$

Solution:
$E_{\text {cell }}=E_{A \mu^{3+} / A L}^o-E_{N e^{2+} / N 6}^o=1.50-(-0.25)=1.75$

Topic: Chemistry, Electrochemistry, JEE Main
Emf of the following cell at 298 K in V is x×10-2

ZnZn2+0.1M‖Ag+0.01MAg

The value of x is ________________ (Rounded off to the nearest integer)

Given: EZn2+/Znθ=-0.76 V; EAg+/Agθ=+0.80 V; 2.303RTF=0.059

Options:

The correct answer is:
147

Solution:
Ecell0=EAg+/Ag0-EZn2+/Zn0

=0.80--0.76

=1.56 V

Ecell=1.56-0.0592logZn2+Ag+2

=1.56-0.0592log0.10.012

=1.56-0.0592×3

=1.56-0.0885

=1.4715

=147.15×10-2

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2021 (26 Feb Shift 2)
EMF of the following cell is 0.6 volt.

AgsAgBrsKBr0.01M||AgNO30.001 M|Ag(s)

Ksp of AgBr is expressed as 1Ã10-x, x is

Take2.303RTF=0.06âV

Options:

The correct answer is:
15.00

Solution:
AgâAg+L+ e-
Ag+0.001 M+e-âAg
Ag+(0.001 M)âAg+(L)
0.6=0-0.061logKsp10-2Ã10-3
10-10=Ksp10-5âKsp=10-15

Topic: Chemistry, Electrochemistry, JEE Main
Empirical formula of a compound is $\mathrm{CH}_{2} \mathrm{O}$ and its molecular mass is 90 , the molecular formula of the compound is

Options:
A $\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{3}$
B $\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}$
C $\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}$
D $\mathrm{CH}_{2} \mathrm{O}$

The correct answer is:
$\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{3}$

Solution:
The empirical formula of compound $=\mathrm{CH}_{2} \mathrm{O}$
Empirical formula mass
$=(1 \times 12)+(2 \times 1)+(1 \times 16)=30$
Molecular mass $=90$
$\therefore n=\frac{\text { Molecular mass }}{\text { Empirical mass }}=\frac{90}{30}=3$
and molecular formula $=3 \times \mathrm{CH}_{2} \mathrm{O}$
$=\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{3}$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
Emulsifier is mixed to

Options:
A Increase the stability of emulsion
B Decrease the stability of emulsion
C Change oil into water like emulsion
D None of these

The correct answer is:
Increase the stability of emulsion

Solution:
Correct option is 1. Increase the stability of emulsion to increase the stability of the emulsion

Topic: Chemistry, Surface Chemistry, JEE Main
End product C in above reaction is

Options:
A
B
C
D

The correct answer is:

Solution:
Here the reaction occurs as follows to give cyclohexenone

Topic: Chemistry, Aldehydes and Ketones, JEE Main
End product of the following reaction is

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
End product of the following reaction is
$$
\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH} \frac{\mathrm{Cl}_{2}}{\text { red } \mathrm{P}} X \stackrel{\mathrm{Alc} \mathrm{KOH}}{\longrightarrow}
$$

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
Endosperm is consumed by developing embryo in the seed of

Options:
A Maize
B Coconut
C Castor
D Pea

The correct answer is:
Pea

Solution:
Other three are endospermic/albuminous seeds.

Topic: Biology, Morphology of Flowering Plants, NEET, NEET 2008 (Mains)
Endosperms in angiosperms is:

Options:
A haustorial
B assimilatory
C protective
D nutritive

The correct answer is:
nutritive

Solution:
Double fertilization is a unique characteristic observed in the Angiosperm group of plants. In this, two fertilizations are observed - 1st between the female and male gametic cells forming zygote and 2nd is between the second male gamete and polar nuclei forming the primary endosperm nucleus. PEN later develops in to the endosperm tissue and its main function is to provide nourishment to the developing embryo.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
Energy is not carried by which of the following waves

Options:
A Stationary
B Progressive
C Transverse
D Electromagnetic

The correct answer is:
Stationary

Solution:
Energy is not carried by stationary waves

Topic: Physics, Waves and Sound, JEE Main
Energy levels A, B and C of a certain atom correspond to increasing values of energy, i.e. EA < EB < EC. If Î» 1, Î» 2 and Î» 3 are the wavelengths of radiations corresponding to transitions C to B, B to A and C to A respectively, which of the following relations is correct?

Options:
A λ 3 = λ 1 + λ 2
B Î» 3 = Î» 1 Î» 2 Î» 1 + Î» 2
C λ3=λ1λ2
D λ3=λ12+λ22

The correct answer is:
λ 3 = λ 1 λ 2 λ 1 + λ 2

Solution:
hcλ1=EC-EB
hcλ2=EB-EA
hcλ3=EC-EA
⇒ 1 λ 3 = 1 λ 1 + 1 λ 2

Topic: Physics, Atomic Physics, JEE Main
Energy needed in breaking a drop of radius $R$ into $n$ drops of radii $r$ is given by

Options:
A $4 \pi T\left(n r^2-R^2\right)$
B $\frac{4}{3} \pi\left(r^3 n-R^2\right)$
C $4 \pi T\left(R^2-n r^2\right)$
D $4 \pi T\left(n r^2+R^2\right)$

The correct answer is:
$4 \pi T\left(n r^2-R^2\right)$

Solution:
Energy needed = Increment in surface energy $=$ (surface energy of $n$ small drops) - (surface energy of one big drop)
$=n 4 \pi r^2 T-4 \pi R^2 T=4 \pi T\left(n r^2-R^2\right)$

Topic: Physics, Mechanical Properties of Fluids, JEE Main
Energy needed in breaking a drop of radius R into n identical drops of radii r is given by

Options:
A 4πT(n r 2 − R 2 )
B 4 3 π( r 3 n− R 2 )
C 4πT( R 2 −n r 2 )
D 4πT(n r 2 + R 2 )

The correct answer is:
4πT(n r 2 − R 2 )

Solution:
Energy needed = Increment in surface energy = (surface energy of n small drops) – (surface energy of one big drop) =n4π r 2 T−4π R 2 T=4πT(n r 2 − R 2 )

Topic: Physics, Mechanical Properties of Fluids, JEE Main
Energy of 10 non rigid diatomic molecules at temperature $T$ is :

Options:
A $70 \mathrm{~K}_{\mathrm{B}} \mathrm{T}$
B $35 \mathrm{~K}_{\mathrm{B}} \mathrm{T}$
C $\frac{7}{2} \mathrm{RT}$
D $35 \mathrm{RT}$

The correct answer is:
$35 \mathrm{~K}_{\mathrm{B}} \mathrm{T}$

Solution:
$\begin{aligned}
& \text { Degree of freedom }(f)=5+2(3 \mathrm{~N}-5) \\
& \qquad f=5+2(3 \times 2-1)=7 \\
& \text { energy of one molecule }=\frac{f}{2} K_B T
\end{aligned}$
energy of 10 molecules
$=10\left(\frac{\mathrm{f}}{2} \mathrm{~K}_{\mathrm{B}} \mathrm{T}\right)=10\left(\frac{7}{2} \mathrm{~K}_{\mathrm{B}} \mathrm{T}\right)=35 \mathrm{~K}_{\mathrm{B}} \mathrm{T}$

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2024 (06 Apr Shift 2)
Energy of an electron is given by E=-2.178×10-18Z2n2 J. Wavelength of light required to excite an electron in a hydrogen atom from level n=1 to n=2 will be :
h=6.62×10-34 Js and c=3.0×108 ms-1

Options:
A 6.500 Ã 1 0 - 7 m
B 8.500 × 1 0 - 7 m
C 1.214 Ã 1 0 - 7 m
D 2.816 Ã 1 0 - 7 m

The correct answer is:
1.214 × 1 0 - 7 m

Solution:
E1=−2.178×10−18×11=−2.178×10−18J
=2.178×10−181.6×10−19=13.6 eV
E2=−2.178×10−8(2)2=−2.178×10−184=0.5448×10−18J
=−3.4 eV
ΔE=E2-E1=-3.4--13.6=10.2 eV
λ in Å=1240010.2≈1214 Å=1214×10−10m=1.214×10−7m { ∴ 1 Å=10 −10 m }

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2013 (07 Apr)
Energy required for the electron excitation in $\mathrm{Li}^{++}$from the first to the third Bohr orbit is :

Options:
A
$36.3 \mathrm{eV}$

B
$108.8 \mathrm{eV}$

C
$122.4 \mathrm{eV}$

D
$12.1 \mathrm{eV}$

The correct answer is:
$108.8 \mathrm{eV}$

Solution:
$E_n=-13.6 \frac{Z^2}{n^2}$
$E_{\mathrm{L}}{ }^{* *}=-13.6 \times \frac{9}{1}=-122.4 \mathrm{eV}$
$E_{\mathrm{L}}{ }^{* *}=-13.6 \times \frac{9}{9}=-13.6 \mathrm{eV}$
$\Delta E=-13.6-(-122.4)$
$=108.8 \mathrm{eV}$

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2011
Energy required to move a body of mass $m$ from an orbit of radius $2 R$ to $3 R$ is

Options:
A $G M m / 12 R^2$
B $G M m / 3 R^2$
C $G M m / 8 R$
D $\mathrm{GMm} / 6 R$

The correct answer is:
$\mathrm{GMm} / 6 R$

Solution:
Change in potential energy in displacing a body from $r_1$ to $r_2$ is given by
$\Delta U=G M m\left[\frac{1}{r_1}-\frac{1}{r_2}\right]=G M m\left(\frac{1}{2 R}-\frac{1}{3 R}\right)=\frac{G M m}{6 R}$

Topic: Physics, Gravitation, JEE Main
Energy required to move a body of mass m from an orbit of radius 2R to 3R is

Options:
A
$\mathrm{GMm} / 12 \mathrm{R}^2$

B
$\mathrm{GMm} / 3 \mathrm{R}^2$

C
$\mathrm{GMm} / 8 \mathrm{R}$

D
$\mathrm{GMm} / 6 \mathrm{R}$

The correct answer is:
$\mathrm{GMm} / 6 \mathrm{R}$

Solution:
$\mathrm{E}=(\mathrm{PE})_{\text {final }}-(\mathrm{PE})_{\text {initial }}=\frac{-\mathrm{GMm}}{3 \mathrm{R}}+\frac{\mathrm{GMm}}{\mathrm{R}}=\frac{\mathrm{GMm}}{6 \mathrm{R}}$

Topic: Physics, Gravitation, JEE Main, JEE Main 2002
Energy required to stop the ejection of electrons from \( \mathrm{Cu}^{-} \)plate is \( 0.27 \mathrm{eV} \). Calculate the work function (in \( \mathrm{eV} \) ) When radiation of \( \lambda=235 \mathrm{~nm} \) strikes the plate.

Options:

The correct answer is:
5

Solution:
Ephoton=work function +12 mv2=work function +eV0↓Stopping potential ....(i)Ephoton=hcλ=6.625×10-34×3×108235×10-9=5.29 eV

from equation i

Ephoton=W0+0.27

W0=Ephoton-0.27

=5.27-0.27=5 eV

Topic: Chemistry, Structure of Atom, JEE Main
Energy transfer in photosynthesis occurs as

Options:
A Phycoerythrin $\rightarrow$ Phycocyanin $\rightarrow$ Carotenoid $\rightarrow$ Chlorophyll $a$
B Chlorophyll $b \rightarrow$ Carotenoid $\rightarrow$ Phycoerythrin $\rightarrow$ Chlorophyll $a$
C Phycocyanin $\rightarrow$ Phycoerythrin $\rightarrow$ Carotenoid $\rightarrow$ Chlorophyll $a$
D Chlorophyll $b \rightarrow$ Carotenoid $\rightarrow$ Phycocyanin $\rightarrow$ Chlorophyll $a$

The correct answer is:
Phycocyanin $\rightarrow$ Phycoerythrin $\rightarrow$ Carotenoid $\rightarrow$ Chlorophyll $a$

Solution:
Photosynthetic pigments mainly occur on thylakoid membranes. These are major pigments like chlorophylls and accessory pigments like phycoerythrin, phycocyanin, and carotenoids. The accessory pigments absorb light energy of various wavelength and get excited and transfer the energy to chlorophyll-a molecule through electron spin resonance. Order of energy transfer in photosynthesis occur as phycocyanin followed by phycoerythrin followed by carotenoids and finally to chlorophylls.

Topic: Biology, Photosynthesis in Higher Plants, JIPMER, JIPMER 2005
Enthalpies of formation of CCl4( g),H2O(g),CO2( g) and HCl are -105,-242,-394 and -92 kJ mol-1 respectively. The magnitude of enthalpy of the reaction given below is kJmol-1. (nearest integer) CCl4( g)+2H2O(g)→CO2( g)+4HCl(g)

Options:

The correct answer is:
173

Solution:
Using Hess's law of constant heat summation:
Enthalpy of reaction = Enthalpy of formation of products - Enthalpy of formation of reactants
∴ΔHf°=ΔHf°CO2, g+4ΔHf°(HCl)-ΔHf°CCl4-2ΔHf°H2O,r

=[-394]+4[-92]+105-2×[242]

=-394-368+105+484

=-173 kJ/mole

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2023 (31 Jan Shift 2)
Enthalpy change for reaction, $\frac{1}{2} \mathrm{H}_2+\frac{1}{2} \mathrm{Cl}_2 \rightarrow \mathrm{HCl}$, is called

Options:
A Enthalpy of combination
B Enthalpy of reaction
C Enthalpy of formation
D Enthalpy of fusion

The correct answer is:
Enthalpy of formation

Solution:
Enthalpy of formation of $\mathrm{HCl}$.

Topic: Chemistry, Thermodynamics (C), JEE Main
Enthalpy $(H)$ is equal to

Options:
A Internal energy (E)
B Product of pressure (P) and volume (V) of gas
C Internal energy (E)+PV
D Work (W) done by a system

The correct answer is:
Internal energy (E)+PV

Solution:
Enthalpy $(H)$ is defined as the sum of internal energy $E+P V, H=E+P V$.

Topic: Chemistry, Thermodynamics (C), JEE Main
Enthalpy is equal to

Options:
A T2∂G/T∂TP
B –T2∂G/T∂TP
C T2∂G/T∂TV
D –T2∂G/T∂TV

The correct answer is:
–T2∂G/T∂TP

Solution:
G=H–TS …(i)
G=U+PV–TS
ΔG=ΔU+PΔV+VΔP-TΔS-SΔT
From the first and second laws,
TΔS=ΔU+PΔV
ΔG=VΔP-SΔT
At constant pressure, ΔP=0
ΔGΔT=–S …(ii)
From equations. (i) and (ii)
G=H+TΔGΔT
or G=H+T∂G∂TP
-HT2=-GT2+1T∂G∂TP=∂G/T∂TP
H=-T2∂G/T∂TP

Topic: Chemistry, Thermodynamics (C), JEE Main
Enthalpy of a reaction at 27oC is 15 kJ mol-1. The reaction will be feasible if entropy is

Options:
A 15 J mol-1 K-1
B -50 J mol-1 K-1
C Greater than 50 J mol-1 K-1
D Less than 50 J mol-1 K-1

The correct answer is:
Greater than 50 J mol-1 K-1

Solution:
Î G < Î H - T Î S for spontaneous reaction
Î G should be - ve
T Î S > Î H
i.e., 3 0 0 Î S > 15000 J
ÎS>50 Jâ mol -1 ââ K -1

Topic: Chemistry, Thermodynamics (C), JEE Main
Epimers of glucose is

Options:
A Fructose
B Galactose
C Talose
D Ribose

The correct answer is:
Galactose

Solution:
Diastereomers having more than two chiral carbon and configuration at only one carbon is different are epimers of each other..

Configuration of chiral carbon at D- glucose is 2R, 3S, 4R, 5R whereas configuration of chiral carbon at D- galactose is 2R, 3S, 4S, 5R.

Galactose is C-4 epimer of glucose because configuration at 4th carbon is different.

Topic: Chemistry, Biomolecules, JEE Main
Epimers of glucose is

Options:
A Fructose
B Galactose
C Talose
D Ribose

The correct answer is:
Galactose

Solution:
Galactose is C-4 epimer of glucose

Topic: Chemistry, Biomolecules, JEE Main
Epipetalous stamens are found in:

Options:
A Trifolium
B Brinjal
C Asparagus
D Indigofera

The correct answer is:
Brinjal

Solution:
When stamen remains attached to petals, this adhesion is known as epipetalous stamens. It is present in members of Solanaceae family. Brinjal is member of Solanaceae family. In Asparagus, member of Liliaceae family stamen remains attached to tepals and this condition is known as epiphyllous condition.

Topic: Biology, Morphology of Flowering Plants, NEET
Equal amount of an alkyl halide (C4H9Cl) is reacted at the same temperature with equal volume of 0Â·2 M and 0Â·4 M solution of KOH, respectively, in two separate experiments. The time taken for the reaction of 50% of (C4H9Cl) was found to be same, the alkyl halide is

Options:
A
B
C
D

The correct answer is:

Solution:
Hydrolysis of 3° RX is independent of nucleophilic concentration SN1reaction.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Equal masses of methane and ethane have their total translational kinetic energy in the ratio 3 : 1, then their temperatures are in the ratio is

Options:
A 5: 8
B 45: 8
C 15: 8
D 8: 5

The correct answer is:
8: 5

Solution:
E1=32×M16RT1
E2=32×M30RT2
⇒E1E2=3016T1T2
⇒31=3016T1T2 ⇒T1T2=85

Topic: Chemistry, Redox Reactions, JEE Main
Equal masses of methane and oxygen are mixed in an empty container at $25^{\circ} \mathrm{C}$. The fraction of the total pressure exerted by oxygen is

Options:
A
$\frac{2}{3}$

B
$\frac{1}{3} \times \frac{273}{298}$

C
$\frac{1}{3}$

D
$\frac{1}{2}$

The correct answer is:
$\frac{1}{3}$

Solution:
Let the mass of methane and oxygen is $w$
$\begin{aligned}
& \text { mole fraction of oxygen }=\frac{\frac{w}{32}}{\frac{w}{32}+\frac{w}{16}} \\
& =\frac{\frac{1}{32}}{\frac{1}{32}+\frac{1}{16}}=\frac{\frac{1}{32}}{\frac{3}{32}}=\frac{1}{3}
\end{aligned}$
Let the total pressure be $\mathrm{P}$
The pressure exerted by oxygen (partial pressure) $=\mathrm{X}_{\mathrm{O}_2} \times \mathrm{P}_{\text {total }}$
$\Rightarrow \mathrm{P} \times \frac{1}{3}$
Hence, (C) is correct.

Topic: Chemistry, States of Matter, JEE Main, JEE Main 2007
Equal masses of water and a liquid of density 2 are mixed together, then the mixture has a density of

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

The correct answer is:
$4 / 3$

Solution:
If two ljquid of equal masses and different densities are mixed together then density of the mixture
$\rho=\frac{2 \rho_1 \rho_2}{\rho_1+\rho_2}=\frac{2 \times 1 \times 2}{1+2}=\frac{4}{3}$

Topic: Physics, Mechanical Properties of Fluids, JEE Main
Equal volumes of 0.1 M NaCl and 0.06 M CaCl2 solutions are separated by a semi- permeable membrane in container. For this system, choose the correct answer

Options:
A Water flows from NaCl solution towards CaCl2 solution
B There is no movement of any solution across the membrane
C Osmotic pressure of 0.1 M NaCl is lower than the osmotic pressure of CaCl2 (Assume complete dissociation)
D Water flows form CaCl2 solution towards NaCl solution

The correct answer is:
Water flows form CaCl2 solution towards NaCl solution

Solution:
Using πV=iCRT
i = 2 for NaCl
i = 3 for CaCl2
πVNaCl=2×0.1RT=0.2RT
πVCaCl2=3×0.06RT=0.18RT
As πVNaCl>πVCaCl2 .
Here water flows from CaCl2 towards NaCl solution.

Topic: Chemistry, Solutions, JEE Main
Equal volumes of monoatomic and diatomic gases at same initial temperature and pressure are mixed. The ratio of specific heats of the mixture $\left(C_p / C_v\right)$ will be

Options:
A 1
B 2
C 1.67
D 1.5

The correct answer is:
1.5

Solution:
$C_v=\frac{3}{2} R T ; C_p=\frac{5}{2} R T$ for monoatomic gas
$C_v=\frac{5}{2} \mathrm{RT} ; C_p=\frac{7}{2} R T$ for diatomic gas
Thus for mixture of 1 mole each, $C_v=\frac{\frac{3}{2} R T+\frac{5}{2} R T}{2}$ and $C_p=\frac{\frac{5}{2} R T+\frac{7}{2} R T}{2}$
Therefore, $C_p / C_v=\frac{3 R T}{2 R T}=1.5$.

Topic: Chemistry, Thermodynamics (C), JEE Main
Equal volumes of three acid solutions of $\mathrm{pH} 3,4$ and 5 are mixed in a vessel. What will be the $\mathrm{H}^{+}$ion concentration in the mixture?

Options:
A $1.11 \times 10^{-3} \mathrm{M}$
B $1.11 \times 10^{-4} \mathrm{M}$
C $3.7 \times 10^{-4} \mathrm{M}$
D $3.7 \times 10^{-3} \mathrm{M}$

The correct answer is:
$3.7 \times 10^{-4} \mathrm{M}$

Solution:
$\mathrm{N}_1 \mathrm{~V}_1+\mathrm{N}_2 \mathrm{~V}_2+\mathrm{N}_3 \mathrm{~V}_3=\mathrm{N}_{\mathrm{m}} \mathrm{V}_{\mathrm{m}}$
or $10^{-3} \times 1+10^{-4} \times 1+10^{-5} \times 1=\mathrm{N}_{\mathrm{m}} \times 3$
or $1.11 \times 10^{-3}=\mathrm{N}_{\mathrm{m}} \times 3$
or $\mathrm{N}_{\mathrm{m}}=0.37 \times 10^{-3}$
$\left[\mathrm{H}^{+}\right]=3.7 \times 10^{-4} \mathrm{M}$

Topic: Chemistry, Ionic Equilibrium, NEET, NEET 2008 (Mains)
Equal volumes of two gases, having their densities in the ratio of $1: 16$ exert equal pressures on the walls of two containers. The ratio of their rms velocities $\left(\frac{c_{1}}{c_{2}}\right)$

Options:
A $1: 8$
B $8: 1$
C $1: 4$
D $4: 1$

The correct answer is:
$4: 1$

Solution:
Densities ratio of two gases,
$$
\rho_{1}: \rho_{2}=1: 16
$$
$$
\Rightarrow \quad \frac{\rho_{1}}{\rho_{2}}=\frac{1}{16}
$$
Since, pressure of both gases are equal.
$\therefore \quad p_{1}=p_{\mathrm{g}}$
$\frac{1}{3} \rho_{1} c_{1}^{2}=\frac{1}{2} \rho_{2} c_{2}^{2}$
$\Rightarrow \quad \frac{c_{1}^{2}}{c_{2}^{2}}=\frac{\rho_{2}}{\rho_{1}}=\frac{16}{1}$
$=\left(\frac{4}{1}\right)^{2}$
$\Rightarrow \quad \frac{c_{1}}{c_{2}}=\frac{4}{1}$
$\Rightarrow \quad c_{1}: c_{2}=4: 1$

Topic: Physics, Kinetic Theory of Gases, JEE Main
Equal volumes of two monoatomic gases, $A$ and $B$, at same temperature and pressure are mixed. The ratio of specific heats $\left(c_p / c_V\right)$ of the mixture will be

Options:
A 0.83
B 1.50
C 3.3
D 1.67

The correct answer is:
1.67

Solution:
For monoatomic gases,
$$
\frac{c_p}{c_V}=\gamma=\frac{\frac{5}{2} R}{\frac{3}{2} R} 1.67
$$
$\because$ Equal volumes of two monoatomic gases are mixed.
$\therefore \gamma$ will remain the same.

Topic: Chemistry, States of Matter, NEET, NEET 2012 (Mains)
Equal weights of ethane and hydrogen are mixed in an empty container at 25 C. The fraction of total pressure exerted by hydrogen is

Options:
A 1: 2
B 1: 1
C 01:16
D 15:16

The correct answer is:
15:16

Solution:
Given, equal weights of ethane and hydrogen are mixed in an empty container at $25^{\circ} \mathrm{C}$
Initial gram weight $=\begin{array}{ll}\mathrm{C}_{2} \mathrm{H}_{6} & \mathrm{H}_{2} \\ \mathrm{Wg} & \mathrm{Wg}\end{array}$
Number of moles $=\frac{w}{30} \quad \frac{w}{2}$ According to Henry's law,
$$
\begin{array}{l}\frac{p_{\mathrm{H}_{2}}}{p_{\text {total }}}=\chi_{\mathrm{H}_{2}}=\frac{n_{\mathrm{H}_{2}}}{n_{\mathrm{H}_{2}}+n_{\mathrm{C}_{2} \mathrm{H}_{6}}}=\frac{\frac{w}{2}}{\frac{w}{2}+\frac{w}{30}} \\ \frac{p_{\mathrm{H}_{2}}}{p_{\text {total }}}=\frac{\frac{w}{2}}{\frac{15 w+w}{30}}=\frac{w}{2} \times \frac{30}{16 w}=\frac{15}{16}\end{array}
$$
So, the fraction of total pressure exerted by
hydrogen is 15: 16

Topic: Chemistry, Solutions, JEE Main
Equal weights of two gases of molecular weight 4 and 40 are mixed. The pressure of the mixture is 1.1 atm. The partial pressure of the light gas in this mixture is

Options:

The correct answer is:
1

Solution:
Number of mol of gas 1=w4
Number of mol of gas 2=w40
Mole fraction of gas 1, X1=w4w4+w40=1011
Partial pressure of gas 1 = mol fraction of gas 1× Total pressure
=1011×1.1 atm = 1 atm

Topic: Chemistry, States of Matter, JEE Main
Equanil acts as alan:

Options:
A Artificial sweetener
B Tranquilizer
C Antihistamine
D Antifertility drug

The correct answer is:
Tranquilizer

Solution:
Equanil is a tramquilizer.
Structure of equanil
Equanil is a tranquilizer used in controlling depression and hypertension

Topic: Chemistry, Chemistry in Everyday Life, JEE Main
Equation of a common tangent to the circle, x2+y2-6x=0 and the parabola, y2=4x is:

Options:
A 23y=-x-12
B 3y=x+3
C 3y=3x+1
D 23y=12x+1

The correct answer is:
3y=x+3

Solution:
Let tangent to parabola y2=4x is
y=mx+1m...i
If equation (i) is tangent to given circle whose centre is (3, 0) and radius is 3 then length of perpendicular from centre of circle to equation (i) is equal to radius of circle.
Hence, 3m+1mm2+1=3
⇒3m2+1=3m1+m2
⇒9m4+6m2+1=9m2+9m4
⇒m=±13
∴ common tangents are y=x3+3 or y=-x3-3
⇒3y=x+3 or 3y=-x-3

Topic: Mathematics, Parabola, JEE Main, JEE Main 2019 (09 Jan Shift 1)
Equation of a common tangent to the parabola $y^{2}=4 x$ and the hyperbola $x y=2$ is :

Options:
A $x+y+1=0$
B $x-2 y+4=0$
C $x+2 y+4=0$
D $4 x+2 y+1=0$

The correct answer is:
$x+2 y+4=0$

Solution:
Equation of a tangent to parabola $y^{2}=4 x$ is:

$y=m x+\frac{1}{m}$

This line is a tangent to $x y=2$ $\therefore \quad x\left(m x+\frac{1}{m}\right)=2 \Rightarrow m x^{2}+\frac{1}{m} x-2=0$

$\because$ Tangent is common for parabola and hyperbola. $\therefore \quad D=\left(\frac{1}{m}\right)^{2}-4 \cdot m \cdot(-2)=0$

$\frac{1}{m^{2}}+8 m=0$

$1+8 m^{3}=0$

$m^{3}=-\frac{1}{8} \Rightarrow m=-\frac{1}{2}$

$\therefore$ Equation of common tangent: $y=-\frac{1}{2} x-2$

$\Rightarrow 2 y=-x-4 \Rightarrow x+2 y+4=0$

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2019 (11 Jan Shift 1)
Equation of a plane at a distance 221 units from the origin, which contains the line of intersection of the planes x-y-z-1=0 and 2x+y-3 z+4=0, is

Options:
A -x+2y+2z-3=0
B 3x-4z+3=0
C 3x-1y-5z+2=0
D 4x-y-5z+2=0

The correct answer is:
4x-y-5z+2=0

Solution:
Required equation of plane

P1+λP2=0

x-y-z-1+λ2x+y-3z+4=0

Given that its distance from origin is 221

Thus,

4λ-12λ+12+λ-12+-3λ-12=221

214λ-12=214λ2+8λ+3

336λ2-168λ+21=28λ2+16λ+6

308λ2-184λ+15=0

308λ2-154λ-30λ+15=0

2λ-1154λ-15=0

λ=12 & 15154

For λ=12, required plane is 4x-y-5z+2=0

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2021 (27 Aug Shift 1)