datavorous/entrance-exam-dataset huggingface dataset:
The eubacterial cellular part that is similar to a eukaryotic cell is ____

Options:
A plasma membrane.
B nucleus.
C ribosome.
D cell wall.

The correct answer is:
plasma membrane.

Solution:
Eubacteria are prokaryotic but are enclosed by plasma membrane like the eukaryotic cells. The composition and architecture of the plasma membrane remain the same for all cells.

Ribosomes are of 70S type in bacteria. The cell wall has peptidoglycan as the main component. The prokaryotes are devoid of the true nucleus.

Topic: Biology, Cell - The Unit of Life, NEET
The eutrophication of water body results in

Options:
A Increase in biodiversity
B Loss in biodiversity
C Break down of organic matter
D decrease in BOD.

The correct answer is:
Loss in biodiversity

Solution:
The addition of phosphates in water enhances algae growth. Such profuse growth of algae, covers the water surface and reduces the oxygen concentration in water. This leads to anaerobic conditions, commonly with accumulation of abnoxious decay and animal death. Thus, bloom-infested water inhibits the growth of other living organisms in the water body. This process in which nutrient enriched water bodies support a dense plant population, which kills animal life by depriving it of oxygen and results in subsequent loss of biodiversity is known as Eutrophication.

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2022 (25 Jun Shift 1)
"The Evil Quartet" is a sobriquet used to describe

Options:
A four major causes of biodiversity losses
B four major causes of forest loss
C four major causes of air pollution
D four major causes of population explosion

The correct answer is:
four major causes of biodiversity losses

Solution:
Presently the rapid rate of species extinction is mainly due to human activities, which ultimately cause biodiversity loss of some species. There are mainly four causes for biodiversity loss, which are collectively called "The Evil Quartet". These are habitat loss and fragmentation, over-exploitation, alien-species invasion and co-extinction.

Topic: Biology, Biodiversity and its Conservation, NEET
The exact order of acidity of the compounds p-nitrophenol, acetic acid, acetylene and ethanol is

Options:
A p-nitrophenol $ < $ acetic acid $ < $ acetylene $ < $ ethanol
B acetic acid $ < \mathrm{p}-$ nitrophenol $ < $ acetylene $ < $ ethanol
C acetylene $ < \mathrm{p}$-nitrophenol $ < $ ethanol $ < $ acetic acid
D acetylene $ < $ ethanol $ < $ p-nitrophenol $ < $ acetic acid

The correct answer is:
acetylene $ < $ ethanol $ < $ p-nitrophenol $ < $ acetic acid

Solution:
Acetic acid is most acidic as its conjugte base has equivalent Resonating structure followed by para-nitrophenol. Ethanol is less acidic then para-nitrophenol as conjugate base of p-nitrophenol is resonance stablized but not ethoxide $\left(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{O}\right)$ Also electronegativity of carbon is lesser than oxygen, hence acetylene is less acidic.

Topic: Chemistry, General Organic Chemistry, JEE Main
The exact volumes of 1 M NaOH solution required to neutralise 50 mL of 1 M H3PO3 solution and 100 mL of 2 M H3PO2 solution, respectively, are :

Options:
A 100 mLand 100 mL
B 100 mL and 50 mL
C 100 mL and 200 mL
D 50 mL and 50 mL

The correct answer is:
100 mL and 200 mL

Solution:
H3PO3+2NaOH→Na2HPO3+2H2O

50 ml1 M

1 M V=?

⇒nNaoHnH3PO3=21

⇒1×V50×1=21⇒VNaOH=100 ml

H3PO2+NaOH→NaH2PO3+H2O

2M 1 M

100 mL V=?

⇒nNaoHnH3PO3=11⇒1×V2×100=11⇒VNaOH=200 ml

Topic: Chemistry, Redox Reactions, JEE Main, JEE Main 2021 (16 Mar Shift 2)
The excess pressure inside a soap bubble is thrice the excess pressure inside a second soap bubble. The ratio between the volume of the first and the second bubble is:

Options:
A $1: 9$
B $1: 3$
C $1: 27$
D $1: 81$

The correct answer is:
$1: 27$

Solution:
$\mathrm{P}_1-\mathrm{P}_0=\frac{4 \mathrm{~T}}{\mathrm{r}_1} \qquad \quad \quad \quad\mathrm{P}_2-\mathrm{P}_0=\frac{4 \mathrm{~T}}{\mathrm{r}_2}$
$\begin{aligned} & \mathrm{P}_1-\mathrm{P}_0=3\left(\mathrm{P}_2-\mathrm{P}_0\right) \\ & \frac{4 \mathrm{~T}}{\mathrm{r}_1}=3 \frac{4 \mathrm{~T}}{\mathrm{r}_2} \\ & \mathrm{r}_2=3 \mathrm{r}_1 \\ & \frac{\mathrm{V}_1}{\mathrm{~V}_2}=\frac{\frac{4}{3} \pi \mathrm{r}_1^3}{\frac{4}{3} \pi \mathrm{r}_2^3}=\frac{1}{27}\end{aligned}$

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2024 (09 Apr Shift 2)
The exchange of gases with the surrounding atmosphere for respiration takes place through the skin in which of the following organisms?

Options:
A Homo sapiens
B Rana tigrina
C Psittacula krameri
D Panthera leo

The correct answer is:
Rana tigrina

Solution:
Frog (Rana tigrina) has lungs as its main respiratory organs, but during hibernation and aestivation and during its habitat in water, it respires through the skin.

Topic: Biology, Breathing and Exchange of Gases, NEET
The existence of a substance in more than one solid modifications is known as or Any compound having more than two crystal structures is called

Options:
A Polymorphism
B Isomorphism
C Allotropy
D Enantiomorphism

The correct answer is:
Polymorphism

Solution:
The existence of a substance in more than one crystalline form is known as polymorphism.

Topic: Chemistry, Solid State, JEE Main
The exothermic formation of $\mathrm{ClF}_3$ is represented by the equation:
$$
\mathrm{Cl}_{2(g)}+3 \mathrm{~F}_{2(g)} \rightleftharpoons 2 \mathrm{ClF}_{3(g)} ; \Delta \mathrm{r} \mathrm{H}=-329 \mathrm{~kJ}
$$
Which of the following will increase the quantity of $\mathrm{ClF}_3$ in an equilibrium mixture of $\mathrm{Cl}_2, \mathrm{~F}_2$ and $\mathrm{ClF}_3$ ?

Options:
A
Increasing the temperature

B
Removing $\mathrm{Cl}_2$

C
Increasing the volume of the container

D
Adding $\mathrm{F}_2$

The correct answer is:
Adding $\mathrm{F}_2$

Solution:
$\mathrm{M}_3 \mathrm{~V}_3=\mathrm{M}_1 \mathrm{~V}_2+\mathrm{M}_2 \mathrm{~V}_2$
$\mathrm{M}=\frac{480(1.5)+520(1.2)}{1000}=1.344 \mathrm{M}$

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2005
The expected graphical representation of the variation of angle of deviation ' δ ' with angle of incidence 'i' in a prism is :

Options:
A
B
C
D

The correct answer is:

Solution:
Standard graph between angle of deviation and
incident angle.

Topic: Physics, Ray Optics, JEE Main, JEE Main 2021 (27 Jul Shift 2)
The experimental data for the reaction $2 A+B_2 \rightarrow 2 A B$ is
\(\begin{array}{|l|l|l|c|}
\hline \text {Exp. } & {[A]_0} & {[B]_0} & \text {Rate }\left(\mathrm{mole} \mathrm{s}^{-1}\right) \\
\hline(1) & 0.50 & 0.50 & 1.6 \times 10^{-4} \\
\hline(2) & 0.50 & 1.00 & 3.2 \times 10^{-4} \\
\hline(3) & 1.00 & 1.00 & 3.2 \times 10^4 \\
\hline
\end{array}\)
The rate equation for the above data is

Options:
A Rate $=k\left[B_2\right]$
B Rate $=k\left[B_2\right]^2$
C Rate $=k[A]^2[B]^2$
D Rate $=k[A]^2[B]$

The correct answer is:
Rate $=k\left[B_2\right]$

Solution:
Rate $=k\left[B_2\right]$

Topic: Chemistry, Chemical Kinetics, JEE Main
The experimental data for the reaction,

2A+B2→2AB is

The rate equation for the above data is

Options:
A Rate =kA2B2
B Rate =kA2B
C Rate =kB2
D Rate =kB22

The correct answer is:
Rate =kB2

Solution:
For the reaction 2A+B2⇌2AB ,
Rate ∝[A]x[B2]y . On substituting the given data, we get
From experimental 1, 1.6×10-4∝0.50x0.50y ...(i)
From experimental 2, 3.2×10-4∝0.50x1.00y ...(ii)
From experimental 3, 3.2×10-4∝1.00x1.00y ...(iii)
1=1.000.50x
or 1=2x or 20=2x or x=0
Now, divide equation (ii) by equation (i) we get,
2=1.000.50y
2=2y
y=1
Thus rate equation is: Rate =kA0B21=kB2 .

Topic: Chemistry, Chemical Kinetics, JEE Main
The experimental value of the dipole moment of HCl is 1.03ââD. The length of the H-Cl bond is 1.275 â« . The percentage of ionic character in HCl is:
[Given: 1D=10-18esu cm]

Options:
A 43
B 21
C 17
D 7

The correct answer is:
17

Solution:
% Ionic character=Experimental value of dipole momentTheoritical value of dipolemomentÃ100
=Î¼expÎ¼theoÃ100
Where:
Î¼exp =1.030 and Î¼theo=qÃd
d=1.275â«=1.275Ã10-8cm
q=4.8Ã10-10 esu
Î¼theo=4.8Ã10-10Ã1.275Ã10-8esu-cm
=4.8Ã1.275Ã10-18esu-cm
=4.8Ã1.275D
1D=10-18esu cm
% Ionic character=1.034.8Ã1.275Ã100=17

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The experimental value of the dipole moment of \( \mathrm{HCl} \) is \( 1.03 \mathrm{D} \). The length of the \( \mathrm{H}-\mathrm{Cl} \) bond is \( 1.275 Å \). The percentage of ionic character in \( \mathrm{HCl} \) is: [Given: \( 1 \mathrm{D}=10^{-18} \) esu cm]

Options:
A \( 43 \% \)
B \( 21 \% \)
C \( 17 \% \)
D \( 7 \% \)

The correct answer is:
\( 17 \% \)

Solution:
%Ionic character=Experimental value of dipole momentTheoretical value of dipole moment×100
=μexpμtheo×100
Where:
μexp =1.03 D and μtheo=q×d
d=1.275Å=1.275×10-8cm
q=4.8×10-10 esu
μtheo=4.8×10-10×1.275×10-8esu cm
=4.8×1.275×10-18esu cm
=4.8×1.275D
1D=10-18esu cm
% Ionic character=1.034.8×1.275×100=17%(approx.)

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The exponent of 7 in 100C50 is

Options:
A 0
B 1
C 2
D 3

The correct answer is:
0

Solution:
We know that, 100C50=100!50!50!
The exponent of 7 in 50! is
507+5072=7+1=8
And the exponent of 7 in 100! is
1007+10072=14+2=16
Thus, the exponent of 7 in 100C50 is 16-2Ã8=0

Topic: Mathematics, Permutation Combination, JEE Main
The exponent of \( 7 \) in \( 100 ! \) is

Options:
A \( 14 \)
B \( 15 \)
C \( 16 \)
D None of these

The correct answer is:
\( 16 \)

Solution:
The exponent of 7 in 100! can be found by dividing 100 by exponents of 7 and adding the quotients.

The required exponent =

1007 + 10049+100343 ...=14+2+0=16

where, . is the greatest integer function.

Topic: Mathematics, Permutation Combination, JEE Main
The expression $\frac{\tan A}{1-\cot A}+\frac{\cot A}{1-\tan A}$ can be written as

Options:
A $\sin A \cos A+1$
B $\sec A \operatorname{cosec} A+1$
C $\tan A+\cot A$
D $\sec A+\operatorname{cosec} A$

The correct answer is:
$\sec A \operatorname{cosec} A+1$

Solution:
$\frac{\tan A}{1-\cot A}+\frac{\cot A}{1-\tan A}=\frac{\sin A}{\cos A} \times \frac{\sin A}{\sin A-\cos A}$
$$
+\frac{\cos A}{\sin A} \times \frac{\cos A}{\cos A-\sin A}
$$
$$
\begin{aligned}
&=\frac{1}{\sin A-\cos A}\left\{\frac{\sin ^{3} A-\cos ^{3} A}{\cos A \sin A}\right\} \\
&=\frac{\sin ^{2} A+\cos ^{2} A+\sin A \cos A}{\cos A \sin A}=\sec A \operatorname{cosec} A+1
\end{aligned}
$$

Topic: Mathematics, Trigonometric Equations, JEE Main
The expression $\frac{2^{2}+1}{2^{2}-1}+\frac{3^{2}+1}{3^{2}-1}+\frac{4^{2}+1}{4^{2}-1}+\ldots \ldots+\frac{(2011)^{2}+1}{(2011)^{2}-1}$ lies in the interval

Options:
A $\left(2010,2010 \frac{1}{2}\right)$
B $\left(2011-\frac{1}{2011}, 2011-\frac{1}{2012}\right)$
C $\left(2011,2011 \frac{1}{2}\right)$
D $\left(2012,2012 \frac{1}{2}\right)$

The correct answer is:
$\left(2011,2011 \frac{1}{2}\right)$

Solution:
$\frac{2^{2}+1}{2^{2}-1}+\frac{3^{2}+1}{3^{2}-1}+\frac{4^{2}+1}{4^{2}-1}+\ldots .+\frac{(2011)^{2}+1}{(2011)^{2}-1}$
$\sum_{r=2}^{2011} \frac{r^{2}+1}{r^{2}-1}=\sum_{r=2}^{2011}\left[1+\frac{2}{(r+1)(r-1)}\right]$
$\begin{array}{l}
=\sum_{r=2}^{2011}\left[1+\frac{1}{r-1}-\frac{1}{r+1}\right] \\
=2010+\left[1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+\ldots .+\frac{1}{2010}-\frac{1}{2012}\right] \\
=2010+1+\frac{1}{2}-\frac{1}{2012}-\frac{1}{2011} \\
=2011+\frac{1}{2}-\left[\frac{1}{2011}+\frac{1}{2012}\right]
\end{array}$
lies between $\left(2011,2011 \frac{1}{2}\right)$

Topic: Mathematics, Sequences and Series, JEE Main
The expression $2 \cos \frac{\pi}{13} \cdot \cos \frac{9 \pi}{13}+\cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13}$ is equal to

Options:
A -1
B 0
C 1
D None of these

The correct answer is:
0

Solution:
$\begin{aligned}
2 \cos \frac{\pi}{13} \cdot \cos \frac{9 \pi}{13} & +\cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13} \\
& =2 \cos \frac{\pi}{13} \cdot \cos \frac{9 \pi}{13}+2 \cos \frac{4 \pi}{13} \cos \frac{\pi}{13} \\
& =2 \cos \frac{\pi}{13}\left[\cos \frac{9 \pi}{13}+\cos \frac{4 \pi}{13}\right] \\
& =2 \cos \frac{\pi}{13}\left[2 \cos \frac{\pi}{2} \cdot \cos \frac{5 \pi}{26}\right]=0,\left[\text { cos } \frac{\pi}{2}=0\right] .
\end{aligned}$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
The expression $\bar{A}(A+B)+(B+A A)(A+\bar{B})$ simplifies to

Options:
A $A+B$
B $A B$
C $\overline{\mathrm{A}+\mathrm{B}}$
D $\bar{A}+\bar{B}$

The correct answer is:
$A+B$

Solution:
$\begin{aligned} & \bar{A}(A+B)+(B+A A)(A+\bar{B}) \\=& \bar{A} A+\bar{A} B+(B+A)(A+\bar{B}) \\=& O+\bar{A} B+A B+A+B \bar{B}+A \bar{B} \\=&(\bar{A}+A) B+A+O+A \bar{B} \\=& 1 . B+A(1+\bar{B}) \\=& A+B \end{aligned}$

Topic: Physics, Mathematics in Physics, JEE Main
The expression for the magnetic energy stored in a solenoid of length $L$, in terms of magnetic field $B$ and area $\mathrm{A}$ is

Options:
A $\frac{1}{2 \mu_0} \mathrm{~B}^2 \mathrm{AL}$
B $\frac{1}{2 \epsilon_0} \mathrm{~B}^2 \mathrm{AL}$
C $\frac{1}{2 \mu_0} \mathrm{BA}^2 \mathrm{~L}$
D $\frac{1}{2 \epsilon_0} \mathrm{BA}^2 \mathrm{~L}$

The correct answer is:
$\frac{1}{2 \mu_0} \mathrm{~B}^2 \mathrm{AL}$

Solution:
The expression for the magnetic energy stored in a solenoid is given by
$$
\begin{aligned}
& \frac{U}{V}=\frac{1}{2} \epsilon_0 E^2 \\
& \frac{U}{A L}=\frac{1}{2} \frac{1}{C^2 \mu_0}(B c)^2\left(\because c^2=\frac{1}{\mu_0 \epsilon_0}, c=\frac{E}{B}\right) \\
& U=\frac{A L}{2 c^2 \mu_0} \times B^2 c^2=\frac{1}{2 \mu_o} B^2 A L
\end{aligned}
$$

Topic: Physics, Magnetic Effects of Current, JEE Main
The expression for the solubility product of $\mathrm{Al}_2\left(\mathrm{SO}_4\right)_3$ is

Options:
A $K_{s p}=\left[\mathrm{Al}^{3+}\right]\left[\mathrm{SO}_4^{2-}\right]$
B $K_{s p}=\left[\mathrm{Al}^{3+}\right]^2\left[\mathrm{SO}_4^{2-}\right]^3$
C $K_{s p}=\left[\mathrm{Al}^{3+}\right]^3\left[\mathrm{SO}_4^{2-}\right]^2$
D $K_{s p}=\left[\mathrm{Al}^{3+}\right]^2\left[\mathrm{SO}_4^{2-}\right]^2$

The correct answer is:
$K_{s p}=\left[\mathrm{Al}^{3+}\right]^2\left[\mathrm{SO}_4^{2-}\right]^3$

Solution:
Solubility of $\mathrm{Al}_2\left(\mathrm{SO}_4\right)_3$
$\begin{aligned} & \mathrm{Al}_2\left(\mathrm{SO}_4\right)_3 \rightleftharpoons 2 \mathrm{Al}^{3+}+3 \mathrm{SO}_4^{2-} \\ & K_{s p}=\left[\mathrm{Al}^{3+}\right]^2\left[\mathrm{SO}_4^{2-}\right]^3\end{aligned}$

Topic: Chemistry, Ionic Equilibrium, JEE Main
The expression of a phenotype

Options:
A depends on the genotype but doesn't depend on the environment
B depends on the environment but doesn't depend on the genotype
C depends on both the environment as well as on the genotype
D doesn't depend on either the environment or the genotype

The correct answer is:
depends on both the environment as well as on the genotype

Solution:
The expression of a phenotypedepends on both the environment as well as on the genotype. Example: A pea plant with TT genotype should grow tall but is the nutrients for growth of the plant in the soil are missing or deficient its growth may remain stunted.

Topic: Biology, Principles of Inheritance and Variation, NEET
The expression ~~p→q is logically equivalent to

Options:
A p∧ ~q
B ~p∧~q
C p∧q
D ~p∧q

The correct answer is:
~p∧~q

Solution:
We know that ~p→q=p∧~q.

Hence, the expression ~~p→q is logically equivalent to ~p∧~q.

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2019 (12 Jan Shift 2)
The expression sin⁡27°cos⁡57°sin⁡87° simplifies to

Options:
A sin⁡9°4
B cos⁡9°4
C sin⁡9°2
D cos⁡9°2

The correct answer is:
cos⁡9°4

Solution:
sin⁡27°cos⁡57°sin⁡90°-3°
=12sin⁡27°2cos⁡57°cos⁡3°
=12sin⁡27°cos⁡60°+cos⁡54°
=12sin⁡27°12+12sin⁡27°sin⁡36°
=14sin⁡27°+142sin⁡27°sin⁡36°
=14sin⁡27°+14cos⁡9°-cos⁡63°
=14sin⁡27°+14cos⁡9°-14sin⁡27°
=14cos⁡9°

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
The expression tanA1-cotA+cotA1-tanA can be written as :

Options:
A tanA+cotA
B secA+cosecA
C sinAcosA+1
D secAcosecA+1

The correct answer is:
secAcosecA+1

Solution:
Given tanA1-cotA+cotA1-tanA
=sin2AcosAsinA-cosA+cos2AsinAcosA-sinA
=sin3A-cos3Asin A cos Asin A-cos A
=sin2A+cos2A+sin A cos Asin A cos A
=secAcosecA+1

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main, JEE Main 2013 (07 Apr)
The expression $x^2+2 b x+c$ has the positive value if

Options:
A $b^2-4 c\gt0$
B $b^2-4 c \lt 0$
C $c^2 \lt b$
D $b^2 \lt c$

The correct answer is:
$b^2 \lt c$

Solution:
$x^2+2 b x+c=(x+b)^2+c-b^2$
$(x+b)^2$ is a perfect square, therefore the given expression is positive if $c-b^2\gt0$ or $b^2 \lt c$

Topic: Mathematics, Quadratic Equation, JEE Main
The expressions below give current $I$ through an electronic component as a function of applied potential $V . I_0$ and $V_0$ are constants having dimensions of current and potential respectively. Which of the following are dimensionally incorrect?
(A) $I=I_0\left(e^{\frac{2 V}{V_0}}+1\right)$
(B) $I=I_0\left(e^{\frac{v}{2 V_0}}-1\right)$
(C) $I=I_0 V_0\left(e^{\frac{V}{V_0}}-1\right)$
(D) $I=I_0\left(\frac{V}{V_0}\right)\left(e^{\frac{V}{V_0}}-1\right)$

Options:
A $\mathrm{A}$
B $\mathrm{B}$
C $\mathrm{C}$
D $\mathrm{D}$

The correct answer is:
$\mathrm{C}$

Solution:
According to the principle of homogeneity, a physical equation will be dimensionally correct only if the dimensions of all the terms occuring on both side of the equation are the same.
Hence, in option (c)
$I=I_0 V_0\left(e^{\frac{v}{V_0}}-1\right)$
Dimension of $I \neq$ dimensions of $I_0 V_0$ Therefore, equation shown in option (c) is not dimensionally correct.

Topic: Physics, Units and Dimensions, JEE Main
The extent of charge of lead accumulator is determined by

Options:
A amount of $\mathrm{PbSO}_4$ in the battery
B amount of $\mathrm{PbO}_2$ in the battery
C specific gravity of $\mathrm{H}_2 \mathrm{SO}_4$ of the battery
D amount of $\mathrm{Pb}$ in the battery

The correct answer is:
specific gravity of $\mathrm{H}_2 \mathrm{SO}_4$ of the battery

Solution:
In a fully charged lead accumulator or lead storage battery, sulphuric acid has a specific gravity (i.e., density that varies from 1.260 to 1.285. But during discharge (i.e., when the battery is in use). $\mathrm{H}_2 \mathrm{SO}_4$ is used up.
$$
\begin{array}{r}
\mathrm{Pb}(s)+\mathrm{PbO}_2(s)+4 \mathrm{H}^{+}(a q)+2 \mathrm{SO}_4^{2-}(a q) \longrightarrow \\
2 \mathrm{PbSO}_4(l)+2 \mathrm{H}_2 \mathrm{O}
\end{array}
$$
As a result, the specific gravity of $\mathrm{H}_2 \mathrm{SO}_4$ falls when it falls below 1.230 the battery needs recharging.

Topic: Chemistry, Electrochemistry, JEE Main
The extra volume of air that can be inspired forcibly after normal inspiration is

Options:
A Tidal volume
B Inspiratory Reserve Volume
C Vital capacity
D Inspiratory Capacity

The correct answer is:
Inspiratory Capacity

Solution:
Respiratory capacities are the volume of air present at each stage of the gas exchange process found in the human body. Total lung capacity is the amount of air one can accommodate ~5000 mL, vital capacity is the volume of air that can forced out ~4500 mL. The amount of air that will remain after normal exhalation is functional residual capacity ~2300 mL, and inspiratory capacity is the extra volume of air that can be inspired forcibly after after a normal inspiration is ~3000 mL.

Topic: Biology, Breathing and Exchange of Gases, NEET
The extraction principle used in extraction of a metal 'M' can be represented as follows

MS + 2 O 2 → MSO 4 MSO 4 + MS → 2 M + 2 SO 2

The metal 'M' is

Options:
A Mg
B Pb
C Sn
D Fe

The correct answer is:
Pb

Solution:
It is a case of self reduction.
The concentrated galena (PbS) is partially oxidised by adjusting the air supply.
PbS + 2 O 2 → PbSO 4
The temperature is raised, air supply reduced and fresh galena (PbS) added.
PbS + PbSO 4 → 2 Pb + 2 SO 2 ↑

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
The extremities of a diagonal of a rectangle are 0,0 and 4,3. The equations of the tangents to the circumcircle of the rectangle which are parallel to the diagonal are

Options:
A 3x-4y=±5
B 3x+4y=Â±10
C 3x+4y=Â±25
D 6x-8y=±25

The correct answer is:
6x-8y=±25

Solution:
Let, ABCD be the rectangle such that the coordinates of the vertices A and C of diagonal AC are A0,0 and C4,3.
The equation of the circumcircle of rectangle ABCD is
x-0x-4+y-0y-3=0
⇒x2+y2-4x-3y=0 … 1
The equation of a line parallel to AC is
3x-4y+λ=0 … 2
If it touches the circle 1, then
λ9+16=4+94
⇒λ5=52
⇒λ=±252
Substituting the value of λ in 2, we obtain that the equations of the required tangents are
3x-4y±252=0

Topic: Mathematics, Circle, JEE Main
The eye can be regarded as a single refracting surface. The radius of curvature of this surface is equal to that of the cornea (7.8 mm). This surface separates two media of refractive indices 1 and 1.34. Calculate the distance from the refracting surface at which a parallel beam of light will come to focus.

Options:
A 4.0cm
B 1cm
C 2 cm
D 3.07 cm

The correct answer is:
3.07 cm

Solution:
We know eye lens is convex refracting surface.
Use expression for refraction from curved surface
μ1→ Refractive index of medium having incident ray
μ2→ Refractive index of a medium having reflected ray
μ2v-μ1u=μ2-μ1R
1.34v-1∞=1.34-17.8
v=30.7 mm
v=3.07 cm

Topic: Physics, Ray Optics, JEE Main, JEE Main 2019 (10 Jan Shift 2)
The eye can be regarded as a single refracting surface. The radius of curvature of this surface is equal to that of the cornea (7.8 mm). This surface separates two media of refractive indices 1 and 1.34. Calculate the distance from the refracting surface at which a parallel beam of light will come to focus.

Options:
A 4.0cm
B 1cm
C 2 cm
D 3.07 cm

The correct answer is:
3.07 cm

Solution:
We know eye lens is convex refracting surface.
Use expression for refraction from curved surface
μ1→ Refractive index of medium having incident ray
μ2→ Refractive index of a medium having reflected ray
μ2v-μ1u=μ2-μ1R
1.34v-1∞=1.34-17.8
v=30.7 mm
v=3.07 cm

Topic: Physics, Ray Optics, JEE Main
The F-Br-F bond angles in BrF5 and the Cl-P-Cl bond angles in PCl5, respectively, are

Options:
A identical in BrF5 but non-identical in PCl5
B identical in BrF5 and identical in PCl5

C non-identical in BrF5 but identical in PCl5
D non-identical in BrF5 and non-identical in PCl5

The correct answer is:
non-identical in BrF5 and non-identical in PCl5

Solution:
The geometry of BrF5 is square pyramidal.

Here, the lone pair occupies the axial position and hence axial bonds will suffer more repulsion than equatorial bonds. Thus, the axial Br-F bond length will be different from equatorial Br-F.

The geometry of PCl5 is triangular pyramidal.

The axial bonds suffer more repulsions than equatorial bonds, so they are larger in bond length.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The 'f' orbitals are half and completely filled, respectively in lanthanide ions [Given : Atomic no. Eu,63:Sm,62:Tm,69;Tb,65;Yb,70;Dy,66]

Options:
A Eu2+ and Tm2+
B Sm2+ and Tm3+
C Tb4+ and Yb2+
D Dy3+ and Yb3+

The correct answer is:
Tb4+ and Yb2+

Solution:
Ion	Electronic configuration

Eu2+	4f7

Tm2+	4f13

Sm2+	4f6

Tm3+	4f12

Tb4+	4f7

Yb2+	4f14

Dy3+	4f9

Yb3+	4f13

Hence, the ions with half and completely filled configurations are Tb4+ and Yb2+

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2022 (27 Jun Shift 2)
The factor which does not affect the rate of alveolar diffusion is

Options:
A Thickness of the membranes
B Pressure gradient
C Concentration gradient
D Reactivity of the gases

The correct answer is:
Reactivity of the gases

Solution:
Alveoli (thin, irregular-walled and vascularized bag-like structure at the end of bronchiole) are the primary sites of exchange of gases. O2 and CO2 are exchanged in these sites by simple diffusion mainly based on pressure/concentration gradient. Solubility of the gases as well as thickness of the membranes involved in diffusion are also some important factors that can affect the rate of diffusion. Reactivity of the gases does not affect the rate of alveolar diffusion.

Topic: Biology, Breathing and Exchange of Gases, NEET
The factor which is not affecting the conductivity of any solution is

Options:
A Dilution
B Nature of electrolyte
C Temperature
D None of these

The correct answer is:
None of these

Solution:
Dilution, temperature and nature of electrolyte affect the conductivity of solution.

Topic: Chemistry, Electrochemistry, JEE Main