datavorous/entrance-exam-dataset huggingface dataset:
Apical dominance in plants is caused by

Options:
A high concentration of auxins in the terminal bud.
B high concentration of gibberellins in the apical bud.
C high concentration of auxins in the lateral bud.
D absence of auxins and gibberellins in the apical bud.

The correct answer is:
high concentration of auxins in the terminal bud.

Solution:
Apical dominance is the phenomenon in which main axis or terminal bud continuous to grow suppressing the growth of lateral bud. High concentration of auxin in the terminal bud promotes its growth and it increases ethylene formation in lateral bud which suppresses its growth.

Topic: Biology, Plant - Growth and Development, NEET
Apparatus used to find out the velocity of sound in gas is

Options:
A Melde's apparatus
B Kundt's tube
C Quincke's tube
D None of these

The correct answer is:
Kundt's tube

Solution:
Kundt's tube

Topic: Physics, Waves and Sound, JEE Main
Appearance of blood red colour, on treatment of the sodium fusion extract of an organic compound with FeSO4 in presence of concentrated H2SO4 indicates the presence of element/s

Options:
A Br
B N
C N and S
D S

The correct answer is:
N and S

Solution:
The sodium fusion test, or Lassaigne's test, is used in elemental analysis for the qualitative determination of the presence of foreign elements, namely halogens, nitrogen, and sulphur, in an organic compound.

In Lassaigne's test, if both N and S are present in the organic compound, they are converted to NaSCN.

If SCN- is added to a solution with iron(III) ions, a blood-red solution forms mainly due to the formation of [Fe(SCN)(H2O)5]2+

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2024 (29 Jan Shift 1)
Appearance of vegetative propagules from the nodes of plants such as sugarcane and ginger is mainly because

Options:
A
nodes are shorter than internodes

B
nodes have meristamatic cells

C
nodes are located near the soil

D
nodes have non-photosynthetic cells

The correct answer is:
nodes have meristamatic cells

Solution:
- The vegetative propagules appears from the nodes of plants such as sugarcane and ginger is mainly because of the nodes having meristematic cells.
These cells are responsible to control the growth and development of tissues and organs in plants. Nodes which are present in the modified stems, when come in contact with damp soil or water, they produce roots and gives rise to new plants.

Topic: Biology, Reproduction in Organisms
Applying the principle of homogeneity of dimensions, determine which one is correct,
where $T$ is time period, $G$ is gravitational constant, $M$ is mass, $r$ is radius of orbit.

Options:
A $T^2=\frac{4 \pi^2 r^2}{G M}$
B $T^2=\frac{4 \pi^2 r}{G M^2}$
C $T^2=\frac{4 \pi^2 r^3}{G M}$
D $T^2=4 \pi^2 r^3$

The correct answer is:
$T^2=\frac{4 \pi^2 r^3}{G M}$

Solution:
According to principle of homogeneity dimension of LHS should be equal to dimensions of RHS so option (3) is correct.
$\begin{aligned}
& \mathrm{T}^2=\frac{4 \pi^2 \mathrm{r}^3}{\mathrm{GM}} \\
& {\left[\mathrm{T}^2\right]=\frac{\left[\mathrm{L}^3\right]}{\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right][\mathrm{M}]}}
\end{aligned}$
(Dimension of $\mathrm{G}$ is $\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]$ )
$\left[\mathrm{T}^2\right]=\frac{\left[\mathrm{L}^3\right]}{\left[\mathrm{L}^3 \mathrm{~T}^{-2}\right]}=\left[\mathrm{T}^2\right]$

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2024 (04 Apr Shift 2)
Aqua regia is used for dissolving noble metals (Au, Pt, etc.). The gas evolved in this process is

Options:
A NO
B N2O5
C N2
D N2O3

The correct answer is:
NO

Solution:
Aqua regia is HNO3 : HCl

1 : 3

HNO3+3HCl → 2H2O+NO+3Cl

(i) Au+3Cl → AuCl3→HClHAuCl4

(ii) Pt+4Cl → PtCl4 →HClH2PtCl6

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2020 (03 Sep Shift 1)
Aqueous solution of Ni2+ contains [Ni(H2O)6]2+ and its magnetic moment is 2.83 BM. When ammonia is added in it, comment on the magnetic moment of solution

Options:
A It will remain same
B It increases from 2.83 BM
C It decreases from 2.83 BM
D It cannot be predicted theoretically

The correct answer is:
It will remain same

Solution:
On adding ammonia in aqueous solution of Ni2+ containing [Ni(H2O)6]2+ a new complex is formed as follows
[Ni(H2O)6]2+sp3d2Hyb.μeff=2.8BM+6NH3⇌[Ni(NH3)6]2+sp3d2Hyb.μeff=2.8BM+6H2O
As there is no change in hybridisation and number of unpaired electrons so magnetic moment remains same.

Topic: Chemistry, Coordination Compounds, NEET
Aqueous solution of Ni2+ contains [Ni (H2O)6]2+ and its magnetic moment is 2.83 BM when ammonia is added in it , commment on the magnetic moment of solution.

Options:
A It decreases from 2.83 BM
B It increases from 2.83 BM
C It will remains same
D Cannot be predicted by given information

The correct answer is:
It will remains same

Solution:
Ammonia is consered a strong ligand generally, but in reality, strength of a ligand depends on the metal it is interacting with. As it stands, NH3 isn't a strong enough ligand to cause pairing up in Ni, hence the magnetic moment remains the same.
Ni H 2 O 6 2 + + 6 NH 3 â Ni NH 3 6 2 + + 6 H 2 O
sp 3 d 2 Hybridisation sp 3 d 2 Hybridisation
Î¼ = 2.8 BM Î¼ = 2.8 BM

Topic: Chemistry, Coordination Compounds, JEE Main
Aqueous solution of orthoboric acid can be titrated against sodium hydroxide using phenolphthalein indicator only in presence of:

Options:
A trans-glycerol
B catechol
C cis-glycerol
D both B and C

The correct answer is:
both B and C

Solution:
It forms a stable cyclic complex with polyhydroxy compounds like cis-glycerol, catechol, etc. This helps in the release of H+as H3O+ therefore, boric acid acts as a strong acid and hence can be titrated with sodium hydroxide using phenolphthalein.

Topic: Chemistry, Redox Reactions, JEE Main
Aqueous solution of which of the following boron compounds will be strongly basic in nature?

Options:
A NaBH4
B LiBH4
C B2H6
D Na2B4O7

The correct answer is:
Na2B4O7

Solution:
Na2B4O7 gives H3BO3 and NaOH (strong base) in water.

So, it is a salt of strong base and weak acid. So, its solution will be strongly basic in water. It will have very high pH.

Lithium borohydride and sodium borohydride reacts with water to produce hydrogen.

And diborane act as lewis acid.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main, JEE Main 2022 (29 Jun Shift 2)
Aqueous solution of which of the following salts will not contain ions with the electronic configuration 1s22s22p63s23p6 ?

Options:
A NaF
B KBr
C NaCl
D CaI2

The correct answer is:
NaF

Solution:
NaF →Na+→1s2 2s2 2p6F-→1s2 2s2 2p6 Not Match
KBr →K+=1s22s22p63s23p6Br-=1s22s22p63s23p64s24p6
NaCl →Na+=1s22s22p6Cl=1s22s22p63s23p6
CaI2→Ca+2=1s22s22p63s23p6I-=Kr4d105s25p6

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2016 (10 Apr Online)
Archaebacteria can survive better than other bacteria in unfavourable conditions because of their

Options:
A Food
B Mesosomes
C Cell wall
D Plasmids

The correct answer is:
Cell wall

Solution:
The cell walls of archaebacteria are distinctive from those of eubacteria. Archaebacterial cell walls are composed of different polysaccharides and proteins, with no peptidoglycan. Many archaebacteria have cell walls made of the polysaccharide pseudomurein. The cell wall of archaea, as of any other prokaryote, is surrounding the cell outside the cytoplasmic membrane and is mediating the interaction with the environment. In this regard, it can be involved in cell shape maintenance, protection against virus, heat, acidity or alkalinity.

Topic: Biology, Biological Classification, NEET
Archaebacteria differ from other bacteria in having a different___________

Options:
A Size
B Shape
C Cell wall
D Ribosome

The correct answer is:
Cell wall

Solution:
Archaebacteria differ from the bacteria in having different cell wall (Pseudopeptidoglycan) which is helpful for survival in extreme condition.

Organism	Composition
Bacteria	Peptidoglycans (N-Acetyl Muramic acid + N-Acetyl Glucosmaine)
Fungi	Chitin (also known as fungal cellulose)
Plants	Cellulose + Hemicellulose + Pectins + Proteins
Algae	Cellulose + Galactans + Mannans + Minerals like calcium carbonate

Topic: Biology, Biological Classification, NEET
________ are important decomposers that cause decay and decomposition of dead bodies of plants and animals.

Options:
A Saprotrophic bacteria
B Saprotrophic fungi
C Earthworms
D Both A and B

The correct answer is:
Both A and B

Solution:
The organic content of the dead and decaying organisms is broken down to inorganic substances by the process called decomposition. It is brought about by decomposers. Decomposition is mainly an oxygen dependent process and its rate is dependent on chemical composition and climatic factors. Decomposers play a vital role in energy flow through the ecosystem and also in making available the essential nutrients to the producers. Saprophytes are organisms that show saprophytic nutrition where the organisms digest decaying organic matter extracellularly. Bacteria and fungi show saprophytic nutrition and participate in the decomposition process in the food chain.

Topic: Biology, Ecosystem, NEET
Are the following set of ordered pairs functions? If so examine whether the mapping is injective or surjective.
(i) $\{(x, y)$ : $x$ is a person, $y$ is the mother of $x\}$.
(ii) $\{(\mathrm{a}, \mathrm{b})$ : a is a person, $\mathrm{b}$ is an ancestor of a $\}$.

Options:

None

Solution:
(i) Given set of ordered pair is $\{(x, y): x$ is a person, $y$ is the mother of $\mathrm{x}\}$.
It represents a function. Here, the image of distinct elements of $\mathrm{x}$ under $\mathrm{f}$ are not distinct, so it is not injective but it is surjective.
(ii) Set of ordered pairs $=\{(\mathrm{a}, \mathrm{b})$ : a is a person, $\mathrm{b}$ is an ancestor of a
Here, each element of domain does not have a unique image. So, it does not represent function.

Topic: Mathematics, Relations and Functions (Class 11)
are

Options:
A Position isomers
B Chain isomers
C Functional isomers
D Metamers

The correct answer is:
Metamers

Solution:
Compounds having the same molecular formula but different number of carbon atoms ( alkyl groups) on either side of functional group ( i.e., -O-,-S-, -NH-, -C(=O)-) are called metamers and the phenomenon is called metamerism. All isomeric esters are metamers.

Topic: Chemistry, General Organic Chemistry, JEE Main
Area between the curve $y=\cos x$ and $x$ - axis when $0 \leq x$ is

Options:
A $2$
B $4$
C $0$
D $3$

The correct answer is:
$4$

Solution:
$\begin{array}{l}y=\cos x \text {, When } x \in\left[0, \frac{\pi}{2}\right], \cos x \geq 0 \\\text { When } x \in\left[\frac{\pi}{2}, \frac{3 \pi}{2}\right], \cos x \leq 0 \\\text { When } x \in\left[\frac{3 \pi}{2}, 2 \pi\right], \cos x \geq 0\end{array}$
Thus required area is given by,
$\begin{aligned}\int_0^{\pi / 2} y d x & =\int_0^{\pi / 2} \cos x d x+\int_{\pi / 2}^{3 \pi / 2}(-\cos x) d x+\int_{3 \pi / 2}^{2 \pi} \cos x d x \\& =1+2+1=4 \text { sq. unit. }\end{aligned}$

Topic: Mathematics, Area Under Curves, JEE Main
Area bounded by curve $x y=c, x$ - axis between $x=1$ and $x=4$, is

Options:
A $c \log 3$ sq. unit
B $2 \log c_{\text {sq. }}$. unit
C $2 c \log 2$ sq. unit
D ${2 c \log 5}$ sq. Unit

The correct answer is:
$2 c \log 2$ sq. unit

Solution:
Required area $=\int_1^4 y d x=c \int_1^4 \frac{1}{x} d x=2 c \log 2$ sq. unit.

Topic: Mathematics, Area Under Curves, JEE Main
Area bounded by curve $y=x^3, x$ - axis and ordinates $x=1$ and $x=4$, is

Options:
A $64$ sq. unit
B $27$ sq. unit
C $\frac{127}{4}$sq. unit
D $\frac{255}{4}$ sq. Unit

The correct answer is:
$\frac{255}{4}$ sq. Unit

Solution:
$\text { Required area }=\int_1^4 x^3 d x=\left[\frac{x^4}{4}\right]_1^4=\frac{255}{4}$ sq. unit.

Topic: Mathematics, Area Under Curves, JEE Main
Area bounded by lines $y=2+x, y=2-x$ and $x=2$ is

Options:
A $3$
B $4$
C $8$
D $16$

The correct answer is:
$4$

Solution:
Obviously, triangle $A C B$ is right angled at $C$.
$\therefore$ Required area $=\frac{1}{2} \times A C \times B C$
$=\frac{1}{2} \times 2 \sqrt{2} \times 2 \sqrt{2}=4$sq. unit.

Topic: Mathematics, Area Under Curves, JEE Main
Area bounded by parabola $y^2=x$ and straight line $2 y=x$ is

Options:
A $\frac{4}{3}$
B $1$
C $\frac{2}{3}$
D $\frac{1}{3}$

The correct answer is:
$\frac{4}{3}$

Solution:
$\begin{aligned}y^2=x \text { and } 2 y=x \Rightarrow y^2 & =2 y \Rightarrow y=0,2 \\\therefore \text { Required area } & =\int_0^2\left(y^2-2 y\right) d y=\left(\frac{y^3}{3}-y^2\right)_0^2=\frac{4}{3}\end{aligned}$

Topic: Mathematics, Area Under Curves, JEE Main
Area bounded by the curve $y=\log x$ and the coordinate axes is

Options:
A 2
B 1
C 5
D $2 \sqrt{2}$

The correct answer is:
$2 \sqrt{2}$

Solution:
Observing the graph of $\log x$, we find that the required area lies below $\mathrm{X}$-axis between $\mathrm{x}=0$ and $x=1$.
$$
\begin{aligned}
& \text { Sorequired area }=\left|\int_0^1 \log x d x\right|=\left.|(x \log x-x)|\right|_0 ^1 \\
& =|-1|=1
\end{aligned}
$$

Topic: Mathematics, Area Under Curves, JEE Main
Area bounded by the curve $y=\log x, x$ - axis and the ordinates $x=1, x=2$ is

Options:
A $\log 4$ sq. unit
B $(\log 4+1)$ sq. unit
C $(\log 4-1)$ sq. unit
D None of these

The correct answer is:
$(\log 4-1)$ sq. unit

Solution:
$\begin{aligned} & \text { Given curve } y=\log x \text { and } x=1, x=2 . \\ & \text { Hence required area }=\int_1^2 \log x d x=(x \log x-x)_1^2 \\ & =2 \log 2-1=(\log 4-1) \text { sq. Unit. }\end{aligned}$

Topic: Mathematics, Area Under Curves, JEE Main
Area bounded by the curve $y=\sin x$ between $x=0$ and $x=2 \pi$ is

Options:
A $2$ sq. unit
B $4$ sq. unit
C $8$ sq. unit
D None of these

The correct answer is:
$4$ sq. unit

Solution:
We have $y=\sin x$
$\begin{array}{|c|c|c|c|c|c|c|} \hline x & 0 & \pi / 6 & \pi / 2 & \pi & 3 \pi / 2 & 2 \pi \\ \hline y & 0 & 0.5 & 1 & 0 & -1 & 0 \\ \hline\end{array}$
Join these points with a free hand to obtain a rough sketch
Required area $=($ area of $O A B)+($ area of $B C D)$
$=\int_0^\pi y d x+\int_\pi^{2 \pi}(-y) d x$
(Area $B C D$ is below $x$-axis)
$=\int_0^\pi \sin x d x-\int_\pi^{2 \pi} \sin x d x=4$ sq. unit.

Topic: Mathematics, Area Under Curves, JEE Main
Area bounded by the curve $y=x e^{x^2}, x$ - axis and the ordinates $x=0, x=a$

Options:
A $\frac{e^{a^2}+1}{2}$sq. unit
B $\frac{e^{a^2}-1}{2}$sq. unit
C $e^{a^2}+1$ sq. unit
D $e^{a^2}-1$ sq. Unit

The correct answer is:
$\frac{e^{a^2}-1}{2}$sq. unit

Solution:
Required area is $\int_0^a y d x=\int_0^a x e^{x^2} d x$
We put $x^2=t \Rightarrow d x=\frac{d t}{2 x}$ as $x=0 \Rightarrow t=0$ and $x=a \Rightarrow t=a^2$, then it reduces to $\frac{1}{2} \int_0^{a^2} e^t d t=\frac{1}{2}\left[e^t\right]_0^{a^2}=\frac{e^{a^2}-1}{2}$sq. unit.

Topic: Mathematics, Area Under Curves, JEE Main
Area bounded by the lines $y=x, x=-1, x=2$ and $x$-axis is

Options:
A $\frac{5}{2}$ sq. unit
B $\frac{3}{2}$ sq. unit
C $\frac{1}{2}$ sq. unit
D None of these

The correct answer is:
$\frac{5}{2}$ sq. unit

Solution:
Required area $\int_{-1}^2 y d x=\int_{-1}^0 y \cdot d x+\int_0^2 y . d x=\frac{5}{2}$ sq. unit.

Topic: Mathematics, Area Under Curves, JEE Main
Area bounded by $y=x \sin x$ and $x$ - axis between $x=0$ and $x=2 \pi$, is

Options:
A $0$
B $2 \pi$ sq. unit
C $\pi$ sq. unit
D $4 \pi$ sq. unit

The correct answer is:
$4 \pi$ sq. unit

Solution:
Required area is
$A_1+A_2=\int_0^\pi y d x+\left|\int_\pi^{2 \pi} y d x\right|=4 \pi \text { sq.unit }$

Topic: Mathematics, Area Under Curves, JEE Main
Area enclosed between the curves \( |\mathrm{y}|=1-\mathrm{x}^{2} \) and \( \mathrm{x}^{2}+\mathrm{y}^{2}=1 \) is

Options:
A \( \frac{3 \pi-8}{3} \) sq. units
B \( \frac{\pi-8}{3} \) sq. units
C \( \frac{2 \pi-8}{3} \) sq. units
D None of these

The correct answer is:
\( \frac{3 \pi-8}{3} \) sq. units

Solution:
The dotted area is
A=∫011-x2dx=x-x3301=1-13=23
Hence, area bounded by circle x2+y2=1 and y=1-x2
=Lined area
=Area of curcle-Area bounded by y=1-x2
=π-4.23=3π-83 sq. units

Topic: Mathematics, Area Under Curves, JEE Main
Area (in sq. units) of the region outside x2+y3=1 and inside the ellipse x24+y29=1 is

Options:
A 6π−2
B 3π-2
C 34-π
D 64−π

The correct answer is:
6π−2

Solution:
Area of ellipse =πab=6π
Area enclosed by |x|2+|y|3=1 is
=4×1223=12
So, required area=6π-12 sq. units

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2020 (02 Sep Shift 1)
Area of retina in human eye containing only cone cells is called

Options:
A optic disc
B tapetum lucidum
C macula lutea
D fovea centralis

The correct answer is:
fovea centralis

Solution:
The innermost layer of the eye made of think, delicate and transparent membrane is called retina. It has three regions - optic disc, macula lutea and peripheral retina. The retina has several layers including the layer of special cells called photoreceptors (rod cells and cone cells). Rod cells help with motion detection and low light vision along with black-white visuals. The photoreceptor cells called as cone cells are mainly concentrated in a small area on retina called as macula. A small depression called fovea centralis is found in the macula region which is concentrated with only cone cells. This section of retina is responsible for maximum visual acuity and color vision.

Topic: Biology, Neural Control and Coordination, NEET
Area of the figure bounded by the parabola $y^2+8 x=16$ and $y^2-24 x=48$ is

Options:
A $\frac{11}{9}$ sq. unit
B $\frac{32}{3} \sqrt{6}$ sq. unit
C $\frac{16}{3}$ sq. unit
D $\frac{24}{5}$ sq. unit

The correct answer is:
$\frac{32}{3} \sqrt{6}$ sq. unit

Solution:
$$
\begin{gathered}
y^2=-8(x-2) \\
y^2=24(x+2) \cdots(2)
\end{gathered}
$$
Solving (1) and $(2) \Rightarrow x=-1$
Area $=2\left[\int_{-2}^{-1} 2 \sqrt{6} \sqrt{x+2} \mathrm{dx}+\int_{-1}^2 2 \sqrt{2} \sqrt{2-x} \mathrm{dx}\right]=\frac{32}{3} \sqrt{6}$ sq. unit

Topic: Mathematics, Area Under Curves, JEE Main
Area of the greatest rectangle that can be inscribed in the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is

Options:
A
$2 a b$

B
$a b$

C
$3a b$

D
$\frac{a}{b}$

The correct answer is:
$2 a b$

Solution:
Area of rectangle $A B C D=(2 a\cos \theta)$
$(2 b \sin \theta)=2 a b \sin 2 \theta$
$\Rightarrow$ Area of greatest rectangle is equal to
$2 a b$
when $\sin 2 \theta=1$.

Topic: Mathematics, Application of Derivatives, JEE Main
Area of the greatest rectangle that can be inscribed in the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is

Options:
A
$2 a b$

B
$a b$

C
$3a b$

D
$\frac{a}{b}$

The correct answer is:
$2 a b$

Solution:
Area of rectangle $A B C D=(2 a\cos \theta)$
$(2 b \sin \theta)=2 a b \sin 2 \theta$
$\Rightarrow$ Area of greatest rectangle is equal to
$2 a b$
when $\sin 2 \theta=1$.

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2005
Area of the region bounded by the curve $y=\tan x$, tangent drawn to the curve at $x=\frac{\pi}{4}$ and the $x$-axis is

Options:
A $\frac{1}{4}$
B $\log \sqrt{2}+\frac{1}{4}$
C $\log \sqrt{2}-\frac{1}{4}$
D None of these

The correct answer is:
None of these

Solution:
$\begin{aligned} & \text { Shaded area }=\int_0^{\pi / 2} \tan x d x=[\log \sec x]_0^{\pi / 4} \\ & =\log \sec (\pi / 4)-\log \sec 0=\log \sqrt{2}-\log 1=\log \sqrt{2}\end{aligned}$

Topic: Mathematics, Area Under Curves, JEE Main
Area of the region x,y:x2+y-22≤4,x2≥2y is

Options:
A π+83
B 2π+163
C π-83
D 2π-163

The correct answer is:
2π-163

Solution:
Given,

x2+y-22≤4 & x2≥2y

Now on solving, x2+y-22=4 & x2=2y we get intersecting point of the curves as ±2,2,

Now plotting the diagram we get,

So, from above diagram area of required region will be,

=214π×22-∫022·ydy

=2π−2⋅y323202

=2π-223·22

=2π-83

=2π-163

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2023 (11 Apr Shift 1)
Area under the curve $y=\sqrt{3 x+4}$ between $x=0$ and $x=4$, is

Options:
A $\frac{56}{9}$sq. unit
B $\frac{64}{9}$ sq. unit
C $8$ sq. unit
D None of these

The correct answer is:
None of these

Solution:
Area $=\int_0^4 \sqrt{3 x+4} d x=\left|\frac{(3 x+4)^{3 / 2}}{3 .(3 / 2)}\right|_0^4$
$=\frac{2}{9} \times 56=\frac{112}{9} s q .$ unit.

Topic: Mathematics, Area Under Curves, JEE Main
Area under the curve $y=\sin 2 x+\cos 2 x$ between $x=0$ and $x=\frac{\pi}{4}$, is

Options:
A 2 sq. unit
B 1 sq. unit
C 3 sq. unit
D 4 sq. unit

The correct answer is:
1 sq. unit

Solution:
$\begin{aligned}& \text { Required area }=\int_0^{\pi / 4}(\sin 2 x+\cos 2 x) d x \\& =\left[-\frac{\cos 2 x}{2}+\frac{\sin 2 x}{2}\right]_0^{\pi / 4} \\& =\frac{1}{2}\left[-\cos \frac{\pi}{2}+\sin \frac{\pi}{2}+\cos 0-\sin 0\right]=1 s q unit.\end{aligned}$

Topic: Mathematics, Area Under Curves, JEE Main
Area under the curve $y=x^2-4 x$ within the $x$-axis and the line $x=2$, is

Options:
A $\frac{16}{3}$ sq.unit
B $-\frac{16}{3}$ sq.unit
C $\frac{4}{7}$ sq.unit
D Cannot be calculated

The correct answer is:
$\frac{16}{3}$ sq.unit

Solution:
$\int_0^2\left(x^2-4 x\right) d x=\left[\frac{x^3}{3}-\frac{4 x^2}{2}\right]_0^2=\frac{16}{3} s q .$unit.

Topic: Mathematics, Area Under Curves, JEE Main
$\operatorname{Arg}\left(\sin \frac{6 \pi}{5}+i\left(1+\cos \frac{6 \pi}{5}\right)\right)=$

Options:
A $\frac{5 \pi}{6}$
B $\frac{6 \pi}{5}$
C $\frac{2 \pi}{5}$
D $\frac{9 \pi}{10}$

The correct answer is:
$\frac{9 \pi}{10}$

Solution:
$\begin{aligned} & \operatorname{Arg}\left(\sin \frac{6 \pi}{5}+i\left(1+\cos \frac{6 \pi}{5}\right)\right) \\ = & \operatorname{Arg}\left(2 \sin \frac{3 \pi}{5} \cdot \cos \frac{3 \pi}{5}+i \cdot 2 \cos ^2 \frac{3 \pi}{5}\right) \\ = & \operatorname{Arg}\left[\left(2 \cos \frac{3 \pi}{5}\right)\left(\sin \frac{3 \pi}{5}+i \cos \frac{3 \pi}{5}\right)\right] \\ = & \operatorname{Arg}\left[\left(2 \cos \frac{3 \pi}{5}\right)\left(\sin \left(\frac{\pi}{2}+\frac{\pi}{10}\right)+i \cos \left(\frac{\pi}{2}+\frac{\pi}{10}\right)\right)\right] \\ = & \operatorname{Arg}\left[\left(2 \cos \frac{3 \pi}{5}\right)\left(\cos \frac{\pi}{10}-i \sin \frac{\pi}{10}\right)\right] \\ \therefore & \quad \operatorname{Arg}\left(Z_1 \cdot Z_2\right)=\arg Z_1+\arg Z_2\end{aligned}$
$=0-\frac{\pi}{10}=\frac{-\pi}{10}$ or $\pi-\frac{\pi}{10}=\frac{9 \pi}{10}$.

Topic: Mathematics, Complex Number, JEE Main
Arginine is nonessential amino acid which is present is human body. How many Nitrogen atoms are present in sp2 hybrid state in Arginine.

Options:

The correct answer is:
3.00

Solution:

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Argon is used

Options:
A in filling airships
B to obtain low temperature
C in high temperature welding
D in radiotherapy for treatment of cancer

The correct answer is:
in high temperature welding

Solution:
Argon is used in high temperature welding and other operations which require a non-oxidising atmosphere and the absence of nitrogen.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Argon is used in arc welding because of its

Options:
A low reactivity with metal
B ability to lower the melting point of metal
C flammability
D high calorific value

The correct answer is:
low reactivity with metal

Solution:
Argon is used mainly to provide an inert atmosphere in high temperature metallurgical (arc welding of metals/alloys) extraction.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
Argument of $\frac{1-i \sqrt{3}}{1+i \sqrt{3}}$ is

Options:
A $210^{\circ}$
B $120^{\circ}$
C $240^{\circ}$
D $60^{\circ}$

The correct answer is:
$240^{\circ}$

Solution:
$z=\frac{1-i \sqrt{3}}{1+i \sqrt{3}} \times \frac{1-i \sqrt{3}}{1-i \sqrt{3}}=\frac{-2-2 \sqrt{3} i}{4}=-\frac{1}{2}-\frac{1}{2} \sqrt{3} i$
$\begin{aligned} & \text { Now Arg }(Z)=\pi+\tan ^{-1}\left|\frac{\operatorname{Im}(Z)}{\operatorname{Re}(Z)}\right| \\ & =180^{\circ}+\tan ^{-1}(\sqrt{3})=180^{\circ}+60^{\circ}=240^{\circ}\end{aligned}$

Topic: Mathematics, Complex Number, JEE Main