datavorous/entrance-exam-dataset huggingface dataset:
An urn contains 25 balls of which 10 balls bear a mark ' $\mathrm{X}$ 'and the remaining 15 bear a mark ' $\mathrm{Y}$ '. A ball is drawn at random from the urn, its mark is noted down and it is replaced. If 6 balls are drawn in this way, find the probability that
(i) all will bear ' $\mathrm{X}$ ' mark.
(ii) not more than 2 will bear ' $Y$ ' mark.
(iii) at least one ball will bear ' $Y$ ' mark.
(iv) the number of balls with ' $\mathrm{X}$ ' mark and ' $\mathrm{Y}$ ' mark will be equal.

Options:

None

Solution:
Total number of balls $=25$,
Number of balls marked ' $\mathrm{X}$ ' $=10$
Let $\mathrm{X}$ denote the event of getting a ball marked $\mathrm{X}, \mathrm{Y}=$ event of getting a ball $Y$
$\therefore \quad \mathrm{P}(\mathrm{X})=\frac{10}{25}=\frac{2}{5}=\mathrm{p}$
$\therefore \quad \mathrm{P}(\mathrm{Y})=1-\frac{2}{5}=\frac{3}{5}=\mathrm{q}$
Now 6 balls are drawn.
(i) $\mathrm{P}($ all the balls are marked $\mathrm{X})=\left(\frac{2}{5}\right)^6$
(ii) Event that not more than 2 will bear ' $\mathrm{Y}$ ' mark)
$\Rightarrow$ (6X, 0Y), $(5 \mathrm{X}, 1 \mathrm{Y}),(4 \mathrm{X}, 2 \mathrm{Y})$
$\therefore \mathrm{P}$ (not more than two will bear ' $\mathrm{Y}$ ' mark)
$=\mathrm{P}(6)+\mathrm{P}(5)+\mathrm{P}(4)$
$=\left(\frac{2}{5}\right)^6+{ }^6 \mathrm{C}_5\left(\frac{3}{5}\right)\left(\frac{2}{5}\right)^5+{ }^6 \mathrm{C}_4\left(\frac{3}{5}\right)^2\left(\frac{2}{5}\right)^4$
$=\left(\frac{2}{5}\right)^4\left[\frac{4}{25}+\frac{6 \times 3 \times 2}{25}+\frac{15 \times 9}{25}\right]$
$$
\begin{aligned}
&=\left(\frac{2}{5}\right)^4\left[\frac{4+36+135}{25}\right] \\
&=\left(\frac{2}{5}\right)^4\left[\frac{175}{25}\right]=7\left(\frac{2}{5}\right)^4
\end{aligned}
$$
(iii) Event that at last 1 ball will bear ' $\mathrm{Y}$ ' $=\{(5 \mathrm{X}, 1 \mathrm{Y}),(4 \mathrm{X}, 2 \mathrm{Y}),(3 \mathrm{X}, 3 \mathrm{Y})$, $(2 \mathrm{X}, 4 \mathrm{Y}),(\mathrm{X}, 5 \mathrm{Y}),(6 \mathrm{X}, 6 \mathrm{Y})\}$
$\mathrm{P}$ (at least 1 ball will bear mark $\mathrm{Y}$ $=1-\mathrm{P}$ (no ball will bear mark $\mathrm{Y})$ $=1-\mathrm{P}($ all balls bear mark $\mathrm{X})=1-\left(\frac{2}{5}\right)^6$
(iv) $\mathrm{P}(3$ balls each are marked $\mathrm{X}$ and $\mathrm{Y})$
$$
\begin{aligned}
P(3) &={ }^6 C_3\left(\frac{3}{5}\right)^3\left(\frac{2}{5}\right)^3 \\
&=\frac{6 \times 5 \times 4}{6} \times \frac{27}{125} \times \frac{8}{125}=\frac{864}{3125}
\end{aligned}
$$

Topic: Mathematics, Probability
An urn contains nine balls of which three are red, four are blue and two are green. Three balls are drawn at random without replacement from the urn. The probability that the three balls have different colour is

Options:
A
$\frac{2}{7}$

B
$\frac{1}{21}$

C
$\frac{2}{23}$

D
$\frac{1}{3}$

The correct answer is:
$\frac{2}{7}$

Solution:
$\mathrm{n}(\mathrm{S})={ }^9 \mathrm{C}_3$
$\mathrm{n}(\mathrm{E})={ }^3 \mathrm{C}_1 \times{ }^4 \mathrm{C}_1 \times{ }^2 \mathrm{C}_1$
Probability $=\frac{3 \times 4 \times 2}{{ }^9 \mathrm{C}_3}=\frac{24 \times 3 !}{9 !} \times 6 !=\frac{24 \times 6}{9 \times 8 \times 7}=\frac{2}{7}$

Topic: Mathematics, Probability, JEE Main, JEE Main 2010
An X-ray tube produces a continuous spectrum of radiation with a cutoff wavelength of 45Ã10-3 nm. The maximum energy of a photon in the radiation (in eV) is
[h=6.62Ã10-34 J s, c=3Ã108 m s-1]

Options:
A 27500
B 22500
C 17500
D 12500

The correct answer is:
27500

Solution:
E=hcÎ»
=6.62Ã10-34Ã3Ã108 45Ã10-12 J
=0.44Ã10-141.6Ã10-19 eV=0.275Ã105 eV=27500 eV

Topic: Physics, Atomic Physics, JEE Main
An X-ray tube is operated at 1.24 million volt. The shortest wavelength of the produced photon will be:

Options:
A 10-3nm
B 10-2nm
C 10-4nm
D 10-1nm

The correct answer is:
10-3nm

Solution:
λmin=1240ΔVnm

=12401.24×106=10-3nm

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2021 (24 Feb Shift 2)
An X-ray tube is operated at $15 \mathrm{kV}$. Calculate the upper limit of the speed of the electrons striking the target.

Options:
A $7.26 \times 10^{7} \mathrm{~m} / \mathrm{s}$
B $7.62 \times 10^{7} \mathrm{~m} / \mathrm{s}$
C $7.62 \times 10^{7} \mathrm{~cm} / \mathrm{s}$
D $7.26 \times 10^{9} \mathrm{~m} / \mathrm{s}$

The correct answer is:
$7.26 \times 10^{7} \mathrm{~m} / \mathrm{s}$

Solution:
The maximum kinetic energy of an electron accelerated through a potential difference of $\mathrm{V}$ volt is $\frac{1}{2} \mathrm{mv}^{2}=\mathrm{eV}$
$$
\begin{array}{l}
\therefore \text { maximum velocity } \mathrm{v}=\sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}} \\
\mathrm{v}=\sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 15000}{9.1 \times 10^{-31}}} \\
\mathrm{v}=7.26 \times 10^{7} \mathrm{~m} / \mathrm{s}
\end{array}
$$

Topic: Physics, Electromagnetic Waves, JEE Main
Analyse the events during every stage of cell cycle and notice how the following two parameters change
Number of chromosomes $(\mathrm{N})$ per cell
ii Amount of DNA content $($ C $)$ per cell

Options:

None

Solution:
$2 \mathrm{~N}$ corresponds to diploid chromosomes where as $2 \mathrm{C}$ corresponds to diploid.
(i) Number of chromosomes doubles after mitotic cell division and becomes half after meiotic cell division.
(ii) During $\mathrm{S}$ phase the DNA content doubles, but number of chromosomes remains the same.

Topic: Biology, Cell Cycle and Cell Division
Analyse the following reaction and identify the nature of $A$ and $B$

Options:
A

B

C

D

The correct answer is:

Solution:
$\xrightarrow{\mathrm{HBr}} A$
Formation of A is a electrophilic addition reaction
Formation of B is a free radical addition reaction

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Analysis show that iron oxide consist of Iron ion with 94% ions having d6 configuration and 6% having d5 configuration. Which amongst the following best represents the formula of the oxide?

Options:
A Fe0.97O
B Fe1.03O
C Fe0.60O
D Fe0.94O0.94

The correct answer is:
Fe0.97O

Solution:
d6→Fe+2,d5→Fe+3
Total charge of Iron
=0.94×2+0.06×3=1.88+0.18=2.06
Number of O–2 ion=2.062=1.03
Formula of solid =FeO1.03=Fe0.97O

Topic: Chemistry, Hydrocarbons, JEE Main
Analysis show that nickel oxide consists of nickel ion with 96% ions having d8 configuration and 4% having of d7 configuration. Which amongst the following best represents the formula of the oxide.

Options:
A Ni1.02O1.00
B Ni0.96O1.00
C Ni0.98O0.98
D Ni0.98O1.00

The correct answer is:
Ni0.98O1.00

Solution:
d8âNi+2, d7âNi+3
Total charge of nickel
0.96Ã2+0.04Ã3=2.04
Number of O-2 ion =2.042=1.02
Formula of solid =NiO1.02=Ni0.98O1.00

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
Analysis shows that nickel oxide has the formula $\mathrm{Ni}_{0.98} \mathrm{O}_{1.00}$. The fractions $\mathrm{Ni}^{2+}$ and $\mathrm{Ni}^{3+}$ ions in the crystal are ......

Options:
A $\mathrm{Ni}^{2+}=98 \%$ and $\mathrm{Ni}^{3+}=2 \%$
B $\mathrm{Ni}^{2+}=2 \%$ and $\mathrm{Ni}^{3+}=98 \%$
C $\mathrm{Ni}^{2+}=4 \%$ and $\mathrm{Ni}^{3+}=96 \%$
D $\mathrm{Ni}^{2+}=96 \%$ and $\mathrm{Ni}^{3+}=4 \%$

The correct answer is:
$\mathrm{Ni}^{2+}=96 \%$ and $\mathrm{Ni}^{3+}=4 \%$

Solution:
It is given that, nickel oxide has the formula as $\mathrm{Ni}_{0.98} \mathrm{O}_{1.00}$.
As per the formula, there are $98 \mathrm{Ni}$ ions for 100 oxide ions. Out of $98 \mathrm{Ni}$ ions, let $x$ ions be in +2 oxidation state.
$(98-x)$ ions will be in +3 oxidation state.
Oxide ion has -2 charge.
To maintain electrical neutrality, total positive charge on cations = total negative charge on anions.
$$
\begin{aligned}
2 x+3(98-x)+100(-2) & =0 \\
x & =94
\end{aligned}
$$
Fraction of $\mathrm{Ni}^{2+}$ ions $=\frac{94}{98}=0.96$
Fraction of $\mathrm{Ni}^{3+}$ ions $=\frac{4}{98}=0.04$
Hence, the fractions of nickel that exists as $\mathrm{Ni}^{2+}$ and $\mathrm{Ni}^{3+}$ are $\mathrm{Ni}^{2+}=96 \%$ and $\mathrm{Ni}^{3+}=4 \%$ respectively.

Topic: Chemistry, Solid State, AP EAMCET, AP EAMCET 2020 (22 Sep Shift 1)
Analyze the events occurring during every stage of the cell cycle, how the amount of DNA content (C) per cell changes and select the correct option.

Options:
A DNA content becomes doubled during S phase of cell cycle
B DNA content is reduced to half during anaphase
C DNA content remain same during meiosis I
D Both (a) and (b)

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

Solution:
Synthesis of DNA occurs by DNA replication and the amount of DNA per cell doubles (C becomes 2C). It is important to note that in spite of the doubling DNA, chromosome number remains the same.
(ii) During anaphase as the division of centromere takes place, the chromatids move to the opposite poles, causing the DNA content to become half (from 2C to C), however, the number of chromosomes remains the same.

Topic: Biology, Cell Cycle and Cell Division, NEET
Analyze the following pairs and identify the correct option given.

I. Chromoplasts - Contain pigments other than chlorophyll
II. Leucoplasts - Devoid of any pigments
III. Amyloplasts - Store proteins
IV. Aleuroplasts - Store oils and fats
V. Elaioplasts - Store carbohydrates

Options:
A II and III are correct
B III and IV are correct
C IV and V are correct
D I and II are correct

The correct answer is:
I and II are correct

Solution:
Plastids are mainly of two types:
(i) Coloured (including chromoplasts containing pigments other than chlorophyll and chloroplast containing green pigment chlorophyll).
(ii) Leucoplasts, which store reserve food material, these are devoid of any pigment and may be carbohydrate storing amyloplast, lipid storing elaioplast or protein storing proteinoplast (aleuroplast).

Topic: Biology, Cell - The Unit of Life, NEET
Analyze the picture given below and fill the labeling (a), (b) and (c).

Options:
A (a) Thinking, memory, behavior, movement, sight
(b) Maintenance of body temperature
(c) Breathing, heart rate
B (a) Breathing, heart rate, temperature
(b) Sight
(c) Thinking, memory, behavior, movement
C

(a) Thinking, memory, behavior, movement

(b) Sight

(c) Breathing, heart rate, temperature

D (a) Breathing, heart rate, thinking, memory, behavior, movement
(b) Maintenance of body temperature
(c) Sight

The correct answer is:

(a) Thinking, memory, behavior, movement

(b) Sight

(c) Breathing, heart rate, temperature

Solution:
The frontal lobe of the cerebrum controls all the voluntary activities of the body. It is the seat of memory, will, intelligence, reasoning, learning, behavior and movement. The temporal lobe is responsible for sight and the brain stem controls the breathing and respiration, regulates the heartbeat and maintains the temperature.

Topic: Biology, Neural Control and Coordination, NEET
Anatomically fairly old dicotyledonous root is distinguished from the dicotyledonous stem by :

Options:
A Position of protoxylem
B absence of secondary xylem
C Absence of secondary phloem
D Presence of cortex

The correct answer is:
Position of protoxylem

Solution:
No Solution Available

Topic: Biology, Anatomy of Flowering Plants, NEET, NEET 2009 (Mains)
Angle between the lines $\frac{x}{a}+\frac{y}{b}=1$ and $\frac{x}{a}-\frac{y}{b}=1$ is

Options:
A $2 \tan ^{-1} \frac{b}{a}$
B $\tan ^{-1} \frac{2 a b}{a^2+b^2}$
C $\tan ^{-1} \frac{a^2-b^2}{a^2+b^2}$
D None of these

The correct answer is:
$2 \tan ^{-1} \frac{b}{a}$

Solution:
The lines are $b x+a y-a b=0$ and $b x-a y-a b=0$
Hence the required angle is
$\tan ^{-1}\left|\frac{a b-(-a b)}{b^2+\left(-a^2\right)}\right|=\tan ^{-1}\left|\frac{2 a b}{b^2-a^2}\right|=2 \tan ^{-1} \frac{b}{a}$

Topic: Mathematics, Straight Lines, JEE Main
Angle between the planes $x+y+2 z=6$ and $2 x-y+z=9$ is

Options:
A $\frac{\pi}{4}$
B $\frac{\pi}{6}$
C $\frac{\pi}{3}$
D $\frac{\pi}{2}$

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

Solution:
Equations of planes are
$x+y+2 z-6=0$
and $\quad 2 x-y+z-9=0$
$\cos \theta=\frac{a_{1} a_{2}+b_{1}b_{2}+c
_{1}c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}$
$=\frac{1 \times 2+1 \times(-1)+2 \times 1}{\sqrt{1+1+4} \sqrt{4+1+1}}$
$=\frac{2-1+2}{\sqrt{6} \times \sqrt{6}}=\frac{3}{6}=\frac{1}{2}$
$\Rightarrow \quad \cos \theta=\cos \left(\frac{\pi}{3}\right)$
$\therefore \quad \theta=\frac{\pi}{3}$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
Angle between the tangents to the curve $y=x^2-5 x+6$ at the points $(2,0)$ and $(3,0)$ is

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

B
$\frac{\pi}{2}$

C
$\frac{\pi}{6}$

D
$\frac{\pi}{4}$

The correct answer is:
$\frac{\pi}{2}$

Solution:
$\frac{d y}{d x}=2 x-5$
$\therefore m_1=(2 x-5)_{(2,0)}=-1, m_2=(2 x-5)_{(3,0)}=1$
$\Rightarrow m_1 m_2=-1$

Topic: Mathematics, Parabola, JEE Main, JEE Main 2006
Angle of deviation $(\delta)$ by a prism (refractive index $=\mu$ and supposing the angle of prism $\mathrm{A}$ to be small) can be given by

Options:
A $\delta=(\mu-1) \mathrm{A}$
B $\delta=(\mu+1) A$
C $\delta=\frac{\sin \frac{A+\delta}{2}}{\sin \frac{A}{2}}$
D $\delta=\frac{\mu-1}{\mu+1} \mathrm{~A}$

The correct answer is:
$\delta=(\mu-1) \mathrm{A}$

Solution:
Correct option is 1. $\delta=(\mu-1) A$
The refractive index of a prism is given by,
$\mu=\frac{\sin \left(\frac{A+\delta}{2}\right)}{\sin \left(\frac{A}{2}\right)}$
For a very thin prism, we can assume that $\sin \theta \approx \theta$
$\begin{aligned} & \therefore \mu=\frac{\frac{A+\delta}{2}}{\frac{A}{2}} \\ & \text { or, } \mu=\frac{A+\delta}{A}\end{aligned}$
or, $\delta=\mu A-A$
or, $\delta=(\mu-1) A$ is the required angle of deviation.

Topic: Physics, Ray Optics, JEE Main
Angle of minimum deviation for a prism of refractive index 1.5 is equal to the angle of prism. The angle of prism is $\left(\cos 41^{\circ}=0.75\right)$

Options:
A $62^{\circ}$
B $41^{\circ}$
C $82^{\circ}$
D $31^{\circ}$

The correct answer is:
$82^{\circ}$

Solution:
By prism formula $n=\frac{\sin \frac{A+A}{2}}{\sin \frac{A}{2}}=\frac{2 \sin \frac{A}{2} \cos \frac{A}{2}}{\sin \frac{A}{2}}$
$\therefore \cos \frac{A}{2}=\frac{n}{2}=\frac{1.5}{2}=0.75=\cos 41^{\circ} \Rightarrow A=82^{\circ}$

Topic: Physics, Ray Optics, JEE Main
Angles of a triangle are in the ratio $4: 1: 1$. Then the ratio of its greatest side to its perimeter is

Options:
A $3:(2+\sqrt{3})$
B $\sqrt{3}:(2+\sqrt{3})$
C $\sqrt{3}:(2-\sqrt{3})$
D $1:(2+\sqrt{3})$

The correct answer is:
$\sqrt{3}:(2+\sqrt{3})$

Solution:
Let the angles of the triangle be $4 x, x$ and $x$.
$$
\therefore 4 x+x+x=180^{\circ} \Rightarrow 6 x=180^{\circ} \Rightarrow x=30^{\circ}
$$
By sine rule,
$$
\begin{aligned}
& \frac{\sin 120^{\circ}}{\mathrm{a}}=\frac{\sin 30^{\circ}}{\mathrm{b}}=\frac{\sin 30^{\circ}}{\mathrm{c}} \\
\therefore \quad & \mathrm{a}:(\mathrm{a}+\mathrm{b}+\mathrm{c}) \\
& =\left(\sin 120^{\circ}\right):\left(\sin 120^{\circ}+\sin 30^{\circ}+\sin 30^{\circ}\right) \\
& =\frac{\sqrt{3}}{2}: \frac{\sqrt{3}+2}{2}=\sqrt{3}: \sqrt{3}+2
\end{aligned}
$$

Topic: Mathematics, Sequences and Series, JEE Main
Angular momentum is conserved

Options:
A always
B never
C when external force is absent
D when external torque is absent

The correct answer is:
when external torque is absent

Solution:
According to the law of conservation of angular momentum, if there is no external torque on the system, then the angular momentum remains constant.

Topic: Physics, Rotational Motion, NEET
Angular momentum of a single particle moving with constant speed along circular path :

Options:
A remains same in magnitude but changes in the direction
B remains same in magnitude and direction
C is zero
D changes in magnitude but remains same in the direction

The correct answer is:
remains same in magnitude and direction

Solution:
The angular momentum is defined as the vector product of linear momentum and perpendicular distance, L→=r→×p→=r→×mv→, And direction will be upward & remain constant.

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2021 (31 Aug Shift 1)
Angular momentum of the particle rotating with a central force is constant due to

Options:
A
Constant Force

B
Constant linear momentum.

C
Zero Torque

D
Constant Torque

The correct answer is:
Zero Torque

Solution:
No Solution Available

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2007
Angular width $(\beta)$ of central maximum of a diffraction pattern on a single slit does not depend upon

Options:
A Distance between slit and source
B Wavelength of light used
C Width of the slit
D Frequency of light used

The correct answer is:
Distance between slit and source

Solution:
For single slit diffraction pattern $d \sin \theta=\lambda \quad$ ( $d=$ slit width)
Angular width $=2 \theta=2 \sin ^{-1}\left(\frac{\lambda}{d}\right)$
It is independent of $D$ i.e. distance between screen and slit

Topic: Physics, Wave Optics, JEE Main
Anhydrous $\mathrm{AlCl}_3$ is obtained from

Options:
A $\mathrm{HCl}$ and aluminium metal
B Aluminium and chlorine gas
C Hydrogen chloride gas and alumminium metal
D None of the above

The correct answer is:
Hydrogen chloride gas and alumminium metal

Solution:
$2 \mathrm{Al}+6 \mathrm{HCl} \rightarrow 2 \mathrm{AlCl}_3+3 \mathrm{H}_2$

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
Anhydrous ferric chloride is prepared by:

Options:
A Heating hydrated ferric chloride at a high temperature in a stream of air.
B Heating metallic iron in a stream of dry chlorine gas.
C Reaction of ferric oxide with HCl (aq).
D Reaction of metallic iron with HCl (aq).

The correct answer is:
Heating metallic iron in a stream of dry chlorine gas.

Solution:
(1)On heating we get FeCl 2 and Cl 2
2  2Fe+ 3Cl 2 → 2FeCl 3 (anhydrous);
3 , 4 do not give anhydrous FeCl3 but give hydrated FeCl3 .

Topic: Chemistry, d and f Block Elements, JEE Main
Aniline and methyl amine can be differentiated by

Options:
A Reaction with chloroform and aqueous solution of $\mathrm{KOH}$
B Diazotinium followed by coupling with phenol
C Reaction with
D None of these

The correct answer is:
Diazotinium followed by coupling with phenol

Solution:
Phenol react with aniline to give diazonium salt by coupling but Methyl amine not react with phenol.

Topic: Chemistry, Amines, JEE Main
Aniline dissolved in dilute HCl is reacted with sodium nitrite at 0°C. This solution was added dropwise to a solution containing an equimolar mixture of aniline and phenol in dilute HCl. The structure of the major product is

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Amines, JEE Main, JEE Main 2019 (09 Apr Shift 1)
Aniline in a set of reactions yielded a product ' $d$ ':
The structure of the product ' $d$ ' would be

Options:
A $\mathrm{C}_6 \mathrm{H}_5 \mathrm{NHOH}$
B $\mathrm{C}_6 \mathrm{H}_5 \mathrm{NHCH}_2 \mathrm{CH}_3$
C $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{NH}_2$
D $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{OH}$

The correct answer is:
$\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{OH}$

Solution:

Topic: Chemistry, Amines, JEE Main
Aniline is prepared in presence of Fe/HCI from

Options:
A Benzene
B Nitrobenzene
C Dinitrobenzene
D Aniline

The correct answer is:
Nitrobenzene

Solution:
Aniline is prepared by the reduction of nitrobenzene in acidic medium.

Topic: Chemistry, Amines, JEE Main
Aniline on treatment with conc. $\mathrm{HNO}_3+$ conc. $\mathrm{H}_2 \mathrm{SO}_4$ mixture yields

Options:
A o- and p-nitroanilines
B $m$-nitroanilines
C A black tarry matter
D No reaction

The correct answer is:
A black tarry matter

Solution:

Topic: Chemistry, Amines, JEE Main
Aniline on treatment with excess of bromine water gives

Options:
A Aniline bromide
B o-bromoaniline
C p-bromoaniline
D 2, 4, 6-tribromoaniline

The correct answer is:
2, 4, 6-tribromoaniline

Solution:

Topic: Chemistry, Amines, JEE Main
Aniline reacts with alkyl halide to give

Options:
A Amino compound
B Tertiary compound
C Quaternary ammonium compound
D Azomethane

The correct answer is:
Quaternary ammonium compound

Solution:

Topic: Chemistry, Amines, JEE Main
Aniline when diazotised in cold and then treated with dimethyl aniline gives a coloured product. Its structure would be:

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Amines, JEE Main
Animal husbandry and plant breeding programmes are the examples of:

Options:
A
development of embryo

B
homologous organs

C
fossils

D
analogous organs

The correct answer is:
fossils

Solution:
Palaeontological evidences for evolution refer to the evidences from fossils. The preserved remains or traces of organisms from the distant past are fossils and its study is called Palaeontology. The other options are not correct because the development patterns of embryo refer to embryological evidences for evolution.
Homologous and analogous organs provide evidences for comparative anatomy and morphology.

Topic: Biology, Evolution
Animal starch is the other name of

Options:
A amylose
B maltose
C glycogen
D amylopectin

The correct answer is:
glycogen

Solution:
Glycogen often referred to as "animal starch" is a multi-branched glucose polysaccharide that is used as a source of energy storage in animals, fungi and bacteria. Glycogen functions as one of two forms of energy reserves, glycogen being stored for the short term and triglyceride being stored in adipose tissue (i.e. body fat) for long-term storage. In humans, glycogen is produced and stored primarily in the liver and skeletal muscle cells.

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2022 (26 Jul Shift 2)
Animals having a built in thermostat to maintain constant body temperature are known as

Options:
A biothermic
B poikilothermic
C oligothermic
D homeothermic

The correct answer is:
homeothermic

Solution:
Homeothermic are the animals having a nearly uniform or constant body temperature.
These animals are known as warm blooded animals, eg, birds, man.Poikilothermic are those, having a variable temperature which fluctuates with that of environment. They are called the cold blooded animals, eg, reptiles, amphibians.

Topic: Biology, Organisms and Populations, JIPMER, JIPMER 2009
Anisole can be prepared by the action of methyl iodide on sodium phenate. The reaction is called

Options:
A Wurtz's reaction
B Williamson's reaction
C Fittig's reaction
D Etard's reaction

The correct answer is:
Williamson's reaction

Solution:
The reaction of alkyl halide with sodium alkoxide to give ether (alkoxy alkane), is known as Williamson's synthesis. In this reaction an ether (anisole) is prepared by the action of alkyl halide
(methyl iodide) on sodium alkoxide (sodium phenate), so it is an example of Williamson's synthesis.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Anomalous behavior of nitrogen is due to

Options:
A small size and high electronegativity
B non-availability of d-orbitals in valency shells
C ease of multiple bond formation
D all of the above

The correct answer is:
all of the above

Solution:
The anomalous behaviour of nitrogen is due to following reasons.

1. Small size and high electronegativity in the group.

2. Non-availability of vacant d-orbitals in valency shells.

3. Ease of multiple bond formation due to effective pπ-pπ overlapping.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main