datavorous/entrance-exam-dataset huggingface dataset:
Name the logic gate equivalent to the diagram attached

Options:
A NAND
B AND
C NOR
D OR

The correct answer is:
NOR

Solution:
Let's assume that when the path is short, the state is called state 1 and when it is open, the state is called state 0.

Depending up on whether the path is short or not in the circuit, the following table can be obtained:

A	B	Out

1	1	0

1	0	0

0	1	0

0	0	1

The above table is similar to the truth table corresponding to NOR gate. Hence, the circuit indicates a NOR logic gate.

Topic: Physics, Semiconductors, JEE Main, JEE Main 2023 (06 Apr Shift 1)
Name the main processes of urine formation.

Options:
A Glomerular Filtration, Reabsorption and tubular secretion
B Tubular reabsorption, Tubular secretion and Tubular Filtration
C Glomerular Filtration, Tubular secretion and Tubular Filtration
D Glomerular Filtration, Glomerular reabsorption and secretion

The correct answer is:
Glomerular Filtration, Reabsorption and tubular secretion

Solution:
The main processes of urine formation are Glomerular Filtration, Reabsorption and tubular secretion.

Topic: Biology, Excretory Products and their Elimination, NEET
Name the moderator used in the nuclear reactor?

Options:
A Plutonium
B Thorium
C Graphite
D Berilium

The correct answer is:
Graphite

Solution:
In nuclear engineering, a neutron moderator is a medium that reduces the speed of fast neutrons, thereby turning them into thermal neutrons capable of sustaining a nuclear chain reaction involving U-235 or a fissile nuclide. Graphite is used as moderator in nuclear reactor.

Topic: Physics, Nuclear Physics, JEE Main
Name the physical quantity whose S.I. unit is \( \mathrm{J} \mathrm{C}^{-1} . \) Is it a scalar or a vector quantity?

Options:
A Current, vector
B Current, scalar
C Electrostatic potential, vector
D Electrostatic potential, scalar

The correct answer is:
Electrostatic potential, scalar

Solution:
The SI unit of electrostatic potential is joule per coulomb or J C−1. It is a scalar quantity.

Topic: Physics, Mathematics in Physics, JEE Main
Name the pulmonary disease in which alveolar surface area involved in gas exchange is drastically reduced due to damage in the alveolar walls.

Options:
A Emphysema
B Pneumonia
C Asthma
D Pleurisy

The correct answer is:
Emphysema

Solution:
Emphysema: The air-pollutants that cause chronic bronchitis, may breakdown the alveoli of the lungs, reducing the surface area for gas exchange. The victim becomes permanently short of breath. It mainly occurs due to smoking.

Topic: Biology, Breathing and Exchange of Gases, NEET, NEET 2015 (Phase 2)
Name three groups of plants that bear archegonia. Briefly describe the life cycle of any one of them.

Options:

None

Solution:
Three groups of plant that bear archegonia are
1. Bryophyta,
2. Pteridophyta and
3. Gymnosperm.
Life cycle of gymnosperms : The gymnosperms are heterosporous, they produce haploid microspores and megaspores. The two kinds of spores are produced within sporangia that are borne on sporophyll which are arranged spirally along an axis to form strobili or cones. The strobili bearing microsporophylls and microsporongia are called microsporangiate or male strobili. The microspores develop into a male gametophytic generation which is highly reduced, and confined to only limited number of cells. The reduced gametophyte is called pollen grain. The development of pollen grain takes place within microsporangia. The cones bearing megaspo-rophylls with
ovules or megasporangia are called macrosporangiate or female strobili. The male or female cones or strobili may be borne on the same tree or different trees. The megaspore mother cell is differentiated from one of the cells of the nucellus. The nucellus is protected by envelopes and the composite structure is called an ovule. The ovules are borne on megasporophylls which may be clustered to form the female cones. The females cones are borne on the main plant body of the sporophyte. The megaspore mother cell divides meiotically to form four megaspore. One of the megaspore enclosed within the megasporangium develops into multicellular female gametophyte that bears 2 or more archegonia or female sex organs. The multicellular female gametophyte is also remains within megasporangium.

Topic: Biology, Plant Kingdom
Name two cell-organelles that are double membrane bound. What are the characteristics of these two organelles? State their functions and draw labelled diagram of both.

Options:

None

Solution:
Two double membrane bound cell organelles:
(a) Mitochondria : It has finger like folds in the inner membrane called cristae. Mitochondria is the place for aerobic respiration.
(b) Chloroplast : Chloroplast is responsible for converting light energy into chemical energy. Chloroplast contains stacked thylakoid in its matrix.
Functions of mitochondria: The double membrane mitochondria are actively associated with aerobic respiration and the release of energy for cellular activity. The biological oxidation of the fats and carbohydrate release much amount of energy which is utilised by mitochondria for ATP synthesis. Required energy is released from ATP molecules for various cell processes in cells so they are termed as "the powerhouse of the cell"
Functions of chloroplast :
(i) Their main function of chloroplast is to trap sun's energy and to convert it into chemical energy of food by photosynthesis.
(ii) Storage of starch.
(iii) Chloroplast in fruits and flowers change into chromoplasts.

Topic: Biology, Cell - The Unit of Life
$\mathrm{NaOH}$ is a strong base. What will be $\mathrm{pH}$ of $5.0 \times 10^{-2} \mathrm{M} \mathrm{NaOH}$ solution ? $(\log 2=0.3)$

Options:
A
$14.00$

B
$13.70$

C
$13.00$

D
$12.70$

The correct answer is:
$12.70$

Solution:
Given $\left[\mathrm{OH}^{-}\right]=5 \times 10^{-2}$
$$
\begin{aligned}
& \therefore \mathrm{pOH}=-\log 5 \times 10^{-2} \\
& \quad \quad=-\log 5+2 \log 10=1.30 \\
& \because \mathrm{pH}+\mathrm{pOH}=14 \\
& \because \mathrm{pH}=14-\mathrm{pOH} \\
& \quad=14-1.30=12.70
\end{aligned}
$$

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2013 (22 Apr Online)
Narcotics are chemical substances which produce sleep and unconsciousness. Morphine diacetate is most widely used analgesic. How many double bond equivalents are present in morphine diacetate?

Options:

The correct answer is:
11

Solution:
One double bond = 1 DBE
One ring = 1 DBE

Topic: Chemistry, Chemistry in Everyday Life, JEE Main
Natalite is a mixture of

Options:
A Acetic acid and diethyl ether
B Ethyl bromide and diethyl ether
C Ethyl alcohol and dimethyl ether
D Diethyl ether and ethyl alcohol

The correct answer is:
Diethyl ether and ethyl alcohol

Solution:
Natalite is a trade name for the mixture of diethyl ether and ethyl alcohol. It is used as petrol substitute. It consist of 54 percent alcohol, 45 percent ether and 1 percent trimethylamine.

Topic: Chemistry, General Organic Chemistry, JEE Main
Natural glucose is termed D-glucose because :

Options:
A
$-\mathrm{OH}$ on the second carbon is on the right side in Fischer projection

B
$-\mathrm{OH}$ on the sixth carbon is on the right side in Fischer projection.

C
$-\mathrm{OH}$ on the fifth carbon is on the right side in Fischer projection.

D
It is dextrorotatory.

The correct answer is:
$-\mathrm{OH}$ on the fifth carbon is on the right side in Fischer projection.

Solution:
Fischer gave the prefix "D" to compounds whose bottom chiral has its $\mathrm{OH}$ to the right. So natural glucose is called D-glucose or dextrose.
Structure of D-Glucose :

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2013 (23 Apr Online)
Natural rubber belongs to which class of polymer

Options:
A Addition polymer
B Condensation polymer
C Co-ordination polymer
D Co-polymer

The correct answer is:
Addition polymer

Solution:
Natural rubber is addition polymer of isoprene (2-methyl-1, 3-butadiene)

Topic: Chemistry, Polymers, JEE Main
n-butane and isobutane are examples of

Options:
A Chain isomers
B Geometrical isomers
C Position isomers
D Tautomers

The correct answer is:
Chain isomers

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Main
n-Butylamine (I), diethylamine (II) and N,N-dimethylethyl amine (III) have the same molar mass. The increasing order of their boiling point is

Options:
A III < II < I
B I < II < III
C II < III < I
D II < I < III

The correct answer is:
III < II < I

Solution:
H3C-CH2-CH2-CH2-NH2n-butylamine primary amine
CH3CH2-NH-CH2-CH3Diethylamine Secondary amine
CH3-N|-C2H5CH3 ethyldimethylamine (tertiary amine)
Primary amines have two hydrogen atoms available for hydrogen bond formation, while 2° amines have only one hydrogen atom for hydrogen bond formation. 3° amines do not have intermolecular association due to absence of hydrogen atom.
Thus, order of boiling point of isomeric amines is as follows
3°<2°<1°
or III<II<I .

Topic: Chemistry, Amines, JEE Main
Nd2+=______

Options:
A 4f26s2
B 4f4
C 4f3
D 4f46s2

The correct answer is:
4f4

Solution:
The atomic number of Neodymium (Nd) is 60. Its electronic configuration is Xe 4f4 6s2.

Now, by removing two electrons from the outermost shell 6s, Nd2+ ion is formed. Hence, electronic configuration of Nd2+ is Xe 4f4.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2023 (31 Jan Shift 1)
Nearly 10% of the power of a 110 W light bulb is converted to visible radiation. The change in average intensities of visible radiation, at a distance of 1 m from the bulb to a distance of 5 m is a×10-2 W m-2. The value of 'a' will be

Options:

The correct answer is:
84

Solution:
Power of visible radiation is P'=10100×110 W =11 W

Change in average intensity of visible radiation is ∆Iradiation=Iradiation1-Iradiation2

⇒I1-I2=P'4πr12-P'4πr22

=114π11-125 =114π×2425

=264π×10-2=84×10-2 W m-2

Hence, the value of a=84.

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2022 (29 Jul Shift 2)
Nearly (X) per cent of CO2 is transported by RBCs whereas nearly (Y) per cent of O2 is transported by RBCs

Options:
A

X	Y
97	97

B

X	Y
25	25

C

X	Y
97	25

D

X	Y

25	97

The correct answer is:

X	Y

25	97

Solution:
Blood is the medium of transport for O2 and CO2. About 97 per cent of O2 is transported by RBCs in the blood. The remaining 3 per cent of O2 is carried in a dissolved state through the plasma. Nearly 20-25 per cent of CO2 is transported by RBCs whereas 70 per cent of it is carried as bicarbonate. About 7 per cent of CO2 is carried in a dissolved state through plasma.

Topic: Biology, Breathing and Exchange of Gases, NEET
Needles $\mathrm{N}_1, \mathrm{~N}_2$ and $\mathrm{N}_3$ are made of a ferromagnetic, a paramagnetic and a diamagnetic substance respectively. A magnet when brought close to them will

Options:
A
attract all three of them

B
attract $\mathrm{N}_1$ and $\mathrm{N}_2$ strongly but repel $\mathrm{N}_3$

C
attract $\mathrm{N}_1$ strongly, $\mathrm{N}_2$ weakly and repel $\mathrm{N}_3$ weakly

D
attract $\mathrm{N}_1$ strongly, but repel $\mathrm{N}_2$ and $\mathrm{N}_3$ weakly

The correct answer is:
attract $\mathrm{N}_1$ strongly, $\mathrm{N}_2$ weakly and repel $\mathrm{N}_3$ weakly

Solution:
attracts $\mathrm{N}_1$ strongly, $\mathrm{N}_2$ weakly and Repel $\mathrm{N}_3$ weakly

Topic: Physics, Magnetic Properties of Matter, JEE Main, JEE Main 2006
Negation is " $2+3=5$ and $8 \lt 10$ " is

Options:
A $2+3 \neq 5$ and $ \lt 10$
B $2+3=5$ and $8 \nless 10$
C $2+3 \neq 5$ or $8 \nless 10$
D None of these

The correct answer is:
$2+3 \neq 5$ or $8 \nless 10$

Solution:
Let $p: 2+3=5, q: 8 \lt 10$
Given proposition is : $p \wedge q$
Its negation is $\sim(p \wedge q)=\sim p \vee \sim q$
$\therefore$ we have $2+3 \neq 5$ or $8 \nless 10$.

Topic: Mathematics, Mathematical Reasoning, JEE Main
Negation of “Monu is in class X or Anu is in class XII” is

Options:
A Monu is not in class X and Anu is in class XII
B Monu is not in class X and Anu is not in class XII
C Either Monu is not in class X or Anu is not in class XII
D None of these

The correct answer is:
Monu is not in class X and Anu is not in class XII

Solution:
Let
p: Monu is in class X
q: Anu is in class XII
Given statement is pâ¨q
â~pâ¨qâ¡~pâ§~q

Topic: Mathematics, Mathematical Reasoning, JEE Main
Negation of p∧q∧~p∧q is

Options:
A ~p∧q∨p
B p∨q
C ~p∨q
D ~p∧q∧q

The correct answer is:
~p∧q∨p

Solution:
Given,

Expression p∧q∧~p∧q

So, negation of the above expression will be,

~p∧q∧~p∧q

Now using demorgan's law ~A∧B≡~A∨~B & ~A∨B≡~A∧~B we get,

~p∧q∧~p∧q

≡~p∧q∧~p∨~q

≡~p∧q∧~p∨q∧~q

≡~p∧q∧~p∨F

≡~p∧q∧~p

≡~p∨~q∧~p

≡~p∨~q∨p

≡~p∨~q∨p

≡~p∧q∨p

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2023 (15 Apr Shift 1)
Negation of p→q→q→p is

Options:
A p~∨p
B q∧(~p)
C ~q∧p
D p∨~q

The correct answer is:
q∧(~p)

Solution:
We know that, A→B≡~A∨B

Now using the above formula we will solve,

(p→q)→(q→p)

≡~~p∨q∨~q∨p

≡p∧~q∨~q∨p

≡~q∨p

Now finding the negation of p∨~q we get,

~p∨~q

≡~p∧q

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2023 (08 Apr Shift 1)
Negation of the Boolean expression p↔q→p is

Options:
A ~p∧q
B p∧~q
C ~p∨~q
D ~p∧~q

The correct answer is:
~p∧~q

Solution:
To find ~p↔q→p

We know ~p↔q≡p∧~q∨q∧~p

So ~p↔q→p

≡p∧~q→p∨q→p∧~p

Now p∧~q→p≡p∧q∧~p≡p∧~p∧q≡c

and q→p∧~p≡~q∨p∧~p≡~p∧~q∨p

≡~p∧~q∨~p∧p=~p∧~q

Hence ~p↔q→p≡c∨~p∧~q≡~p∧~q

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2022 (26 Jul Shift 2)
Negation of the Boolean statement p∨q⇒~r∨p is equivalent to:

Options:
A p∧~q∧r
B ~p∧~q∧r
C ~p∧q∧r
D p∧q∧~r

The correct answer is:
~p∧q∧r

Solution:
Given,

p∨q⇒~r∨p

≡~p∨q∨~r∨p

≡~p∧~q∨p∨~r

≡~p∨p∧~q∨p∨~r

≡~q∨p∨~r

So negation of~q∨p∨~r is ~p∧q∧r.

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2022 (29 Jun Shift 2)
Negation of the conditional : "If it rains, I shall go to school" is

Options:
A It rains and I shall go to school
B It rains and I shall not go to school
C It does not raihs and I shall go to school
D None of these

The correct answer is:
It rains and I shall not go to school

Solution:
The correct option is B It rains and I shall not go to school
Let p : It rains, q : I shall go to school
Thus, we have \(\mathrm{p} \Rightarrow \mathrm{q}\)
Its negative is \(\sim(p \Rightarrow q)\),i.e., \(p \wedge \sim q\)
i.e., it rains and I shall not go to school.

Topic: Mathematics, Mathematical Reasoning, JEE Main
Negation of the conditional, "If it rains, I shall go to school" is

Options:
A It rains and I shall go to school
B It rains and I shall not go to school
C It does not rains and I shall go to school
D None of the above

The correct answer is:
It rains and I shall not go to school

Solution:
Given a circuit

The circuit pq is true when p is true or q is true.

Hence, the circuit pq is represented by p∨q.

Similarly, the circuit pr is true when p is true or r is true.

Hence, the circuit pr is represented by p∨r.

Now, the whole circuit is true only when both pq and pr is true.

Thus, the whole circuit is represented by p∨q∧p∨r.

According to the distributive law p∨q∧p∨r=p∨(q∧r).

Topic: Mathematics, Mathematical Reasoning, JEE Main
Negation of the statement: 5 is an integer or 5 is irrational is:

Options:
A 5 is not an integer 5 is not irrational
B 5 is not an integer and 5 is not irrational
C 5 is irrational or 5 is an integer
D 5 is an integer and 5 irrational

The correct answer is:
5 is not an integer and 5 is not irrational

Solution:
Using~p∨q=~p∧~q, we get negation of given statement as 5 is not an integer and 5 is not irrational Number.

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2020 (09 Jan Shift 1)
Negation of the statement, “I will work hard and party” is

Options:
A I will not work hard and party
B I will not work hard or I will party
C I will not work hard or I will not party
D I will not work hard and I will not party

The correct answer is:
I will not work hard or I will not party

Solution:
~p∧q≡~p∨~q
Therefore, negation of the statement, “I will work hard and party” is "I will not work hard or I will not party".

Topic: Mathematics, Mathematical Reasoning, JEE Main
Negation of the statement (p∨r)⇒(q∨r) is :

Options:
A ~p∧q∧~r
B ~p∧q∧r
C p∧~q∧~r
D p∧q∧r

The correct answer is:
p∧~q∧~r

Solution:
The negation of the statement (p∨r)⇒(q∨r) is given by

~p∨r⇒q∨r

≡~~p∨r∨q∨r

≡p∨r∧~q∨r

≡p∨r∧~q∧~r

≡p∨r∧~r∧~q

≡p∧~r∨r∧~r∧~q

≡((p∧~r)∨ϕ)∧(~q)

≡(p∧~r)∧(~q)

≡p∧~q∧~r

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2021 (31 Aug Shift 2)
Neopentyl bromide, undergoes dehydrohalogenation to give alkenes even though it has no β−H . This is due to

Options:
A E 2 mechanism
B Rearrangement of carbocations by E 1 mechanism
C E 1 cB mechanism
D E 4 mechanism

The correct answer is:
Rearrangement of carbocations by E 1 mechanism

Solution:
Neopentyl bromide give a carbocation as intermediate which undergo for rearrangement and show E 1 mechanism (even though it has no β−H atom).
Thus Rearrangement of carbocations by E 1 mechanism is the correct option.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Nesseler's reagent is

Options:
A $\mathrm{K}_2 \mathrm{HgI}_4$
B $\mathrm{K}_2 \mathrm{HgI}_4+\mathrm{KOH}$
C $\mathrm{K}_2 \mathrm{HgI}_2+\mathrm{KOH}$
D $\mathrm{K}_2 \mathrm{Hgl}_4+\mathrm{Hg}$

The correct answer is:
$\mathrm{K}_2 \mathrm{HgI}_4+\mathrm{KOH}$

Solution:
$2 \mathrm{KI}+\mathrm{HgI}_2 \rightarrow \mathrm{K}_2 \mathrm{HgI}_4+\mathrm{KOH}$

Topic: Chemistry, d and f Block Elements, JEE Main
Newton's first law of motion describes the following

Options:
A Energy
B Work
C Inertia
D Moment of inertia

The correct answer is:
Inertia

Solution:
Newton's first law of motion defines the inertia of body. It states that every body has a tendency to remain in its state (either rest or motion) due to its inerta.

Topic: Physics, Laws of Motion, JEE Main
NH4HS (s)âNH3 (g)+H2S (g)
In the above reaction, if the pressure at equilibrium and at 300 K is 100 atm then what will be the equilibrium constant Kp?

Options:
A 2500 atm2
B 50Â atm2
C 100Â atm2
D 200Â atm2

The correct answer is:
2500 atm2

Solution:
Given,
Total pressure at equilibrium =P+P=100 atm
â2P=100 atm
âP=50 atm
KP = PNH3ÃPH2S
KP=50Ã50=2500 atm2

Topic: Chemistry, Chemical Equilibrium, JEE Main
Nickel shows ferromagnetic property at room temperature. If the temperature is increased beyond Curie temperature, then it will show

Options:
A paramagnetism
B anti-ferromagnetism
C no magnetic property
D diamagnetism

The correct answer is:
paramagnetism

Solution:
Nickel exhibits ferromagnetism because of a quantum physical effect called exchange coupling in which the electron spins of one atom interact with those of neighbouring atoms. The result is alignment of the magnetic dipole moments of the atoms, inspite of the randomizing tendency of atomic collisions. This persistent alignment is what gives ferromagnetic materials their permanent magnetism. If the temperature of a ferromagnetic material is raised above a certain critical value, called the Curie temperature, the exchange coupling ceases to be effective. Most such materials then become simply paramagnetic; that is, the dipoles still tend to align with an external field but much more weakly, and thermal agitation can now more easily disrupt the alignment.

Topic: Physics, Magnetic Properties of Matter, JEE Main
Nickel (Zâ=â28) combines with a uninegative monodentate ligand to form a diamagnetic complex NiL 4 2â . The hybridisation involved and the number of unpaired electrons present in the complex are respectively:

Options:
A dsp 2 , zero
B sp 3 , zero
C dsp 2 ,, one
D sp 3 , two

The correct answer is:
dsp 2 , zero

Solution:
​ 28 Ni has +2 oxidation state and since the complex is Diamagnetic, all electrons are paired in inner 3d, orbital to provide the electronic configuration as under:
L − might have been a strong field like CN − . Hybridisation is dsp 2 . Number of unpaired electrons = 0

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2014 (19 Apr Online)
Nickel (Zâ=â28) combines with a uninegative monodentate ligand to form a diamagnetic complex NiL 4 2â . The hybridisation involved and the number of unpaired electrons present in the complex are respectively:

Options:
A dsp 2 , zero
B sp 3 , zero
C dsp 2 ,, one
D sp 3 , two

The correct answer is:
dsp 2 , zero

Solution:
â 28 Ni has +2 oxidation state and since the complex is Diamagnetic, all electrons are paired in inner 3d, orbital to provide the electronic configuration as under:
L − might have been a strong field like CN − . Hybridisation is dsp 2 . Number of unpaired electrons = 0

Topic: Chemistry, Coordination Compounds, JEE Main
Nickel $(Z=28)$ combines with a uninegative monodentate ligand $\mathrm{X}^{-}$to form a paramagnetic complex $\left[\mathrm{NiX}_4\right]^{2-}$. The number of unpaired electron(s) in the nickel and geometry of this complex ion are, respectively

Options:
A
one, tetrahedral

B
two, tetrahedral

C
one, square planar

D
two, square planar

The correct answer is:
two, tetrahedral

Solution:
${ }_{28} \mathrm{Ni}: \ldots \ldots \ldots \mathrm{s}^2, 3 \mathrm{p}^6, 3 \mathrm{~d}^8, 4 \mathrm{~s}^2$
$\mathrm{Ni}^{2+}: 3 \mathrm{~s}^2, 3 \mathrm{p}^6, 3 \mathrm{~d}^8$

Topic: Chemistry, Coordination Compounds, JEE Main
Nickel $(Z=28)$ combines with a uninegative monodentate ligand $\mathrm{X}^{-}$to form a paramagnetic complex $\left[\mathrm{NiX}_4\right]^{2-}$. The number of unpaired electron(s) in the nickel and geometry of this complex ion are, respectively

Options:
A
one, tetrahedral

B
two, tetrahedral

C
one, square planar

D
two, square planar

The correct answer is:
two, tetrahedral

Solution:
${ }_{28} \mathrm{Ni}: \ldots \ldots \ldots \mathrm{s}^2, 3 \mathrm{p}^6, 3 \mathrm{~d}^8, 4 \mathrm{~s}^2$
$\mathrm{Ni}^{2+}: 3 \mathrm{~s}^2, 3 \mathrm{p}^6, 3 \mathrm{~d}^8$

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2006
Night blindness is due to the deficiency of

Options:
A vitamin A
B vitamin B
C vitamin C
D Vitamin D

The correct answer is:
vitamin A

Solution:
A vitamin is an organic molecule that is an essential micronutrient which an organism needs in small quantities for the proper functioning of its metabolism. Deficiency of vitamin A causes, hardening of cornea of eye and night blindness.

Topic: Chemistry, Biomolecules, JEE Main
Nissl bodies are mainly composed of

Options:
A Nucleic acids and SER
B DNA and RNA
C Proteins and lipids
D Free ribosomes and RER

The correct answer is:
Free ribosomes and RER

Solution:
Nissl granules are found in the cell body of neurons. These granules are composed of Rough Endoplasmic Reticulum (RER) that bears free ribosomes. They act as the site of protein synthesis. These granules were named after its discoverer Franz Nissl.

Topic: Biology, Neural Control and Coordination, NEET, NEET 2018
Nitrate is confirmed by ring test. The brown colour of the ring is due to the formation of

Options:
A Ferrous nitrite
B $\mathrm{FeSO}_4 \mathrm{NO}$
C $\mathrm{FeSO}_4 \mathrm{NO}_2$
D Ferrous nitrate

The correct answer is:
$\mathrm{FeSO}_4 \mathrm{NO}$

Solution:
When is $\mathrm{NO}^{3-}$ passed through $\mathrm{FeSO}_4$ (neutral) and then few drops of conc. $\mathrm{H}_2 \mathrm{SO}_4$ is added then the brown ring is obtained
$\begin{aligned} & \mathrm{FeSO}_4+\mathrm{NO} \rightarrow \mathrm{Fe}(\mathrm{NO}) \mathrm{SO}_4 \\ & \text { (Brown ring) } \\ & \text { Witroso Femousuly hate } \\ & \end{aligned}$

Topic: Chemistry, Practical Chemistry, JEE Main
Nitrates of all the metals are

Options:
A Coloured
B Unstable
C Soluble in water
D Insoluble in water

The correct answer is:
Soluble in water

Solution:
Nitrates and acetates of all metals are water soluble

Topic: Chemistry, Practical Chemistry, JEE Main
Nitration of aniline in strong acidic medium also gives m-nitroaniline because:

Options:
A In absence of substituents nitro group always goes to m-position.
B In electrophilic substitution reactions amino group is meta directive.
C In spite of substituents nitro group always goes to only m-position.
D In acidic (strong) medium aniline is present as anilinium ion.

The correct answer is:
In acidic (strong) medium aniline is present as anilinium ion.

Solution:
In acidic medium aniline is protonated to form anilinium ion which is metadirecting (due to +I effect).

Topic: Chemistry, General Organic Chemistry, JEE Main