datavorous/entrance-exam-dataset huggingface dataset:
The moment of inertia of a thin uniform rod of mass $M$ and length $L$ about an axis passing through its mid-point and perpendicular to its length is $I_0$. Its moment of inertia about an axis passing through one of its ends and perpendicular to its length is

Options:
A $I_0+M L^2 / 4$
B $I_0+2 M L^2$
C $I_0+M L^2$
D $I_0+M L^2 / 2$

The correct answer is:
$I_0+M L^2 / 4$

Solution:
The theorem of parallel axis for moment of inertia.
$$
\begin{aligned}
& I=I_{\mathrm{CM}}+M h^2 \\
& I=I_0+M\left(\frac{L}{2}\right)^2
\end{aligned}
$$
$$
I=I_0+\frac{M L^2}{4}
$$

Topic: Physics, Rotational Motion, JEE Main
The moment of inertia of a uniform circular disc is maximum about an axis perpendicular to the disc and passing through

Options:
A $B$
B $C$
C $D$
D $A$

The correct answer is:
$B$

Solution:
Moment of inertia \(\mathrm{I}=\mathrm{I}_{\mathrm{cm}}+\mathrm{md}^2\)
d is maximum for B
\(\therefore \mathrm{I}\) is maximum for B.

Topic: Physics, Rotational Motion, NEET, NEET 2012 (Mains)
The moment of inertia of a uniform circular disc of radius $R$ and mass $M$ about an axis passing from the edge of the disc and normal to the disc is:

Options:
A $M R^2$
B $\frac{1}{2} M R^2$
C $\frac{3}{2} M R^2$
D $\frac{7}{2} M R^2$

The correct answer is:
$\frac{3}{2} M R^2$

Solution:
M.I. of disc about its normal $=\frac{1}{2} M R^2$
M.I. about its one edge $=M R^2+$ $\frac{M R^2}{2}$
$$
\therefore \quad \text { M.I. }=\frac{3}{2} M R^2
$$

Topic: Physics, Rotational Motion, JEE Main
The moment of inertia of a uniform circular disc of radius ' $R$ ' and mass ' $M$ ' about an axis passing from the edge of the disc and normal to the disc is

Options:
A $M R^2$
B $\frac{1}{2} M R^2$
C $\frac{3}{2} M R^2$
D $\frac{1}{2} M R^2$

The correct answer is:
$\frac{3}{2} M R^2$

Solution:
[c] M.I. of a uniform circular disc of radius 'R' and mass 'M' about an axis passing through C.M and normal to the disc is
$\left.I_{C . M .}=\frac{1}{2} M R^2 \right\rvert\,$ From parallel axis theorem $\left.I_T=I_{C . M .}+M R^2=\frac{1}{2} M R^2+M R^2=\frac{3}{2} M R^2 \right\rvert\,$

Topic: Physics, Rotational Motion, JEE Main
The moment of inertia of a uniform circular disc of radius $R$ and mass $M$ about an axis touching the disc at its diameter and normal to the disc:

Options:
A $\frac{1}{2} M R^2$
B $M R^2$
C $\frac{2}{5} M R^2$
D $\frac{3}{2} M R^2$

The correct answer is:
$\frac{3}{2} M R^2$

Solution:
M. I. of uniform circular disc about an axis and perpendicular to the plane is:
$$
I_C=\frac{1}{2} M R^2
$$
Using the theorem of parallel axis.
M. I. of uniform circular disc about an axis touching the disc at it diameter is:
$$
\begin{aligned}
I & =I_C+M R^2 \\
& =\frac{1}{2} M R^2+M R^2 \\
& =\frac{3}{2} M R^2 .
\end{aligned}
$$

Topic: Physics, Rotational Motion, JEE Main
The moment of inertia of a uniform circular disc of radius $R$ and mass $M$ about an axis touching the disc at its diameter and normal to the disc is

Options:
A $M R^{2}$
B $\frac{2}{5} M R^{2}$
C $\frac{3}{2} M R^{2}$
D $\frac{1}{2} M R^{2}$

The correct answer is:
$\frac{3}{2} M R^{2}$

Solution:
The moment of inertia about an axis passing through centre of mass of disc and perpendicular to its plane is
$$
I_{\mathrm{CM}}=\frac{1}{2} M R^{2}
$$
Where $M$ is the mass of disc and $R$ its radius According to theorem of parallel axis, moment of inertia of circular disc about an axis
touching the disc at its diameter and normal to the disc is
$$
\begin{aligned}
I &=I_{\mathrm{CM}}+M R^{2} \\
&=\frac{1}{2} M R^{2}+M R^{2} \\
&=\frac{3}{2} M R^{2}
\end{aligned}
$$

Topic: Physics, Rotational Motion, JEE Main
The moment of inertia of a uniform cylinder of length l and radius R about its perpendicular bisector is I. What is the ratio l/R such that the moment of inertia is minimum?

Options:
A 32
B 32
C 32
D 1

The correct answer is:
32

Solution:
m=ρπR2l ...(1)
dm=ρπ2R.dRl+R2dl
0=ρπ2RldR+R2dl
2Rl dR=-R2dl ...(2)
I=MR24+Ml212
I=M12 3R2+l2
dI=M126RdR+2l dl=0 ... (3)
Putting (1), (2) in (3)
6R-R2dl2Rl+2l dl=0
-3R2l+2ldl=0
-3R2+2l2=0
2l2=3R2
lR= 32

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2017 (02 Apr)
The moment of inertia of a uniform semicircular disc of mass $M$ and radius $r$ about a line perpendicular to the plane of the disc through the centre is

Options:
A
$\frac{1}{4} \mathrm{Mr}^2$

B
$\frac{2}{5} \mathrm{Mr}^2$

C
$\mathrm{Mr}^2$

D
$\frac{1}{2} \mathrm{Mr}^2$

The correct answer is:
$\frac{1}{2} \mathrm{Mr}^2$

Solution:
$2 \mathrm{I}=2 \mathrm{M} \frac{\mathrm{R}^2}{2}$
$\therefore \mathrm{I}=\frac{\mathrm{MR}^2}{2}$

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2005
The moment of inertia of a uniform thin rod about a perpendicular axis passing through one end is I1. The same rod is bent into a ring and its moment of inertia about a diameter is I2. If I1I2 is xπ23, then the value of x will be ______.

Options:

The correct answer is:
8

Solution:
Moment of inertia of rod rotating about an axis passing through one end is I1=ml23 ⋯1

Moment of inertia of ring rotating about one of its diameters is I2=mR22.

Since the rod is bent into a ring, the circumference of the ring and the length of the rod are equal, l=2πR

Then, I2=mR22=ml28π2

Thus, the ratio I1I2=8π23, so x=8.

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2022 (29 Jun Shift 2)
The moment of inertia of the body about an axis is \( 1.2 \mathrm{~kg}-\mathrm{m}^{2} \). Initially the body is at rest. In order to produce a rotational kinetic energy of \( 1500 \mathrm{~J} \), an angular acceleration of \( 25 \mathrm{rad} / \mathrm{s}^{2} \) must be applied about the axis for the duration of

Options:
A \( 2 \mathrm{~s} \)
B \( 4 \mathrm{~s} \)
C \( 8 \mathrm{~s} \)
D \( 10 \mathrm{~s} \)

The correct answer is:
\( 2 \mathrm{~s} \)

Solution:
Rotational kinetic energy = 1 2 Iâ¡ ω 2
According to question
1 2 Iâ¡ ω 2 = 1 5 0 0
1 2 Iâ¡ α t 2 = 1 5 0 0
1.2 × 2 5 2 × t 2 = 3 0 0 0
1.2 × 6 2 5 × t 2 = 3 0 0 0
t 2 = 3 0 0 0 1.2 × 6 2 5 = 4
t = 2 s

Topic: Physics, Rotational Motion, JEE Main
The moment of the force, F→=4i^+5j^−6k^ at ( 2,0,−3 ), about the point ( 2,−2,−2 ), is given by

Options:
A \(-7 \hat{\mathrm{i}}-8 \hat{\mathrm{j}}-4 \widehat{\mathrm{k}}\)
B \(-4 \hat{\mathrm{i}}-\hat{\mathrm{j}}-8 \widehat{\mathrm{k}}\)
C \(-8 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}-7 \hat{\mathrm{k}}\)
D \(-7 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}-8 \widehat{\mathrm{k}}\)

The correct answer is:
\(-7 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}-8 \widehat{\mathrm{k}}\)

Solution:
Fâ=4i^+5j^-6k^
râ=2-2i^+0--2j^+-3--2k^=2j^-1k^
râÃFâ=2j^-1k^Ã4k^+5j^-6k^
= i^j^k^02-145-6=-7i^-4j^-8k^

Topic: Physics, Mathematics in Physics, JEE Main
The moments of inertia of a non-uniform circular disc (of mass $\mathrm{M}$ and radius $\mathrm{R}$ ) about four mutually perpendicular tangents $\mathrm{AB}, \mathrm{BC} \mathrm{CD}, \mathrm{DA}$ are $I_{1}, I_{2}, I_{3}$ and $I_{4}$ respectively (the square ABCD circumscribes the circle.) The distance of the center of mass of the disc from its geometrical center is given by.

Options:
A $\frac{1}{4 M R} \sqrt{\left(I_{3}-I_{3}\right)^{2}+\left(I_{2}-I_{4}\right)^{2}}$
B $\frac{1}{12 M R} \sqrt{\left(I_{3}-I_{3}\right)^{2}+\left(I_{2}-I_{4}\right)^{2}}$
C $\frac{1}{3 M R} \sqrt{\left(I_{1}-I_{2}\right)^{2}+\left(I_{3}-I_{4}\right)^{2}}$
D $\frac{1}{2 M R} \sqrt{\left(I_{1}+I_{3}\right)^{2}+\left(I_{2}+I_{4}\right)^{2}}$

The correct answer is:
$\frac{1}{4 M R} \sqrt{\left(I_{3}-I_{3}\right)^{2}+\left(I_{2}-I_{4}\right)^{2}}$

Solution:
$$
\begin{array}{l}
I_{1}=I_{c m}+M(R-y)^{2} \\
I_{3}=I_{c m}+M(R+y)^{2}
\end{array}
$$
So $I_{3}-I_{1}=M\left[(R+y)^{2}-(R-y)^{2}\right\rfloor$
same as $I_{4}-I_{2}=M\left[(R+x)^{2}-(R-x)^{2}\right]$
Ву
Distance of $\mathrm{CM}$ from center of disc $=\sqrt{x^{2}+y^{2}}=\frac{1}{4 M R} \sqrt{\left(I_{3}-I_{3}\right)^{2}+\left(I_{2}-I_{4}\right)^{2}}$

Topic: Physics, Rotational Motion, JEE Main
The momentum of a body is increased by 50%. The percentage increase in the kinetic energy of the body is________%.

Options:

The correct answer is:
125

Solution:
The relation of kinetic energy and momentum is given by

K=p22m

The new value of momentum is p'=1.5p

The new value of kinetic energy is

⇒K'=1.5p22m=2.25K

Hence, the percentage increase in kinetic energy is

K'-KK×100=2.25K-KK=125%

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2023 (08 Apr Shift 1)
The momentum of a system is conserved

Options:
A Always
B Never
C In the absence of an external force on the system
D None of the above

The correct answer is:
In the absence of an external force on the system

Solution:
Correct option is C. In the absence of an external force on the system The law of conservation of momentum says that the momentum of a closed system is constant in time (conserved). A closed (or isolated) system is defined to be one for which the mass remains constant, and the external force is zero. The total momentum of a system is conserved when the system is closed.

Topic: Physics, Laws of Motion, JEE Main
The momentum of an electron revolving in nth orbit is given by : (Symbols have their usual meanings)

Options:
A nh2πr
B nh2r
C nh2π
D 2πrnh

The correct answer is:
nh2πr

Solution:
According to Bohr's postulates, angular momentum of an electron in an orbit is given by,

L=mvr=nh2π

Therefore, linear momentum p=Lr=nh2πr

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2022 (25 Jul Shift 1)
The momentum of electrons having a wavelength $2 Å$ (given, $h=6.626 \times 10^{-34} \mathrm{~J}$-s, $\left.m=9.1 \times 10^{-35} \mathrm{~kg}\right)$ is

Options:
A $6.313 \times 10^{-24} \mathrm{~kg}^{-\mathrm{m} \mathrm{s}^{-1}}$
B $3.313 \times 10^{-24} \mathrm{~kg}-\mathrm{m} \mathrm{s}^{-1}$
C $9.313 \times 10^{-25} \mathrm{~kg}-\mathrm{m} \mathrm{s}^{-1}$
D $12.313 \times 10^{-24} \mathrm{~kg}-\mathrm{m} \mathrm{s}^{-1}$

The correct answer is:
$3.313 \times 10^{-24} \mathrm{~kg}-\mathrm{m} \mathrm{s}^{-1}$

Solution:
Given, $h=6.626 \times 10^{-34} \mathrm{~J}-\mathrm{s}$
$m=9.1 \times 10^{-35} \mathrm{~kg}$ Wavelength, $\lambda=2 Å$ $=2 \times 10^{-10} \mathrm{~m}$ $\therefore$ Momentum of electron is given as $$ \begin{aligned} p &=\frac{h}{\lambda} \\ &=\frac{6.626 \times 10^{-34}}{2 \times 10^{-10}}=3.313 \times 10^{-24} \mathrm{~kg}-\mathrm{ms}^{-1} \end{aligned} $$

Topic: Physics, Dual Nature of Matter, JEE Main
The Mond process is used for the:

Options:
A Purification of Zr and Ti
B Purification of Ni
C Extraction of Zn
D Extraction of Mo

The correct answer is:
Purification of Ni

Solution:
Mond's Process: In this process, nickel is heated in a stream of carbon monoxide forming a volatile complex, nickel tetracarbonyl:

Nis+4COg→330-350 KNiCO4g

The carbonyl is subjected to a higher temperature so that it is decomposed giving the pure metal:

NiCO4 →450-470 KNi+4CO

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2019 (08 Apr Shift 2)
The moon's radius is $1 / 4$ that of the earth and its mass is $1 / 80$ times that of the earth. If $g$ represents the acceleration due to gravity on the surface of the earth, that on the surface of the moon is

Options:
A $g / 4$
B $g / 5$
C $g / 6$
D $g / 8$

The correct answer is:
$g / 5$

Solution:
Using $g=\frac{G M}{R^2}$ we get $g_m=g / 5$

Topic: Physics, Gravitation, JEE Main
The most abundant element on earth crust is

Options:
A Hydrogen
B Oxygen
C Silicon
D Carbon

The correct answer is:
Oxygen

Solution:
$\begin{array}{lc}
\text { Element } & \text { % abundance by weight } \\
\mathrm{O} & 46.6 \\
\mathrm{Si} & 27.7 \\
\mathrm{Al} & 8.3 \\
\mathrm{Fe} & 5.1 \\
\mathrm{Ca} & 3.6
\end{array}$

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
The most abundant elements by mass in the body of a healthy human adult are:
Oxygen (61.4%); Carbon (22.9%), Hydrogen (10.0%); and Nitrogen (2.6%). The weight which a 75kg person would gain if all 1H atoms are replaced by 2H atoms is:

Options:
A 37.5 kg
B 7.5 kg
C 10 kg
D 15 kg

The correct answer is:
7.5 kg

Solution:
Weight of Hydrogen in body of 75 kg person =75×10100=7.5kg
On replacing 1H by 2H , the new weight of H=7.5×2=15kg
Weight increase = 7.5kg

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2017 (02 Apr)
The most abundant metal ion present in the human body is

Options:
A $\mathrm{Zn}^{2+}$
B $\mathrm{Ca}^{2+}$
C $\mathrm{Na}^{+}$
D $\mathrm{Fe}^{2+}$

The correct answer is:
$\mathrm{Ca}^{2+}$

Solution:
Calcium Mainly present in Bones all other are required in lower amount

Topic: Chemistry, s Block Elements, KVPY, KVPY 2017 (19 Nov SB/SX)
The most acidic oxide for nitrogen is

Options:
A NO2
B N2O
C NO
D N2O5

The correct answer is:
N2O5

Solution:
The acidic nature of the oxides of nitrogen can be determined by the strength of the acid formed by the nitrogen oxide in water.

More higher be the (+) ve oxidation number of N-atom in its oxide more acidic be the nature of oxide. Thus N2O5, is most acidic oxide.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
The most active phagocytic white blood cells are

Options:
A Neutrophils and monocytes
B Neutrophils and eosinophils
C Lymphocytes and macrophages
D Eosinophils and lymphocytes

The correct answer is:
Neutrophils and monocytes

Solution:
Neutrophils and monocytes have phagocytic function.

Topic: Biology, Body Fluids and Circulation, NEET, NEET 2008 (Mains)
The most apparent change during the evolutionary history of Homo sapiens is traced in

Options:
A loss of body hair
B walking upright
C shortening of the jaws
D remarkable increase in the brain size

The correct answer is:
remarkable increase in the brain size

Solution:
Rounding and enlargement of cranium, ie, gradual increase in the brain size is the most apparent change during the evolutionary history of Homo sapiens.

Topic: Biology, Evolution, NEET, NEET 2010 (Mains)
The most appropriate method of making egg-albumin sol is:

Options:
A Break an egg carefully and transfer the transparent part of the content to 100 mL of 5% w/V saline solution and stir well
B Keep the egg in boiling water for 10 minutes. After removing the shell, transfer the yellow part of the content to 100 mL of 5% w/V saline solution and homogenize with a mechanical shaker
C Keep the egg in boiling water for 10 minutes. After removing the shell, transfer the white part of the content to 100 mL of 5% w/V saline solution and homogenize with a mechanical shaker
D Break an egg carefully and transfer only the yellow part of the content to 100 mL of 5% w/V saline solution and stir well

The correct answer is:
Break an egg carefully and transfer the transparent part of the content to 100 mL of 5% w/V saline solution and stir well

Solution:
Preparation of egg albumin sol. is done following two steps.
(i) Prepare100 mL of 5% (w/V) solution of NaCl in water in a 250 mL beaker.
(ii) Break one egg in a porcelain dish and pipette out the albumin and pour it in sodium chloride solution. Stir well to ensure the preparation of sol.

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2016 (09 Apr Online)
The most appropriate reagent for conversion of C2H5CN into CH3CH2CH2NH2 is :

Options:
A NaBH4
B CaH2
C LiAlH4
D Na(CN)BH3

The correct answer is:
LiAlH4

Solution:
CH3CH2−C≡N→LiAlH4CH3CH3−CH2−NH2

NaBH4 does not reduce R-CN.

Topic: Chemistry, Amines, JEE Main, JEE Main 2020 (05 Sep Shift 1)
The most appropriate reagent to distinguish between acetaldehyde and formaldehyde is

Options:
A Fehling's solution
B Tollen's reagent
C Schiff's reagent
D lodine in presence of base

The correct answer is:
lodine in presence of base

Solution:
Acetaldehyde have $\mathrm{CH}_3 \mathrm{CO}-$ group so it give positive iodoform test with $I_2$ and $\mathrm{NaOH}$ while formaldehyde does not have $-\mathrm{CH}_3 \mathrm{CO}$ group so it will not give the positive haloform test. Option D is correct.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
The most appropriate sequence of the reactions for carrying out the following conversion is

Options:
A (i) peracid (ii) H+ (iii) Zn/dil. HCl
B (i) Alkaline KMnO4 (ii) NaIO4 (iii) N2H4/KOH
C (i) Alkaline KMnO4 (ii) H+ (iii) Zn/dil. HCl
D (i) O3/Me2S (ii) NaOEt (iii) N2H4/KOH

The correct answer is:
(i) O3/Me2S (ii) NaOEt (iii) N2H4/KOH

Solution:
It involves ozonolysis followed by intramolecular aldol condensation

Topic: Chemistry, Hydrocarbons, JEE Main
The most basic compound among the following is

Options:
A
Acetanilide

B
Benzylamine

C
$p$-Nitro aniline

D
Aniline

The correct answer is:
Benzylamine

Solution:
< smiles>NCc1ccccc1 < /smiles>
basic due to presence of localized lone pair of electron on nitrogen atom while other compounds have delocalized lone pair of electron.

Topic: Chemistry, Amines, JEE Main, JEE Main 2012 (07 May Online)
The most basic nitrogen in the following compound is

Options:
A I
B II
C III
D IV

The correct answer is:
III

Solution:
As lone pair of N III is localised therefore it is most basic.

Topic: Chemistry, General Organic Chemistry, JEE Main
The most common or abundant connective tissue of body is

Options:
A Areolar connective tissue
B Fluid connective tissue
C Dense connective tissue
D Adipose tissue

The correct answer is:
Areolar connective tissue

Solution:
The areolar connective tissue is a tissue with 'small open spaces' (areola) and with small airy pockets between the network of cells and fibers. Within this tissue there are three main types of cells (fibroblasts, white blood cells, and mast cells) and three types of fibers (collagen fibers, elastin fibers, and reticular fibers), a fluid matrix called ground substance.

Topic: Biology, Structural Organisation in Animals, NEET
The most common oxidation state of Lanthanoid elements is +3. Which of the following is likely to deviate easily from +3 oxidation state?

Options:
A CeAt. No,58
B LaAt. No,57
C LuAt. No,71
D GdAt. No,64

The correct answer is:
CeAt. No,58

Solution:
Electronic configuration of Ce=Xe4f15d16s2⏟→−4e−Xe⏟

Cerium in +4 oxidation state due to stable xenon configuration. La, Lu and Gd are stable in +3 oxidation state.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2022 (26 Jun Shift 2)
The most commonly used food preservative is:

Options:
A Sodium cyclamate
B Sodium benzoate
C Sodium acetate
D Vanillin

The correct answer is:
Sodium benzoate

Solution:
Sodium benzoate is a preservative. As a food additive, sodium benzoate has the E number E-211. It is bacteriostatic and fungistatic under acidic conditions.
It is most widely used in acidic foods such as salad dressings (vinegar), carbonated drinks (carbonic acid), jams and fruit juices (citric acid), pickles (vinegar), and condiments.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main
The most delicate stage in the life cycle of an angiosperm where maximum mortality observed is the

Options:
A seed stage
B flowering stage
C fruiting stage
D seedling stage

The correct answer is:
seedling stage

Solution:
Seed germination is the process where seed develops into a new plant by producing the shoot and root on germination. The germinated seed forms a young sporophyte called a seedling. It has plumule and radicle, where plumule forms the shoot bearing leaves and flower while the radicle forms the root part. Seedling stage is the most delicate stage of angiosperms and has high mortality rate due to drought condition been one of the main cause as roots are not able to maintain contact with soil sufficiently to prevent water stress.

Topic: Biology, Plant - Growth and Development, NEET
The most electropositive amongst the alkaline earth metals is

Options:
A Beryllium
B Magnesium
C Calcium
D Barium

The correct answer is:
Barium

Solution:
As we go down the group electropositive character increases because I.E. decreases. $\mathrm{Ba}$ is most electropositive element in the group.

Topic: Chemistry, s Block Elements, JEE Main
The most important force which pulls water up in tall trees is :

Options:
A imbibition force
B osmotic force
C cohesive force
D electromagnetic force

The correct answer is:
cohesive force

Solution:
The most important force which pulls water up in the tall trees is the cohesive (inter-molecular) force between $\mathrm{H}_2 \mathrm{O}$ molecules. It also helps in maintaining the continuity of water column in tall trees.

Topic: Biology, Transport in Plants, JIPMER, JIPMER 2006
The most polar compound among the following is:

Options:
A
B
C
D

The correct answer is:

Solution:
Polar compounds are chemical compounds that are held together by polar covalent bonds. The term ‘polar compound’ can be defined as a chemical species which consists of two or more atoms that are held together by covalent bonds that are polar in nature due to the unequal sharing of electrons. When two atoms are bound together via a covalent bond the differences in the electronegativities of the bonded atoms may cause the bond pair of electrons to shift closer to the more electronegative atom. This results in the accumulation of a partial positive charge at the location of the more electropositive atom and the accumulation of a partial negative charge at the location of the more electronegative atom. Polar compounds are chemical compounds that are held together by such bonds.

In the bond dipole vector of C - F bond is not subtractive.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2018 (16 Apr Online)
The most popularly known blood grouping is the $\mathrm{ABO}$ grouping. It is named $\mathrm{ABO}$ and not $\mathrm{ABC}$, because "O" in it refers to having :

Options:
A No antigens A and B on RBCs
B Other antigens besides A and B on RBCs
C Overdominance of this type on the genes for $A$ and $B$ types
D One antibody only-either anti-A and anti-B on the RBCs

The correct answer is:
No antigens A and B on RBCs

Solution:
No Solution Available

Topic: Biology, Body Fluids and Circulation, NEET, NEET 2009 (Mains)
The most precise measurement among the following is

Options:
A \( 5.00 \mathrm{~mm} \)
B \( 5.00 \mathrm{~cm} \)
C \( 5.00 \mathrm{~m} \)
D \( 5.00 \mathrm{~km} \)

The correct answer is:
\( 5.00 \mathrm{~mm} \)

Solution:
All given measurements are correct up to two decimal places.

As here 5.00 mm has the smallest unit and the error in 5.00 mm is least (commonly taken as 0.01 mm if not specified), hence, 5.00 mm is most precise.

Topic: Physics, Mathematics in Physics, JEE Main
The most predominantly ionic compounds will be obtained from the combination of elements belonging to

Options:
A 1 and 7 groups
B 2 and 6 groups
C 3 and 5 groups
D 0 and 7 groups

The correct answer is:
1 and 7 groups

Solution:
Ionic bond is formed when there is large difference of electro-negativities between the atoms.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
The most stable carbocation for the following is:

Options:
A c
B d
C b
D a

The correct answer is:
c

Solution:
Maximum resonating structure are possible in (c) structure. In (a) structure the carbocation is stabilised by lone pair on nitrogen as well as two double bonds in the ring. In (b) and (d) structure lone pair of nitrogen is not involved in conjugation to stabilise positive charge. So, NH2 group will show -I effect.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2023 (30 Jan Shift 2)
The most stable carbonium ion among the following is

Options:
A $\mathrm{C}_6 \mathrm{H}_5 \stackrel{+}{\mathrm{C}} \mathrm{HC}_6 \mathrm{H}_5$
B $\mathrm{C}_6 \mathrm{H}_5 \stackrel{+}{\mathrm{C}} \mathrm{H}_2$
C $\mathrm{CH}_3 \mathrm{C}_2$
D $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{C}_2 \mathrm{H}_2$

The correct answer is:
$\mathrm{C}_6 \mathrm{H}_5 \stackrel{+}{\mathrm{C}} \mathrm{HC}_6 \mathrm{H}_5$

Solution:
$\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH} \mathrm{C}_6 \mathrm{H}_5$ is the most stable carbonium ion due to resonance.

Topic: Chemistry, Hydrocarbons, JEE Main
The most stable conformation of 2,3 -dibromobutane is $-$

Options:
A
B
C
D

The correct answer is:

Solution:
No Solution

Topic: Chemistry, General Organic Chemistry, JEE Main