datavorous/entrance-exam-dataset huggingface dataset:
Arrange the following in increasing order of their covalent character.

(A) CaF2

(B) CaCl2

(C) CaBr2

(D) CaI2

Choose the correct answer from the options given below.

Options:
A B<A<C<D
B A<B<C<D
C A<B<D<C
D A<C<B<D

The correct answer is:
A<B<C<D

Solution:
Cation is same in all case while anions are different.

According to Fajan's rule, Larger the anion, less is the effective nuclear charge that holds the valence electron of the ion in place. Since the last electron is loosely bound in large anions, it can easily be polarised by a cation, thereby making the compound more covalent.

Hence, the larger the size of the anion, greater is the covalent character of the ionic bond.

CaF2<CaCl2<CaBr2<CaI2

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2022 (26 Jul Shift 2)
Arrange the following in the ascending order of wavelength:
A. Gamma rays $\left(\lambda_1\right)$
B. $x$ - rays $\left(\lambda_2\right)$
C. Infrared waves $\left(\lambda_3\right)$
D. Microwaves $\left(\lambda_4\right)$
Choose the most appropriate answer from the options given below

Options:
A $\lambda_4 < \lambda_3 < \lambda_1 < \lambda_2$
B $\lambda_2 < \lambda_1 < \lambda_4 < \lambda_3$
C $\lambda_1 < \lambda_2 < \lambda_3 < \lambda_4$
D $\lambda_4 < \lambda_3 < \lambda_2 < \lambda_1$

The correct answer is:
$\lambda_1 < \lambda_2 < \lambda_3 < \lambda_4$

Solution:
$\lambda_1 < \lambda_2 < \lambda_3 < \lambda_4$

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2024 (04 Apr Shift 2)
Arrange the following in the correct order of Lewis acidity: BF3, BCl3, BBr3.

Options:
A BF3 > BBr3 > BCl3
B BF3 > BCl3 > BBr3
C BF3 < BCl3 < BBr3
D BBr3 < BF3 < BCl3

The correct answer is:
BF3 < BCl3 < BBr3

Solution:
In BF3, BCl3 and BBr3 each will shown p Ï - p Ï back bonding as follows.
Extant of overlapping as the 2pâ2p>2pâ3p>2pâ4p difference between the size of p orbital increases the extent of overlapping decreases the availability of vacant orbital increases and the acidic character increases.
Hence correct order of Lewis acidity represented by option (BF3 < BCl3 < BBr3) .

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Arrange the following in the decreasing order (most important first) of their importance, for the welfare of human society. Give reasons for your answer.
Biogas, Citric acid, Penicillin and Curd

Options:

None

Solution:
Curd : Less important for society - as it depends on individual use and has only nutritions value.
Citric acid: Industrial use, not for dissipation in community.
Penicillin : Medicinal use of microbes, good for health of society, commercially more usable.
Biogas: Most important for community welfare as
(i) it reduces excreta, waste from community.
(ii) it produces inflammable gases, can be used as energy source.
(iii) it is a renewable source.
(iv) it has multidimensional utility.
(v) it is easily maintained and dissipated for community purpose.
So, Penicillin $>$ Biogas $>$ Curd $>$ Citric acid.

Topic: Biology, Microbes in Human Welfare
Arrange the following in the decreasing order of their covalent character :

(A) LiCl

(B) NaCl

(C) KCl

(D) CsCl

Choose the most appropriate answer from the options given below

Options:
A A>C>B>D
B B>A>C>D
C A>B>C>D
D A>B>D>C

The correct answer is:
A>B>C>D

Solution:
Lithium chloride is the alkali metal chloride which is most covalent in nature. This is due to the small size and high polarising power of lithium. That is, smaller the cation, greater is the polarising power and hence, greater is the covalent character.

So order of covalent character will be CsCl<RbCl<NaCl<LiCl

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2022 (29 Jun Shift 1)
Arrange the following in the increasing order of their basic strengths $\mathrm{CH}_{3} \mathrm{NH}_{2},\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}$, $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N}, \mathrm{NH}_{3}$

Options:
A $\mathrm{NH}_{3} < \left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N} < \left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH} < \mathrm{CH}_{3} \mathrm{NH}_{2}$
B $\mathrm{NH}_{3} < \left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N} < \mathrm{CH}_{3} \mathrm{NH}_{2} < \left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}$
C $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N} < \mathrm{NH}_{3} < \mathrm{CH}_{3} \mathrm{NH}_{2} < \left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}$
D $\mathrm{CH}_{3} \mathrm{NH}_{2} < \left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH} < \left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N} < \mathrm{NH}_{3}$

The correct answer is:
$\mathrm{NH}_{3} < \left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N} < \mathrm{CH}_{3} \mathrm{NH}_{2} < \left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}$

Solution:
Aliphatic amines are more basic than $\mathrm{NH}_{3}$ due to $+I$ effect of alkyl groups. In aqueous medium tertiary amine is less basic than secondary amine because the cation formed by protonation of tertiary amine is less solvated as compared to the cation formed by protonation of secondary amine. Hence, the increasing order of their basic strength is as:
$$
\mathrm{NH}_{3} < \left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N} < \mathrm{CH}_{3} \mathrm{NH}_{2} < \left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}
$$

Topic: Chemistry, Amines, JEE Main
Arrange the following ions in the order of decreasing $\mathrm{X}-\mathrm{O}$ bond length, where $\mathrm{X}$ is the central atom

Options:
A $\mathrm{ClO}_{4}^{-}, \mathrm{SO}_{4}^{2-}, \mathrm{PO}_{4}^{3-}, \mathrm{SiO}_{4}^{-}$
B $\mathrm{SiO}_{4}^{4-}, \mathrm{PO}_{4}^{3-}, \mathrm{SO}_{4}^{2-}, \mathrm{ClO}_{4}^{-}$
C $\mathrm{SiO}_{4}^{4-}, \mathrm{PO}_{4}^{3-}, \mathrm{ClO}_{4}^{-}, \mathrm{SO}_{4}^{2-}$
D $\mathrm{SiO}_{4}^{4-}, \mathrm{SO}_{4}^{2-}, \mathrm{PO}_{4}^{3-}, \mathrm{ClO}_{4}^{-}$

The correct answer is:
$\mathrm{SiO}_{4}^{4-}, \mathrm{PO}_{4}^{3-}, \mathrm{SO}_{4}^{2-}, \mathrm{ClO}_{4}^{-}$

Solution:
More will be the electronegativity of $\mathrm{X}$ lesser will be the bond length of $\underline{X}-\mathrm{O}$ bond.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
Arrange the following labelled hydrogens in decreasing order of acidity:

Options:
A b>a>c>d
B c>b>a>d
C b>c>d>a
D c>b>d>a

The correct answer is:
b>c>d>a

Solution:
Acidic strength∝ Stability of conjugate base General Order of acidic strength:
R-COOH>Ph-OH>R-C≡CH
'C' is more acidic due to-M effect of-NO2 .

Hence , order of acidic strength is: b>c>d>a

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2020 (02 Sep Shift 2)
Arrange the following molecules from most to least polar.

Options:
A III > V > II > I = IV
B II > IV > III > IV > I
C III > II > V > IV > I
D V > III > II > IV = I

The correct answer is:
III > V > II > I = IV

Solution:
(i) lnCH4 and CCl4 being tetrahedral (symmetrical molecules), bond dipoles cancel one other resulting in zero dipole moment.
(ii) As electronegativity difference between two bonded atoms increases, the bond polarity increases resulting into the increase in dipole moment.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Arrange the following orbitals in decreasing order of energy.

A. n=3, l=0, m=0
B. n=4, l=0, m=0
C. n=3, l=1, m=0
D. n=3, l=2, m=1

The correct option for the order is:

Options:
A D > B > C > A
B B > D > C > A
C A > C > B > D
D D > B > A > C

The correct answer is:
D > B > C > A

Solution:
In multi-electronic species, energy is decided on the basis of (n+l) rule.

3s = 3 + 0 + 0= 3

4s = 4 + 0 +0 = 4

3p = 3 + 1 +0 = 4

3d = 3 + 2 + 1 = 6

For 4s and 3p, an orbital with a higher value of n has higher energy.

So increasing the order of energy is 3d>4s>3p>3s.

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2023 (31 Jan Shift 2)
Arrange the following polymers in an increasing order of intermolecular forces; fibre, plastic, elastomer.

Options:
A Elastomer < Fibre < Plastic
B Elastomer < Plastic < Fibre
C Plastic < Elastomer < Fibore
D Fibre < Elastomer < Plastic

The correct answer is:
Elastomer < Plastic < Fibre

Solution:
Elastomers or rubbers have the weakest intermolecular forces of attraction followed by plastics while fibres have the strongest forces of attraction. Thus, the increasing intermolecular forces of attraction follows the order
elastomer < plastic < fibre.

Topic: Chemistry, Polymers, JEE Main
Arrange the following solutions in the decreasing order of pOH :
(A) 0.01 MHCl
(B) 0.01 MNaOH
(C) 0.01 MCH3COONa
(D) 0.01 MNaCl

Options:
A (A)>(C)>(D)>(B)
B (A)>(D)>(C)>(B)
C (B)>(C)>(D)>(A)
D (B)>(D)>(C)>(A)

The correct answer is:
(A)>(D)>(C)>(B)

Solution:
pH 0.01MHCl=2
pH of 0.01MNaOH=12
pH of 0.01MCH3COONa>7
pH of 0.01MNaCl=7
So the order of pH:B>C>D>A

So the order of pOH: B<C<D<A

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2020 (06 Sep Shift 1)
Arrange the formation of the following cells in the ovary in the sequential order:

i) Ovum
ii) Primary follicle
iii) Primary oocytes
iv) Graafian follicle

Options:
A ii), i), iv), iii)
B iii), ii), i), iv)
C iii), ii), iv), i)
D i), ii), iii), iv)

The correct answer is:
iii), ii), iv), i)

Solution:
In an ovary, there are millions of oocytes. An oocyte is not the same thing as a germ cell. The germ cells develop in the fetal ovary and before birth begin meiosis, and then freeze right at the start of the first meiotic division. These cells are stopped in prophase I of meiosis I, and are called primary oocytes. They lie within the ovaries surrounded by other cells, the follicular cells. The primary oocyte plus its follicular cells comprise the primary follicle.
At puberty, each oocyte enters meiosis again, just before it is ready to be ovulated. This means some oocytes can remain dormant for up to 50 years. FSH and LH act on a primary follicle in the ovaries, promoting its development into a secondary follicle.
The secondary follicle will continue to develop, and eventually forms a mature ovarian follicle called a tertiary, or Graafian, follicle. Once this mature follicle has developed, the further rise in oestrogen leads to a surge in LH (and some FSH), which stimulates ovulation.
The ovum (surrounded by its zona pellucida and cumulus of granulosa cells) will burst out of the follicle, ovulated and free to travel towards the uterus, hoping to meet a sperm cell on the way.

Topic: Biology, Human Reproduction, NEET
Arrange the given species F, F-, O and O-2 in the order of their radii

Options:
A O2->F->O>F
B O2->F->F>O
C F->O2->F>O
D O2->O>F->F

The correct answer is:
O2->F->O>F

Solution:
Atomic radius decreases on going from left to right in a period. Thus size of O>F and size of anion is larger then their atoms. As O2- and F- are isoelectronic, therefore, size of O2->F-.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
Arrange the main steps of plant breeding sequentially:

I. Cross-hybridisation among the selected parents
II. Evaluation and selection of parents
III. Selection and testing of superior recombinants
IV. Testing, release and commercialization of new cultivars
V. Collection of variability

Options:
A II V I IV III
B V II III I IV
C V II I III IV
D II I V III IV

The correct answer is:
V II I III IV

Solution:
Collection of variability is the entire collection of seeds or genes of a crop plant. Success of any plant breeding programme depends on the germplasm collection. Parents are evaluated and selected, followed by cross hybridisation among the selected parents, than testing and selection of superior recombinants. This is followed by testing the new cultivar in farmers field for three successive growing season.

Topic: Biology, Strategies for Enhancement in Food Production, NEET
Arrange the resonating structures of CH 2 N 2 in increasing order of stability.

Options:
A II > I > IV > III
B I > II > IV > III
C III > II > IV > I
D IV > II > III > I

The correct answer is:
II > I > IV > III

Solution:
(ii) Octet complete according to charge dispersion rule
(i) Octet complete but −ve charge present at less electronegative atom
(iv) −ve charge stable at more electronegative
(iii) −ve charge less stable at less electronegative atom

Topic: Chemistry, General Organic Chemistry, JEE Main
Arrange the various respiratory volumes in ascending order.

Options:
A Tidal Volume → Residual Volume → Expiratory Reserve Volume → Inspiratory Reserve Volume
B Tidal Volume → Expiratory Reserve Volume → Residual Volume → Inspiratory Reserve Volume
C Tidal Volume → Residual Volume → Inspiratory Reserve Volume → Expiratory Reserve Volume
D Residual Volume → Tidal Volume → Inspiratory Reserve Volume → Expiratory Reserve Volume

The correct answer is:
Tidal Volume → Expiratory Reserve Volume → Residual Volume → Inspiratory Reserve Volume

Solution:
Tidal Volume (TV): Volume of air inspired or expired during a normal respiration. It is approx. 500 mL., i.e., a healthy man can inspire or expire approximately 6000 to 8000 mL of air per minute. Inspiratory Reserve Volume (IRV): Additional volume of air, a person can inspire by a forcible inspiration. This averages 2500 mL to 3000 mL. Expiratory Reserve Volume (ERV): Additional volume of air, a person can expire by a forcible expiration. This averages 1000 mL to 1100 mL. Residual Volume (RV): Volume of air remaining in the lungs even after a forcible expiration. This averages 1100 mL to 1200 mL.

Topic: Biology, Breathing and Exchange of Gases, NEET
Arrange Yb+3 , Ce+3,La+3,Pm+3 ions in increasing order of their ionic radii

Options:
A Ce+3<Yb+3<Pm+3<La+3
B Yb+3<Pm+3<La+3<Ce+3
C Yb+3<Pm+3<Ce+3<La+3
D Pm+3<La+3<Ce+3<Yb+3

The correct answer is:
Yb+3<Pm+3<Ce+3<La+3

Solution:
Due to Lanthanoid contraction their is greater decrease in ionic radii of elements in lanthanide series because of poor shielding of nuclear charge by 4f electrons, hence the order will be Yb+3<Pm3+<Ce+3<La3+

Topic: Chemistry, d and f Block Elements, JEE Main
Arrangement of $\left(\mathrm{CH}_3\right)_3-\mathrm{C}-,\left(\mathrm{CH}_3\right)_2-\mathrm{CH}-, \mathrm{CH}_3-\mathrm{CH}_2$ - when attached to benzyl or an unsaturated group in increasing order of inductive effect is

Options:
A
$\left(\mathrm{CH}_3\right)_3-\mathrm{C}- < \left(\mathrm{CH}_3\right)_2-\mathrm{CH}- < \mathrm{CH}_3-\mathrm{CH}_2$
B
$\mathrm{CH}_3-\mathrm{CH}_2- < \left(\mathrm{CH}_3\right)_2- < \mathrm{CH}- < \left(\mathrm{CH}_3\right)_3-\mathrm{C}-$
C
$\left(\mathrm{CH}_3\right)_2-\mathrm{CH}- < \left(\mathrm{CH}_3\right)_3-\mathrm{C}- < \mathrm{CH}_3,-\mathrm{CH}_2$
D
$\left(\mathrm{CH}_3\right)_3-\mathrm{C}- < \mathrm{CH}_3-\mathrm{CH}_2-\left(\mathrm{CH}_3\right)_2-\mathrm{CH}-$

The correct answer is:
$\mathrm{CH}_3-\mathrm{CH}_2- < \left(\mathrm{CH}_3\right)_2- < \mathrm{CH}- < \left(\mathrm{CH}_3\right)_3-\mathrm{C}-$

Solution:
$\mathrm{CH}_3$ group has $+\mathrm{I}$ effect, as number of $-\mathrm{CH}_3$ group increases, the inductive effect increases.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2002
Arrhenius equation k=Ae-Ea/RT
If the activation energy of the reaction is found to be equal to RT, then [given : 1e=0.3679]

Options:
A the rate of reaction does not depend upon initial concentration
B the rate constant becomes about 37% of the Arrhenius constant A
C the rate constant becomes equal to 73% of the Arrhenius constant A
D the rate of the reaction becomes infinite of zero

The correct answer is:
the rate constant becomes about 37% of the Arrhenius constant A

Solution:
k=Ae-Ea/RT
When, Ea=RT
kA=1e=0.3679=0.37

Topic: Chemistry, Chemical Kinetics, JEE Main
Arsenic drugs are mainly used in the treatment of

Options:
A Jaundice
B Typhoid
C Syphilis
D Cholera.

The correct answer is:
Syphilis

Solution:
Arsenic drugs such as salvarsan is used for treatment of syphilis.

Topic: Chemistry, Chemistry in Everyday Life, AIIMS, AIIMS 2016
Arsenic sulphide sol is prepared by passing H2S through As2O3 solution. The charge developed on sol particles is due to the adsorption of

Options:
A H+
B S2-
C OH-
D O2-

The correct answer is:
S2-

Solution:
As2O3+3H2S→As2S3(sol)+3H2O
H2S molecules get adsorbed on As2S3 sol particles. Enthalpy of adsorption dissociates H2S to form H+ and S2- . H+ ions are lost and S2- ions remain adsorbed on the surface sol particles.

Topic: Chemistry, Surface Chemistry, NEET
Artificial ecosystem are

Options:
A The ecosystems that are created and maintained by man.
B The ecosystem made by the addition of energy and planned manipulation with a natural balance.
C Corp lands, orchards, gardens, aquariums, greenhouses etc., are examples for artificial ecosystems.
D All of these

The correct answer is:
All of these

Solution:
An ecosystem is a functional unit of nature, where living organisms interact among themselves and also with the surrounding physical environment. Ecosystem varies greatly in size from a small pond to a large forest or a sea. Forest, grassland and desert are some examples of terrestrial ecosystems; pond, lake, wetland, river and estuary are some examples of aquatic ecosystems. Crop fields and an aquarium may also be considered as man-made ecosystems or artificial ecosystems. These are the ecosystems that are maintained by the man with certain manipulation.

Topic: Biology, Ecosystem, NEET
Artificial light, extended work-time and reduced sleep-time disrupt the activity of

Options:
A thymus gland
B pineal gland
C adrenal gland
D posterior pituitary gland

The correct answer is:
pineal gland

Solution:
Artificial light, extended work time and reduced sleep time disrupt the activity of pineal gland. It is a small pea-shaped gland in the brain. It produces melatonin, which helps maintain circadian rhythm and regulate reproductive hormones.

Topic: Biology, Chemical Coordination and Integration, NEET, NEET 2019 (Odisha)
Aryl fluoride may be prepared from arene diazonium chloride using :

Options:
A
$\mathrm{HBF}_4 / \Delta$

B
$\mathrm{HBF}_4 / \mathrm{NaNO}_2, \mathrm{Cu}, \Delta$

C
$\mathrm{CuF} / \mathrm{HF}$

D
$\mathrm{Cu} / \mathrm{HF}$
0

The correct answer is:
$\mathrm{HBF}_4 / \Delta$

Solution:

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2013 (09 Apr Online)
As a wave propagates

Options:
A The wave intensity remains constant for a plane wave
B The wave intensity decreases as the inverse of the distance from the source for a spherical wave
C The wave intensity decreases as the inverse square of the distance from the
source for a spherical wave
D Total intensity of the spherical wave over the spherical surface centered at the source remains constant at all times

The correct answer is:
The wave intensity decreases as the inverse of the distance from the source for a spherical wave

Solution:
For a travelling wave, the intensity of wave remains constant if it is a plane wave
Intensity of spherical wave on the spherical surface centred at source always
remains same. Here total intensity means power $P$

Topic: Physics, Waves and Sound, JEE Main
As an electron makes a transition from an excited state to the ground state of a hydrogen-like atom/ion

Options:
A kinetic energy and total energy decrease but potential energy increases
B its kinetic energy increases but potential energy and total energy decrease
C kinetic energy, potential energy and total energy decrease
D kinetic energy decreases, potential energy increases but total energy remains same

The correct answer is:
its kinetic energy increases but potential energy and total energy decrease

Solution:
U=-e24πε0r, U= potential energy, K=e24πε0r
K=kinetic energy. E=U+K=-e28πε0r, E=total energy∴ As electron de-excites from an excited state to ground state, K increases, but U and E decrease.

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2015 (04 Apr)
As an electron makes a transition from an excited state to the ground state of a hydrogen-like atom/ion

Options:
A kinetic energy and total energy decrease but potential energy increases
B its kinetic energy increases but potential energy and total energy decrease
C kinetic energy, potential energy and total energy decrease
D kinetic energy decreases, potential energy increases but total energy remains same

The correct answer is:
its kinetic energy increases but potential energy and total energy decrease

Solution:
U=-e24πε0r
U= potential energy
K=e24πε0r
K=kinetic energy
E=U+K= -e28πε0r
E=total energy
∴ As electron de-excites from an excited state to ground state, K increases but U and E decrease.

Topic: Physics, Atomic Physics, NEET
As compared to a dicot root, a monocot root has

Options:
A more abundant secondary xylem
B many xylem bundles
C inconspicuous annual rings
D relatively thicker periderm

The correct answer is:
many xylem bundles

Solution:
The vasculature is radial in dicot root i.e., xylem bundles alternate with phloem bundles. Usually vascular bundles are $2-6$ in number while in monocot root the vascular strand consists of several ( 8 or more) alternate radial xylem and phloem bundles. Secondary growth takes place in dicot root while absent in monocot roots and annual rings do not form.

Topic: Biology, Anatomy of Flowering Plants, NEET, NEET 2012 (Mains)
As compared to lithium, sodium reacts quickly with water because

Options:
A Its molecular weight is less
B It is stronger electronegative
C It is stronger electropositive
D It is a metal

The correct answer is:
It is stronger electropositive

Solution:
On moving down the group electropositive character increases.

Topic: Chemistry, s Block Elements, JEE Main
As compared to potassium, sodium has

Options:
A Lower electronegativity
B Higher ionization potential
C Greater atomic radius
D Lower melting point

The correct answer is:
Higher ionization potential

Solution:
\(\begin{array} \text{Element} & \text{Na} & \text{K} \\ \text{IE}_1 & 496 & 419 \\ \text{IE}_2 & 4562 & 305 \\ \end{array}\)

Topic: Chemistry, s Block Elements, JEE Main
As each cell of the sporogenous tissue is capable of giving rise to a microspore tetrad. Each one is a potential

Options:
A Pollen grain
B Microspore
C Male gametophyte
D Pollen mother cell

The correct answer is:
Pollen mother cell

Solution:
The primary sporogenous cells divide to produce a mass of sporogenous cells or microsporocytes. Each microspore mother cell divides meiotically to form four haploid microspores or pollen grains and remains arranged in tetrads.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
As in figure shown, if a capacitor $C$ is charged by connecting it with resistance $R$, then energy is given by the battery will be

Options:
A $\frac{1}{2} C V^2$
B More than $\frac{1}{2} \mathrm{CV}^2$
C Less than $\frac{1}{2} \mathrm{CV}^2$
D Zero

The correct answer is:
Less than $\frac{1}{2} \mathrm{CV}^2$

Solution:
The maximum stared by the capacitor is given by,
\(\mathrm{u}_{\max }=\frac{1}{2} \mathrm{CV}^2\)
But when capacitor is charging with a resistor attached to it, a little extra energy will be required than \(u_{\max }\) to full charge the capacitor because resister is providing a resistance which make less in energy in form of heat.

Topic: Physics, Capacitance, JEE Main