datavorous/entrance-exam-dataset huggingface dataset:
Blood and bone marrow tests are mainly done for analysis of

Options:
A Leukemia
B Gastric carcinoma
C Skin carcinoma
D Brain tumour

The correct answer is:
Leukemia

Solution:
Cancer detection and diagnosis:
A. Biopsy and histopathological studies of the tissue. In a biopsy, a piece of the suspected tissue is taken out, cut into thin sections, stained, and examined under a microscope (histopathological studies) by a pathologist for a proper suggestion.
B. Blood and bone marrow tests: These analyze the increased cell counts in the case of leukemias.
C. Cancers of the internal organs are mainly detected by the use of techniques like radiography (use of X-rays), CT (computed tomography), and MRI (magnetic resonance imaging).
CT: It uses X-rays to generate a three-dimensional image of the internals of an object.
MRI: It uses strong magnetic fields and non-ionizing radiations to detect the pathological/physiological changes in the living tissue.
D. Detection of Antibodies against cancer-specific antigens.
E. Techniques of Molecular biology to detect genes in individuals with inherited susceptibility to certain cancers.

Topic: Biology, Human Health and Diseases, NEET
Blood calcium level is a resultant of how much dietary calcium is absorbed, how much calcium is lost in the urine, how much bone dissolves releasing calcium into blood and how much calcium from blood enters tissues. A number of factor play an important role in these processes. Mark the one which has no role.

Options:
A
Vitamin-D

B
Parathyroid hormone

C
Thyrocalcitonin

D
Thymosin

The correct answer is:
Thymosin

Solution:
Thymosin hormone has no role in regulation of blood calcium homeostasis. It plays a major role in the differentiation of T-lymphocytes, which provides cell mediated immunity, vitamin-D, thyrocalcitonin and parathyroid hormone play a vital role in maintaining calcium homeostasis by regulating blood calcium levels.

Topic: Biology, Chemical Coordination and Integration
Bob of a simple pendulum of length $l$ is made of iron. The pendulum is oscillating over a horizontal coil carrying direct current. If the time period of the pendulum is $\mathrm{T}$ then :

Options:
A
$T < 2 \pi \sqrt{\frac{l}{g}}$ and damping is smaller than in air alone.
B
$T=2 \pi \sqrt{\frac{l}{g}}$ and damping is larger than in air alone.

C
$T>2 \pi \sqrt{\frac{l}{g}}$ and damping is smaller than in air alone.

D
$T < 2 \pi \sqrt{\frac{l}{g}}$ and damping is larger than in air alone.

The correct answer is:
$T < 2 \pi \sqrt{\frac{l}{g}}$ and damping is larger than in air alone.

Solution:
When the pendulum is oscillating over a current carrying coil, and when the direction of oscillating pendulum bob is opposite to the direction of current. Its instantaneous acceleration increases.
Hence time period $\mathrm{T} < 2 \pi \sqrt{\frac{\ell}{\mathrm{g}}}$ and damping is larger than in air alone due energy dissipation.

Topic: Physics, Oscillations, JEE Main, JEE Main 2013 (23 Apr Online)
BOD stands for:

Options:
A Bacterial Oxidation Demand
B Biological oxygen demand
C Biochemical oxygen demand
D Both A and B

The correct answer is:
Both A and B

Solution:
Biochemical Oxygen Demand (BOD) is also called Biological Oxygen Demand.

It is the amount of dissolved oxygen needed (i.e. demanded) by aerobic biological organisms to break down organic material present in a given water sample at certain temperature over a specific time period

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2016 (09 Apr Online)
BOD values (in ppm) for clean water (A) and polluted water (B) are expected respectively as:

Options:
A A<5, B>17
B A>50, B<27
C A>15, B>47
D A>25, B<17

The correct answer is:
A<5, B>17

Solution:
Clean water would have BOD a value of less than, 5 ppm whereas highly polluted water could have a BOD value of 17 ppm or more.

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2021 (31 Aug Shift 1)
B.O.D values of four sample of water \( A, B, C \) and \( D \) is given below
A. \( 160 \mathrm{ppm} \)
B. \( 35 \mathrm{ppm} \)
C. \( 180 \mathrm{ppm} \)
D. \( 25 \mathrm{ppm} \)
The decreasing order of extent of pollution in water is

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

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

Solution:
Less polluted water has lower B.O.D value.

Topic: Chemistry, Environmental Chemistry, JEE Main
Body A of mass 4m moving with speed u collides with another body B of mass 2m at rest. The collision is head on and elastic in nature. After the collision, the fraction of energy lost by the colliding body A is

Options:
A 19
B 89
C 49
D 59

The correct answer is:
89

Solution:
4mu+0=4mv1+2mv2
⇒2u=2v1+v2 …(A)
For elastic collision,
e=1
⇒u=v2-v1 …(B)

From A and B,
4u=3v2⇒v2=43u
and v1=u3
Fraction of energy lost by (A),
=124mu2 - 124mv12124mu2
=u2 - u29u2=89

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Bohr's atom model assumes____

Options:
A The nucleus is of infinite mass and is at rest.
B Electrons in a quantised orbit will not radiate energy.
C mass of the electron remains constant.
D All the above conditions.

The correct answer is:
All the above conditions.

Solution:
In 1913, Niels Bohr proposed a model for the hydrogen atom. The Bohr model was based on the following assumptions.
- The electron in a hydrogen atom travels around the nucleus in a circular orbit
- The nucleus is of infinite mass and is rest
- Electrons in a quantized orbit will not radiate energy.
- Mass of electron remains constant.

Topic: Physics, Atomic Physics, JEE Main
Bohr's atom model assumes

Options:
A the nucleus is of infinite mass and is at rest
B electrons in a quantised orbit will not radiate energy
C mass of the electron remains constant
D All of the above

The correct answer is:
All of the above

Solution:
Bohr assumed one of the postulates of Rutherford's atomic model. That is the mass of the nucleus is very large compared to that of the electrons and hence assumed to be infinite. Electrons in quantised orbit does not radiate energy. Mass of electron remains constant.

Topic: Physics, Atomic Physics, COMEDK, COMEDK 2017
Bohr's radius can have

Options:
A Discrete values
B +ve values
C -ve values
D Fractional values

The correct answer is:
+ve values

Solution:
Bohr's radius $=\frac{n^2 h^2}{4 \pi^2 m e^2 z}$. Which is a positive quantity.

Topic: Chemistry, Structure of Atom, JEE Main
Boiling of hard water is helpful in removing the temporary hardness by converting calcium hydrogen carbonate and magnesium hydrogen carbonate to

Options:
A CaCO3 and MgOH2
B

CaCO3 and MgCO3

C CaOH2 and MgCO3
D CaOH2 and MgOH2

The correct answer is:
CaCO3 and MgOH2

Solution:
During boiling MgHCO32 is converted to insoluble MgOH2

MgHCO32→BoilingMgOH2↓+2CO2

CaHCO32→boilingCaCO3↓+H2O+CO2

CaHCO32 changes to insoluble CaCO3

Mg(OH)2â is less soluble than MgCO3. Hence, the products formed are CaCO3 and Mg(OH)2.â

Topic: Chemistry, Hydrogen, JEE Main, JEE Main 2022 (26 Jun Shift 2)
Boiling point of a 2% aqueous solution of a nonvolatile solute A is equal to the boiling point of 8% aqueous solution of a non-volatile solute B. The relation between molecular weights of A and B is.

Options:
A MA=4MB
B MB=4MA
C MA=8MB
D MB=8MA

The correct answer is:
MB=4MA

Solution:
For A : 100gm solution →2gm solute A ∴ Molality =2/MA0.098

For B : 100gm solution →8gm solute B ∴ Molality =8/MB0.092

∵ ΔTBA=ΔTBB

∴ Molality of A= Molality of B

∴20.098MA=80.092MB

298×928=MAMB

14.261=MAMB

∴ MB=4.261×MA

Topic: Chemistry, Solutions, JEE Main, JEE Main 2022 (27 Jul Shift 1)
Boiling point of \( \mathrm{H}_{2} \mathrm{O} \) is higher than that of \( \mathrm{H}_{2} \mathrm{~S} \), because the former

Options:
A is capable of forming \( \mathrm{H} \) - bonds
B has higher molecular mass
C has relatively strong covalent bonds
D is capable of forming co-ordinate bonds with \( \mathrm{H}^{+} \)ions

The correct answer is:
is capable of forming \( \mathrm{H} \) - bonds

Solution:
In H2OH-bond is present

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Bond angles of \( \mathrm{NH}_{3}, \mathrm{PH}_{3}, \mathrm{AsH}_{3} \) and \( \mathrm{SbH}_{3} \) are in the order

Options:
A \( \mathrm{PH}_{3}>\mathrm{AsH}_{3}>\mathrm{SbH}_{3}>\mathrm{NH}_{3} \)
B \( \mathrm{SbH}_{3}>\mathrm{AsH}_{3}>\mathrm{PH}_{3}>\mathrm{NH}_{3} \)
C \( \mathrm{SbH}_{3}>\mathrm{AsH}_{3}>\mathrm{NH}_{3}>\mathrm{PH}_{3} \)
D \( \mathrm{NH}_{3}>\mathrm{PH}_{3}>\mathrm{AsH}_{3}>\mathrm{SbH}_{3} \)

The correct answer is:
\( \mathrm{NH}_{3}>\mathrm{PH}_{3}>\mathrm{AsH}_{3}>\mathrm{SbH}_{3} \)

Solution:
According to Drago's rule, elements belonging to the third period or below do not show hybridisation. This is accompanied by a decrease in bond angle to nearly 90°.

However, in the case of ammonia, the bond angle is approximately 107°.
NH3>PH3>AsH3>SbH3

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Bond dissociation energy of E-H bond of the "H2E" hydrides of group 16 elements (given below), follows order.
(A) O
(B) S
(C) Se
(D) Te

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

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

Solution:
Bond strength is inversely proportional to the bond length of E - H bond of hydrides of 16th group hydrides. The bond length increases down the group. Hence, bond dissociation enthalpy decreases down the group.

Hence, bond dissociation energy of E-H bond in hydrides of group 16 follows the order

H2O > H2 S > H2Se > H2Te.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2023 (30 Jan Shift 2)
Bond dissociation enthalpy of $\mathrm{H}_2, \mathrm{Cl}_2$ and $\mathrm{HCl}$ are 434,242 and $431 \mathrm{kJmol}^{-}$ ${ }^1$ respectively. Enthalpy of formation of $\mathrm{HCl}$ is

Options:
A $245 \mathrm{kJmol}^{-1}$
B $93 \mathrm{kJmol}^{-1}$
C $-245 \mathrm{kJmol}^{-1}$
D $-93 \mathrm{kJmol}^{-1}$

The correct answer is:
$-93 \mathrm{kJmol}^{-1}$

Solution:
$$
\begin{aligned}
& 2[\mathrm{H}]+2[\mathrm{Cl}] \longrightarrow 2 \mathrm{HCl}, \Delta \mathrm{H}=-862 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \mathrm{H}_2 \longrightarrow 2[\mathrm{H}], \Delta \mathrm{H}=434 \mathrm{~kJ} \\
& \mathrm{Cl}_2 \longrightarrow 2[\mathrm{Cl}], \Delta \mathrm{H}=242 \mathrm{~kJ} \\
& \mathrm{H}_2+\mathrm{Cl}_2 \longrightarrow 2 \mathrm{HCl}, \Delta \mathrm{H}=-186 \mathrm{~kJ} \\
& \therefore \Delta \mathrm{H}_4 \text { of } \mathrm{HCl}=\frac{-186}{2}=-93 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}
$$

Topic: Chemistry, Thermodynamics (C), NEET, NEET 2008 (Mains)
Bond dissociation on energy of Cl2 is 240 kJ/mol . The longest wavelength of photon that can break this bond would be NA=6×1023, h=6.6×10-34 J/s

Options:
A 4.95×10-7m
B 9.9×10-7m
C 4.95×10-6m
D 9.9×10-6m

The correct answer is:
4.95×10-7m

Solution:
E=NA×hCλ
240×103=6×1023×6⋅6×10-34×3×108λ
λ=4.95×10-7 m

Topic: Chemistry, Structure of Atom, JEE Main
Bond distance in $\mathrm{HF}$ is $9.17 \times 10^{-11} \mathrm{~m}$. Dipole moment of $\mathrm{HF}$ is $6.104 \times 10^{-30} \mathrm{Cm}$. The percentage ionic character in $\mathrm{HF}$ will be : (electron charge $=1.60 \times 10^{-19} \mathrm{C}$ )

Options:
A
$61.0 \%$

B
$38.0 \%$

C
$35.5 \%$

D
$41.5 \%$

The correct answer is:
$41.5 \%$

Solution:
Given $e=1.60 \times 10^{-19} \mathrm{C}$
$$
d=9.17 \times 10^{-11} \mathrm{~m}
$$
From $\mu=e \times d$
$$
\begin{aligned}
\mu & =1.60 \times 10^{-19} \times 9.17 \times 10^{-11} \\
& =14.672 \times 10^{-30}
\end{aligned}
$$
$\%$ ionic character
$=\frac{\text { Observed dipole moment }}{\text { Dipole moment for } 100 \%}$
$$
\begin{aligned}
& =\frac{6.104 \times 10^{-30}}{14.672 \times 10^{-30}} \times 100 \\
& =41.5 \%
\end{aligned}
$$

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2013 (23 Apr Online)
Bond energies in $\mathrm{NO}, \mathrm{NO}^{+}$and $\mathrm{NO}^{-}$are such as

Options:
A $\mathrm{NO}^{-}\gt\mathrm{NO}\gt\mathrm{NO}^{+}$
B $\quad \mathrm{NO}\gt\mathrm{NO}^{-}\gt\mathrm{NO}^{+}$
C $\mathrm{NO}^{+}\gt\mathrm{NO}\gt\mathrm{NO}^{-}$
D $\quad \mathrm{NO}^{+}\gt\mathrm{NO}^{-}\gt\mathrm{NO}$

The correct answer is:
$\mathrm{NO}^{+}\gt\mathrm{NO}\gt\mathrm{NO}^{-}$

Solution:
Bond order of $\mathrm{NO}^{+}, \mathrm{NO}$ and $\mathrm{NO}^{-}$are $3,2.5$ and 2 respectively, bond energy $\propto$ bond order

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Bond enthalpies of $\mathrm{H}_{2}, \mathrm{X}_{2}$ and $\mathrm{HX}$ are in the ratio 2: 1: 2 . If enthalpy of formation of $\mathrm{HX}$ is $-50 \mathrm{~kJ}$ $\mathrm{mol}^{-1},$ the bond enthalpy of $\mathrm{X}_{2}$ is

Options:
A $100 \mathrm{~kJ} \mathrm{~mol}^{-1}$
B $300 \mathrm{~kJ} \mathrm{~mol}^{-1}$
C $200 \mathrm{~kJ} \mathrm{~mol}^{-1}$
D $400 \mathrm{~kJ} \mathrm{~mol}^{-1}$

The correct answer is:
$100 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Solution:
$\frac{1}{2} \mathrm{H}_{2}+\frac{1}{2} \mathrm{X}_{2} \longrightarrow \mathrm{HX}$
Let the bond enthalpy of $\mathrm{X}-\mathrm{X}$ bond be $x$.
$\Delta \mathrm{H}_{\mathrm{f}}(\mathrm{HX})=-50$
$=\frac{1}{2} \Delta \mathrm{H}_{\mathrm{H}-\mathrm{H}}+\frac{1}{2} \Delta \mathrm{H}_{\mathrm{X}-\mathrm{X}}-\Delta \mathrm{H}_{\mathrm{H}-\mathrm{X}}$
$=\frac{1}{2} 2 x+\frac{1}{2} x-2 x=\frac{-x}{2}$
$\therefore x=50 \times 2=100 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Bond enthalpies of \( \mathrm{A}_{2}, \mathrm{~B}_{2} \) and \( \mathrm{AB} \) are in the ratio \( 2: 1: 2 \). If bond enthalpy of formation of \( \mathrm{AB} \) is \( -100 \mathrm{KJ} \mathrm{mol}^{-1} \). The bond enthalpy of \( \mathrm{B}_{2} \) is

Options:
A \( 100 \mathrm{KJmol}^{-1} \)
B \( 50 \mathrm{KJmol}^{-1} \)
C \( 200 \mathrm{KJmol}^{-1} \)
D \( 150 \mathrm{KJmol}^{-1} \)

The correct answer is:
\( 200 \mathrm{KJmol}^{-1} \)

Solution:
Assume bond strength of A2=2X, then,

B2=X, AB=2X

ΔrHΘ=∑ΔdissHΘ-(R)∑ΔdissHΘ(P)

12 A2+12 B2→AB

X+X2 - 2X = -100 kJ mol-1X = 200 kJ mol-1

Topic: Chemistry, Thermodynamics (C), JEE Main
Bond line formula of HOCH(CN)2 is:

Options:
A
B
C
D

The correct answer is:

Solution:
A bond line structural formula is another method of structural representation of organic compounds. Here, every bond is represented as a line in a zigzag manner. The carbon and hydrogen atoms are not written and do not appear in the structure. Only, the bonds present between the carbon atoms are shown.

In the bond-line formula for HOCHCN2 is

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2024 (27 Jan Shift 2)
Bond order normally gives idea of stability of a molecular species. All the molecules viz. $\mathrm{H}_2, \mathrm{Li}_2$ and $\mathrm{B}_2$ have the same bond order yet they are not equally stable. Their stability order is :

Options:
A
$\mathrm{H}_2>\mathrm{B}_2>\mathrm{Li}_2$

B
$\mathrm{Li}_2>\mathrm{H}_2>\mathrm{B}_2$

C
$\mathrm{Li}_2>\mathrm{B}_2>\mathrm{H}_2$

D
None of these

The correct answer is:
None of these

Solution:
The molecular orbital configuration of the given molecules is
$\mathrm{H}_2=\sigma 1 s^2$ (no electron anti-bonding)
$\mathrm{Li}_2=\sigma 1 s^2 \sigma^* 1 s^2 \sigma 2 s^2$ (two anti-bonding electrons)
$$
\mathrm{B}_2=\sigma 1 s^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2
$$
$$
\left\{\pi 2 p_y^1=\pi 2 p_z^1\right\}
$$
(4 anti-bonding electrons)
Though the bond order of all the species are same $(\mathrm{B} . \mathrm{O}=1)$ but stability is different. This is due to difference in the presence of no. of anti-bonding electron.
Higher the no. of anti-bonding electron lower is the stability hence the correct order is
$$
\mathrm{H}_2>\mathrm{Li}_2>\mathrm{B}_2
$$

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2013 (22 Apr Online)
Bonding in which of the following diatomic molecule(s) become(s) stronger, on the basis of MO Theory, by removal of an electron?

(A) NO

(B) N2

(C) O2

(D) C2

(E) B2

Choose the most appropriate answer from the options given below :

Options:
A A,B,C only
B B,C,E only
C A,C only
D D only

The correct answer is:
A,C only

Solution:
In all these molecules O2​ and NO, is stabilized by removal of electrons because it is having electrons in the anti-bonding orbital. And thus removal of electron from the anti bonding orbital, increases the bond order of O2 and NOâ. And thus is becomes more stable after removal of electrons. Bond order of O2+​is 2.5 and it is more than that of O2â which is having bond order as 2. Similarly Bond order of NO is 2.5 and that of NO+ is 3.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2022 (25 Jun Shift 1)
Boolean relation at the output stage- Y for the following circuit is:

Options:
A A-+B-
B A+B
C A∙B
D A-∙B-

The correct answer is:
A-∙B-

Solution:
First part of figure shown is OR gate and
Second part of figure shown is NOT gate
So YP=OR+NOT=NOR gate
Y= A+B ¯ = A ¯ . B ¯

Topic: Physics, Semiconductors, JEE Main, JEE Main 2020 (08 Jan Shift 1)
Boric acid in solid, whereas BF3 is gas at room temperature because of

Options:
A Strong ionic bond in Boric acid
B Strong van der Waal's interaction in Boric acid
C Strong hydrogen bond in Boric acid
D Strong covalent bond in BF3

The correct answer is:
Strong hydrogen bond in Boric acid

Solution:
Inter-molecular structure of Boric acid(H3BO3) in solid phase:
Boric acid has strong hydrogen bonding as we can see in the above figure, while BF3 does not. Therefore boric acid is solid.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main, JEE Main 2023 (30 Jan Shift 2)
Boric acid is an acid because its molecule

Options:
A contains replaceable \( \mathrm{H}^{+} \)ion.
B gives up a proton.
C accepts \( \mathrm{OH}^{-} \)from water releasing proton.
D combines with proton from water molecule

The correct answer is:
accepts \( \mathrm{OH}^{-} \)from water releasing proton.

Solution:
Boric acid can be considered as an acid because its molecule accepts OH- from water and releases proton. The structure of the boric acid is given as below,

The reaction can be represented as follows:

H3BO3+H2O⇌BOH-4+H+.

Topic: Chemistry, Ionic Equilibrium, JEE Main
Born-Haber cycle is used to determine

Options:
A Crystal energy
B Electron affinity
C Lattice energy
D All of these

The correct answer is:
All of these

Solution:
Born Haber cycle is used to determine the electron affinity, lattice energy and crystal energy.
According to the above discussion, we conclude Born Haber cycle is used to determine the electron affinity, lattice energy and crystal energy.
Hence, the option D is the correct answer. Because all the options match with the question.

Topic: Chemistry, Thermodynamics (C), JEE Main
Boron and silicon of very high purity can be obtained through :

Options:
A liquation
B zone refining
C vapour phase refining
D electrolytic refining

The correct answer is:
zone refining

Solution:
Germanium, Silicon, Boron, Gallium and Indium can be obtained in pure state by zone refining process.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2020 (05 Sep Shift 2)
Boron cannot form which one of the following anions?

Options:
A
$\mathrm{BH}_4^{-}$

B
$\mathrm{B}(\mathrm{OH})_4^{-}$

C
$\mathrm{BO}_2^{-}$

D
$\mathrm{BF}_6^{3-}$

The correct answer is:
$\mathrm{BF}_6^{3-}$

Solution:
As Boron has only four orbitals in the valence shell ( i.e. $2 s, 2 p_x, 2 p_y \& 2 p_z$ ) it can show a maximum valency of four only.
Therefore $\left[\mathrm{BF}_6\right]^{3-}$ is not possible

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main, JEE Main 2011
Boron form covalent compound due to

Options:
A Higher ionization energy
B Lower ionization energy
C Small size
D Both (1) and (3)

The correct answer is:
Both (1) and (3)

Solution:
As boron is predominantly a non-metal it hardly loses the electrons and forms ionic bonds. It always forms covalent bonds due to following reasons:
1) Small size
2) High ionisation energy

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
Boron halides behave as Lewis acids because of their nature.

Options:
A proton donor
B covalent
C electron deficient
D ionising

The correct answer is:
electron deficient

Solution:
According to Lewis, the compound which can accept a lone pair of electron, are called acids. Boron halides, being electron deficient compounds, can accept a lone pair of electrons, so termed as Lewis acid.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
Boron in green plants assists in:

Options:
A activation of enzymes
B acting as enzyme co-factor
C photosynthesis
D sugar transport

The correct answer is:
sugar transport

Solution:
Boron in green plants assist sugar transport.
Related Theory
Boron is a micronutrient which performs a variety of functions such as pollination, seed set, cell wall formation, maintenance of the structural and functional integrity of plant cells and movement of sugar or energy in the growing parts of the plants.

Topic: Biology, Mineral Nutrition, NEET, NEET 2003
Both an electron and a photon have same de-Broglie wavelength of $1.2 Å$. The ratio of their energies is nearly

Options:
A 1 : 100
B 1 : 10
C 1 : 1000
D 1 : 1

The correct answer is:
1 : 100

Solution:
The de-Broglie wavelength of electron
$$
\lambda_e=\frac{h}{\sqrt{2 m K_e}}
$$
$K_e=$ Kinetic energy of electron
$$
\lambda_e^2=\frac{h^2}{2 m_e K_e} \Rightarrow K_e=\frac{h^2}{\lambda_e^2 \cdot 2 m_e}
$$
De-Broglie wavelength of photon is $\lambda_p$.
Kinetic energy of photon
$$
\begin{aligned}
& K_P=h v=\frac{h c}{\lambda_P} \\
& \therefore \quad \frac{K_e}{K_P}=\frac{h^2 / \lambda_e^2 \cdot 2 m_e}{h c / \lambda_P}
\end{aligned}
$$
As, $\lambda_e=\lambda_p$
$$
\begin{aligned}
& \text { Then, } \frac{K_e}{K_P}=\frac{h}{c \lambda_e \times 2 m_e} \\
& =\frac{6.66 \times 10^{-34}}{3 \times 10^8 \times 1.2 \times 10^{-10} \times 2 \times 9.1 \times 10^{-31}} \cong \frac{1}{100} \\
& \therefore \quad K_e: K_P:: 1: 100
\end{aligned}
$$

Topic: Physics, Dual Nature of Matter, JEE Main
Both Co3+ and Pt4+ have a coordination number of six. Which of the following pair of complexes will show approximately the same electrical conductance for their 0.001 M aqueous solutions? (assume 100% ionisation)

Options:
A CoCl3.4NH3Â andÂ PtCl4.4NH3
B CoCl3.3NH3Â andÂ PtCl4.5NH3
C CoCl3.6NH3 and PtCl4.5NH3
D CoCl3.6NH3Â andÂ PtCl4.3NH3

The correct answer is:
CoCl3.6NH3 and PtCl4.5NH3

Solution:
CoCl3.6NH3→CoNH36Cl3
Number of ion = 4
PtCl4.5NH3→PtNH35ClCl3
Number of ion = 4

Topic: Chemistry, Coordination Compounds, JEE Main
Both corpus luteum and macula lutea are

Options:
A found in human ovaries
B a source of hormones
C characterized by a yellow colour
D contributory in maintaining pregnancy.

The correct answer is:
characterized by a yellow colour

Solution:
Both corpus luteum and macula lutea are characterised by a yellow colour. After an egg is released from the ovary, the cells that line the ovarian follicle change shape and colour to become a corpus luteum (yellow body). Macula lutea is the yellowish spot present at the posterior pole of eye. Lutein, zeaxanthin, $\alpha$-carotene and $\beta$-cryptoxanthin are responsible for its yellow colouration.

Topic: Biology, Human Reproduction, AIIMS, AIIMS 2003
Both, hydrarch and xerarch successions lead to

Options:
A medium water conditions
B xeric conditions
C highly dry conditions
D excessive wet conditions

The correct answer is:
medium water conditions

Solution:
Hydrarch succession takes place in water areas and the successional series progress from hydric to the mesic conditions. As against this, xerarch succession takes place in dry areas and series progress from xeric to mesic conditions. Hence, both hydrarch and xerarch successions lead to medium water conditions (mesic)-neither too dry (xeric) nor too wet (hydric).

Topic: Biology, Ecosystem, NEET, NEET 2011 (Mains)