datavorous/entrance-exam-dataset huggingface dataset:
Which of the following are not electromagnetic waves?

Options:
A
cosmic rays

B
gamma rays

C
$\beta$-rays

D
X- rays

The correct answer is:
$\beta$-rays

Solution:
$\beta$ - rays are the beam of fast moving electrons

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2002
Which of the following are not membrane-bound?

Options:
A Ribosomes
B Lysosomes
C Mesosomes
D Vacuoles

The correct answer is:
Ribosomes

Solution:
Ribosomes were first observed under an electron microscope by George Palade (1953). Ribosomes are composed of RNA and proteins.

STRUCTURE:
The eukaryotic ribosomes are 80S, while the prokaryotic ribosomes are 70S. In this case, ‘S’ stands for sedimentation coefficient or Svedberg’s Unit. It is an indirect measure of the density and size of the ribosome. Both 70S and 80S ribosomes are composed of two subunits. It has two subunits. Small sub-unit and large sub-unit.
Prokaryotic Ribosome: 70S; Small subunit: 30S; Large subunit: 50S.
Eukaryotic Ribosome: 80S; Small sub-unit: 40S; Large subunit: 60S.

FUNCTIONS:
It is mainly involved in protein synthesis. Ribosomes are known as protein factories of the cell.

Topic: Biology, Cell - The Unit of Life, NEET, NEET 2015 (Phase 2)
Which of the following are not more basic than AlOH3 ?

Options:
A CaOH2, CeOH3
B YbOH3, LuOH3
C BOH3, BeOH2
D CeOH3, LuOH3 .

The correct answer is:
BOH3, BeOH2

Solution:
The basic strength of the hydroxides decreases as we move from cerium to lutetium Also the lanthanoid hydroxides are more basic than AlOH3 but less basic than CaOH2 .
BOH3 is more covalent than AlOH3.BeOH2 is as covalent as AlOH3 .
BOH3 is acidic while BeOH2 and AIOH3 are amphoteric while all others are more basic than AIOH3 .
Hence choice B is correct while A,B and D are incorrect.

Topic: Chemistry, d and f Block Elements, JEE Main
Which of the following are not state functions?
(I) $q+W$
(II) $q$
(III) $W$
(IV) $H-T S$

Options:
A (I) and (IV)
B (II), (III) and (IV)
C (I,) (II) and (III)
D (II) and (III)

The correct answer is:
(II) and (III)

Solution:
Key Idea : State function is the property of the system whose value depends only on the initial and final state of the system and is independent of the path.
$\because$ Internal energy $(\Delta E)=q+W$
It is a state function because it is independent of the path. It is an extensive property.
$\because$ Gibbs energy $(G)=H-T S$
It is also a state function because it is independent of the path. It is also an extensive property.
Heat $(q)$ and work $(W)$ are not state functions being path dependent.

Topic: Chemistry, Structure of Atom, JEE Main
Which of the following are not state functions?
(I) $q+w$
(II) $q$
(III) $\mathrm{w}$
(IV) H-TS

Options:
A (II) and (III)
B (I) and (IV)
C (II), (III) and (IV)
D (I), (II) and (III)

The correct answer is:
(II) and (III)

Solution:
$$
\begin{gathered}
\Delta U=q+w \\
\Delta G=\Delta H-T \Delta S
\end{gathered}
$$
$\Delta \mathrm{U}$ and $\Delta \mathrm{G}$ are state functions but $\mathrm{q}$ and w are not state functions.

Topic: Chemistry, Thermodynamics (C), NEET, NEET 2008 (Mains)
Which of the following are NOT under the control of thyroid hormone?

A. Maintenance of water and electrolyte balance

B. Regulation of basal metabolic rate

C. Normal rhythm of sleep-wake cycle

D. Development of immune system

E. Support the process of R.B.C.s formation

Choose the correct answer from the given options:

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

The correct answer is:
C and D only

Solution:
Development of immune system is a function of Thymosins secreted by Thymus glands.

Melatonin, secreted from the Pineal gland maintains the normal rhythms of sleep-wake cycle.

Thyroid hormones

Play an important role in the regulation of the basal metabolic rate.

These hormones also support the process of red blood cell formation.

Thyroid hormones control the metabolism of carbohydrates, proteins and fats.

Calcitonin from thyroid glands regulate the blood calcium levels.

Maintenance of water and electrolyte balance is also influenced by thyroid hormones.

Topic: Biology, Chemical Coordination and Integration, NEET, NEET 2023 (All India)
Which of the following are real life examples of reflective property of a parabola?

Options:
A Satellite dish
B Headlight
C Fountains
D Plane mirror

The correct answers are:
Satellite dish, Headlight, Plane mirror

Solution:
While the parabola has many beautiful geometric properties as we have seen, it also has a remarkable property known as the reflection property, which is used in such diverse places as car headlights and satellite dishes. A car headlight emits light from one source, but is able to focus that light into a beam so that the light does not move off in all directions - as lamplight does - but is focused directly ahead. Satellite dishes do the opposite. They collect electromagnetic rays as they arrive and concentrate them at one point, thereby obtaining a strong signal.
Headlight reflectors and satellite dishes are designed so their cross-sections are parabolic in shape and the point of collection or emission is the focus of the parabola.

Topic: Mathematics, Parabola, JEE Main
Which of the following are the correct representations of a zero order reaction, where $A$ represents the reactant?

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

The correct answer is:
i, ii, iv

Solution:
(i), (ii), (iv)
In zero order $R^{\prime} n$
$$
t_{\frac{1}{2}} \propto \frac{[A]_0}{2 K}
$$

Topic: Chemistry, Chemical Kinetics, JEE Main
Which of the following are the example of double salt?

(A) FeSO4·NH42SO4·6H2O

(B) CuSO4·4NH3·H2O

(C) K2SO4·Al2SO43·24H2O

(D) Fe(CN)2·4KCN

Choose the correct answer.

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

The correct answer is:
A and C only

Solution:
A double salt is a salt that contains two or more different cations or anions.

(A) FeSO4 + NH42SO4 + 6H2O → FeSO4·NH42SO4·6H2O(Mohr's salt)

Mohr’s salt called double salt which contains more than one simple salt. It undergoes complete dissociation in aqueous solutions.

(B) CuSO4·4NH3·H2O

Ammonia solution in water gives a blue precipitate when it combines with a solution of copper salt. The pale blue precipitate of copper hydroxide dissolves in excess of ammonium hydroxide forming tetraamine copper[II] sulphate, an azure blue (deep blue) soluble complex salt.

(C) K2SO4·Al2SO43·24H2O (Potash alum)

It is a double salt. Basically, it is comprised of two simple salts potassium sulphate and aluminium sulphate. Potash alum will not be wet in its solid form because the water of crystallisation is not free.

(D) Fe(CN)2·4KCN

It is complex salt. It does not dissociate completely into individual ions.

So, the correct answer is A and C only.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2023 (01 Feb Shift 1)
Which of the following are the Green house gases?

A. Water vapour

B.Ozone

C.I2

D.Molecular hydrogen

Choose the most appropriate answer from the options given below :

Options:
A

A and D only

B B and C only
C A and B only
D C and D only

The correct answer is:
A and B only

Solution:
The greenhouse effect is a process that occurs when gases in Earth's atmosphere trap the Sun's heat. This process makes Earth much warmer than it would be without an atmosphere. The main greenhouse gases whose concentrations are rising are carbon dioxide, methane, nitrous oxide, hydrochlorofluorocarbons (HCFCs), hydrofluorocarbons (HFCs) and ozone in the lower atmosphere. I2 and H2 are not green house gases.

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2023 (13 Apr Shift 2)
Which of the following are the transition metal ions responsible for colour in ruby and emerald respectively?

Options:
A Co3+ and Cr3+
B Co3+ and Co3+
C Cr3+ and Cr3+
D Cr3+ and Ag+

The correct answer is:
Cr3+ and Cr3+

Solution:
Ruby is Al2O3 in which Red colour is obtained when Cr+3 is replacing Al+3 ions in octahedral sites. Emerald is Be3Al2SiO36 in which green colour is obtained when Cr+3 is replacing Al+3 in octahedral site.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2016 (10 Apr Online)
Which of the following are true?

A. Speed of light in vacuum is dependent on the direction of propagation.

B. Speed of light in a medium is independent of the wavelength of light.

C. The speed of light is independent of the motion of the source.

D. The speed of light in a medium is independent of intensity.

Choose the correct answer from the question given below :

Options:
A A and C only
B B and D only
C B and C only
D C and D only

The correct answer is:
C and D only

Solution:
Speed of light in vaccum is independent of the frequency, wavelength, nature of the source, direction of propagation, motion of the source and intensity of the wave. But it depends on the refractive index of the medium.

Therefore, the speed of light in vacuum is a universal constant.

Hence, C and D are only correct.

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2023 (29 Jan Shift 1)
Which of the following arrangement does not represent the correct order of the property stated against it?

Options:
A V2+<Cr2+<Mn2+<Fe2+: paramagnetic behaviour
B Sc<Ti<Cr<Mn: number of oxidation states
C Co3+<Fe3+<Cr3+<Sc3+: stability in aqueous solution
D Ni2+<Co2+<Fe2+<Mn2+: ionic size

The correct answer is:
V2+<Cr2+<Mn2+<Fe2+: paramagnetic behaviour

Solution:
A. V2+ has 3 unpaired electrons
Cr2+ has 4 unpaired electrons
Mn2+ has 5 unpaired electrons
Fe2+ has 4 unpaired electrons
Hence the order of paramagentic behaviour should be
V2+<Cr2+=Fe2+<Mn2+
B. Ionic size decreases from left to right in same period.
C. As per data, we have
Co 3+ | Co 2+ =1.97;  Fe 3+ | Fe 2+ =0.77;  Cr 3+ | Cr 2+ =−0.41
Sc3+ is highly stable, because it does not show +2.
D. The oxidation state increases as we go from group 3 to group 7 in same period.

Topic: Chemistry, d and f Block Elements, JEE Main
Which of the following arrangement for the three halogens $\mathrm{Cl}, \mathrm{Br}$ and I when placed in the order of their increasing electron affinity is correct

Options:
A $\mathrm{Cl} \gt \mathrm{Br} \gt \mathrm{I}$
B $\mathrm{Br} \gt \mathrm{Cl} \gt \mathrm{I}$
C $I \gt B r \gt C l$
D $1 \gt \mathrm{Cl} \gt \mathrm{Br}$

The correct answer is:
$\mathrm{Cl} \gt \mathrm{Br} \gt \mathrm{I}$

Solution:
Electron affinity order of $\mathrm{Cl}, \mathrm{Br} ~\&~ \mathrm{I}$ is $\mathrm{Cl} \gt \mathrm{Br} \gt \mathrm{I}$. Hence, Option "1" is the correct answer.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Which of the following arrangements does not represent the correct order of the property stated against it?

Options:
A Co3+ < Fe3+ < Cr3+ < Sc3+ :â Stability in aqueous solution
B Sc < Ti < Cr < Mn :â Number of oxidation states
C V2+< Cr2+ < Mn2+ < Fe2+:ââ Paramagnetic behaviour
D Ni2+ < Co2+ < Fe2+ < Mn2+ :â Ionic size

The correct answer is:
V2+< Cr2+ < Mn2+ < Fe2+:ââ Paramagnetic behaviour

Solution:
No. of unpaired electrons
23V2+18Ar 4S0 3d3 = 324Cr2+18Ar 4S0 3d4 = 425Mn2+18Ar 4S0 3d5 = 526Fe2+18Ar 4S0 3d6 = 4

On moving left to right in the periodic table, ionic size decreases.

Sc shows only one oxidation state, and it is +3 but Mn shows variable oxidation state from +2 to +7.

Topic: Chemistry, d and f Block Elements, JEE Main
Which of the following arrangements does not represent the correct order of the property stated against it?

Options:
A $\mathrm{V}^{2+}<\mathrm{Cr}^{2+}<\mathrm{Mn}^{2+}<\mathrm{Fe}^{2+}$ :paramagnetic

behaviour
B $\mathrm{Ni}^{2+}<\mathrm{Co}^{2+}<\mathrm{Fe}^{2+}<\mathrm{Mn}^{2+}:$ ionic size
C $\mathrm{Co}^{3+}<\mathrm{Fe}^{3+}<\mathrm{Cr}^{3+}<\mathrm{Sc}^{3+}:$ stability in

aqueous solution
D $\mathrm{Sc}<\mathrm{Ti}<\mathrm{Cr}<\mathrm{Mn}:$ number of oxidation states

The correct answer is:
$\mathrm{V}^{2+}<\mathrm{Cr}^{2+}<\mathrm{Mn}^{2+}<\mathrm{Fe}^{2+}$ :paramagnetic

behaviour

Solution:
No Solution

Topic: Chemistry, Coordination Compounds, JEE Main
Which of the following arrangements does not represent the correct order of the property stated against it?

Options:
A Co3+ < Fe3+ < Cr3+ < Sc3+ :â Stability in aqueous solution
B Sc < Ti < Cr < Mn :â Number of oxidation states
C V2+< Cr2+ < Mn2+ < Fe2+:ââ Paramagnetic behaviour
D Ni2+ < Co2+ < Fe2+ < Mn2+ :â Ionic size

The correct answer is:
V2+< Cr2+ < Mn2+ < Fe2+:ââ Paramagnetic behaviour

Solution:
No. of unpaired electrons
23V2+18Ar 4S0 3d3 = 324Cr2+18Ar 4S0 3d4 = 425Mn2+18Ar 4S0 3d5 = 526Fe2+18Ar 4S0 3d6 = 4

On moving left to right in the periodic table, ionic size decreases.

Sc shows only one oxidation state, and it is +3 but Mn shows variable oxidation state from +2 to +7.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2013 (07 Apr)
Which of the following arrangements is correct in respect of solubility in water?

Options:
A $\mathrm{CaSO}_{4}>\mathrm{BaSO}_{4}>\mathrm{BeSO}_{4}>\mathrm{MgSO}_{4}>\mathrm{SrSO}_{4}$
B $\mathrm{BeSO}_{4}>\mathrm{MgSO}_{4}>\mathrm{CaSO}_{4}>\mathrm{SrSO}_{4}>\mathrm{BaSO}_{4}$
C $\mathrm{BaSO}_{4}>\mathrm{Sr} \mathrm{SO}_{4}>\mathrm{CaSO}_{4}>\mathrm{MgSO}_{4}>\mathrm{BeSO}_{4}$
D $\mathrm{BeSO}_{4}>\mathrm{CaSO}_{4}>\mathrm{MgSO}_{4}>\mathrm{SrSO}_{4}>\mathrm{BaSO}_{4}$

The correct answer is:
$\mathrm{BeSO}_{4}>\mathrm{MgSO}_{4}>\mathrm{CaSO}_{4}>\mathrm{SrSO}_{4}>\mathrm{BaSO}_{4}$

Solution:
The correct order of solubility of sulphates in water is
$\mathrm{BeSO}_{4}>\mathrm{MgSO}_{4}>\mathrm{CaSO}_{4}>\mathrm{SrSO}_{4}>\mathrm{BaSO}_{4}$
Solublity of 2nd group sulphates decreases as we down the group due to less release of hydration energy. $\mathrm{Be}^{2+} < \mathrm{Mg}^{2+} < \mathrm{Ca}^{2+} < \mathrm{Sr}^{2+} < \mathrm{Ba}^{24}$ (lonic Size)
As hydration energy decreases more rapidly than latice energy, the solubility decreases down the group. Hydration Energy $=\frac{\text { Charge }}{\text { Size }}$
While lathice energy almost remains constant. Hence, solubility decreases.

Topic: Chemistry, Practical Chemistry, WBJEE, WBJEE 2016
Which of the following arrangements represents the increasing order (smallest to largest) of ionic radii of the given species $\mathrm{O}^{2-}, \mathrm{S}^{2-}, \mathrm{N}^{3-}, \mathrm{P}^{3-}$ ?

Options:
A
$\mathrm{O}^{2-} < \mathrm{N}^{3-} < \mathrm{S}^{2-} < \mathrm{P}^{3-}$
B
$\mathrm{O}^{2-} < \mathrm{P}^{3-} < \mathrm{N}^{3-} < \mathrm{S}^{2-}$
C
$\mathrm{N}^3 < \mathrm{O}^{2-} < \mathrm{P}^{3-} < \mathrm{S}^{2-}$
D
$\mathrm{N}^{3-} < \mathrm{S}^{2-} < \mathrm{O}^{2-} < \mathrm{P}^{3-}$

The correct answer is:
$\mathrm{O}^{2-} < \mathrm{N}^{3-} < \mathrm{S}^{2-} < \mathrm{P}^{3-}$

Solution:
For isoelectronic species ionic radii decreases as the charge on ion decreases. Further on moving down in a group ionic radii increases. Hence the correct order is $\mathrm{O}^{2-} < \mathrm{N}^{3-} < \mathrm{S}^{2-} < \mathrm{P}^{3-}$

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2014 (12 Apr Online)
Which of the following arrangements show schematic alignment of magnetic moments of antiferromagnetic substances?

Options:
A
$\uparrow \uparrow \uparrow(\uparrow \uparrow \uparrow$

B
$\downarrow \emptyset \emptyset \emptyset \emptyset \emptyset$

C
$\uparrow \uparrow \uparrow \uparrow \uparrow \uparrow$

D
$\uparrow \uparrow \uparrow \uparrow \uparrow \uparrow$

The correct answer is:
$\uparrow \uparrow \uparrow \uparrow \uparrow \uparrow$

Solution:
The substances having domain structure which are opposite and cancel each other's magnetic field are known as antiferromagnetic substances.

Topic: Chemistry, Solid State
Which of the following arrangements shows the schematic alignment of magnetic moments of antiferromagnetic substance?

Options:
A

B

C

D

The correct answer is:

Solution:
When the magnetic moments align in a regular pattern with neighbouring spins pointing in opposite directions (as in d), the substance shows antiferromagnetism.

Topic: Chemistry, Solid State, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
Which of the following arrangements shows the schematic alignment of magnetic moments of antiferromagnetic substance?

Options:
A
B
C
D

The correct answer is:

Solution:
Substances which are expected to possess para-magnetic or ferromagnetism on the basis of unpaired electrons, but actually they possess zero net magnetic moment are called anti-ferromagnetic substance.

Topic: Chemistry, Solid State, JEE Main, JEE Main 2018 (15 Apr)
Which of the following arrangements shows the schematic alignment of magnetic moments of anti-ferromagnetic substance?

Options:
A
B
C
D

The correct answer is:

Solution:
In anti-ferromagnetic substances, the magnetic dipoles are oppositely oriented and cancel out each other's magnetic moment.

Topic: Chemistry, Solid State, JEE Main
Which of the following artificial sweeteners has the highest sweetness value in comparison to cane sugar?

Options:
A Aspartame

B Sucralose

C Alitame

D Saccharin

The correct answer is:
Alitame

Solution:
Saccharin, also called ortho-sulphobenzimide, is the first popular artificial sweetening agent. It is about 550 times as sweet as cane sugar. Its use is of great value to diabetic persons and people who need to control their intake of calories.

Aspartame is the most successful and widely used artificial sweetener. It is roughly 100 times as sweet as cane sugar. It is the methyl ester of dipeptide formed from aspartic acid and phenylalanine. Use of aspartame is limited to cold foods and soft drinks because it is unstable at cooking temperature.

Alitame is a high-potency sweetener, and it is roughly 2000 times as sweet as cane sugar.

Sucralose is trichloro derivative of sucrose. Its appearance and taste are like sugar. It is roughly 600 times as sweet as cane sugar.

So, sweetness value order with respect to cane sugar is:

Alitame > Sucralose > Saccharin > Aspartame

Topic: Chemistry, Chemistry in Everyday Life, JEE Main, JEE Main 2023 (31 Jan Shift 1)
Which of the following atomic and physical properties of hydrogen is false?

Options:
A Hydrogen \( > \) Deuterium \( > \) Tritium; (melting point \( / \mathrm{K} \) )
B Hydrogen \( < \) Deuterium \( < \) Tritium; (boiling point \( / \mathrm{K} \) )
C Hydrogen \( < \) Deuterium \( < \) Tritium; (density/ \( g L^{-1} \) )
D Hydrogen \( > \) Deuterium \( > \) Tritium; (\% relative abundance)

The correct answer is:
Hydrogen \( > \) Deuterium \( > \) Tritium; (melting point \( / \mathrm{K} \) )

Solution:
PropertyTDH
Melting point (K) 20.62 > 18.5 > 13.9
Boiling point (K) 24.9 > 23.5 > 20.4
Density (g/L) 0.27 > 0.18 > 0.09
Relative abundance (%) 10-15 < 0.016 < 99.984

Topic: Chemistry, Hydrogen, JEE Main
Which of the following atoms has the highest first ionization energy?

Options:
A K
B Sc
C Rb
D Na

The correct answer is:
Sc

Solution:
Ionisation energy decreases with increase in the size of atom. Na, K, Rb belongs to alkali metal and down the group ionization energy decreases as size of atom increases down the group. Among the given elements, Sc being small has the highest ionization energy (Na= 496, K=419; Sc =631 kJ mol-1)

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2016 (03 Apr)
Which of the following atoms has the highest first ionization energy?

Options:
A Sc
B Rb
C Na
D K

The correct answer is:
Sc

Solution:
Due to poor shielding of d-electrons in Sc, Zeff of Sc becomes more so that ionisation energy of Sc is more than Na, K and Rb.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
Which of the following atoms has the lowest ionization potential?

Options:
A
${ }_7^{14} \mathrm{~N}$

B
${ }_{55}^{133} \mathrm{Cs}$

C
${ }_{18}^{40} \mathrm{Ar}$

D
${ }_8^{16} \mathrm{~O}$

The correct answer is:
${ }_{55}^{133} \mathrm{Cs}$

Solution:
No Solution Available

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2003
Which of the following atoms should have the highest $1^{\prime \prime}$ electron affinity?

Options:
A F
B 0
C $\mathrm{N}$
D C

The correct answer is:
F

Solution:
$\because$ F-atom has 2 nd highest electron affinity
(Cl has highest) in the periodic table
$\therefore$ Among the given options F-has highest electron gain affinity

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
Which of the following attacks glass

Options:
A $\mathrm{HCl}$
B $H F$
C $\mathrm{HI}$
D $\mathrm{HBr}$

The correct answer is:
$H F$

Solution:
Glass being a mixture of sodium and calcium silicates reacts with hydrofluoric acid forming sodium and calcium fluorosilicates respectively.
$\begin{aligned} & \mathrm{Na}_2 \mathrm{SiO}_3+3 \mathrm{H}_2 F_2 \rightarrow \mathrm{Na}_2 \mathrm{SiF}_4+3 \mathrm{H}_2 \mathrm{O} \\ & \mathrm{CaSlO}_3+3 \mathrm{H}_2 F_2 \rightarrow \mathrm{CaSiF}_4+3 \mathrm{H}_2 \mathrm{O}\end{aligned}$

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
Which of the following azeotropes is not correctly matched?

Options:
A HNO3 (68%) +H2O (32%): Maximum boiling azeotrope, boiling point = 393.5 K
B H2O (43%) + HI (57%): Minimum boiling azeotrope, boiling point = 290 K
C C2H5OH (95.5%) +H2O (4.5%): Minimum boiling azeotrope, boiling point = 351.15 K
D Chloroform (93.2%) +C2H5OH (6.8%): Minimum boiling azeotrope, boiling point = 332.3 K

The correct answer is:
H2O (43%) + HI (57%): Minimum boiling azeotrope, boiling point = 290 K

Solution:
H2O (43%) and HI (57%) make maximum boiling azeotropic mixture on boiling at 400 K.

Topic: Chemistry, Solutions, JEE Main
Which of the following best describes how the capacitance changes when a slab of material with a dielectric constant K, having the same area as the plates of a parallel-plate capacitor but with a thickness of (3/4)d (where d is the separation of the plates), is inserted between the plates?

Options:
A The capacitance increases by a factor of 4K/(3+K).
B The capacitance decreases by a factor of (K+4)/3K.
C The capacitance remains the same.
D The capacitance increases by a factor of (K+3)/4K.

The correct answer is:
The capacitance increases by a factor of 4K/(3+K).

Solution:
Correct Option is : (A)
The capacitance increases by a factor of 4K/(3+K).

Topic: Chemistry, Capacitance, NEET
Which of the following best describes the concept of orthogonal trajectories in the context of differential equations?

Options:
A Solutions that intersect at right angles.
B Solutions that never intersect.
C Solutions that are parallel to each other.
D Solutions that have the same slope at every point.

The correct answer is:
Solutions that intersect at right angles.

Solution:
The correct answer is:
1- Solutions that intersect at right angles.

Topic: Mathematics, Differential Equations, JEE Main
Which of the following best describes the diagram of molecular orbital?

Options:
A
A bonding $\pi$ orbital

B
A non-bonding orbital

C
An antibonding $\sigma$ orbital

D
An antibonding $\pi$ orbital

The correct answer is:
An antibonding $\pi$ orbital

Solution:
An antibonding $\pi$ orbital best describes the diagram of a molecular orbital. Two orbitals laterally overlap to form pi bond. Out of phase combination, of these two $p$ orbitals, give $\pi^*$ MO.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
Which of the following best estimates the amount of ice remaining in a cubical thermacole icebox with 30 cm sides and 5.0 cm thickness, after 6 hours, when initially filled with 4.0 kg of ice, and exposed to an outside temperature of 45°C, considering a coefficient of thermal conductivity of 0.01 J${{s}^{-1}}{{m}^{-1}}$ ${{K}^{-1}}$ for thermacole and a heat of fusion of water equal to 335 ×${{10}^{3}}$ J $k{{g}^{-1}}$ ?

Options:
A Approximately 2.67 kg of ice remaining
B Approximately 3.68 kg of ice remaining
C Approximately 3.98 kg of ice remaining
D Approximately 4.63 kg of ice remaining

The correct answer is:
Approximately 3.68 kg of ice remaining

Solution:
Correct Option is : (B)
Approximately 3.68 kg of ice remaining

Topic: Chemistry, Thermal Properties of Matter, NEET