datavorous/entrance-exam-dataset huggingface dataset:
Mercury is a liquid metal because

Options:
A it has a completely filled $s$-orbital
B it has a small atomic size
C it has a completely filled $d$-orbital that prevents $d-d$ overlapping of orbitals
D it has a completely filled $d$-orbital that causes d-(d) overlapping

The correct answer is:
it has a completely filled $s$-orbital

Solution:
Mercury is a liquid metal because it has completely filled $6 s$-orbital. The inner $f$-orbital poorly screens nuclear charge that stabilise $6 s$-orbital. Thus, it is not possible to form any bond with other elements.

Topic: Chemistry, d and f Block Elements, JEE Main
Mercury is filled in a tube of radius $2 \mathrm{~cm}$ up to a height of $30 \mathrm{~cm}$. The force exerted by mercury on the bottom of the tube is _____ N.
(Given, atmospheric pressure $=10^5 \mathrm{Nm}^{-2}$, density of mercury $=1.36 \times 10^4 \mathrm{~kg} \mathrm{~m}^{-}$ $\left.{ }^3, \mathrm{~g}=10 \mathrm{~m} \mathrm{~s}^{-2}, \pi=\frac{22}{7}\right)$

Options:

The correct answer is:
177

Solution:
$\begin{aligned} & \mathrm{F}=\mathrm{P}_0 \mathrm{~A}+\rho_{\mathrm{m}} \mathrm{ghA} \\ & =10^5 \times \frac{22}{7} \times\left(2 \times 10^{-2}\right)^2 \\ & +1.36 \times 10^4 \times 10 \times\left(30 \times 10^{-2}\right)\left(\frac{22}{7} \times\left(2 \times 10^{-2}\right)^2\right) \\ & \mathrm{F}=51.29+125.71=177 \mathrm{~N}\end{aligned}$

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2024 (04 Apr Shift 2)
Mercury on heating with aqua-regia gives :

Options:
A Hg NO 3 2
B HgCl2
C Hg(NO2)2
D Hg2Cl2

The correct answer is:
HgCl2

Solution:
Aqua regia is a mtextxture of nitric acid and hydrochloric acid, optimally in a molar ratio of 1:3
3Hg+2HNO3+6HCl
â 3HgCl2+2NO+4H2O

Topic: Chemistry, d and f Block Elements, JEE Main
Mesityl oxide is a common name of:

Options:
A 2, 4 -Dimethylpentan-3-one
B 3-Methylcyclohexanecarbaldehyde
C 2-Methylcyclohexanone
D 4-Methylpent-3-en-2-one

The correct answer is:
4-Methylpent-3-en-2-one

Solution:
IUPAC [4-Methylpent-3-en-2-one]

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2021 (17 Mar Shift 1)
Meso-tartaric acid is

Options:
A Optically inactive
B Optically active because of molecular symmetry
C Optically inactive due to external compensation
D Optically active because of asymmetric carbon atom

The correct answer is:
Optically inactive

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Main
Metagenesis refers to:

Options:
A Alternation of generation between asexual and sexual phases of an organism
B Occurrence a drastic change in form during post-embryonic development
C Presence of a segmented body and parthenogenetic mode of reproduction
D Presence of different morphic forms

The correct answer is:
Alternation of generation between asexual and sexual phases of an organism

Solution:
Body forms: Two basic body forms are found in cnidarians which differ in structure and function from one another. These are called polyp and medusa. A polyp is a cylindrical form that remains attached to the substratum (hence sessile) whereas a medusa is an umbrella-shaped and free-swimming form. Some cnidarians exist only in polyp form, e.g., Hydra, Adamsia, while some have medusa as the dominant form in their life cycle (polyp may be reduced or absent), e.g., Aurelia (jellyfish). Some other cnidarians exist in both forms, i.e., polyp and medusa, e.g., Obelia. Those cnidarians which exist in both forms show alternation of generation in their life (metagenesis).

Topic: Biology, Animal Kingdom, NEET, NEET 2015 (Phase 2)
Metal alloys are used for making standard resistance coils because

Options:
A They have high thermal conductivity.
B Their resistance depend weakly on temperature.
C They have low thermal conductivity.
D Their resistance depend strongly on temperature.

The correct answer is:
Their resistance depend weakly on temperature.

Solution:
Metal alloys are used for standard resistance coils because their resistance depends weakly on temperature.

Topic: Physics, Current Electricity, JEE Main
Metal M crystallizes into a FCC lattice with the edge length of 4.0×10-8 cm. The atomic mass of the metal is g/mol.

(Use : NA=6.02×1023 mol-1, density of metal, M=9.03 g cm-3)

Options:

The correct answer is:
87

Solution:
Given ,a=4×10-8 cm

d=9.03 g/ml

d=ZMNAa3

Volume of a unit cell = a3, where a is the edge length

d is the density of the solid substance and M is the molar mass.

Mass of the unit cell = number of atoms in unit cell × mass of each atom =Z×m

(Here z is the number of atoms present in one unit cell and m is the mass of a single atom)M=9.03×6.02×1023×64×10-244=86.97g/mol

Topic: Chemistry, Solid State, JEE Main, JEE Main 2022 (28 Jul Shift 2)
Metal 'M' forms a carbonlyl compound in which it is present in its lower valance state. Which of the following bonding is possible in this metal carbonyl?

Options:
A
B
C
D

The correct answer is:

Solution:
Bonding in metal carbonyl
π -bond → filled d-orbitel of metal and empty π* molecular orbital of CO take part.
σ -bond → Empty hybrid orbital of metal and filled molecular orbital of carbon mono oxide.

Topic: Chemistry, Coordination Compounds, JEE Main
Metal occur in the native from because of their

Options:
A High electronegativity
B High reactivity
C Low reactivity
D Low density

The correct answer is:
Low reactivity

Solution:
Metal occur in the native from because of their less reactive metals are found in native state (free state).

Less reactive metal means it has low ionisation energy that means it can easily give electrons to show its reactivity.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
Metal which can be extracted from all the three dolomite, magnesite and carnallite is

Options:
A Na
B K
C Mg
D Ca

The correct answer is:
Mg

Solution:
Dolamite	MgCO 3 . CaCO 3

Magnesite	MgCO 3

Carnallite	KCl .MgCl 2 .6H 2 O

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
Metals can be prevented from rusting by -

Options:
A Connecting iron to more electropositive metal - cathodic protection
B Connecting iron to more electropositive metal - anodic protection
C Connecting iron to less electropositive metal - anodic protection
D Connecting iron to less electropositive metal - cathodic protection

The correct answer is:
Connecting iron to more electropositive metal - cathodic protection

Solution:
The process is called cathodic protection where iron acts as cathode and thus, not oxidised.

Topic: Chemistry, Electrochemistry, JEE Main
Metals generally melt at very high temperatures, Among the following which one has the highest melting point?

Options:
A Ga
B Ag
C Hg
D Cs

The correct answer is:
Ag

Solution:
Transition metals have high melting point due to their high enthalpy of atomisation. They form strong metallic bonds with unpaired electrons.

Melting point of Ag=961.8°C

Melting point of Cs=28.5°C

Melting point of Ga=30°C

Melting point of Hg=-38.83°C

Cs,Ga and Hg are liquid at or near room temperature.

Among the given options Ag has the maximum Melting point, as it belongs to transition metal series.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2022 (24 Jun Shift 2)
Metals of groups 7,8 and 9 do not form metallic hydrides. This is termed as

Options:
A Hydride gap
B Hydride shift
C Anhydride
D Dehydride

The correct answer is:
Hydride gap

Solution:
Hydride gap

Topic: Chemistry, Hydrogen, JEE Main
Metals which will not evolve H2 gas with dil. HCl are

Options:
A Cu, Ag, Au
B Cu, Zn, Al
C Fe, Ag, Pt
D Hg, Mg, Pt

The correct answer is:
Cu, Ag, Au

Solution:
The metals whose reduction potentials are greater than the reduction potential of Hydrogen do not liberate H2 gas.
H + + e â â 1 2 H 2 ;E Î¿ =0.00 V
Au 3+ + 3â e â â Au E o = + 1.50âV
Ag + + e â â Ag E o = + 0.799 V
Cu 2+ + 2  e – → Cu E o = + 0.34 V

Topic: Chemistry, Electrochemistry, JEE Main
Methane is a greenhouse gas because,

Options:
A it absorbs longer wavelengths of the electromagnetic spectrum while transmitting shorter wavelengths.
B it absorbs shorter wavelengths of the electromagnetic spectrum while transmitting longer wavelengths.
C it absorbs all wavelengths of the electromagnetic spectrum.
D it transmits all wavelengths of the electromagnetic spectrum.

The correct answer is:
it absorbs longer wavelengths of the electromagnetic spectrum while transmitting shorter wavelengths.

Solution:
There are various types of rays in sunlight like visible, infrared and may be UV. Out of these, infrared is responsible for heating effects on earth, and it has longer wavelength compared to others, i.e., approximately 8000 A0 and the molecular structure is such that it only absorbs longer wavelength and release shorter wavelengths. So, it is more responsible for global warming than other gases like CO2.

Topic: Physics, Electromagnetic Waves, JEE Main
Method used for extraction of Pb and Sn from their chief ore

Options:
A Carbon reduction and self reduction.
B Self reduction and carbon reduction.
C Electrolysis and self reduction.
D Self reduction and electrolysis.

The correct answer is:
Self reduction and carbon reduction.

Solution:
PbO & PbSO4 get reduced by PbS itself which is already present in mixture so because the reduction took place by mixture itself, hence is known as self reduction.
2PbO+PbS→Δ3Pb+SO2↑
PbSO4+PbS→Δ2Pb+2SO2↑
While Sn is obtained by carbon reduction from SnO2 .
SnO2+2C→ΔSn+2CO

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
Methods used for purification of organic compounds are based on :

Options:
A nature of compound and presence of impurity.
B neither on nature of compound nor on the impurity present.
C nature of compound only.
D presence of impurity only.

The correct answer is:
nature of compound and presence of impurity.

Solution:
Organic compounds are purified based on their nature and impruity present in it.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2024 (09 Apr Shift 1)
Methoxybenzene on treatment with $\mathrm{HI}$ produces

Options:
A iodobenzene and methanol
B phenol and methyl iodide
C iodobenzene and methyl iodide
D phenol and methanol

The correct answer is:
phenol and methyl iodide

Solution:
Methoxybenzene on treatment with HI produces phenol and methyl iodide. This is due the more stable aryl oxygen bond. Here. methyl phenyl oxonium ion is formed by the protonation of ether. The bond between $\mathrm{O}-\mathrm{CH}_{3}$ is weaker than the bond between $\mathrm{O}-\mathrm{C}_{6} \mathrm{H}_{5}$ because the carbon of phenyl group is $s p^{2}$ -hybridised and there is a partial double bond character. Thus, the reaction yields phenol and alkyl halide.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Methyl alcohol can be distinguished from ethyl alcohol using

Options:
A Fehling solution
B Schiff's reagent
C Sodium hydroxide and iodine
D Phthalein fusion test

The correct answer is:
Sodium hydroxide and iodine

Solution:
$\begin{aligned} & \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH} \underset{\text { lodoform test }}{\stackrel{\mathrm{NaOH} / \mathrm{I}_2}{\rightarrow}} \mathrm{CHI}_3 \text { yellow ppt } \mathrm{CH}_3 \mathrm{OH} \xrightarrow{\mathrm{NaOH} / \mathrm{I}_2} \text { No ppt } \end{aligned}$

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Methyl amine reacts with methyl iodide. For completion of reaction, how many moles of methyl iodide are required?

Options:
A \( 1 \)
B \( 2 \)
C \( 3 \)
D \( 4 \)

The correct answer is:
\( 3 \)

Solution:
therefore, 3 moles of CH3I (methyl iodide) are required for reaction with methyl amine.

Topic: Chemistry, Amines, JEE Main
Methyl ethyl ketone is prepared by the oxidation of

Options:
A 2-propanol
B 1-butanol
C 2-butanol
D t-butyl alcohol

The correct answer is:
2-butanol

Solution:
\(\underset{\text{2-butanol}}{\mathrm{CH}_3-\underset{\substack{|~~~~ \\ \Large \mathrm{OH}}}{\mathrm{CH}}-\mathrm{CH}_2-\mathrm{CH}_3} \xrightarrow[\text { Oxidation }]{\mathrm{KMnO}_4} \underset{\text{ethyl methylketone}}{\mathrm{CH}_3-\stackrel{\substack{\Large \mathrm{O} \\ ||}}{\mathrm{C}}-\mathrm{CH}_2 \mathrm{CH}_3}\)

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Methylation of 10 g of benzene gave 9.2 g of toluene. Calculate the percentage yield of toluene (Nearest integer)

Options:

The correct answer is:
78

Solution:
C6H6+CH3Cl⟶C6H5CH3+HCl 10 gm 1078×92gm
Percentage yield of toluene = Actual yieldTheoretical yield×100

=9.2920×78×100=78%

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2021 (22 Jul Shift 1)
MF + XeF 4 → 'A' (M + = Alkali metal cation)

The state of hybridisation of the central atom 'A' and shape of the species are :

Options:
A sp3d , TBP
B sp3d3 , distorted octahedral
C sp3d3 , pentagonal planar
D No compound formed at all

The correct answer is:
sp3d3 , pentagonal planar

Solution:
MF + XeF 4 → M + XeF 5 -
sp3d3 , Pentagonal planar

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
MF +  XeF 4 →A ( M + = Alkali metal cation). The state of hybridisation of the central atom in A and shape of the species are

Options:
A sp3d, TBP
B sp3d3, distorted octahedral
C sp3d3, pentagonal planar
D No compound formed at all

The correct answer is:
sp3d3, pentagonal planar

Solution:
MF + XeF 4 ⟶ M + XeF 5 -
Pentagonal planar sp 3 d 3

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
$\mathrm{Mg}^{2+}$ is isoelectronic with

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

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

Solution:
Isoelectronic species have same number of electron. $\mathrm{Mg}^{2+}$ and $\mathrm{Na}^{+}$both have 10 electrons hence, they are isoelectronic species.

Topic: Chemistry, Structure of Atom, JEE Main
Mg → Heat Air X + Y → H 2 O Z → Colourless gas H 2 O Solution → CuSO 4 (A) Blue coloured solution
Substances X, Y, Z and A are solution respectively

Options:
A Mg 3 N 2 , MgO,  NH 3 ,  CuSO 4  ⋅  5H 2 O
B Mg(NO 3 ) 2 , MgO,  H 2 ,  CuSO 4 ⋅  5H 2 O
C Mg3N2,MgO,NH3,CuNH34SO4
D Mg(NO 3 ) 2 ,  MgO 2 ,  H 2 O 2 ,  CuSO 4 ⋅ 5H 2 O

The correct answer is:
Mg3N2,MgO,NH3,CuNH34SO4

Solution:
3Mg+N2→Mg3N2X

Mg3N2+6H2O→3MgOH2+2NH3Z

2Mg+O2→MgOY;

NH3+H2O→NH4OHSolution

CuSO4+4NH4OH→CuNSO4+4H2O

Topic: Chemistry, s Block Elements, JEE Main
Micelles are formed only

Options:
A below the CMC and the Kraft temperature.
B above the CMC and below the Kraft temperature.
C above the CMC and above the Kraft temperature,
D below the CMC and above the Kraft temperature

The correct answer is:
above the CMC and above the Kraft temperature,

Solution:
The formation of micelles takes place only above a particular temperature called Kraft temperature Tk and above a particular concentration called critical micelle concentration CMC. On dilution, these colloids revert to individual ions.

Topic: Chemistry, Surface Chemistry, JEE Main
Microbes can be used to decrease the use of chemical fertilizers and pesticides. Explain how this can be accomplished.

Options:

None

Solution:
Microbes can be used to decrease the use of chemical fertilizers and pesticides. This can be accomplished by the use of biofertilizers. The main source of biofertilizers are bacteria, fungi and cyanobacteria. They can provide plant nutrients at cheaper cost when compared with the chemical fertilizers. e.g.,
(i) Use of leguminous plant with Rhizobium.
(ii) Use of sulphur fixing bacteria.
(iii) Use of hydrogen fixing bacterium.

Topic: Biology, Microbes in Human Welfare
Microscope is an optical instrument which

Options:
A Enlarges the object
B Increases the visual angle formed by the object at the eye
C Decreases the visual angle formed by the object at the eye
D Brings the object nearer

The correct answer is:
Increases the visual angle formed by the object at the eye

Solution:
In microscope final image formed is enlarged which in turn increases the visual angle.

Topic: Physics, Ray Optics, JEE Main
Microwave oven acts on the principle of:

Options:
A giving rotational energy to water molecules
B giving translational energy to water molecules
C giving vibrational energy to water molecules
D transferring electrons from lower to higher energy levels in water molecule

The correct answer is:
giving vibrational energy to water molecules

Solution:
The food having more content of water cooks faster in the microwave oven. Energy of microwaves lie in the range of vibration energy of water molecules.

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2016 (09 Apr Online)
Midbrain is located between

Options:
A Thalamus/hypothalamus of forebrain and pons varolii of hindbrain
B Thalamus/hypothalamus of forebrain and medulla of hindbrain
C Olfactory lobe of forebrain and pons varolii of hindbrain
D Olfactory lobe of forebrain and medulla of hindbrain

The correct answer is:
Thalamus/hypothalamus of forebrain and pons varolii of hindbrain

Solution:
The brain stem consists of midbrain and hindbrain. Midbrain is located in between the thalamus/hypothalamus (of the forebrain) and pons (of hindbrain). Cerebral aqueduct passes through the midbrain. The dorsal portion of the midbrain mainly consists of 4 round swellings (lobes). These are termed as Corpora quadrigemina. Midbrain receives tactile and visual responses.

Topic: Biology, Neural Control and Coordination, NEET
Middle lamella is composed mainly of :

Options:
A Phosphoglycerides
B Hemicellulose
C Muramic acid
D Calcium pectate

The correct answer is:
Calcium pectate

Solution:
No Solution Available

Topic: Biology, Cell - The Unit of Life, NEET, NEET 2009 (Mains)
Middle lamella is mainly composed of

Options:
A hemicellulose
B muramic acid
C calcium pectate
D phosphoglycerides

The correct answer is:
calcium pectate

Solution:
The middle lamella is cementing layer between the cells. It is made up of $\mathrm{Ca}$ and $\mathrm{Mg}$ pectates. The basic chemical unit of pectin is galacturonic acid, which have the capability of salt formation with calcium and magnesium (an acid base reaction).

Topic: Biology, Cell - The Unit of Life, NEET, NEET 2009 (Screening)
Middle lamella is

Options:
A plasma membrane covering the two cells together.
B made of calcium pectate or magnesium pectate.
C a pore in the plasma membrane of cell.
D a cell wall between two adjacent cells.

The correct answer is:
made of calcium pectate or magnesium pectate.

Solution:
Cell wall is the outermost layer in the eukaryotic cells (absent in animals). It is a non -living and rigid layer. The main function is to protect the cells and also to give shape to the cells. It also prevents entry of unimportant substances. Young cells contain three layers: Primary wall, middle lamella and Secondary wall. Primary wall is responsible for growth and hence it is not seen in adult cells. Middle lamella is seen in all types of cells. This contains calcium pectate which helps to stick to the neighbouring cells. Plasmodesmata connects the middle lamella and cytoplasm of neighbouring cells.

Topic: Biology, Cell - The Unit of Life, NEET
Middle point of the chord of the circle $x^2+y^2=25$ intercepted on the line $x-2 y=2$ is

Options:
A $\left(\frac{3}{5}, \frac{4}{5}\right)$
B $(-2,-2)$
C $\left(\frac{2}{5},-\frac{4}{5}\right)$
D $\left(\frac{8}{3}, \frac{1}{3}\right)$

The correct answer is:
$\left(\frac{2}{5},-\frac{4}{5}\right)$

Solution:
Here the intersection point of chord and circle can be found by solving the equation of circle with the equation of given line, therefore, the points of intersection are $(-4,-3)$ and $\left(\frac{24}{5}, \frac{7}{5}\right)$. Hence the midpoint is $\left(\frac{-4+\frac{24}{5}}{2}, \frac{-3+\frac{7}{5}}{2}\right)=\left(\frac{2}{5},-\frac{4}{5}\right)$

Topic: Mathematics, Circle, JEE Main
Millimoles of calcium hydroxyide required to produce 100mL of the aqueous solution of pH 12 is x×10-1. The value of x is _____ (Nearest integer).
Assume complete dissociation.

Options:

The correct answer is:
5

Solution:
We know,
pH = -log[H+] , pH = -log[OH-], pH + pOH = 14
∵ pH=12

∴ H+=10-12M

∴ OH-=10-2M
We also know,
Ca(OH)2 ⇌ 2OH- + Ca2+

∴ Ca(OH)2=5×10-3M

5×10-3=milli moles of CaOH2 100mL

milli moles of CaOH2=5×10-1

Ans. =5

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2023 (29 Jan Shift 1)
Minamata disease is caused due to the consumption of

Options:
A sea food containing lot of cadmium
B fish contaminated with mercury
C oysters with lot of pesticide
D sea food contaminated with selenium

The correct answer is:
fish contaminated with mercury

Solution:
Minamata disease is a neurological syndrome. It is caused by mercury poisoning. The symptoms seen are Ataxia, general muscle weakness, damage to hearing and speech etc. It is so named because it was first detected in Minamata city of Japan. The main reason was release of methylmercury from industrial waste water. This highly toxic chemical accumulated in water bodies and thus was consumed by fishes which when consumed by local humans resulted in the condition.

Topic: Biology, Environmental Issues, NEET
Minamata disease is due to pollution of

Options:
A Organic waste into drinking water
B Oil spill in water
C Industrial waste mercury into fishing water
D Arsenic into the atmosphere

The correct answer is:
Industrial waste mercury into fishing water

Solution:
Mercury poisoning often produces a crippling and fatal disease called Minamata disease.

Topic: Chemistry, Environmental Chemistry, JEE Main
Minerals are present in the soil in sufficient amounts. Do plants need to adjust the type of solutes that reach the xylem? Which molecules help to adjust this? How do plants regulate the type and quantity of solutes that reach xylem?

Options:

None

Solution:
Plants do need to adjust the type and quantity of solutes that reach the xylem. The transport of proteins in endodermal cells help in maintaining and adjusting solute movement. The minerals are present in soil as charged particles with a very low concentration compared to that of roots, and thus cannot be completely transported passively across the cell membranes of roots hairs.
Minerals are thus transported both by active and passive processes, to the xylem. Upon reaching xylem, they are further transported, upwards to sinks through transpiration stream. At the sink regions mineral ions are unloaded through diffusion and active uptake by receptor cells. The mineral ions moving frequently through xylem include.
(i) Sulphur and Phosphorus in small amounts are carried in organic forms.
(ii) Nitrogen travels in plants as inorganic ions $\mathrm{NO}_2$ and $\mathrm{NO}_3$ but much of the nitrogen moves in the form of amino acids and related organic compounds.
(iii) Mineral ions are frequently remobilised particularly from older senescing parts. Older dying leaves export much of their mineral content to younger leaves. Similarly, before leaf fall in deciduous plants, minerals are removed to other parts.
The most readily mobilised elements are phosphorus, sulphur, nitrogen and potassium. Structural components elements like calcium are not remobilised.

Topic: Biology, Transport in Plants
Minimum area of the triangle formed by any tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with co- ordinate axes is

Options:
A $\frac{a^2+b^2}{2}$
B $\frac{(a+b)^2}{2}$
C ab
D $\frac{(a-b)^2}{2}$

The correct answer is:
ab

Solution:
Let equation of tangent be $\frac{\mathrm{x} \cos \phi}{\mathrm{a}}+\frac{\mathrm{y} \sin \phi}{\mathrm{b}}=1$
Area of Triangle formed $=\frac{1 \mathrm{a}}{2 \cos \phi} \times \frac{\mathrm{b}}{\sin \phi}$
$=\frac{\mathrm{ab}}{\sin 2 \phi}$
$\sin 2 \phi$ has maximum value 1 .
$\therefore$ Minimum area $=\mathrm{ab}$

Topic: Mathematics, Ellipse, JEE Main
Minimum deviation is observed with a prism having angle of prism $A$, angle of deviation $\delta$, angle of incidence $i$ and angle of emergence $e$. We then have generally

Options:
A $i \gt e$
B $\underset{\sim}{i \lt e}$
C $i=e$
D $i=e=\delta$

The correct answer is:
$i=e$

Solution:
In minimum deviation condition $\angle i=\angle e, \angle r_1=\angle r_2$

Topic: Physics, Ray Optics, JEE Main