datavorous/entrance-exam-dataset huggingface dataset:
In Youngâs double-slit experiment, the distance between the two identical slits is 6.1 times larger than the slit width. Then the number of intensity maxima observed within the central maximum of the single-slit diffraction pattern is :

Options:
A 3
B 6
C 24
D 12

The correct answer is:
12

Solution:
dâ= separation between 2 slits
aâ= slit width
Î² y = fringe width in YDSE & Î² D = width of central maxima
Î²y=Î»Dd=Î»D6.1a;Î²D=2Î»Da
Î² D =2Ã6.1â Î² y
=12.2â Î² y

Topic: Physics, Wave Optics, JEE Main
In Young's double-slit experiment, the intensity at a point P on the screen is half the maximum intensity in the interference pattern. If the wavelength of light used is λ and d is the distance between the slits, the angular separation between point P and the center of the screen is

Options:
A sin-1⁡λd
B sin-1⁡λ2d
C sin-1⁡λ3d
D sin-1⁡λ4d

The correct answer is:
sin-1⁡λ4d

Solution:
If δ is the phase difference between the interfering waves at point P, then the intensity at point P is given by
cos2⁡δ2
Given I=Imax2. Hence
cos2δ2=12, which gives δ2=π4
Or δ=π2
The angular separation θ between points P and O is given by tan⁡θ=yD . Since θ≃sin⁡θ. Hence
sin⁡θ=yD ...... (i)
If β is the fringe width, then
yβ=π/22π=14 ...... (ii)
This is because the phase difference δ between two consecutive maxima is 2π . Now β=λDd. using this in Equation. (ii), we get
ydλd=14
Or yD=λ4d .....(iii)
Using Equation. (i) in Equation. (iii), we have
sin⁡θ=λ4d or θ=sin-1⁡λ4d

Topic: Physics, Wave Optics, JEE Main
In Young's double-slit experiment, the ratio of intensities of a bright band and a dark band is 16:1. The ratio of amplitudes of interfering waves will be

Options:
A 16:1
B 4:1
C 3:1
D 5:3

The correct answer is:
5:3

Solution:
Since Imax.Imin.=161
( I 1 + I 2 ) 2 ( I 1 â I 2 ) 2 = 16 1
I 1 + I 2 =4 I 1 â4 I 2
A 1 + A 2 =4A 1 â4 A 2
3 A 1 =5 A 2
â A 1 A 2 = 5 3

Topic: Physics, Wave Optics, JEE Main
In Young's double-slit experiment, the separation between the slits is d=0.25 cm and the distance of the screen from the slits is D=100 cm. When the wavelength of light used in the experiment is Î»=6000 Ã, the intensity at a distance x=4Ã10-5 m from the central maximum is pqI0, where I0 is the intensity of central maxima and p and q are the smallest integers. What is the value of p+q?

Options:

The correct answer is:
7

Solution:
Path diff.=xdD

⇒ Path diff.=4×10-5×0.25×10-21
Path diff. = 1 × 1 0 - 7
Phase diff.=path diff.λ×2π

⇒ Phase diff.=1×10-76×10-7×2π
Phase diff.=2π6 ⇒ Phase diff.=π3 ⇒ ϕ=60∘
IR=I1+I2+2I1I2cos60∘ ⇒ IR=I1+I2+2I×12
IR=3I ⇒ IR=3I04

⇒p=3, q=4⇒p+q=7

Topic: Physics, Wave Optics, JEE Main
In Young’s double-slit experiment, the separation between the slits is halved and the distance between the slits and the screen is doubled. The fringe width will

Options:
A remain the same
B be halved
C be doubled
D be quadrupled

The correct answer is:
be quadrupled

Solution:
The fringe width in Youngâs double-slit experiment
Î²=Î»Dd
where Î» is the wavelength of light used
D is the distance between slit and screen.
d is the distance between the slit.
â´âââââÎ²'=Î»(2D)d2=4Î»Dd=4Î²

Topic: Physics, Wave Optics, NEET
In Youngâs double-slit experiment, the slit separation is 1 mm and the screen is at a distance of 1 m from the slit. For a monochromatic light of wavelength 500 nm, the distance of 3rd minima from the central maxima is

Options:
A 0.50 mm
B 1.25 mm
C 1.50 mm
D 1.75 mm

The correct answer is:
1.25 mm

Solution:
Distance of nth minima from central bright fringe
xn=2n-1λD2d
For n=3 i.e., 3rd minima
x3=2×3-1×500×10-9×12×1×10-3
=5×500×10-62=1.25×10-3 m=1.25 mm

Topic: Physics, Wave Optics, JEE Main
In Youngâs double-slit experiment, the two slits, 0.12 mm apart, are illuminated by monochromatic light of wavelength 420 nm. If the screen is 1.0 m away from the slits, then the distance of the second bright fringe from the central maximum is

Options:
A 7 mm
B 5.25 mm
C 6 mm
D 10 mm

The correct answer is:
7 mm

Solution:
Here d=0.12 mm =0.12× 10 −3 m =12× 10 −5 m

λ=420 nm =420× 10 −9 m, D=1.0 m

The distance of the second bright fringe

x 2 = 2λD d                                                                [ ∵ x n = nλD d ]

= 2×420× 10 −9 ×1.0 12× 10 −5 =70× 10 −4 =7× 10 −3 m=7 mm

Topic: Physics, Wave Optics, JEE Main
In Young's experiment, light of wavelength $4000 Å$ is used to produce bright fringes of width $0.6 \mathrm{~mm}$, at a distance of 2 meters. If the whole apparatus is dipped in a liquid of refractive index 1.5 , then fringe width will be

Options:
A $0.2 \mathrm{~mm}$
B $0.3 \mathrm{~mm}$
C $0.4 \mathrm{~mm}$
D $1.2 \mathrm{~mm}$

The correct answer is:
$0.4 \mathrm{~mm}$

Solution:
$\beta_{\text {medium }}=\frac{\beta_{\text {air }}}{\mu}=\frac{0.6}{1.5}=0.4 \mathrm{~mm}$

Topic: Physics, Wave Optics, JEE Main
In Young's experiment, the distance between slits is $0.28 \mathrm{~mm}$ and distance between slits and screen is 1.4 $\mathrm{m}$. Distance between central bright fringe and third bright fringe is $0.9 \mathrm{~cm}$. What is the wavelength of used light

Options:
A $5000 Å Å$
B $6000 Å$
C $7000 Å$
D $9000 Å$

The correct answer is:
$6000 Å$

Solution:
Position of $3^{\text {rd }}$ bright fringe $x_3=\frac{3 D \lambda}{d}$
$\Rightarrow \lambda=\frac{x_3 d}{3 D}=\frac{\left(0.9 \times 10^{-2}\right) \times\left(0.28 \times 10^{-3}\right)}{3 \times 1.4}=6000 Å$

Topic: Physics, Wave Optics, JEE Main
In zinc blende structure the coordination number of $\mathrm{Zn}^{2+}$ ion is

Options:
A $2$
B $4$
C $6$
D $8$

The correct answer is:
$4$

Solution:
Zinc blende (ZnS) has fcc structure and is an ionic crystal having $4: 4$ co-ordination number.

Topic: Chemistry, Solid State, JEE Main
Incandescent bulbs are designed by keeping in mind that the resistance of their filament increases with the increase in temperature. If at room temperature, $100 \mathrm{~W}, 60 \mathrm{~W}$ and $40 \mathrm{~W}$ bulbs have filament resistances $R_{100}, R_{60}$ and $R_{40}$, respectively, the relation between these resistances is

Options:
A
$\frac{1}{R_{100}}=\frac{1}{R_{40}}+\frac{1}{R_{60}}$

B
$R_{100}=R_{40}+R_{60}$

C
$R_{100}>R_{60}>R_{40}$

D
$\frac{1}{R_{100}}>\frac{1}{R_{60}}>\frac{1}{R_{40}}$

The correct answer is:
$\frac{1}{R_{100}}>\frac{1}{R_{60}}>\frac{1}{R_{40}}$

Solution:
$R=\frac{V^2}{P}$ or $R \propto \frac{1}{P}$ $\therefore \quad \frac{1}{R_{100}}>\frac{1}{R_{60}}>\frac{1}{R_{40}}$
Hence, the correct option is (d).

Topic: Physics, Current Electricity, JEE Main
Incandescent bulbs are designed keeping in mind that the resistance of their filament increases with the increase in temperature. If at the room temperature 100 W, 60 W and 40ââW bulbs have filament resistances R100, R60 and R40 respectively, then the relation between these resistances is

Options:
A 1 R 1 0 0 = 1 R 4 0 + 1 R 6 0
B R 1 0 0 = R 4 0 + R 6 0
C R 1 0 0 > R 4 0 > R 6 0
D 1 R 1 0 0 > 1 R 6 0 > 1 R 4 0

The correct answer is:
1 R 1 0 0 > 1 R 6 0 > 1 R 4 0

Solution:
R = V 2 P or R ∝ 1 P
∴ 1 R 1 0 0 > 1 R 6 0 > 1 R 4 0

Topic: Physics, Current Electricity, JEE Main
Incomplete division of egg during cleavage is known as

Options:
A meroblastic
B holoblastic
C meridional
D spiral

The correct answer is:
meroblastic

Solution:
In meroblastic cleavage, division occurs only in small amount of metabolically active cytoplasm, confined to the animal pole or peripheral region of egg. The yolk remains undivided. It is found in macrolecithal eggs. e.g. birds

Topic: Biology, Human Reproduction, JIPMER, JIPMER 2004
Incorrect match among the following is:

Options:
A Vitamin B12-Cu
B Cis-platin - Pt
C Wilkinson catalyst - Rh
D Chlorophyll - Mg

The correct answer is:
Vitamin B12-Cu

Solution:
Vitamin B12 contains Co.

Topic: Chemistry, Coordination Compounds, JEE Main
Incorrect matches is

Options:
A COCl2 - phosgene
B SO 2 Cl 2 −Thionyl chloride
C ClCH2CH2SCH2CH2Cl - mustard gas
D H 2 SO 5  − Caro's acid

The correct answer is:
SO 2 Cl 2 −Thionyl chloride

Solution:
SOCl 2 =Thionyl chloride
SO 2 Cl 2 =Sulfuryl chloride

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Incorrect method of preparation for alcohols from the following is:

Options:
A Reaction of Ketone with RMgBr followed by hydrolysis.
B Reaction of alkyl halide with aqueous NaOH.
C Hydroboration-oxidation of alkene.
D Ozonolysis of alkene

The correct answer is:
Ozonolysis of alkene

Solution:
Reacting a Grignard reagent with a ketone gives a tertiary alcohol. When alkyl halide reacts with aqueous sodium hydroxide to give alcohol, the reaction is called nucleophilic substitution. The Hydroboration Oxidation reaction is an organic chemical reaction which is employed for the conversion of alkenes into alcohols that are neutral. Reductive ozonolysis of alkenes will lead to formation of aldehyde or ketones, oxidative ozonolysis of alkenes will lead to formation of carboxylic acids or ketones.

So, alcohol is not formed by ozonolysis of alkenes.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2023 (10 Apr Shift 2)
Incorrect order of properties given in parenthesis are

Options:
A Cl>S>F>O (Electron affinity)
B O>F>B>C IE2
C N>O>P>S IE1
D O>C>B>N (Electron affinity)

The correct answer is:
Cl>S>F>O (Electron affinity)

Solution:
Correct order is:
Cl>F>S>O

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
Incorrect order of radius is

Options:
A Sr2+<Rb+<Br-<Se2-
B Nb5+<Zr4+<Y3+
C Co>Co2+>Co3+>Co4+
D Ba 2+  >  Cs +  >  Se 2−  >  As 3−

The correct answer is:
Ba 2+  >  Cs +  >  Se 2−  >  As 3−

Solution:
In A and B use (z/e) concept for isoelectronic species.
In C size of neutral atom is greater than its cation.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
Incorrect set of quantum numbers from the following is

Options:
A $\mathrm{n}=4, \mathrm{l}=2, \mathrm{~m}_{\mathrm{l}}=-2,-1,0,+1,+2, \mathrm{~m}_{\mathrm{s}}=-1 / 2$
B $\mathrm{n}=5, \mathrm{l}=3, \mathrm{~m}_{\mathrm{l}}=-3,-2,-1,0,+1,+2,+3, \mathrm{~m}_{\mathrm{s}}=+1 / 2$
C $\mathrm{n}=4, \mathrm{l}=3, \mathrm{~m}_{\mathrm{l}}=-3,-2,-1,0,+1,+2,+3, \mathrm{~m}_{\mathrm{s}}=-1 / 2$
D $\mathrm{n}=5, \mathrm{l}=2, \mathrm{~m}_{\mathrm{l}}=-2,-1,+1,+2, \mathrm{~m}_{\mathrm{s}}=+1 / 2$

The correct answer is:
$\mathrm{n}=5, \mathrm{l}=2, \mathrm{~m}_{\mathrm{l}}=-2,-1,+1,+2, \mathrm{~m}_{\mathrm{s}}=+1 / 2$

Solution:
For any value of $\mathrm{I}, m_1$ value will be $-\mathrm{l}$ to $+\mathrm{I}$
So for $\mathrm{n}=5$ and $\mathrm{I}=2$ : $\mathrm{m}_{\mathrm{l}}=-2,-1,0,+1,+2$ and $\mathrm{m}_{\mathrm{s}}=+\frac{1}{2}$ or $-\frac{1}{2}$
Hence, the set $n=5, l=2, m_l=-2,-1,+1,+2, m_s=+\frac{1}{2}$ is incorrect

Topic: Chemistry, Structure of Atom, JEE Main
Incorrect statement about carbon monoxide is: -

Options:
A It is highly soluble in water
B It burns in oxygen to produce considerable amount of heat
C It is toxic having bond order =3
D It is found in coal gas, water gas and produces gas

The correct answer is:
It is highly soluble in water

Solution:
CO is sparingly soluble in water

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
Incorrect statement about Mn2O7 is:

Options:
A It is a covalent oxide
B Mn - O - Mn linkage is present
C It is purple in colour
D Mn is tetrahedrally surrounded by O atoms

The correct answer is:
It is purple in colour

Solution:
Mn2O7 is green in colour

Topic: Chemistry, d and f Block Elements, JEE Main
Incorrect statement among the following is:

Options:
A Oxidation state of chromium in chromate and dichromate is same
B Oxidation of manganese is different in manganate and permanganate
C Colour of chromate and dichromate is orange
D Chromate ion gets converted into dichromate ion in acidic medium

The correct answer is:
Colour of chromate and dichromate is orange

Solution:
Chromate is yellow and dichromate is orange.

Topic: Chemistry, d and f Block Elements, JEE Main
Incorrect statement for the use of indicator in acid-base titration is:

Options:
A Methyl orange may be used for a weak acid vs weak bas titration.
B Phenolphthalein may be used for a strong acid vs strong base titration.
C Methyl orange is a suitable indicator for a strong acid vs weak base titration.
D Phenolphthalein is a suitable indicator for a weak acid vs strong base titration.

The correct answer is:
Methyl orange may be used for a weak acid vs weak bas titration.

Solution:
Methyl orange and phenolphthalein indicators work with the change in pH and are termed as acid–base indicators.

Methy lorange is sutable indicator for strong acid vs strong base and strong acid vs weak base titration.

Phenolphthalein is sutable indicator for strong acid vs strong base and weak acid vs strong base titration.

There is no suitable indicator that can be used in the titration of weak acid and weak base. So, the correct option is A.

Topic: Chemistry, Ionic Equilibrium, JEE Main
Incorrect statement for the use of indicator in acid-base titration is:

Options:
A Methyl orange may be used for a weak acid vs weak bas titration.
B Phenolphthalein may be used for a strong acid vs strong base titration.
C Methyl orange is a suitable indicator for a strong acid vs weak base titration.
D Phenolphthalein is a suitable indicator for a weak acid vs strong base titration.

The correct answer is:
Methyl orange may be used for a weak acid vs weak bas titration.

Solution:
Methyl orange and phenolphthalein indicators work with the change in pH and are termed as acid–base indicators.

Methy lorange is sutable indicator for strong acid vs strong base and strong acid vs weak base titration.

Phenolphthalein is sutable indicator for strong acid vs strong base and weak acid vs strong base titration.

There is no suitable indicator that can be used in the titration of weak acid and weak base. So, the correct option is A.

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2023 (31 Jan Shift 2)
Incorrect statement for Tyndall effect is

Options:
A The refractive indices of the dispersed phase and the dispersion medium differ greatly in magnitude.
B The diameter of the dispersed particles is much smaller than the wavelength of the light used.
C It is used to distinguish a true solution from a colloidal solution.
D During projection of movies in the cinemas hall, Tyndall effect is noticed.

The correct answer is:
The diameter of the dispersed particles is much smaller than the wavelength of the light used.

Solution:
Tyndall effect is observed only when the following two conditions are satisfied.

(i) The diameter of the dispersed particles is not much smaller than the wavelength of the light used; and

(ii) The refractive indices of the dispersed phase and the dispersion medium differ greatly in magnitude.

This is a characteristic property of the colloid solutions.

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2022 (25 Jun Shift 1)
Incorrect statement is

Options:
A MgO>AlF3>MgF2 : Lattice energy
B Li>Na>Mg : Electron affinity
C SF6>PF5>SiF4 : Lewis acidic character
D SiCl4>SiBr4>SiI4 : Decreasing order of electronegativity of Si

The correct answer is:
SF6>PF5>SiF4 : Lewis acidic character

Solution:
Here the incorrect statement is option C as the correct order of Lewis acidic character: PF5>SiF4>SF6 . Although, S has vaccant 3d-orbital but it cannot accept co-ordinate bond from Lewis base due to steric crowding factor as S-atom is already bonded to six F-atoms.
Rest other statements are correct here.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Incorrect statement regarding following reaction is

Options:
A 'X' is explosive
B 'Y' is an oxyacid of xenon
C Both are examples of non-redox reaction
D XeF6 can undergoes partial hydrolysis

The correct answer is:
'Y' is an oxyacid of xenon

Solution:
Here statement B is incorrect as here Y is not an oxyacid of xenon but an oxy Fluoride.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Incorrect statement related to extraction of copper from copper pyrite is:

Options:
A Iron silicate is obtained as slag
B Copper matte in the form of CuS+FeS is obtained
C Copper is obtained by self reduction
D Blister copper is obtained after reduction process

The correct answer is:
Copper matte in the form of CuS+FeS is obtained

Solution:
Copper Matte is Cu2S+FeS

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
Increase in temperature of the material causes
in elasticity.

Options:
A Decrease
B Increase
C No effect
D None of these

The correct answer is:
Decrease

Solution:
With increase in temperature, the average intermolecular distance increases and hence the intermolecular force decreases. This results in decrease of elasticity.

Topic: Physics, Mechanical Properties of Solids, JEE Main
Increasing order of acid strength of halogen acid is

Options:
A $\mathrm{HF} \lt \mathrm{HCl} \lt \mathrm{HBr} \lt \mathrm{HI}$
B $\mathrm{HCl} \lt \mathrm{HBr} \lt \mathrm{HI} \lt \mathrm{HF}$
C $\mathrm{HF} \lt \mathrm{HI} \lt \mathrm{HBr} \lt \mathrm{HCl}$
D None of these

The correct answer is:
$\mathrm{HF} \lt \mathrm{HCl} \lt \mathrm{HBr} \lt \mathrm{HI}$

Solution:
As going down the group size increases, an liberation of $\mathrm{H}^{+}$ion becomes easy. So the order of acidity is: $\mathrm{HI}\gt\mathrm{HBr}\gt\mathrm{HCl}\gt\mathrm{HF}$

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
Increasing order of basic strength among given compounds is-

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

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

Solution:
Its conjugate acid do not undergo proper solvation and hence its stability decreases. Hence,basic strength decreases.
⇒ Lone pair is in resonance.
⇒
Steric inhibition(two methyl groups present at ortho position pushes nitrogen out of the plane of the benzene ring and prevents resonance of lone pair) takes place in resonance. (Basic strength increases).

Topic: Chemistry, General Organic Chemistry, JEE Main
Increasing order of density is

Options:
A Li < K < Na < Rb < Cs
B LiÂ <Â NaÂ <Â KÂ <Â RbÂ <Â Cs
C LiÂ <Â NaÂ <Â KÂ <Â CsÂ <Â Rb
D KÂ <Â LiÂ <Â NaÂ <Â RbÂ <Â Cs

The correct answer is:
Li < K < Na < Rb < Cs

Solution:
As per Lother-Meyer's curve, the atomic volume of Cs is maximum among s-block elements. But atomic masses increases more rapidly than atomic volume, hence density increases along the group

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
Increasing order of reactivity of the following compounds for SN1 substitution is:

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

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

Solution:
SN1 reactivity order ∝ stability of carbocation

SN1 is unimolecular because the slow step of the reaction involves only the substrate, the reaction is unimolecular. Because only the substrate is present in the transition state, the rate of the reaction depends only on its concentration and not on the concentration of the nucleophile.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2019 (09 Apr Shift 2)
Increasing order of stability among the three main conformations (i.e. Eclipse, Anti, Gauche) of 2-fluoroethanol is

Options:
A
Eclipse, Gauche, Anti

B
Gauche, Eclipse, Anti

C
Eclipse, Anti, Gauche

D
Anti, Gauche, Eclipse

The correct answer is:
Eclipse, Anti, Gauche

Solution:
No Solution Available

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2006
Increasing order of stability of the resonance structure is:

Options:
A C,D,B,A
B C,A,B,D
C D,C,A,B
D D,C,B,A

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

Solution:
In the molecule C, positive charge is present on the electronegative oxygen atom and is not having complete octet. Hence, the structure C is the least stable resonance structure. In the structure A, the delocalisation is taken place away from the electron withdrawing group, which is not favourable delocalisation. In the structure D, more number of covalent bonds are there and having all atoms' octet.

Order should be : C<A<B<D

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2023 (24 Jan Shift 1)
Increasing rate of SN1 reaction in the following compounds is:
(I)
(II)
(III)
(IV)

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

The correct answer is:
(II)<(I)< (III)<(IV)

Solution:
The rate of SN1 reaction ∝ stability of carbocation
stability of carbocation ∝+ve charge delocation.
∝+M, +HC, +I
Order of effect effectiveness
+M>+HC>+I (+M )> (-I)
D

C

A

B
So, order of stability B<A<C<D

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2019 (10 Apr Shift 1)
Increasing the temperature of an aqueous solution will cause

Options:
A Decrease in molality
B Decrease in molarity
C Decrease in mole fraction
D Decrease in $\% w / w$

The correct answer is:
Decrease in molarity

Solution:
An increases in temperature increase the volume of the solution and thus decreases its molarity.

Topic: Chemistry, Solutions, JEE Main
Indentify 'M' in the following sequence of reactions

Options:
A
B
C
D

The correct answer is:

Solution:
and

Topic: Chemistry, Amines, JEE Main
India's Mangalyan was sent to the Mars by launching it into a transfer orbit EOM around the sun. It leaves the earth at E and meets Mars at M. If the semi-major axis of Earth's orbit is a e = 1.5 × 1 0 1 1 m , that of Mar's orbit a m = 2.28 × 1 0 1 1 m , taking Kepler's laws, give the estimate of time for Mangalyan to reach Mars from Earth.

Options:
A 220 days.
B 500 days.
C 260 days.
D 320 days.

The correct answer is:
260 days.

Solution:
Semi major axis of mangalyan,
a Mn = a e + a m 2
= 1.89 × 1 0 1 1 m
From Kepler's law,
T 2 ∝ R 3
∴ T Mn Te 2 = R M R e 3
∴ TMn≈518 days
∴ Required time = 5 1 8 2 = 2 6 0 days .

Topic: Physics, Gravitation, JEE Main, JEE Main 2014 (09 Apr Online)
India's wheat yield revolution in the 1960s was possible primarily due to:

Options:
A hybrid seeds
B increased chlorophyll content
C mutations resulting in plant height reduction
D acquired trait mutations.

The correct answer is:
mutations resulting in plant height reduction

Solution:
Through selective breeding, Norman Borlaug, an American biologist, created a dwarf variety of wheat that put most of its energy into edible kernels rather than long, inedible stems.

Topic: Biology, Capacitance, JEE Main
Indicate the complex/complex ion which did not show any geometrical isomerism :

Options:
A CoNH34Cl2+
B CoNH33NO23
C Co(CN)5(NC)3-
D CoCl2(en)2

The correct answer is:
Co(CN)5(NC)3-

Solution:
1CoNH34Cl2+show cis-trans isomerism.
2CoCl2(en)2 show cis-trans isomerism.
3CoNH33NO23 show fac-mer isomerism.
4[Co(CN)5(NC)]3-can't show geometrical isomerism due to the presence of plane of symmetry.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2021 (26 Aug Shift 2)
Indicate the correct IUPAC name of the co-ordination compound shown in the figure.

Options:
A Cis-dichlorotetraminochromium (III) chloride
B Trans-dichlorotetraminochromium (III) chloride
C Trans-tetraminedichlorochromium (III) chloride
D Cis-tetraamminedichlorochromium (III) chloride

The correct answer is:
Cis-tetraamminedichlorochromium (III) chloride

Solution:
Hint:

Topic: Chemistry, Coordination Compounds, JEE Main
Indicate the incorrect statement for a 1-L sample of $\mathrm{N}_2(\mathrm{~g})$ and $\mathrm{CO}_2(\mathrm{~g})$ at $298 \mathrm{~K}$ and $1 \mathrm{~atm}$ pressure

Options:
A The average translational $\mathrm{KE}$ per molecule is the same in $\mathrm{N}_2$ and $\mathrm{CO}_2$
B The rms speed remains constant for both $\mathrm{N}_2$ and $\mathrm{CO}_2$
C The density of $\mathrm{N}_2$ is less than that of $\mathrm{CO}_2$
D The total translational $\mathrm{KE}$ of both $\mathrm{N}_2$ and $\mathrm{CO}_2$ is the same

The correct answer is:
The rms speed remains constant for both $\mathrm{N}_2$ and $\mathrm{CO}_2$

Solution:
Kinetic energies per molecule will be same because it is proportional to absolute temperature only.
$\frac{d_{N_2}}{d_{C O_2}}=\frac{M_{N_2}}{M_{\mathrm{CO}_2}}=\frac{28}{44}$ i.e. $d N_2 \lt d \mathrm{CO}_2$
Total translational kinetic energy will also be same because at same temperature \& pressure number of molecules present in same volume would be same (according to Avogadro's Law)

Topic: Chemistry, States of Matter, JEE Main