datavorous/entrance-exam-dataset huggingface dataset:
Originally the radioactive beta decay was thought as a decay of a nucleus with the emission of electrons only (Case I) . However, in addition to the electron, another (nearly) massless and electrically neutral particle is also emitted (Case II). Based on the figure below, which of the following is correct?

Options:
A (a) in both cases I and II
B (a) in case I and (b) in case II
C (a) in case II and (b) in case I
D (b) in both cases I and II

The correct answer is:
(a) in case II and (b) in case I

Solution:
β- decay occurs as

XZA⟶YZ+1A+β-10+v¯

Most of the decay energy is carried by both emitted electron β- and antineutrino v¯. So, β-- particles emerges with varrying energies and energy spectrum of emitted β- is as shown;

If there are no antineutrinos, energy is carried mainly by β-- particles, so energy spectrum will be

Topic: Physics, Nuclear Physics, KVPY, KVPY 2019 (SB/SX)
Orlon fibres are made up of:

Options:
A Polyacrylonitrile
B Polyesters
C Polyamide
D Cellulose

The correct answer is:
Polyacrylonitrile

Solution:
Orlon fibres are made up of Polyacrylonitrile.

Topic: Chemistry, Polymers, JEE Main, JEE Main 2021 (20 Jul Shift 1)
Orlon fibres are made up of

Options:
A Polyacrylonitrile
B Polyesters
C Polyamide
D Cellulose

The correct answer is:
Polyacrylonitrile

Solution:
→orlon fibers are made up of Polyacrylonitrile

Topic: Chemistry, Polymers, JEE Main
Orthoboric acid in aqueous medium is

Options:
A Monobasic
B Dibasic
C Tribasic
D All are correct

The correct answer is:
Monobasic

Solution:
$\mathrm{H}_3 \mathrm{BO}_3+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \left[\mathrm{B}(\mathrm{OH})_4\right]^{-}+\mathrm{H}^{+}$
$\mathrm{H}_3 \mathrm{BO}_3$ is a weak monobasic acid if does not act as a $\mathrm{H}^{+}$ donor but behaves as a Lewis acid

Topic: Chemistry, Ionic Equilibrium, JEE Main
Ortho-Nitrophenol is less soluble in water than $\mathrm{p}-$ and $\mathrm{m}-$ Nitrophenols because :

Options:
A
$\mathrm{o-}$Nitrophenol is more volatile in steam than those of $\mathrm{m}-$ and $\mathrm{p}-$ isomers

B
$\mathrm{o-}$Nitrophenol shows Intramolecular $\mathrm{H}-$ bonding

C
$\mathrm{o-}$ Nitrophenol shows Intermolecular $\mathrm{H}-$ bonding

D
Melting point of $\mathrm{o-}$Nitrophenol is lower than those of $\mathrm{m}-$ and $\mathrm{p}-$ isomers.

The correct answer is:
$\mathrm{o-}$Nitrophenol shows Intramolecular $\mathrm{H}-$ bonding

Solution:
Intramolecular $\mathrm{H}$-bonding decreases water solubility.

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2012 (Offline)
Ortho-nitrophenol is steam volatile whereas para-nitrophenol is not. This is due to

Options:
A Intramolecular hydrogen bonding present in ortho-nitrophenol
B IIntermolecular hydrogen bonding
C Intramolecular hydrogen bonding present in para-nitrophenol
D None of these

The correct answer is:
Intramolecular hydrogen bonding present in ortho-nitrophenol

Solution:
Intramolecular hydrogen bonding present in ortho-nitrophenol

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Orthophosphoric acid represents the molaysis condition due to

Options:
A Hydrogen bonding
B Phosphorous group
C Maximum oxygen group
D Tribasicity

The correct answer is:
Phosphorous group

Solution:
Orthophosphoric acid represents the molaysis condition due to Phosphorous group. Hence, Option "2" is the correct answer.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Ortho-xylene on ozonolysis will give

Options:
A
B
C
D

The correct answer is:

Solution:
Ortho-xylene on ozonolysis gives different products as follows

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Osazone formation involves only 2 carbon atoms of glucose because of

Options:
A Chelation
B Oxidation
C Reduction
D Hydrolysis

The correct answer is:
Oxidation

Solution:
Glucose reacts with one molecule of phenyl hydrazine to form glucose phenylhydrazine and the second molecule of phenyl hydrazine oxidises second Carbon atom to form keto group, the third molecule of phenyl hydrazine then reacts with this Carbon atom.

Topic: Chemistry, Biomolecules, JEE Main
Osmoreceptors in the body are activated by changes in

Options:
A blood volume
B body fluid volume
C ionic concentration
D all of these

The correct answer is:
all of these

Solution:
Sensory receptors that detect changes in osmotic pressure and assist in maintaining fluid balance in the body are called osmoreceptors. These are spread out in the kidneys and hypothalamus of the body. These play important role in homeostasis. Changes in the blood volume, ion concentration and body fluid level trigger these receptors for response.

Topic: Biology, Excretory Products and their Elimination, NEET
Osmotic pressure observed when benzoic acid is dissolved in benzene is less than that expected from theoretical considerations. This is because

Options:
A benzoic acid is an organic solute
B benzoic acid has higher molar mass than benzene
C benzoic acid gets associated in benzene
D benzoic acid gets dissociated in benzene

The correct answer is:
benzoic acid gets associated in benzene

Solution:
Osmotic pressure is a colligative property. It depends upon the number of solute particles. Benzoic acid undergoes dimerisation.
$\therefore$ Number of particles reduces $(i < 1)$ Hence, osmotic pressure observed is less than the theoretical consideration.

Topic: Chemistry, Solutions, JEE Main
Osmotic pressure of a sugar solution at 24oC is 2.5 atmospheres. Determine the concentration of the solution in mole per litre.

Options:
A 0.0821 moles/litre
B 1.082 moles/litre
C 0.1025 moles/litre
D 0.0827 moles/litre

The correct answer is:
0.1025 moles/litre

Solution:
Here it is given that
π=2.5 atm, T=24+273=297 K, S=0.0821 lit. atm  K −1  mo 1 −1 C = ?
We know that Ï= CST or C=ÏST=2.50.0821Ã297=0.1025 moles/litre

Topic: Chemistry, Solutions, JEE Main
o-Toluic acid on reaction with $\mathrm{Br}_2+\mathrm{Fe}$, gives

Options:
A

B

C

D

The correct answer is:

Solution:
$-\mathrm{COOH}$ is meta directing group

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
Out of 5 apples, 10 mangoes and 15 oranges, any 15 fruits distributed among two persons. The total number of ways of distribution

Options:
A $66$
B $36$
C $60$
D None of these

The correct answer is:
$66$

Solution:
Number of ways $=$ coefficient of $x^{15}$ in the expansion $\left(1+x+x^2+x^3+x^4+x^5\right)$
$\left(1+x+x^2+\ldots \ldots . . x^{10}\right)$
$\left(1+x+x^2+\ldots \ldots+x^{15}\right)$
$\left(1+x+x^2+x^3+x^4+x^5\right)\left(1+x+x^2+\ldots .+x^{10}\right)$
$\left(1+x+x^2+\ldots+x^{15}\right)=\left(1-x^6-x^{11}\right)\left(1+{ }^3 C_1 x+{ }^4 C_2 x^2\right.$
$\left.+\ldots \ldots+{ }^6 C_4 x^4+{ }^{11} C_9 x^9+{ }^{17} C_{15} x^{15}+\ldots \ldots \ldots . ..\right)$
$=\ldots \ldots . .+\ldots \ldots .+x{ }^{15}\left(-{ }^{11} C_9-{ }^6 C_4+{ }^{17} C_{15}\right)$
$=\ldots \ldots . .+\ldots \ldots+x^{15}(-55-15+136)=x^{15} \times 66$
$\therefore$ Coefficient of $x^{15}=66$.

Topic: Mathematics, Permutation Combination, JEE Main
Out of 6 books, in how many ways can a set of one or more books be chosen

Options:
A $64$
B $63$
C $62$
D $65$

The correct answer is:
$63$

Solution:
Required number of ways
${ }_6 C_{1+}{ }_6 C_{2+}{ }_6 C_{3+} C_{4+} C_{5+} C_{6+}=2^{6} -1=63$

Topic: Mathematics, Permutation Combination, JEE Main
Out of 10 white, 8 black and 6 red balls, the number of ways in which one or more balls can be selected is (assuming balls of the same colour are identical)

Options:
A 681
B 691
C 679
D 692

The correct answer is:
692

Solution:
Required number of ways
=10+18+16+1-1
=11×9×7-1
=693-1
=692

Topic: Mathematics, Permutation Combination, JEE Main
Out of 10 white, 9 black and 7 red balls, the number of ways in which selection of one or more balls can be made, is

Options:
A $881$
B $891$
C $879$
D $892$

The correct answer is:
$879$

Solution:
The required number of ways are
$(10+1)(9+1)(7+1)-1=11^* 10^* 8-1=879 \text {. }$

Topic: Mathematics, Permutation Combination, JEE Main
Out of 11 consecutive natural number if three numbers are selected at random (without repetition), then the probability that they are in A.P. with positive common difference is :

Options:
A 15101
B 5101
C 533
D 1099

The correct answer is:
533

Solution:
2b=a+c⇒sum of a & c is even and b depends on a & c

Thus, we can not choose all three independently, but we can choose a & c

Moreover, a and c should be both either even or odd as their sum is even.

Out of 11 consecutive natural numbers, either five are even and six are odd or six are even and five are odd.

Hence, P=12C25+C26C311+12C26+C25C311=533

Topic: Mathematics, Probability, JEE Main, JEE Main 2020 (06 Sep Shift 1)
Out of 60% female and 40% male candidates appearing in an exam, 60% candidates qualify it. The number of females qualifying the exam is twice the number of males qualifying it. A candidate is randomly chosen from the qualified candidates. The probability, that the chosen candidate is a female, is

Options:
A 23
B 1116
C 2332
D 1316

The correct answer is:
23

Solution:
Given,

Out of 60% female and 40% male candidates appearing in an exam, 60% candidates qualify it. The number of females qualifying the exam is twice the number of males qualifying it. A candidate is randomly chosen from the qualified candidates. The probability, that the chosen candidate is a female, is

Now plotting the given value we have,

Now probability that chosen candidate is female =4060=23

Note: we have modified the option, as this question of 28th july 2022 jee mains was a bonus.

Topic: Mathematics, Probability, JEE Main, JEE Main 2022 (28 Jul Shift 1)
Out of 64 codons, 61 codons code for 20 types of amino acid, it is called:

Options:
A degeneracy of genetic code
B overlapping of gene
C wobbling of codon
D universality of codons

The correct answer is:
degeneracy of genetic code

Solution:
Degeneracy of codons is the redundancy of the genetic code, exhibited as the multiplicity of three-base pair codon combinations that specify an amino acid. The exceptions are methionine (AUG) and tryptophan (UGG).
Related Theory
The genetic code is degenerate mainly at the third codon position. The degeneracy of the genetic code made it possible for organisms to prosper on Earth. Organisms, which did not use a degenerate genetic code, would extinguish from this planet. This is one significant point of the genetic code.

Topic: Biology, Molecular Basis of Inheritance, NEET, NEET 2002
Out of \( 64 \) students, the number of students taking Mathematics is \( 45 \) and the number of students taking both Mathematics and Biology is \( 10 \). Then, the number of students taking the only Biology is

Options:
A \( 18 \)
B \( 19 \)
C \( 20 \)
D \( 17 \)

The correct answer is:
\( 19 \)

Solution:
Let M & B denote the sets of students taking Mathematics and Biology respectively.

Given nM=45, nM∩B=10, nM∪B=64

∴nB=nM∪B-nM+nM∩B

⇒nB=64-45+10=29

∴nonly B=nB-A=nB-nM∩B=29-10=19

Drawing the Venn diagram :

Therefore, the number of students taking only Biology =19.

Topic: Mathematics, Sets and Relations, JEE Main
Out of \( 1000 \) boys in a school, \( 300 \) played cricket, \( 380 \) played hockey and \( 420 \) played basketball. Of the total, \( 120 \) played both basketball and hockey, \( 100 \) played cricket and basketball, \( 70 \) played cricket and hockey and \( 56 \) played all the three games. If the probability of the number of boys who did not play any game is \( k \), then \( 200 k \) is equal to

Options:

The correct answer is:
26.8

Solution:
nC=300,nH=380,nB=420

nH∩B=120,nB∩C=100

nH∩C=70,nC∩H∩B=56

nCc∩Hc∩Bc=nC∪H∪Bc

=nU-nC∪H∪B=1000-nc+nH+nB-nH∩C-nH∩B-nC∩B+nCC∩H∩B

=1000-300+380+420-120-100-70+56

=134
Required probability =1341000

Topic: Mathematics, Probability, JEE Main
Out of 1000 boys in a school, 300 played cricket, 380 played hockey and 420 played basketball. Of the total, 120 played both basketball and hockey, 100 played cricket and basketball, 70 played cricket and hockey and 56 played all the three games. If the probability of the number of boys who did not play any game is k, then 200k is equal to

Options:

The correct answer is:
26.8

Solution:
nC=300,nH=380,nB=420
nH∩B=120,nB∩C=100
nH∩C=70,nC∩H∩B=56
nCc∩Hc∩Bc=nC∪H∪Bc
=nU-nC∪H∪B
=1000-[n(c)+n(H)+n(B)-nH∩C-nH∩B-nC∩B+nCC∩H∩B
=1000-300+380+420-120-100-70+56
=134
Required probability =1341000

Topic: Mathematics, Probability, JEE Main
Out of all the patients in a hospital 89% are found to be suffering from heart ailment and 98% are suffering from lungs infection. If K% of them are suffering from both ailments, then K can not belong to the set:

Options:
A {79,81,83,85}
B {84,87,90,93}
C {80,83,86,89}
D {84,86,88,90}

The correct answer is:
{79,81,83,85}

Solution:
Let H and L is set of people suffered from heart ailment and lungs infections respectively.

nH=89%

nL=98%

nH∩L=K%

maxn(H), nL≤nH∪L≤nU

Where, U is the universal set.

⇒98≤nH∪L≤100

⇒98≤nH+nL-nH∩L≤100

⇒98≤89+98-k≤100

⇒0≤89-k≤2

⇒-89≤-k≤-87

⇒87≤K≤89

Topic: Mathematics, Sets and Relations, JEE Main, JEE Main 2021 (26 Aug Shift 1)
Out of all the possible isomers of C5H11Cl, how many are secondary in nature? (Exclude stereoisomers if any.)

Options:

The correct answer is:
3

Solution:

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Out of aphelion and perihelion, where is the speed of the earth more and why?

Options:

None

Solution:
According to Kepler's first law, Earth revolve around the Sun in an elliptical orbit and sun remains at its one focus. Aphelion is the location of the earth where it is at the greatest distance from the sun and perihelion is the location of the earth where it is at the nearest distance from the sun that shown in figure:
Now, according to the Kepler's second law the areal velocity of the earth around the sun is constant.
$$
v=\frac{d A}{d t}=\frac{L}{2 m}=\frac{r \times p}{2 m}=\frac{r \times m v}{2 m}=\frac{1}{2} r \times v
$$
Hence if $r$ increase, the speed of the earth is more at the perihelion than at the aphelion.

Topic: Physics, Gravitation
Out of following isomeric forms of uracil, which one is present in RNA?

Options:
A
B
C
D

The correct answer is:

Solution:
Uracil is one of the four nucleo bases in the nucleic acid RNA .

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2021 (27 Aug Shift 1)
Out of $\mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}, \mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{4}, \mathrm{H}_{2} \mathrm{SO}_{5}$ and $\mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{8}$ peroxy acids are

Options:
A $\mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}, \mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{8}$
B $\mathrm{H}_{2} \mathrm{SO}_{5}, \mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{8}$
C $\mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{4}, \mathrm{H}_{2} \mathrm{SO}_{5}$
D $\mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}, \mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{4}$

The correct answer is:
$\mathrm{H}_{2} \mathrm{SO}_{5}, \mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{8}$

Solution:
Peroxy acids contain $-\mathrm{O}-\mathrm{O}-$ linkage.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Out of \(\mathrm{C}\) and \(\mathrm{CO}\), which is a better reducing agent for \(\mathrm{ZnO}\) ?

Options:

None

Solution:
The two reduction reactions are :
\(\mathrm{ZnO}(s)+\mathrm{C}(s) \longrightarrow \mathrm{Zn}(s)+\mathrm{CO}(g)\) ...(i)
\(\mathrm{ZnO}(s)+\mathrm{CO}(g) \longrightarrow \mathrm{Zn}(s)+\mathrm{CO}_2(g)\) ...(ii)
In the first case, there is increase in the magnitude of \(\Delta \mathrm{S}^{\circ}\) while in the second case, it almost remains the same. In other words \(\Delta G^{\circ}\) will have more negative value in the first case when \(\mathrm{C}(s)\) is used as the reducing agent than in the second case when \(\mathrm{CO}(g)\) acts as the reducing agent. Therefore, \(\mathrm{C}(s)\) is a better reducing agent.

Topic: Chemistry, General Principles and Processes of Isolation of Metals
Out of nutrient cycles of sulphur-a; phosphorus-b; carbon-c; and nitrogen-d; the sedimentary cycle is/are:

Options:
A a only
B b and (d) only
C c only
D aÂ andÂ b only

The correct answer is:
a and b only

Solution:
The movement and exchange of organic and inorganic matter between living and non-living component in a cyclical manner refers to nutrient cycle. These are also called as biogeochemical cycles as there is involvement of biotic and abiotic factors and chemical reactions. There are many important cycles which help in maintaining the biosphere like carbon cycle, nitrogen cycle etc. Calcium cycle, sodium cycle, phosphorous cycle, iron cycle, magnesium cycle, sulfur cycle are sedimentary cycles. The nitrogen, carbon and oxygen are found in the gaseous form in the atmosphere and have gaseous phase in their cycles.

Topic: Biology, Ecosystem, NEET
Out of the compounds below the vapour pressure of (B) at a particular temperature is

Options:
A higher than that of (A)
B lower than that of (A)
C higher or lower than (A), depending on the size of the vessel
D same as that of (A)

The correct answer is:
higher than that of (A)

Solution:
In (A) para-nitro phenol intermolecular (between two molecules) H-bonding exists while in $(B)$ ortho- nitro phenol, intramolecular H-bonding exists.
Because of the presence of intramolecular H-bonding, the boiling point of $(B)$ is lower as compared to $(A)$ and thus, $(B)$ is more volatile, ie, has higher vapour pressure as compared to (A).

Topic: Chemistry, Solutions, JEE Main
Out of the following a skew-symmetric matrix is

Options:
A $\left[\begin{array}{ccc}0 & 4 & 5 \\ -4 & 0 & -6 \\ -5 & 6 & 0\end{array}\right]$
B $\left[\begin{array}{ccc}1 & 4 & 5 \\ -4 & 1 & -6 \\ -5 & 6 & 1\end{array}\right]$
C $\left[\begin{array}{ccc}1 & 4 & 5 \\ -4 & 2 & -6 \\ -5 & 6 & 3\end{array}\right]$
D $\left[\begin{array}{ccc}i+1 & 4 & 5 \\ -4 & i & -6 \\ -5 & 6 & i\end{array}\right]$

The correct answer is:
$\left[\begin{array}{ccc}0 & 4 & 5 \\ -4 & 0 & -6 \\ -5 & 6 & 0\end{array}\right]$

Solution:
In a skew-symmetrix matrix $a_{i j}=-a_{j i}$ where $i, j=1,2,3$
for $j=i, a_{i i}=-a_{j i} \Rightarrow$ each $a_{i i}=0$.Hence the matrix $\left[\begin{array}{ccc}0 & 4 & 5 \\ -4 & 0 & -6 \\ -5 & 6 & 0\end{array}\right]$ is skew-symmetric.

Topic: Mathematics, Matrices, JEE Main
Out of the following functions representing motion of a particle which represents SHM
I. $y=\sin \omega t-\cos \omega t$
II. $y=\sin ^3 \omega t$
III. $y=5 \cos \left(\frac{3 \pi}{4}-3 \omega t\right)$
IV. $y=1+\omega t+\omega^2 t^2$

Options:
A Only (IV) does not represent SHM
B (I) and (III)
C (I) and (II)
D Only (I)

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

Solution:
For a simple harmonic motion
$$
\frac{d^2 y}{d t^2} \propto-y
$$
Hence, equation $y=\sin \omega t-\cos \omega t$ and $y=5 \cos \left(\frac{3 \pi}{4}-3 \omega t\right)$ are satisfying this condition and equation $y=1+\omega t+\omega^2 t^2$ is not periodic and $y=\sin ^3 \omega t$ is periodic but not SHM.
Option (3) is correct

Topic: Physics, Oscillations, JEE Main
Out of the following ions Ti3+, V3+, Cu+, Sc3+, Mn2+, Fe3+ and Co2+ the colourless ions will be

Options:
A Cu+, Sc3+
B Ti3+, V3+
C Cu+, Co2+
D Sc3+, Fe3+

The correct answer is:
Cu+, Sc3+

Solution:
Only those ions will be coloured which have incomplete d - orbitals. Ions which has complete or vacant d - orbitals are colourless.

Ti3+ = [Ar] 3d1	Purple

V3+ = [Ar] 3d2	Green

Cu+ = [Ar] 3d10	Colourless

Sc3+ = [Ar]	Colourless

Mn2+ = [Ar] 3d5	Pink

Fe3+ = [Ar] 3d5	Brown

Co2+ = [Ar] 3d7	Pink

As Sc3+ and Cu+ have 3d0 and 3d10 configuration in their valence shell so their aqueous solutions are colourless. All others, i.e., Ti3+, V3+, Mn2+, Fe3+ and Co2+ are coloured in aqueous medium.

Topic: Chemistry, d and f Block Elements, JEE Main
Out of the following options which one can be used to produce a propagating electromagnetic wave?

Options:
A A charge moving at constant velocity
B A stationary charge
C A chargeless particle
D An accelerating charge

The correct answer is:
An accelerating charge

Solution:
An accelerating charge can produce electromagnetic wave.

Topic: Physics, Electromagnetic Waves, JEE Main
Out of the following outer electronic configurations of atoms, the highest oxidation state is achieved by which one?

Options:
A $(n-1) d^{8} n s^{2}$
B $(n-1) d^{5} n s^{2}$
C $(n-1) d^{3} n s^{2}$
D $(n-1) d^{5} n s^{1}$

The correct answer is:
$(n-1) d^{5} n s^{2}$

Solution:
Out of the given outer electronic configuration of atoms, the highest oxidation state is achieved by $(n-1) d^{5} n s^{2}$ i.e. 7
A large number of oxidation state is due to the fact the $(n-1) d$ -electrons may get involved along with $n s$ electrons in bonding as electrons in $(n-1) d-$ orbitals are in an energy state comparable to $n s$ -electrons. Oxidation state of other options are as follows:

Topic: Chemistry, Redox Reactions, JEE Main
Out of the following pair, which one does NOT have identical dimensions is

Options:
A
angular momentum and Planck's constant

B
impulse and momentum

C
moment of inertia and moment of a force

D
work and torque

The correct answer is:
moment of inertia and moment of a force

Solution:
Using dimension

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2005
Out of the following statements which is not correct

Options:
A When unpolarised light passes through a Nicol's prism, the emergent light is elliptically polarised
B Nicol's prism works on the principle of double refraction and total internal reflection
C Nicol's prism can be used to produce and analyse polarised light
D Calcite and Quartz are both doubly refracting crystals

The correct answer is:
When unpolarised light passes through a Nicol's prism, the emergent light is elliptically polarised

Solution:
It magnitude of light vector varies periodically during it's rotation, the tip of vector traces an ellipse and light is said to be elliptically polarised. This is not in nicol prism.

Topic: Physics, Wave Optics, JEE Main
Out of the following, the alkene that exhibits optical isomerism is

Options:
A
3-methyl-2-pentene

B
4-methyl-1-pentene

C
3-methyl-1-pentene

D
2-methyl-2-pentene

The correct answer is:
3-methyl-1-pentene

Solution:
only 3-methyl-1-pentene has a chiral carbon

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2010
Out of the following which has smallest bond length

Options:
A $\mathrm{O}_2$
B $\mathrm{O}_2^{+}$
C $\mathrm{O}_2^{-}$
D $\mathrm{O}_2^{2-}$

The correct answer is:
$\mathrm{O}_2^{+}$

Solution:
Bond order of $O_2^{+}$is highest so its bond length is smallest.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Out of the following, which is the correct match for the radial probability of finding the electron for 2s orbital

Options:
A A - H, B - He+ , C - Li2+
B A - He+ , B - H, C - Li2+
C Can't say
D A - Li2+ , B - He+ , C - H

The correct answer is:
A - Li2+ , B - He+ , C - H

Solution:
Radial nodes for 2s orbital is at distance 2 a 0 Z
For (A) 2 a 0 Z is least Z = 3 for Li2+
For (B) 2 a 0 Z is intermediate Z = 2 for He+
For (C) 2 a 0 Z is maximum Z = 1 for H.
A - Li2+ , B - He+ , C - H.

Topic: Chemistry, Structure of Atom, JEE Main
Out of the following which one is not a possible energy for a photon to be emitted by hydrogen atom according to Bohr's atomic model?

Options:
A $1.9 \mathrm{eV}$
B $11.1 \mathrm{eV}$
C $13.6 \mathrm{eV}$
D $0.65 \mathrm{eV}$

The correct answer is:
$11.1 \mathrm{eV}$

Solution:
Obviously difference of $11.1 \mathrm{eV}$ is not possible.

Topic: Physics, Atomic Physics, NEET, NEET 2011 (Mains)
Out of the following, which pair of solutions is not a buffer solution

Options:
A $\mathrm{NH}_4 \mathrm{Cl}+\mathrm{NH}_4 \mathrm{OH}$
B $\mathrm{NaCl}+\mathrm{NaOH}$
C $\mathrm{Na}_2 \mathrm{HPO}_4+\mathrm{Na}_3 \mathrm{PO}_4$
D $\mathrm{CH}_3 \mathrm{COOH}+\mathrm{CH}_3 \mathrm{COONa}$

The correct answer is:
$\mathrm{NaCl}+\mathrm{NaOH}$

Solution:
Buffer solution is a combination of weak acid and conjugate base. $\mathrm{NaCl}$ is a salt and $\mathrm{NaOH}$ is the base.

Topic: Chemistry, Ionic Equilibrium, JEE Main