datavorous/entrance-exam-dataset huggingface dataset:
The IUPAC name of the following compound is:

Options:
A 2-Ethyl-1, 1-dimethyl cyclohexane
B 1, 1-Dimethyl-2-ethyl cyclohexane
C 2, 2-Dimethyl-1-1-ethyl cyclohexane
D 1-Ethyl-2, 2-dimethyl cyclohexane

The correct answer is:
2-Ethyl-1, 1-dimethyl cyclohexane

Solution:
Six membered cyclic ring will be the root chain. It is an alkane with only single bonds so suffix "ane" will be used. The IUPAC name of the compound will be.

2-Ethyl-1, 1-dimethyl cyclohexane

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2017 (08 Apr Online)
The IUPAC name of the following compound is :

Options:
A
$3-$ethyl$-4-$methylhex$-4-$ene

B
$4,4-$diethyl$-3-$methylbut$-2-$ene

C
$4-$methyl$-3-$ethylhex$-4-$ene

D
$4-$ethyl$-3-$methylhex$-2-$ene

The correct answer is:
$4-$ethyl$-3-$methylhex$-2-$ene

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
The IUPAC name of the following compound is

Options:
A 3-ethyl-4-methylhex-4-ene
B 4,4-diethyl-3-methylbut-2-ene
C 4-methyl-3-ethylhex-4-ene
D 4-ethyl-3-methylhex-2-ene

The correct answer is:
4-ethyl-3-methylhex-2-ene

Solution:
4-Ethyl-3-methylhex-2-ene

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2018 (15 Apr)
The IUPAC name of the following compound is:

Options:
A 5-Bromo-3-methylcyclopentanoic acid
B 4-Bromo-2-methylcyclopentane carboxylic acid
C 3-Bromo-5-methylcyclopentanoic acid
D 3-Bromo-5-methylcyclopentanoic carboxylic acid

The correct answer is:
4-Bromo-2-methylcyclopentane carboxylic acid

Solution:
4-Bromo-2-methylcyclopentane carboxylic acid

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2020 (04 Sep Shift 1)
The IUPAC name of the following compound
is

Options:
A trans-2-chloro-3-iodo-2-pentene
B cis-3-iodo-4-chloro-3-pentene
C trans-3-iodo-4-chloro-3-pentene
D cis-2-chloro-3-iodo-2-pentene

The correct answer is:
trans-2-chloro-3-iodo-2-pentene

Solution:

Topic: Chemistry, General Organic Chemistry, NEET, NEET 2011 (Mains)
The IUPAC name of the following structure is

Options:
A 1-hydroxy 4-methyl 3-pentanone
B 2-methyl 5-hydroxy 3-pentanone
C 4-methyl 3-oxo 1-pentanol
D Hexanol-1, one-3

The correct answer is:
1-hydroxy 4-methyl 3-pentanone

Solution:
1-hydroxy 4-methyl 3-pentanone

Topic: Chemistry, Aldehydes and Ketones, JEE Main
The IUPAC name of the given structure

is-

Options:
A 4-keto-3-bromo-2-iodopentanoic acid
B 2-iodo-3-bromo-4-keto-pentanoic acid
C 3-bromo-2-iodo-4-ketopentanoic acid
D 4-carboxy-3-bromo-4-iodopentanoic acid

The correct answer is:
3-bromo-2-iodo-4-ketopentanoic acid

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Main
The IUPAC name of the product obtained by the oxidation of phenol with the help of chromic acid is

Options:
A cyclohexa-2, 4-diene-1, 4-diol
B cyclohexa-2, 4diene-1, 4-dione
C cyclohexa-2, 5-diene-1, 4-diol
D cyclohexa-2, 5-diene-1, 4-dione

The correct answer is:
cyclohexa-2, 5-diene-1, 4-dione

Solution:
cyclohexa-2, 5-diene-1, 4-dione

Topic: Chemistry, General Organic Chemistry, JEE Main
The IUPAC nomenclature of an element with electronic configuration Rn5f146d17s2 is

Options:
A Unnilunium
B Unnilbium
C Unniltrium
D Unnilquadium

The correct answer is:
Unniltrium

Solution:
IUPAC nomenclature has been recommended for all the elements with Z>100. It was decided by IUPAC that the names of elements beyond atomic number 100 should use Latin words for their numbers. The names of these elements are derived from their numerical roots.

Numerical	0	1	2	3	4	5	6	7	8	9

roots	nil	un	bi	tri	quad	pent	hex	sept	oct	en

The atomic mass of given molecule [Rn]5f146d17s2 = 86+14+1+2 = 103

Name of element with this atomic number = un+nil+tri+ium = unniltrium

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2022 (25 Jul Shift 1)
The IUPAC symbol for the element with atomic number 119 would be:

Options:
A Unh
B Uun
C Une
D Uue

The correct answer is:
Uue

Solution:
The element with atomic number 119 is called ununennium (Uue)

Symbol	Atomic number

Unh	106

Uun	110

Une	109

Uue	119

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2019 (08 Apr Shift 2)
The Keq values in HCN addition to following aldehydes are in the order

Options:
A i > ii > iii
B ii > iii > i
C iii > i > ii
D ii > i > iii

The correct answer is:
ii > i > iii

Solution:
Rate of nucleophilic addition reaction in carbonyl compound ∝ magnitude of the charge on carbonyl C-atom

Topic: Chemistry, Aldehydes and Ketones, JEE Main
The KÎ± line obtained for molybdenum (Z=42) is 0.71 Aâ. Then, the wavelength of the KÎ± line of copper (Z=29) is

Options:
A 2.14 Aâ
B 1.52 A∘
C 1.04 A∘
D 0.71 A∘

The correct answer is:
1.52 A∘

Solution:
λmin∝1Z-12
∴λCuλMo=ZMo-12ZCu-12=412282=2.14
∴ λCu=2.14×0.71A∘=1.52 A∘

Topic: Physics, Atomic Physics, JEE Main
The K Î± line obtained for molybdenum (Z=42) is 0.71Ã. Then, the wavelength of the K Î± line of copper (Z=29) is

Options:
A 2.14 Ã
B 1.52 Ã
C 1.04 Ã
D 0.71 Ã

The correct answer is:
1.52 Ã

Solution:
Î»minâ1Z-12
â´ Î» Cu Î» Mo = Z Mo - 1 2 Z Cu - 1 2 = 4 1 2 2 8 2 = 2.14
â´ Î»Cu=2.14Ã0.71Ã=1.52 Ã

Topic: Physics, Atomic Physics, JEE Main
The Ksp for bismuth sulphide Bi2S3 is 1.08×10-73. The solubility of Bi2S3 in molL-1 at 298 K is

Options:
A 1.0×10-15
B 2.7×10-12
C 3.2×10-10
D 4.2×10-8

The correct answer is:
1.0×10-15

Solution:
Bi2S3 will get dissociated as:

Bi2S3⇌2Bi+3+3S-2 2s 3s

Here,

Ksp=2s23s3=108 s5s=Ksp10815s=1.08×10-7310815s=10-15 mol L-1

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2022 (25 Jun Shift 2)
The $K_{S P}$ of $A g I$ is $1.5 \times 10^{-16}$. On mixing equal volumes of the following solutions, precipitation will occur only with

Options:
A $10^{-7} \mathrm{M} \mathrm{Ag}^{+}$and $10^{-19} \mathrm{M}^{-}$
B $10^{-8} \mathrm{M} \mathrm{Ag}^{+}$and $10^{-8} \mathrm{MI}^{-}$
C $10^{-16} \mathrm{M} \mathrm{Ag}^{+}$and ${ }^{10^{-16} \mathrm{M} \mathrm{I}^{-}}$
D $10^{-9} \mathrm{M} \mathrm{Ag}^{+}$and $10^{-9} \mathrm{MI}^{-}$

The correct answer is:
$10^{-8} \mathrm{M} \mathrm{Ag}^{+}$and $10^{-8} \mathrm{MI}^{-}$

Solution:
$K_{s p}$ of $\mathrm{AgI}=1.5 \times 10^{-16}$
$\begin{aligned} & 10^{-8} \mathrm{MAg}^{+} \text {and } 10^{-8} \mathrm{MI}^{-} \\ & \text {Ionic product }=10^{-16} \\ & K_{s p}=\text { Ionic product }\end{aligned}$

Topic: Chemistry, Ionic Equilibrium, JEE Main
The K sp of FeS=4Ã 10 â19 at 298âK. The minimum concentration of H + ions required to prevent the precipitation of FeS from a 0.01âM solution Fe 2+ salt by passing H 2 S

(Given H 2 S k a 1 Ã k b 1 = 10 â21 â )

Options:
A 1.6 × 10–3 M
B 2.5 × 10–4 M
C 2.0 × 10–2 M
D 1.2 × 10–4 M

The correct answer is:
1.6 × 10–3 M

Solution:
Fe2+ S–2=4 × 10–19⇒ S–2=4×10–191×10–2= 4×10–17 M
In order to precipitate FeS, S–2 required is 4 × 10–17 M from 0.01 M Fe2+ salts.
Now H+24×10–170.1= 1 × 10–21
⇒ H+2 =2.5 × 10–6
⇒ H+=1.6 × 10–3

Topic: Chemistry, Ionic Equilibrium, JEE Main
The Ksp of Mg (OH)2 is 1×10-12. A 0.01 M MgCl2 solution will precipitate at what limiting pH value?

Options:

The correct answer is:
9

Solution:
Ksp=Mg2+ OH-2
OH-=1×10-120.01
OH-=10-5
pOH=5
pH=9

Topic: Chemistry, Ionic Equilibrium, JEE Main
The Kα X-ray of molybdenum has wavelength 0.071 nm. If the energy of a molybdenum atom with a K electron knocked out is 27.5 keV, the energy of this atom when an L electron is knocked out will be keV. (Round off to the nearest integer ) h=4.14×10-15 eV s, c=3×108 m s-1

Options:

The correct answer is:
10

Solution:
Ekα=Ek-EL

hcλkα=Ek-EL

EL=Ek-hcλkα

=27.5 KeV-12.42×10-7eV m0.071×10-9 m

EL=(27.5-17.5) keV

=10 keV

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2021 (27 Jul Shift 2)
The key concept of Darwin’s theory of evolution is

Options:
A natural selection.
B branching descent.
C saltation.
D Both (A) & (B).

The correct answer is:
Both (A) & (B).

Solution:
To explain evolution, Darwin talked about two things

1) Branching descent
2) Natural selection

Branching descent explains the âpatternâ of evolution and natural selection explains the âmechanismâ of evolution. Branching descent and natural selection are the two key concepts of the Darwinian theory of evolution.
Branching descent: He said that evolution is a branching process. With time, new life forms appear and these new life forms arise as âbranchesâ from the previously existing life forms. Hence, branching descent points towards the concept of common ancestry.
Natural selection: He emphasized that the selection of âuseful variationsâ by nature is the main (but not exclusive) mechanism for evolution to occur.

Topic: Biology, Evolution, NEET
The key step in cannizzaro's reaction is the intermolecular shift of

Options:
A Proton
B Hydride ion
C Hydronium ion
D Hydrogen bond

The correct answer is:
Hydride ion

Solution:

Topic: Chemistry, Aldehydes and Ketones, JEE Main
The kind of tissue that forms the supportive structure in our pinna (external ears) is also found in -

Options:
A tip of the nose
B vertebrae
C nails
D ear ossicles

The correct answer is:
tip of the nose

Solution:
No Solution Available

Topic: Biology, Structural Organisation in Animals, NEET, NEET 2009 (Mains)
The kinetic e nergy acquired by a body of mass 'M' in travelling a certain distance 'd:
starting from rest, under the action of constant force is

Options:
A inversely proportional to $\sqrt{\mathrm{M}}$.
B directly proportional to M.
C independent of M.
D directly proportional to $\sqrt{M}$.

The correct answer is:
independent of M.

Solution:
Work done by external forces $=$ Change in kinetic energy
Since particle starts from rest, its initial Kinetic Energy is o.
Therefore, K.E.=Fd
It is given that force is constant.
We see that Kinetic energy is independent of mass.

Topic: Physics, Work Power Energy, JEE Main
The kinetic energy and potential energy of a particle executing simple harmonic motion will be equal, when displacement (amplitude $=a$) is

Options:
A $\frac{a}{2}$
B $a \sqrt{2}$
C $\frac{a}{\sqrt{2}}$
D $\frac{a \sqrt{2}}{3}$

The correct answer is:
$\frac{a}{\sqrt{2}}$

Solution:
Suppose at displacement $y$ from mean position potential energy = kinetic energy
$\begin{aligned} & \Rightarrow \frac{1}{2} m\left(a^2-y^2\right) \omega^2=\frac{1}{2} m \omega^2 y^2 \\ & \Rightarrow a^2=2 y^2 \Rightarrow y=\frac{a}{\sqrt{2}}\end{aligned}$

Topic: Physics, Oscillations, JEE Main
The kinetic energy K of a particle moving along a circle of radius R depends on the distance covered s, as K=as2, where, a is constant. Then the force acting on the particle is

Options:
A 2as2R
B 2as1+s2R212
C 2as
D 2aR2s

The correct answer is:
2as1+s2R212

Solution:
According to given problem, 12mv2=as2
âv=s2am
So, ac=v2R=2as2mR ... (i)
Further more as
at=dvdt=dvds.dsdt=vÃdvds ... (ii)
[by chain rule]
Which in light of Equation. (i) i.e. v=s2am yields
at=s2am2am=2asm ... (iii)
So that a=aR2+at2
=2as2mR2+2âasm2
Hence, a=2asm1+sR2
â´ Force acting on the particle F=ma=2as1+sR2

Topic: Physics, Oscillations, JEE Main
The kinetic energy K of a particle moving along a circle of radius R depends upon the distance s , as K=as2 . The force acting on the particle is-

Options:
A 2as2R
B 2as1+s2R212
C 2as
D 2a

The correct answer is:
2as1+s2R212

Solution:
K=as2

⇒v=2ams

⇒dvdt=2amdsdt=2ams

The net force on the particle is

F=mv2R2+mdvdt2=2as2R2+(2as)2

=2as1+s2R21/2

Topic: Physics, Laws of Motion, JEE Main
The kinetic energy needed to project a body of mass \(m\) from the earth surface (radius \(\mathrm{R}\)) to infinity is

Options:
A \(2 \mathrm{mgR}\)
B \(\mathrm{mgR} / 4\)
C \(\mathrm{mgR} / 2\)
D \(\mathrm{mgR}\)

The correct answer is:
\(\mathrm{mgR} / 4\)

Solution:
If any object through with escaping velocity then the object will never come back at the ground, So we can assume it will reach at infinite distance.
\(\mathrm{KE}=\frac{1}{2} \mathrm{mv} v_{\text {esc }}^2=\frac{1}{2} \mathrm{~m}(\sqrt{2 \mathrm{gR}})^2=\mathrm{mgR}\)

Topic: Physics, Gravitation, JEE Main
The kinetic energy needed to project a body of mass $m$ from the earth surface (radius $R$ ) to infinity is

Options:
A
$\mathrm{mgR} / 2$

B
$2 \mathrm{mgR}$

C
$\mathrm{mgR}$

D
$\mathrm{mgR} / 4$

The correct answer is:
$\mathrm{mgR}$

Solution:
$\mathrm{KE}=\frac{1}{2} \mathrm{mv}_{\mathrm{esc}}^2=\frac{1}{2} \mathrm{~m}(\sqrt{2 \mathrm{gR}})^2=\mathrm{mgR}$

Topic: Physics, Gravitation, JEE Main, JEE Main 2002
The kinetic energy of a body becomes four times its initial value. The new linear momentum will be

Options:
A same as the initial value
B four times of the initial value
C twice of the initial value
D eight times of the initial value.

The correct answer is:
twice of the initial value

Solution:
Initial kinetic energy $\left(E_1\right)=E$ and final kinetic energy $\left(E_2\right)=4 E$.
Kinetic energy of a body is $\frac{1}{2} m v^2$
Since the value of $m$ remains constant, therefore for the kinetic energy to be 4 times, the new value of velocity (v) should be 2 times the initial value. Initial linear momentum $\left(p_1\right)=m v$.
Therefore new linear momentum $\left(p_2\right)=m \times 2 v$ $=2 m v=2 p_1$

Topic: Physics, Center of Mass Momentum and Collision, AIIMS, AIIMS 2002
The kinetic energy of a body is increased by 4 times. Its momentum will

Options:
A increase twice
B increase four times
C decrease twice
D remains constant

The correct answer is:
increase twice

Solution:
Given, initial kinetic energy, \(K_1=K\)
Final kinetic energy, \(K_2=4 K\)
Momentum and kinetic energy are related as,
\(\begin{aligned}
& \qquad K=\frac{p^2}{2 m} \Rightarrow p=\sqrt{2 m K} \\
& \begin{aligned}
\therefore \text {Initial momentum, } p_1 & =\sqrt{2 m K_1} \\
\Rightarrow & =\sqrt{2 m K} \quad \ldots (i) \\
\text {Final momentum, } p_2 & =\sqrt{2 m K_2}=\sqrt{2 m \cdot 4 K} \\
\Rightarrow & \ldots \text { (i) } \\
& =2 \sqrt{2 m K}=2 p_1 \quad \text { [from Eq. (i)] }
\end{aligned}
\end{aligned}\)

Topic: Physics, Center of Mass Momentum and Collision, AP EAMCET, AP EAMCET 2020 (18 Sep Shift 2)
The kinetic energy of a light body and a heavy body is same. Which one of the following statements is CORRECT?

Options:
A The light body has greater momentum.
B A body having high velocity has greater momentum.
C Both bodies have same momentum.
D The heavy body has greater momentum.

The correct answer is:
The heavy body has greater momentum.

Solution:
Kinetic energy,
$\mathrm{k}=\frac{1}{2} \mathrm{my}^{2}=\frac{(\mathrm{mv})^{2}}{2 \mathrm{~m}}=\frac{\mathrm{p}^{2}}{2 \mathrm{~m}}$
or $\mathrm{p}=\sqrt{2 \mathrm{mk}}$ i.e., $\mathrm{p} \propto=\sqrt{\mathrm{m}}$
Thus the heavier body has a greater momentum than the lighter one.

Topic: Physics, Work Power Energy, JEE Main
The kinetic energy of a particle executing S.H.M. is 16 J when it is at its mean position. If the amplitude of oscillations is 25 cm, and the mass of the particle is 5.12 kg, then the time period of the oscillation is

Options:
A 20π s
B 2π s
C π/5 s
D 5π s

The correct answer is:
π/5 s

Solution:
When the particle is in its mean position, the kinetic energy will be maximum,
Kmax=12mω2a2 and ω=2πT
⇒16=12×5.12×4π2T2×0.252
⇒T2=2.56×4π2×0.254×0.25
⇒T=1.6×2π×0.254=π5 s

Topic: Physics, Oscillations, JEE Main
The kinetic energy of an electron having de-Broglie wavelength Î» is ( h= Planck's constant, m= mass of electron)

Options:
A h2mλ
B h22mλ2
C h22m2λ2
D h22m2λ

The correct answer is:
h22mλ2

Solution:
Î»= h 2mK.E
Î» 2 = h 2 2m( K.E )
( K.E )= h 2 2m Î» 2

Topic: Physics, Atomic Physics, JEE Main
The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is $\left[a_{0}\right.$ is Bohr radius $]$ :

Options:
A $\frac{h^{2}}{4 \pi^{2} m a_{0}^{2}}$
B $\frac{h^{2}}{16 \pi^{2} m a_{0}^{2}}$
C $\frac{h^{2}}{32 \pi^{2} m a_{0}^{2}}$
D $\frac{h^{2}}{64 \pi^{2} m a_{0}^{2}}$

The correct answer is:
$\frac{h^{2}}{32 \pi^{2} m a_{0}^{2}}$

Solution:
As per Bohr's postulate,

$\begin{array}{l}

m v r=\frac{n h}{2 \pi} \quad \text { So, } v=\frac{n h}{2 \pi m r} \\

\mathrm{KE}=\frac{1}{2} m v^{2} \quad \text { So, } \mathrm{KE}=\frac{1}{2} m\left(\frac{n h}{2 \pi m r}\right)^{2}

\end{array}$

$\text { Since, } r=\frac{a_{\mathrm{o}} \times n^{2}}{z}$

So, for $2^{\text {nd }}$ Bohr orbit

$\begin{array}{l}

r=\frac{a_{\mathrm{o}} \times 2^{2}}{1}=4 a_{\mathrm{o}} \\

\mathrm{KE}=\frac{1}{2} m\left(\frac{2^{2} h^{2}}{4 \pi^{2} m^{2} \times\left(4 a_{\mathrm{o}}\right)^{2}}\right)=\frac{h^{2}}{32 \pi^{2} m a_{\mathrm{o}}^{2}}

\end{array}$

Topic: Chemistry, Structure of Atom, JEE Main
The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is [ a0 is Bohr radius of first shell of hydrogen atom]

Options:
A h264π2ma02
B h232π2ma02
C h216π2ma02
D h24π2ma02

The correct answer is:
h232π2ma02

Solution:
K.E.=12mv2 ...(i)
mvr=nh2π
v=nh2πmr ...(ii)
Putting (ii) and (i)
K.E.=12mn2h24π2m2r2=n2h28π2r2m
Now r=4a0 (since n =2 ) and a0= Bohr's radius
n2h28π2×16a02m=n2h2128π2a02m
Now n=2
K.E.=4h2128π2ma02=h232π2ma02

Topic: Chemistry, Structure of Atom, JEE Main
The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is equal to h2x ma02. The value of 10 x is (a0 is radius of Bohr's orbit)

(Nearest integer)

[Given: π=3.14]

Options:

The correct answer is:
3155

Solution:
x=8π2×164=32π2

=32(3.14)2

=315.5072

10x=315.5072

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2021 (27 Aug Shift 1)
The kinetic energy of an electron, α-particle and a proton are given as 4K, 2K and K respectively. The de-Broglie wavelength associated with electron (λe), α-particle (λα) and the proton (λp) are as follows:

Options:
A λα=λp>λe
B λα<λp<λe
C λα=λp<λe
D λα>λp>λe

The correct answer is:
λα<λp<λe

Solution:
The formula to calculate the de Broglie wavelength of a particle is given by

λ=h2mK ...1

Now, the masses of the given particles are interrelated by the following relations:

me≈mp1837mα=4mp

The de Broglie wavelength for electron is given by

λe=h2me4K= h81837mpK ...2

The de Broglie wavelength for proton is given by

λp=h2mp2K= h4mpK ...3

And, the de Broglie wavelength of alpha particle is given by

λα=h2mαK= h24mpK= h8mpK ...4

From equation (2), (3) and (4), it can be concluded that λe>λp>λα.

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2023 (06 Apr Shift 1)
The kinetic energy of the photoelectrons increases by $0.52 \mathrm{eV}$ when the wavelength of incident light is changed from $500 \mathrm{~nm}$ to another wavelength which is approximately

Options:
A $400 \mathrm{~nm}$
B $1250 \mathrm{~nm}$
C $1000 \mathrm{~nm}$
D $700 \mathrm{~nm}$

The correct answer is:
$400 \mathrm{~nm}$

Solution:
Given, change in kinetic energy of photoelectrons,
$$
\begin{aligned}
\Delta K & =0.52 \mathrm{eV} \\
\lambda_1 & =500 \mathrm{~nm}, \lambda_2=?
\end{aligned}
$$
We know that, kinetic energy of emitted photoelectron (using Einstein's equation)
$$
\begin{gathered}
K=\frac{h c}{\lambda}-\phi \\
\therefore \text { At wavelength } \lambda_1 \\
K_1=\frac{h c}{\lambda_1}-\phi
\end{gathered}
$$
At wavelength $\lambda_{2^{\prime}}$
$$
K_2=\frac{h c}{\lambda_2}-\phi
$$
From Eqs. (i) and (ii), we have
$$
\begin{array}{cc}
& K_2-K_1=\frac{h c}{\lambda_2}-\frac{h c}{\lambda_1} \\
\Rightarrow \quad \Delta K=h c\left(\frac{1}{\lambda_2}-\frac{1}{\lambda_1}\right) \\
\Rightarrow \quad 0.52=1242\left(\frac{1}{\lambda_2}-\frac{1}{500}\right)
\end{array}
$$
$$
\begin{array}{rlrl}
\Rightarrow & & \frac{0.52}{1242} & =\frac{1}{\lambda_2}-\frac{1}{500} \\
\Rightarrow & \frac{1}{\lambda_2} & =\frac{0.52}{1242}+\frac{1}{500} \\
\Rightarrow & & \lambda_2 & \approx 400 \mathrm{~nm}
\end{array}
$$

Topic: Physics, Dual Nature of Matter, JEE Main
The kinetic energy possessed by a body of mass $m$ moving with a velocity $v$ is equal to $\frac{1}{2} m v^2$, provided

Options:
A The body moves with velocities comparable to that of light
B The body moves with velocities negligible compared to the speed of light
C The body moves with velocities greater than that of light
D None of the above statement is correct

The correct answer is:
The body moves with velocities negligible compared to the speed of light

Solution:
Due to theory of relativity.

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
The Kirchhoff's first law $\left(\sum i=0\right)$ and second law $\left(\sum i R=\sum E\right)$, where the symbols have their usual meanings, are respectively based on

Options:
A
conservation of charge, conservation of energy

B
conservation of charge, conservation of momentum

C
conservation of energy, conservation of charge

D
conservation of momentum, conservation of charge

The correct answer is:
conservation of charge, conservation of energy

Solution:
Conservation of charge, conservation of energy

Topic: Physics, Current Electricity, JEE Main, JEE Main 2006
The Kjeldahl method of Nitrogen estimation fails for which of the following reaction products?

Options:
A (c) and (d)
B (a) and (d)
C (a), (c) and (d)
D (b) and (c)

The correct answer is:
(c) and (d)

Solution:
(a)	(N- present in product so will show Kjeldhal Test)

(b)	(-N2+ Never show Kjeldhal Test)

(c)	(N- absent, wo will not show Kjeldhal Test)

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2020 (03 Sep Shift 1)
The lac operon consists of

Options:
A four regulatory genes only
B one regulatory gene and three structural genes
C two regulatory genes and two structural genes
D three regulatory genes and three structural genes

The correct answer is:
one regulatory gene and three structural genes

Solution:
Lac operon is the operon that regulates lactose metabolism in the bacterium Escherichia coli. Its form was first postulated in 1961 by Francois Jacob and Jacques Monod to explain control of $\beta$-galactosidase synthesis and for this work they were awarded Nobel Prize. Lac operon system or inducible operon system consists of regulator gene (i), promotor gene (p), operator gene (o), three structural genes ( $\mathrm{z}, \mathrm{y}$ and $\mathrm{a})$, repressor protein and inducer.

Topic: Biology, Molecular Basis of Inheritance, NEET, NEET 2010 (Mains)
The lanthanide contraction is responsible for the fact that:

Options:
A \( \mathrm{Zr} \) and \( \mathrm{Y} \) have about the same radius
B Zr and Nb have similar oxidation state
C \( \mathrm{Zr} \) and \( \mathrm{Zn} \) have the same oxidation state
D \( \mathrm{Zr} \) and \( \mathrm{Hf} \) have about the same radius

The correct answer is:
\( \mathrm{Zr} \) and \( \mathrm{Hf} \) have about the same radius

Solution:
Zr and Hf belong to 4d & 5d series in same group have same atomic size due to Lanthanide Contraction.

The major cause for lanthanide contraction is due to the inappropriate shielding of the 4f electrons due to the improper shape of the f-orbitals. As the atomic number increases in the lanthanide series, for every proton in the nucleus the extra electron goes to fill the 4f-orbitals.

Topic: Chemistry, d and f Block Elements, JEE Main
The lanthanide contraction is responsible for the fact that

Options:
A
Zr and Y have about the same radius

B
Zr and Nb have similar oxidation state

C
Zr and Hf have about the same radius

D
Zr and Zn have the same oxidation

The correct answer is:
Zr and Hf have about the same radius

Solution:
Due to Lanthanide contraction.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2005
The lanthanide contraction is responsible for the fact that

Options:
A $Z r$ and $Y$ have about the same radius
B $\mathrm{Zr}$ and $\mathrm{Nb}$ have similar oxidation state
C $\mathrm{Zr}$ and $H f$ have about the same radius
D $Z r$ and $Z n$ have the same oxidation state

The correct answer is:
$\mathrm{Zr}$ and $H f$ have about the same radius

Solution:
$\mathrm{Zr}$ and $\mathrm{Hf}$ have about the same radius

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main