datavorous/entrance-exam-dataset huggingface dataset:
AB is a light rigid rod, which is rotating about a vertical axis passing through end A. A spring of force constant k and natural length l is attached at A and its other end is attached to a small bead of mass m. The bead can slide without friction on the rod. At the initial moment, the bead is at rest (with respect to the rod) and the spring is unstretched. The maximum speed of the bead with respect to the rod during the course of motion is

Options:
A Vmax=mω2l2k-mω2
B Vmax=mÏ4l2k-mÏ2
C Vmax=mω4l2mω2-k
D Vmax=mω2l2mω2-k

The correct answer is:
Vmax=mω4l2k-mω2

Solution:
Velocity will be maximum at the equilibrium position
âkx=mÏ2l+x
â x=mÏ2lk-mÏ2
Now, using work-energy theorem

âKE= Work done by all the forces
12mVmax2=â«0xmÏ2l+xdx-12kx2
â Vmax2=2mÏ2lx+mÏ2x2-kx2m
Vmax2=mÏ2l+mÏ2l+x-kxxm
=mÏ2lxm
â Vmax2=Ï2lx=mÏ4l2k-mÏ2
Vmax=mÏ4l2k-mÏ2

Topic: Physics, Laws of Motion, JEE Main
AB is a light rigid rod, which is rotating about a vertical axis passing through end A . A spring of force constant k and natural length l is attached at A and its other end is attached to a small bead of mass m. The bead can slide without friction on the rod. At the initial moment, the bead is at rest (w.r.t the rod) and the spring is unstretched. Select incorrect options :

Options:
A Vmax=mω2l2k-mω2
B Vmax=mω4l2k-mω2
C Vmax=mω4l2mω2-k
D Vmax=mω2l2mω2-k

The correct answer is:
Vmax=mω4l2k-mω2

Solution:
Velocity will be maximum at the equilibrium position
⇒kx=mω2l+x
⇒ x=mω2lk-mω2
Now, using work-energy theorem

∆KE= Work done by all the forces
12mVmax2=∫0xmω2l+xdx-12kx2
⇒ Vmax2=2mω2lx+mω2x2-kx2m
Vmax2=mω2l+mω2l+x-kxxm
=mω2lxm
⇒ Vmax2=ω2lx=mω4l2k-mω2
Vmax=mω4l2k-mω2

Topic: Physics, Laws of Motion, JEE Main
ABA is involved in

Options:
A Dormancy of seeds
B Root elongation
C Shoot elongation
D Increased cell division

The correct answer is:
Dormancy of seeds

Solution:
Abscisic acid acts as a growth inhibitor and induces dormancy of buds towards the approach of winter. Dormancy of seeds is mainly caused by abscisic acid. Because of its action in inducing dormancy abscisic acid (ABA) is also called dormin. The buds, as well as seeds, sprout only when abscisic acid is overcome by gibberellins.

Topic: Biology, Plant - Growth and Development, NEET
ABC is a triangle. Forces $\vec{P}, \vec{Q}, \vec{R}$ acting along $I A, I B$ and $I C$ respectively are in equilibrium, where $I$ is the incentre of $\triangle A B C$. Then $P: Q: R$ is

Options:
A
$\sin A: \sin B: \sin C$

B
$\sin \frac{A}{2}: \sin \frac{B}{2}: \sin \frac{C}{2}$

C
$\cos \frac{\mathrm{A}}{2}: \cos \frac{\mathrm{B}}{2}: \cos \frac{\mathrm{C}}{2}$

D
$\cos A: \cos B: \cos C$

The correct answer is:
$\cos \frac{\mathrm{A}}{2}: \cos \frac{\mathrm{B}}{2}: \cos \frac{\mathrm{C}}{2}$

Solution:
Using Lami's Theorem
$$
\therefore P: Q: R=\cos \frac{A}{2}: \cos \frac{B}{2}: \cos \frac{C}{2} \text {. }
$$

Topic: Mathematics, Properties of Triangles, JEE Main, JEE Main 2005
$\mathrm{ABC}$ is an equilateral triangle. Charges ${ }^{+} q$ are placed at each corner. The electric intensity at 0 will be

Options:
A $\frac{1}{4 \pi \varepsilon_0} \frac{q}{r^2}$
B $\frac{1}{4 \pi \varepsilon_0} \frac{q}{r}$
C Zero
D $\frac{1}{4 \pi \varepsilon_0} \frac{3 q}{r^2}$

The correct answer is:
Zero

Solution:

Topic: Physics, Electrostatics, JEE Main
ABO blood grouping is controlled by gene I, which has three alleles and show codominance. There are six genotypes. How many phenotypes in all are possible?

Options:
A Six
B Three
C Four
D Five

The correct answer is:
Four

Solution:
In $\mathrm{ABO}$ blood group system, inheritance of grouping is controlled by a single autosomal gene on chromosome-9 with three major alleles $\mathrm{A}, \mathrm{B}$ and $\mathrm{O}\left(\mathrm{I}^{\mathrm{A}}, \mathrm{I}^{\mathrm{B}}\right.$ and $\left.\mathrm{I}^{\mathrm{O}}\right)$. The $\mathrm{ABO}$ blood group system has at least 6 genotypes. On the basis of presence or absence of antigens and antibodies, 4 blood groups (phenotypes) have been differentiated, ie, $A, B, A B$ and $O$ blood groups.

Topic: Biology, Principles of Inheritance and Variation, NEET, NEET 2010 (Mains)
About $70 \%$ of total global carbon is found in

Options:
A Forests
B Grasslands
C Agroecosystems
D Oceans

The correct answer is:
Oceans

Solution:
Oceans : For the C, cycling pool consists of $6 \times 10^{14} \mathrm{~kg}(29 \%)$ of free $\mathrm{CO}_2$ in the atmosphere and $1.45 \times 10^{15} \mathrm{~kg}(71 \%)$ dissolved $\mathrm{CO}_2$ in hydrosphere.

Topic: Biology, Organisms and Populations, NEET, NEET 2008 (Mains)
About which day in a normal human menstrual cycle does rapid secretion of LH (popularly called LH surge) normally occurs?

Options:
A $14^{\text {th }}$ day
B $20^{\text {th }}$ day
C $5^{\text {th }}$ day
D $11^{\text {th }}$ day

The correct answer is:
$14^{\text {th }}$ day

Solution:
Rapid secretion of LH leading to its maximum level during the mid-cycle, $i e$, about the $14^{\text {th }}$ day of menstrual cycle, popularly called LH surge induces rupture of Graafian follicle and thereby the release of ovum (ovulation).

Topic: Biology, Human Reproduction, NEET, NEET 2011 (Mains)
Above curie temperature:

Options:
A a paramagnetic substance becomes ferromagnetic
B a ferromagnetic substance becomes paramagnetic
C a paramagnetic substance becomes diamagnetic
D a diamagnetic substance becomes paramagnetic

The correct answer is:
a ferromagnetic substance becomes paramagnetic

Solution:
Curie temperature \(\left(\mathrm{T}_{\mathrm{C}}\right)\) is the temperature at which certain materials lose their permanent magnetic properties, to be replaced by induced magnetism.
Above the Curie temperature, the material is paramagnetic, as the atoms lose their ordered magnetic moments when the material undergoes a phase transition.
The ordered magnetic moments change and become disordered at the Curie temperature. Higher temperatures make magnets weaker, as spontaneous magnetism only occurs below the Curie temperature.

Topic: Physics, Magnetic Properties of Matter, JEE Main
Above diagram represents

Options:
A anaphase-I
B metaphase-I
C telophase-I
D prophase-।

The correct answer is:
anaphase-I

Solution:
The given figure is showing anaphase-I of meiosis. It begins when the two chromosomes of each bivalent separate and start moving towards opposite poles of the cell. Unlike the anaphase of mitosis, the sister chromatids remain attached at their centromeres and move together toward the poles.

Topic: Biology, Cell Cycle and Cell Division, JIPMER, JIPMER 2018
Absolute alcohol (100\% alcohol) is prepared by distilling rectified spirit over

Options:
A $\mathrm{Na}$
B $\mathrm{CaCl}_2$
C $\mathrm{Mg}$
D $\mathrm{Mg}\left(\mathrm{OC}_2 \mathrm{H}_5\right)_2$

The correct answer is:
$\mathrm{Mg}\left(\mathrm{OC}_2 \mathrm{H}_5\right)_2$

Solution:
Rectified spirit ( $95.6 \%$ alcohol), on distillation, either with magnesium ethoxide or quick lime gives absolute alcohol.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Acceleration velocity graph of a particle moving in a straight line is as shown in figure. The slope of velocity-displacement
graph

Options:
A increases linearly
B decreases linearly
C is constant
D increases parabolically

The correct answer is:
is constant

Solution:
From the acceleration-velocity graph, we have a=k v where k is a constant which represents the slope of the given line.
As, a=vdvds, so vdvds=kv
or dvds=k=a constant
Thus, the slope of the velocity-displacement graph is the same as that of acceleration-velocity. Which is constant.

Topic: Physics, Motion In One Dimension, JEE Main
Accessory sexual character in female is promoted by

Options:
A androgen
B progesterone
C estrogen
D testosterone

The correct answer is:
estrogen

Solution:
Estrogens are secreted by cells of Graafian follicles. Estradiol is the principal feminizing estrogen. It stimulates the development of female secondary sex characteristics during puberty and maintains them through the reproductive years of adult life.

Topic: Biology, Cell - The Unit of Life, JIPMER, JIPMER 2008
According to Allen's rule, the mammals from colder climates have

Options:
A
shorter ears and longer limbs

B
longer ears and shorter limbs

C
longer ears and longer limbs

D
shorter ears and shorter limbs

The correct answer is:
shorter ears and shorter limbs

Solution:
, According to Allen's rule, the mammals (endothermal animals) from colder climates or areas show shorter extremities like ears and limbs as compared to the mammals of warm region. The shorter extremities of mammals in colder region helps them. To minimise heat loss and maintain homeostasis.
Rest of the other options does not follow Allen's rule.

Topic: Biology, Organisms and Populations
According to Arrhenius equation, the rate constant $(\mathrm{k})$ is related to temperature $(\mathrm{T})$ as

Options:
A $\operatorname{In} \frac{\mathrm{k}_{2}}{\mathrm{k}_{1}}=\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{R}}\left[\frac{1}{\mathrm{~T}_{1}}-\frac{1}{\mathrm{~T}_{2}}\right]$
B In $\frac{\mathrm{k}_{2}}{\mathrm{k}_{1}}=-\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{R}}\left[\frac{1}{\mathrm{~T}_{1}}-\frac{1}{\mathrm{~T}_{2}}\right]$
C In $\frac{\mathrm{k}_{2}}{\mathrm{k}_{1}}=\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{R}}\left[\frac{1}{\mathrm{~T}_{1}}+\frac{1}{\mathrm{~T}_{2}}\right]$
D $\operatorname{In} \frac{\mathrm{k}_{2}}{\mathrm{k}_{1}}=-\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{R}}\left[\frac{1}{\mathrm{~T}_{1}}+\frac{1}{\mathrm{~T}_{2}}\right]$

The correct answer is:
$\operatorname{In} \frac{\mathrm{k}_{2}}{\mathrm{k}_{1}}=\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{R}}\left[\frac{1}{\mathrm{~T}_{1}}-\frac{1}{\mathrm{~T}_{2}}\right]$

Solution:
$\ln \frac{k_{2}}{k_{1}}=\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R}}\left[\frac{1}{\mathrm{~T}_{1}}-\frac{1}{\mathrm{~T}_{2}}\right]$

Topic: Chemistry, Chemical Kinetics, JEE Main
According to Arrhenius theory, the activation energy is

Options:
A The energy it should possess so that it can enter into an effective collision
B The energy which the molecule should possess in order to undergo reaction
C The energy it has to acquire further so that it can enter into a effective collision
D The energy gained by the molecules on colliding with another molecule

The correct answer is:
The energy it has to acquire further so that it can enter into a effective collision

Solution:
The definition of activation energy.

Topic: Chemistry, Chemical Kinetics, JEE Main
According to Aufbau's principle, which of the three $4 d, 5 p$ and $5 s_{\text {will be filled with electrons first }}$

Options:
A $4 d$
B $5 p$
C $5 \mathrm{~s}$
D $4 d$ and $5 s$ will be filled simultaneously

The correct answer is:
$5 \mathrm{~s}$

Solution:
According to the Auffau's principle, electron will be first enters in those orbital which have least energy.

Topic: Chemistry, Structure of Atom, JEE Main
According to Bemoulli's equation $\frac{P}{\rho g}+h+\frac{1}{2} \frac{v^2}{g}=$ constant. The terms $A, B$ and $C$ are generally called

Options:
A Gravitational head, pressure head and velocity head
B Gravity, gravitational head and velocity head
C Pressure head, gravitational head and velocity head
D Gravity, pressure and velocity head

The correct answer is:
Pressure head, gravitational head and velocity head

Solution:
According to bernoulli's formula it is sum of pressure head, gravitational head velocity head.

Topic: Physics, Mechanical Properties of Fluids, JEE Main
According to Bohr atom model, in which of the following transitions will the frequency be maximum?

Options:
A n=3 to n=2
B n=5 to n=4
C n=4 to n=3
D n=2 to n=1

The correct answer is:
n=2 to n=1

Solution:
ΔE=13.61n12-1n22=hv

It is maximum if n1=1 and n2=2

n=5………………………………..-0.544eV

n=4 .................................................... -0.850 eV

n=3…………………………………..-1.511eV

n=2…………………………………-3.4eV

n=1………………………………….-13.6eV

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2021 (24 Feb Shift 2)
According to Bohr, the angular momentum of an electron, in any orbit, should be:

Options:
A \(h / 2 \pi\)
B \(\mathrm{nh} / 2 \pi\)
C \(\mathrm{h} / \mathrm{mv}\)
D \(2 \pi / n h\)

The correct answer is:
\(\mathrm{nh} / 2 \pi\)

Solution:
According to Bohr's theory, angular momentum of electrons is quantized. It is an integral multiple of \(\frac{h}{2 \pi}\)
where \(n=1,2,3 \ldots\)

Topic: Physics, Atomic Physics, JEE Main
According to Bohr's atomic model, the radius of the orbit is directly proportional to

Options:
A $n^2$
B $1 / n^2$
C $1 / n$
D $n$

The correct answer is:
$n^2$

Solution:
The radius of $n$th orbit is $r_n=0.53 \times \frac{n^2}{Z} A^{\circ}$. Here, $Z$ is atomic number which is constant for atom.Hence, the radius of stable orbits is proportional to $\mathrm{n}^2$

Topic: Physics, Atomic Physics, JEE Main
According to Bohr's model, the minimum energy (in eV) required to remove an electron from the ground state of a doubly ionized Li atom Z=3 is

Options:
A 40.8
B 13
C 122.4
D 1.51

The correct answer is:
122.4

Solution:
The doubly ionized Li atom is H-like atom
En=-13.6Z2n2 eV
The ground state energy of the doubly ionised Li atom
(Z = 3, n = 1) will be
E1=-13.6×91=-122.4 eV
Hence, the ionisation energy for an electron in the ground state of doubly ionised Li atom =122.4 eV

Topic: Physics, Atomic Physics, JEE Main
According to Bohr's principle, which is the correct relation between principle quantum number (n) and radius of orbit (r)?

Options:
A r∝n
B r∝n2
C r∝1n
D r∝1n2

The correct answer is:
r∝n2

Solution:
Bohr suggest a formulae to calculate the radius and energy of each orbit and gave the following formulae
rn=n2h24π2kme2Z
Where except n2 , all other unit are constant so rn∝n2 .

Topic: Chemistry, Structure of Atom, JEE Main
According to Bohr's theory of hydrogen atom, the total energy of the electron in the $n^{\text {th }}$ stationary orbit is

Options:
A directly proportional to $\mathrm{n}$
B inversely proportional to $\mathrm{n}$
C directly proportional to $\mathrm{n}^2$
D inversely proportional to $\mathrm{n}^2$

The correct answer is:
inversely proportional to $\mathrm{n}^2$

Solution:
According to Bohr's theory of hydrogen atom, the equation for total energy of the electron in the $\mathrm{n}^{\mathrm{th}}$ stationary orbit is,
$\begin{aligned}
& \mathrm{E}_{\mathrm{n}}=\frac{-\mathrm{mZ}^2 \mathrm{e}^4}{8 \varepsilon_0{ }^2 \mathrm{~h}^2 \mathrm{n}^2} \\
\therefore \quad & \mathrm{E}_{\mathrm{n}} \propto \frac{1}{\mathrm{n}^2}
\end{aligned}$

Topic: Physics, Atomic Physics, JEE Main
According to Bohr's theory, the angular momentum of an electron in $5^{\text {th }}$ orbit is

Options:
A
$25 \frac{\mathrm{h}}{\pi}$

B
$1.0 \frac{\mathrm{h}}{\pi}$

C
$10 \frac{\mathrm{h}}{\pi}$

D
$2.5 \frac{\mathrm{h}}{\pi}$

The correct answer is:
$2.5 \frac{\mathrm{h}}{\pi}$

Solution:
$\begin{aligned} \mathrm{mvr} & =\frac{\mathrm{nh}}{2 \pi} \\ & =\frac{5 \mathrm{~h}}{2 \pi}=2.5 \frac{\mathrm{h}}{\pi}\end{aligned}$

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2006
According to Bohr's theory, the angular momentum of an electron revolving in second orbit of hydrogen atom will be

Options:
A \( 2 \pi h \)
B \( \pi h \)
C \( \frac{h}{\pi} \)
D \( \frac{2 h}{\pi} \)

The correct answer is:
\( \frac{h}{\pi} \)

Solution:
Angular momentum
=nh2π=2h2π=hπ

Topic: Physics, Atomic Physics, JEE Main
According to Bohr's theory, the moment of momentum of an electron revolving in $4^{\text {th }}$ orbit of hydrogen atom is:

Options:
A $\frac{h}{\pi}$
B $\frac{h}{2 \pi}$
C $8 \frac{h}{\pi}$
D $2 \frac{h}{\pi}$

The correct answer is:
$2 \frac{h}{\pi}$

Solution:
Moment of momentum is $\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{P}}$
$\begin{aligned}
& \overrightarrow{\mathrm{L}}=\overrightarrow{\mathrm{r}} \times \mathrm{m} \overrightarrow{\mathrm{v}} \\
& \mathrm{L}=\mathrm{mvr}=\frac{\mathrm{nh}}{2 \pi}=\frac{4 \mathrm{~h}}{2 \pi}=\frac{2 \mathrm{~h}}{\pi}
\end{aligned}$

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2024 (04 Apr Shift 2)
According to Bohr's theory, the time averaged magnetic field at the centre (i.e., nucleus) of a hydrogen atom due to the motion of electrons in the nth orbit is proportional to: (n= principal quantum number)

Options:
A n-3
B n-2
C n-4
D n-5

The correct answer is:
n-5

Solution:
v∝ z n
r∝ n 2 z
B= μI 2R
I= e T = e ( 2πR v ) = e 2π v R
B= μ 0 2 e 2π v R 2
B= μ 0 e 4π z n ( n 2 z ) 2
B=μ0e4πz3n5
B∝ n −5 .

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2017 (08 Apr Online)
According to Boyle's law the product $P V$ remain constant. The dimension of $P V$ is same as

Options:
A Energy
B Force
C Impulse
D Momentum

The correct answer is:
Energy

Solution:
$[P V] \equiv\left[\frac{\text { Force }}{\text { Area }} \times\right.$ Volume $]=\left[M^1 L T^{-2} \frac{L^3}{L^2}\right]=\left[M^1 L T^{-2} L\right] \equiv[$ Force $\times$ Length $] \equiv$ Energy

Topic: Physics, Units and Dimensions, MHT CET, MHT CET 2022 (11 Aug Shift 1)
According to Boyle's law, the product PV remains constant. The unit of PV is same as that of

Options:
A energy
B force
C impulse
D momentum

The correct answer is:
energy

Solution:
The units of PV can be calculated as follows:
The unit of pressure is $\mathrm{Kgm}^{-1} \mathrm{~s}^{-7}$
The unit of volume is $\mathrm{m}^3$
$\therefore \quad$ The unit of $\mathrm{PV}$ is
Unit $=\mathrm{Kgm}^{-1} \mathrm{~s}^{-2} \mathrm{~m}^3=\mathrm{Kgm}^2 \mathrm{~s}^{-2}$
This is a unit of energy.

Topic: Physics, Kinetic Theory of Gases, MHT CET, MHT CET 2023 (11 May Shift 2)
According to Bronsted-Lowry concept, base is a substance which is

Options:
A A proton donor
B An electron pair acceptor
C A proton acceptor
D An electron pair donor

The correct answer is:
A proton acceptor

Solution:
According to Bronsted concept, base is a substance which accepts proton.
Hence, option 3 is correct.

Topic: Chemistry, Ionic Equilibrium, JEE Main
According to Cahn-Ingold-Prelog sequence rules, the correct order of priority for the given groups is

Options:
A $-\mathrm{COOH}>-\mathrm{CH}_2 \mathrm{OH}>-\mathrm{OH}>-\mathrm{CHO}$
B $-\mathrm{COOH}>-\mathrm{CHO}>-\mathrm{CH}_2 \mathrm{OH}>-\mathrm{OH}$
C $\mathrm{Ni} / 140^{\circ} \mathrm{C}$ and $\mathrm{Na} / \mathrm{NH}_3$ (liq. )
D $\mathrm{Pd} / \mathrm{BaSO}_4+\mathrm{H}_2$ and $\mathrm{Na} / \mathrm{NH}_3$ (liq.)

The correct answer is:
$\mathrm{Pd} / \mathrm{BaSO}_4+\mathrm{H}_2$ and $\mathrm{Na} / \mathrm{NH}_3$ (liq.)

Solution:
According to Cahn-Ingold-Prelog sequence rules, the priority of groups is decided by the atomic number of their atoms. When the atom (which is directly attached to the asymmetric carbon atom) of a group has higher atomic number, then the group gets higher priority. Groups with atoms of comparable atomic number having double or triple bond, have high priority than those have single bond.
Hence, the order of priority of groups is
$$
-\mathrm{OH}>-\mathrm{COOH}>-\mathrm{CHO}>-\mathrm{CH}_2 \mathrm{OH}
$$

Topic: Chemistry, General Organic Chemistry, JEE Main
According to Central Pollution Control Board (CPCB), which particulate size in diameter (in micrometers) of the air pollutants is responsible for greatest harm to human health?

Options:
A $5.2-2.5$
B 2.5 or less
C 1.5 or less
D 1.0 or less

The correct answer is:
2.5 or less

Solution:
This is the report of Central Pollution Control Board.

Topic: Biology, Environmental Issues, NEET, NEET 2008 (Mains)
According to Curie's law, the magnetic susceptibility of a substance at an absolute temperature $T$ is proportional to:

Options:
A $\frac{1}{T}$
B $T$
C $\frac{1}{T^2}$
D $T^2$

The correct answer is:
$\frac{1}{T}$

Solution:
From Curie Law $X_m \propto \frac{1}{T}$

Topic: Physics, Magnetic Properties of Matter, JEE Main
According to Darwin, organic evolution proceeds in which of the following orders?

Options:
A Overproduction, variations, constancy of population size, natural selection
B Variations, constancy of population size, overproduction, natural selection
C Overproduction, constancy of population size, variations, natural selection
D Variations, natural selection, overproduction, constancy of population size

The correct answer is:
Overproduction, constancy of population size, variations, natural selection

Solution:
The main features of theory of Natural Selection are as follows
1 Over production (High rate of Reproduction): All organisms have capability to produce enormous number of offspring, organisms multiply in geometric ratio. 2 Struggle for existence: Every individual competes with other of the same and other species for basic necessities like space, shelter and food. It is called struggle for existence and it continues for the whole life from zygote stage to its natural death. 3 Variations and heredity: According to Darwin the variations are continuous and those which are helpful in the adaptation of an organism towards its surroundings would be passed on to the next generation, while the others will disappear. 4 Survival of the fittest or natural selection: According to Darwin most suitable and fit individuals are successful in struggle for existence. The individuals with most favourable adaptations are able to lead most successful life and are able to win over their mating partners. In the struggle for existence only those members survive which possess useful variations means nature selects fit individuals. This was called Natural Selection.

Topic: Biology, Evolution, NEET
According to de-Broglie hypothesis, the wavelength associated with moving electron of mass m is Î»e. Using mass energy relation and Planck's quantum theory, the wavelength associated with photon is Î»p. If the energy (E) of electron and photon is same then relation between Î»e and Î»p is

Options:
A λp α λe
B λp α λe2
C λp αλe
D λp α1λe

The correct answer is:
λp α λe

Solution:
For photon: E=hcλp
∴ λp=hcE ..........(i)
For electron: E=mc2=pc
∴ p=EC
λe=hp=hcE ........(ii)
By Equations. (i) and (ii),
λp ∝ λe

Topic: Physics, Dual Nature of Matter, JEE Main
According to Einstein's photoelectric equation, the graph between the kinetic energy of photoelectrons elected and the frequency of incident radiations is-

Options:
A
B
C
D

The correct answer is:

Solution:
KE = hv- hv0
for v >v0, KE∝ v
(-ve) intercept on y axis = h ν 0

Topic: Chemistry, Structure of Atom, JEE Main
According to EInstein's photoelectric equation, the plot of the kinetic energy of the emitted photo electrons from a metal $\mathrm{V}_{\mathrm{s}}$ the frequency, of the incident radiation gives straight line whose slope

Options:
A
depends on the nature of the metal used

B
depends on the intensity of the radiation

C
depends both on the intensity of the radiation and the metal used

D
is the same for all metals and independent of the intensity of the radiation.

The correct answer is:
is the same for all metals and independent of the intensity of the radiation.

Solution:

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2004
According to Fajan's rule, covalent bond is favoured by

Options:
A Large cation and small anion
B Large cation and large anion
C Small cation and large anion
D Small cation and small anion

The correct answer is:
Small cation and large anion

Solution:
Small cation and large anion
According to Fajan's rule, covalent bond is favoured by small cation and large anion.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
According to Faraday's law of electromagnetic induction

Options:
A The direction of induced current is such that it opposes the cause producing it
B The magnitude of induced e.m.f. produced in a coil is directly proportional to the rate of change of magnetic flux
C The direction of induced e.m.f. is such that it opposes the cause producing it
D None of the above

The correct answer is:
The magnitude of induced e.m.f. produced in a coil is directly proportional to the rate of change of magnetic flux

Solution:
The magnitude of induced e.m.f. produced in a coil is directly proportional to the rate of change of magnetic flux

Topic: Physics, Electromagnetic Induction, JEE Main