datavorous/entrance-exam-dataset huggingface dataset:
Excitation energy of a hydrogen like ion in its first excitation state is $40.8 \mathrm{eV}$. Energy needed to remove the electron from the ion in ground state is

Options:
A $54.4 \mathrm{eV}$
B $13.6 \mathrm{eV}$
C $40.8 \mathrm{eV}$
D $27.2 \mathrm{eV}$

The correct answer is:
$54.4 \mathrm{eV}$

Solution:
Excitation energy of hydrogen like ion in its first excitation state, $\Delta E=40.8 \mathrm{eV}$
For the hydrogen like ion, energy in the ground state $(n=1)$,
$$
E=\frac{-13.6 Z^{2}}{n^{2}} \mathrm{eV}=-13.6 Z^{2} \mathrm{eV}
$$
Energy in the first excited state, $E_{2}=\frac{-13.6 Z^{2}}{n^{2}}$
$$
=\frac{-13.6 Z^{2}}{4}
$$
Excitation energy, $\Delta E=E_{2}-E_{1}$
$$
\begin{aligned}
&=\left(\frac{-13.6 Z^{2}}{4}+13.6 Z^{2}\right) \mathrm{eV} \\
&=(-3.4+13.6) Z^{2} \mathrm{eV} \\
\Rightarrow \quad 40.8 \mathrm{eV} &=10.2 Z^{2} \mathrm{eV} \\
\Rightarrow \quad \quad Z^{2} &=\frac{40.8}{10.2} \\
\Rightarrow \quad \quad Z^{2} &=4 \\
\Rightarrow \quad \quad Z &=2
\end{aligned}
$$
Thus, energy of the electron in the ground state is given as
$$
\begin{array}{ll}
\Rightarrow & E_{1}=-13.6(2)^{2} \mathrm{eV} \\
\Rightarrow & E_{1}=-54.4 \mathrm{eV}
\end{array}
$$
Hence, the energy needed to remove the electron from the ion in ground state (ionisation energy)
$$
=-(-54.4) \mathrm{eV}=54.4 \mathrm{eV}
$$

Topic: Physics, Atomic Physics, JEE Main
Excited hydrogen atom emits light in the ultraviolet region at $2.47 \times 10^{15} \mathrm{~Hz}$. With this frequency, the energy of a single photon is: $\left(\mathrm{h}=6.63 \times 10^{-34} \mathrm{Js}\right)$

Options:
A
$8.041 \times 10^{-40} \mathrm{~J}$

B
$2.680 \times 10^{-19} \mathrm{~J}$

C
$1.640 \times 10^{-18} \mathrm{~J}$

D
$6.111 \times 10^{-17} \mathrm{~J}$

The correct answer is:
$1.640 \times 10^{-18} \mathrm{~J}$

Solution:
$\mathrm{E}=\mathrm{hv}$
$$
\begin{aligned}
&=6.63 \times 10^{-34} \times 2.47 \times 10^{15} \\
&=1.640 \times 10^{-18} \mathrm{~J}
\end{aligned}
$$

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2014 (12 Apr Online)
Excretion in cockroach takes place by

Options:
A nephridium
B parotid gland
C coxal glands
D malpighian tubules.

The correct answer is:
malpighian tubules.

Solution:
In cockroach, excretion is mainly performed by one or two pairs of malpighian tubules. One of this pair opens into the middle region of intestine and other pair of coxal glands are located in prosoma and opens at bases of fifth walking legs.

Topic: Biology, Structural Organisation in Animals, AIIMS, AIIMS 2002
Experimental evidence for the existence of the atomic nucleus comes from

Options:
A Millikan's oil drop experiment
B Atomic emission spectroscopy
C The magnetic bending of cathode rays
D Alpha scattering by a thin metal foil

The correct answer is:
Alpha scattering by a thin metal foil

Solution:
Rutherford $\alpha$-Scattering experiment.

Topic: Chemistry, Structure of Atom, JEE Main
Experimentally it is found that 12.8 eV energy is required to separate a hydrogen atom into a proton and an electron. So the orbital radius of the electron in a hydrogen atom is 9x×10-10 m.

The value of the x is: _____.

(1 eV=1.6×10-19 J, 14πε0=9×109 Nm2C2 and electronic charge =1.6×10-19 C)

Options:

The correct answer is:
16

Solution:
The given data is

q=1.6×10-19 C14πε0=9×109 N m2 C-2

The formula for the value of energy in a hydrogen atom is given by

E=q28πε0r⇒E=q22(4πε0)r

⇒r=9×109×1.6×10-1912.8×2

⇒r=916×10-10 m

It is given that r=9x×10-10 m⇒916×10-10=9x×10-10 ⇒x=16

Topic: Physics, Electrostatics, JEE Main, JEE Main 2023 (06 Apr Shift 2)
Experimentally it was found that a metal oxide has formula M0.98O. Metal M, is present as M2+ and M3+ in its oxide. Fraction of the metal which exists as M3+ would be:

Options:
A 6.05%
B 5.08%
C 7.01%
D 4.08%

The correct answer is:
4.08%

Solution:
Consider one mole of the oxide, then
Moles of M = 0.98; Moles of O2- = 1
Let 'x' moles of M are in +3 oxidation state, then Moles of M2+ = 0.98-x
On doing charge balancing we get
(0.98 - x)2 + 3x = 2
Or 1.96 - 2x + 3x - 2 = 0
Or x = 0.04
∴ %of M3+=0.040.98×100=4.08%

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2013 (07 Apr)
Experimentally it was found that a metal oxide has formula M0.98O. Metal M, present as M2+ and M3+ in its oxide. Fraction of the metal which exists as M3+ would be nearly
[Report your answer by rounding it upto nearset whole number]

Options:

The correct answer is:
4

Solution:
For a one mole of the oxide
Moles of M=0.98, Moles of O2-=1
Let moles of M3+=x
Moles of M2+=0.98-x
on balancing charge
0.98-x×2+3x-2=0
x=0.04
% of M3+=0.040.98×100=4.08%≈4%

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
Experimentally reducing a functional group cannot be done by which one of the following reagents ?

Options:
A Zn/H2O
B Pt-C/H2
C Pd-C/H2
D Na/H2

The correct answer is:
Na/H2

Solution:
Sodium reacts with Hydrogen to form sodium hydride.

2Na + H2 →2NaH

So Na/H2 cannot be used for reduction of functional groups.

All other reagents are used as reducing agent.

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2021 (01 Sep Shift 2)
Explain how does the- $\mathrm{OH}$ group attached to a carbon of benzene ring activate it towards electrophilic substitution?

Options:

None

Solution:
Phenol may be regarded as a resonance hybrid of structures I-V, shown below.
As a result of $+\mathrm{R}$ effect of the $-\mathrm{OH}$ group, the electron density in the benzene ring increases thereby facilitating the attack of an electrophile. In other words, presence of $-\mathrm{OH}$ group, activates the benzene ring towards electrophilic substitution reactions. Further, since the electron density is relatively higher at the two $o$-and one $p$-position, therefore electrophilic substitution occurs mainly at $o$-and $p$-positions.

Topic: Chemistry, Alcohols Phenols and Ethers
Explain how during light reaction of photosynthesis, ATP Synthesis is a chemiosmotic phenomenon.

Options:

None

Solution:
In light reaction plants solar radiation is trapped by photosynthetic pigments, which converts light energy into chemical energy. Photophosphorylation is the main event of light reaction i.e., formation of ATP from ADP + Pi by using energy of excited electron movement through electron transport chain, that is present in thylakoid membrane.
The movement of ions across a selectively permeable membrane, down the electrochemical/proton gradient is known as chemiosmosis.
Chemiosmosis hypothesis of ATP formation was first proposed by Mitchell (1961), according to ATP generated by enzyme ATP synthase via a membrane, proton pump and proton gradient. ATP synthase allows ions $\mathrm{O}_2$ protons to pass through membrane and proton pump.
Which creates a high concentration of protons $\left(\mathrm{H}^{+}\right)$in the lumen and hence diffuses across the membrane to activate ATPase, releasing ATP molecules. One molecule of ATP is released for every two $\left(\mathrm{H}^{+}\right)$ions passing through ATPase.

Topic: Biology, Photosynthesis in Higher Plants
Explain how Newton's corpuscular theory predicts that the speed of light in a medium, say water, is greater than the speed of light in vacuum. Is the prediction confirmed by experimental determination of the speed of light in water? If not, which alternative picture of light is consistent with experiment?

Options:

None

Solution:
In Newton's corpuscular (particle) picture of refraction, particles of light incident from a rarer to a denser medium experience a force of attraction normal to the surface.
This results in an increase in the normal component of velocity but the component along the surface remains unchanged.
Considering a ray of light going from a rarer medium (air) to a denser medium (water).
Let $c=$ speed of light in vacuum (or air),
$v=$ speed of light in water,
$i=$ angle of incidence, and $r=$ angle of refraction
Then according to Newton's corpuscular theory,
Component of velocity $c$ along surface of separation =
Component of velocity
$v$ along the surface of separation
$\therefore c \sin i=v \sin r$ or $\frac{v}{c}=\frac{\sin i}{\sin r}={ }^a \mu_\omega$
$\operatorname{As}^a \mu_\omega>1$, therefore $v>c$.
So, according to Newtons corpuscular theory the speed of light in medium is larger than speed of light in air $v>c$ but in fact the experimental observation shows that speed of light is smaller in denser medium as compared to rare medium.

Topic: Physics, Wave Optics
Explain how vacancies are introduced in an ionic solid when a cation of higher valence is added as an impurity in it.

Options:

None

Solution:
Let us take an example where \(\mathrm{NaCl}\) is doped with \(\mathrm{SrCl}_2\) impurity. When \(\mathrm{SrCl}_2\) is added to \(\mathrm{NaCl}\) solid as an impurity, two \(\mathrm{Na}^{+}\)ions will be replaced and one of their sites will be occupied by \(\mathrm{Sr}^{2+}\) while the other will remain vacant. Thus, we can say that when a cation of higher valence is added as an impurity to an ionic solid, two or more cations of lower valency are replaced by a cation of higher valency to maintain electrical neutrality. Hence, some cationic vacancies are created.

Topic: Chemistry, Solid State
Explain the bonding in coordination compounds in terms of Werner's postulates.

Options:

None

Solution:
The main postulates of Werner's theory of coordination compounds are as follows:
(a) Metals possess two types of valencies called
(i) primary valency which are ionisable;
(ii) secondary valency which are nonionisable
(b) Primary valency is satisfied by the negative ions and it is that which a metal exhibits in the formation of its simple salts.
(c) Secondary valencies are satisfied by neutral ligand or negative ligand and are those which metal exercises in the formation of its complex ions. Every cation has a fixed number of secondary valencies which are directed in space about central metal ion in certain fixed directions, e.g., In \(\mathrm{CoCl}_3 \cdot 6 \mathrm{NH}_3\), valencies between \(\mathrm{Co}\) and \(\mathrm{Cl}\) are primary valencies and valencies between \(\mathrm{Co}\) and \(\mathrm{NH}_3\) are secondary. In \(\mathrm{CoCl}_3 \cdot 6 \mathrm{NH}_3\), six ammonia molecules linked to \(\mathrm{Co}\) by secondary valencies are directed to six corners of a regular octahedron and thus account for structure of \(\mathrm{CoCl}_3 \cdot 6 \mathrm{NH}_3\) as follows:
In modern theory, it is now referred as coordination number of central metal atom or ion.

Topic: Chemistry, Coordination Compounds
Explain the fact that in aryl alkyl ethers
(i) the alkoxy group activates the benzene ring towards electrophilic substitution and
(ii) it directs the incoming substituents to ortho and para positions in benzene ring.

Options:

None

Solution:
In aryl alkyl ethers, the $+$ R-effect of the alkoxy (OR) group increases the electron density in the benzene ring, thereby activating the benzene ring towards electrophilic substitution reaction.
Since the electron density increases more at the two ortho and one para position as compared to meta position therefore, electrophilic substitution reactions mainly occur at $o$-and $p$-positions.

Topic: Chemistry, Alcohols Phenols and Ethers
Explain the fact that in aryl alkyl ethers the alkoxy group activates the benzene ring towards electrophilic substitution and it directs the incoming substituents to ortho and para positions in benzene ring

Options:
A In aryl alkyl ethers, +R effect of the alkoxy group (-OR) increases the electron density in the benzene ring thereby activating the benzene ring towards electrophilic substitution reactions.Electron density is more at o- and p- positions so o- and p-products are mainly formed during electrophilic substitution reactions.
B The alkoxy group is an electron-withdrawing group that decreases electron density on the benzene ring, making it less reactive towards electrophilic substitution. The alkoxy group directs incoming substituents to meta positions through inductive effects, leading to substitution at the meta position.
C The alkoxy group is inert and does not affect the reactivity of the benzene ring towards electrophilic substitution. It also does not direct the incoming substituents to any specific positions on the benzene ring.
D The alkoxy group is an electron-donating group that decreases electron density on the benzene ring, making it less reactive towards electrophilic substitution. The alkoxy group directs incoming substituents to ortho and para positions through inductive effects, leading to substitution at these positions.

The correct answer is:
In aryl alkyl ethers, +R effect of the alkoxy group (-OR) increases the electron density in the benzene ring thereby activating the benzene ring towards electrophilic substitution reactions.Electron density is more at o- and p- positions so o- and p-products are mainly formed during electrophilic substitution reactions.

Solution:
Correct Option is : (A)
In aryl alkyl ethers, +R effect of the alkoxy group (-OR) increases the electron density in the benzene ring thereby activating the benzene ring towards electrophilic substitution reactions.Electron density is more at o- and p- positions so o- and p-products are mainly formed during electrophilic substitution reactions.

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
Explain the Law of Dominance using a monohybrid cross.

Options:

None

Solution:
When two different factors (genes) or a pair of contrasting forms of a character are present in an organism, only one expresses itself in the $\mathrm{F}_1$ generation and is termed as dominant while the other remains unexpressed and called recessive factors (gene).
A tall (TT) true breeding plant is crossed with a dwarf(tt) plant. The character of height is represented by ' $\mathrm{T}$ ' for tall ' $t$ ' for dwarf are the alternate form as character of height. The $\mathrm{F}_1$ hybrid ' $\mathrm{Tt}$ ' is Tall, showing that tall is dominant over dwarf while dwarf remains unexpressed in $\mathrm{F}_1$ offspring due to phenomenon of dominance by tall factor or gene.
In this $\mathrm{Tt}$ heterozygous has tall phenotype showing $\mathrm{T}$ is dominant over t allele.

Topic: Biology, Principles of Inheritance and Variation
Explain the mechanism of formation of concentrated urine in mammals.

Options:

None

Solution:
Mammals have the ability to produce concentrated urine. The loop of Henle and vasa recta play important role in it, which is discussed as follows:
(i) The proximity between the Henle's loop and vasa recta, as well as the counter current that is formed due to the flow of filtrate in two limb's of Henle's loop in opposite direction and help in maintaining an increasing osmolality towards the inner medullary interstitium, i.e., from $300 \mathrm{mOsmoL}^{-1}$ in the cortex to about $1200 \mathrm{mOsmol}^{-1}$ in the inner medulla.
(ii) This gradient is caused mainly due to $\mathrm{NaCl}$ and urea ascending limb of Henle's loop transports $\mathrm{NaCl}$, that is exchanged with the descending limb of vasa recta.
(iii) Through the ascending portion of vasa recta. $\mathrm{NaCl}$ is returned to the interstitium.
(iv) Similarly, a small amount of urea enters the thin segment of the ascending limb of Henle's loop, which is transported back to the intersitium by the collecting tubule.
(v) This special arrangement of Henle's loop, and vasa recta, is called the counter current mechanism.
(vi) The rate of dissipation is reduced by the counter current exchange. This in turn, reduces the rate at which the current must pump $\mathrm{Na}^{+}$to maintain any given gradient.
(vii) Presence of such interstitial garden helps in an easy passage of water from the collecting tubule thereby concentrating the filtrate (urine).
(viii) Human kidneys produces urine nearly four times concentrated than the initial filtrate formed.

Topic: Biology, Excretory Products and their Elimination
Explain the process of the transport and release of neurotransmitter with the help of a labelled diagram showing a complete neuron, axon terminal and synapse.

Options:

None

Solution:
The three main parts of a neuron include the
(i) Cell body
(ii) Axon
(iii) Dendrites
Stimulus or nerve impulse of any kind passes from one neuron to another via axon. This nerve impulse is wave of bioelectric/electrochemical disturbance that passes along the neuron during conduction of an excitation.
Within a synapse transport and release of a neuro transmiter occurs. At a chemical synapse, the membranes of the pre- and post-synaptic neurons are separated by a fluid-filled space called synaptic cleft. Chemicals called neurotransmitters are involved in the transmission of impulses at these synapses. The axon terminals contain vesicles filled with these neurotransmitters.
Upon arrival of an impulse (action potential) at the axon terminal, it stimulates the movement of the synaptic vesciles towards the membrane, where they fuse with the plasma membrane and release their neurotransmitters in the synaptic cleft.
The released neurotransmitters bind to their specific receptors, present on the post-synaptic membrane. This binding opens ion channels allowing the entry of ions, that can generate a new action potential in the postsynaptic neuron.

Topic: Biology, Neural Control and Coordination
Explain the significance of sodium, potassium, magnesium and calcium in biological fluids.

Options:

None

Solution:
Sodium (Na): They are found in our blood plasma and the interstitial fluids around the cells. They help in
(a) Transmission of nerve signals.
(b) They regulate the flow of water across the membranes of the neighbouring cells.
(c) Transport sugars and amino acids from and to cells. Potassium (K): They are found mostly in the cell fluids in greater quantities. They help in
(a) Activating enzymes
(b) Oxidising glucose to form ATP
(c) Transmitting nerve signals.
Magneslum (Mg) and calcium (Ca): They are also called as macro-minerals named so because of their abundance in out body. Mg helps in
(a) Relaxing nerves and muscles
(b) Building and strengthening bones
(c) Maintaining blood circulation in our body Ca helps in
(a) coagulation of blood
(b) Maintaining homeostasis.

Topic: Chemistry, s Block Elements
Explain the structure of middle and internal ear with the help of diagram.

Options:

None

Solution:
Ears are a part of statoacoustic organ meant for balancing and hearing the external ear in most mammals is a heap of tissue also called pinna. It is a part of auditory system. The human ear consists of three main parts external ear, middle, ear and internal ear.
Structure of Middle Ear
The middle ear consists of three bones or ossicles-the malleus (hammer), incus (anvil and stapes (stir-up). These bones are attached to one another in a chain-like manner. The malleus is attached to the tympanic membrane and the stapes is attached to the oval window (a membrane beneath the stapes) of cochlea. These three ossicles increase the efficiency of transmission of sound waves to the inner ear. The middle ear also opens into the eustachian tube, which connects with the pharynx and maintains the pressure between the middle ear and the outside atmosphere.
Structure of Internal Ear
The inner ear consists of a labyrinth of chambers filled with fluid within temporal bone of the skull. The labyrinth consists of two parts the bony and membranous labyrinth. The bony labyrinth is a series of channels. Membranous labyrinth lies inside these channels which is surrounded by a fluid called perilymph. The membranous labyrinth is filled with a fluid called endolymph. The coiled portion of the labyrinth is called cochlea.
The cochlea has two large canal separated by a small cochlear duct (scala media). An upper vestibular canal (scala vestibuli) and a lower tympanic canal (scala tympani). The vestibular and tympanic canals contain perilymph and the cochlear duct is filled with endolymph.
The wall of membranous labyrinth comes in contact with the fenestra ovalis at the base of scale vestibuli while the fenestra rotunda.

Topic: Biology, Neural Control and Coordination
Explain this common observation clearly: If you look out of the window of a fast moving train, the nearby trees, houses etc., seem to move rapidly in a direction opposite to the train's motion, but the distant objects (hill tops, the Moon, the stars etc.) seem to be stationary. (In fact, since you are aware that you are moving, these distant objects seem to move with you).

Options:

None

Solution:
The line joining a given object to our eye is known as the line of sight. When a train moves rapidly, the line of sight of a passenger sitting in the train for nearby trees' changes its direction rapidly. As a result, the nearby trees and other objects appear to run in a direction opposite to the train's motion. However, the line of sight of distant and large size objects e.g., hill tops, the Moon, the stars etc., almost remains unchanged (or changes by an extremely small angle). As a result, the distant object seems to be stationary.

Topic: Physics, Units and Dimensions
Explain tropospheric pollution in 100 words.

Options:

None

Solution:
Tropospheric pollution occurs due to the presence of undesirable gases and the solid particles in the air. The major gaseous and the particulate pollutants present in the troposphere are as follows:
(i) Gaseous air pollutants. These include mainly oxides of sulphur \(\left(\mathrm{SO}_2, \mathrm{SO}_3\right)\), oxides of nitrogen \(\left(\mathrm{NO}, \mathrm{NO}_2\right)\) and oxides of carbon \(\left(\mathrm{CO}, \mathrm{CO}_2\right)\) in addition to hydrogen sulphide \(\left(\mathrm{H}_2 \mathrm{~S}\right)\), hydrocarbons, ozone and other oxidants.
(ii) Particulate pollutants. These include dust, mist, fumes, smoke, smog, etc.

Topic: Chemistry, Environmental Chemistry
Explain what would happen if in the capacitor given in question 2.8, a $3 \mathrm{~mm}$ thick mica sheet (of dielectric constant $=6$ ) were inserted between the plates,
(a) while the voltage supply remained connected and,
(b) after the supply was disconnected.

Options:

None

Solution:
Given, $\mathrm{C}_0=17.7 \mathrm{pF}, \mathrm{C}_{\mathrm{K}}=\mathrm{KC}_0=6 \mathrm{C}_0$;
(a) While the supply remained connected; voltage will remain constant
$$
\begin{aligned}
\mathrm{q}_{\mathrm{K}} &=\mathrm{C}_{\mathrm{K}} \mathrm{V}=6 \mathrm{C}_0 \mathrm{~V}=6 \times 17.7 \times 10^{-10} \mathrm{C} \\
&=1.06 \times 10^{-8} \mathrm{C} .
\end{aligned}
$$
Charge on capacitor becomes six times.
(b) When supply is disconnected; Charge will remain constant
$$
\begin{aligned}
&\mathrm{q}=\mathrm{C}_{\mathrm{K}} \mathrm{V}_{\mathrm{K}} \\
&\mathrm{V}_{\mathrm{K}}=\frac{\mathrm{q}}{\mathrm{C}_{\mathrm{k}}}=\frac{\mathrm{q}}{\mathrm{KC}_0}=16.7 \mathrm{~V}
\end{aligned}
$$
Potential difference between plates becomes one-sixth.

Topic: Physics, Electrostatic Potential and Capacitance