datavorous/entrance-exam-dataset huggingface dataset:
Consider the following four statements (1-4), select the correct option stating which ones are true (T) and which ones are false (F)

1 An organism at each trophic level depends on those at a higher trophic level for their energy demands.
2 Each trophic level has a certain mass of living material at a particular time called the standing crop.
3 Each trophic level has a certain mass of non-living material at a particular time called standing state.
4 The number of trophic levels in a grazing food chain is not restricted.

Options:
A 1-T, 2-T, 3-F, 4-F
B 1-F, 2-T, 3-F, 4-T
C 1-T, 2-F, 3-T, 4-F
D 1-F, 2-T, 3-T, 4-F

The correct answer is:
1-F, 2-T, 3-T, 4-F

Solution:
Standing crop is the total amount of living material in a specified population at a particular time, generally expressed as biomass. The standing crop may show variation at different times of the year. The availability of the inorganic content available to an organism at given time in the ecosystem is referred to as standing state. Nutrients, such as, carbon, nitrogen, phosphorus, calcium, etc. form part of the inorganic content.
The sequential transfer of nutrients and energy in the form of food from one organism to next. The position of organism in a food chain is termed as trophic level. In a typical food chain, the energy that is transferred from one trophic level to the next is always unidirectional and never forms a cyclical pattern. The main source of energy is the sun. As energy transfer in each successive trophic level reduces is only 10% of the previous, number of trophic levels in a food chain is restricted. After 5-6 trophic levels, very low amount of energy will be transferred making it nearly impossible to survive. Food chain starting with a producer is referred to as grazing food chain.

Topic: Biology, Ecosystem, NEET
Consider the following four statements A, B, C and D select the right option for two correct statements. Statements :
(A) In vexillary aestivation, the large posteior petal is called-standard, two lateral ones are wings and two small anteior petals are termed keel.
(B) The floral formula for Litiaceae is
(C) In pea flower the stamens are monadelphous
(D) The floral formula for Solanaceae is
The Correct statements are

Options:
A A and C
B A and $B$
C B and C
D C and D

The correct answer is:
A and $B$

Solution:
In vexillary aestivation (eg, papilionaceous plants), the large posterior petal is called standard, two lateral ones are wings and two small anterior petals are termed keel.
General floral formula for family-Liliaceae is $\mathrm{Br} \oplus \overbrace{}^2 \mathrm{P}_{3+3} \mathrm{~A}_{3+3} \mathrm{G}_{(3)}$
In pea flower, the stamens are diadelphous, ie, anthers remain free, while filaments unite to form two separate bundles.
The general floral formula for family-Solanaceae is

Topic: Biology, Morphology of Flowering Plants, NEET, NEET 2010 (Mains)
Consider the following four statements $(\mathrm{a}-\mathrm{d})$ about certain desert animals such as kangaroo rat.
(a) They have dark colour and high rate of reproduction and excrete solid urine
(b) They do not drink water, breathe at a slow rate to conserve water and have their body covered with thick hairs
(c) They feed on dry seeds and do not require drinking water
(d) They excrete very concentrated urine and do not use water to regulate body temperature
Which two of the above statements for such animals are true?

Options:
A a and b
B $\mathrm{c}$ and $\mathrm{d}$
C $b$ and $c$
D c and a

The correct answer is:
$\mathrm{c}$ and $\mathrm{d}$

Solution:
Kangaroo rat is the only mammal which does not ingest any water and reamains only on metabolic water which it can utilize due to nasal counter current mechanism. It feeds on dry castor seeds. Sweat glands are absent and secrete concentrated urine with nitrogenous waste as urea.

Topic: Biology, Organisms and Populations, NEET, NEET 2008 (Mains)
Consider the following four statements (a-d) and select the option which includes all the correct ones only.
(a) Single cell Spirulina can produce large quantities of food rich in protein, minerals, vitamins etc.
(b) Body weight-wise the microorganism Methylophilus methylotrophus may be able to produce several times more proteins than the cows per day
(c) Common button mushrooms are a very rich source of vitamin C
(d) A rice variety has been developed which is very rich in calcium.

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

The correct answer is:
Statements (a), (b)

Solution:
Spirulina is SCP rich in protein, vitamins & minerals & rice variety rich in iron content. 250 grambiomass of Methylophilus methylotrophus produce 25 tonn protein/day while cow of 250 Kg . produces only 200 gm. protein/day.

Topic: Biology, Biological Classification, NEET, NEET 2012 (Mains)
Consider the following four statements (I-IV) related to the common frog Rana tigrina, and select the correct option stating which ones are true (T) and which ones are false $(\mathrm{F})$.
I. On dry land it would die due to lack of $\mathrm{O}_2$ its mouth is forcibly kept closed for a few days.
II. It has four-chambered heart.
III. On day land it turns uricotelic from ureotelic.
IV. Its life-history is carried out in pond water.

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

The correct answer is:
(a)

Solution:
Frogs respire on land and in the water by two different methods. In water, skin acts as aquatic respiratory organ (cutaneous respiration). Dissolved $\mathrm{O}_2$ in the water is exchanged through the skin by diffusion. On land, the buccal cavity, skin and lungs act as the respiratory organs. The respiration by lungs is called is called pulmonary respiration.
In frog, heart has three-chambers, two atria and one ventricle, and is covered by a membrane called pericardium.
The frog excretes urea and thus is a ureotelic animal. It never becomes uricotelic.
In frog, fertilization is external and takes place in water.

Topic: Biology, Structural Organisation in Animals, NEET, NEET 2011 (Mains)
Consider the following gaseous equilibria with equilibrium constants K1 and K2 respectively.

SO2g+12O2(g)âSO3(g)

2SO3gâ2SO2g+O2(g)

The equilibrium constants are related as

Options:
A 2K1=K22
B K12=1K2
C K22=1K1
D K2=2K12

The correct answer is:
K12=1K2

Solution:
For the reaction,
SO2g+12O2(g)âSO3(g)
Equilibrium constant, K1=[SO3]SO2O21/2 ...(i)
For the reaction,
2SO3gâ2SO2g+O2(g)
Equilibrium constant, K2=SO22[O2]SO32 ...(ii)
On squaring both sides in Eq.(i), we get
K12=SO32SO22[O2] ...(iii)
Eq. (ii) Ã Eq (iii), we get
K12ÃK2=1
or K2=1K12
or K12=1K2

Topic: Chemistry, Chemical Equilibrium, JEE Main
Consider the following gaseous reaction where initial pressure of A(g) is 200 torr
A(g)→3B(g) + C(g)
After time t, the pressure of system increased to 650 torr. Calculate the value of t (in seconds)
(Given: rate constant of reaction, k = 4.606 ×  10 −3   sec −1 , log 2 = 0.3 )

Options:
A 600
B 300
C 50
D 5

The correct answer is:
300

Solution:
200âpâ+â3pâ+âpâ=â650
pâ=150âtorr
t=1kÃ2.303Ãlogâ¡200200-p
= 1âÃâ2.303âÃlogâ4 4.606âÃâ 10 â3 â=â300 sec

Topic: Chemistry, Chemical Kinetics, JEE Main
Consider the following graph:

X, Y and Z can be respectively

Options:
A Ne, Ar and Xe
B Ar, Xe and He
C Kr, Ar and Ne
D Ar, He and Ne

The correct answer is:
Kr, Ar and Ne

Solution:
Most probable speeds are inversely proportional to the atomic mass.

Topic: Chemistry, States of Matter, JEE Main
Consider the following ionization enthalpies of two elements 'A' and 'B'.

Element	Ionization enthalphy kJ/mol

1st	2nd	3rd

A	899	1757	14847

B	737	1450	7731

Which of the following statements is correct?

Options:
A Both 'A' and 'B' belong to group-1 where 'A' comes below 'B'.
B Both 'A' and 'B' belong to group-2 where 'A' comes below 'B'.
C Both 'A' and 'B' belong to group-1 where 'B' comes below 'A'.
D Both 'A' and 'B' belong to group-2 where 'B' comes below 'A'.

The correct answer is:
Both 'A' and 'B' belong to group-2 where 'B' comes below 'A'.

Solution:
Both 'A' and 'B' belong to group-2, because the 3rd ionization energy is very high, it means 3rd electron gets removed from internal shell and where 'B' comes below 'A', because on moving from top to bottom in group the ionization energy decreases.

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2017 (08 Apr Online)
Consider the following ions

1. Ni2+

2. Co2+

3. Cr2+

4. Fe3+

Atomic number: Cr = 24, Fe = 26, Co = 27, Ni = 28.

The correct sequence of increasing order of the number of unpaired electrons in these ions is

Options:
A 1, 2, 3, 4
B 4, 2, 3, 1
C 1, 3, 2, 4
D 3, 4, 2, 1

The correct answer is:
1, 2, 3, 4

Solution:

Topic: Chemistry, Structure of Atom, JEE Main
Consider the following ligands NH2-,acac,SH-,Gly,O2-,Phen,DMG,NO2-,CO32-,I-,CH3COO-,en,SO42-. Then calculate value of
"P + Q – R – S"
Here, P: Total number of ligands which act as bridging as well as monodentate only.
Q: Total number of flexidentate ligands
R: Total number of bidentate ligands only
S: Total number of unsymmetrical bidentate ligands

Options:

The correct answer is:
1

Solution:
P=NH2-,O2-,SH-,I-=4
Q=SO42-,NO2-,CO32-,CH3COO-=4
R=en,acac,Phen,DMG,Gly=5
S=DMG,Gly=2
4 + 4 – 5 – 2 = 1.

Topic: Chemistry, Coordination Compounds, JEE Main
Consider the following match of hormones and select the incorrect one:

Options:
A Cortisol: Anti-inflammatory, immuno-suppressive and stimulate RBC production.
B Thyroxine: Regulate BMR, maintain water-electrolyte balance and support RBC formation.
C Adrenaline: Prepare body for emergency or '3 F' conditions.
D Calcitonin: Decreases bone calcium level in human body.

The correct answer is:
Calcitonin: Decreases bone calcium level in human body.

Solution:
Glucocorticoids, mainly cortisol hormone, produce anti-inflammatory reactions and suppress the immune response. They also stimulate RBC production. Adrenaline or epinephrine is secreted from the adrenal medulla in response to stress or emergency or fight, flight and fright conditions. Thyroxine regulates BMR, maintains water-electrolyte balance and supports RBC formation. Thyrocalcitonin (TCT) is a long peptide hormone secreted by parafollicular cells of the thyroid gland (C-cells). Its secretion is regulated by an increased plasma level of calcium by a feedback mechanism. TCT lowers calcium level in the blood to normal by increasing calcium deposition in the bones, so it checks osteoporosis and stimulates excretion of calcium in the urine.

Topic: Biology, Chemical Coordination and Integration, NEET
Consider the following metallurgical processes
(I) Heating impure metal with CO and distilling the resulting volatile carbonyl (boiling point. 43°C ) and finally decomposition at 150°-200°C to get the pure metal
(II) Heating the sulphide ore in air until a part is converted to oxide and then further heating in the absence of air to let the oxide react with unchanged metal sulphide
(III) Electrolysis of the molten electrolyte containing approximately equal amounts of the metal chloride and NaCl to obtained the metal
The processes used for obtaining magnesium, nickel and copper are respectively

Options:
A (I), (II) and (III)
B (II), (III) and (I)
C (III), (I) and (II)
D (II), (I) and (III)

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

Solution:
(I) Ni(impure)+4CO→50°CNiCO4volatile→230°CNipure+4CO↑
(II) Cu2S+32O2→ΔCu2O+SO2
Cu2S+2Cu2O→hightemp.6Cu+SO2
(III) MgCl2s→lectrolysisMg2+l+2Cl-l
At cathode Mg2++2e-→Mgs
At anode 2Cl-l→Cl2g+2e-

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
Consider the following molecule

If π - electron cloud of C1-C2 is present in the plane of paper then which of the following is/are correct

Options:
A fluorine is perpendicular to the plane of paper
B chlorine is present in the plane of paper
C σ -bond of C2-C3 is perpendicular to the plane of paper
D π -electron cloud of C2-C3 bond and Cl is present in same plane

The correct answer is:
π -electron cloud of C2-C3 bond and Cl is present in same plane

Solution:
Here π -electron cloud of C2-C3 bond and Cl are present in same plane

Topic: Chemistry, General Organic Chemistry, JEE Main
Consider the following molecules and statements related to them:

(a) (B) is more likely to be crystaline than (A)

(b) (B) has higher boiling point than (A)

(c) (B) dissolves more readily than (A) in water

Identify the correct option from below:

Options:
A (a) and (b) are true
B (a) and (c) are true
C only (a) is true
D (b) and (c) are true

The correct answer is:
(b) and (c) are true

Solution:
The molecules of B are joined by intermolecular hydrogen bonding and molecules of A associated through dipole-dipole interactions.Hence, the B is more likely to be crystalline than A. Due to inter molecularH-Bonding in B, than A, B is more soluble and having more boiling point than A.

*Note: All the three statements given in the question are correct but the given answer according to NTA, is statements (b) and (c).

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2020 (03 Sep Shift 2)
Consider the following molecules

O2I,O2AsF6II,KO2III

Choose the correct answer regarding O-O bond from the following.

Options:
A The correct decreasing bond order is I > III > II
B The correct decreasing order of bond length is III > I > II
C The bond strength of I is less than that of III
D Bond dissociation energy is highest in case of III

The correct answer is:
The correct decreasing bond order is I > III > II

Solution:
In O2AsF6,O2+ has bond order 2.5.
O2 has bond order 2.0.
In KO2,O2- has bond order 1.5.
Higher the bond order, smaller is the bond length.
Hence, the order is O2AsF6<O2<KO2 .

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Consider the following nuclear reactions
$$
\begin{aligned}
& { }_{92}^{238} \mathrm{M} \rightarrow{ }_{\mathrm{y}}^{\mathrm{x}} \mathrm{N}+{ }_2^4 \mathrm{He} \\
& { }_{\mathrm{y}}^{\mathrm{N}} \mathrm{N} \rightarrow{ }_{\mathrm{B}}^{\mathrm{A}} \mathrm{L}+2 \beta^{+}
\end{aligned}
$$ The number of neutrons in the element $L$ is

Options:
A
142

B
146

C
140

D
144

The correct answer is:
144

Solution:
No Solution Available

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2004
Consider the following oxides
1. OF2
2. Cl2O
3. Br2O
The correct sequence of X-O-X bond angle is

Options:
A 1 > 2 > 3
B 3 > 2 > 1
C 2 > 1 > 3
D 1 > 3 > 2

The correct answer is:
3 > 2 > 1

Solution:
The above compounds are the dihalogen oxides as the size of the Halogen atom increases the X-O-X bond of
Here bond angle decrease as follows

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Consider the following pairs of compounds

Of these pairs

Options:
A both X and Y are resonance forms
B both X and Y are tautomeric forms
C X is a resonance pair, while Y is a pair of tautomers
D X is a pair of tautomers; while Y is a resonance pair

The correct answer is:
X is a resonance pair, while Y is a pair of tautomers

Solution:
In 'X' pair of compounds, delocalization of electrons occurs without the change in the relative position of bonded atoms (or nuclei). Hence, X is a resonance pair.
In Y pair of compounds, shifting of proton H+ occurs from N atom to O atom. Hence, 'Y' is a pair of tautomers.

Topic: Chemistry, General Organic Chemistry, JEE Main
Consider the following pairs of electrons

(A) (a) n=3, l=1, ml=1,ms=+12
(b) n=3, l=2,ml=1,ms=+12
(B) (a) n=3, l=2,ml=-2,ms=-12
(b) n=3, l=2,ml=-1,ms=-12
(C) (a) n=4, l=2,ml=2,ms=+12
(b) n=3,l=2,ml=2,ms=+12

The pairs of electrons present in degenerate orbitals is/are

Options:
A Only A
B Only B
C Only C
D B and C both

The correct answer is:
Only B

Solution:
The orbitals which have same energy are called degenerate orbitals. Energy or an orbital depends on two quantum numbers, principle quantum number (n) and azimuthal quantum number (l).

The orbitals with similar values of n & l but with different value of m will be degenerate orbitals.

Among the given set or orbital only option (B) fulfil this condition.

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2022 (24 Jun Shift 1)
Consider the following pairs of solution which will be isotonic at the same temperature. The number of pairs of solutions is/ are _____

A. 1 M aq. NaCl and 2 M aq. urea

B. 1 M aq. CaCl2 and 1.5 M aq. KCl

C. 1.5 M aq. AlCl3 and 2 M aq. Na2SO4

D. 2.5 M aq. KCl and 1 M aq. Al2(SO4)3

Options:

The correct answer is:
4

Solution:
A solution having the same solute and solvent concentration as a cell across the semipermeable membrane is referred to as isotonic solution.

The effective concentration of each solution can be calculated as follows,

Effective concentration =i(Von't Hoff factor)×C

1M NaCl has effective concentration =2×1=2M and hence it is isotonic with two molar urea solution.

1M CaCl2 has effective concentration = 3M and 1.5M KCl has also 3M, hence, both are isotonic.

1.5M AlCl3≅6M and 2M Na2SO4≅6M, Hence both are isotonic.

2.5M KCl≅5M and 1M Al2SO43≅5M, Hence both are isotonic.

Topic: Chemistry, Solutions, JEE Main, JEE Main 2023 (06 Apr Shift 2)
Consider the following parallel reactions being given by A (t1/2 = 1.386 Ã 102 hours), each path being 1st order.

If distribution of A in the product mixture is 50%, calculate partial half-life of A for the conversion into B.

Options:
A 346.5 h
B 231 h
C 154 h
D 92.4 h

The correct answer is:
231 h

Solution:
From question
2 k 1 2 k 1 + 3 k 2 = 0.5 and k 1 + k 2 = 0.693 1.386 × 1 0 2 = 5 × 1 0 - 3 h -1
Solving k 1 k 2 = 3 2 and k 1 = 3 × 1 0 - 3 h -1 and k 2 = 2 × 1 0 - 3 h -1
t 1 / 2 A → B = 0.693 k 1 = 0.693 3 × 1 0 - 3 = 2 3 1 h

Topic: Chemistry, Chemical Kinetics, JEE Main
Consider the following parametric equation of a curve ;
$\mathrm{x}(\theta)=|\cos 4 \theta| \cos \theta$ $\mathrm{y}(\theta)=|\cos 4 \theta| \sin \theta$
for $0 \leq \theta \leq 2 \pi$
Which one of the following graphs represents the curve ?

Options:
A
B
C
D

The correct answer is:

Solution:
Make graph and observe yourself

Topic: Mathematics, Area Under Curves, JEE Main
Consider the following perhalate ion in acidic medium
ClO4-I,BrO4-II,IO4-III
Arrange these in the decreasing order of oxidizing power.

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

The correct answer is:
II>III>I

Solution:
Standard reduction potential (in volts) of the perhalate ions are gives as belowâ
ClO 4 â â+â 2H + â+â 2e â â ClO 3 â â+â H 2 O;â E o â=â1.19âV
BrO 4 â â+â 2H + â+â 2e â â BrO 3 â â+â H 2 O;â E o â=â1.74âV
IO 4 â â+â 2H + â+â 2e â â IO 3 â â+â H 2 O; E o â=â+1.65âV
More the standard reduction potential, more is the tendency of perhalite ion to undergo reduction and stronger will be an oxidizing agent.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Consider the following planes
$$
\begin{aligned}
& P: x+y-2 z+7=0 \\
& Q: x+y+2 z+2=0 \\
& R: 3 x+3 y-6 z-11=0
\end{aligned}
$$

Options:
A
$P$ and $R$ are perpendicular

B
$Q$ and $R$ are perpendicular

C
$P$ and $Q$ are parallel

D
$P$ and $R$ are parallel

The correct answer is:
$P$ and $R$ are parallel

Solution:
Given planes are
$$
\begin{aligned}
& P: x+y-2 z+7=0 \\
& Q: x+y+2 z+2=0
\end{aligned}
$$
and $R: 3 x+3 y-6 z-11=0$
Consider Plane $P$ and $R$.
Here $a_1=1, b_1=1, c_1=-2$ and $a_2=3, b_2=3, c_2=-6$
Since, $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}=\frac{1}{3}$ therefore $P$ and $R$ are parallel.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2012 (26 May Online)
Consider the following plots of rate constant versus 1T for four different reactions. Which of the following orders is correct for the activation energies of these reactions?

Options:
A Eb>Ea>Ed>Ec
B Ea>Ec>Ed>Eb
C Ec>Ea>Ed>Eb
D Eb>Ed>Ec>Ea

The correct answer is:
Ec>Ea>Ed>Eb

Solution:
logk=logA-Ea2.303RT

Slope =-Ea2.303RT⟹Ec>Ea>Ed>Eb

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2020 (08 Jan Shift 2)
Consider the following plots of rate constant versus 1T for four different reactions. which of the following orders is correct for the activation energies of these reactions?

Options:
A Ec>Ea>Ed>Eb
B Eb>Ea>Ed>Ec
C Ea>Ec>Ed>Eb
D Eb>Ed>Ec>Ea

The correct answer is:
Ec>Ea>Ed>Eb

Solution:
As we know that Arhenious equation is given as
K=Ae-EaRT
logâ¡K=logâ¡A-Ea2.303 RT
Slope=-Ea2.303 RT
Slope c>a>d>b.
Ec>Ea>Ed>Eb

Topic: Chemistry, Chemical Kinetics, JEE Main
Consider the following processes
$\Delta H(\mathrm{~kJ} / \mathrm{mol})$
$\begin{array}{ll}1 / 2 A \rightarrow B & +150 \\ 3 B \rightarrow 2 C+D & -125 \\ E+A \rightarrow 2 D & +350\end{array}$
For $B+D \rightarrow E+2 C, \Delta H$ will be

Options:
A $525 \mathrm{~kJ} / \mathrm{mol}$
B $-175 \mathrm{~kJ} / \mathrm{mol}$
C $-325 \mathrm{~kJ} / \mathrm{mol}$
D $325 \mathrm{~kJ} / \mathrm{mol}$

The correct answer is:
$-175 \mathrm{~kJ} / \mathrm{mol}$

Solution:
$\begin{aligned}
& \frac{1}{2} A \longrightarrow B ; \quad \Delta H=150 \mathrm{~kJ} / \mathrm{mol} \\
& 3 B \longrightarrow 2 C+D ; \Delta H=-125 \mathrm{~kJ} / \mathrm{mol} \\
& E+A \longrightarrow 2 D ; \Delta H=+350 \mathrm{~kJ} / \mathrm{mol} \\
& \hline \text { By }[2 \times(\mathrm{i})+(\mathrm{ii})]-(\mathrm{iii}), \text { we have } \\
& B+D \longrightarrow E+2 C \\
& \therefore \quad \Delta H=150 \times 2+(-125)-350 \\
& \quad=-175 \mathrm{~kJ} / \mathrm{mol} .
\end{aligned}$

Topic: Chemistry, Thermodynamics (C), NEET, NEET 2011 (Mains)
Consider the following radial distribution function diagrams. Which of the following has the correct matching of curve and orbital ?

Options:
A I(3s), II(3p), III(3d)
B I(3d), II(3p), III(3s)
C I(3s), II(3d), III(3p)
D I(3p), II(3d), III(3s)

The correct answer is:
I(3p), II(3d), III(3s)

Solution:
No of radial node = n - â - 1 .
I(3p) ---- n = 3 and â =1. One radial node

II(3d) ---- n = 3 and â =2. No radial node
III(3s) ---- n = 3 and â =0. Two radial nodes.
All the orbitals provided in choices belong to match. I(3p), II(3d), III(3s)

Topic: Chemistry, Structure of Atom, JEE Main
Consider the following radioactive decay process:

A84218⟶αA1⟶β-A2⟶γA3⟶αA4⟶β+A5⟶γA6

The mass number and the atomic number of A6 are given by:

Options:
A 210 and 82
B 210 and 84
C 210 and 80
D 211 and 80

The correct answer is:
210 and 80

Solution:
When an alpha particle is emitted from nucleus the nucleus loses two proton and two neutrons. This means the atomic mass number decreases by 4 and the atomic number decreases by 2.

In beta decays, the mass number of the atoms remains the same, but the atomic number changes: the atomic number increases in the negative beta decay and decreases in the positive beta decay by 1, respectively.

In gamma decay, mass number and atomic number remains unchanged.

Step by step variation is given below:

84218A⟶α 82214A1⟶β- 83214A2⟶γ 83214A3⟶α 81210A4⟶β+ 80210A5⟶γ 80210A6

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2023 (24 Jan Shift 1)
Consider the following radioactive decays

I. U92⟶-α90Th and

II. 90Th⟶-αRa88

In which case group of parent and daughter elements remains unchanged.

Options:
A In (I)
B In (II)
C Both in (I) and (II)
D None of the above

The correct answer is:
In (I)

Solution:
For the following radioactive decays

I. U92⟶-α90Th and
II. Th90⟶-α88Ra

In both Uranium and Thorium belongs to IIIB group.

∵ Elements with atomic number 90 to 103 belongs to same group, i.e. III-B 3rd group (i.e. actinoids).

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
Consider the following reaction
2XS+3O2âÎ2XO+2SO2
2XO+XS → Δ 3X+SO 2
The 'X' cannot be

Options:
A Hg
B Pb
C Zn
D Cu

The correct answer is:
Zn

Solution:
Here X cannot be Zn as Zn is extracted from its sulphide ore by roasting followed by carbon reduction Hg, Pb and Cu are extracted by self reduction from their sulphide ores.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main