datavorous/entrance-exam-dataset huggingface dataset:
In the group $\{1,2,3,4,5,6\}$ under multiplication $\bmod 7,2^{-1} X_{7}^{4}=$

Options:
A 1
B 4
C 2
D 3

The correct answer is:
2

Solution:
Under Multiplication Modulo 7 , inverse of 2 is $2 \times 4=1(\bmod 7)$
Hence, $\quad 2^{-1}=4$
Now, $2^{-1} \times 4=4 \times 4=16$
Now, $\quad 16=2 \times 7+2$
Hence dividing 16 by 7 gives a remainder 2 ,
Therefore, value of above multiplication mod is 2.

Topic: Mathematics, Mathematical Reasoning, COMEDK, COMEDK 2018
In the group $G=\{1,5,7,11\}$ under multiplication modulo 12, the solution of $7^{-1} \underset{12}{\times}(x \underset{12}{12} 11)=5$ is $x=$

Options:
A 11
B 7
C 1
D 5

The correct answer is:
1

Solution:
Given that, $G=\{1,5,7,11\}$ is a group under multiplication module 12 .
$\because \quad 7^{-1}=7 \quad\left[\because 7^{-1} \otimes_{12} 7=1\right]$
$\begin{array}{ll}\text { Now, } & 7^{-1} \otimes_{12}\left(x \otimes_{12} 11\right)=5 \\ \Rightarrow & 7 \otimes_{12}\left(11 \otimes_{12} x\right)=5 \\ \Rightarrow & \left(7 \otimes_{12} 11\right) \otimes_{12} x=5 \\ \Rightarrow & \left\{7 \otimes_{12} 11=\text { remainder after dividing } 77 \text { from } 12\right\} \\ \Rightarrow & 5 \otimes_{12} x=5 \\ \Rightarrow & x=1\end{array}$

Topic: Mathematics, Mathematical Induction, COMEDK, COMEDK 2016
In the group III radicals, in place of $\mathrm{NH}_4 \mathrm{Cl}$ which of the following can be used

Options:
A $\mathrm{NH}_4 \mathrm{NO}_3$
B $\left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4$
C $\left(\mathrm{NH}_4\right)_2 \mathrm{CO}_3$
D $\mathrm{NaCl}$

The correct answer is:
$\mathrm{NH}_4 \mathrm{NO}_3$

Solution:
$\mathrm{NH}_4$ ions are required to supress the ionization of $\mathrm{NH}_4 \mathrm{OH}$ but sufficient to precipitate the III group radicals. $\left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4$ can't be used as $\mathrm{SO}_4^{2-}$ ions will precipitate $\mathrm{Ba}^{2+}$ as $\mathrm{BaSO}_4$

Topic: Chemistry, Practical Chemistry, JEE Main
In the group the carbonyl carbon is joined to other atoms by:

Options:
A Two sigma and one pi bonds
B Three sigma and one pi bonds
C One sigma and two pi bonds
D Two sigma and two pi bonds

The correct answer is:
Three sigma and one pi bonds

Solution:

Topic: Chemistry, Aldehydes and Ketones, JEE Main
In the half wave rectifier circuit operating from \(50 \mathrm{~Hz}\) mains frequency, the fundamental frequency in the ripple would be

Options:
A \(25 \mathrm{~Hz}\)
B \(50 \mathrm{~Hz}\)
C \(70.7 \mathrm{~Hz}\)
D \(100 \mathrm{~Hz}\)

The correct answer is:
\(50 \mathrm{~Hz}\)

Solution:
In full wave a.c. or half wave rectifier, the time interval between two peaks are the same. But in half wave rectification, the negative wave part is cut off. But as for frequency or the period, they will be the same as seen from the figures. Therefore, this also will be 50 cycles/second.

Topic: Physics, Semiconductors, AIIMS, AIIMS 2007
In the half wave rectifier circuit operating with $50 \mathrm{~Hz}$ mains frequency. The fundamental frequency in the ripple will be

Options:
A $100 \mathrm{~Hz}$
B $20 \mathrm{~Hz}$
C $50 \mathrm{~Hz}$
D $25 \mathrm{~Hz}$

The correct answer is:
$50 \mathrm{~Hz}$

Solution:
The output is obtained for half cycle only in half wave rectifier. Therefore, frequency of the ripple is same as that of the input i.e. $50 \mathrm{~Hz}$.

Topic: Physics, Semiconductors, VITEEE, VITEEE 2015
In the Hall-Heroult process, aluminium is formed at the cathode. The cathode is made out of:

Options:
A Platinum
B Carbon
C Copper
D Pure aluminium

The correct answer is:
Carbon

Solution:
The electrolytic cell which is used for extraction of aluminium is a steel vessel. The vessel is lined with carbon, which acts as cathode and graphite is used at the anode.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2019 (12 Jan Shift 1)
In the Hatch and Slack pathway, malic acid is formed in

Options:
A Mesophyll cell of maize leaf
B Mesophyll cell of tomato
C Mesophyll cell of banana
D Mesophyll of cactus

The correct answer is:
Mesophyll cell of maize leaf

Solution:
Plants that are adapted to dry tropical regions have the C4 pathway. Though these plants have the oxaloacetic acid as the first CO2 fixation product, they use the C3 pathway or the Calvin cycle as the main biosynthetic pathway. Since this is discovered by Hatch and Slack, it is known as Hatch and slack pathway.

Topic: Biology, Photosynthesis in Higher Plants, NEET
In the Henderson-Hasselbalch equation for a buffer solution, what happens to the pH when the concentration of the conjugate base ([A-]) is equal to the concentration of the weak acid ([HA])?

Options:
A The pH increases
B The pH decreases
C The pH remains unchanged
D The pH becomes neutral

The correct answer is:
The pH remains unchanged

Solution:
In the Henderson-Hasselbalch equation, when the concentration of the conjugate base ([A–]) is equal to the concentration of the weak acid ([HA]), the pH remains unchanged.

Topic: Chemistry, Ionic Equilibrium, NEET
In the Henderson-Hasselbalch equation for a buffer solution, why does the pH remain unchanged upon dilution?

Options:
A Dilution increases the concentration of the conjugate base.
B Dilution decreases the concentration of the weak acid.
C Dilution affects the pKa value.
D Dilution does not impact the ratio of [Conjugate acid, BH+] to [Base, B].

The correct answer is:
Dilution does not impact the ratio of [Conjugate acid, BH+] to [Base, B].

Solution:
In the Henderson-Hasselbalch equation, the pH remains unchanged upon dilution because dilution does not impact the ratio of [Conjugate acid, BH+] to [Base, B].

Topic: Chemistry, Ionic Equilibrium, NEET
In the Hoffmann bromamide degradation reaction, the number of moles of NaOH and Br2 used per mole of amine produced are:

Options:
A Two moles of NaOH and two moles of Br2
B Four moles of NaOH and one mole of Br2
C One mole of NaOH and one mole of Br2
D Four moles of NaOH and two moles of Br2

The correct answer is:
Four moles of NaOH and one mole of Br2

Solution:
To find number of moles of NaOH and Br2 used per mole of amine produced. The complete balanced equation is:

Topic: Chemistry, Amines, JEE Main, JEE Main 2016 (03 Apr)
In the Hoop’s process for refining of aluminum, the fused materials form three different layers and they remain separated during electrolysis also. This is because

Options:
A There is special arrangement in the cell to keep the layers separate
B The three layers have different densities
C The three layers are maintained at different temperatures.
D The upper layer is kept attracted by the cathode and the lower layer is kept attracted by the anode.

The correct answer is:
The three layers have different densities

Solution:
In the Hoop’s process for refining of aluminum, the fused materials form three different layers and they remain separated during electrolysis also. This is because the three layers have different densities

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
In the human brain, the cerebral aqueduct passes through:

Options:
A Corpus callosum
B Cerebral hemispheres
C Midbrain
D Spinal cord

The correct answer is:
Midbrain

Solution:
Location: between the thalamus/hypothalamus (of the forebrain) and pons (of hindbrain).
Cerebral aqueduct passes through the midbrain. The dorsal portion of the midbrain mainly consists of 4 round swellings (lobes). These are termed as corpora quadrigemina. Midbrain receives tactile and visual responses.

Topic: Biology, Neural Control and Coordination, NEET
In the human female, menstruation can be deferred by the administration of:

Options:
A combination of $\mathrm{FSH}$ and $\mathrm{LH}$
B combination of estrogen and progesterone
C FSH only
D LH only.

The correct answer is:
combination of estrogen and progesterone

Solution:
The progesterone and oestradiol continue the hypertrophy of endometrial lining in the uterus and fallopian tubes and maintain it throughout pregnancy. progesterone is also necessary for the proper implantation of the foetus in the uterine wall. It stimulates the endometrial glands to secrete a nutrient fluid for the foetus, hence the term secretory phase. The progesterone inhibits the release of FSH so that it may not develop additional follicles and eggs.
Related Theory
There is no menstrual cycle during pregnancy due to the secretion of progesterone hormone by the corpus luteum. The progesterone hormone inhibits the production of the gonadotropic hormones from the pituitary and prevents the uterine lining from sloughing and supports pregnancy.

Topic: Biology, Human Reproduction, NEET, NEET 2007
In the hydroboration - oxidation reaction of propene with diborane, H2O2 and NaOH, the organic compound formed is:

Options:
A CH3CH2CH2OH
B CH3CH2OH
C CH3CHOHCH3
D CH33COH

The correct answer is:
CH3CH2CH2OH

Solution:
According to the question, The hydroboration oxidation reaction is an organic chemical reaction which is employed for the conversion of alkenes into alcohols that are neutral. This is done via a two-step process which includes a hydroboration step and an oxidation step.

Hydroboration oxidation reaction is a two-step reaction in which alkene reacts with BH3, THF (tetrahydrofuran) and hydrogen peroxide in a basic medium to give alcohol. This is done by a net addition (across the entire double bond) of water.

Hydroboration oxidation reaction mechanism can be considered as an anti-markovnikov reaction where a hydroxyl group attaches itself to the carbon which is less substituted.

So, according to this reaction, propene give n-propyl-alcohol.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2014 (09 Apr Online)
In the hydroboration - oxidation reaction of propene with diborane, H 2 O 2 and NaOH, the organic compound formed is:

Options:
A CH 3 CH 2 CH 2 OH
B CH 3 CH 2 OH
C CH 3 CHOHCH 3
D CH 3 3 COH

The correct answer is:
CH 3 CH 2 CH 2 OH

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
In the hydrocarbon
$$
\underset{8}{\mathrm{CH}_3}-\mathrm{CH}=\underset{4}{\mathrm{CH}}=\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{C} \equiv \mathrm{CH}_1
$$
The state of hybridization of carbons 1,3 and 5 are in the following sequence

Options:
A $\mathrm{sp}^3, \mathrm{sp}^2, \mathrm{sp}$
B $\mathrm{sp}^2, \mathrm{sp}, \mathrm{sp}^3$
C $\mathrm{sp}, \mathrm{sp}^3, \mathrm{sp}_3^2$
D $\mathrm{sp}, \mathrm{sp}^2, \mathrm{sp}^3$

The correct answer is:
$\mathrm{sp}, \mathrm{sp}^3, \mathrm{sp}_3^2$

Solution:
No Solution Available

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET, NEET 2008 (Mains)
In the hydrogen atom spectrum, Î»3-1 and Î»2-1 represent wavelengths emitted due to transition from second and first excited states to the ground state respectively. The ratio Î»3-1Î»2-1=pq, where p and q are the smallest positive integers. What is the value of p+q?

Options:

The correct answer is:
59

Solution:
1Î»3-1=R112-132=8R9
1Î»2-1=R112-122=3R4
â Î»3-1Î»2-1=2732

Topic: Physics, Atomic Physics, JEE Main
In the hydrogen atom spectrum, the emission of the least energetic photon takes place during the transition from n=6 energy level to n=.…… energy level.

Options:
A 1
B 3
C 5
D 4

The correct answer is:
5

Solution:
According to the relation
ÎEâ1Î»âÏâ1nL2-1nu2

nL= lower energy level

nu= higher energy level
In hydrogen atom the emission of the least energetic photon take place only when change in energy is minimum. This is possible for an electron that is present in n=6 and jumps to n=5 .

Topic: Chemistry, Structure of Atom, JEE Main
In the hydrogen spectrum the wavelength of H Î» line is 656 nm whereas, in the spectrum of a distant galaxy the H Î» line has a wavelength of 706 nm. The estimated speed of the galaxy with respect to the earth is

Options:
A 1.2×107 m s-1
B 2.2Ã107 m s-1
C 3×107 m s-1
D 1.9×107 m s-1

The correct answer is:
2.2Ã107 m s-1

Solution:
Since, the wavelength observed is more than the true wavelength, we can say that the galaxy is receding. Doppler effect in this case gives us
Î»'=Î»c+vc-v
706 = 656 c + v c - v
or, c+vc-v=7066562=1.16
or, c+v=1.16c-1.16v
v=0.16c2.16=0.16Ã3.0Ã1082.16=0.22Ã108 m s-1
v=2.2Ã107 m s-1

Topic: Physics, Wave Optics, JEE Main
In the hydrolysis of an organic chloride in presence of large excess of water; $\mathrm{RCI}+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{ROH}+\mathrm{HCl}$

Options:
A Molecularity and order of reaction both are 2
B Molecularity is 2 but order of reaction is 1
C Molecularity is 1 but order of reaction is 2
D Molecularity is 1 and order of reaction is also 1

The correct answer is:
Molecularity is 2 but order of reaction is 1

Solution:
Hints : As water used is in large excess.

Topic: Chemistry, Chemical Kinetics, JEE Main
In the hysteresis curve the value of magnetization (B) which will be present in a substance when value of magnetizing force $(\mathrm{H})$ is made zero $(\mathrm{H}=0)$ is called as

Options:
A coercivity
B retentivity
C domain
D saturation

The correct answer is:
retentivity

Solution:
Retentivity or remanence is the ability of a magnetic substance to retain magnetism even in the absence of a magnetising field.
$\therefore \quad$ Retentivity corresponds to the value of $\mathrm{B}$ when value of $\mathrm{H}$ is zero.

Topic: Physics, Magnetic Properties of Matter, MHT CET, MHT CET 2023 (14 May Shift 2)
In the ideal double-slit experiment, when a glass-plate (refractive index 1.5) of thickness t is introduced in the path of one of the interfering beams (wavelength Î»), the intensity at the position where the central maximum occurred previously remains unchanged. The minimum thickness of the glass-plate is

Options:
A 2 Î»
B 2 λ 3
C λ 3
D λ

The correct answer is:
2 λ

Solution:
For no change of Intensity at central maxima there should again be maxima.
â´ Path difference due to slab should be integral multiple of Î»
or Îx=nÎ»
or Î¼ - 1 t = Î» n = 1 , 2 , 3 ,........
or t = n Î» Î¼ - 1
For minimum value of t , n = 1
t = Î» Î¼ - 1 = Î» 1 Â· 5 - 1 = 2 Î»

Topic: Physics, Wave Optics, NEET
In the ideal double-slit experiment, when a glass-plate (refractive index 1.5) of thickness t is introduced in the path of one of the interfering beams (wavelength λ), the intensity at the position where the central maximum occurred previously remains unchanged. The minimum thickness of the glass-plate is

Options:
A 2 Î»
B 2 λ 3
C λ 3
D λ

The correct answer is:
2 λ

Solution:
For no change of Intensity at central maxima there should again be maxima.
â´ Path difference due to slab should be integral multiple of Î»
or Îx=nÎ»
or Î¼ - 1 t = Î» n = 1 , 2 , 3 ,........
or t = n Î» Î¼ - 1
For minimum value of t , n = 1
t = Î» Î¼ - 1 = Î» 1 Â· 5 - 1 = 2 Î»

Topic: Physics, Wave Optics, NEET
In the increasing order of their sizes, arrange the following ions \( \mathrm{N}^{3-}, \mathrm{Na}^{+}, \mathrm{F}^{-}, \mathrm{Mg}^{2+} \) and \( \mathrm{O}^{2-} \)

Options:
A \( \mathrm{Mg}^{2+}>\mathrm{Na}^{+}>\mathrm{F}>\mathrm{O}^{2-} < \mathrm{N}^{3-} \)
B \( \mathrm{N}^{3-} < \mathrm{F}>\mathrm{O}^{2}>\mathrm{Na}^{+}>\mathrm{Mg}^{2+} \)
C \( \mathrm{Mg}^{2+} < \mathrm{Na}^{+} < \mathrm{F} < \mathrm{O}^{2-} < \mathrm{N}^{3-} \)
D \( \mathrm{N}^{3}>\mathrm{O}^{2}>\mathrm{F}>\mathrm{Na}^{+} < \mathrm{Mg}^{2+} \)

The correct answer is:
\( \mathrm{Mg}^{2+} < \mathrm{Na}^{+} < \mathrm{F} < \mathrm{O}^{2-} < \mathrm{N}^{3-} \)

Solution:
Mg2+<Na+<F<O2-<N3-

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
In the industrial production of which of the following, molecular hydrogen is obtained as a bye product.

Options:
A Na2CO3
B NaOH
C Na metal
D NaHCO3

The correct answer is:
NaOH

Solution:
NaOH is produced in Nelson cell using Brine solution.

Reaction taking place will be

Cathode: 2H2O+2e-→H2+2OH-

Anode: 2Cl-→Cl2+2e-

In solution Na++OH-→NaOH

So, H2 gas is released at cathode during preparation of NaOH.

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2022 (24 Jun Shift 2)
In the interval 0,2Ï , sum of all the roots of the equation sinÏ log 3 1 x = 0 is

Options:
A 3 2
B 4
C 9 2
D 13 3

The correct answer is:
9 2

Solution:
π log 3 1 x = k π , k ∈ I
log 3 1 x = k ⇒ x = 3 - k
Possible values of k are – 1, 0, 1, 2, 3, .......
S = 3 + 1 + 1 3 + 1 3 2 + 1 3 3 + ..... ∞
= 4 + 1 / 3 1 - 1 / 3 = 4 + 1 2 = 9 2

Topic: Mathematics, Trigonometric Equations, JEE Main
In the interval -π2,π2, the equation logsin⁡θ⁡cos⁡2θ=2 has

Options:
A no solution
B a unique solution
C two solutions
D infinite many solutions

The correct answer is:
a unique solution

Solution:
Since, -π2≤θ≤π2⇒sin⁡θ∈-1,1
Here sin⁡θ∈0,1 for logsin⁡θ⁡cos⁡2θ=2 to be defined.
⇒cos⁡2θ=sin2⁡θ
⇒1-2sin2⁡θ=sin2⁡θ
⇒sin2⁡θ=13⇒sin⁡θ=13 (as sin⁡θ∈0,1 )
⇒The equation has a unique solution.

Topic: Mathematics, Trigonometric Equations, JEE Main
In the interval 0,2, on which of the following function Lagrange's mean value theorem is not applicable?

Options:
A fx=sin⁡xxx≠01x=0
B fx=1-xx<11-x2x≥1
C fx=x2x
D fx=ex-1

The correct answer is:
fx=1-xx<11-x2x≥1

Solution:
Note that LMVT is applicable in a,b if fx is continuous in a,b and differentiable in a,b

Since fx=1-xx<11-x2x≥1 is non-differentiable at x=1, hence LMVT cannot be applied to it in 0,2

Topic: Mathematics, Application of Derivatives, JEE Main
In the ionic equation
$-\mathrm{BiO}_{3}^{-}+6 \mathrm{H}^{+}+x e^{-} \longrightarrow \mathrm{Bi}^{3+}+3 \mathrm{H}_{2} \mathrm{O}, \quad$ the values of $x$ is

Options:
A 6
B 2
C 4
D 3

The correct answer is:
2

Solution:
$\mathrm{BiO}_{3}^{-}+6 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Bi}^{3+}+3 \mathrm{H}_{2} \mathrm{O}$

Topic: Chemistry, Redox Reactions, JEE Main
In the isolation of metals, calcination process usually results in:

Options:
A Metal oxide
B Metal carbonate
C Metal sulphide
D Metal hydroxide

The correct answer is:
Metal oxide

Solution:
Calcination is used for decomposition of metal carbonates.
MCO3 →∆MO+CO2↑

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2015 (10 Apr Online)
In the isothermal expansion of $10 \mathrm{~g}$ of gas from volume $\mathrm{V}$ to $2 \mathrm{~V}$ the work done by the gas is $575 \mathrm{~J}$.
What is the root mean square speed of the molecules of the gas at that temperature?

Options:
A
$398 \mathrm{~m} / \mathrm{s}$

B
$520 \mathrm{~m} / \mathrm{s}$

C
$499 \mathrm{~m} / \mathrm{s}$

D
$532 \mathrm{~m} / \mathrm{s}$

The correct answer is:
$499 \mathrm{~m} / \mathrm{s}$

Solution:
$\mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{3 \rho \mathrm{v}}{\text { mass of the gas }}}$

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2013 (25 Apr Online)
In the Kjeldahl's method for estimation of nitrogen present in a soil sample, ammonia evolved from 0.75 g of sample neutralized 10 mL of 1M H2SO4. The percentage of nitrogen in the soil is
[Report your answer upto two decimal places]

Options:

The correct answer is:
37.33

Solution:
In 0.75 g of sample the percentage of nitrogen can be find out as follows
N%=14×N×VWO.C.×100
Here N×V of H2SO4=n×M×V
=2×1×10×10-3=20×10-3
% Nitrogen =14×20×10-30.75×100=37.33%

Topic: Chemistry, Hydrocarbons, JEE Main
In the laboratory, manganese (II) salt is oxidise to permanganate ion in aqueous solution by

Options:
A hydrogen peroxide
B conc. nitric acid
C peroxy disulphate
D dichromate

The correct answer is:
peroxy disulphate

Solution:
In laboratory, manganese (II) ion salt is oxidised to permanganate ion in aqueous solution by peroxodisulphate.
$2 \mathrm{Mn}^{2+}+\underset{\text{peroxodisulphate ion}}{5 \mathrm{~S}_2 \mathrm{O}_8^{2-}}+8 \mathrm{H}_2 \mathrm{O} \rightarrow$ $2 \mathrm{MnO}_4^{-}+10 \mathrm{SO}_4^{2-}+16 \mathrm{H}^{+}$

Topic: Chemistry, d and f Block Elements, JEE Main
In the LC circuit shown below, the current is in direction as shown and the charges on the capacitor plates have the sign as shown. At this instant

Options:
A I is increasing Q is increasing
B I is increasing Q is decreasing
C I is decreasing Q is increasing
D I is decreasing and Q is decreasing

The correct answer is:
I is decreasing Q is increasing

Solution:
As clearly ‘Q’ is increasing so electric field energy in capacitor must be increasing. And hence, magnetic field energy in inductor must be decreasing. So I is decreasing.

Topic: Physics, Electromagnetic Induction, JEE Main
In the leaching method, bauxite ore is digested with a concentrated solution of $\mathrm{NaOH}$ that produces ' $\mathrm{X}$ '. When $\mathrm{CO}_2$ gas is passed through the aqueous solution of ' $X^{\prime}$, a hydrated compound ' $\mathrm{Y}^{\prime}$ ' is precipitated. ' $\mathrm{X}$ ' and ' $\mathrm{Y}$ ' respectively are.

Options:
A
$\mathrm{Na}\left[\mathrm{Al}(\mathrm{OH})_4\right]$ and $\mathrm{Al}_2\left(\mathrm{CO}_3\right)_3 \cdot \mathrm{xH}_2 \mathrm{O}$

B
$\mathrm{Al}(\mathrm{OH})_3$ and $\mathrm{Al}_2 \mathrm{O}_3 \cdot \mathrm{xH}_2 \mathrm{O}$

C
$\mathrm{NaAlO}_2$ and $\mathrm{Al}_2\left(\mathrm{CO}_3\right)_3 \cdot \mathrm{xH}_2 \mathrm{O}$

D
$\mathrm{Na}\left[\mathrm{Al}(\mathrm{OH})_4\right]$ and $\mathrm{Al}_2 \mathrm{O}_3 \cdot \mathrm{xH}_2 \mathrm{O}$

The correct answer is:
$\mathrm{Na}\left[\mathrm{Al}(\mathrm{OH})_4\right]$ and $\mathrm{Al}_2 \mathrm{O}_3 \cdot \mathrm{xH}_2 \mathrm{O}$

Solution:
$\mathrm{Al}_2 \mathrm{O}_3(\mathrm{~s})+2 \mathrm{NaOH}(\mathrm{aq})+3 \mathrm{H}_2 \mathrm{O}$ (l)
Alumina

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
In the leaching of alumina from bauxite, the ore expected to leach out in the process by reacting with NaOH is :

Options:
A TiO2
B Fe2O3
C ZnO
D SiO2

The correct answer is:
SiO2

Solution:
In bauxite impurities of Fe2O3,TiO2 and SiO2 are present, Fe2O3 and TiO2 are basic oxides therefore does not reacts with or dissolve in NaOH whereas SiO2 is acidic oxide it gets dissolve in NaOH, hence leach out

SiO2+2NaOH→Na2SiO3aq.+H2O

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2021 (25 Jul Shift 1)
In the lewis dot structure for \(\mathrm{NO}_2^{-}\), total number of valence electrons around nitrogen is _______

Options:

The correct answer is:
8

Solution:
Number of valence $\mathrm{e}^{-}$around $\mathrm{N}$-atom $=8$

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2024 (05 Apr Shift 1)
In the light of recent classification of living organisms into three domains of life (bacteria, archaea and eukarya), which one of the following statements is true about archaea?

Options:
A Archaea completely differ from prokaryotes
B Archaea resemble eukarya in all respects
C Archaea have some novel features that are absent in other prokaryotes and eukaryotes
D Archaea completely differ from both prokaryotes and eukaryotes

The correct answer is:
Archaea have some novel features that are absent in other prokaryotes and eukaryotes

Solution:
Archea have primitive forms with histones, no organized nucleus, membrane bound organelles absent and proteinous and noncellulosic carbohydrate nature of cell wall.

Topic: Biology, Biological Classification, NEET, NEET 2008 (Mains)
In the light of recent classification of living organisms into three domains of life (bacteria, archaea and eukarya), which one of the following statement is true about archaea?

Options:
A Archaea resemble eukarya in all respects
B Archaea have some novel features that are absent in other prokaryotes and eukaryotes
C Archaea completely differ from both prokaryotes and eukaryotes
D Archaea completely differ from prokaryotes

The correct answer is:
Archaea have some novel features that are absent in other prokaryotes and eukaryotes

Solution:
The term archaebacteria refers to the ancient origin of this group of bacteria, which seems to have diverged very early from the eubacteria. All archaebacteria share certain key characteristics.
(i) Their cell wall lack peptidoglycan (important component of cell wall of eubacteria).
(ii) Lipids in cell membrane of archaebacteria have different structure than those in all other organisms.
(iii) Archaebacteria have distinct ribosomal RNA sequence.
(iv) Some genes of archaebacteria possess introns unlike those of other bacteria.

Topic: Biology, Biological Classification, NEET, NEET 2008 (Screening)