datavorous/entrance-exam-dataset huggingface dataset:
According to Werner, which type of valences are satisfied by neutral molecules or negative ions?

Options:
A Primary valences
B Secondary valences
C Tertiary valences
D Quaternary valences

The correct answer is:
Secondary valences

Solution:
Secondary valences, as proposed by Werner, are satisfied by neutral molecules or negative ions.

Topic: Chemistry, Coordination Compounds, NEET
According to Werner's concept, what is the term used for the number of groups bound directly to the metal ion?

Options:
A Primary valence
B Secondary valence
C Tertiary valence
D Quaternary valence

The correct answer is:
Secondary valence

Solution:
Werner used the term "secondary valence" for the number of groups bound directly to the metal ion.

Topic: Chemistry, Coordination Compounds, NEET
According to Werner's theory, how many types of linkages (valences) do metals show in coordination compounds?

Options:
A One
B Two
C Three
D Four

The correct answer is:
Two

Solution:
Werner's theory states that metals show two types of linkages (valences) in coordination compounds - primary and secondary.

Topic: Chemistry, Coordination Compounds, NEET
According to Werner's theory of coordination compounds,

Options:
A primary valency is ionisable
B secondary valency is ionisable
C primary and secondary valencies are ionisable
D neither primary nor secondary valency is ionisable

The correct answer is:
primary valency is ionisable

Solution:
Primary valency is ionisable while secondary valency is non-ionisable according to Werner's theory of coordination compounds.

Topic: Chemistry, Coordination Compounds, NEET
According to Werner’s theory the geometry of the complex is determined by

Options:
A Only from the primary valence is space
B Number and position of the primary valences in space
C Number and position of the secondary valences in space
D Only from the position of secondary valence in space

The correct answer is:
Number and position of the secondary valences in space

Solution:
Werner’s theory was used to describe the structure and formation of complex compounds or coordination compounds. According to this theory the primary valency gives the oxidation number and the secondary valency gives the coordination number. Also, the geometry of the complex is determined by number and position of secondary valences is space as the ligand satisfying secondary valences are always directed towards fixed position in space.

Topic: Chemistry, Coordination Compounds, MHT CET, MHT CET 2019 (Shift 2)
According to Werner's theory, the number of groups bonded to the central metal atom/ion in a coordination complex represent.

Options:
A Oxidation state
B Primary Valency
C Secondary Valency
D Polyhedron

The correct answer is:
Secondary Valency

Solution:
The secondary valency of a central metal atom/ ion in a complex represents the number of neutral or negatively charged groups called ligands.
The primary valency represents the oxidation state of the metal.

Topic: Chemistry, Coordination Compounds, TS EAMCET, TS EAMCET 2023 (13 May Shift 1)
Account for the following statement- Gabriel phthalimide synthesis is preferred for synthesizing primary amines.

Options:
A because Gabriel phthalimide synthesis is less expensive than other methods.
B because Gabriel phthalimide synthesis provides a high yield of primary amines.
C because Gabriel phthalimide synthesis results in the formation of only 10amines. Secondary and tertiary amines cannot be formed through this process.
D because Gabriel phthalimide synthesis is less sensitive to reaction conditions and is therefore more reliable for primary amine synthesis.

The correct answer is:
because Gabriel phthalimide synthesis results in the formation of only 10amines. Secondary and tertiary amines cannot be formed through this process.

Solution:
Correct Option is : (C)
because Gabriel phthalimide synthesis results in the formation of only 10amines. Secondary and tertiary amines cannot be formed through this process.

Topic: Chemistry, Amines, NEET
Accumulation of which one of the following acids results in the closure of stomata.

Options:
A Malic acid
B Aspartic acid
C PEP
D Oxaloacetic acid.

The correct answer is:
Malic acid

Solution:
According to K+-H+ pump theory during the closure of stomata, ABA stops the energy dependant proton pump in guard cell, thus K+ diffuses out of the cell and H+ enters inside, which combines with malate ion to form malic acid. OP of guard cells decreases thus stomata close.

Topic: Biology, Transport in Plants, NEET
Acid hydrolysis of ester is first order reaction and rate constant is given by $k=\frac{2.303}{t} \log \frac{V_{\infty}-V_{0}}{V_{\infty}-V_{t}}$ where, $V_{0}, V_{t}$ and $V_{\infty}$ are the volumle of standard $\mathrm{NaOH}$ required to neutralise acid present at a given time, if ester is $50 \%$ neutralised then

Options:
A $V_{\infty}=V_{t}$
B $V_{\infty}=\left(V_{t}-V_{0}\right)$
C $V_{\infty}=2 V_{t}-V_{0}$
D $V_{\infty}=2 V_{t}+V_{0}$

The correct answer is:
$V_{\infty}=2 V_{t}-V_{0}$

Solution:
$\mathrm{RCOOR}^{\prime}+\mathrm{H}_{2} \mathrm{O} \stackrel{\mathrm{H}^{+}}{\longrightarrow} \mathrm{RCOOH}+\mathrm{R}^{\prime} \mathrm{OH}$
$\begin{array}{lccc}\text { At } \mathrm{t}=0, & \text { a } & 0 & 0 \\ \text { At timet, } & \mathrm{a}-\mathrm{x} & \mathrm{x} & \mathrm{x} \\ \text { At time } \infty, & \mathrm{a}-\mathrm{a} & \mathrm{a} & \mathrm{a} \\ \text { At } \mathrm{t}=0, \mathrm{~V}_{0}=\text { volume } & \mathrm{Va}_{\mathrm{t}}=\mathrm{x}+\mathrm{V}_{0} & & \text { (catalyst) }\end{array}$
$\mathrm{V}_{\infty}=\mathrm{a}+\mathrm{V}_{0}$
If ester is $50 \%$ hydrolysed then, $x=\frac{a}{2}$
$\begin{array}{l}
\text { or } \mathrm{V}_{\mathrm{t}}=\frac{\mathrm{a}}{2}+\mathrm{V}_{0} \\
\text { or } \mathrm{a}=2 \mathrm{~V}_{\mathrm{t}}-2 \mathrm{~V}_{0} \\
\therefore \mathrm{V}_{\infty}=2 \mathrm{~V}_{\mathrm{t}}-2 \mathrm{~V}_{0}+\mathrm{V}_{0} \\
=2 \mathrm{~V}_{\mathrm{t}}-\mathrm{V}_{0}
\end{array}$

Topic: Chemistry, Chemical Kinetics, VITEEE, VITEEE 2013
Acoin is tossed If a head is observed, a number is randomly selected from the set $\{1,2,3\}$ and if a tail is observed. number is randomly selected from the set $\{2,3,4,5\}$. If the selected number be denoted by $X$, what is the probability that $X=3$ ?

Options:
A 27
B $1 / 5$
C $1 / 6$
D $7 / 2 A$

The correct answer is:
$7 / 2 A$

Solution:
Probability that $(\mathrm{X}=3)=\frac{1}{2} \times \frac{1}{3}+\frac{1}{2} \times \frac{1}{4}=\frac{7}{24}$

Topic: Mathematics, Probability, NDA, NDA 2008 (Phase 2)
Activated sludge is

A. formed after sequential filtration and sedimentation in primary setting tank.

B. bacterial flocs settling after secondary treatment.

C. passed into anaerobic sludge digester.

Options:
A Only (A) is incorrect
B Only (A) and (C) are incorrect
C Only (B) and (C) are incorrect
D (A), (B), and (C) are incorrect

The correct answer is:
Only (A) is incorrect

Solution:
The effluent from the primary settling tank is taken for secondary treatment, where the primary effluent is passed into large-aeration tanks with constant agitation, and the air is pumped into it. Aeration tanks show vigorous growth of useful aerobic heterotrophic microbes into flocs (mass of bacterial and fungal filaments). The microbes consume a significant part of the organic matter and significantly reduce the BOD of the effluent. The effluent is passed into a secondary settling tank where bacterial flocs are allowed to sediment to form activated sludge.

A part of activated sludge is fed into anaerobic sludge digesters, where the anaerobic methanogenic bacteria digest the organic matter present in sludge to produce biogas, used as sources of energy.

Topic: Biology, Microbes in Human Welfare, NEET
Activation energy of a chemical reaction can be determined by ...... .

Options:
A
determining the rate constant at standard temperature

B
determining the rate constant at two temperatures

C
determining probability of collision

D
using catalyst

The correct answer is:
determining the rate constant at two temperatures

Solution:
Arrhenius equation relates activation energy of a chemical reaction with rate constant of a reaction at two different temperatures.
$\ln \left(\frac{k_1}{k_2}\right)=\frac{E_a}{R}\left[\frac{1}{T_1}-\frac{1}{T_2}\right]$

Topic: Chemistry, Chemical Kinetics
Activation energy of any chemical reaction can be calculated if one knows the value of

Options:
A probability of collision
B orientation of reactant molecules during collision
C rate constant at two different temperatures
D rate constant at standard temperature

The correct answer is:
rate constant at two different temperatures

Solution:
To calculate value of $E_a$
Equation used is
$\log \left(\frac{k_2}{k_1}\right)=\frac{E_a}{2.303 R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$
Hence $E_a$ can be calculated if value of rate constant $k$ is known at two different temperatures $T_1$ and $\mathrm{T}_2$.

Topic: Chemistry, Chemical Kinetics, NEET, NEET 2024
Aerobic bacteria are found in:

Options:
A Activated sludge sediment in settlement tanks of sewage treatment plant
B Stomach of ruminants and produce methane
C Biogas plant and produce biogas
D Both (A) and (B)

The correct answer is:
Activated sludge sediment in settlement tanks of sewage treatment plant

Solution:
In the secondary treatment of waste, flocs are used which are a mesh formed due to the association of fungal filament with the aerobic bacteria. These flocs for sediments are the treatment called as activated sludge. This sludge can be said to be rich with aerobic bacteria as it is formed of flocs. This activated sludge is later used in anaerobic sludge digestor. Methanogens present in stomach of ruminants and biogas plants are anaerobic bacteria. Mushrooms and truffles are fungi

Topic: Biology, Microbes in Human Welfare, NEET
Aflower in the form of a sector has been fenced by a wire of $40 \mathrm{~m}$ length. If the flower-bed has the greatest possible area, then what is the radius of the sector? $\quad$

Options:
A $25 \mathrm{~m}$
B $20 \mathrm{~m}$
C $10 \mathrm{~m}$
D $5 \mathrm{~m}$

The correct answer is:
$10 \mathrm{~m}$

Solution:
$\quad \ell+2 r=40$
Area is maximum, when $\ell+2 r=20 \Rightarrow r=10$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2018 (Phase 2)
After double fertilisation, a mature ovule has

Options:
A 1 diploid and 1 haploid cell
B 1 diploid and 1 triploid cell
C 2 haploid and 1 triploid cell
D 1 haploid and 1 triploid cell.

The correct answer is:
1 diploid and 1 triploid cell

Solution:
Double fertilisation is found only in angiosperms. In this group of plants, the pollen tube on entering the embryo sac bursts open in one of the two synergids to release the two male gametes. One male gamete fuses with the egg or oosphere to form a diploid zygote or oospore. It is called generative fertilisation. The second male gamete descends down and fuses with the diploid secondary nucleus of the central cell to form a triploid primary endosperm cell. It is known as vegetative fertilisation. Thus, after double fertilisation, a mature angiospermous ovule contains one diploid cell (zygote) and one triploid cell (primary endosperm cell). The haploid cells of the ovule such as antipodals and synergids degenerate after fertilisation.

Topic: Biology, Sexual Reproduction in Flowering Plants, AIIMS, AIIMS 2017
After double fertilization, a mature ovule has

Options:
A 1 diploid and 1 haploid cell
B 1 diploid and 1 triploid cell
C 2 haploid and 1 triploid cell
D 1 haploid and 1 triploid cell

The correct answer is:
1 diploid and 1 triploid cell

Solution:
Double fertilization is the fusion of two male gametes brought about by a pollen tube fusing to two different cells of the same female gametophyte in order to produce two different structures. It is found only in angiosperms. In angiosperms, the pollen tube bursts open in one of the two synergids to release the two male gametes. One male gamete fuses with the egg or oosphere to form a diploid zygote or oospore. It is called generative fertilization. The second male gamete descends down and fuses with the diploid secondary nucleus of the central cell to form a triploid primary endosperm cell. It is known as vegetative fertilization. Thus, after double fertilization, a mature angiospermous ovule contains one diploid cell (zygote) and one triploid cell (endosperm). The haploid cells of the ovule such as antipodals and synergids degenerate after fertilization.

Topic: Biology, Sexual Reproduction in Flowering Plants, AIIMS, AIIMS 2013
After fertilization, in the embryo sac, which will grow faster?

Options:
A Nucellus
B Endosperm
C Pollen tube
D Synergids

The correct answer is:
Endosperm

Solution:
Double fertilization occurs in Angiosperm. In this fertilization, one male gamete enters to the ovule and fuse with the egg cell result into the formation of diploid zygote while another male gamete fuse with the polar nuclei (2n) or secondary nucleus present in the center results into the formation of primary endosperm nucleus(3n) and that will be triploid in nature. Endosperm will form from the primary endosperm nucleus. They provide nourishment to the developing embryo. So the correct answer is B. Embryo will be able to survive when endosperm grows faster. Synergids guide the pollen tube and it degenerates after fertilization.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
After paying 30 out of 40 installments of a debt of Rs. 3600 , one third of the debt is unpaid. If the installments are forming an arithmetic series, then what is the first instalment?

Options:
A $\operatorname{Rs} 50$
B $\operatorname{Rs} 51$
C $\operatorname{Rs} 105$
D $\operatorname{Rs} 110$

The correct answer is:
$\operatorname{Rs} 51$

Solution:
Let first instalement be Rs. $x$ and difference of consecutive instalments be Rs. $\mathrm{d}$.
$\Rightarrow \frac{30}{2}[2 \mathrm{x}+29 \mathrm{~d}]=\frac{3600 \times 2}{3}$
$\left(\because 1 / 3^{\mathrm{rd}}\right.$ amount is unpaid, $2 / 3^{\mathrm{rd}}$ amount is paid)
$\Rightarrow 2 x+29 d=\frac{2400}{15}$
$\Rightarrow 2 x+29 d=160$ ...(i)
Since total amount was 3600 and it was to be paid $1 n 40$ installment.
$\Rightarrow \frac{40}{2}[2 \mathrm{x}+39 \mathrm{~d}]=3600$
$\Rightarrow 2 \mathrm{x}+39 \mathrm{~d}=180$ ...(ii)
On solving eqs. (1) and (2), we get $\mathrm{x}=51$ and $\mathrm{d}=2$
First instalment $=$ Rs. 51

Topic: Mathematics, Sequences and Series, NDA, NDA 2008 (Phase 1)
After simplification, what is the number of terms in the expansion of $\left[(3 x+y)^{5}\right]^{4}-\left[(3 x-y)^{4}\right]^{5} ?$

Options:
A 4
B 5
C 10
D 11

The correct answer is:
10

Solution:
Given expression is:
$\left[(3 x+y)^{5}\right]^{4}-\left[(3 x-y)^{4}\right]^{5}=[(3 x+y)]^{20}-[(3 x-y)]^{20}$
First and second expansion will have 21 terms each but odd terms in second expansion be Ist, $3 \mathrm{rd}$, 5 th.....21st will be equal and opposite to those of first expansion. Thus, the number of terms in the expansion of above expression is 10

Topic: Mathematics, Binomial Theorem, NDA, NDA 2007 (Phase 2)
After subtending an angle of $1000^{\circ}$ from its inital position, the revolving line will be situated in which one of the following quadrants?

Options:
A First quadrant
B Second quadrant
C Third quadrant
D Fourth quadrant

The correct answer is:
Fourth quadrant

Solution:
$1000^{\circ}=2 \times 360^{\circ}+280^{\circ}$
It is clear that the revolving live will be in fourth quadrant.

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2009 (Phase 1)
Air column in two identical tubes is vibrating. Tube A has one end closed and tube B has both ends open. Neglecting end correction, the ratio of the fundamental frequency of air column in tube A to that in tube B is

Options:
A $1: 4$
B $4: 1$
C $1: 2$
D $2: 2$

The correct answer is:
$1: 2$

Solution:
The correct option is (C).
Consider the following diagram:
Velocity of a wave $v$ in a given medium is fixed, as $v=\lambda \mathrm{f}$, where $\lambda$ is the wavelength and $f$ is the frequency.
Therefore,
$f \propto \frac{1}{\lambda}$
The ratio of frequencies in case A and case B is given by,
$\frac{\mathrm{f}_1}{\mathrm{f}_2}=\frac{\lambda_2}{\lambda_1}=\frac{1}{2}$

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2022 (05 Aug Shift 1)
Air pollutants that produce photochemical smog include

Options:
A CO2, CO, SO2
B nitrous oxide and fluorocarbons
C ozone, PAN and nitrogen oxides
D O2, Cl, SO2

The correct answer is:
ozone, PAN and nitrogen oxides

Solution:
Smog is a mixture of fog with dust particles and smoke in the air of the environment. When the UV rays react with the nitrous acid to form smog that is called as photochemical smog. It is observed as a brown haze and very prominent during morning and afternoon. It also contains PAN and other secondary pollutants.

Topic: Biology, Environmental Issues, NEET
Alcohols react with active metals e.g. Na, K etc. to give corresponding alkoxides. What is the decreasing order of reactivity of sodium metal towards primary, secondary and tertiary alcohols?

Options:
A Decreasing order of reactivity of sodium metal is : 2° > 3° > 1°
B Decreasing order of reactivity of sodium metal is : 1° > 2° > 3°
C Decreasing order of reactivity of sodium metal is : 1° > 3° > 2°
D Decreasing order of reactivity of sodium metal is : 2° > 1° > 3°

The correct answer is:
Decreasing order of reactivity of sodium metal is : 1° > 2° > 3°

Solution:
Correct Option is : (B)
Decreasing order of reactivity of sodium metal is : 1° > 2° > 3°

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
Alcohols react with active metals e.g., $\mathrm{Na}, \mathrm{K}$ etc., to give corresponding alkoxides. Write down the decreasing order of reactivity of sodium metal towards primary, secondary and tertiary alcohols.

Options:

None

Solution:
Alcohols are acidic in nature while sodium metal is basic in nature. The reactivity of alcohol with sodium metal decreases with decrease in acidic strength of alcohol. The order of acidic strength of alcohol is as follows
$1^{\circ}$ alcohol $>2^{\circ}$ alcohol $>3^{\circ}$ alcohol
Therefore, the correct order of reactivity of sodium metal towards these alcohols will be as follows:
$1^{\circ}$ alcohol $>2^{\circ}$ alcohol $>3^{\circ}$ alcohol

Topic: Chemistry, Alcohols Phenols and Ethers
Alkali metals have negative reduction potential and hence they behave as

Options:
A oxidising agents
B Lewis bases
C reducing agents
D electrolytes

The correct answer is:
reducing agents

Solution:
Standard reduction potential is a measure of tendency of an element to lose electron in aqueous solution. More negative is the reduction potential, higher is the ability of element to lose electrons and hence stronger is the reducing character. Therefore, alkali metals behave as reducing agents.

Topic: Chemistry, Electrochemistry, KCET, KCET 2013
Alkyl cyanides undergo Stephen reduction to produce

Options:
A aldehyde
B secondary amine
C primary amine
D amide

The correct answer is:
aldehyde

Solution:
Stephen reduction is partial reduction of alkyl or aryl cyanides to give corresponding aldehydes with a suspension of anhydrous stannous chloride in ether.
$\mathrm{SnCl}_{2}+2 \mathrm{HCl} \longrightarrow \mathrm{SnCl}_{4}+2[\mathrm{H}]$

Topic: Chemistry, Aldehydes and Ketones, VITEEE, VITEEE 2008
Alkyl cyanides undergo Stephen reduction to produce

Options:
A Aldehyde
B Secondary amine
C Primary amine
D Amide

The correct answer is:
Aldehyde

Solution:
An alkyl cyanide is dissolved in ether or better in ethyl formate or ethyl acetate, and reduced with SnCl 2 and HCl and then steam distilled. The whole process is called Stephen reaction. In this process, alkyl cyanide is reduced to aldehyde.
There is no analogous method for the preparation of ketones.

Topic: Chemistry, Amines, NEET
All of the following statements are correct about active immunity, except

Options:
A It is slow and takes time to give its full effective response.
B Memory cells are formed.
C Antibodies are not formed during primary response.
D Exposure of antigen is a pre-requisite for the development of active immunity.

The correct answer is:
Antibodies are not formed during primary response.

Solution:
It is developed when a host is exposed to antigens (living or dead microbes/proteins) and antibodies are produced in the host body (both in primary and secondary immune response). Active immunity is slow and takes time to give its full effective response. It can be developed by injecting the microbes or by the natural entry of infectious organisms into the body.
Example: During Vaccination and Immunisation.

Topic: Biology, Human Health and Diseases, NEET
All of the following structures of the human respiratory system contains incomplete cartilaginous rings, except:

Options:
A Secondary bronchi and tertiary bronchi
B Trachea and primary bronchi
C Initial bronchioles and tertiary bronchi
D Distal bronchioles and terminal bronchioles

The correct answer is:
Distal bronchioles and terminal bronchioles

Solution:
The trachea, primary bronchi, secondary bronchi, tertiary bronchi and initial bronchioles are supported by incomplete cartilaginous rings.

Topic: Biology, Breathing and Exchange of Gases, NEET
All the given properties are found in the meristematic cell except

Options:
A It shows the ability to actively divide.
B Vacuole is very large and maybe more than one.
C It shows prominent nucleus and abundant cytoplasm.
D Meristematic cell has a thin cell wall of cellulose.

The correct answer is:
Vacuole is very large and maybe more than one.

Solution:
The meristematic cell has certain set of characteristic features. It is generally isodiametric in shape, has a prominent nucleus and abundant cytoplasm. Generally, these cells do not possess vacuoles. Their cell walls are composed of cellulose and are thin to facilitate cell division. The meristematic tissue contains compactly arranged cells.

Topic: Biology, Anatomy of Flowering Plants, NEET
All these statements about trophic levels and ecosystems are false except

Options:
A Trophic level represents a functional level, not a species as such
B A given species never occupies more than one trophic level in the same ecosystem at the same time
C In most ecosystems, producers are less in number and biomass than the herbivores
D Pyramid of energy can never be upright.

The correct answer is:
Trophic level represents a functional level, not a species as such

Solution:
Organisms occupy a specific position in an ecosystem based on their source of food which is known as trophic levels. In trophic levels, the first level is occupied by producers second by herbivores and third level by carnivores. These levels do not depict the species but their functional level. A sparrow becomes a primary consumer when it eats seeds and secondary consumers when it feeds on worms. Hence a given specie can occupy more than one trophic level at the same time in the ecosystem. In the most ecosystem, the pyramid of energy and biomass is upright as producers more in number and biomass than herbivores and herbivores are more in number and biomass than carnivores.

Topic: Biology, Ecosystem, NEET