datavorous/entrance-exam-dataset huggingface dataset:
Mean and standard deviation of 100 observations were found to be 40 and 10 , respectively. If at the time of calculation two observations were wrongly taken as 30 and 70 in place of 3 and 27 respectively, find the correct standard deviation.

Options:

None

Solution:
Here, $n=100, \bar{x}=40$, S.D. $=10$
$$
\begin{aligned}
&\Rightarrow \quad 40=\bar{x}=\frac{\Sigma x}{100} \Rightarrow \Sigma x=4000(\text { Incorrect }) \\
&\therefore \quad \text { Correct } \Sigma x=4000-30-70+3+27=3930 \\
&\therefore \quad \text { Correct Mean }=\frac{3930}{100}=39.3 \\
&\Rightarrow \quad \sigma=\sqrt{\frac{\sum x^2}{n}-\left(\frac{\Sigma x}{n}\right)^2} \\
&\Rightarrow \quad 10=\sqrt{\frac{\Sigma x^2}{100}-(40)^2} \\
&\Rightarrow \quad \Sigma x^2=170000 \\
&\therefore \text { Correct } \Sigma x^2=170000-(30)^2-(70)^2+(3)^2+(27)^2 \\
&\quad=164938 \\
&\therefore \text { Correct S.D. }
\end{aligned}
$$
$=\sqrt{\frac{164938}{100}-(39.3)^2}=\sqrt{104.89}=10.24$

Topic: Mathematics, Statistics
Mean and standard deviation of marks obtained in some particular subject by four classes are given below. Report the class with best performance

Options:
A 80,18
B 75,5
C 80,21
D 76,7

The correct answer is:
75,5

Solution:
Performance of the class will be best if mean of the marks obtained is maximum but standard deviation of the marks obtained is minimum.
Hence the class which has mean and standard deviation of the marks obtained as 75 and 5 respectively performs best.

Topic: Mathematics, Matrices, VITEEE, VITEEE 2008
Megaspore mother cell originates from

Options:
A superficial layer of nucellus
B hypodermal cell of nucellus
C inner layer of integument
D outer layer of integument

The correct answer is:
hypodermal cell of nucellus

Solution:
Megaspore mother cell is the cell that divides meiotically and forms the female gametophyte. The end result is the 7 celled and 8 nuclei embryo sac.
The ovule is surrounded by two layers of integuments and encloses a mass of cell called the nucellus tissue. It has abundant reserve of food materials. The megaspore mother cell which develops from the hypodermal cell of nucellus which is at the micropylar region of the ovule.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
Meiotic division of the secondary oocyte is completed

Options:
A At the time of copulation
B After zygote formation
C At the time of fusion of a sperm with an ovum
D Prior to ovulation

The correct answer is:
At the time of fusion of a sperm with an ovum

Solution:
The entry of sperm into the ovum induces completion of the meiotic division of the secondary oocyte. The second meiotic division is also unequal and results in the formation of a second polar body and a haploid ovum (ootid). Soon the haploid nucleus of the ovum fuses together to form a diploid zygote.

Topic: Biology, Human Reproduction, NEET, NEET 2020 (Phase 1)
Melting points are normally the highest for

Options:
A tertiary amides
B secondary amides
C primary amides
D amines.

The correct answer is:
primary amides

Solution:
Melting points are normally higher for primary amides due to presence of hydrogen bonding.

Topic: Chemistry, Amines, AIIMS, AIIMS 2004
Members of kingdom Protista are primarily ____.

Options:
A photosynthetic
B unicellular eukaryotes
C aquatic
D Both (B) and (C)

The correct answer is:
Both (B) and (C)

Solution:
All single-celled eukaryotes are placed under Protista, but the boundaries of this kingdom are not well defined. It includes Chrysophytes, Dinoflagellates, Euglenoids, Slime moulds and Protozoans under Protista. Members of Protista are primarily aquatic. This kingdom forms a link with the others dealing with plants, animals and fungi.

Topic: Biology, Biological Classification, NEET
Meristematic tissue in vascular bundle is

Options:
A phellem
B procambium
C interfascicular cambium
D fascicular cambium

The correct answer is:
fascicular cambium

Solution:
The cambium present between the xylem and phloem of vascular bundle is called vascular cambium. It is a lateral meristem that gives rise to secondary xylem and phloem and occurs in the form of a thin strip. Vascular cambium is also called fascicular cambium.

Topic: Biology, Anatomy of Flowering Plants, JIPMER, JIPMER 2008
Methanogens do not produce

Options:
A
oxygen

B
methane

C
hydrogen sulphide

D
carbon dioxide

The correct answer is:
oxygen

Solution:
. During secondary treatment of waste water in the digesters, heterotrophic microbes (methanogens) anaerobically digest bacteria and fungi in sludge producing mixture of gases such as methane, hydrogen sulphide and $\mathrm{CO}_2$ which form the biogas. The $\mathrm{O}_2$ is not produced by methanogens.

Topic: Biology, Microbes in Human Welfare
Milk-water mixture when viewed by the reflected light appears in which colour?

Options:
A Yellow
B Green
C Blue
D Red

The correct answer is:
Blue

Solution:
Tyndall Effect - scattering phenomenon.

Topic: Chemistry, Surface Chemistry, AP EAMCET, AP EAMCET 2022 (06 Jul Shift 1)
Molar volume is the volume occupied by 1 mole of any ideal gas at standard temperature and pressure (STP) which are 1 atmospheric pressure and $0^{\circ} \mathrm{C}$. Show that it is $22.4$ litres.

Options:

None

Solution:
For one mole of an ideal gas, $P V=R T$ $\therefore V=\frac{R T}{P}$ [Where, $R=8.31 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$, $T=273 \mathrm{~K}, P=1$ atmosphere $\left.=1.013 \times 10^5 \mathrm{Nm}^{-2}\right]$ $=\frac{8.31 \times 273}{1.013 \times 10^5}=0.0224 \mathrm{~m}^3=22.4$ litres.

Topic: Physics, Kinetic Theory of Gases
Most animals in deep oceanic water are

Options:
A Primary consumers
B Secondary consumers
C Tertiary consumers
D Detrivores

The correct answer is:
Detrivores

Solution:
The animals like crustaceans, polychaetes, and microorganisms that are living in deep oceanic waters are called benthos and are also called as scavengers or detrivores.

Topic: Biology, Ecosystem, NEET
Most animals that live in deep oceanic waters are

Options:
A tertiary consumers
B detritivores
C primary consumers
D secondary consumers.

The correct answer is:
detritivores

Solution:
Benthos is the community of organisms which live on, in, or near the sea bed, also known as the benthic zone. As no light is available in this zone of sea, the energy source for deep benthic ecosystem is often organic matter from higher water column which drifts down to the depths. This dead and decaying matter sustains the benthic organisms, and therefore, most organisms in benthic zone i.e., in deep oceanic waters, are scavengers or detritivores.

Topic: Biology, Ecosystem, AIIMS, AIIMS 2016
Most common isotopes of hydrogen (non-radioactive) is

Options:
A protium
B deuterium
C titrium
D All of these

The correct answer is:
protium

Solution:
Protium $\left({ }_1^1 H\right)$ deuterium or heavy hydrogen $\left({ }_1^2 \mathrm{H}\right.$ or D$)$ and tritium $\left({ }_1^3 \mathrm{H}\right.$ or T $)$ are the isotopes of hydrogen.
Protium or ordinary hydrogen has one proton and no neutron in the nucleus and one electron revolves around the nucleus. It is the most abundant and common form of hydrogen (approximately $99.98 \%$ ).
${ }_1^2 \mathrm{H}+{ }_{\mathrm{l}}^2 \mathrm{H} \longrightarrow{ }_1^3 \mathrm{H}+{ }_1^1 \mathrm{H}$

Topic: Chemistry, Hydrogen, JIPMER, JIPMER 2019
Most of the biological activities of proteins are due to:

Options:
A Primary configuration
B Secondary configuration
C Tertiary configuration
D More than one option is correct

The correct answer is:
Tertiary configuration

Solution:
Primary proteins are linear cahins of amino acids, in secondary proteins coiling or folding occurs. In tertiary proteins complex coiling and folding occurs which makes the structure more stable. Most enzymatic proteins have tertiary structure. In quaternary structures more than 2 polypeptide are coiled together.

Topic: Chemistry, Biomolecules, NEET
$n_{1}$ is the frequency of the pipe closed at one end and $n_{2}$ is the frequency of the pipe open at both ends. If both are joined end to end, find the fundamental frequency of closed pipe so formed

Options:
A $\frac{n_{1} n_{2}}{n_{2}+2 n_{1}}$
B $\frac{n_{1} n_{2}}{2 n_{2}+n_{1}}$
C $\frac{n_{1}+2 n_{2}}{n_{2} n_{1}}$
D $\frac{2 n_{1}+n_{2}}{n_{2} n_{1}}$

The correct answer is:
$\frac{n_{1} n_{2}}{n_{2}+2 n_{1}}$

Solution:
Frequency of closed pipe, $n_{1}=\frac{v}{4 l_{1}} \Rightarrow l_{1}=\frac{v}{4 n_{1}}$
Frequency of open pipe, $n_{2}=\frac{v}{2 l_{1}} \Rightarrow l_{2}=\frac{v}{2 n_{2}}$
when both pipe are joined then length of closed pipe
$$
\begin{array}{c}
l=l_{1}+l_{2} \\
\frac{v}{4 n}=\frac{v}{4 n_{1}}+\frac{v}{2 n_{2}}
\end{array}
$$
or $\quad \frac{1}{2 n}=\frac{1}{2 n_{1}}+\frac{1}{n_{2}}$
or
$$
\frac{1}{2 n}=\frac{n_{2}+2 n_{1}}{2 n_{1} n_{2}}
$$
or
$$
n=\frac{n_{1} n_{2}}{n_{2}+2 n_{1}}
$$

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2010
$$
\underset{\mathrm{N}}{\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}} \frac{\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}}{\mathrm{H}_{2} \mathrm{SO}_{4}} \mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O} \cdot \frac{\mathrm{I}_{2} / \mathrm{NaOH}}{\mathrm{Warm}}, \mathrm{CHI}_{3}
$$ Here, N is

Options:
A
B
C
D

The correct answer is:

Solution:
Given sequence of chemical equation is
$\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O} \frac{\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}}{\mathrm{H}_{2} \mathrm{SO}_{4}}-\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O} \frac{\mathrm{I}_{2} / \mathrm{NaOH}}{\text { Warm }} \mathrm{CHI}_{3}$
$(\mathrm{N})$
Here N will be $=$ Butan- $2 \cdot 0$ l) Here,
Step I It involves the oxidation of secondary alcohol to corresponding alcohol in the presence of $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ and $\mathrm{H}_{2} \mathrm{SO}_{4}$

Step II It involves the lodoform reaction. Ketone formed in step 1 have one methyl group linked to the carbonyl carbon atom are oxidised by sodium hypohalite to sodium salts of corresponding carboxylic acids having one carbon atom less than that of carbonyl compound.

Topic: Chemistry, Alcohols Phenols and Ethers, WBJEE, WBJEE 2018
Name an organism found as secondary carnivore in an aquatic ecosystem.

Options:

None

Solution:
The food chain in an aquatic ecosystem can be drawn as follows: Producer (Phytoplankton) $\rightarrow$ Primary consumer(Zooplankton) $\rightarrow$ Secondary consumer (small fish and water beetle) $\rightarrow$ Tertiary consumer (large fish, water birds like fowl and duck). The large fish and water birds like ducks and water fowls which are the tertiary consumers in an aquatic ecosystem occupy the position of secondary carnivore.

Topic: Biology, Ecosystem
Name the following halide according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides: (CH3)2CHCH(Cl)CH3

Options:
A 2-chloro-2-methylbutane and it is a primary alkyl halide
B 2-chloro-3-methylbutane and it is a primary alkyl halide
C 2-chloro-2-methylbutane and it is a secondary alkyl halide
D 2-chloro-3-methylbutane and it is a secondary alkyl halide

The correct answer is:
2-chloro-3-methylbutane and it is a secondary alkyl halide

Solution:
Correct Option is : (D)
2-chloro-3-methylbutane and it is a secondary alkyl halide

Topic: Chemistry, Haloalkanes and Haloarenes, NEET
Name the following halide according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides: (CH3)3CCH2CH(Br)C6H5

Options:
A 1-Bromo-3,3-dimethyl-1-phenylbutane and it is a tertiary benzyl halide
B 1-Bromo-2,3-dimethyl-1-phenylbutane and it is a secondary benzyl halide
C 1-Bromo-3,3-dimethyl-1-phenylbutane and it is a secondary benzyl halide
D 1-Bromo-2,3-dimethyl-1-phenylbutane and it is a primary benzyl halide

The correct answer is:
1-Bromo-3,3-dimethyl-1-phenylbutane and it is a secondary benzyl halide

Solution:
Correct Option is : (C)
1-Bromo-3,3-dimethyl-1-phenylbutane and it is a secondary benzyl halide

Topic: Chemistry, Haloalkanes and Haloarenes, NEET
Name the following halide according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides: CH3C(C2H5)2CH2Br

Options:
A 1-bromo-2-ethyl-2-methylbutane and it is a primary allyl halide
B 1-bromo-2-ethyl-2-methylbutane and it is a primary alkyl halide
C 1-bromo-1-ethyl-2-methylbutane and it is a primary alkyl halide
D 1-bromo-2-ethyl-2-methylbutane and it is a secondary alkyl halide

The correct answer is:
1-bromo-2-ethyl-2-methylbutane and it is a primary alkyl halide

Solution:
Correct Option is : (B)
1-bromo-2-ethyl-2-methylbutane and it is a primary alkyl halide

Topic: Chemistry, Haloalkanes and Haloarenes, NEET
Name the following halide according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides: CH3C(Cl)(C2H5)CH2CH3

Options:
A 2-chloro-3-methylpentane and it is a tertiary alkyl halide
B 3-chloro-3-methylpentane and it is a secondary alkyl halide
C 2-chloro-3-methylpentane and it is a secondary alkyl halide
D 3-chloro-3-methylpentane and it is a tertiary alkyl halide

The correct answer is:
3-chloro-3-methylpentane and it is a tertiary alkyl halide

Solution:
Correct Option is : (D)
3-chloro-3-methylpentane and it is a tertiary alkyl halide

Topic: Chemistry, Haloalkanes and Haloarenes, NEET
Name the following halide according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides: CH3CH2C(CH3)2CH2I

Options:
A 1-Iodo-2,2-dimethylbutane and it is a secondary alkyl halide
B 1-Iodo-2,2-dimethylbutane and it is a primary alkyl halide
C 1-Iodo-1,2-dimethylbutane and it is a secondary alkyl halide
D 1-Iodo-1,2-dimethylbutane and it is a primary alkyl halide

The correct answer is:
1-Iodo-2,2-dimethylbutane and it is a primary alkyl halide

Solution:
Correct Option is : (B)
1-Iodo-2,2-dimethylbutane and it is a primary alkyl halide

Topic: Chemistry, Haloalkanes and Haloarenes, NEET
Name the following halide according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides: CH3CH2CH(CH3)CH(C2H5)Cl

Options:
A 3-chloro-4-methyl-hexane and it is a secondary alkyl halide
B 3-chloro-3-methyl-hexane and it is a secondary alkyl halide
C 3-chloro-4-methyl-hexane and it is a primary alkyl halide
D 3-chloro-3-methyl-hexane and it is a primary alkyl halide

The correct answer is:
3-chloro-4-methyl-hexane and it is a secondary alkyl halide

Solution:
Correct Option is : (A)
3-chloro-4-methyl-hexane and it is a secondary alkyl halide

Topic: Chemistry, Haloalkanes and Haloarenes, NEET
Name the following halide according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides: CH3CH=C(Cl)CH2CH(CH3)2

Options:
A 3-chloro-5-methylhex-2-ene and it is a aryl halide
B 4-chloro-5-methylhex-2-ene and it is a vinyl halide
C 3-chloro-5-methylhex-2-ene and it is a vinyl halide
D 3-chloro-4-methylhex-2-ene and it is a aryl halide

The correct answer is:
3-chloro-5-methylhex-2-ene and it is a vinyl halide

Solution:
Correct Option is : (C)
3-chloro-5-methylhex-2-ene and it is a vinyl halide

Topic: Chemistry, Haloalkanes and Haloarenes, NEET
Name the following halide according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides: CH3CH(CH3)CH(Br)CH3

Options:
A 2-bromo-3-methyl-butane and it is a secondary alkyl halide
B 2-bromo-2-methyl-butane and it is a secondary alkyl halide
C 2-bromo-3-methyl-butane and it is a primary alkyl halide
D 2-bromo-3-methyl-butane and it is a tertiary alkyl halide

The correct answer is:
2-bromo-3-methyl-butane and it is a secondary alkyl halide

Solution:
Correct Option is : (A)
2-bromo-3-methyl-butane and it is a secondary alkyl halide

Topic: Chemistry, Haloalkanes and Haloarenes, NEET
Name the following halide according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides: CH3CH=CHC(Br)(CH3)2

Options:
A 4-bromo-4-methylpent-2-ene and it is a allyl halide
B 3-bromo-3-methylpent-2-ene and it is a allyl halide
C 4-bromo-4-methylpent-2-ene and it is a vinyl halide
D 4-bromo-3-methylpent-2-ene and it is a vinyl halide

The correct answer is:
4-bromo-4-methylpent-2-ene and it is a allyl halide

Solution:
Correct Option is : (A)
4-bromo-4-methylpent-2-ene and it is a allyl halide

Topic: Chemistry, Haloalkanes and Haloarenes, NEET
Name the following halide according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides: m-ClCH2C6H4CH2C(CH3)3

Options:
A 1-chloromethyl-3-(2,2-dimethylpropyl)-benzene and it is a secondary benzyl halide
B 1-chloromethyl-3-(2,2-dimethylpropyl)-benzene and it is a primary benzyl halide
C 1-chloromethyl-3-(2,2-dimethylpropyl)-benzene and it is a tertiary benzyl halide
D 1-chloromethyl-3-(2,4-dimethylpropyl)-benzene and it is a primary benzyl halide

The correct answer is:
1-chloromethyl-3-(2,2-dimethylpropyl)-benzene and it is a primary benzyl halide

Solution:
Correct Option is : (B)
1-chloromethyl-3-(2,2-dimethylpropyl)-benzene and it is a primary benzyl halide

Topic: Chemistry, Haloalkanes and Haloarenes, NEET
Name the following halide according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides: o-Br-C6H4CH(CH3)CH2CH3

Options:
A 1-bromo-2-(3-methylpropyl)-benzene and it is a vinyl halide
B 1-bromo-2-(1-methylpropyl)-benzene and it is a vinyl halide
C 1-bromo-2-(2-methylpropyl)-benzene and it is a aryl halide
D 1-bromo-2-(1-methylpropyl)-benzene and it is a aryl halide

The correct answer is:
1-bromo-2-(1-methylpropyl)-benzene and it is a aryl halide

Solution:
Correct Option is : (D)
1-bromo-2-(1-methylpropyl)-benzene and it is a aryl halide

Topic: Chemistry, Haloalkanes and Haloarenes, NEET
Name the following halide according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides: p-ClC6H4CH2CH(CH3)2

Options:
A 1-chloro-4-(2-methylpropyl)-benzene and it is a paimary alkyl halide
B 1-chloro-3-(2-methylpropyl)-benzene and it is a aryl halide
C 1-chloro-4-(2-methylpropyl)-benzene and it is a aryl halide
D 1-chloro-2-(2-methylpropyl)-benzene and it is a vinyl halide

The correct answer is:
1-chloro-4-(2-methylpropyl)-benzene and it is a aryl halide

Solution:
Correct Option is : (C)
1-chloro-4-(2-methylpropyl)-benzene and it is a aryl halide

Topic: Chemistry, Haloalkanes and Haloarenes, NEET
Name the following halides according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides:
(i) \(\left(\mathrm{CH}_3\right)_2 \mathrm{CHCH}_{(\mathrm{Cl}) \mathrm{CH}_3}\)
(ii) \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}\left(\mathrm{CH}_3\right) \mathrm{CH}\left(\mathrm{C}_2 \mathrm{H}_5\right) \mathrm{Cl}\)
(iii) \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{C}\left(\mathrm{CH}_3\right)_2 \mathrm{CH}_2 \mathrm{I}\)
(iv) \(\left(\mathrm{CH}_3\right)_3 \mathrm{CCH}_2 \mathrm{CH}(\mathrm{Br}) \mathrm{C}_6 \mathrm{H}_5\)
(v) \(\mathrm{CH}_3 \mathrm{CH}\left(\mathrm{CH}_3\right) \mathrm{CH}(\mathrm{Br}) \mathrm{CH}_3\)
(vi) \(\mathrm{CH}_3 \mathrm{C}\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{CH}_2 \mathrm{Br}\)
(vii) \(\mathrm{CH}_3 \mathrm{C}(\mathrm{Cl})\left(\mathrm{C}_2 \mathrm{H}_5\right) \mathrm{CH}_2 \mathrm{CH}_3\)
(viii) \(\mathrm{CH}_3 \mathrm{CH}=\mathrm{C}\left(\mathrm{Cl}_2 \mathrm{CH}_2 \mathrm{CH}\left(\mathrm{CH}_3\right)_2\right.\)
(ix) \(\mathrm{CH}_3 \mathrm{CH}=\mathrm{CHC}(\mathrm{Br})\left(\mathrm{CH}_3\right)_2\)
(x) \(\mathrm{p}-\mathrm{ClC}_6 \mathrm{H}_4 \mathrm{CH}_2 \mathrm{CH}\left(\mathrm{CH}_3\right)_2\)
(xi) \(\mathrm{m}-\mathrm{ClCH}_2 \mathrm{C}_6 \mathrm{H}_4 \mathrm{CH}_2 \mathrm{C}\left(\mathrm{CH}_3\right)_3\)
(xii) \(o-\mathrm{Br}-\mathrm{C}_6 \mathrm{H}_4 \mathrm{CH}\left(\mathrm{CH}_3\right) \mathrm{CH}_2 \mathrm{CH}_3\)

Options:

None

Solution:
(i) 2-Chloro-3methylbutane, \(2^{\circ}\) alkyl halide
(ii) 3-Chloro-4methyl hexane, \(2^{\circ}\) alkyl halide
(iii) 1-Iodo-2,2-dimethylbutane, \(1^{\circ}\) alkyl halide
(iv) 1-Bromo-3,3-dimethyl-1-phenylbutane, \(2^{\circ}\) benzylic halide
(v) 2-Bromo-3-methylbutane, \(2^{\circ}\) alkyl halide
(vi) 1-Bromo-2-ethyl-2-methylbutane, \(1^{\circ}\) alkyl halide
(vii) 3-Chloro-3-methylpentane, \(3^{\circ}\) alkyl halide
(viii) 3-Chloro-5-methylhex-2-ene, vinylic halide
(ix) 4-Bromo-4-methylpent-2-ene, allylic halide
(x) 1-Chloro-4-(2-methylpropyl) benzene, aryl halide
(xi) 1-Chloromethyl-3- (2,2-dimethylpropyl) benzene, \(1^{\circ}\) benzylic halide.
(xii) 1-Bromo-2-(1-methylpropyl) benzene, aryl halide.

Topic: Chemistry, Haloalkanes and Haloarenes
Name the primary and secondary lymphoid organs.

Options:

None

Solution:
Primary lymphoid organs - Bone marrow and thymus.
Secondary lymhoid organs $-$ Spleen, lymph nodes, tonsils.

Topic: Biology, Human Health and Diseases
Name the three basic tissue systems in the flowering plants. Give the tissue names under each system.

Options:

None

Solution:
There are three basic tissue systems viz., epidermal, ground and vascular tissue system.
(i) Epidermal tissue system-Epidermis, stomata and epidermal appendages.
(ii) Fundamental or ground tissue system - Cortex (hypodermis, general cortex and endodermis), pericycle and pith.
(iii) Vascular tissue system - Xylem and phloem.

Topic: Biology, Anatomy of Flowering Plants
Natural abundances of ${ }^{12} \mathrm{C}$ and ${ }^{13} \mathrm{C}$ isotopes of carbon are $99 \%$ and $1 \%$, respectively. Assuming they only contributes to the mol. wt. of $\mathrm{C}_{2} \mathrm{F}_{4}$, the percentage of $\mathrm{C}_{2} \mathrm{F}_{4}$ having a molecular mass of $101$ is

Options:
A $1.98$
B $98$
C $0.198$
D $99$

The correct answer is:
$1.98$

Solution:
Percentage of $\mathrm{C}_{2} \mathrm{F}_{4}$ of Molar mass $100=\frac{1}{100} \times \frac{1}{100} \times 100=0.01 \%$
Percentage of $\mathrm{C}_{2} \mathrm{F}_{4}$ of Molar mass $102=\frac{99}{100} \times \frac{99}{100} \times 100=98.01 \%$
Percentage of $\mathrm{C}_{2} \mathrm{F}_{4}$ of Molar mass $101=100-(0.01+98.01)=1.98 \%$

Topic: Chemistry, Some Basic Concepts of Chemistry, KVPY, KVPY 2013 (SA)
Natural numbers are divided into groups as $(1),(2,3)$ $(4,5,6),(7,8,9,10)$ and so on. What is the sum of the numbers in the 11 th group?

Options:
A 605
B 615
C 67
D 693

The correct answer is:
67

Solution:
The 11 th term of the group is $(56,57,58,59,60,61,62,63,64,65,66)$
$\mathrm{sum}=\frac{n\left(n^{2}+1\right)}{2}=\frac{11(121+1)}{2}=11 \times 61=671$
Their sum is 671

Topic: Mathematics, Sets and Relations, NDA, NDA 2009 (Phase 2)
Nitroalkanes are obtained in laboratory from primary or secondary alkyl halides by the action of

Options:
A AgNO2
B NaNO3
C AgNO3
D HNO3

The correct answer is:
AgNO2

Solution:
RX+AgNO2→RNO2NitroalkaneMajorproduct+RONOAlkylNitriteMinorproduct+AgX

Topic: Chemistry, Haloalkanes and Haloarenes, MHT CET, MHT CET 2019 (02 May Shift 1)
Nitroglycerine MW=227.1 detonates according to the following equations,

2C3H5NO33l⟶3 N2g+12O2g+6CO2g+5H2Og

The standard molar enthalpies of formation, ΔHf° for all the compounds are given below.

ΔHf°C3H5NO33=-364 kJ/mol
ΔHf°CO2g=-393.5 kJ/mol
ΔHf°H2Og=-241.8 kJ/mol
ΔHf°N2g=0 kJ/mol
ΔHf°O2g=0 kJ/mol

The enthalpy change when 10 g of nitroglycerine is detonated is

Options:
A -100.5 kJ/mol
B -62.5 kJ/mol
C -80.3 kJ/mol
D -74.9 kJ/mol

The correct answer is:
-62.5 kJ/mol

Solution:
ΣHReaction°=ΣHfP°-ΣHf(R)°

ΔHReaction°=3×0+12×0+6×-393.5+5×-241.8-2×-364

=-2361-1209+728

=-2842 kJ/mol

For 2 moles of nitroglycerine enthalpy
-2842 kJ/mol

∴ For 1 mole of nitroglycerine enthalpy
=-28422=-1421 kJ/mol

227.1 g of C3H5NO33 has enthalpy -1421 kJ/mol.

∴10 g of C3H5NO33 has enthalpy

=-1421227.1×10

=-62.5 kJ/mol

Topic: Chemistry, Thermodynamics (C), KVPY, KVPY 2018 (SB/SX)