datavorous/entrance-exam-dataset huggingface dataset:
\begin{array}{|c|c|c|c|c|c|c|c|}
\hline \multirow{2}{*} { Year } & \multicolumn{3}{|c|} { Male } & \multicolumn{2}{|c|} { Female } & \multirow{2}{*} { Total } \\
\cline { 2 - 7 } & & Urban & Rural & Total & Urban & Rural & Total & \\
\hline 1995 & 280 & 350 & & & 310 & & 1350 \\
\hline 1996 & 370 & & 670 & 180 & & 450 & \\
\hline 1997 & & 130 & 440 & & 190 & & \\
\hline 1998 & 400 & 280 & & 290 & & & \\
\hline Total & & & & 1060 & 850 & & \\
\hline
\end{array}
What is the urban population in the year 1997?

Options:
A 400
B 460
C 490
D 510

The correct answer is:
490

Solution:
Complete table is
\begin{array}{|c|c|c|c|c|c|c|c|}
\hline \multirow{2}{*} { Year } & \multicolumn{3}{|c|} { Male } & \multicolumn{3}{|c|} { Female } & \multirow{2}{*} { Total } \\
\cline { 2 - 8 } & Urban & Rural & Total & Urban & Rural & Total & \\
\hline 1995 & 280 & 350 & 630 & 410 & 310 & 720 & 1350 \\
\hline 1996 & 370 & 300 & 670 & 180 & 270 & 450 & 1120 \\
\hline 1997 & 310 & 130 & 440 & 180 & 190 & 370 & 810 \\
\hline 1998 & 400 & 280 & 680 & 290 & 80 & 370 & 1050 \\
\hline Total & 1360 & 1060 & 2420 & 1060 & 850 & 1910 & 4330 \\
\hline
\end{array}
Urban population in the yean $1997=310+180=490 .$

Topic: Mathematics, Statistics, NDA, NDA 2011 (Phase 2)
\begin{array}{|c|c|c|c|c|c|c|c|}
\hline \multirow{2}{*} { Year } & \multicolumn{3}{|c|} { Male } & \multicolumn{2}{|c|} { Female } & \multirow{2}{*} { Total } \\
\cline { 2 - 7 } & & Urban & Rural & Total & Urban & Rural & Total & \\
\hline 1995 & 280 & 350 & & & 310 & & 1350 \\
\hline 1996 & 370 & & 670 & 180 & & 450 & \\
\hline 1997 & & 130 & 440 & & 190 & & \\
\hline 1998 & 400 & 280 & & 290 & & & \\
\hline Total & & & & 1060 & 850 & & \\
\hline
\end{array}
What is the percentage of rural male population (over the whole population) in the year $1998 ?$

Options:
A $\frac{80}{3} \%$
B $\frac{100}{3} \%$
C $35 \%$
D $40 \%$

The correct answer is:
$\frac{80}{3} \%$

Solution:
Complete table is
\begin{array}{|c|c|c|c|c|c|c|c|}
\hline \multirow{2}{*} { Year } & \multicolumn{3}{|c|} { Male } & \multicolumn{3}{|c|} { Female } & \multirow{2}{*} { Total } \\
\cline { 2 - 8 } & Urban & Rural & Total & Urban & Rural & Total & \\
\hline 1995 & 280 & 350 & 630 & 410 & 310 & 720 & 1350 \\
\hline 1996 & 370 & 300 & 670 & 180 & 270 & 450 & 1120 \\
\hline 1997 & 310 & 130 & 440 & 180 & 190 & 370 & 810 \\
\hline 1998 & 400 & 280 & 680 & 290 & 80 & 370 & 1050 \\
\hline Total & 1360 & 1060 & 2420 & 1060 & 850 & 1910 & 4330 \\
\hline
\end{array}
Total rural male population in the year $1998=280$ Required $\%=\frac{280}{1050} \times 100 \%=\frac{28}{105} \times 100 \%$
$=\frac{28 \times 20}{21} \%=\frac{4 \times 20}{3} \%=\frac{80}{3} \%$

Topic: Mathematics, Statistics, NDA, NDA 2011 (Phase 2)
\begin{array}{|c|c|c|c|c|c|c|c|}
\hline \multirow{2}{*} { Year } & \multicolumn{3}{|c|} { Male } & \multicolumn{2}{|c|} { Female } & \multirow{2}{*} { Total } \\
\cline { 2 - 7 } & & Urban & Rural & Total & Urban & Rural & Total & \\
\hline 1995 & 280 & 350 & & & 310 & & 1350 \\
\hline 1996 & 370 & & 670 & 180 & & 450 & \\
\hline 1997 & & 130 & 440 & & 190 & & \\
\hline 1998 & 400 & 280 & & 290 & & & \\
\hline Total & & & & 1060 & 850 & & \\
\hline
\end{array}
In which year is the female population maximum?

Options:
A 1995
B 1996
C 1997
D 1998

The correct answer is:
1995

Solution:
Complete table is
\begin{array}{|c|c|c|c|c|c|c|c|}
\hline \multirow{2}{*} { Year } & \multicolumn{3}{|c|} { Male } & \multicolumn{3}{|c|} { Female } & \multirow{2}{*} { Total } \\
\cline { 2 - 8 } & Urban & Rural & Total & Urban & Rural & Total & \\
\hline 1995 & 280 & 350 & 630 & 410 & 310 & 720 & 1350 \\
\hline 1996 & 370 & 300 & 670 & 180 & 270 & 450 & 1120 \\
\hline 1997 & 310 & 130 & 440 & 180 & 190 & 370 & 810 \\
\hline 1998 & 400 & 280 & 680 & 290 & 80 & 370 & 1050 \\
\hline Total & 1360 & 1060 & 2420 & 1060 & 850 & 1910 & 4330 \\
\hline
\end{array}
In 1995 , the female population is maximum.

Topic: Mathematics, Statistics, NDA, NDA 2011 (Phase 2)
\begin{array}{|c|c|c|c|c|c|c|c|}
\hline \multirow{2}{*} { Year } & \multicolumn{3}{|c|} { Male } & \multicolumn{2}{|c|} { Female } & \multirow{2}{*} { Total } \\
\cline { 2 - 7 } & & Urban & Rural & Total & Urban & Rural & Total & \\
\hline 1995 & 280 & 350 & & & 310 & & 1350 \\
\hline 1996 & 370 & & 670 & 180 & & 450 & \\
\hline 1997 & & 130 & 440 & & 190 & & \\
\hline 1998 & 400 & 280 & & 290 & & & \\
\hline Total & & & & 1060 & 850 & & \\
\hline
\end{array}
What is the total population in the year $1998 ? \quad$

Options:
A 1000
B 1020
C 1040
D 1050

The correct answer is:
1050

Solution:
Complete table is
\begin{array}{|c|c|c|c|c|c|c|c|}
\hline \multirow{2}{*} { Year } & \multicolumn{3}{|c|} { Male } & \multicolumn{3}{|c|} { Female } & \multirow{2}{*} { Total } \\
\cline { 2 - 8 } & Urban & Rural & Total & Urban & Rural & Total & \\
\hline 1995 & 280 & 350 & 630 & 410 & 310 & 720 & 1350 \\
\hline 1996 & 370 & 300 & 670 & 180 & 270 & 450 & 1120 \\
\hline 1997 & 310 & 130 & 440 & 180 & 190 & 370 & 810 \\
\hline 1998 & 400 & 280 & 680 & 290 & 80 & 370 & 1050 \\
\hline Total & 1360 & 1060 & 2420 & 1060 & 850 & 1910 & 4330 \\
\hline
\end{array}
Total population in the year $1998=1050$.

Topic: Mathematics, Statistics, NDA, NDA 2011 (Phase 2)
\[
\begin{array}{l}
M n O_{2}+H C l-\Delta \\
A_{(g)}+F_{2}(\text { excess })_{--}>->B_{(g)} \\
B_{(l)}+U_{(s)}-\cdots C_{(g)}+D_{(g)}
\end{array}
\]
The gases \( A, B, C \) and \( D \) are respectively

Options:
A \( \mathrm{Cl}_{2}, \mathrm{ClF}_{3}, \mathrm{UF}_{6}, \mathrm{ClF} \)
B \( \mathrm{O}_{2}, \mathrm{O}_{2} \mathrm{~F}_{2}, \mathrm{U}_{2} \mathrm{O}_{3}, \mathrm{OF}_{2} \)
C \( \mathrm{Cl}_{2}, \mathrm{ClF}, U \mathrm{~F}_{6}, \mathrm{ClF}_{3} \)
D \( \mathrm{O}_{2}, \mathrm{OF}_{2}, \mathrm{U}_{2} \mathrm{O}_{3}, \mathrm{O}_{2} \mathrm{~F}_{2} \)

The correct answer is:
\( \mathrm{Cl}_{2}, \mathrm{ClF}_{3}, \mathrm{UF}_{6}, \mathrm{ClF} \)

Solution:
\( \mathrm{MnO}_{2}+4 \mathrm{HCl}_{-}>\underset{(A)}{\Delta} \mathrm{Cl}_{2}(g)+\mathrm{MnCl}_{2}+2 \mathrm{H}_{2} \mathrm{O} \)
\( \mathrm{Cl}_{2}(g)+3 \mathrm{~F}_{2} \) (excess) \( \quad{ }_{573 \mathrm{~K}}^{(A)} \rightarrow 2 \mathrm{ClF}_{3}(g) \)

Topic: Chemistry, Electrochemistry, KCET, KCET 2014
Below is a list of plant fibres. From which part of the plant these are obtained.
(a) Coir
(b) Hemp
(c) Cotton
(d) Jute

Options:

None

Solution:
(a) Coir is a natural fibre obtained from coconut husk. It is the fibrous mesoderm of the fruit of Cocos nucifera (coconut).
(b) Hemp fibre is obtained from the stems of Cannabis sativa. It is the bast fibre (soft or stem fibre) obtained from secondary phloem.
(c) Cotton fibre is the epidermal growth in cotton (Gossypium hirsutum) seed. It is an elongated structure made up of cellulose.
(d) Jute is a natural bast fibre made up of cellulose and lignin obtained from Corchorus capsularis.

Topic: Biology, Anatomy of Flowering Plants
Benzene reacts with n-propyl chloride in the presence of anhydrous AlCl3 to give predominantly

Options:
A n-Propyl benzene
B Isopropyl benzene
C 3-Propyl-1-chloro benzene
D 1-Chloro-3-n-propyl benzene

The correct answer is:
Isopropyl benzene

Solution:
This reaction is known as Friedel Craft's alkylation reaction. in presence of anhydrous AlCl3, n-propyl chloride introduce an alkylating agent. It is an electrophilic substitution reaction.

Reaction occurs in following step.

Generation of electrophile.

CH3CH2CH2Cl+AlCl3→CH3CH2CH2++AlCl4-

Carbocation formed in step 1 rearranges to form secondary carbocation.

CH3CH2CH2+↔ CH3CH+CH3 1° 2°

This secondary carbocation attack on benzene ring.

Next step is the abstraction of proton from conjugate base

Hence, the product is isopropyl benzene.

Topic: Chemistry, Hydrocarbons, TS EAMCET, TS EAMCET 2021 (05 Aug Shift 1)
Benzenesulfonyl chloride (Hinsberg's reagent) reacts with which type of amines to form sulphonamides?

Options:
A Primary and secondary amines
B Primary amines only
C Secondary amines only
D Tertiary amines only

The correct answer is:
Primary and secondary amines

Solution:
Benzenesulfonyl chloride reacts with both primary and secondary amines to form sulphonamides.

Topic: Chemistry, Amines, NEET
Besides Annelida and Arthropoda, true metamerism is found in

Options:
A Ctenophora
B Mollusca
C Chordata
D Porifera

The correct answer is:
Chordata

Solution:
True segmentation metamerism is found in Chordata, Annelida and Arthropoda. Metamerism is the phenomenon of having a linear series of body segments fundamentally similar in structure, though not all such structures are entirely alike in any single life form because some of them perform special functions. In animals, metameric segments are referred to as somites or metameres.

Topic: Biology, Animal Kingdom, NEET
Between the primary and secondary rainbows, there is a dark band known as Alexandar's dark band. This is because

Options:
A
light scattered into this region interfere destructively

B
there is no light scattered into this region

C
light is absorbed in this region

D
angle made at the eye by the scattered rays with respect to the incident light of the sun lies between approximately $42^{\circ}$ and $50^{\circ}$

The correct answers are:
light scattered into this region interfere destructively
,
angle made at the eye by the scattered rays with respect to the incident light of the sun lies between approximately $42^{\circ}$ and $50^{\circ}$

Solution:
Between the primary and secondary rainbows, there is a dark band known as Alexandar's dark, this forms due to light scattered into this region interfere destructively. Since, primary and secondary rainbows subtends an angle nearly $41^{\circ}$ to $42^{\circ}$ and $51^{\circ}$ to $54^{\circ}$ respectively at observer's eye w.r.t. incident light ray. Then, the scattered rays with respect to the incident ray of the sun lies between approximately $42^{\circ}$ and $50^{\circ}$.

Topic: Physics, Ray Optics
Bilateral symmetry, metameric segmentation, coelom and open circulatory system are the features of

Options:
A Annelida
B Arthropoda
C Mollusca
D Echinodermata

The correct answer is:
Arthropoda

Solution:
The members of phylum-Arthropoda show bilateral symmetry, three germ layers in body wall, external metamerism, jointed and paired appendages, haemocoel and open type of circulatory system with dorsal heart.

Topic: Biology, Animal Kingdom, JIPMER, JIPMER 2012
Bilateral symmetry, metameric segmentation, coelom and open circulatory system characterizes which of the following phylum?

Options:
A Annelida
B Mollusca
C Arthropoda
D Echinodermata

The correct answer is:
Arthropoda

Solution:
The phylum Arthropoda is the largest and most varied in the animal kingdom. It includes insects, arachnids, myriapods and crustaceans. They have bilateral symmetry as their body can be divided vertically into two mirror images. They have metameric segmentation in which body organs and appendages were serially repeated in each segment. They are eucoelomate with an open circulatory system.

Topic: Biology, Animal Kingdom, NEET
Biodiversity of a geographical region represents

Options:
A endangered species found in the region.
B the diversity in the organisms living in the region.
C genetic diversity in the dominant species of the region.
D species endemic to the region.

The correct answer is:
the diversity in the organisms living in the region.

Solution:
Biodiversity (biological diversity) is the existence of a wide variety of species (species diversity) or other taxa of plants, animals and microorganisms in a natural community or habitat, or of communities within a particular environment (ecological diversity), or of genetic variation within a species (genetic diversity).

Topic: Biology, Biodiversity and its Conservation, NEET
Biodiversity of a geographical region represents

Options:
A Endangered species found in the region.
B The diversity in the organisms living in the region.
C Genetic diversity present in the dominant species of the region.
D Species endemic to the region.

The correct answer is:
The diversity in the organisms living in the region.

Solution:
Biodiversity is the number of variety of organism found within a specified geographical region. In our biosphere, immense diversity (or heterogeneity) exists not only at the species level but at all levels of biological organisation ranging from macromolecules within cells to biomes.

Topic: Biology, Biodiversity and its Conservation, NEET
Biome distribution with respect to annual temperature and precipitation is given below. Identify grassland and coniferous forest by selecting the CORRECT option:

Options:
A A and B
B B and E
C B and D
D C and D

The correct answer is:
B and E

Solution:
Ecology is the study of interactions between biotic and abiotic factors in a given environment and is concerned with basically four levels of biological organization - organisms, populations, community and biomes. Biome is a community of fauna having similar features for the environment they live in.
Vegetation that is dominated by the continuous cover of grasses is said to be grasslands. The region is covered by small shrubs and herbs rather than tall plants. The mean annual temperature and precipitation in this region is about 0oC to 25oC and approximately 25-100 cm.
Forest region having an abundance of cone-bearing needle leaf evergreen trees with long winter and moderate to high precipitation is considered under coniferous forest. They are found in the region above the tropical region. The mean annual temperature is around -1oC to 13oC and mean annual precipitation is approximately 50-250 cm.
Below image describes the requirement of MAP and MAT of different forest types.

Topic: Biology, Organisms and Populations, NEET
Blue colour of the sky is due to

Options:
A reflection of light
B refraction of light
C diffraction of light
D scattering of light

The correct answer is:
scattering of light

Solution:
Blue colour of the sky is due to scattering of light. Scattering $\propto \frac{1}{\lambda^4}$ and $\lambda$ is least for blue colour among the fundamental colour red, green and blue. So blue colour of light scattered the most. Therefore sky appear blue.

Topic: Physics, Ray Optics, AP EAMCET, AP EAMCET 2023 (15 May Shift 1)
Both ethanal and propanone will react with which of the following?

Options:
A Tollen reagent
B Schiff reagent
C Fehling reagent
D Grignard reagent

The correct answer is:
Grignard reagent

Solution:
Acetaldehyde and acetone both react with a Grignard reagent to give secondary and tertiary alcohols respectively.
While Fehling's reagent, Tollen's reagent and Schiff's reagent are reduced by only aldehydes (i.e., acetaldehyde). Hence, the correct option is (4).

Topic: Chemistry, Aldehydes and Ketones, AP EAMCET, AP EAMCET 2020 (22 Sep Shift 2)
Briefly describe the cell theory.

Options:

None

Solution:
Schleiden and Schwann formulated the cell theory, in 1938-39 which stated
(i) All living beings are made up of cells and products formed by the cells.
(ii) Cells are the structural and functional units of life
The cell theory stated by Schleiden and Schwann failed to explain the question of origin of cells. A major expansion of the cell theory was expressed by Virchow in his statment 'Omnis cellula e cellula' (all cells arise from pre-existing cells) in 1855. This concept, was the actual idea of Nagelli (1846), which later on was elaborated by Virchow, along with considerable evidences in its support. The work of Nagelli and Virchow established cell division as the central pehnomenon in the continuity of life.
The modern cell theory is thus based on two facts
(i) All living organisms are composed of cells and products of cells.
(ii) Cells are the basic structural and functional units of life.
(iii) All cells arise from pre-existing cells.
Viruses are exception to cell theory as they are not composed of cell. They consist of a nucleic acid (DNA or RNA) surrounded by a protein sheet and are incapable of independant existence, self regulation and self reproduction.

Topic: Biology, Cell - The Unit of Life
Briefly describe the followings
(a) anaemia
(b) angina pectoris
(c) atherosclerosis
(d) hypertension
(e) heart failure
(f) erythroblastosis foetalis

Options:

None

Solution:
(a) Anaemia This is the most common disorder of the blood, which is caused due to decrease in the number of RBC than the normal amount and also due to less quantity of haemoglobin than the normal value in blood. This is the most common disorder of the blood.
(b) Angina Pectoris When there is blockage in coronary artery, thus insufficient supply of blood reaches to heart muscles. That results in chest pain, fear, anxiety, pale skin, profuse sweating and vomiting. The pain usually starts in the centre of the chest spreads down to the left arm which last for only few second.
(c) Atherosclerosis It refers the deposition of cholesterol or fatty substance in the inner lining of arteries called atherosclerotic plaque. Sometimes arteries get completely blocked, this may result in stroke of heart attack.
(d) Hypertension It is sometimes also known as arterial hypertension. The blood pressure in the arteries gets elevated. It could be primary or secondary hypertension which are caused by various conditions which affect kidneys, arteries heart or endocrine system.
(e) Heart Failure It is the state of heart, causes in when heart does not pump blood effectively enough to meet the requirement of the body.
(f) Erythroblastosis foetalis It is a haemolytic disease causes in new borns, which is an allo-immune condition that develops in foetus, when igG molecules produced by mother pass through placenta and attack RBC. It causes reticulocytosis and anaemia. It develops due to $\mathrm{Rh}$ incompatibility between the couples.
In a man with $\mathrm{RH}^{+}$blood and women with $\mathrm{Rh}^{-}$blood, the second pregnancy foetus may have this problem due to $\operatorname{lgG}$ accumulation in women during first child development and delivery.

Topic: Biology, Body Fluids and Circulation
Brown haze produced in photochemical smog is due to:

Options:
A NO
B NO2
C CH3-CO||OO-NO2
D CH2=CH-CH=O

The correct answer is:
NO2

Solution:
Photochemical smog is initiated by the photochemical dissociation of NO2 and the resulting secondary reactions involving unsaturated hydrocarbons, other organic compounds and free radicals, leading to the formation of organic peroxides and ozone.
Hydrocarbons +O3, O2, O,NO2, NO; peroxides, peroxyacetyl nitrate, formaldehyde, ozone aldehyde, acrolein, etc. oxidised hydrocarbons and ozone in the presence of humidity cause photochemical smog, which dissipates at night.

Topic: Chemistry, Environmental Chemistry, NEET
Brunner's glands are

Options:
A present in the ileum and is a secondary lymphatic tissue storing lymphocytes.
B present in the duodenum and is a secondary lymphatic tissue storing lymphocytes.
C present in the ileum and secrete an alkaline mucous fluid.
D present in the duodenum and secrete an alkaline mucous fluid.

The correct answer is:
present in the duodenum and secrete an alkaline mucous fluid.

Solution:
Brunner's glands are found in the submucosa of the duodenum and secrete an alkaline fluid, having mucin in it. These glands play an important function in protecting the duodenal epithelium from the acidic chyme of the stomach and its gastric juices, by coating it.

Topic: Biology, Digestion and Absorption, NEET
By convention which of the following is not characteristic for having Zero water potential?

Options:
A Atmospheric pressure is taken to be 1 atm
B Water is not under any pressure
C Water is at standard temperature
D Pure water is taken

The correct answer is:
Atmospheric pressure is taken to be 1 atm

Solution:
By convention, the water potential of pure water at standard temperatures, which is not under any pressure, is taken to be zero.

Topic: Biology, Transport in Plants, NEET
ByBaye's theorem, which one of the following probabilities is calculated?

Options:
A Prior probability
B Likelihood probability
C Posterior probability
D Conditional probability

The correct answer is:
Conditional probability

Solution:
By Baye's theorem, we know that, conditional probability is calculated.

Topic: Mathematics, Probability, NDA, NDA 2009 (Phase 1)
$\mathrm{C}_3 \mathrm{H}_9 \mathrm{~N}$ reacts with Hinsberg reagent and the product is insoluble in alkali but soluble in ether. This nitrogen containing compound is

Options:
A primary amine
B secondary amine
C tertiary amine
D methyl isocyanide

The correct answer is:
secondary amine

Solution:
The compound with molecular formula $\mathrm{C}_3 \mathrm{H}_9 \mathrm{~N}$ is a secondary amine. When it reacts with Hinsberg reagent, the product obtained is soluble in ether.

Topic: Chemistry, Amines, JIPMER, JIPMER 2016
$$
\mathrm{C}_6 \mathrm{H}_5-\mathrm{O}-\mathrm{CH}_2 \mathrm{CH}_3 \underset{\Delta}{\stackrel{\mathrm{HI}}{\longrightarrow}} Y+Z
$$
Identify $Y$ and $Z$ in the above reaction.
\begin{array}{l|l}
X & Y
\end{array}

Options:
A \begin{array}{l|l}
\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH} & \mathrm{H}_3 \mathrm{CCH}_3
\end{array}
B \begin{array}{l|l}
\mathrm{C}_2 \mathrm{H}_5 \mathrm{I} & \mathrm{C}_6 \mathrm{H}_5 \mathrm{CHO}
\end{array}
C \begin{array}{l|l}
\mathrm{C}_6 \mathrm{H}_5 \mathrm{I} & \mathrm{H}_3 \mathrm{CCH}_2 \mathrm{OH}
\end{array}
D $\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH} \quad \mathrm{H}_3 \mathrm{CCH}_2$

The correct answer is:
$\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH} \quad \mathrm{H}_3 \mathrm{CCH}_2$

Solution:
When an alkyl aryl ether is reacted with halogen acid $(H X)$, the products are always phenol and an alkyl halide.
$$
\begin{aligned}
\mathrm{C}_6 \mathrm{H}_5-\mathrm{O}-\mathrm{CH}_2- & \mathrm{CH}_3+\mathrm{HI} \stackrel{100 \mathrm{C}}{\longrightarrow} \\
& \mathrm{C}_6 \mathrm{H}_5 \mathrm{OH}+\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{I}
\end{aligned}
$$

Topic: Chemistry, Alcohols Phenols and Ethers, AP EAMCET, AP EAMCET 2014
$$
\mathrm{C}_6 \mathrm{H}_5-\mathrm{O}-\mathrm{CH}_2 \mathrm{CH}_3 \underset{\Delta}{\stackrel{\mathrm{HI}}{\longrightarrow}} Y+Z
$$
Identify $Y$ and $Z$ in the above reaction.
\begin{array}{l|l}
X & Y
\end{array}

Options:
A \begin{array}{l|l}
\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH} & \mathrm{H}_3 \mathrm{CCH}_3
\end{array}
B \begin{array}{l|l}
\mathrm{C}_2 \mathrm{H}_5 \mathrm{I} & \mathrm{C}_6 \mathrm{H}_5 \mathrm{CHO}
\end{array}
C \begin{array}{l|l}
\mathrm{C}_6 \mathrm{H}_5 \mathrm{I} & \mathrm{H}_3 \mathrm{CCH}_2 \mathrm{OH}
\end{array}
D $\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH} \quad \mathrm{H}_3 \mathrm{CCH}_2$

The correct answer is:
$\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH} \quad \mathrm{H}_3 \mathrm{CCH}_2$

Solution:
When an alkyl aryl ether is reacted with halogen acid $(H X)$, the products are always phenol and an alkyl halide.
$$
\begin{aligned}
\mathrm{C}_6 \mathrm{H}_5-\mathrm{O}-\mathrm{CH}_2- & \mathrm{CH}_3+\mathrm{HI} \stackrel{100 \mathrm{C}}{\longrightarrow} \\
& \mathrm{C}_6 \mathrm{H}_5 \mathrm{OH}+\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{I}
\end{aligned}
$$

Topic: Chemistry, Alcohols Phenols and Ethers, TS EAMCET, TS EAMCET 2014
C is associated with

Options:
A $1,4,6,7,9,11$
B $2,4,8,9$
C $1,4,6,7,9$
D None of these

The correct answer is:
$1,4,6,7,9$

Solution:
$\tan x$ is a periodic function with period $\pi$ and is
discontinuous at $x=\frac{m \pi}{2}$. Also, tan $x$ is not differentiable at every point on $(-\infty, \infty)$ and increases on $\left(0, \frac{\pi}{2}\right)$ and increases on $\left(\frac{\pi}{2}, \pi\right)$. Hence, option (c) is correct.

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2010 (Phase 1)
Calculate standard cell potential for a cell having following reaction,
$\begin{aligned} & 2 \mathrm{Al}_{(\mathrm{s})}+3 \mathrm{Ni}^{2+} \rightarrow 2 \mathrm{Al}^{3+}+3 \mathrm{Ni}(\mathrm{s}) \\ & \left(\mathrm{E}_{\mathrm{Ni}}^{\circ}=-0.25 \mathrm{~V}, \mathrm{E}_{\mathrm{A} 1}^{\circ}=-1.66 \mathrm{~V}\right)\end{aligned}$

Options:
A $0.50 \mathrm{~V}$
B $1.41 \mathrm{~V}$
C $-0.50 \mathrm{~V}$
D $0.41 \mathrm{~V}$

The correct answer is:
$1.41 \mathrm{~V}$

Solution:
$\begin{aligned} & \mathrm{E}_{\text {cell }}^{\circ}=\mathrm{E}_{\mathrm{Ni}}^{\circ}-\mathrm{E}_{\mathrm{Ai}}^{\circ} \\ & =-0.25-(-1.66) \\ & =1.41 \mathrm{~V}\end{aligned}$

Topic: Chemistry, Electrochemistry, MHT CET, MHT CET 2022 (05 Aug Shift 2)