datavorous/entrance-exam-dataset huggingface dataset:
Arrange given metals in order of their reducing powers. The standard reduction electrode potentials of three metals A, B and C are respectively are +0.5V , –3.0V and –1.2V . The

Options:
A B>C>A
B A>B>C
C C>B>A
D A>C>B

The correct answer is:
B>C>A

Solution:
More is reduction potential, more is the power to get itself reduced or lesser is reducing power or greater is oxidising power.

Topic: Chemistry, Electrochemistry, NEET
Arrange in the order of stability of enol form of the compounds:

Options:
A iii > ii > i
B i > ii > iii
C iiÂ >Â iÂ >Â iii
D iiÂ >Â iiiÂ >Â i

The correct answer is:
i > ii > iii

Solution:
In 1,2,3 the acidic -H order is 1,>2,>3
1, 3-Diketones are prone to exist as enols because of conjugation of the enol or enolate with the other carbonyl group, and the stability gained in forming a six-membered ring.

Topic: Chemistry, General Organic Chemistry, NEET
Arrange reactivity of given compounds in decreasing order with benzaldehyde in the presence of secondary amine.
1.CH2(COOC2H5)2
2. CH2(CN)2
3. CH2(NO2)2
4.

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

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

Solution:
Reaction is nucleophilic addition reaction of benzaldehyde. More is the nucleophilicity of anion of these compounds, more is the reactivity.

Topic: Chemistry, Aldehydes and Ketones, NEET
Arrange the below four layers in proper order starting from the outermost to the innermost layer after secondary growth has already occurred in a dicot stem:

I. Phellem
II. Wood
III. Phelloderm
IV. Vascular cambium

Options:
A I → II → III → IV
B I → III → II → IV
C IV → III → II → I
D III → II → I → IV

The correct answer is:
I → III → II → IV

Solution:
The meristematic tissue called cork cambium or phellogen develops, usually in the cortex region. Phellogen is a couple of layers thick. It is made of narrow, thin-walled, and nearly rectangular cells. Phellogen cuts off cells on both sides. The outer cells differentiate into cork or phellem while the inner cells differentiate into secondary cortex or phelloderm. Internal to the secondary cortex lies the xylem (wood), then the vascular cambium and then the phloem is the innermost.

Topic: Biology, Anatomy of Flowering Plants, NEET
Arrange the elements in the increasing order of their energy of atomization.

Options:
A V
B Fe
C Mn
D Cr

The correct answer is:
Mn

Solution:
Correct Option is : 3
Mn

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Topic: Chemistry, d and f Block Elements, NEET
Arrange the following alcohols in order of increasing reactivity towards sodium metal.
(i) (CH3)3C-OH
(ii) (CH3)2CH-OH
(iii) CH3CH2OH

Options:
A (iii) < (ii) < (i)
B (ii) < (i) < (iii)
C (i) < (ii) < (iii)
D (iii) < (i) < (ii)

The correct answer is:
(i) < (ii) < (iii)

Solution:
More the number of methyl groups (electron releasing) lesser will be the acidity. Hence the reactivity order towards sodium metal will be as follows
tertiary (i) < secondary (ii) < primary (iii).

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
Arrange the following elements in the increasing order of metallic character: Si, Be, Mg, Na, P.

Options:
A Be
B P
C Na
D Si

The correct answer is:
P

Solution:
Correct Option is : (B)
P

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Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET
Arrange the following events occurring in the ovary in the CORRECT sequence

a. The antrum is formed.

b. Formation of a first tiny polar body.

c. The primary follicle get surrounded by more and more layers of granulosa cells.

d. The primary oocyte is arrested in prophase-I of the meiotic division.

Options:
A d,Â a,Â b and c
B a,Â c,Â d and b
C d,Â a,Â c and b
D d, c, a and b

The correct answer is:
d, c, a and b

Solution:
Oogenesis involves the development of oocyte and also folliculogenesis, which takes place in the Ovary. The primary oocyte develops first and begins its meiosis-I, which gets arrested at prophase-I. This stage is followed by the development of granulosa cells around it, known as primary follicle formation. The primary follicle then develops into the secondary follicle, which is characterized by the formation of the theca layer around granulosa cells. This then develops into the tertiary follicle. The differentiating feature is the formation of Antrum, which is initially crescent-shaped and later becomes large. This tertiary follicle still shows the presence of primary oocyte in it. These two then undergo further development. Primary oocyte completes its meiosis-I and forms two unequal cells: Large secondary oocyte and a small first polar body. Tertiary follicle further develops into Graffian follicle and secondary oocyte starts it meiosis-II.

Topic: Biology, Human Reproduction, NEET
Arrange the following formed elements in the decreasing order of their abundance in blood in humans :
(a) Platelets
(b) Neutrophils
(c) Erythrocytes
(d) Eosinophils
(e) Monocytes
Choose the most appropriate answer from the options given below :

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

The correct answer is:
(c), (a), (b), (e), (d)

Solution:
Number or abundance of blood cells can be arranged in decreasing order as follows:
RBCs $>$ Platelets $>$ Neutrophills $>$ Monocytes $>$ Eosinophills $>$ Platelets

Topic: Biology, Body Fluids and Circulation, NEET, NEET 2022 (Phase 2)
Arrange the following in decreasing order of their boiling points.

Options:
A $A>B>D>C$
B $A>C>D>B$
C $B>C>D>A$
D $C>A>B>D$

The correct answer is:
$A>B>D>C$

Solution:
The boiling point of $(\mathrm{A})$ alcohol is highest due to $\mathrm{H}$-bonding.
Out of primary (B), secondary (D) and tertiary amines (C), the boiling point in decreasing order is
Primary amine $>$ Secondary amine $>$ Tertiary amine
as intermolecular association decreases in same order, being highest in primary amine and least in tertiary amine due to steric hindrance.
$\therefore$ The correct decreasing order of boiling point in given compound is
A $>$ B $>$ D $>$ C.

Topic: Chemistry, Amines, AP EAMCET, AP EAMCET 2022 (05 Jul Shift 1)
Arrange the following in decreasing order of their pKb values

Options:
A d>a>c>b
B a>b>d>c
C d>c>b>a
D a>c>d>b

The correct answer is:
d>c>b>a

Solution:
The reaction of amines with mineral acids to form ammonium salts shows that these are basic in nature. Amines have an unshared pair of electrons on nitrogen atom due to which they behave as Lewis base

Basic character of amines can be better understood in terms of their Kb and pKb, Larger the value of Kb or smaller the value of pKb , stronger is the base.

The order of basic strength in case of methyl substituted amines and in aqueous solution depends on inductive effect ,steric effect and solvation effect.

Hence, the correct basicity order is

secondary amines >primary amines> tertiary amines

Hence, CH33N is less basic than CH3NH2.

Therefore, pKb value is more for CH33N.

C and D are less basic than A and B due to the presence of phenyl group in C and D which causes -I effect.

But C is more basic than D due to lesser steric hindrance.Hence, the pKb value is more D

So, the correct order of pKb values is

Topic: Chemistry, General Organic Chemistry, AP EAMCET, AP EAMCET 2022 (04 Jul Shift 1)
Arrange the following in the increasing order of their basic strengths $\mathrm{CH}_{3} \mathrm{NH}_{2},\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}$, $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N}, \mathrm{NH}_{3}$

Options:
A $\mathrm{NH}_{3} < \left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N} < \left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH} < \mathrm{CH}_{3} \mathrm{NH}_{2}$
B $\mathrm{NH}_{3} < \left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N} < \mathrm{CH}_{3} \mathrm{NH}_{2} < \left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}$
C $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N} < \mathrm{NH}_{3} < \mathrm{CH}_{3} \mathrm{NH}_{2} < \left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}$
D $\mathrm{CH}_{3} \mathrm{NH}_{2} < \left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH} < \left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N} < \mathrm{NH}_{3}$

The correct answer is:
$\mathrm{NH}_{3} < \left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N} < \mathrm{CH}_{3} \mathrm{NH}_{2} < \left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}$

Solution:
Aliphatic amines are more basic than $\mathrm{NH}_{3}$ due to $+I$ effect of alkyl groups. In aqueous medium tertiary amine is less basic than secondary amine because the cation formed by protonation of tertiary amine is less solvated as compared to the cation formed by protonation of secondary amine. Hence, the increasing order of their basic strength is as:
$$
\mathrm{NH}_{3} < \left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N} < \mathrm{CH}_{3} \mathrm{NH}_{2} < \left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}
$$

Topic: Chemistry, Amines, KCET, KCET 2009
Arrange the following in the order of decreasing basicity
$\begin{array}{ccc}\mathrm{RN}=\mathrm{CHR}^1 & \mathrm{RC} \equiv \mathrm{N} & \mathrm{RNH}_2 \\ \text { I } & \text { II } & \text { III }\end{array}$

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

The correct answer is:
III $>$ I $>$ II

Solution:
In $\mathrm{RNH}_2, \mathrm{~N}$ atom is $\mathrm{sp}^3$ hybridised. In $\mathrm{RN}=\mathrm{CHR}$, $\mathrm{N}$ atom is $\mathrm{sp}^2$ hybridised whereas in $\mathrm{RC} \equiv \mathrm{N}, \mathrm{N}$ atom is sp hybridised. The sp hybridised $\mathrm{N}$ atom is most electronegative due to maximum s character. More electronegative element has less tendency to donate electrons and therefore is least basic.
The order of basicity of aliphatic amines should be primary $>$ secondary $>$ tertiary.
Thus, the order of basicity will be:

Topic: Chemistry, Amines, TS EAMCET, TS EAMCET 2023 (14 May Shift 2)
Arrange the following in the sequence you would find them in a plant starting from the periphery-phellem, phellogen, phelloderm.

Options:

None

Solution:
The outer most layer is phellem or cork followed by phellogen (cork cambium) which in turn is followed by phelloderm (secondary cortex.

Topic: Biology, Anatomy of Flowering Plants
Arrange the following structures based on their location from the periphery to centre with respect to the anatomy of a dicotyledonous root.
I. Pericycle
II. Casparian strips
III. Epiblema
IV. Conjunctive tissue

Options:
A IV, I, II, III
B IV, II, III,I
C I, II, III, IV
D III, II, I, IV

The correct answer is:
III, II, I, IV

Solution:
The outermost layer is epiblema. Many of the cells of epiblema protrude in the form of unicellular root hairs. The cortex consists of several layers of thin-walled parenchyma cells, with intercellular spaces. The innermost layer of the cortex is called endodermis. It comprises a single layer of barrel-shaped cells without any intercellular spaces. The tangential, as well as radial walls of the endodermal cells, have a deposition of water-impermeable, waxy material suberin in the form of casparian strips. Next to endodermis lies a few layers of thick-walled parenchymatous cells referred to as pericycle. Initiation of lateral roots and vascular cambium during the secondary growth takes place in these cells. The pith is small or inconspicuous. The parenchymatous cells which lie between the xylem and the phloem are called conjunctive tissue.

Topic: Biology, Anatomy of Flowering Plants, NEET
As a tree grows very old, which of the following increases more rapidly in thickness?

Options:
A Cortex
B Heartwood
C Phloem
D Pericycle

The correct answer is:
Heartwood

Solution:
As a tree matures, secondary growth results in more amount of secondary xylem as compared to secondary phloem. This secondary xylem gradually becomes non functional and forms heart wood.

Topic: Biology, Anatomy of Flowering Plants, NEET
As compared to a dicot root, a monocot root has

Options:
A Many xylem bundles.
B Relatively thicker periderm.
C Inconspicuous annual rings.
D More abundant secondary xylem.

The correct answer is:
Many xylem bundles.

Solution:
Vascular bundles in monocot root are arranged radially having eight bundles each of xylem and phloem. This type of arrangement is basically called as polyarch with xylem in exarch condition.

Topic: Biology, Anatomy of Flowering Plants, NEET
As shown in the figure, two spherical cavities are made in the uniform solid sphere of radius $R$. The boundaries of the cavities touch at the centre of the sphere. The centers of the cavities and the sphere lie on the $X$-axis. The mass of the solid sphere before the cavities were created was $M$. The gravitational force on a point mass $m$ at a distance $d$ away from the centre of the solid sphere is

Options:
A $\frac{G M m}{d^2}\left[1-\frac{1}{8} \frac{1}{\left(1+\frac{R}{2 d}\right)^2}-\frac{1}{8} \frac{1}{\left(1-\frac{R}{2 d}\right)^2}\right]$
B $\frac{G M m}{d^2}\left[1-\frac{1}{8} \frac{1}{\left(1+\frac{R}{d}\right)^2}-\frac{1}{8} \frac{1}{\left(1-\frac{R}{d}\right)^2}\right]$
C $\frac{G M m}{d^2}\left[1-\frac{1}{8} \frac{1}{\left(1+\frac{d}{R}\right)^2}-\frac{1}{8} \frac{1}{\left(1-\frac{d}{R}\right)^2}\right]$
D $\frac{G M m}{d^2}\left[1-\frac{1}{8} \frac{1}{\left(1+\frac{d}{R}\right)^2}+\frac{1}{8} \frac{1}{\left(1-\frac{d}{R}\right)^2}\right]$

The correct answer is:
$\frac{G M m}{d^2}\left[1-\frac{1}{8} \frac{1}{\left(1+\frac{R}{2 d}\right)^2}-\frac{1}{8} \frac{1}{\left(1-\frac{R}{2 d}\right)^2}\right]$

Solution:
Given situation is as shown
Radius of sphere given $=R$
Mass of sphere $=M$
Density of sphere, $d=\frac{M}{\frac{4}{3} \pi R^3}$
Radius of each of cavity $=R / 2$
Mass of each of portion removed to create a cavity $=$ density $\times$ volume
$$
=\frac{M}{\frac{4}{3} \pi R^3} \times \frac{4}{3} \pi\left(\frac{R}{2}\right)^3=\frac{M}{8}
$$
Now, force of gravity on $m$,
$F$ = force due to complete sphere of mass $M$ - force of mass of cavity of sphere centre at $A$ - force of mass of cavity of sphere centre at $B$.
$\begin{aligned} & =\frac{G M m}{d^2}=\frac{G M^{\prime} m}{\left(d+\frac{R}{2}\right)^2}-\frac{G M^{\prime} m}{\left(d-\frac{R}{2}\right)^2} \\ & \Rightarrow F=\frac{G M m}{d^2}-\frac{G M m}{8\left(d+\frac{R}{2}\right)^2}-\frac{G M m}{8\left(d-\frac{R}{2}\right)^2} \\ & =G M m\left(\frac{1}{d^2}-\frac{1}{8\left(d+\frac{R}{2}\right)^2}-\frac{1}{8\left(d-\frac{R}{2}\right)^2}\right) \\ & =\frac{G M m}{d^2}\left(1-\frac{1}{8\left(1+\frac{R}{2 d}\right)^2}-\frac{1}{8\left(1-\frac{R}{2 d}\right)^2}\right)\end{aligned}$

Topic: Physics, Gravitation, AP EAMCET, AP EAMCET 2022 (07 Jul Shift 2)
As we go from periphery to the centre of a dicot root, the correct order of tissues that we will encounter is

Options:
A Epiblema - Cortex - Endodermis - Metaxylem - Protoxylem - Pith
B Cortex - Epiblema - Endodermis - Metaxylem - Protoxylem - Pith
C Cortex - Epiblema - Endodermis - Protoxylem - Metaxylem - Pith
D Epiblema - Cortex - Endodermis - Protoxylem - Metaxylem - Pith

The correct answer is:
Epiblema - Cortex - Endodermis - Protoxylem - Metaxylem - Pith

Solution:
The outermost layer is epiblema. Many of the cells of epiblema protrude in the form of unicellular root hairs. The cortex consists of several layers of thin-walled parenchyma cells with intercellular spaces. The innermost layer of the cortex is called endodermis. It comprises a single layer of barrel-shaped cells without any intercellular spaces. The tangential, as well as radial walls of the endodermal cells, have a deposition of water-impermeable, waxy material suberin in the form of Casparian strips. Next to endodermis lies a few layers of thick-walled parenchyomatous cells referred to as pericycle. The initiation of lateral roots and vascular cambium during secondary growth takes place in these cells. The pith is small or inconspicuous. The parenchymatous cells which lie between the xylem and the phloem are called conjunctive tissue. There are usually two to four xylem and phloem patches. Later, a cambium ring develops between the xylem and phloem. All tissues on the inner side of the endodermis such as pericycle, vascular bundles, and pith constitute the stele.

Topic: Biology, Anatomy of Flowering Plants, NEET
Assertion $(\mathbf{A}):\left(\frac{-1+\sqrt{-3}}{2}\right)^{29}+\left(\frac{-1-\sqrt{-3}}{2}\right)^{29}=-1$
Reason $(\mathbf{R}): \omega^{2}=-1$

Options:
A Both A and $\mathrm{R}$ are true and $\mathrm{R}$ is the correct explanation of A
B Both A and $\mathrm{R}$ are true but $\mathrm{R}$ is not the correct explanation of A
C $\mathrm{A}$ is true but $\mathrm{R}$ is false
D A is false but $\mathrm{R}$ is true

The correct answer is:
$\mathrm{A}$ is true but $\mathrm{R}$ is false

Solution:
(A) Consider $\left[\frac{-1+\sqrt{-3}}{2}\right]^{29}+\left[\frac{-1-\sqrt{-3}}{2}\right]^{29}$
$=\left[\frac{-1+\sqrt{3 i}}{2}\right]^{29}+\left[\frac{-1-\sqrt{3 i}}{2}\right]^{29}$
$=(\omega)^{29}+\left(\omega^{2}\right)^{29}=\omega^{27} \cdot \omega^{2}+\left(\omega^{3}\right)^{19} \cdot \omega$
$=\left(\omega^{3}\right)^{9} \omega^{2}+\left(\omega^{3}\right)^{19} \cdot \omega$
$=1 . \omega^{2}+1 \cdot \omega \quad\left(\because \omega^{3}=1\right)$
$=\omega^{2}+\omega=-1$
(R) $\omega^{2} \neq-1$
A is true but $\mathrm{R}$ is false.

Topic: Mathematics, Complex Number, NDA, NDA 2009 (Phase 1)
Assertion (A) $: f(x)=x \sin \left(\frac{1}{x}\right)$ is differentiable at $x=0$.
Reason $(\mathbf{R}): \mathrm{f}(\mathrm{x})$ is continuous at $\mathrm{x}=0$

Options:
A Both $\mathbf{A}$ and $\mathbf{R}$ are individually true, and $\mathbf{R}$ is the correct explanation of $\mathbf{A}$
B Both $\mathbf{A}$ and $\mathbf{R}$ are individually true but $\mathbf{R}$ is not the correct explanation of $\mathbf{A}$.
C A is true but $\mathbf{R}$ is false.
D A is false but $\mathbf{R}$

The correct answer is:
A is false but $\mathbf{R}$

Solution:
Given function : $\mathrm{f}(\mathrm{x})=\mathrm{x} \sin \left(\frac{1}{\mathrm{x}}\right)$
For differentiability at $\mathrm{x}=0 ; \mathrm{LHD}=\mathrm{RHD}$ at $\mathrm{x}=0$
$\begin{aligned} \mathrm{LHD} &=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{f}(0-\mathrm{h})-\mathrm{f}(0)}{-\mathrm{h}} \\ &=\lim _{\mathrm{h} \rightarrow 0} \frac{(-\mathrm{h}) \sin \left(-\frac{1}{\mathrm{~h}}\right)}{-\mathrm{h}}=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{h} \sin \left(\frac{1}{\mathrm{~h}}\right)}{-\mathrm{h}} \end{aligned}$
$=\lim _{h \rightarrow 0} \sin \left(\frac{1}{h}\right)=a$ finite value lies between $-1$
and 1 which cannot be qualified exactly.
$\mathrm{RHD}=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{f}(0+\mathrm{h})-\mathrm{f}(0)}{\mathrm{h}}$
$=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{h} \sin \left(\frac{1}{\mathrm{~h}}\right)}{\mathrm{h}}=\lim _{\mathrm{h} \rightarrow 0} \sin \left(\frac{1}{\mathrm{~h}}\right)$
$=$ a finite value lies between $-1$ and 1 which cannot be qualified exactly. $\mathrm{LHD} \neq \mathrm{RHD} \neq$ a definite value.
Hence, $\mathrm{f}(\mathrm{x})$ is not differentiable at $\mathrm{x}=0$. For continuity at $\mathrm{x}=0$ :
$\lim _{x \rightarrow 0}$ LHL $=\lim _{x \rightarrow 0} \mathrm{RHL}=\mathrm{V} . \mathrm{F}$. at $\mathrm{x}=0$
$\mathrm{LHL}=\lim _{\mathrm{x} \rightarrow 0} \mathrm{f}(0-\mathrm{h})=\lim _{\mathrm{h} \rightarrow 0}-\mathrm{h} \sin \left(-\frac{1}{\mathrm{~h}}\right)$
$=\lim _{h \rightarrow 0} \mathrm{~h} \sin \frac{1}{\mathrm{~h}}=0$
$\mathrm{RHL}=\lim _{\mathrm{x} \rightarrow 0^{\prime}} \mathrm{f}(0+\mathrm{h})=\lim _{\mathrm{h} \rightarrow 0} \mathrm{~h} \sin \frac{1}{\mathrm{~h}}=0$
$\mathrm{f}(0)=0$
$\Rightarrow \mathrm{LHL}=\mathrm{RHL}=\mathrm{f}(0)$
Hence, $\mathrm{f}(\mathrm{x})$ is continuous at $\mathrm{x}=0$ Thus, (A) is false but $(\mathrm{R})$ is true.

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2008 (Phase 1)
Assertion (A) : For every chemical reaction at equilibrium, standard Gibbs energy change of the reaction is zero
Reason (R) : At constant temperature and pressure chemical reactions are spontaneous in the direction of decreasing Gibbs energy.

Options:
A Both (A) and (R) are correct and (R) is the correct explanation of (A)
B Both (A) and (R) are correct but (R) is not the correct explanation of (A)
C (A) is correct and (R) is not correct
D (A) is not correct but (R) is correct

The correct answer is:
(A) is not correct but (R) is correct

Solution:
No solution. Refer to answer key.

Topic: Chemistry, Thermodynamics (C), AP EAMCET, AP EAMCET 2017 (26 Apr Shift 2)
Assertion (A) $:$ If $f(x)=\log x$, then $f(x)>0$ for all $x>0$.
Reason $(\mathbf{R}): \mathrm{f}(\mathrm{x})=\log \mathrm{x}$, is defined for all $\mathrm{x}>0$

Options:
A Both $\mathbf{A}$ and $\mathbf{R}$ are individually true, and $\mathbf{R}$ is the correct explanation of $\mathbf{A}$
B Both $\mathbf{A}$ and $\mathbf{R}$ are individually true but $\mathbf{R}$ is not the correct explanation of $\mathbf{A}$.
C A is true but $\mathbf{R}$ is false.
D A is false but $\mathbf{R}$

The correct answer is:
A is false but $\mathbf{R}$

Solution:
$(A): f(x)=\log x$ for $x=1, f(x)=\log 1=0$
$(\mathrm{R}): \mathrm{f}(\mathrm{x})=\log \mathrm{x}$
and $\mathrm{f}(\mathrm{x}) \geq 0 \forall \mathrm{x}>0$
Thus, (A) is false but $(\mathrm{R})$ is true.

Topic: Mathematics, Functions, NDA, NDA 2008 (Phase 1)
Assertion (A) : In aqueous solution, α-amino acids exists as internal salt called as zwitter ion.

Reason (R) : Proline is a natural amino acid having a secondary amino group.
The correct option among the following is

Options:
A (A) is true, (R) is true and (R) is the correct explanation for (A)
B (A) is true, (R) is true but (R) is not the correct explanation for (A)
C (A) is true but (R) is false
D (A) is false but (R) is true

The correct answer is:
(A) is true, (R) is true but (R) is not the correct explanation for (A)

Solution:
In aqueous solution, the carboxyl group can lose a proton and amino group can accept a proton, giving rise to a dipolar ion known as zwitter ion. This is neutral but contains both positive and negative charges.

In zwitter ionic form, amino acids show amphoteric behaviour as they react both with acids and bases.

Proline has a secondary amine group

Topic: Chemistry, Biomolecules, TS EAMCET, TS EAMCET 2021 (05 Aug Shift 1)
Assertion : A middle aged woman is reported to have small breasts and undersized uterus.Reason : Her genotypic analysis shows XO condition of allosomes.

Options:
A If both assertion and reason are true and reason is the correct explanation of assertion
B If both assertion and reason are true but reason is not the correct explanation of assertion
C If assertion is true but reason is false
D If both assertion and reason are false

The correct answer is:
If both assertion and reason are true and reason is the correct explanation of assertion

Solution:
Females having 45 chromosomes $(2 \mathrm{~A}+\mathrm{X} 0)$ are affected with Turner's syndrome. Individuals having Turner's syndrome have female sexual differentiation but ovaries are rudimentary. Other associated phenotypes of this condition are short stature, webbed-neck, broad chest, lack of secondary sexual characteristics and sterility. Thus, any imbalance in the copies of the sex chromosomes may disrupt the genetic information necessary for normal sexual development.

Topic: Biology, Principles of Inheritance and Variation, AIIMS, AIIMS 2012