datavorous/entrance-exam-dataset huggingface dataset:
Davisson and Germer studied the diffraction from crystal of beams of

Options:
A Electrons
B Photons
C Alpha particles
D Protons

The correct answer is:
Electrons

Solution:
Davisson and Germer's actual objective was to study the surface of a piece of nickel by directing a beam of electrons at the surface and observing how many electrons bounced off at various angles. They expected that because of the small size of electrons, even the smoothest crystal surface would be too rough and thus the electron beam would experience diffuse reflection. During the experiment an accident occurred and air entered the chamber, producing an oxide film on the nickel surface. To remove the oxide, Davisson and Germer heated the specimen in a high temperature oven, not knowing that this affected the formerly polycrystalline structure of the nickel to form large single crystal areas with crystal planes continuous over the width of the electron beam.
When they started the experiment again and the electrons hit the surface, they were scattered by atoms which originated from crystal planes inside the nickel crystal. In 1925, they generated a diffraction pattern with unexpected peaks.
So, the answer is option (1).

Topic: Physics, Dual Nature of Matter, JEE Main
DDT residues are rapidly passed through food chain causing biomagnification because DDT is :

Options:
A Water soluble
B Lipo soluble
C Moderately toxic
D Non-toxic to aquatic animals

The correct answer is:
Lipo soluble

Solution:
No Solution Available

Topic: Biology, Environmental Issues, NEET, NEET 2009 (Mains)
De-Broglie equation states the

Options:
A dual nature
B particle nature
C wave nature
D More than one of the above

The correct answer is:
dual nature

Solution:
In Bohr's theory, electron is treated as a particle. But according to de-Broglie's theory electron has a dual character, both as a material particle and as a wave. The energy $(\mathrm{E})$, and mass $(\mathrm{m})$ of matter are related through the relationship.
$E=m c^2$

Topic: Physics, Dual Nature of Matter, JEE Main
De-Broglie wavelength associated with an electron accelerated through a potential difference ' $\mathrm{V}$ ' is ' $\lambda$ '. When the accelerating potential is increased to ' $4 \mathrm{~V}$ ', de-Broglie wavelength.

Options:
A reduces to half
B remains the same
C reduces to $(1 / 4)^{\text {th }}$
D increases by $25 \%$

The correct answer is:
reduces to half

Solution:
De-Broglie wavelength, $\lambda=\frac{1.228}{\sqrt{V}}(\mathrm{~nm})$
If $\mathrm{V}$ is increased to $4 \mathrm{~V}$, then $\lambda$ will become $\frac{1}{\sqrt{4}}$ times or $\frac{1}{2}$ times.

Topic: Physics, Dual Nature of Matter, MHT CET, MHT CET 2021 (24 Sep Shift 1)
De-Broglie wavelength of an electron accelerated by a voltage of 50âV is close to e=1.6Ã10-19 C, me=9.1Ã10-31 kg, h=6.6Ã10-34 J s

Options:
A 0.5 â«
B 1.2Â â«
C 1.7 â«
D 2.4Â â«

The correct answer is:
1.7 â«

Solution:
De-Broglie wavelength λ in given by
λ=hmv=h2meV
and for electron λ=12.27 A°V
=12.27 A°50
=1.7 A°

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2015 (10 Apr Online)
De-Broglie wavelength of an electron accelerated by a voltage of 50âV is close to e=1.6Ã10-19 C, me=9.1Ã10-31 kg, h=6.6Ã10-34 J s

Options:
A 0.5 â«
B 1.2Â â«
C 1.7 â«
D 2.4Â â«

The correct answer is:
1.7 â«

Solution:
De-Broglie wavelength λ in given by
λ=hmv=h2meV
and for electron λ=12.27 A°V
=12.27 A°50
=1.7 A°

Topic: Physics, Dual Nature of Matter, JEE Main
Decarboxylation of all six possible forms of diaminobenzoic acids C6H3NH22COOH yields three products A,B and C. Three acids give a product 'A', two acids gives a product 'B' and one acid give a product 'C'. The melting point of product 'C' is

Options:
A 63 °C
B 90 °C
C 104 °C
D 142 °C

The correct answer is:
142 °C

Solution:
C is a para isomer.

The melting point of para-isomer is comparatively higher than the corresponding ortho and meta-isomers , because of more symmetrical nature; it fits into crystal lattice more readily and so, close packing of molecules in crystal lattice increases the intermolecular force of attraction between molecules.

Hence C has highest melting point that is 142 °C

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main, JEE Main 2022 (27 Jun Shift 2)
Decarboxylation of malonic acid gives

Options:
A $\mathrm{CH}_{4}$
B $\mathrm{CH}_{3} \mathrm{COOH}$
C Both (a) and (b)
D None of the above

The correct answer is:
$\mathrm{CH}_{3} \mathrm{COOH}$

Solution:
Decarboxylation of malonic acid gives acetic acid and $\mathrm{CO}_{2}$

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
Decay constant of a reaction is $1.1 \times 10^{-9} / \mathrm{sec}$, then the half life of the reaction is

Options:
A $1.2 \times 10^8$
B $6.3 \times 10^8$
C $3.3 \times 10^8$
D $2.1 \times 10^8$

The correct answer is:
$6.3 \times 10^8$

Solution:
$t_{1 / 2}=\frac{0.693}{k}, \frac{0.693}{1.1 \times 10^{-9}}=6.3 \times 10^8$ sec.

Topic: Chemistry, Chemical Kinetics, JEE Main
Decay of ${ }_{92} U^{235}$ is .....order reaction

Options:
A Zero
B First
C Second
D Third

The correct answer is:
First

Solution:
The radioactive disintegration reactions are of first order because this rate of disintegration depends on the concentration term of radioactive material only.

Topic: Chemistry, Chemical Kinetics, JEE Main
Decibel is unit of

Options:
A Intensity of light
B X-rays radiation capacity
C Sound loudness
D Energy of radiation

The correct answer is:
Sound loudness

Solution:
Decibel (dB), unit for expressing the ratio between two physical quantities, usually amounts of acoustic or electric power, or for measuring the relative loudness of sounds

Topic: Physics, Waves and Sound, JEE Main
Decomposers like fungi and bacteria are
(i) autotrophs
(ii) heterotrophs
(iii) saprotrophs
(iv) chemo-autotrophs
Choose the correct answer

Options:
A
(i) and (iii)

B
(i) and (iv)

C
(ii) and (iii)

D
(i) and (ii)

The correct answer is:
(ii) and (iii)

Solution:
Those organisms which are able to synthesise their own food are known as autotrophs. e.g., plants while the organisms which derive their food, from autotrophs or plants are termed as heterotrophs. These may be herbivores, carnivores and omnivores.
Saprotrophs are the organisms, which obtain their nourishment from dead organic matter or decayed remains of both plants and animals. They are also known as decomposer. They play an important role in recycling of minerals by decomposing dead organic matter.

Topic: Biology, Ecosystem
Decomposition of H2O2 follows a first order reaction. In fifty minutes the concentration of H 2 O 2 decreases from 0.5 to 0.125 M in one such decomposition. When the concentration of H2O2 reaches 0.05 M , the rate of formation of O2 will be:

Options:
A 2.66Â Lmin-1â¡Â atÂ STP
B 1.34Ã10-2Â molÂ min-1
C 6.93Ã10-2Â molÂ min-1
D 6.93Ã10-4 mol min-1

The correct answer is:
6.93Ã10-4 mol min-1

Solution:
H 2 O 2 → H 2 O+ 1 2 O 2
-d[H2O2]dt=d[H2O]dt=2d[O2]dt
t12=25
t 1 2 = 0.69314 k
k =  0.69314 25
rate of reaction = k[H2O2]
∴ d[O2]dt=12×k[H2O2]
=12×0.6931425×0.05
=6.93×10-4 mol min-1

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2016 (03 Apr)
Decomposition of H2O2 follows a first order reaction. In fifty minutes the concentration of H 2 O 2 decreases from 0.5 to 0.125 M in one such decomposition. When the concentration of H2O2 reaches 0.05 M , the rate of formation of O2 will be:

Options:
A 2.66Â Lmin-1â¡Â atÂ STP
B 1.34Ã10-2Â molÂ min-1
C 6.93Ã10-2Â molÂ min-1
D 6.93Ã10-4 mol min-1

The correct answer is:
6.93Ã10-4 mol min-1

Solution:
H 2 O 2 → H 2 O+ 1 2 O 2
-d[H2O2]dt=d[H2O]dt=2d[O2]dt
t12=25
t 1 2 = 0.69314 k
kâ=â 0.69314 25
rate of reaction = k[H2O2]
â´ d[O2]dt=12Ãk[H2O2]
=12Ã0.6931425Ã0.05
=6.93Ã10-4 mol min-1

Topic: Chemistry, Chemical Kinetics, JEE Main
Decomposition of H2O2 follows a first order reaction.In fifty minutes the concentration of H2O2 decreases from 0.5 to 0.125 M in one such decomposition. When the concentration of H2O2 reaches 0.05 M, the rate of formation of O2 will be

Options:
A 1.34×10-2molmin-1⁡
B 6.93×10-2molmin-1⁡
C 6.93×10-4molmin-1⁡
D 2.66 L min-1⁡ at STP

The correct answer is:
6.93×10-4molmin-1⁡

Solution:
H2O2aq→H2Oaq+12O2g
k=1tln⁡a0a1
=150ln⁡0.50.125
=150ln⁡4min-1⁡
Rate of disappearance of H2O21=Rate of appearance of O212
RateO2=12×RateH2O2
=12kH2O2
=12×150×ln⁡4×0.05
=6.93×10-4Mmin-1⁡

Topic: Chemistry, Chemical Kinetics, JEE Main
Decomposition of $\mathrm{H}_2 \mathrm{O}_2$ is prevented by

Options:
A $\mathrm{NaOH}$
B $\mathrm{MnO}_2$
C Acetanilide
D Oxalic acid

The correct answer is:
Acetanilide

Solution:
Acetanilide: Acetanilide acts as a stabilizer which, when added to the solutions, retards the decomposition

Topic: Chemistry, Hydrogen, JEE Main
Decomposition of $\mathrm{NH}_{3}$ on the surface of platinum has a rate constant of $2.5 \times 10^{-4} \mathrm{~mol} \mathrm{dm^{-3 } \mathrm { s } ^ { - 1 } \text { at } 3 5 0 \mathrm { K } \text { . The order of }}$ the reaction is

Options:
A 0
B 1
C $1 / 2$
D 2

The correct answer is:
1

Solution:
Unit of rate constant of $n$th order
$$
(k)=\left(\frac{\mathrm{mol}}{\mathrm{dm}^{3}}\right)^{1-n} \mathrm{~s}^{-1}
$$
for zero order $(n)=0$
$$
\therefore \quad k=\operatorname{mol~dm}{ }^{-3} \mathrm{~s}^{-1}
$$
$\therefore$ The order of reaction is 1st.

Topic: Chemistry, Chemical Kinetics, JEE Main
Decomposition of X exhibits a rate constant of 0.05 μg/year. How many years are required for the decomposition of 5 μg of X into 2.5 μg?

Options:
A 20
B 40
C 25
D 50

The correct answer is:
50

Solution:
According to unit of rate constant, it is a zero order reaction. Then, half life of reaction will be:
t12=C02k=5μg2×0.05μg/year=50 year

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2019 (12 Jan Shift 1)
Decreasing order of basicity of the given compound is which of the following?
(1) $\mathrm{CH}_3 \mathrm{CONH}_2$
(2) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{NH}_2$
(3) $\mathrm{PhCH}_2 \mathrm{CONH}_2$

Options:
A $1\gt2\gt3$
B $2\gt1\gt3$
C $3\gt2\gt1$
D $1\gt3\gt2$

The correct answer is:
$2\gt1\gt3$

Solution:
Explanation: Comparing the basicity of alkyl amines to amides. With an alkyl amine the lone pair electron is localized on the nitrogen. However, the lone pair electron on an amide are delocalized between the nitrogen and the oxygen through resonance. This makes amides much less basic compared to alkyl amines. And more the stabilization in amide less will be basic, here $\mathrm{PhCH}_2 \mathrm{CONH}_2$ will be stabilizes by resonance.

Topic: Chemistry, Amines, JEE Main
Decreasing order of dehydration of the following alcohols is

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

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

Solution:
The dehydration of alcohol takes place via the formation of a carbocation. Hence, the rate of reaction depends on the stability of carbocation.

No dehydration is possible in phenol because carbocation formation is difficult in case of phenol.

Option (d) fast due to loss of H2O and conjugation increasing stable of compound as compared to (c). Option(b) is very fast because stable compound formed.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2023 (01 Feb Shift 1)
Decreasing order of reactivity towards electrophilic substitution for the following compounds is:

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

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

Solution:
Reactivity of aromatic compounds towards electrophilic substitution depends on the electron density on benzene ring. Higher the electron density on benzene ring, more be the reactivity towards electrophilic substitution reaction. Hence, the correct decreasing order is

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2023 (15 Apr Shift 1)
Decreasing order of stability of following alkenes is

(i) CH3-CH=CH2

(ii) CH3-CH=CH-CH3

(iii) CH 3 −C = CH− CH 3           |          CH 3

(iv)           CH 3 CH 3            |     | CH 3 −C = C− CH 3

Options:
A (i) > (ii) > (iii) > (iv)
B (iv) > (iii) > (ii) > (i)
C (iii) > (ii) > (i) > (iv)
D (ii) > (iii) > (iv) > (i)

The correct answer is:
(iv) > (iii) > (ii) > (i)

Solution:
Stability ∝ No. of hyper conjugating structures ∝ No. of α -hydrogens.

Topic: Chemistry, General Organic Chemistry, JEE Main
Decreasing order of the hydrogen bonding in following forms of water is correctly represented by
A. Liquid water
B. Ice
C. Impure water

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

The correct answer is:
B>A>C

Solution:
The extra hydrogen bonds that occur when water freezes increase the space between molecules, causing a decrease in overall density. In fact, each water ice molecule forms hydrogen bonds with four other molecules, while water molecules only form hydrogen bonds with an average of 3.4 other molecules.

Due to impurity extent of H -Bonding decreases.

Ice > Liquid water > Impure water

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2023 (24 Jan Shift 1)
Decreasing order towards SN1 reaction for the following compounds is:

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

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

Solution:
The relative reactivity of alkyl halides in the SN1 reaction depends on the stability of carbocation. Electron-withdrawing groups decrease the stability and electron-releasing groups increase the stability of carbocations. -NO2 group is an electron-withdrawing group by -M effect. -OMe is electron-releasing groups by +M effect. Hence, the overall order is

∴ Reactivity order (b) > (d) > (c) > (a)

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2023 (30 Jan Shift 2)
Deep pink colour is given to flame by the salts of

Options:
A Strontium
B Potassium
C Zinc
D Barium

The correct answer is:
Strontium

Solution:
Strontium Crimson or pink colour

Topic: Chemistry, s Block Elements, JEE Main
Deep sea divers used to respirate a mixture of

Options:
A Oxygen and argon
B Oxygen and helium
C Oxygen and nitrogen
D Oxygen and hydrogen

The correct answer is:
Oxygen and helium

Solution:
An oxygen- helium mixture is used for artificial respiration in deep sea diving instead of air because nitrogen present in air dissolves in blood under high pressure when sea diver goes into deep sea. When he comes to the surface nitrogen bubbles out of the blood due to decrease in pressure, causing pain. This disease is called "bends".

Topic: Chemistry, Solutions, JEE Main
Deficiency of vitamin K causes :

Options:
A Increase in blood clotting time
B Increase in fragility of RBC 's
C Cheilosis
D Decrease in blood clotting time

The correct answer is:
Increase in blood clotting time

Solution:
Vitamin K refers to structurally similar, fat-soluble vitamins found in foods and marketed as dietary supplements. The human body requires vitamin K for post-synthesis modification of certain proteins that are required for blood coagulation or for controlling binding of calcium in bones and other tissues. Due to deficiency of vitamin K causes increases in blood clotting time.

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2021 (18 Mar Shift 2)
Define a function $f: R \rightarrow R$ by $f(x)=\max \{|x|,|x-1|, \ldots \ldots .|x-2 n|\}$
Where $\mathrm{n}$ is a fixed natural number, Then $\int_{0}^{2 n} f(x) d x$ is

Options:
A $\mathrm{n}$
B $\mathrm{n}^{2}$
C $3 \mathrm{n}$
D $3 \mathrm{n}^{2}$

The correct answer is:
$3 \mathrm{n}^{2}$

Solution:
$\int_{0}^{2 n} f(x) d x=\int_{0}^{n}|x-2 n| d x+\int_{n}^{2 n}|x| d x$
$\int_{0}^{2 n}(2 n-x) d x+\int_{n}^{2 n} x d x$
$=\left(2 n x-\frac{x^{2}}{2}\right)_{0}^{n}+\left(\frac{x^{2}}{2}\right)_{n}^{2 n}=3 n^{2}$

Topic: Mathematics, Definite Integration, JEE Main
Define a relation $R$ on $A=\{1,2,3,4\}$ as $x R y$ if $x$ divides $y . R$ is

Options:
A reflexive and transitive
B reflexive and symmetric
C symmetric and transitive
D equivalence

The correct answer is:
reflexive and transitive

Solution:
Given set $A=\{1,2,3,4\}$ and relation, $x R y$ if $x$ divides $y$.
$\Rightarrow$ Relation
$=\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,3),$, $(1,4),(2,4)\}$
Reflexive We have, $x R y \Leftrightarrow y / x$ for $x, y \in A$
For any $x \in A$, we have $x / x \Rightarrow x R x$.
Thus, $x R x$ for all $x \in A$. So, $R$ is reflexive on $A$.
Symmetry $R$ ia not symmetry because, if $y / x$, then $x$ may not divide $y$. For example $4 / 2$ but $2 / 4$.
Transitive, Let $x, y, z \in A$, such that $x R y$ and yRz.
Then, $x R y$ and $y R z \Rightarrow \frac{y}{x}$ and $\frac{z}{y} \Rightarrow \frac{z}{x}$. So, $R$ is a transitive relation on $A$.

Topic: Mathematics, Sets and Relations, JEE Main
Define a relation $R$ on the set $N$ of natural numbers by $R=\{(x, y): y=x+5, x$ is a natural number less than $4, x$, $y \in N\}$. Depict this relationship using (i) roster form (ii) an arrow diagram. Write down the domain and the range.

Options:

None

Solution:
Here, $R=\{(x, y): y=x+5, x$ is a natural number less than 4 , $x, y \in \mathrm{N}\}$.
(i) In roster form, $R=\{(1,6),(2,7),(3,8)\}$.
(ii) The arrow diagram is :
Domain is the set of first elements i.e., $\{1,2,3\}$. Range is the set of second elements i.e., $\{6,7,8\}$.

Topic: Mathematics, Relations and Functions (Class 11)
Define $f: R \rightarrow R$ by $f(x)=\max \{x+1,1-x, 2\}$. Then, $f$ is

Options:
A one-one but not onto
B onto but not one-one
C neither one-one nor onto
D both one-one and onto

The correct answer is:
neither one-one nor onto

Solution:
$f: R \rightarrow R$
$f(x)=\max \{x+1,1-x, 2\}$
Since, $f(x)$ contains 2 in range when domain is $[-1,1]$
Hence, it is not one-one.
Since, $f(x)$ does not take any value less than 2 in it's domain.
Hence, Codomain $\neq$ Range
$\Rightarrow f(x)$ is not onto.
Hence, $f(x)$ is neither one-one nor onto.

Topic: Mathematics, Functions, AP EAMCET, AP EAMCET 2021 (23 Aug Shift 2)
Define $f: R \rightarrow R$ by
$f(x)= \begin{cases}(x-a) \frac{e^{\frac{1}{(x-a)}}-1}{\frac{1}{(x-a)}}, 1 & \text { for } x \neq a \\ 0, & \text { at } x=a\end{cases}$
Then which one of the following is true?

Options:
A Left and right limits of $f$ at $x=a$ are equal and they are not equal to $f(a)$
B Both left and right limits of $f$ at $x=a$ exist and are not equal
C The function $f(x)$ is continuous at $x=a$
D The function $f(x)$ has a simple discontinuity at a point other than a

The correct answer is:
The function $f(x)$ is continuous at $x=a$

Solution:
We have,
$f(x)=\left\{\begin{array}{cc}
(x-a) \frac{e^{\frac{1}{x-a}}-1}{\frac{1}{e^{x-a}}+1}, & x \neq a \\
0, & x=a
\end{array}\right.$
$\begin{aligned}
\lim _{x \rightarrow a^{-}} f(x) & =\lim _{x \rightarrow a^{-}}(x-a) \frac{e^{\frac{1}{x-a}}-1}{e^{\frac{1}{x-a}}+1}=\lim _{h \rightarrow 0}-h \frac{e^{-1 / h}-1}{e^{-1 / h}+1}=0 \\
\lim _{x \rightarrow a+} f(x) & =\lim _{x \rightarrow a+}(x-a) \frac{e^{\frac{1}{x-a}}-1}{\frac{e^{x-a}}{x}} \\
& =\lim _{h \rightarrow 0} \frac{h\left(1-e^{-1 / h}\right)}{1+e^{-1 / h}}=0
\end{aligned}$
$\because f(x)$ is continuous at $x=a$

Topic: Mathematics, Limits, JEE Main
Define $f(x)=\left\{\begin{array}{cc}x^2+b x+c, & x < 1 \\ x, & x \geq 1\end{array}\right.$. If $f(x)$ is differentiable at $x=1$, then $(b-c)$ is equal to

Options:
A $-2$
B $0$
C $1$
D $2$

The correct answer is:
$-2$

Solution:
Given function,
$\begin{aligned}
& f(x)=\left\{\begin{array}{cc}
x^2+b x+c, & x < 1 \\
x, & x \geq 1
\end{array}\right. \\
& f^{\prime}(x)=\left\{\begin{array}{cc}
2 x+b, & x < 1 \\
1, & x \geq 1
\end{array}\right.
\end{aligned}$
Since, $f(x)$ is differentiable at $x=1$.
$\begin{aligned}
& \lim _{x \rightarrow 1^{+}} f^{\prime}(x)=\lim _{x \rightarrow 1^{+}} f^{\prime}(x) \\
& \lim _{x \rightarrow 1^{+}}(2 x+b)=\lim _{x \rightarrow 1^{+}} 1 \\
& 2+b=1 \Rightarrow b=-1
\end{aligned}$
As form is differentiable at $n=1$ So, it will be continuous at $x=1$ also.
$\begin{aligned}
& \Rightarrow \quad \lim _{x \rightarrow 1^{+}} f(x)=\lim _{\substack{x \rightarrow 1^{+}}} f^{\prime}(x)=f(1) \\
& \Rightarrow \quad \lim _{x \rightarrow 1^{+}} x^2+b x+c=\lim _{x \rightarrow 1^{+}} x=1 \\
& \Rightarrow \quad 1+b+c=1 \\
& \Rightarrow \quad 1-1+c=1 \Rightarrow c=1
\end{aligned}$
Hence, $b-c=-1-1=-2$

Topic: Mathematics, Continuity and Differentiability, JEE Main
Define $\mathrm{g}(\mathrm{x})=\int_{-3}^{3} \mathrm{f}(\mathrm{x}-\mathrm{y}) \mathrm{f}(\mathrm{y}) \mathrm{dy}$, for all real $\mathrm{x}$, where $f(t)=\left\{\begin{array}{ll}1, & 0 \leq t \leq 1 \\ 0, & \text { elsewhere }\end{array}\right.$
Then

Options:
A $\mathrm{g}(\mathrm{x})$ is not continuous everywhere
B $\mathrm{g}(\mathrm{x})$ is continuous everywhere but differentiable nowhere
C $\mathrm{g}(\mathrm{x})$ is continuous everywhere and differentiable everywhere except at $\mathrm{x}=0,1$
D $g(x)$ is continuous everywhere and differentiable everywhere except at $\mathrm{x}=0,1,2$

The correct answer is:
$g(x)$ is continuous everywhere and differentiable everywhere except at $\mathrm{x}=0,1,2$

Solution:
Definition can be break as
$\begin{array}{l}
g(x)=\int_{0}^{1} f(x-y) d y \\
x-y=t ;-d y \cdot d t \\
g(x)=\int_{x-1}^{x} f(t) d t \\
g(x)=\left\{\begin{array}{cc}
0 & x \leq 0 \\
x & 0 < x < 1 \\
2-x & 1 \leq x \leq 2 \\
0 & x>2
\end{array}\right.
\end{array}$
Now, check yourself

Topic: Mathematics, Continuity and Differentiability, JEE Main
Define self sustainability.

Options:

None

Solution:
The maintenance of an ecosystem itself or naturally is self-sustainability.

Topic: Biology, Ecosystem
Define the following as related to proteins:
(i) Peptide linkage
(ii) Primary structure
(iii) Denaturation

Options:

None

Solution:
(i) Peptide bond: Proteins are condensation polymers of \(\alpha\)-amino acids in which the same or different \(\alpha\)-amino acids are joined by peptide bonds. Chemically, a peptide bond is an amide linkage formed between - COOH group of one \(\alpha\)-amino acid and \(-\mathrm{NH}_2\) group of the other \(\alpha\)-amino acid by loss of a molecule of water. For example,
(ii) Primary structure: Proteins may contain one or more polypeptide chains. Each polypeptide chain has a large number of \(\alpha\)-amino acids which are linked to one another in a specific manner. The specific sequence in which the various amino acids present in a protein linked to one another is called its primary structure. Any change in the sequence of \(\alpha\)-amino acids creates a different protein.
(iii) Denaturation : Each protein in the biological system has a unique three-dimensional structure and has specific biological activity. This is called native form of a protein. When a protein in its native form is subjected to a physical change such as change in temperature or a chemical change like change in \(\mathrm{pH}\), etc., hydrogen bonds gets broken. As a result, soluble forms of proteins such as globular proteins undergo coagulation or precipitation to give fibrous proteins which are insoluble in water. This coagulation also results in loss of biological activity of the proteins and this loss in biological activity is called denaturation. During denaturation, \(2^{\circ}\) and \(3^{\circ}\) structures of proteins are destroyed but \(1^{\circ}\) structure remains intact.
The most common example of denaturation of proteins is the coagulation of albumin present in the white of an egg. When the egg is boiled hard, the soluble globular protein present in it is denatured and is converted into insoluble fibrous protein.

Topic: Chemistry, Biomolecules
Define the following terms?
(a) Tidal volume
(b) Residual volume
(c) Asthma

Options:

None

Solution:
(a) Tidal Volume (TV) is volume of air inspired or expired during a normal respiration. It is approx. $500 \mathrm{ml}$.i.e., a healthy man can inspire or expire approximately 6000 to $8000 \mathrm{~mL}$ of air per minute.
(b) Residual Volume : (RV) is volume of air remaining in the lungs even after a forcible expiration. This averages $1100 \mathrm{~mL}$ to $1200 \mathrm{~mL}$.
(c) Asthma is an allergic reaction that causes constriction of the bronchiole muscles, thereby reducing the air passage thus the amount of the air that can get to the alveoli.

Topic: Biology, Breathing and Exchange of Gases