datavorous/entrance-exam-dataset huggingface dataset:
Class XII students were asked to prepare one litre of buffer solution of pH 8.26 by their chemistry teacher. The amount of ammonium chloride to be dissolved by the student in 0.2M ammonia solution to make one litre of the buffer is (Given pKbNH3=4.74; Molar mass of NH3=17 g mol-1. Molar mass of NH4Cl=53.5 g mol-1)

Options:
A 53.5 g
B 72.3 g
C 107 g
D 126g

The correct answer is:
107 g

Solution:
A mixture of NH3 and NH4Cl will form a basic buffer.

pH of final solution =8.26

pOH of final solution=14-8.26 =5.74

Let the number of moles of NH4Cl added to 1 L solution of be x.

Using Henderson’s equation.

pOH=pKb+logNH4ClNH3

5.74=4.74+logx0.2

∴ x=2

The amount of ammonium chloride to be dissolved by the student in 0.2M ammonia solution to make one litre of buffer=2×53.5=107g

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2022 (26 Jul Shift 2)
Classically, an electron can be in any orbit around the nucleus of an atom. Then what determines the typical atomic size? Why is an atom not, say, thousand times bigger than its typical size? The question had greatly puzzled Bohr before he arrived at his famous model of the atom that you have learnt in the text. To simulate what he might well have done before his discovery, let us play as follows with the basic constants of nature and see if we can get a quantity with the dimensions of length that isroughly equal totheknown size of an atom $\left(\sim 10^{-}\right.$ 10 m).
(a) Construct a quantity with the dimensions of length from the fundamental constants $e, m_e$, and c. Determine its numerical value.
(b) You will find that the length obtained in (a) is many orders of magnitude smaller than the atomic dimensions. Further, it involves c. But energies of atoms are mostly in non-relativistic domain where c is not expected to play any role. This is what may have suggested Bohr to discard $\mathrm{c}$ and look for 'something else' to get the right atomic size. Now, the Planck's constant h had already made its appearance elsewhere. Bohr's great insight lay in recognising that $h, \mathrm{~m}_{\mathrm{e}}$ and e will yield the right atomic size. Construct a quantity with the dimension of length from $h, m_e$ and e and confirm that its numerical value has indeed the correct order of magnitude.

Options:

None

Solution:
(a) The quantity $\left(\frac{\mathrm{e}^2}{\varepsilon_0 \mathrm{mc}^2}\right)$ has dimensions of length. Its value $\simeq 10^{-15} \mathrm{~m}$ is much smaller than atomic size.
(b) The quantity $\left(\frac{\varepsilon_0(h)^2}{m e^2}\right)$ has dimensions of length. Its value $\approx 10^{-10} \mathrm{~m}$ of the order of atomic size. However constants like $2 \pi$ or $4 \pi$ have to be added to arrive at correct formula.

Topic: Physics, Atomic Physics
$\lambda_{\mathrm{ClCH}_2 \mathrm{COONa}}=224 \mathrm{ohm}^{-1} \mathrm{~cm}^2 \mathrm{gmeq}^{-1}$
$\lambda_{\mathrm{NaCl}}=38.2 \mathrm{ohm}^{-1} \mathrm{~cm}^2 \mathrm{gmeq}^{-1}$
$\lambda_{H C l}=2030 \mathrm{hm}^{-1} \mathrm{~cm}^2 \mathrm{gmeq}^{-1}$
What is the value of $\lambda_{\mathrm{ClCH}_2 \mathrm{COOH}}$

Options:
A $288.5 \mathrm{ohm}^{-1} \mathrm{~cm}^2 \mathrm{gmeq}^{-1}$
B $289.5 \mathrm{ohm}^{-1} \mathrm{~cm}^2 \mathrm{gmeq}^{-1}$
C $388.5 \mathrm{ohm}^{-1} \mathrm{~cm}^2 \mathrm{gmeq}^{-1}$
D $59.5 \mathrm{ohm}^{-1} \mathrm{~cm}^2 \mathrm{gmeq}^{-1}$

The correct answer is:
$388.5 \mathrm{ohm}^{-1} \mathrm{~cm}^2 \mathrm{gmeq}^{-1}$

Solution:
$\mathrm{ClCH}_2 \mathrm{COONa}^{\mathrm{HCl}} \rightarrow \mathrm{ClCH}_2 \mathrm{COOH}+\mathrm{NaCl}$
$\lambda_{\mathrm{ClCH}_2 \mathrm{COONa}}+\lambda_{\mathrm{HCl}}=\lambda_{\mathrm{ClCH}_2 \mathrm{COOH}}+\lambda_{\mathrm{NaCl}}$
$224+203=\lambda_{\mathrm{ClCH}_2 \mathrm{COOH}}+38.2$
$\lambda_{\mathrm{ClCH}_2 \mathrm{COOH}}=427-38.2=388.8 \mathrm{ohm}^{-1} \mathrm{~cm}^2 \mathrm{gmeq}^{-1}$

Topic: Chemistry, Electrochemistry, JEE Main
Clemmensen reduction of a ketone is carried out in the presence of :

Options:
A
$\mathrm{LiAlH}_4$

B
$\mathrm{Zn}-\mathrm{Hg}$ with $\mathrm{HCl}$

C
Glycol with $\mathrm{KOH}$

D
$\mathrm{H}_2$ with $\mathrm{Pt}$ as catalyst

The correct answer is:
$\mathrm{Zn}-\mathrm{Hg}$ with $\mathrm{HCl}$

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2013 (22 Apr Online)
ClF5 at room temperature is a

Options:
A

Colourless liquid with trigonal bipyramidal geometry

B

Colourless gas with square pyramidal geometry

C

Colourless gas with trigonal bipyramidal geometry

D

Colourless liquid with square pyramidal geometry

The correct answer is:

Colourless liquid with square pyramidal geometry

Solution:
Chlorine pentafluoride is an inter halogen compound. ClF5 is colourless liquid. The central chlorine atom is having five bond pairs and one lone pair on it. Hence, according to VSEPR theory it has square pyramidal geometry. It is toxic by inhalation and an irritant to skin, eyes and mucus membranes.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2023 (13 Apr Shift 1)
ClO 2 is an/a

Options:
A anhydride of HClO 2
B anhydride of HClO 3
C mixed anhydride of HClO 2 and HClO 3
D mixed anhydride of HClO 3 and HClO 4

The correct answer is:
mixed anhydride of HClO 2 and HClO 3

Solution:
CIO 2 reacts KOH forming KClO 3 and KCIO 2
2âKOHâ+â 2CIO 2 â KCIO 2 â+â KCIO 3 â+â H 2 O
This reaction indicates that CIO 2 is a mixed anhydride of HClO 2 and HCIO 3 .

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Closeness of two or more measured values is called as

Options:
A accuracy
B precision
C error
D approximation

The correct answer is:
precision

Solution:
Closeness of two or more measured values is called as precision. The precision reflects how reproducible measurements are, even if they are far from the accepted value. As much the measured values are close, more the measurement is accurate.

Topic: Physics, Mathematics in Physics, JEE Main
CN − solution used in extraction of which metal?

Options:
A Ag
B Ti
C Zn
D Sn

The correct answer is:
Ag

Solution:
CN − solution used in extraction of Ag metal in the cyanide process.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
CNG is an important transportation fuel. When 100 gCNG is mixed with 208 g oxygen in vehicles, it leads to the formation of CO2 and H2O and produces large quantity of heat during this combustion, then the amount of carbon dioxide, produced in grams is [nearest integer] [Assume CNG to be methane]

Options:

The correct answer is:
143

Solution:
CH4+ 2O2⟶ CO2+2H2O

10016 20832

Number of molesstoichiometric coefficient for CH4=6.251 Number of molesstoichiometric coefficient for O2=6.52=3.25

The reagent with lower ratio of number of moles to stoichiometric coefficient is called limiting reagent.

Hence, O2 is limiting reagent

Limiting reagent decides the amount of product,

As per the equation 2 moles of O2 gives rise to 1 mole of CO2, So 6.5 moles of O2 gives 6.52 moles of CO2

⇒nCO2=6.52

So mass of CO2=6.52×44=143gm

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2022 (26 Jun Shift 2)
C+O 2 â CO 2 ;ââÎH=X

CO+12O2âCO2 ;Î H = Y

Then the heat of formation of CO is

Options:
A X-Y
B Y-2X
C X+Y
D 2X-Y

The correct answer is:
X-Y

Solution:
Câ+â O 2 â CO 2 ; âÎHâ=âX ...(i)
COâ+â 1 2 â O 2 â CO 2 ; âÎHâ=âY ...(ii)
Eqs. (i) -(ii)
C +  1 2   O 2 →CO;  ΔH = X−Y

Topic: Chemistry, Thermodynamics (C), JEE Main
Co2CO8 displays

Options:
A no Co-Co bonds, six terminal CO and two bridging CO.
B no Co-Co bonds, four terminal CO and four bridging CO.
C one Co-Co bonds, six terminal CO and two bridging CO.
D one Co-Co bonds, four terminal CO and four bridging CO.

The correct answer is:
one Co-Co bonds, six terminal CO and two bridging CO.

Solution:
There is 1 metal-metal bond.

The structure of Co2(CO)8 is octahedral.

There are 6 ligands in the structure.

The structure has two tautomeric forms.

In Co2(CO)8, 1 Co-Co bond, 6 terminal Co and 2 bridging Co is present.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2017 (09 Apr Online)
CO2 gas adsorbs on charcoal following Freundlich adsorption isotherm. For a given amount of charcoal, the mass of CO2 adsorbed becomes 64 times when the pressure of CO2 is doubled.

The value of n in the Freundlich isotherm equation is ____ ×102. (Round off to the Nearest Integer)

Options:

The correct answer is:
17

Solution:
Freundlich isotherm. ;
xm=k·p1n
Substituting values ; 641=(2)1n⇒n=16=0.166
≅17×10-2

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2021 (27 Jul Shift 1)
$\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{NO}_{2}$ and $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{ClNO}_{2}\right] \mathrm{Cl}$ exhibit which type of isomerism?

Options:
A Geometrical
B Optical
C Linkage
D Ionisation

The correct answer is:
Ionisation

Solution:
On ionisation they give different ions
$$
\begin{array}{c}
{\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{NO}_{2} \longrightarrow}
{\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right]^{+}+\mathrm{NO}_{2}^{-}} \\
{\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{ClNO}_{2}\right] \mathrm{Cl} \longrightarrow}
{\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{CINO}_{2}\right]^{+}+\mathrm{Cl}^{-}}
\end{array}
$$
So, they show ionisation isomerism.

Topic: Chemistry, Coordination Compounds, JEE Main
Coagulating value of the electrolytes AlCl3 and NaCl tor As2 S3 are 0.09 and 50.04 respectively. The coagulating power of AlCl3 is x times the coagulating power of NaCl. The value of x is

Options:

The correct answer is:
556

Solution:
The coagulating power of an electrolyte is a measure of its ability to cause the coagulation or precipitation of a colloidal solution.

Coagulation power α1Coagulation value

Coagulating value of AlCl3 for As2 S3=0.09
Coagulating value of NaCl for As2 S3=50.04

Then,

(C.V)AlCl3 (C.V)NaCl =(C.P)NaCl (C.P)AlCl30.0950.04=(C.P)NaCl (C.P)AlCl3 (C.P)AlCl3=556(C.P)NaCl
So the answer is 556.

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2023 (08 Apr Shift 2)
Coagulation of egg, on heating is because of :

Options:
A The secondary structure of protein remains unchanged
B Denaturation of protein occurs
C Biological property of protein remains unchanged
D Breaking of the peptide linkage in the primary structure of protein occurs

The correct answer is:
Denaturation of protein occurs

Solution:
Coagulation of egg give primary structure of protein, which is known as denaturation of protein

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2024 (05 Apr Shift 2)
Cobalt (60) isotope is used in the treatment of :

Options:
A Heart diseases
B Skin diseases
C Diabetes
D Cancer

The correct answer is:
Cancer

Solution:
Cobalt ( 60 ) isotope is used in the treatment of cancer.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
Cobalt chloride when dissolved in water forms pink colored complex X which has octahedral geometry. This solution on treating with cone HCl forms deep blue complex, Y which has a Z geometry. X, Y and Z, respectively, are

Options:
A X=CoH2O62+, Y=CoCl42-, Z=Tetrahedral
B X=CoH2O62+, Y=CoCl63-, Z=Octahedral
C X=CoH2O63+, Y=CoCl63-, Z=Octahedral
D X=CoH2O4Cl2+, Y=CoCl42-, Z=Tetrahedral

The correct answer is:
X=CoH2O62+, Y=CoCl42-, Z=Tetrahedral

Solution:
Cobalt chloride when dissolved in water forms pink coloured isCoH2O6Cl2 . This is due to the electronic transition of an electron from t2g to eg energy level of CoH2O62+ complex. When excess of HCl is added to the solution CoH2O62+ is transformed into deep blue colour complex [CoCl4]2−.

△t= 49 △o

△t < △o

Tetrahedral complexes have smaller crystal field splitting than octahedral complexes.

CoCl2+6H2O⟶CoH2O6Cl2

CoH2O62+ →Conc. HCl [CoCl4]2-

Hence, X=CoH2O62+, Y=CoCl42-, Z=Tetrahedral

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2023 (31 Jan Shift 1)
Cobalt-57 is radioactive, emitting Î² -particles. The half-life for this is 270 days. If 100 mg of this is kept in an open container, then what mass (in mg) of Cobalt-57 will remain after 540 days?

Options:

The correct answer is:
25

Solution:
Here, T12=270 days.
After 540 days, it means two half lives have been completed i.e., n=2
If NN0=12n
â mm0=12n [ âµ from Equation. (i)]
â m100=14
â m=25 mg

Topic: Physics, Nuclear Physics, JEE Main
Coefficient of kinetic friction between 3 kg and 2 kg block is 0.25. The horizontal table surface is smooth. Find the acceleration (in S.I unit) of the block of mass 10 kg.

Options:

The correct answer is:
6

Solution:
100-T2=10a

T2-T1-0.25×20=3a

T1-0.25×20=2a

Adding the above three equations and solving for acceleration we get

a=100-0.25×20-0.25×2015=6 m s-2

Topic: Physics, Laws of Motion, JEE Main
Coefficient of linear expansion of brass and steel rods are α1 and α2. Lengths of brass and steel rods are l1 and l2, respectively. If l2-l1 is maintained the same at all temperatures, which one of the following relations holds good?

Options:
A α1l2=α22l1
B α1l22=α2l12
C α12l2=α22l1
D α1l1=α2l2

The correct answer is:
α1l1=α2l2

Solution:
The difference in length is the same, so the increase in length is equal,

∆l1=∆l2

l1α1∆T=l2α2∆T

⇒l1α1=l2α2.

Topic: Physics, Thermal Properties of Matter, JEE Main
Coefficient of linear expansion of brass and steel rods are α1 and α2. Lengths of brass and steel rods are l1 and l2, respectively. If l2-l1 is maintained the same at all temperatures, which one of the following relations holds good?

Options:
A α1l2=α22l1
B α1l22=α2l12
C α12l2=α22l1
D α1l1=α2l2

The correct answer is:
α1l1=α2l2

Solution:
The difference in length is the same, so the increase in length is equal,

∆l1=∆l2

l1α1∆T=l2α2∆T

⇒l1α1=l2α2.

Topic: Physics, Thermal Properties of Matter, NEET, NEET 2016 (Phase 1)
Coefficient of αt in the expansion of α+pm-1+α+pm-2α+q+α+pm-3α+q2+.………+α+qm-1, where α≠-q and p≠q is

Options:
A Cmtpt-qtp-q
B Cmtpm-t-qm-tp-q
C Cmtpt+qtp-q
D Cmtpm-t+qm-tp-q

The correct answer is:
Cmtpm-t-qm-tp-q

Solution:
The terms are in GP with a=α+pm-1,r=α+qα+p and number of terms =m.
S=α+pm−1α+qα+pm−1α+qα+p−1=α+qm-α+pmq−p
Coefficient of αt in 1q-pα+qm-α+pm
=1q-pCtm qm-t-Ctm pm-t
=Ctpm-t-qm-tmp-q

Topic: Mathematics, Binomial Theorem, JEE Main
Coefficient of thermal conductivity is the product of heat, distance and reciprocal of (area Ã difference in temperature Ã time). The new value of a unit of coefficient of thermal conductivity, if fundamental units are 21.6kg, 1 decimetre, 4 K and 1 minute will be ______ Ã10-6 new units. If your answer is K, mark 2K as answer after rounding off to nearest integer.

Options:

The correct answer is:
5

Solution:
Coefficient of thermal conductivity, k=QÃlAÃÎÎ¸Ãt
â K=ml2t2Ãll2ÃÎÎ¸Ãt
â K=mÃlÎÎ¸Ãt3=21.6Ã0.14Ã603
=2.5Ã10-6 new units

Topic: Physics, Units and Dimensions, JEE Main
Coefficient of thermal conductivity is the product of heat, distance and reciprocal of (area Ã difference in temperature Ã time). The new value of a unit of coefficient of thermal conductivity, if fundamental units are 21.6kg, 1 decimetre, 4 K and 1 minute will be ______ Ã10-6 new units.

Options:

The correct answer is:
02.50

Solution:
Coefficient of thermal conductivity, k=QÃlAÃÎÎ¸Ãt
â K=ml2t2Ãll2ÃÎÎ¸Ãt
â K=mÃlÎÎ¸Ãt3=21.6Ã0.14Ã603
=2.5Ã10-6 new units

Topic: Physics, Units and Dimensions, JEE Main
Coefficient of variation of two distributions are 50 and 60 and their arithmetic means are 30 and 25, respectively. Difference of their standard deviation is

Options:
A 0
B 1
C 1.3
D 2.5

The correct answer is:
0

Solution:
Given, coefficient of variation C1=50

and coefficient of variation C2=60.

We have, x¯1=30 and x¯2=25

∵ C=σx¯×100

50=σ130×100

⇒ σ1=15

and 60=σ225×100

⇒ σ2=15

∴ Required difference σ1-σ2=15-15=0

Topic: Mathematics, Statistics, JEE Main
Coefficient of variation of two distributions are 60% and 75%, and their standard deviation are 18 and 15 respectively, then their arithmetic means respectively are

Options:
A 30,30
B 30,20
C 20,30
D 20,20

The correct answer is:
30,20

Solution:
Given, CV1=60,CV2=75,σ1=18 and σ2=15
Let x-1 and x-2 be the means of 1st and 2nd distribution respectively.
Then, CV1=σ1x-1×100⇒x-1=18×10060=30
CV2=σ2x-2×100⇒x-2=15×10075=20
Hence, x-1=30 and x-2=20

Topic: Mathematics, Statistics, JEE Main
Coefficient of x48 in ∑r=050C50rx-2r350-r

Options:
A C502
B C482
C 348
D 248

The correct answer is:
C502

Solution:
∑r=050C50r350-rx-2r
=3+x-250
=x+150
Coefficient of x48 in 1+x50 is C5048=C502

Topic: Mathematics, Binomial Theorem, JEE Main
Coefficient of $x$ in the expansion of $\left(x^2+\frac{a}{x}\right)^5$ is

Options:
A $9 a^2$
B $10 a^3$
C $10 a^2$
D $10 a$

The correct answer is:
$10 a^3$

Solution:
In the expansion of $\left(x^2+\frac{a}{x}\right)^5$
the general term is $T_{r+1}={ }^5 C_r\left(x^2\right)^{5-r}\left(\frac{a}{x}\right)^r={ }^5 C_r a^r x^{10-3 r}$
Here, exponent of $x$ is $10-3 r=1 \Rightarrow r=3$
$\therefore \quad T_{2+1}={ }^5 C_3 a^3 x=10 a^3 \cdot x$
Hence coefficient of $x$ is $10 a^3$

Topic: Mathematics, Binomial Theorem, JEE Main
Coherent sources are characterized by the same

Options:
A Phase and phase velocity
B Wavelength, amplitude and phase velocity
C Wavelength
D Wavelength and phase

The correct answer is:
Wavelength, amplitude and phase velocity

Solution:
Because in general phase velocity = wave velocity. But in case of complex waves (many waves together) phase velocity $\neq$ wave velocity

Topic: Physics, Waves and Sound, JEE Main
Coherent sources are those sources for which

Options:
A Phase difference remain constant
B Frequency remains constant
C Both phase difference and frequency remains constant
D None of these

The correct answer is:
Both phase difference and frequency remains constant

Solution:
Both phase difference and frequency remains constant

Topic: Physics, Wave Optics, JEE Main
Cohesion, Adhesion and Surface tension, gives water-

Options:
A High Tensile strength and low capillarity
B Low tensile strength and low capillarity
C High tensile strength and high capillarity
D Low tensile strength and high capillarity

The correct answer is:
High tensile strength and high capillarity

Solution:
The transpiration driven ascent of xylem sap depends mainly on the

Topic: Biology, Transport in Plants, NEET
â âColdââdil.âalk .KMnO 4 â X â â HIO 4 â Y â â OH â â Z

In the above sequence of reactions, Z is

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
Cold-blooded animals fall under the category of

Options:
A Ectotherms
B Psychrotherms
C Endotherms
D Thermophiles

The correct answer is:
Ectotherms

Solution:
Ectotherms (poikilotherms) are the organisms that maintain body temperature by absorbing heat from the surrounding environment. All animals except mammals and birds are ectotherms. They are often described as being cold-blooded and are unable to regulate their body temperature metabolically.

Topic: Biology, Animal Kingdom, NEET
Collenchyma differs from sclerenchyma in

Options:
A Having thick walls
B Retaining protoplasm at maturity
C Being meristematic
D Having a wide lumen

The correct answer is:
Retaining protoplasm at maturity

Solution:
Collenchyma are living mechanical cells, they have non uniformly thick wall which is mainly made up of pectin. They retain protoplasm at maturity and if chloroplast is present they carry out photosynthesis.

Topic: Biology, Anatomy of Flowering Plants, NEET
Colloidal solutions can be purified by

Options:
A
emulsification

B
electrodialysis

C
peptization

D
using Tyndall effect

The correct answer is:
electrodialysis

Solution:
No Solution Available

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2012 (26 May Online)
Colloidal solutions of metals like gold, silver and platinum are generally prepared by using

Options:
A Peptization
B Bredig's arc method
C Exchange of solvent
D Oxidation method

The correct answer is:
Bredig's arc method

Solution:
The correct option is $\mathbf{B}$ bredig's arc method Colloidal solutions of metals are generally prepared by Bredig's arc method in which an arc is struck between the metal electrodes under the surface of water Containing some stabilizing agents such as a track of $\mathrm{KOH}$.

Topic: Chemistry, Surface Chemistry, JEE Main
Colloidal sulphur is obtained when

Options:
A sulphur is heated gradually to high temperature
B sulphur is heated with dilute sulphuric acid
C hydrogen sulphide is passed through an aqueous solution of nitric acid
D sulphur is warned with carbon disulphide

The correct answer is:
hydrogen sulphide is passed through an aqueous solution of nitric acid

Solution:
HNO3 oxideses H2S to S , which is obtained in colloidal form
H 2 S+2HN O 3 →S+2N O 2 +2 H 2 O

Topic: Chemistry, Surface Chemistry, JEE Main
Colour of the sky is blue due to

Options:
A Scattering of light
B Total internal reflection
C Total emission
D None of the above

The correct answer is:
Scattering of light

Solution:
Intensity of scattered light $I \propto \frac{1}{\lambda^4}$, since $\lambda_{\text {blue }}$ is least that's why sky looks blue.

Topic: Physics, Ray Optics, JEE Main
Colouration of $\mathrm{Br}_2 / \mathrm{CCl}_4$ will be discharged by

Options:
A Cinnamic acid
B Benzoic acid
C o-phthalic acid
D Acetophenone

The correct answer is:
Cinnamic acid

Solution:
Cinnamic acid reacts with bromine in carbon tetrachloride to give dibremecipnamic acid.

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
Column I list the parts of the human brain and column II lists the functions. Match the two columns and identify the CORRECT choice from those given.

Column I	Column II

A. Cerebrum	p. Controls the pituitary

B. Cerebellum	q. Controls vision and hearing

C. Hypothalamus	r. Controls the rate of heartbeat

D. Midbrain	s. Seat of intelligence

t. Maintains body posture

Options:
A A = t, B= s, C = q, D = p
B A = s, B= t, C = q, D = p
C A = t, B= s, C = p, D = q
D A = s, B= t, C = p, D = q

The correct answer is:
A = s, B= t, C = p, D = q

Solution:
The largest part of the brain located in the front of the skull is the cerebrum. It is divided into two hemisphere and is responsible integration of complex sensory and neural functions. It is the main centre for storage of information, memories and intelligence.
One of the major structures of the hindbrain is the cerebellum (little brain). It is responsible for the voluntary motor movements in a co-ordinated manner and also involved in the balance and posture of an individual. The information received by the sensory system is processed by the cerebellum to give the motor output like balancing.
Mesencephalon, composed of the tectum and tegmentum of the developing vertebrate brain, is also called as midbrain. This part of the brain plays an important role in movement functions, particularly movements of the eye. Also, helps in auditory and visual processing.
A small section located at the base of brain near pituitary is called as hypothalamus. Hypothalamus plays a crucial role in various important functions. It controls the activity of pituitary gland, regulates body temperature.

Topic: Biology, Neural Control and Coordination, NEET
Column-I lists the parts of the human brain and column-II lists the functions. Match the two columns and identify the correct choice from those given.

Column I	Column II

A.	Cerebrum	p.	Seat of intelligence

B.	Cerebellum	q.	Receive visual and auditory inputs

C.	Hypothalamus	r.	Controls the rate of heartbeat

D.	Midbrain	s.	Controls the pituitary

t.	Maintains body posture

Options:
A A = t; B = r; C = q; D = p
B A = s; B = t; C = q; D = p
C A = t; B = s; C = p; D = r
D A = p; B = t; C = s; D = q

The correct answer is:
A = p; B = t; C = s; D = q

Solution:
Cerebrum	The cerebrum is the largest part of the human brain. It is the seat of intelligence

Cerebellum	The cerebellum is the part of the hindbrain responsible for maintaining body Posture.

Hypothalamus	Hypothalamus is located near the pituitary gland, helps to control the pituitary.

Midbrain	Midbrain receives visual and auditory inputs.

Medulla oblongata	It controls the rate of heartbeat.

Topic: Biology, Neural Control and Coordination, NEET