datavorous/entrance-exam-dataset huggingface dataset:
Vectors $\vec{A}, \vec{B}$, and $\vec{C}$ are such $\vec{A} \cdot \vec{B}=0$ $\vec{A} \cdot \vec{C}=0$. Then the vector parallel is $\vec{A}$ is:

Options:
A $\vec{A} \times \vec{B}$
B $\vec{B} \times \vec{C}$
C $\vec{B} \times \vec{C}$
D $\vec{B}$ and $\vec{C}$

The correct answer is:
$\vec{B} \times \vec{C}$

Solution:
From vector produce
$$
\vec{A} \times(\vec{B} \times \vec{C})=(\vec{B} \cdot \vec{C}) \vec{B}-(\vec{A} \cdot \vec{B}) \vec{C}
$$
Given that $\vec{A} \cdot \vec{B}=0, \vec{A} \cdot \vec{C}=\mathrm{C}$ then $\vec{A} \times(\vec{B} \times \vec{C})=0$ thus vector $A$ is paralles to $\vec{B} \times \vec{C}$

Topic: Physics, Mathematics in Physics, JEE Main
Vectors ai^+bj^+k^ and 2i^-3j^+4k^ are perpendicular to each other when 3a+2b=7, the ratio of a to b is x2. The value of x is _____.

Options:

The correct answer is:
1

Solution:
Dot product of two perpendicular vectors will be zero. Therefore,

ai^+bj^+k^·2i^-3j^+4k^=0

⇒2a-3b+4=0 ...Eq(1)

As given in the question,

3a+2b=7 ...Eq(2)

From both equations, we get a=1 & b=2

Then,

ab=x2⇒x=2ab=2×12

⇒ x=1

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2023 (24 Jan Shift 1)
Vegetative propagation by runner or modified runner takes place in

Options:
A Eichhornia
B Strawberry
C Both A and B
D None of these

The correct answer is:
Both A and B

Solution:
Vegetative propagules are the plant parts or organs that are involved in the process of reproduction even though their primary function is not it. The reproductive process is called as vegetative propagation as the non-reproductive parts/organs (leaf/stem/roots) are involved.
Stolon, modified runner, is a slender lateral branch that arises from the base of main axis. It is a modified sub-aerial stem observed in water hyacinth (Eichhornia crassipes), strawberry. Initially, stolon grows upwards and then bends down and touches the soil where its terminal end gives rise to a new shoot and roots.

Topic: Biology, Reproduction in Organisms, NEET
Vegetative propagation in mint occurs by :

Options:
A Sucker
B Runner
C Offset
D Rhizome

The correct answer is:
Sucker

Solution:
No Solution Available

Topic: Biology, Reproduction in Organisms, NEET, NEET 2009 (Mains)
Vegetative propagation in Pistia is

Options:
A Stolon
B Offset
C Runner
D Sucker

The correct answer is:
Offset

Solution:
Pistia is a floating aquatic plant. Vegetative propagation in Pistia occurs by the formation of offset. An offset or offshoot is a new shoot, branch, or stem or a whole young rooted plant produced by the main stem of the parents, usually starting from an axillary-bud at the plant base or from a rhizome or tubers.

Topic: Biology, Reproduction in Organisms, NEET
Vegetative propagation in Pistia occurs by

Options:
A stolen
B offset
C runner
D sucker

The correct answer is:
offset

Solution:
In water lattuce (Pistia), vegetative propagation occurs by offsets. Offsets are more shorter and thicker small runners growing in the axil of leaf. It bears cluster of leaves in rosette manner above the water level and adventitious roots below the water level arising from all nodes.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET, NEET 2010 (Mains)
Velocities (V) and accelerations (a) in two systems of units 1 and 2 are related as $V_2=\frac{n}{m^2} V_1$ and $a_2=\frac{a_1}{m n}$ respectively. Here $\mathrm{m}$ and $\mathrm{n}$ are constants. Dimensionally relations between distances $\left(\mathrm{S}_1\right.$ and $\left.\mathrm{S}_2\right)$ and times $\left(\mathrm{t}_1\right.$ and $t_2$ ) in the two systems are respectively.

Options:
A $S_2=\left(\frac{n}{m}\right)^3 S_1$ and $t_2=\frac{n^2}{m} t_1$
B $\mathrm{S}_2=\left(\frac{\mathrm{n}}{\mathrm{m}}\right)^3 \mathrm{~S}_1$ and $\mathrm{t}_2=\frac{\mathrm{m}}{\mathrm{n}^2} \mathrm{t}_1$
C $S_2=\frac{m}{n^2} S_1$ and $t_2=\frac{m^2}{n^4} t_1$
D $\mathrm{S}_2=\frac{\mathrm{n}^2}{\mathrm{~m}} \mathrm{~S}_1$ and $\mathrm{t}_2=\frac{\mathrm{m}^2}{\mathrm{n}^4} \mathrm{t}_1$

The correct answer is:
$S_2=\left(\frac{n}{m}\right)^3 S_1$ and $t_2=\frac{n^2}{m} t_1$

Solution:
Given, $v_2=\frac{n}{m^2} v_1$ and $a_2=\frac{a_1}{m n}$
$\begin{aligned}
& \frac{v_2}{v_1}=\frac{n}{m^2} \quad \frac{a_2}{a_1}=\frac{1}{m n} \\
& a=\frac{v}{t} \\
& \frac{a_2}{a_1}=\frac{v_2}{t_2} \times \frac{t_1}{v_1} \Rightarrow \frac{1}{m n}=\frac{n}{m^2} \times \frac{t_1}{t_2} \\
& t_2=\frac{n^2}{m} t_1 \\
& \frac{v_2}{v_1}=\frac{S_2}{t_2} \times \frac{t_1}{S_1} \Rightarrow \frac{n}{m^2}=\frac{S_2}{S_1} \times \frac{m}{n^2} \\
& \frac{S_2}{S_1}=\frac{n^3}{m^3} \Rightarrow S_2=\left(\frac{n}{m}\right)^3 S_1
\end{aligned}$

Topic: Physics, Units and Dimensions, JEE Main
Velocity and acceleration vectors of charged particle moving perpendicular to the direction of a magnetic field at a given instant of time are $\overrightarrow{\mathbf{v}}=2 \hat{\mathbf{i}}+c \hat{\mathbf{j}}$ and $\overrightarrow{\mathbf{a}}=3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}$ respectively. Then the value of $c$ is

Options:
A $3$
B $1.5$
C $-1.5$
D $-3$

The correct answer is:
$-1.5$

Solution:
When a charged particle moves perpendicular to the magnetic field, it follows a circular path. On the circular path its velocity and acceleration will be at right angle to each other.
Given, $\overrightarrow{\mathbf{v}}=2 \hat{\mathbf{i}}+c \hat{\mathbf{j}}, \quad \overrightarrow{\mathbf{a}}=3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}$
As $\overrightarrow{\mathbf{v}}$ and $\overrightarrow{\mathbf{a}}$ are perpendicular, hence
$\overrightarrow{\mathbf{v}} \cdot \overrightarrow{\mathbf{a}}=0$
$(2 \hat{\mathbf{i}}+c \hat{\mathbf{j}}) \cdot(3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}})=0 \Rightarrow 6+4 c=0 \Rightarrow c=-1.5$

Topic: Physics, Magnetic Effects of Current, JEE Main
Velocity constant of a reaction at $290 \mathrm{~K}$ was found to be $3.2 \times 10^{-3}$. At $300 \mathrm{~K}$ it will be

Options:
A $1.28 \times 10^{-2}$
B $9.6 \times 10^{-3}$
C $6.4 \times 10^{-3}$
D $3.2 \times 10^{-4}$

The correct answer is:
$6.4 \times 10^{-3}$

Solution:
The velocity constant doubles for every $10^{\circ} \mathrm{C}$ rise in temperature.

Topic: Chemistry, Chemical Kinetics, JEE Main
Velocity of a particle moving in a straight line varies with its displacement as v = 4 + 4 s m/s . Displacement of particle at time t = 0 is s = 0. Find displacement of particle at time t = 2 s.

Options:
A 8 m
B 6 m
C 4 m
D 10 m

The correct answer is:
8 m

Solution:
as a=vdvds=4+4s124+4sÃ4=2,
which is constant so we can use equation of motion now Squaring the given equation, we get
v2 = 4 + 4s
Now, comparing it with v2 = u2 + 2as
we get, u = 2 m/s and a = 2 m/s2
â´ Displacement at t = 2s is
s = ut + 1 2 at 2 or s = 2 2 + 1 2 2 2 2
or s = 8 m

Topic: Physics, Motion In Two Dimensions, JEE Main
Velocity of sound in air
I. Increases with temperature
II. Decreases with temperature
III. Increase with pressure
IV. Is independent of pressure
V. Is independent of temperature
Choose the correct answer.

Options:
A Only I and II are true
B Only I and III are true
C Only II and III are true
D Only I and IV are true

The correct answer is:
Only I and IV are true

Solution:
Speed of sound $v \propto \sqrt{T}$ and it is independent of pressure.

Topic: Physics, Waves and Sound, JEE Main
Velocity of sound in air is

Options:
A Faster in dry air than in moist air
B Directly proportional to pressure
C Directly proportional to temperature
D Independent of pressure of air

The correct answer is:
Independent of pressure of air

Solution:
$v=\sqrt{\frac{\gamma P}{\rho}}$ as $P$ changes, $\rho$ also changes. Hence $\frac{P}{\rho}$ remains constant so speed remains constant.

Topic: Physics, Waves and Sound, JEE Main
Velocity of water in a river is

Options:
A more in the middle and less near its banks
B less in the middle and more near its banks
C same everywhere
D increase from one bank to other bank

The correct answer is:
more in the middle and less near its banks

Solution:
The correct option is 1 More in middle and less near its bank The velocity is more in the middle as compare to its bank because while flowing it collide with near its banks which results in lesser velocity while in the middle there is no obstruction so the velocity is more in the middle.
The correct option is 1 More in middle and less near its bank
The velocity is more in the middle as compare to its bank because while flowing it collide with near its banks which results in lesser velocity while in the middle there is no obstruction so the velocity is more in the middle.

Topic: Physics, Mechanical Properties of Fluids, JEE Main
Velocity v and acceleration a in two systems of units 1 and 2 are related as v2=nm2v1 and a2=a1mn respectively. Here m and n are constants. The relations for distance and time in two systems respectively are

Options:
A n3 m3 L1=L2 and n2 m T1=T2
B L1=n4 m2 L2 and T1=n2 m T2
C L1=n2mL2 and T1=n4m2T2
D n2 m L1=L2 and n4 m2 T1=T2

The correct answer is:
n3 m3 L1=L2 and n2 m T1=T2

Solution:
Distance is stated as [L]=v2[a]

⇒v22a2=nm2v12a1mn2=n3m3v12a1

Or, we have L2=n3m3L1

Similarly, time T=va

⇒v2a2=nm2v1a1mn=n2mv1a1

So, T2=n2mT1

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2022 (28 Jun Shift 2)
Verify Rolle's theorem for the function
$f(x)=x^2+2 x-8, x \in[-4,2]$

Options:

None

Solution:
Now $\mathrm{f}(\mathrm{x})=\mathrm{x}^2+2 \mathrm{x}-8$ is a polynomial
$\therefore$ it is continuous and derivable in its domain $\mathrm{x} \in \mathrm{R}$.
Hence it is continuous in the interval $[-4,2]$ and derivable in the interval $(-4,2)$
$f(-4)=(-4)^2+2(-4)-8=16-8-8=0$
$f(2)=2^2+4-8=8-8=0$
Conditions of Rolle's theorem are satisfied.
$\begin{aligned}
&\mathrm{f}^{\prime}(\mathrm{x})=2 \mathrm{x}+2 \\
&\therefore \mathrm{f}^{\prime}(\mathrm{c})=2 \mathrm{c}+2=0 \\
&\text { or } \quad \mathrm{c}=-1, \quad \mathrm{c}=-1 \in[-4,2]
\end{aligned}$
Thus $\mathrm{f}^{\prime}(\mathrm{c})=0$ at $\mathrm{c}=-1$.

Topic: Mathematics, Continuity and Differentiability
Vernalization stimulates flowering in

Options:
A zamikand
B turmeric
C carrot
D ginger

The correct answer is:
carrot

Solution:
Vernalization is a process of shortening of juvenile or vegetative phase and hastening flowering by a previous cold treatment. Example-henbane, carrot, cabbage.

Topic: Biology, Plant - Growth and Development, NEET, NEET 2012 (Mains)
Vertex A of the acute angled triangle ABC is equidistant from its circumcentre O and orthocentre H, then the possible value of ∠A is

Options:
A 300
B 600
C 750
D 900

The correct answer is:
600

Solution:
Let, the position of circumcentre(O) be origin, position vector of orthocentre(H) is xâ and position vectors of A,B,C are aâ,bâ,câ respectively
Centroid of ÎABC G=13aâ+bâ+câ
xâ+0â3=aâ+bâ+câ3âxâ=aâ+bâ+câ
OAâ=HAâ (given)
aâ=bâ+câ aâ=bâ=câ=R
aâ2=bâ2+câ2+2bâcâcosâ¡Î¸
âR2=R2+R2+2R2cosâ¡Î¸
âcosâ¡Î¸=-12âÎ¸=120o

From diagram, â A=60o

Topic: Mathematics, Vector Algebra, JEE Main
Vertex of the parabola $9 x^2-6 x+36 y+9=0$ is

Options:
A $(1 / 3,-2 / 9)$
B $(-1 / 3,-1 / 2)$
C $(-1 / 3,1 / 2)$
D $(1 / 3,1 / 2)$

The correct answer is:
$(1 / 3,-2 / 9)$

Solution:
The equation can be written as $(3 x-1)^2=-4(9 y+2)$.
Hence the vertex is $\left(\frac{1}{3},-\frac{2}{9}\right)$

Topic: Mathematics, Parabola, JEE Main
Vertex of the parabola $y^2+2 y+x=0$ lies in the quadrant

Options:
A First
B Second
C Third
D Fourth

The correct answer is:
Fourth

Solution:
Given parabola can be written as $(y+1)^2=-(x-1)$
Hence vertex is $(1,-1)$, which lies in IV quadrant.

Topic: Mathematics, Parabola, JEE Main
Vertex of the parabola $y^2+2 y+x=0$ lies in the quadrant

Options:
A First
B Second
C Third
D Fourth

The correct answer is:
Fourth

Solution:
Given parabola can be written as $(y+1)^2=-(x-1)$.
Hence vertex is $(1,-1)$, which lies in IV quadrant.

Topic: Mathematics, Parabola, JEE Main
Very pure hydrogen $(99.9 \%)$ can be made by which of the following processes?

Options:
A
Reaction of methane with steam

B
Mixing natural hydrocarbons of high molecular weight

C
Electrolysis of water

D
Reaction of salt like hydrides with water

The correct answer is:
Electrolysis of water

Solution:
Highly pure hydrogen is obtained by the electrolysis of water.

Topic: Chemistry, Hydrogen, JEE Main, JEE Main 2012 (Offline)
Very pure hydrogen gas is prepared by the action of pure dilute \( \mathrm{H}_{2} \mathrm{SO}_{4} \) on

Options:
A water
B sodium hydride
C magnesium ribbon
D aluminium

The correct answer is:
magnesium ribbon

Solution:
Very pure hydrogen gas is prepared by the action of pure dilute H2SO4 with magnesium.

Topic: Chemistry, Hydrogen, JEE Main
Viable material of endangered species can be preserved by

Options:
A gene bank.
B gene library.
C herbarium.
D gene pool.

The correct answer is:
gene bank.

Solution:
Viable material of endangered species can be preserved by the gene bank. Gene bank is an institute that maintains stocks of viable seeds (seed banks), live growing plants (orchards), tissue culture and frozen germplasm with the whole range of genetic variability.

Topic: Biology, Biodiversity and its Conservation, NEET
Vicinal and geminal dihalides can be distinguished by

Options:
A KOH (aqueous).
B KOH (alcohol).
C Zn dust.
D None of these.

The correct answer is:
KOH (aqueous).

Solution:
KOH is alkaline in nature and it abstracts proton from water and attacking species OH- is formed, which works as a nucleophile.

From vicinal dihalide, it will form diols:

CH2ClCH2Cl→KOHaqueousHOCH2CH2OHEthane-1, 2-diol

From gem dihalide, gem diol is formed, which converts into carbonyl compound after dehydration.

CH3CHCl2→KOH aqueous gem diol →‐H2OCH3CHOEthanal

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Vidicon works on the principle of

Options:
A electrical conductivity
B photoconductivity
C thermal conductivity
D SONAR

The correct answer is:
photoconductivity

Solution:
The vidicon is a storage-type camera tube in which a charge-density pattern is formed by the imaged scene radiation on a photoconductive surface which is then scanned by a beam of low-velocity electrons. The fluctuating voltage coupled out to a video amplifier can be used to reproduce the scene being imaged. The electrical charge produced by an image will remain in the face plate until it is scanned or until the charge dissipates.

Topic: Physics, Atomic Physics, VITEEE, VITEEE 2006
Vinegar is obtained by double fermentation of starch or sugary material. The first step fermentation is achieved by addition of Saccharomyces cerevisiae and second step fermentation by the addition of

Options:
A Acetobacter aceti
B Lactobacillus lactis
C Penicillium notatum
D Escherichia coli

The correct answer is:
Acetobacter aceti

Solution:
Vinegar is a liquid consisting mainly of acetic acid (CH3COOH) and water. The microbial activity, the sugar is fermented by yeasts or Saccharomyces cerevisiae, followed by bacterial oxidation of the ethanol to acetic acid (acetification by Acetobacter aceti).

Topic: Biology, Microbes in Human Welfare, NEET
Vinegar is obtained by double fermentation of starch or sugary material. The first step of fermentation is achieved by addition of Saccharomyces cerevisiae and second step is achieved by the addition of

Options:
A Acetobacter aceti
B Lactobacillus lactis
C Penicillium notatum
D E. coli

The correct answer is:
Acetobacter aceti

Solution:
When a natural solution of fermentable sugars is exposed to indigenous microbial activities is an alcoholic fermentation by yeasts (Saccharomyces cerevisiae), followed by bacterial oxidation of the ethanol to acetic acid (acetification by Acetobacter aceti). If sufficient acetic acid is produced it may be classified as vinegar. (Vinegar is a liquid consisting mainly of acetic acid (CH3COOH) and water).

Topic: Biology, Microbes in Human Welfare, NEET
Vinyl chloride reacts with $\mathrm{HCl}$ to form

Options:
A 1, 1- dichloro ethane
B 1, 2- dichloro ethane
C Tetrachloro ethylene
D Mixture of 1, 2 and 1, 1 – dichloro ethane

The correct answer is:
1, 1- dichloro ethane

Solution:
$\mathrm{CH}_2=\mathrm{CH}-\mathrm{Cl}+\mathrm{HCl} \rightarrow \mathrm{CH}_3-\mathrm{CHCl}_2$
1,1-dichloro ethane

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Vipin and Shubham are playing a game with a coin, that comes up heads with a probability p. They take turns flipping the coin until one of them wins, with Vipin going first. Vipin wins if he flips a heads and Shubham wins if he flips a tails. Given that the probability of Vipin winning the game is 12, then the value of p is ksin2162o. The value of k10 is equal to

Options:

The correct answer is:
0.4

Solution:
PH=p,  PT=1-p
Probability of Vipin winning game is PH+PTPHPH+PTPHPTPHPH+.......infinite series
=PH1-PHP(T)=p1-p(1-p)=p1-p+p2

⇒p1-p+p2=12 given
⇒2p=1-p+p2
⇒p2-3p+1=0

⇒p=3±52 ∴p=3+52>1
⇒p=3-52=5-142×4=4sin2162o
⇒k=4

Topic: Mathematics, Probability, JEE Main
Viral infection is usually absent in

Options:
A Phloem cells
B Xylem cells
C Pith cells
D Apical meristem

The correct answer is:
Apical meristem

Solution:
The apical meristems are present in the apices of primary and secondary shoots and roots of the plant. The cells of apical meristem are in an active stage of division, have dense cytoplasm, thin cell wall, and remain virus-free. Due to this reason, meristematic culture is carried out to obtain virus-free plants.

Topic: Biology, Anatomy of Flowering Plants, NEET
Visible light of wavelength 6000×10-8cm falls normally on a single slit and produces a diffraction pattern. It is found that the second diffraction minimum is at 60o from the central maximum. If the first minimum is produced at θ1, then θ1 is close to

Options:
A 20o
B 30o
C 25o
D 45o

The correct answer is:
25o

Solution:
For 2nd minima,
dsin⁡θ=2λ
sin⁡θ=32 , (given)
⇒λd=34 ...(1)
So, for 1st minima
dsinθ=λ
⇒sinθ=λd=34
⇒θ=25.65o
⇒θ≈25o

Topic: Physics, Wave Optics, JEE Main, JEE Main 2020 (07 Jan Shift 1)
Vitamin B1 is

Options:
A Riboflavin
B Cobalamin
C Thiamine
D Pvridoxine

The correct answer is:
Thiamine

Solution:
Vitamins are the group of biomolecules which are used in the normal metabolic process and for life growth and health of human beings and animals. Vitamin B1 is soluble in water. It is destroyed by the heat. There mainly sources of vitamin B1 are pulses, nuts, milk, fruits and green vegetables. Beriberi diseases is due to the deficiency of vitamin B1. Hence, vitamin B1 is thiamine.

Topic: Chemistry, Biomolecules, AP EAMCET, AP EAMCET 2021 (20 Aug Shift 1)
Vitamin-A is called

Options:
A ascorbic acid
B retinol
C calciferol
D tocopherol

The correct answer is:
retinol

Solution:
Vitamin-A helps to maintain healthy teeth and skeletal skin. It is also known as retinol because it produces the pigments in the retina of the eye.

Topic: Chemistry, Biomolecules, AP EAMCET, AP EAMCET 2021 (24 Aug Shift 2)
Vitamins \( \mathrm{A}, \mathrm{D}, \mathrm{E} ,\mathrm{~K} \) are

Options:
A water soluble
B oil or fat soluble
C both (A) and (B)
D insoluble

The correct answer is:
oil or fat soluble

Solution:
These are fat soluble vitamins. The fat-soluble vitamins, A,D,E & K, are stored in the body for long periods of time and generally pose a greater risk for toxicity than water-soluble vitamins when consumed in excess.

Topic: Chemistry, Biomolecules, JEE Main
Voltage rating of a parallel plate capacitor is 500 V. Its dielectric can withstand a maximum electric field of 106V/m. The plate area is 10-4 m2 . What is the dielectric constant if the capacitance is 15 pF ?
given ϵ0=8.86×10-12C2/Nm2

Options:
A 3.8
B 8.5
C 4.5
D 6.2

The correct answer is:
8.5

Solution:
Electric field inside capacitor =Vd
=VdAkε0Akε0=VCAkε0∵kAε0d=C
106=500×15×10-1210-4×k×8.85×10-12
k=15×5008.85×10-4×106=758.85
k=8.47≈8.5

Topic: Physics, Capacitance, JEE Main, JEE Main 2019 (08 Apr Shift 1)
Volume occupied by one molecule of water (density $\left.=1 \mathrm{~g} \mathrm{~cm}^{-3}\right)$ is

Options:
A $5.5 \times 10^{-23} \mathrm{~cm}^3$
B $9.0 \times 10^{-23} \mathrm{~cm}^3$
C $6.023 \times 10^{-23} \mathrm{~cm}^3$
D $3.0 \times 10^{-23} \mathrm{~cm}^3$

The correct answer is:
$3.0 \times 10^{-23} \mathrm{~cm}^3$

Solution:
Volume of 1 molecule of $\mathrm{H}_2 \mathrm{O}=\frac{18 \mathrm{~g}}{6.02 \times 10^{23} \times 1 \mathrm{~g} / \mathrm{cc}}$
$$
\cong 3.0 \times 10^{-23} \mathrm{~cm}^3
$$

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET, NEET 2008 (Mains)
Volume occupied by single \( \mathrm{CsCl} \) ion pair in a crystal is \( 7.014 \times 10^{-23} \mathrm{~cm}^{3} \). The smallest \( \mathrm{Cs}- \)
Cs inter-nuclear distance is equal to length of the side of the cube corresponding to volume of
one \( \mathrm{CsCl} \) ion pair. The smallest \( \mathrm{Cs}-\mathrm{Cs} \) inter-nuclear distance is nearly

Options:
A \( 4.3 Å \)
B \( 4.5 Å \)
C \( 4.4 Å \)
D None of the above

The correct answer is:
None of the above

Solution:
Volume of the unit cell $=7.014 \times 10^{-23}$
$\Rightarrow a^{3}=7.014 \times 10^{-23}$ (a is the smallest Cs-Cs distance)
$a=\sqrt{7.014 \times 10^{-23}}$
$a=4.12 Å$

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
Volume of 0.1 M K2Cr2O7 required to oxidize 35 mL of 0.5 M FeSO4 solution is

Options:
A 29.2 mL
B 175 mL
C 185 mL
D 145 mL

The correct answer is:
29.2 mL

Solution:
Cr2O72-+14H++6e-→2Cr3++7H2O
a1=6
a2=1
a1=n -factor of K2Cr2O7
a2=n -factor of FeSO4
Fe2+→Fe3++e-
a1M1V1K2Cr2O7a1=6=a2M2V2FeSO4a2=1
M1=0.1M
M2=0.5 M
V1=?
V2=35 mL
So V1=1×0.5×35mL6×0.1
=29.2 mL.

Topic: Chemistry, Redox Reactions, JEE Main
Volume of 3M NaOH (formula weight 40 g mol-1 ) which can be prepared from 84 g of NaOH is ____×10-1dm3.

Options:

The correct answer is:
7

Solution:
Number of moles = Mass / Molar mass​

Molar mass of NaOH=40 g/mol

By Using the definition of molarity to find the volume of the solution:

Molarity= Number of moles of soluteVolume of solution (in liters)

M=nNaOHVsol( in L)⇒3=(84/40)V⇒V=0.7 L=7×10-1 L

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2024 (27 Jan Shift 2)
Volume of $3 \%$ solution of sodium carbonate necessary to neutralise a litre of $0.1 \mathrm{~N}$ sulphuric acid

Options:
A $176.66 \mathrm{ml}$
B $156.6 \mathrm{ml}$
C $116.0 \mathrm{ml}$
D $196.1 \mathrm{ml}$

The correct answer is:
$176.66 \mathrm{ml}$

Solution:
Normality of $3 \% \mathrm{Na}_{2} \mathrm{CO}_{3}$

$\mathrm{N}=\frac{3 \times 1000}{53 \times 100}=0.566 \mathrm{~N}$

For $\mathrm{H}_{2} \mathrm{SO}_{4}$ sol. $\mathrm{N}_{1}=0.1, \mathrm{~V}_{1}=100 \mathrm{~mL}$

For $\mathrm{Na}_{2} \mathrm{CO}_{3}$ sol. $\mathrm{N}_{2}=0.566$

Now apply $\mathrm{N}_{1} \mathrm{~V}_{1}=\mathrm{N}_{2} \mathrm{~V}_{2}$

$\mathrm{V}_{2}=\frac{\mathrm{N}_{1} \mathrm{~V}_{1}}{\mathrm{~N}_{2}}=\frac{0.1 \times 1000 \mathrm{~mL}}{0.566}=176.66 \mathrm{~mL}$

Topic: Chemistry, Redox Reactions, JEE Main
Vulcanization of rubber is carried out by heating a mixture of

Options:
A isoprene and styrene
B neoprene and sulphur
C isoprene and sulphur
D neoprene and styrene

The correct answer is:
isoprene and sulphur

Solution:
Natural rubber may be considered as a linear polymer of isoprene (2-methyl-1, 3-butadiene) and is also called as cis - 1, 4 - polyisoprene.

To improve the physical properties, a process of vulcanisation is carried out. This process consists of heating a mixture of raw rubber with sulphur and an appropriate additive at a temperature range between 373 K to 415 K.

In the manufacture of tyre rubber, 5% of sulphur is used as a cross linking agent. The probable structures of vulcanised rubber molecules are depicted below:

Topic: Chemistry, Polymers, JEE Main, JEE Main 2022 (26 Jul Shift 2)
War gas is formed from

Options:
A $\mathrm{PH}_3$
B $\mathrm{C}_2 \mathrm{H}_2$
C Zinc phosphate
D Chloropicrin

The correct answer is:
Chloropicrin

Solution:
$\mathrm{CHCl}_3+\mathrm{HO}-\mathrm{NO}_2 \rightarrow \underset{\text { Chloropichn (wargas) }}{\mathrm{CCl}_3 \mathrm{NO}_2}+\mathrm{H}_2 \mathrm{O}$

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Washing soda is Na2CO3·xH2O. The value of 'x' is ______.

Options:

The correct answer is:
10

Solution:
Washing soda is decahydrate of sodium carbonate Na2CO3.10H2O

Topic: Chemistry, s Block Elements, JEE Main