datavorous/entrance-exam-dataset huggingface dataset:
The two ends of a rod of length $L$ and a uniform cross-sectional area $A$ are kept at two temperatures $\mathrm{T}_1$ and $\mathrm{T}_2\left(\mathrm{~T}_1>\mathrm{T}_2\right)$. The rate of heat transfer, $\frac{\mathrm{dQ}}{\mathrm{dt}}$, through the rod in a steady state is given by

Options:
A $\frac{\mathrm{dQ}}{\mathrm{dt}}=\frac{\mathrm{KL}\left(\mathrm{T}_1-\mathrm{T}_2\right)}{\mathrm{A}}$
B $\frac{\mathrm{dQ}}{\mathrm{dt}}=\frac{\mathrm{K}\left(\mathrm{T}_1-\mathrm{T}_2\right)}{\mathrm{LA}}$
C $\frac{\mathrm{dQ}}{\mathrm{dt}}=\mathrm{KLA}\left(\mathrm{T}_1-\mathrm{T}_2\right)$
D $\frac{\mathrm{dQ}}{\mathrm{dt}}=\frac{\mathrm{KA}\left(\mathrm{T}_1-\mathrm{T}_2\right)}{\mathrm{L}}$

The correct answer is:
$\frac{\mathrm{dQ}}{\mathrm{dt}}=\frac{\mathrm{KA}\left(\mathrm{T}_1-\mathrm{T}_2\right)}{\mathrm{L}}$

Solution:
For a rod of length $L$ and area of cross-section $A$ whose faces are maintained at temperature $\mathrm{T}_1$ and $\mathrm{T}_2$ respectively.
Then in steady state the rate of heat flowing from one face to the other face in time $t$ is given by
$$
\frac{\mathrm{dQ}}{\mathrm{dt}}=\frac{\mathrm{KA}\left(\mathrm{T}_1-\mathrm{T}_2\right)}{\mathrm{L}}
$$
The curved surface of rod is kept insulated from surrounding to avoid leakage of heat.

Topic: Physics, Thermal Properties of Matter, NEET, NEET 2009 (Screening)
The two ends of a rod of length $\mathrm{L}$ and a uniform cross-sectional area A are kept at two temperatures $T_1$ and $T_2\left(T_1>T_2\right)$. The rate of heat transfer, $\frac{\mathrm{dQ}}{\mathrm{dt}}$, through the rod in a steady state is given by :

Options:
A $\frac{\mathrm{dQ}}{\mathrm{dt}}=\frac{\mathrm{kA}\left(\mathrm{T}_1-\mathrm{T}_2\right)}{\mathrm{L}}$
B $\frac{\mathrm{dQ}}{\mathrm{dt}}=\frac{\mathrm{kL}\left(\mathrm{T}_1-\mathrm{T}_2\right)}{\mathrm{A}}$
C $\frac{\mathrm{dQ}}{\mathrm{dt}}=\frac{\mathrm{k}\left(\mathrm{T}_1-\mathrm{T}_2\right)}{\mathrm{LA}}$
D $\frac{\mathrm{dQ}}{\mathrm{dt}}=\mathrm{kLA}\left(\mathrm{T}_1-\mathrm{T}_2\right)$

The correct answer is:
$\frac{\mathrm{dQ}}{\mathrm{dt}}=\frac{\mathrm{kA}\left(\mathrm{T}_1-\mathrm{T}_2\right)}{\mathrm{L}}$

Solution:
For steady state $\frac{\mathrm{dQ}}{\mathrm{dt}}=\frac{\mathrm{kA}\left(\mathrm{T}_1-\mathrm{T}_2\right)}{\mathrm{L}}$

Topic: Physics, Thermal Properties of Matter, JEE Main
The two events $A$ and $B$ have probabilities 0.25 and 0.50 respectively. The probability that both $A$ and $B$ occur simultaneously is 0.14 . Then the probability that neither $A$ nor $B$ occurs is

Options:
A $0.39$
B $0.25$
C $0.904$
D None of these

The correct answer is:
$0.39$

Solution:
$P\left(A^c \cap B^c\right)=1-P(A \cup B)$
$=1-[0.25+0.5-0.14]=0.39$.

Topic: Mathematics, Probability, JEE Main
The two forms of D-glucopyranose obtained from the solution of D-glucose are called

Options:
A Isomers
B Anomers
C Epimers
D Enantiomers

The correct answer is:
Anomers

Solution:
Two forms of D-glucopyranose are α-D-+-glucopyranose and β-D-+-glucopyranose. These are anomers (a pair of stereoisomers which differ in configuration only around first-carbon atom are called anomers).

Topic: Chemistry, Biomolecules, JEE Main
The two functional groups present in a typical carbohydrate are :

Options:
A
$-\mathrm{OH}$ and $-\mathrm{COOH}$

B
$-\mathrm{CHO}$ and $-\mathrm{COOH}$

C
$>\mathrm{C}=\mathrm{O}$ and $-\mathrm{OH}$

D
$-\mathrm{OH}$ and $-\mathrm{CHO}$

The correct answer is:
$>\mathrm{C}=\mathrm{O}$ and $-\mathrm{OH}$

Solution:
Carbohydrates are polyhydroxy carbonyl compounds.

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2009
The two geometric means between the number 1 and 64 are

Options:
A $1$ and $64$
B $4$ and $16$
C $2$ and $16$
D $8$ and $16$

The correct answer is:
$4$ and $16$

Solution:
Let $1, a, b, 64 \Rightarrow a^2=b$ and $b^2=64 a$
$\Rightarrow a=4$ and $b=16$

Topic: Mathematics, Sequences and Series, JEE Main
The two light beams having intensities I and 9I interfere to produce a fringe pattern on a screen. The phase difference between the beams is π2 at point P and π at point Q. Then the difference between the resultant intensities at P and Q will be :

Options:
A 2I
B 6I
C 5I
D 7I

The correct answer is:
6I

Solution:
Resultant intensity is given by, Iresultant=I1+I2+2I1I2cosϕ

At P, IP=9I+I+29I×Icosπ2=10I

At Q, IQ=9I+I+29I×Icosπ=10I-2×3I=4I

Now, difference between intensities is 10I-4I=6I

Topic: Physics, Wave Optics, JEE Main, JEE Main 2022 (25 Jun Shift 1)
The two lines $x=a y+b, z=c y+d ;$ and $x=a^{\prime} y+b^{\prime}, z=c^{\prime} y+d^{\prime}$ are perpendicular to each other if

Options:
A
$\mathrm{aa}^{\prime}+\mathrm{cc}^{\prime}=-1$

B
$\mathrm{aa}^{\prime}+\mathrm{cc}^{\prime}=1$

C
$\frac{a}{a^{\prime}}+\frac{c}{c^{\prime}}=-1$

D
$\frac{a}{a^{\prime}}+\frac{c}{c^{\prime}}=1$

The correct answer is:
$\mathrm{aa}^{\prime}+\mathrm{cc}^{\prime}=-1$

Solution:
Equation of lines $\frac{x-b}{a}=y=\frac{z-d}{c}$
$\frac{x-b^{\prime}}{a^{\prime}}=y=\frac{z-d^{\prime}}{c^{\prime}}$
Lines are perpendicular $\Rightarrow a a^{\prime}+1+c c^{\prime}=0$

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2006
The two lines $x=a y+b, z=c y+d$ and $x=a^{\prime} y+b^{\prime} z=c^{\prime} y+d^{\prime}$ will be perpendicular, if and only if

Options:
A
$a a^{\prime}+c c^{\prime}+1=0$

B
$\mathrm{a} \mathrm{a}^{\prime}+\mathrm{b} \mathrm{b}^{\prime}+\mathrm{c} \mathrm{c}^{\prime}+1=0$

C
$\mathrm{a}^{\prime}+\mathrm{b} \mathrm{b}^{\prime}+\mathrm{c} \mathrm{c}^{\prime}=0$

D
$\left(a+a^{\prime}\right)\left(b+b^{\prime}\right)+\left(c+c^{\prime}\right)=0$

The correct answer is:
$a a^{\prime}+c c^{\prime}+1=0$

Solution:
$\frac{\mathrm{x}-\mathrm{b}}{\mathrm{a}}=\frac{\mathrm{y}}{1}=\frac{3-\mathrm{d}}{\mathrm{c}} ; \frac{\mathrm{x}-\mathrm{b}^{\prime}}{\mathrm{a}^{\prime}}=\frac{\mathrm{y}}{1}=\frac{3-\mathrm{d}^{\prime}}{\mathrm{c}^{\prime}}$
For perpendicular $\mathrm{aa}^{\prime}+1+\mathrm{cc}^{\prime}=0$

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2003
The two monomers for the synthesis of nylon 6, 6 are

Options:
A HOOHCH24COOH, H2NCH24NH2
B HOOCCH24COOH, H2NCH26NH2
C HOOCCH26COOH, H2NCH24NH2
D HOOCCH26COOH, H2NCH26NH2

The correct answer is:
HOOCCH24COOH, H2NCH26NH2

Solution:
Nylon 6, 6 is made up of hexamethylenediamine H2NCH26NH2 and adipic acid HCOO-CH24-COOH.
NH2CH26NH2+HCOOCH24COOH

Topic: Chemistry, Polymers, JEE Main, JEE Main 2019 (12 Jan Shift 2)
The two nearest harmonics of a tube closed at one end and open at other end are 220 Hz and 260 Hz. What is the fundamental frequency of the system?

Options:
A 10 Hz
B 20 Hz
C 30 Hz
D 40 Hz

The correct answer is:
20 Hz

Solution:
Two successive frequencies of closed pipe
nv4l=220 ....(i)
n+2v4l=260 ......(ii)
Dividing (ii) by (i), we get
n+2n=260220=1311
11n+22=13n
n=11
So, 11v4l=220
v4l=20
So fundamental frequency is 20 Hz.

Topic: Physics, Waves and Sound, JEE Main
The two statements I and II are regarding cell theory. Identify the statements that were given by which of the following scientists?

I. The bodies of animals and plants are composed of cells and products of cells.
II. All cells arise from pre-existing cells.

Options:
A

I	II

Schleiden and Schwann	Rudolf Vichow

B

I	II

Rudolf Virchow	Schleiden and Schwann

C

I	II

Robert Brown	Schleiden and Schwann

D

I	II

Robert Brown	Robert Hooke

The correct answer is:

I	II

Schleiden and Schwann	Rudolf Vichow

Solution:
The cell theory was formulated by Schleiden and Schwann in 1839. The main feature is as follows:

A cell is a basic unit of life.

All organisms are made up of cells and the product of cells.

Rudolf Virchow expands the cell theory and stated "Omnis cellula E Cellula" means all cells arise from pre-existing cells.

Topic: Biology, Cell - The Unit of Life, NEET
The two subunits of ribosome remain united at a critical ion level of

Options:
A copper
B manganese
C magnesium
D calcium

The correct answer is:
magnesium

Solution:
Magnesium is required in leaves, growing areas of root and stem protein synthesis hence, withdrawn from ageing. It is constituent of chlorophyll, middle lamella, and connected with phosphate transfer in respiration, binding of ribosomes and DNA and RNA synthesis.
Manganese activates enzymes of respiration, photosynthesis and nitrogen metabolism performing oxidation, reduction, decarboxylation, photolysis of water etc.
Copper is activator of plastocyanin, cytochrome oxidase, RuBP carboxylase and many other enzymes. It functions in electron transfer, maintenance of carbohydrate/nitrogen balance, chlorophyll synthesis etc.
Calcium is constituent of middle lamella, activator of enzymes connected with chromosome formation and many aspects of metabolism.

Topic: Biology, Mineral Nutrition, NEET, NEET 2008 (Screening)
The two sub-units of ribosome remain united at a critical ion level of

Options:
A Calcium
B Copper
C Manganese
D Magnesium

The correct answer is:
Magnesium

Solution:
Mangesium concentration is $0.0001 \mathrm{M}$.

Topic: Biology, Cell - The Unit of Life, NEET, NEET 2008 (Mains)
The two surfaces of a biconvex lens has same radii of curvatures. This lens is made of glass of refractive index 1.5 and has a focal length $10 \mathrm{~cm}$ in air. The lens is cut into two equal halves along a plane perpendicular to its principal axis to yield two plano-convex lenses. The two pieces are glued such that the convex surfaces touch each other. If this combination lens is immersed in water (refractive index $=$ $4 / 3$ ), its focal length (in $\mathrm{cm}$ ) is :

Options:
A 5
B 10
C 20
D 40

The correct answer is:
40

Solution:
If $a$ lens of focal length $f$ is divided into two equal parts as shown in figure (1) and each has a focal length $f^{\prime}$ then
$\frac{1}{f}=\frac{1}{f^{\prime}}+\frac{1}{f^{\prime}} \quad \text { i.e., } f^{\prime}=2 f$
i.e., each part will have focal length $2 f$.
Now if these parts are put in contact as in figure (2), then resultant focal length of the combination will be '
$\frac{1}{F}=\frac{1}{2 f}+\frac{1}{2 f} \text { i.e., } F=f \quad \text { (initial value) }$
For this combination,
$\frac{1}{F}=\left({ }_a \mu_g-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$ $\ldots$ (i)
Now, if this combination is immersed in liquid, then
$\frac{1}{F^{\prime}}=\left({ }_l \mu_g-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$ $\ldots$ (ii)
$\therefore \frac{F^{\prime}}{f}=\frac{\left({ }_a \mu_g-1\right)}{\left.{ }_l \mu_g-1\right)}=\frac{(1.5-1)}{\left(\frac{\frac{3}{2}}{\frac{4}{3}}-1\right)}$
or $\frac{F^{\prime}}{f}=\frac{0.5}{\left(\frac{9}{8}-1\right)}=0.5 \times 8$
$\therefore F^{\prime}=0.5 \times 8 \times 10=40 \mathrm{~cm}$

Topic: Physics, Ray Optics, JEE Main
The two tangents to a circle from an external point are always

Options:
A Equal
B Perpendicular to each other
C Parallel to each other
D None of these

The correct answer is:
Equal

Solution:
$\angle A=\angle B=90^{\circ}$
$O C$ is common, $A C=B C$.

Topic: Mathematics, Circle, JEE Main
The two vectors $\overrightarrow{\mathrm{A}}$ and $\overrightarrow{\mathrm{B}}$ are drawn from a common point and $\overrightarrow{\mathrm{C}}=\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}},$ then angle
between $\overrightarrow{\mathrm{A}}$ and $\overrightarrow{\mathrm{B}}$ is -
(1) $90^{\circ}$ if $\mathrm{C}^{2}=\mathrm{A}^{2}+\mathrm{B}^{2}$
(2) greater than $90^{\circ}$ if $\mathrm{C}^{2}<\mathrm{A}^{2}+\mathrm{B}^{2}$
(3) greater than $90^{\circ}$ if $\mathrm{C}^{2}>\mathrm{A}^{2}+\mathrm{B}^{2}$
(4) less than $90^{\circ}$ if $\mathrm{C}^{2}>\mathrm{A}^{2}+\mathrm{B}^{2}$
Correct options are -

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

The correct answer is:
1,2,4

Solution:
No Solution available

Topic: Physics, Mathematics in Physics, JEE Main
The Tyndall effect is observed only when the following conditions are satisfied,
i The diameter of the dispersed particles is much smaller than the wavelength of the light used.
ii The diameter of the dispersed particle is not much smaller than the wavelength of the light used.
iii The refractive indices of the dispersed phase and dispersion medium are almost similar in magnitude.
iv The refractive indices of the dispersed phase and dispersion medium differ greatly in magnitude.

Options:
A ii and iv
B i and iii
C ii and iii
D i and iv

The correct answer is:
ii and iv

Solution:
The Tyndall effect is an easy way of determining whether a mixture is colloidal or not. When light is shined through a true solution, the light passes cleanly through the solution, however when light is passed through a colloidal solution, the substance in the dispersed phases scatters the light in all directions, making it readily seen.
The Tyndall effect is observed only when,
ii The diameter of the dispersed particle is not much smaller than the wavelength of the light used.
iv The refractive indices of the dispersed phase and dispersion medium differ greatly in magnitude.

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2017 (02 Apr)
The Tyndall effect is observed only when the following conditions are satisfied,
(i) The diameter of the dispersed particles is much smaller than the wavelength of the light used.
(ii) The diameter of the dispersed particle is not much smaller than the wavelength of the light used.
(iii) The refractive indices of the dispersed phase and dispersion medium are almost similar in magnitude.
(iv) The refractive indices of the dispersed phase and dispersion medium differ greatly in magnitude.

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

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

Solution:
The Tyndall effect is an easy way of determining whether a mixture is colloidal or not. When light is shined through a true solution, the light passes cleanly through the solution, however when light is passed through a colloidal solution, the substance in the dispersed phases scatters the light in all directions, making it readily seen.
The Tyndall effect is observed only when,
ii The diameter of the dispersed particle is not much smaller than the wavelength of the light used.
iv The refractive indices of the dispersed phase and dispersion medium differ greatly in magnitude.

Topic: Chemistry, Surface Chemistry, JEE Main
The type of a battery, in which the reaction occurs only once and then battery becomes dead over a period of time is known as

Options:
A primary cell
B secondary cell
C lead storage battery
D nickel cadmium cell

The correct answer is:
primary cell

Solution:
By definition:

The type of a battery, in which the reaction occurs only once and then battery becomes dead over a period of time is known as primary cell/batteries.

eg. clock cell, remote cell, etc.

Topic: Chemistry, Electrochemistry, JEE Main
The type of bond that is most important in maintaining secondary structure of a protein is

Options:
A Disulphide bridges
B Hydrogen bonding within the backbone
C Hydrogen bonding between R group
D Salt bridges

The correct answer is:
Hydrogen bonding within the backbone

Solution:
Secondary structure of a protein is generally maintained by hydrogen bonding between the -NH group of one amide and the C=O group of another amide.

Topic: Chemistry, Biomolecules, NEET
The type of hybrid orbitals used by chlorine atom in $\mathrm{ClO}_2^{-}$is

Options:
A $S P^3$
B $S P^2$
C $S P$
D None of these

The correct answer is:
$S P^3$

Solution:
$\mathrm{ClO}_2^{-}$has $\mathrm{SP}^3$-hybridization and two lone pairs on halogen which produces V-shape Bent structure

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
The type of hybridisation of boron in diborane is

Options:
A $s p$-hybridisation
B $\quad s p^2$ - hybridisation
C $s p^3-$ hybridisation
D $s p^3 d^2-$ hybridisation

The correct answer is:
$s p^3-$ hybridisation

Solution:
Boron has three valence electrons, so it is supposed to make 3 bonds in a molecule with hybridization, $\mathrm{sp}^2$ as only $\mathrm{s}$ and two p orbitals are used in hybridization and last p orbitalis vacant.
But diborane, $\mathrm{B}_2 \mathrm{H}_6$ contains two electrons each, three centred bonds. Each Boron atom is in a link with four hydrogen atoms. This makes tetrahedral geometry.
Hence, each Boron atom is $\mathrm{sp}^3$ - hybridized. The correct option is $3$.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
The type of hybridization and no. of lone pair(s) of electron of Xe in XeOF4 , respectively, are:

Options:
A sp3d2  and  1
B sp3d2  and  2
C sp3d  and  2
D sp3d  and  1

The correct answer is:
sp3d2  and  1

Solution:
Steric no. =6, hybridisation =sp3d2
No. of lone pairs on Xe=1

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2019 (10 Jan Shift 1)
The type of immunity that is present at the time of birth in humans is:

Options:
A None of these
B Both of these
C Acquired immunity
D Innate immunity

The correct answer is:
Innate immunity

Solution:
The innate immune system, also known as non-specific immune system, is the first line of defence. It comprises the cells and mechanisms that defend the host from infection by other organisms in a non-specific manner. This means that the cells of the innate system recognize and respond to pathogens in a generic way. Innate Immunity is present at the time of birth. Innate immunity consists of four types of barriers.
â¢ Physical barriers: Skin on our body is the main barrier which prevents entry of the micro-organisms. Mucus coating of the epithelium lining the respiratory, gastrointestinal and urogenital tracts also help in trapping microbes entering our body.
â¢ Physiological barriers: Acid in stomach, saliva in the mouth, tears from eyes all prevent microbial growth.
â¢ Cellular barriers: Certain types of leukocytes (WBC) of our body such as polymorph nuclear leukocytes (PMNL-neutrophils) and monocytes and natural killer cells in the blood as well as macrophages in tissues can phagocytise and destroy microbes.
â¢ Cytokine barriers: Virus-infected cells secrete proteins called interferon which protect non-infected cells from further viral infection.

Topic: Biology, Human Health and Diseases, NEET
The type of isomerism present in nitropentamine chromium (III) chloride is

Options:
A Optical
B Linkage
C Ionization
D Polymerisation

The correct answer is:
Linkage

Solution:
Due to the presence of $NO_2$, it develops linkage isomerism.

Topic: Chemistry, Coordination Compounds, JEE Main
The type of isomerism present in nitropentamine chromium (III) chloride is

Options:
A
optical

B
linkage

C
ionization

D
polymerisation

The correct answer is:
linkage

Solution:
The nitro group can attach to metal through nitrogen as (-NO2) or through oxygen as nitrito (-ONO)

Topic: Chemistry, Coordination Compounds, JEE Main
The type of isomerism present in nitropentamine chromium (III) chloride is

Options:
A
optical

B
linkage

C
ionization

D
polymerisation

The correct answer is:
linkage

Solution:
The nitro group can attach to metal through nitrogen as (-NO2) or through oxygen as nitrito (-ONO)

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2002
The type of muscles present in our

Options:
A heart are involuntary and unstriated smooth muscles
B intestine are striated and involuntary
C thigh are striated and voluntary
D upper arm are smooth muscle fibres fusiform in shape

The correct answer is:
thigh are striated and voluntary

Solution:
Thigh muscles are skeletal muscle that are striated and voluntary skeletal muscle tissue is closely attached to skeletal bones.

Topic: Biology, Locomotion and Movement, NEET, NEET 2011 (Mains)
The type of pollution that gets increased during the daytime and in the presence of O3 is :

Options:
A Reducing smog
B Oxidising smog
C Global warming
D Acid rain

The correct answer is:
Oxidising smog

Solution:
In presence of ozone O3, oxidising smog gets increased during the daytime because automobiles and factories produce main components of the photochemical smog (oxidising smog) results from the action of sunlight on unsaturated hydrocarbon and nitrogen oxide.

Ozone is a strong oxidising agent and can react with the unburnt hydrocarbons in the polluted air to produce chemicals.

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2021 (16 Mar Shift 1)
The type of waves that can be propagated through solid is

Options:
A Transverse
B Longitudinal
C Both (1) and (2)
D None of these

The correct answer is:
Both (1) and (2)

Solution:
Since solid has both the properties (rigidity and elasticity)

Topic: Physics, Waves and Sound, JEE Main
The tyre of a motor car contains air at $15^{\circ} \mathrm{C}$. If the temperature increases to $35^{\circ} \mathrm{C}$, the approximate percentage increase in pressure is (ignore to expansion of tyre)

Options:
A 7
B 9
C 11
D 13

The correct answer is:
7

Solution:
$\begin{aligned}
T_1=15^{\circ} \mathrm{C} & =15+273=288 \mathrm{~K} \\
T_2 & =35^{\circ} \mathrm{C}=35+273=308 \mathrm{~K}
\end{aligned}$
Volume remains constant.
So,
$\begin{aligned}
& \frac{p_1}{T_1}=\frac{p_2}{T_2} \\
& \frac{p_1}{P_2}=\frac{T_1}{T_2} \Rightarrow \frac{p_1}{P_2}=\frac{288}{308} \\
& \frac{p_2}{p_1}=\frac{308}{288}
\end{aligned}$
$\begin{aligned}
\% \text { increase in pressure } & =\frac{p_2-p_1}{p_1} \times 100 \\
& =\frac{308-288}{288} \times 100 \\
& \approx 7 \%
\end{aligned}$

Topic: Physics, Kinetic Theory of Gases, AP EAMCET, AP EAMCET 2005
The tyre of a motor car contains air at $15^{\circ} \mathrm{C}$. If the temperature increases to $35^{\circ} \mathrm{C}$, the approximate percentage increase in pressure is (ignore to expansion of tyre)

Options:
A 7
B 9
C 11
D 13

The correct answer is:
7

Solution:
$\begin{aligned}
T_1=15^{\circ} \mathrm{C} & =15+273=288 \mathrm{~K} \\
T_2 & =35^{\circ} \mathrm{C}=35+273=308 \mathrm{~K}
\end{aligned}$
Volume remains constant.
So,
$\begin{aligned}
& \frac{p_1}{T_1}=\frac{p_2}{T_2} \\
& \frac{p_1}{P_2}=\frac{T_1}{T_2} \Rightarrow \frac{p_1}{P_2}=\frac{288}{308} \\
& \frac{p_2}{p_1}=\frac{308}{288}
\end{aligned}$
$\begin{aligned}
\% \text { increase in pressure } & =\frac{p_2-p_1}{p_1} \times 100 \\
& =\frac{308-288}{288} \times 100 \\
& \approx 7 \%
\end{aligned}$

Topic: Physics, Kinetic Theory of Gases, TS EAMCET, TS EAMCET 2005
The U - tube shown has a uniform cross-section. A liquid is filled in the two arms up to heights h1 and h2, and then the liquid is allowed to move. Neglect viscosity and surface tension. When the levels equalize in the two arms, the liquid will

Options:
A Be at rest
B Be moving with an acceleration of g h 1 - h 2 h 1 + h 2 + h
C Be moving with a velocity of h 1 - h 2 g 2 h 1 + h 2 + h
D Exert a net force to the right on the tube.

The correct answer is:
Be moving with a velocity of h 1 - h 2 g 2 h 1 + h 2 + h

Solution:
When the levels equalize, the height of the liquid in each arm = h 1 + h 2 2
We may then visualize that a length h 1 - h 1 + h 2 2 = h 1 - h 2 2 of the liquid has been transferred from the left arm to the right arm.
Then, the mass of this liquid = h 1 - h 2 2 A ρ
where, A = area of tube, ρ = density of the liquid
Distance through which it moves down = h 1 - h 2 2
∴ Loss in the gravitational potential energy = h 1 - h 2 2 2 A ρg
The mass of the entire liquid = h 1 + h 2 + h A ρ
If this moves with a velocity v, its kinetic energy = 1 2 h 1 + h 2 + h A ρ v 2
Equating energies, we get v=h1-h2g2h1+h2+h

Topic: Physics, Mechanical Properties of Fluids, JEE Main
The unbalanced equation for the reaction of P4S3 with nitrate in aqueous acidic medium is given below.
P4S3+NO3-→H3PO4+SO42-+NO
The number of moles of water required per mol of P4S3 is x3, the value of x is

Options:

The correct answer is:
8

Solution:
P4S3+NO3- âH3PO4+SO42-+NO
( P 4 S 3 â+â28â H 2 Oâ4â H 3 PO 4 â+â3â SO 4 2â â+â44â H + +38â e â )âÃâ3
( NO 3 â â+â4â H + â+â3â e â âNOâ+â2â H 2 O)âÃâ38
â´ 3â P 4 S 3 â+â38â NO 3 â â+â8â H 2 Oâ+â20â H + â12â H 3 PO 4 â+â9â SO 4 2â â+â38âNO

Topic: Chemistry, Redox Reactions, JEE Main