datavorous/entrance-exam-dataset huggingface dataset:
In the solar system, which is conserved

Options:
A Total Energy
B K.E.
C Angular Velocity
D Linear Momentum

The correct answer is:
Total Energy

Solution:
In the solar system, the orbit of a planet is an ellipse with the sun at one of the two foci.
velocity of the planet at every point on orbit will not be same, so momentum and Kinetic energy can not be conserved. The angular velocity will also vary. But the total energy $(K E+P E)$ will remain conserved for the system.

Topic: Physics, Gravitation, JEE Main
In the somatic cell cycle:

Options:
A in $\mathrm{G}_1$ phase DNA content is double the amount of DNA present in the original cell
B DNA replication takes place in S-phase
C a short inter phase is followed by a long mitotic phase.
D $\mathrm{G}_2$ phase follows mitotic phase.

The correct answer is:
DNA replication takes place in S-phase

Solution:
Cell cycle is the series of events in which cells replicate and make two new cells. Cell cycle consists of stages- $\mathrm{G}_1, \mathrm{~S}, \mathrm{G}_2$, and $\mathrm{M}$ (karyokinesis). $\mathrm{G}_1$ is the stage where the cell is preparing to divide. It then moves into the $\mathrm{S}$ phase where the cell copies all the DNA. S stands for synthesis phase (of DNA). After the DNA is copied and there's a complete extra set of all the genetic material, the cell moves into the $\mathrm{G}_2$ stage, where it organizes and condenses the genetic material, or starts to condense the genetic material, and prepares to divide. The next stage is M. M stands for mitosis. It is further divides into four stages,
Related Theory
Some cells enter $G_0$ temporarily until an external signal triggers the onset of $G_r$. Other cells that never or rarely divide, such as mature cardiac muscle and nerve cells, remain in $G_0$ permanently.
A Caution
There is a confusion in the events of Gap-1 and Gap-2 phases. [Mnemonic: $G_2$ phase two - twins - Protwins (Proteins)]

Topic: Biology, Cell Cycle and Cell Division, NEET, NEET 2004
In the spectrum of hydrogen atom, the ratio of the longest wavelength in Lyman series to the longest wavelength in the Balmer series is

Options:
A 527
B 193
C 49
D 32

The correct answer is:
527

Solution:
When an atom comes down from some higher energy level to the first energy level then emitted lines form of Lyman series.
1Î»L = R112-1n2
where R is Rydberg's constant.
When an atom comes from higher energy level to the second level, then Balmer series are obtained.
1Î»B = R122-1n2
For maximum wavelength
nâ¡ = 2 ,1Î»L = R1-1(2)2 = R1-14=3R4 â¦. (i)
n=3,1Î»B = R1(2)2-1(3)2 = R536 â¦(ii)
Dividing Eq. (ii) by Eq. (i), we get
Î»LÎ»B=527

Topic: Physics, Atomic Physics, JEE Main
In the spectrum of the hydrogen atom, the ratio of the wavelengths of the longest wavelength in Lyman series to the longest wavelength in the Balmer series is

Options:
A 5 27
B 1 93
C 4 9
D 3 2

The correct answer is:
5 27

Solution:
When an electron jumps down from any higher energy level to the first energy level, then the emitted lines form the Lyman series

1λL=R112-1n2, where R is the Rydberg constant.

Similarly, when an electron jumps down from any higher energy level to the second energy level, then the emitted lines form the Balmer series

1λB=R122-1n2
For maximum wavelength, the energy gap should be the smallest

n=2, 1λL=R1-122=R1-14=3R4

n=3, 1λB=R122-132=5R36

λLλB=527

Topic: Physics, Atomic Physics, JEE Main
In the structure of SF4, the lone pair of electrons on S is in.

Options:
A Equatorial position and there are two lone pair - bond pair repulsions at 90°.
B equatorial position and there are three lone pair - bond pair repulsions at 90°.
C axial position and there are three lone pair - bond pair repulsion at 90°.
D axial position and there are two lone pair - bond pair repulsion at 90°.

The correct answer is:
Equatorial position and there are two lone pair - bond pair repulsions at 90°.

Solution:
In SF4, The lone pair is in an equatorial position, and there are two lp—bp repulsions. Hence, the following arrangement shown below is more stable.

The shape shown in the following figure is described as a distorted tetrahedron, a folded square or a see-saw.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2022 (28 Jun Shift 2)
In the structure of the dichromate ion, there is a :

Options:
A linear symmetrical Cr-O-Cr bond.
B non-linear unsymmetrical Cr-O-Cr bond.
C non-linear symmetrical Cr-O-Cr bond.
D linear unsymmetrical Cr-O-Cr bond.

The correct answer is:
non-linear symmetrical Cr-O-Cr bond.

Solution:
Non-linear, Symmetrical Cr-O-Cr bond

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2021 (31 Aug Shift 1)
In the study of transistor as an amplifier, if $\alpha=I_c / I_e$ and $\beta=I_c / I_b$, where $I_c, I_b$ and $I_c$ are the collector, base and emitter currents, then

Options:
A $\beta=\frac{1-\alpha}{\alpha}$
B $\beta=\frac{\alpha}{1-\alpha}$
C $\beta=\frac{\alpha}{1+\alpha}$
D $\beta=\frac{1+\alpha}{\alpha}$

The correct answer is:
$\beta=\frac{\alpha}{1-\alpha}$

Solution:
As we know $i_E=i_C+i_B$
$\Rightarrow \frac{i_e}{i_c}=1+\frac{i_b}{i_c} \Rightarrow \frac{1}{\alpha}=1+\frac{1}{\beta} \Rightarrow \beta=\frac{\alpha}{1-\alpha}$

Topic: Physics, Semiconductors, JEE Main
In the sulphur estimation, 0.471 g of an organic compound gave 1.44 g of barium sulfate. The percentage of sulphur in the compound is. (Nearest integer) (Atomic Mass of Ba=137u )

Options:

The correct answer is:
42

Solution:
233 g of BaSO4 contains 32 g sulphur.

Amount of S in 1.44 g of BaSO4=32233×1.44

So percentage of S in 0.471 of organic compound =32233×1.440.471×100=42%

Topic: Chemistry, Chemical Equilibrium, JEE Main, JEE Main 2021 (26 Aug Shift 2)
In the synthesis of glycerol from propene, the intermediates involved are:

Options:
A Allyl chloride, gamma allyl alcohol and Î²-monochlorohydrin
B Glycerol trichloride and glycerol α-chlorohydrin
C Allyl alcohol and Î±-chlorohydrin
D Allyl alcohol and monosodium glycerolate

The correct answer is:
Allyl chloride, gamma allyl alcohol and β-monochlorohydrin

Solution:
Synthesis of glycerol from propene involves following steps

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
In the test of sulphate radical, the white precipitate of sulphate is soluble in

Options:
A Conc. $\mathrm{HCl}$
B Conc. $\mathrm{H}_2 \mathrm{SO}_4$
C Conc. $\mathrm{HNO}_3$
D None of these

The correct answer is:
None of these

Solution:
As the sulphate radical is a strong oxidising agent, it is insoluble in acids and so detection of sulphate radicle requires no other reagent.

Topic: Chemistry, Practical Chemistry, JEE Main
In the thermal decomposition of potassium chlorate given as $2 \mathrm{KClO}_3 \longrightarrow 2 \mathrm{KCl}+3 \mathrm{O}_2$, law of mass action

Options:
A Cannot be applied
B Can be applied
C Can be applied at low temperature
D Can be applied at high temp. and pressure

The correct answer is:
Cannot be applied

Solution:
The thermal decomposition of potassium chlorate is an irreversible reaction which takes place only in the forward direction. Law of mass action can only be applied for reversible reactions.
Hence, the law of mass action is not applicable in this case.

Topic: Chemistry, Chemical Equilibrium, JEE Main
In the titration of KMnO4 and oxalic acid in acidic medium, the change in oxidation number of carbon at the end point is

Options:

The correct answer is:
1

Solution:
The balanced equation of potassium permanganate and oxalic acid reaction in acidic medium is shown below.

2KMn+7O4+5H2C2+3O4+3H2SO4dil.→K2SO4+2M+2nSO4+10C+4O2+8H2O

The initial oxidation state of carbon in the oxalic acid is +3. The oxidation state of carbon at the end point is +4(in carbon dioxide). Hence, the change in oxidation state of carbon at the end point is 1.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2022 (27 Jul Shift 1)
In the transformation of ${ }_{92}^{238} \mathrm{U}$ to ${ }_{92}^{234} \mathrm{U}$, if one emission is an $\alpha$-particle, what should be the other emission(s)?

Options:
A
Two $\beta^{-}$

B
Two $\beta^{-}$and one $\beta^{+}$

C
One $\beta^{-}$and one $\gamma$

D
One $\beta^{+}$and one $\beta^{-}$

The correct answer is:
Two $\beta^{-}$

Solution:
${ }_{92}^{238} \mathrm{U} \longrightarrow{ }_{92}^{234} \mathrm{U}+{ }_2^4 \mathrm{He}+2{ }_{-1}^0 \mathrm{e}$

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2006
In the triangle ABC, vertices A,B,C are 1,2,3,1,-1,6 respectively. If the internal angle bisector of â BAC meets BC at D, then the coordinates of D are

Options:
A 53,83
B 1,72
C 13,133
D 2,34

The correct answer is:
53,83

Solution:
AB=5, AC=20=25

Since AD is the internal angle bisector,
so D divides BC in the ratio of AB:ACâ¡1:2

Now, applying section formula, we get,
The coordinates of D as 6-12+1,2+62+1â¡53,83

Topic: Mathematics, Straight Lines, JEE Main
In the UHF band, frequencies often spread using the:

Options:
A Ground waves
B Skywaves
C Surface waves
D Space waves

The correct answer is:
Space waves

Solution:
Due to their extremely high frequency, ultra-high frequency (UHF) waves are incapable of being reflected by the ionosphere or following the earth's curvature. Line-of-sight communication, which simply means that the waves move in a straight line, is how UHF signals are transmitted.

Topic: Physics, Communication System, JEE Main
In the Uranium radioactive series, the initial nucleus is \({ }_{92}^{238} \mathrm{U}\) and final nucleus is \({ }_{82}^{206} \mathrm{~Pb}\). When the Uranium nucleus decays to lead, the number of \(\alpha\)-particles emitted is .......... and the number of \(\beta\)-particles emitted is ____.

Options:
A 6,8
B 8,6
C 16,6
D 32,2

The correct answer is:
8,6

Solution:
According to question,
Initial nuclei \(={ }_{92}^{238} \mathrm{U}\)
Final nuclei \(={ }_{82}^{206} \mathrm{~Pb}\)
Change in atomic number \(=92-82=10\)
When an \(\alpha\)-particle emits from a nuclei, then its atomic number is decreased by 2 units and atomic mass is decreased by 4 units. When a \(\beta\)-particle emits from a nuclei, then atomic number of parent nuclei is increased by 1 unit whereas its atomic mass remains same.
Change in atomic mass \(=238-206=32\)
Since, atomic mass changes only due to emission \(\alpha\)-particle, hence number of emitted \(\alpha\)-particle
\(=\frac{32}{4}=8\)
Number of emitted \(\beta\)-particles = total change in atomic number due to emission of \(\alpha\)-particles total change in atomic number
\(=8 \times 2-10=6\)

Topic: Physics, Nuclear Physics, AP EAMCET, AP EAMCET 2020 (18 Sep Shift 2)
In the $V-T$ diagram shown in adjoining figure, what is the relation between $p_{1}$ and $p_{2}$ ?

Options:
A $p_{2}=p_{1}$
B $p_{2} < p_{1}$
C $p_{2}>p_{l}$
D insufficient data

The correct answer is:
$p_{2} < p_{1}$

Solution:
In an isobaric process, $\mathrm{p}=$ constant Hence, $\mathrm{V} \propto \mathrm{T}$
i.e., $\mathrm{V}=\left(\frac{\mathrm{nR}}{\mathrm{P}}\right) \mathrm{T}$
$\therefore$ V-T graph is astraight linewith slope $\propto \frac{1}{\mathrm{p}}$
(slope) $_{2}>$ (slope) $_{1}$ $\therefore \mathrm{p}_{2} < \mathrm{p}_{1}$

Topic: Physics, Thermodynamics, JEE Main
In the Victor-Meyer's test, the colour given by $1^{\circ}, 2^{\circ}$ and $3^{\circ}$ alcohols are respectively:

Options:
A
Red, colourless, blue

B
Red, blue, colourless

C
Colourless, red, blue

D
Red, blue, violet

The correct answer is:
Red, blue, colourless

Solution:
$$
\text { Victor Meyer's test: The various steps involved are }
$$
(i)
(ii)
(iii)

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2014 (12 Apr Online)
In the wave equation y=0.5sin2πλ400t-xm the velocity of the wave will be :

Options:
A 200 m s-1
B 2002 m s-1
C 400 m s-1
D 4002 m s-1

The correct answer is:
400 m s-1

Solution:
Given equation of the wave is y=0.5sin2πλ400t-2πλx

Comparing it with the standard equation of motion y=Asinωt-kx we get,

ω=2πλ400 and k=2πλ

Now the velocity of the wave will be,

v=ωk ⇒v=400 m s-1

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2022 (28 Jul Shift 1)
In the wet tests for detection of various cations by precipitation, Ba2+ cations are detected by obtaining precipitate of

Options:
A Ba(ox) : Barium oxalate
B BaCO3
C Ba(OAc)2
D BaSO4

The correct answer is:
BaCO3

Solution:
Ba2+ belongs to Vth Group and hence they are precipitated with CO32- to give ppt of BaCO3. In Wet tests for cations, (NH4)2CO3 is used as a group reagent for fifth group cations(Ba2+,Ca2+,Sr2+).

Ba2++(NH4)2CO3→BaCO3+NH4⊕

Topic: Chemistry, Practical Chemistry, JEE Main, JEE Main 2023 (13 Apr Shift 2)
In the wet tests for identification of various cations by precipitation, which transition element cation doesn't belong to group IV in qualitative inorganic analysis?

Options:
A Fe3+
B Zn2+
C Co2+
D Ni2+

The correct answer is:
Fe3+

Solution:
Cations are segregated in 5 groups depending upon their solubility with the help of various precipitating reagents. As per the segregation, Zn2+, Co2+, Ni2+ belongs to the 4th group and Fe3+ belongs to the 3rd group.

Topic: Chemistry, Practical Chemistry, JEE Main, JEE Main 2023 (30 Jan Shift 1)
In the Wheatstone's bridge (shown in figure) $X=Y$ and $A \gt B$. The direction of the current between $a b$ will be

Options:
A From $a$ to $b$
B From $b$ to $a$
C From $b$ to $a$ through $c$
D From $a$ to $b$ through $c$

The correct answer is:
From $b$ to $a$

Solution:
In the part cbd,
\(V_c-V_b=V_b-V_d \Rightarrow V_b=\frac{V_c+V_d}{2}\)
In the part cad,
\(\mathrm{V}_{\mathrm{c}}-\mathrm{V}_{\mathrm{a}}>\mathrm{V}_{\mathrm{a}}-\mathrm{V}_{\mathrm{d}} \Rightarrow \frac{\mathrm{V}_{\mathrm{c}}+\mathrm{V}_{\mathrm{d}}}{2}>\mathrm{V}_{\mathrm{a}} \Rightarrow \mathrm{V}_{\mathrm{b}}>\mathrm{V}_{\mathrm{a}}\)

Topic: Physics, Current Electricity, JEE Main
In the Wheatstone's bridge shown, $P=2 \Omega, Q=3 \Omega, R=6 \Omega$ and $S=8 \Omega$. In order to obtain balance, shunt resistance across ' $S$ ' must be

Options:
A $2 \Omega$
B $3 \Omega$
C $6 \Omega$
D $8 \Omega$

The correct answer is:
$8 \Omega$

Solution:
Let the value of shunt be $r$. Hence the equivalent resistance of branch containing $S$ will be $\frac{S r}{S+r}$
In balance condition, $\frac{P}{Q}=\frac{S r /(S+r)}{R}$. This gives $r=8 \Omega$

Topic: Physics, Current Electricity, JEE Main
In the $x y$-plane, three distinct lines $l_{1}, l_{2}, l_{3}$ concur at a point $(\lambda, 0)$. Further the lines $l_{1}, l_{2}, l_{3}$ are normals to the parabola $y^{2}=6 x$ at the points $A=\left(x_{1}, y_{1}\right), B=\left(x_{2}, y_{2}\right), C=\left(x_{3}, y_{3}\right)$ respectively. Then we have -

Options:
A $\lambda < -5$
B $\lambda>3$
C $-5 < \lambda < -3$
D $0 < \lambda < 3$

The correct answer is:
$\lambda>3$

Solution:
Any normal
$\mathrm{y}=\mathrm{mx}-2 \mathrm{am}-\mathrm{am}^{3}$ Here $\mathrm{a}=3 / 2$
through $(\lambda, 0)$
$\begin{array}{l}
0=\mathrm{m} \lambda-2 \mathrm{am}-\mathrm{am}^{3} \\
\mathrm{~m}=0, \quad \lambda=2 \mathrm{a}+\mathrm{am}^{2} \\
\mathrm{~m}^{2}=\frac{\lambda}{\mathrm{a}}-2>0 \\
\lambda>2 \mathrm{a} \Rightarrow \lambda>3
\end{array}$

Topic: Mathematics, Parabola, JEE Main
In the xy - plane , the region y>0 has a uniform magnetic field B1k^ and the region y<0 has another uniform magnetic field B2k^. A positively charged particle is projected from the origin along the positive yâaxis with speed v 0 =Ï m s â1 at t=0, as shown in the figure. Neglect gravity in this problem. Let t=T be the time when the particle crosses the xâaxis from below for the first time. If B 2 =4 B 1, the average speed of the particle, in m s â1, along the xâaxis in the time interval T is ________.

Options:

The correct answer is:
2

Solution:
(i) Average speed along x−axis
vx=â«vxâdtâ«dt=d1+d2t1+t2
(ii) We have,
r1=mvqB1,r2=mvqB2
Since, B1=B24
â´ r1=4r2
Time in B1âÏmqB1=t1
Time in B2âÏmqB2=t2
Total distance along x - axis d1+d2=2r1+2r2=2r1+r2=25r2
Total time T=t1+t2=5t2
â´ Average speed =10r25t2=2mvqB2ÃqB2Ïm=2

Topic: Physics, Magnetic Effects of Current, JEE Main
In the YDSE arrangement shown here, the intensity on the screen due to slit-2 is four times that of slit-1. If resultant intensity at the position of central maxima O is I, the resultant intensity at point P, where the phase difference between two waves coming from two slits is cos -1 1 4 is

Options:
A I 3
B 2 I 3
C I 4
D I 2

The correct answer is:
2 I 3

Solution:
I = I 0 + 4 I 0 2 = 9 I 0
I 0 = I g
I ′ = I 0 + 4 I 0 + 2 I 0 . 4 I 0 × 1 4
= 6 I 0
= 2 3 I

Topic: Physics, Wave Optics, JEE Main
In the YDSE shown below, two slits are covered with thin sheets having a thickness t and 2t and refractive indices 2Î¼ and Î¼ respectively. The position of central maxima is shifted by a distance of N (in cm) from the central position of the screen. Find the value of N.
[d = 1 mm; D = 1 m; t = 2 Ã 10-2 mm, Î¼=32 ]

Options:

The correct answer is:
2

Solution:
d sinθ = (2μt + t) – 2μt
⇒ dyD=t ⇒ y=Dtd
⇒ y=1×2×10-51×10-3=2×10-2 m

Topic: Physics, Wave Optics, JEE Main
In the Young's double slit experiment, the resultant intensity at a point on the screen is $75 \%$ of the maximum intensity of the bright fringe. Then the phase difference between the two interfering rays at that point is

Options:
A $\frac{\pi}{6}$
B $\frac{\pi}{4}$
C $\frac{\pi}{3}$
D $\frac{\pi}{2}$

The correct answer is:
$\frac{\pi}{3}$

Solution:
$\begin{aligned} & \text { Given, } I_R=75 \% \text { of } I_{\max } \\ &=\frac{3}{4} I_{\max } \\ &=\frac{3}{4}\left(4 a^2\right)=3 a^2 \\ & \Rightarrow \quad 4 a^2 \cos ^2 \frac{\phi}{2}=3 a^2 \\ & \cos ^2 \frac{\phi}{2}=\frac{3}{4} \text { or } \cos \frac{\phi}{2}=\frac{\sqrt{3}}{2} \\ & \Rightarrow \quad \frac{\phi}{2}=\frac{\pi}{6} \text { or } \phi=\frac{\pi}{3}\end{aligned}$

Topic: Physics, Wave Optics, JEE Main
In the Young's double slit experiment, the spacing between two slits is $0.1 \mathrm{~mm}$. If the screen is kept at a distance of $1.0 \mathrm{~m}$ from the slits and the wavelength of light is $5000 Å$, then the fringe width is

Options:
A $1.0 \mathrm{~cm}$
B $1.5 \mathrm{~cm}$
C $0.5 \mathrm{~cm}$
D $2.0 \mathrm{~cm}$

The correct answer is:
$0.5 \mathrm{~cm}$

Solution:
$\beta=\frac{\lambda D}{d}=\frac{5000 \times 10^{-10} \times 1}{0.1 \times 10^{-3}} \mathrm{~m}=5 \times 10^{-3} \mathrm{~m}=0.5 \mathrm{~cm}$

Topic: Physics, Wave Optics, JEE Main
In thermal equilibrium, the average velocity of gas molecules is

Options:
A Proportional to $\sqrt{T}$
B Proportional to $T^2$
C Proportional to $T^{-1}$
D Zero

The correct answer is:
Proportional to $\sqrt{T}$

Solution:
$v_{\text {average }}=\sqrt{\frac{8 R T}{\pi M}} \Rightarrow v_{a v} \propto \sqrt{T}$

Topic: Physics, Kinetic Theory of Gases, JEE Main
In thermodynamic process, 200 Joules of heat is given to a gas and 100 Joules of work is also done on it. The change in internal energy of the gas

Options:
A $100 \mathrm{~J}$
B $300 \mathrm{~J}$
C $419 \mathrm{~J}$
D $24 \mathrm{~J}$

The correct answer is:
$300 \mathrm{~J}$

Solution:
$\begin{aligned} & \Delta Q=\Delta U+\Delta W ; \Delta Q=200 \mathrm{~J} \text { and } \Delta W=-100 \mathrm{~J} \\ & \Rightarrow \Delta U=\Delta Q-\Delta W=200-(-100)=300 \mathrm{~J}\end{aligned}$

Topic: Physics, Thermodynamics, JEE Main
In thermodynamic processes which of the following statements is not true?

Options:
A In an adiabatic process the system is insulated from the surroundings
B In an isochoric process pressure remains constant
C In an isothermal process the temperature remains constant
D In an adiabatic process $\mathrm{pV}^\gamma=$ constant

The correct answer is:
In an isochoric process pressure remains constant

Solution:
The correct answer is option 1) i.e. In an isochoric process pressure remains constant.
CONCEPT:
- Thermodynamic process: It is the change in a system from its initial state of thermodynamic equilibrium to the final state of thermodynamic equilibrium.
- The various types of thermodynamic processes and their relation with pressure( \((\mathrm{P})\) and volume(V) of gas, is as follows:
$\begin{array}{|c|c|}
\hline \begin{array}{l}
\text{Isobaric process}
\end{array} & \begin{array}{l}
\text{- It is a process that takes place under constant pressure.}
\end{array} \\
\hline \begin{array}{l}
\text{Isochoric process}
\end{array} & \begin{array}{l}
\text{- It is a process where the change in volume is zero.} \\
\text{- As the change in volume is zero, the work done is zero.}
\end{array} \\
\hline \begin{array}{l}
\text{Isothermal process}
\end{array} & \begin{array}{l}
\text{- It is a process that takes place under constant temperature.}
\end{array} \\
\hline \begin{array}{l}
\text{Adiabatic process}
\end{array} & \begin{array}{l}
\text{- It is a process in which no heat is exchanged between the} \\
\text{system and the surrounding i.e. the system is insulated.} \\
\text{- It is represented mathematically as } \mathrm{PV}= \text{ constant}
\end{array} \\
\hline
\end{array}$

EXPLANATION:
- An isochoric process is a thermodynamic process that occurs when the change in volume is zero.
- Hence the statement 'In an isochoric process pressure remains constant is not true.

Topic: Physics, Thermodynamics, NEET, NEET 2009 (Screening)
In thermodynamic processes which of the following statements is not true?

Options:
A In an adiabatic process $\mathrm{PV}^\gamma=$ constant
B In an adiabatic process the system is insulated from the surroundings
C In an isochoric process pressure remains constant
D In an isothermal process the temperature remains constant

The correct answer is:
In an isochoric process pressure remains constant

Solution:
Isochoric $\rightarrow$ Volume constant

Topic: Physics, Thermodynamics, NEET, NEET 2009 (Mains)
In thermodynamics, for an isochoric process, which one of the following statement is INCORRECT?

Options:
A Energy exchanged is used to do work and also to change internal energy
B No work is done in the process
C It is a constant volume process
D Temperature of the system change during the process

The correct answer is:
Energy exchanged is used to do work and also to change internal energy

Solution:
In isochoric process, volume remains constant and hence no work is done. Energy exchanged is used to change in internal energy only. Hence statement (A) is incorrect.

Topic: Physics, Thermodynamics, MHT CET, MHT CET 2021 (21 Sep Shift 2)
In thermodynamics which one of the following is not an intensive property

Options:
A Pressure
B Density
C Volume
D Temperature

The correct answer is:
Volume

Solution:
Volume is not an intensive property.

Topic: Chemistry, Thermodynamics (C), JEE Main
In thermodynamics, why is it important to have a boundary that separates the system from the surroundings?

Options:
A To control all movements of matter and energy
B To allow unlimited exchange of matter and energy
C To minimize the effect of the surroundings on the system
D To ensure that the room temperature remains constant

The correct answer is:
To control all movements of matter and energy

Solution:
The boundary is designed to enable control and monitoring of all movements of matter and energy between the system and surroundings.

Topic: Chemistry, Thermodynamics (C), NEET
In this figure, an ideal liquid flows through the tube, which is of uniform cross-section. The liquid has velocities $v_A$ and $V_B$, and pressure $P_A$ and $P_B$ at points $A$ and $B$ respectively

Options:
A $v_A=v_B$
B $v_B \gt v_A$
C $P_A=P_B$
D $P_B \gt P_A$

The correct answers are:
$v_A=v_B$, $P_B \gt P_A$

Solution:
As per the continuity equation, $\mathrm{Av}=$ constant. As the crosssectional area remains same we get $\mathrm{v}_{\mathrm{A}}=\mathrm{v}_{\mathrm{B}}$.
And as point $B$ is at the lower level, by Bernoulli's theorem $P_B+\frac{1}{2} \rho v^2=P_A+h+\frac{1}{2} \rho v^2$ hence $P_B \gt P_A$.

Topic: Physics, Mechanical Properties of Fluids, JEE Main