datavorous/entrance-exam-dataset huggingface dataset:
The potential difference that must be applied to stop the fastest photoelectrons emitted by a nickel surface, having work function $5.01 \mathrm{eV}$, when ultraviolet light of $200 \mathrm{~nm}$ falls on it, must be

Options:
A $2.4 \mathrm{~V}$
B $-1.2 \mathrm{~V}$
C $-2.4 \mathrm{~V}$
D $1.2 \mathrm{~V}$

The correct answer is:
$1.2 \mathrm{~V}$

Solution:
Energy of incident light $\mathrm{E}(\mathrm{eV})=\frac{12375}{2000}$ $=6.2 \mathrm{eV}(200 \mathrm{~nm}=2000 Å)$
According to the relation $\mathrm{E}=\mathrm{W}_0+\mathrm{eV}_0$
$$
\begin{aligned}
\Rightarrow \quad \mathrm{V}_0 & =\frac{\mathrm{E}-\mathrm{W}_0}{\mathrm{e}} \\
& =\frac{(6.2-5.01) \mathrm{e}}{\mathrm{e}} \\
& =1.2 \mathrm{~V}
\end{aligned}
$$

Topic: Physics, Dual Nature of Matter, NEET, NEET 2010 (Screening)
The potential difference that must be applied to stop the fastest photoelectrons emitted by a nickel surface, having work function 5.01 eV, when ultraviolet light of 200 nm falls on it, must be

Options:
A 2.4 V
B -1.2 V
C -2.4 V
D 1.2 V

The correct answer is:
1.2 V

Solution:
Kmax=hcλ-W=hcλ-5.01
=12375λin Å-5.01
=123752000-5.01=6.1875-5.01=1.17775
=1.2 V

Topic: Physics, Dual Nature of Matter, NEET
The potential difference VA-VB between the points A and B in the given figure is

Options:
A -3 V
B +3 V
C +6 V
D +9 V

The correct answer is:
+9 V

Solution:
Complicated circuits with multiple potential points and batteries can be easily solved using Kirchhoff's laws.

Consider applying Kirchhoff's voltage law (KVL) to the given circuit.

VA-VB=2×2+3+2×1

=4+3+2

=9 V.

Topic: Physics, Current Electricity, NEET, NEET 2016 (Phase 2)
The potential differences across the resistance, capacitance, and inductance are 80 V, 40 V and 100 V, respectively in an LCR circuit. The power factor of this circuit is

Options:
A 0.4
B 0.5
C 0.8
D 1.0

The correct answer is:
0.8

Solution:
Pac=VrmsIrmscosϕ,

where, cosϕ is the power factor.

cosϕ=Rz=VRV=80802+100-402

=80100=0.8.

where, Z is the impedence of the circuit.

Topic: Physics, Alternating Current, NEET, NEET 2016 (Phase 2)
The potential energy for a conservative system is given by

U = ax2âbx

Where a and b are positive constants. The law of the force governing the system is

Options:
A F = constant
B F=bx–2a

C F=b–2ax

D F=2ax

The correct answer is:
F=b–2ax

Solution:
In a conservative fieldF=-dUdr
∴F=-ddxax2-bx=b-2ax

Topic: Physics, Work Power Energy, NEET
The potential energy for a force field Fâ is given by U(x, y)=cos (x + y). The force acting on a particle at the position given by coordinates 0 , Ï / 4 is

Options:
A - 1 2 i ^ + j ^
B 1 2 i ^ + j ^
C 1 2 i ^ + 3 2 j ^
D 1 2 i ^ - 3 2 j ^

The correct answer is:
1 2 i ^ + j ^

Solution:
F x = - ∂ U ∂ x = sin x + y , F y = - ∂ U ∂ x = sin x + y
F x = sin x + y ] 0 , π / 4 = 1 2 ,
F y = sin x + y ] 0 , π / 4 = 1 2
∴ F = 1 2 i ^ + j ^

Topic: Physics, Work Power Energy, NEET
The potential energy for a force field Fâ is given by U(x, y)=cos (x + y). The force acting on a particle at the position given by coordinates 0 , Ï / 4 is

Options:
A -12i^+j^
B 12i^+j^
C 12i^+32j^
D 12i^-32j^

The correct answer is:
12i^+j^

Solution:
F x = - ∂ U ∂ x = sin x + y , F y = - ∂ U ∂ x = sin x + y
Fx=sinx+y0, π/4=12,
Fy=sinx+y0, π/4=12
∴ F=12i^+j^

Topic: Physics, Work Power Energy, NEET
The potential energy of a body of mass 3 kg on the surface of a planet is 54 joule. The escape velocity will be

Options:
A 18 m s-1
B 162 m s-1
C 36 m s-1
D 6 m s-1

The correct answer is:
6 m s-1

Solution:
Uinitial +Kinitial=UFinal+KFinal
12mve2=GMmR
12(3)ve2â¡=54
ve=6 m/s

Topic: Physics, Gravitation, NEET
The potential energy of a body of mass m is U=ax+by. The magnitude of the acceleration of the body will be

Options:
A abm
B a+bm
C a2+b2m
D a2+b2m

The correct answer is:
a2+b2m

Solution:
We know that U=ax+by
∂U∂x=∂∂xax+by
∂U∂x=a
∂U∂y=∂∂yax+by
∂U∂y=b
F→=∂U∂xi^+∂U∂yj^
F→=-ai^-bj^
F=a2+b2

Topic: Physics, Work Power Energy, NEET
The potential energy of a long spring when stretched by 2 cm is U. If the spring is stretched by 8 cm, potential energy stored in it will be:

Options:
A 16U
B 2U
C 4U
D 8U

The correct answer is:
16U

Solution:
Potential energy stored in spring is given by,

U=12Kx2.

First case:

U=12K2×10-22 ...Eq1

Second case:

U'=12K8×10-22 ...Eq2

From both equations, we get

UU'=116⇒U'=16U

Topic: Physics, Work Power Energy, NEET, NEET 2023 (All India)
The potential energy of a long spring when stretched by $2 \mathrm{~cm}$ is $U$. If the spring is stretched by $8 \mathrm{~cm}$ the potential energy stored in it is:

Options:
A $U / 4$
B $4 U$
C $8 U$
D $16 U$

The correct answer is:
$16 U$

Solution:
Potential energy of a spring
$\begin{aligned}
& \begin{aligned}
&= \frac{1}{2} \times \text { force constant } \\
& \times(\text { extension })^2 \\
& \Rightarrow \text { Potential energy } \propto(\text { extension })^2
\end{aligned} \\
& \text {Hence } \frac{U_1}{V_2}=\left(\frac{x_1}{x_2}\right)^2 \\
& \Rightarrow \frac{V_1}{V_2}=\left(\frac{2}{8}\right)^2 \\
& \Rightarrow \frac{U_1}{U_2}- \frac{1}{16}\left(\because U_1=V\right)
\end{aligned}$

Topic: Physics, Work Power Energy, NEET, NEET 2006
The potential energy of a particle executing S.H.M is 2.5 J, when its displacement is half of its amplitude. The total energy of the particle is

Options:
A 2.5 J
B 10 J
C 12 J
D 20 J

The correct answer is:
10 J

Solution:
P.E.=12 mω2x2=2.5 J
As x=A2
12mω2 A22=2.5
∴12mω2A24=2.5
∴12mω2A2=10 J
Total energy of system =12mω2A2=10 J

Topic: Physics, Oscillations, NEET
The potential energy of a particle in a force field is $U=\frac{A}{r^2}-\frac{B}{r}$, where $A$ and $B$ are positive constants and $r$ is the distance of particle from the centre of the field. For stable equilibrium, the distance of the particle is

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

The correct answer is:
$2 A / B$

Solution:
Given, the potential energy of a particle in a force field $U=\frac{A}{r^2}-\frac{B}{r^1}$
For stable equilibrium, $F=-\frac{d U}{d r}=0$
$0=-\frac{2 A}{r^3}+\frac{B}{r^2}$
or $\frac{2 A}{r}=B$
The distance of particle from the centre of the field
$r=\frac{2 A}{B}$

Topic: Physics, Work Power Energy, NEET, NEET 2012 (Screening)
The potential energy of a particle moving along $x$-direction varies as $V=\frac{A x^2}{\sqrt{x}+B}$. The dimensions of $\frac{A^2}{B}$ are:

Options:
A $\left[\mathrm{M}^{3 / 2} \mathrm{~L}^{1 / 2} \mathrm{~T}^{-3}\right]$
B $\left[M^{1 / 2} L^{-3}\right]$
C $\left[\mathrm{M}^2 \mathrm{~L}^{1 / 2} \mathrm{~T}^{-4}\right]$
D $\left[\mathrm{ML}^2 \mathrm{~T}^{-4}\right]$

The correct answer is:
$\left[\mathrm{M}^2 \mathrm{~L}^{1 / 2} \mathrm{~T}^{-4}\right]$

Solution:
$V=\frac{A x^2}{\sqrt{x}+B}$
As per homogeneous rule $B=\sqrt{L}$
$\mathrm{ML}^2 \mathrm{~T}^{-2}=\frac{A \mathrm{~L}^2}{\mathrm{~L}^{1 / 2}}$
$A=\mathrm{ML}^{1 / 2} \mathrm{~T}^{-2}$
$\frac{A^2}{B}=\frac{\mathrm{M}^2 L T^{-4}}{\mathrm{~L}^{1 / 2}}=\mathrm{M}^2 \mathrm{~L}^{1 / 2} \mathrm{~T}^{-4}$

Topic: Physics, Units and Dimensions, NEET, NEET 2024 (Re-NEET)
The potential energy of a particle of mass 5 kg moving in the x-y plane is given by U=-7x+24y J, where x and y are in meters. If t=0, the particle starts from rest from the origin, then speed of the particle at t=2 s is

Options:
A 5Â m/s
B 10 m/s
C 14Â m/s
D 17.5Â m/s

The correct answer is:
10 m/s

Solution:
F→=-∂U∂xi^-∂U∂yj^
F→=-∂∂x7x+24yi^+∂∂y-7x+24yj^
F→=7i^-24j^
ma→=7i^-24j^
a→=7i^-24j^5
a=255=5 m/s2
v=u+at
v=0+5×2
v=10 m/s

Topic: Physics, Work Power Energy, NEET
The potential energy of a particle of mass 5 kg moving in the x-y plane is given by U=-7x+24y J, where x and y are given in metre. If the particle starts from rest, from the origin, then the speed of the particle at t=2 s is

Options:
A 5 m/s
B 14 m/s
C 17.5 m/s
D 10 m/s

The correct answer is:
10 m/s

Solution:
F = - â U â x i ^ - â U â y j ^ = 7 i ^ - 24 j ^
â´ a x = F x mâ¡ = 7 5 = 1.4 m/s 2
a y = F y mâ¡ = - 24 5
=4.8 m/s 2 along negative yâaxis
â â´ vâ¡ x = a x tâ¡ = 1.4 Ã 2
=â2.8âm/s along negative yâaxis
and vâ¡ y = 4.8 Ã 2 = 9.6 m/s
â´ vâ¡ = vâ¡ x 2 + vâ¡ y 2 = 10 m/s

Topic: Physics, Work Power Energy, NEET
The potential energy of a particle of mass 5 kg moving in the x- y plane is given by, U=-7x+24y J
where x and y being in metre. If the particle starts from rest from the origin, then the speed of the particle at t=2 s is :

Options:
A 5 m s-1
B 14 m s-1
C 17 m s-1
D 10 m s-1

The correct answer is:
10 m s-1

Solution:
∵F→=-ΔUΔxi^+ΔUΔyj^=--7 i^+24 j^
∴F=72+242=25 N⇒ a=F/m=25/5=5 m s-2
∵v=u+at
=0+5 × 2=10 m s-1

Topic: Physics, Work Power Energy, NEET
The potential energy of a particle under a conservative force is given by Ux=x2-3x J. The equilibrium position of the particle is at

Options:
A x=1.5 m
B x=2 m
C x=2.5 m
D x=3 m

The correct answer is:
x=1.5 m

Solution:
Ux=x2−3xJ
For a conservative field, Force, F⁡=-d⁡U⁡d⁡x
∴F⁡=-d⁡d⁡xx2-3x=-2x-3=-2x+3
At equilibrium position, F=0
-2x+3=0⇒x=32m⁡=1·5m

Topic: Physics, Work Power Energy, NEET
The potential energy of a particle varies with position x according to the relation Ux=2x4-27x. The point x=32 is point of

Options:
A unstable equilibrium
B stable equilibrium
C neutral equilibrium
D none of these

The correct answer is:
stable equilibrium

Solution:
U=2x4-27x
dUdx=8x3-27
At x=32,dUdx=0 , therefore it is an equilibrium position.
d2Udx2=24x2
At x=32,d2Udx2>0 , therefore it is a stable equilibrium position.

Topic: Physics, Work Power Energy, NEET
The potential energy of a simple harmonic oscillator of mass =2 kg in its mean position is 5 J . If its total energy is 9 J and its amplitude is 0.01 m, its time period would be:

Options:
A π10 s
B π20 s
C π50 s
D π100 s

The correct answer is:
π100 s

Solution:
P.E.=5 J,  T.E.=9 J
Using mechanical energy conservation
T.E. =P.E. +K.E.
9=5+K.E.
⇒K.E.=4 (Maximum)
K.E.max at mean position =12ω2A2m
4=12×2×2πt2×1100×1100
T2=π2100×100
T=π100 s

Topic: Physics, Oscillations, NEET
The potential energy of a simple harmonic oscillator when the particle is half way to its end point is:
where $E$ energy is the total energy

Options:
A $\frac{2}{3} E$
B $\frac{1}{8} E$
C $\frac{1}{4} E$
D $\frac{1}{2} E$

The correct answer is:
$\frac{1}{4} E$

Solution:
P.E. $=\frac{1}{2} K x^2$
$=\frac{1}{2} K\left(\frac{a}{2}\right)^2=\frac{E}{4}$

Topic: Physics, Oscillations, NEET, NEET 2003
The potential energy of a system increase if work is done

Options:
A by the system against a conservative force
B by the system against a nonconservative force
C upon the system by a conservative force
D upon the system by a nonconservative force

The correct answer is:
by the system against a conservative force

Solution:
The potential energy of a system increases if work is done by the system against a conservative force.
$$
-\Delta U=W_{\text {conservative face }}
$$

Topic: Physics, Work Power Energy, NEET, NEET 2011 (Screening)
The potential energy of a system increases if work is done

Options:
A Upon the system by a non conservative force
B By the system against a conservative force
C By the system against a non conservative force
D Upon the system by a conservative force

The correct answer is:
Upon the system by a conservative force

Solution:
When work is done upon the system by a conservative force then its potential energy increases.

Topic: Physics, Work Power Energy, NEET
The potential energy of the electron present in the ground state of Li2+ ion is represented by

Options:
A +3e24πε0r
B -3e4πε0r
C -3e24πε0r2
D -3e24πε0r

The correct answer is:
-3e24πε0r

Solution:
Potential energy of the electron present in ground state is given as
Potential energy =-Ze24πε0r=-3e24πε0r

Topic: Chemistry, Structure of Atom, NEET
The potential energy U between two atoms in a diatomic molecule as a function of the distance x between atom has been shown in the adjoining figure. The atoms are

Options:
A Attracted when x lies between A and B and are repelled when x lies between B and C
B Attracted when x lies between B and C and are repelled when x lies between A and B
C Attracted when they reach B
D repelled when they reach B

The correct answer is:
Attracted when x lies between B and C and are repelled when x lies between A and B

Solution:
Negative value of potential energy decreases from C to B and there after increases. It shows that kinetic energy increases from C to B and there after it decreases. So, there is attraction from C to B and after which repulsion results.

Topic: Physics, Kinetic Theory of Gases, NEET
The potential energy (y) curve for H2 formation as a function of internuclear distance (x) of the H atoms is shown below.

The bond energy of H2 is :

Options:
A (b-a)
B (c-a)2
C (b-a)2
D (c-a)

The correct answer is:
(b-a)

Solution:
As per the given curve bond energy is the amount of energy is released during the bond formation is i.e. =Final-initial

So, Potential energy of two H atoms at infinite distance (initial) \(=\mathrm{a}\) and
Potential energy of two H atoms at distance equal to bond length (final) \(=\mathrm{b}\) Then,
\(\text {B.E. }=(b-a)\)

Topic: Chemistry, Structure of Atom, NEET, NEET 2020 (Phase 2)
The potentiometer wire is of length 1200 cm and it carries a current of 60 mA. For a cell of emf 5 V and internal resistance of 20 Î©, the null point on it is found to be at 1000 cm. The resistance of potentiometer wire is

Options:
A 60 Î©
B 120 Ω
C 100 Î©
D 80 Î©

The correct answer is:
100 Î©

Solution:
Potential gradient =51000=Vp1200RP=VPI=660Ã10-3=100 Î©

Topic: Physics, Current Electricity, NEET
The powdered ore is agitated with water or washed with running steam of water. The heavy ore particles and lighter impurities are separated. This method of concentration is known as

Options:
A metallurgy
B leaching
C gravity separation
D froth floatation process

The correct answer is:
gravity separation

Solution:
Gravity separation is based on the difference in the specific gravities of gangue and ore particles. In the stream of water, heavy ore particles settle down while the lighter gangue particles are washed away.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, NEET
The power consumed by the 4 V battery is

Options:
A 4 W
B 8Â W
C 7Â W
D cannot be calculated

The correct answer is:
4 W

Solution:
Current in the circuit,
I=EnetRnet
⇒I=10-46=1 A
∴ Power consumed by battery
P=V×I
⇒P=1×4=4 W

Topic: Physics, Current Electricity, NEET
The power dissipated in the adjacent circuit is -

Options:
A V02R
B V025R
C V025 R
D V0210R

The correct answer is:
V0210R

Solution:
Z=(Xc−XL)2+R2
=4R2+R2=5 R
Pav=(VrmsZ)2R
=(V025R)2R
Pav=V0210R

Topic: Physics, Alternating Current, NEET
The power factor of the following circuit will be

Options:
A 0.2
B 0.4
C 0.6
D 0.8

The correct answer is:
0.8

Solution:
Z=R2+(XL⁡-Xc⁡)2
⇒Z=(40+40)2+(100-40)2=100 Ω
⇒cos⁡ϕ=RZ=80100=0.8

Topic: Physics, Alternating Current, NEET
The power obtained in a reactor using $\mathrm{U}^{235}$ disintegration is $1000 \mathrm{~kW}$. The mass decay of $\mathrm{U}^{235}$ per hour is

Options:
A $20 \mu \mathrm{g}$
B $40 \mu \mathrm{g}$
C $1 \mu g$
D $10 \mu \mathrm{g}$

The correct answer is:
$40 \mu \mathrm{g}$

Solution:
Let assume power $P=1000 \mathrm{~W}$
Energy per hour $=1000 \times 3600 \mathrm{~J}$
Energy per fission $=200 \mathrm{MeV}$
$=200 \times 1.6 \times 10^{-13} \mathrm{~J}$
$\therefore$ Number of fission per hour
$n=\frac{1000 \times 3600}{200 \times 1.6 \times 10^{-13}}$
$\begin{aligned}
& \text {Number of mole per hour }=\frac{n}{N} \\
& \therefore \text { Mass per hour }=\frac{n}{N} \times 235
\end{aligned}$
$\begin{aligned}
& =\frac{1000 \times 3600 \times 235}{200 \times 1.6 \times 10^{-13} \times 6.02 \times 10^{23}} \\
& =43.9 \times 10^{-6} \mathrm{~g}
\end{aligned}$
This $43.9 \times 10^{-6} \mathrm{~g}$ is nearest value of 40 micro gram so option (b) is correct.

Topic: Physics, Nuclear Physics, NEET, NEET 2011 (Screening)
The power of a biconvex lens is 10 dioptre and the radius of curvature of each surface is 10 cm. Then the refractive index of the material of the lens is,

Options:
A 43
B 98
C 53
D 32

The correct answer is:
32

Solution:
Recall the proper relation between power and focal length, P=100f⇒f=100P=10010=10 cm, now the relation between focal length and radius of curvature and refractive index, of the spherical surface lens,

f=R2(μ-1) (for equiconvex lens)

⇒10=102(μ-1)⇒(μ-1)=12⇒μ=12+1=32.

Topic: Physics, Ray Optics, NEET, NEET 2020 (Phase 2)
The power of sound from the speaker of radio is 10 W. Now, the power of sound from the speaker of the radio is increased to 400 W by increasing the volume of the radio. The power increased in dB as compared to original power is nearly

Options:
A 8
B 12
C 13
D 16

The correct answer is:
16

Solution:
P1=10log1010-12
P1=130 dB
P2=10log40010-12
=10log4×1014
=140+10log4
P2=146 dB
∆P=(146-130)=16 dB

Topic: Physics, Waves and Sound, NEET
The power radiated by a black body is P and it radiates maximum energy at wavelength, λ0. If the temperature of the black body is now changed so that it radiates maximum energy at wavelength 34λ0, the power radiated by it becomes nP. The value of n is

Options:
A 25681
B 43
C 34
D 81256

The correct answer is:
25681

Solution:
By Stefan's law,
P=σAT4⇒P∝T4
According to Wien's law T∝1λm
⇒ P∝1λm4⇒P2P1=λm1λm24
⇒ P2P1=λ034λ04⇒nPP=25681⇒n=25681

Topic: Physics, Thermal Properties of Matter, NEET, NEET 2018
The precursor of mRNA is transcribed by

Options:
A RNA polymerase I
B RNA polymerase II
C RNA polymerase III
D Topoisomerase​

The correct answer is:
RNA polymerase II

Solution:
The precursor mRNA is hnRNA (heterogeneous nuclear RNA). It contains intron along with exons. It is transcribed by RNA polymerase II.

Topic: Biology, Molecular Basis of Inheritance, NEET
The presence of a chromogenic substrate gives blue coloured colonies of bacteria. Insertional inactivation would stop production of which of the following enzyme such that colonies do not produce any colour.

Options:
A Beta-galactosidase
B Alpha galactosidase
C Alpha lactalbumin
D All of the above

The correct answer is:
Beta-galactosidase

Solution:
In PUC plasmid, Lac Z gene is used as a selectable marker. If a transformant is grown in a medium having chromogenic substrate it produces enzyme Beta-galactosidase due to which blue coloured colonies are produced. Insertion of foreign DNA segment in Lac Z gene fails to produces enzyme Beta-galactosidase as a result colourless colonies are produced.

Topic: Biology, Biotechnology - Principles and Processes, NEET
The presence of a physical barrier between stigma and stamen prevents autogamy. This phenomenon is called as

Options:
A Herkogamy
B Heterostyly
C Dichogamy
D Cleistogamy

The correct answer is:
Herkogamy

Solution:
Herkogamy is a contrivance for cross pollination. In a bisexual flower, self-pollination can be prevented by presence of a physical barrier between style and stigma. This is herkogamy. It is seen in pansy where there is hood covering the stigma. Heterostyly is the presence of different lengths of styles in different flowers of the same species. This adaptation also helps in cross pollination. Dichogamy is the phenomenon where male and female parts of the flowers mature at different times thus causing cross pollination. Cleistogamy is the phenomenon in which the flowers always stay in bud condition and never bloom. This ensures self pollination

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
The presence of continuous phenotypic variation in F2 generation suggests that the character shows

Options:
A gene linkage.
B epistasis.
C polygenic inheritance.
D recombination.

The correct answer is:
polygenic inheritance.

Solution:
When inheritance of single character is controlled by more than one gene present at different gene locus it is called as polygenic inheritance. It produces continuous phenotypic variation as each dominant allele contributes equally to the phenotype, and recessive allele are non contributing allele.

Topic: Biology, Principles of Inheritance and Variation, NEET
The presence of which of the following in drinking water is responsible for mottling of teeth?

Options:
A Mercury
B Iodine
C Chlorine
D Fluorine

The correct answer is:
Fluorine

Solution:
Mottling of teeth is due to the presence of fluorine in drinking water (Fluorosis).

Topic: Chemistry, Environmental Chemistry, NEET
The present mass extinction is

Options:
A 3rd
B 4th
C 5th
D 6th

The correct answer is:
6th

Solution:
During the long period since the origin and diversification of life on earth there were five episodes of mass extinction of species. The current species extinction rates are estimated to be 100 to 1000 times faster than in the pre historic times due to human activities.

Topic: Biology, Biodiversity and its Conservation, NEET
The pressure and density of a diatomic gas change from (p, ρ) to ( p' ,ρ ' ) during an adiabatic change. If ρ′ρ=32, value of p′p is

Options:
A 32
B 1/32
C 128
D 1/128

The correct answer is:
128

Solution:
From p1V1γ=p1V2γ
p2p1=V1V2γ=ρ2ρ1γ
∴ P2P1=P P=128

Topic: Physics, Thermodynamics, NEET
The pressure and volume of a given mass of gas at a given temperature are P and V respectively. Keeping the temperature constant, the pressure is increased by 10% and the system is allowed to achieve a steady-state, then the pressure is decreased by 10%. What can be said about the final volume?

Options:
A less than V
B more than V
C equal to V
D less than V for diatomic and more than V for monoatomic

The correct answer is:
more than V

Solution:
When pressure is increased by 10%
P1=P 1+10100= 1110 P
Then presser is decreased by 10%
P2=P1 1-10100= 910 P1
P2=910 1110 P= 99100 P
Then temperature is constant
PV=P1 V1= P2 V2
PV= 99100PV2
V2= 10099 V

Topic: Physics, Thermodynamics, NEET
The pressure and volume of an ideal gas are related as p Î±1v2 for process AâB as shown is figure. The pressure and volume at A are 3p0 and v0 respectively and pressure at B is p0 . The work done in the process AâB is found to be x-x p0v0 find value of x.

Options:
A 3
B 6
C 4
D 5

The correct answer is:
3

Solution:
P=K V2
W= ∫dW=∫PdV
= ∫V1V2K V2 dV
K-1 VV1V2
=P1V1-P2V2
=3P0V0-P03V0
=3-3P0V0
∴ X=3

Topic: Physics, Thermodynamics, NEET
The pressure at the bottom of a liquid tank is not proportional to the

Options:
A Acceleration due to gravity
B Density of the liquid
C Height of the liquid
D Area of the liquid surface

The correct answer is:
Area of the liquid surface

Solution:
p=hρg
p∝h
p∝ρ
p∝g, p does not depend area of the liquid.

Topic: Physics, Mechanical Properties of Fluids, NEET
The pressure at the bottom of a tank of liquid is not proportional to

Options:
A The density of the liquid
B The area of the liquid surface
C The height of the liquid
D The acceleration

The correct answer is:
The area of the liquid surface

Solution:
As we know, P = Patm + ρgh
From this equation, pressure is independent of the area of the liquid surface.
Hence, the area of the liquid surface is correct.

Topic: Physics, Mechanical Properties of Fluids, NEET
The pressure exerted by an ideal gas on the walls of a container is

Options:
A 13rd of kinetic energy per unit volume of a gas
B 23rd of kinetic energy per unit volume of a gas
C 34th of kinetic energy per unit volume of a gas
D 32Ã kinetic energy per unit volume of a gas

The correct answer is:
23rd of kinetic energy per unit volume of a gas

Solution:
The pressure exerted by the gas on the walls of container is P= 1 3 Ï v 2
P =13Ïv2=13mV.v2Ã22=23 KE Volume

Topic: Physics, Thermodynamics, NEET
The pressure inside two soap bubbles is 1.01 and 1.02 atm respectively. The ratio of their respective volumes is

Options:
A 8 : 1

B 2 : 1
C 102 : 101
D (102)3 : (101)3

The correct answer is:
8 : 1

Solution:
Excess pressure, ΔP1=1.01 atm-1 atm=0.01 atm and similarly, ΔP2=0.02 atm and

The volume of the air bubble V=43πr3

Therefore, V∝r3∝[1(ΔP)3]

As ΔP∝(1r) or r∝(1ΔP)

V1V2 = ΔP2ΔP13=0.020.013 = 8

Topic: Physics, Mechanical Properties of Fluids, NEET