datavorous/entrance-exam-dataset huggingface dataset:
An example of halide ore is

Options:
A Galena
B Bauxite
C Cinnabar
D Cryolite

The correct answer is:
Cryolite

Solution:
Cryolite Na3AlF6âHalide ore
Galena PbSCinnabar HgSSulphide ore
Bauxite Al2O3.2H2OâOxide ore

Topic: Chemistry, General Principles and Processes of Isolation of Metals, NEET
An example of plant growth-promoting Gram-negative bacterium, which is pleiomorphic in nature is

Options:
A Pseudomonos
B Rhizobium
C Aspergillus
D Azospirillum

The correct answer is:
Rhizobium

Solution:
Rhizobium is the soil bacterium which is Gram-negative in nature and shows pleiomorphic structure. This is a rhizosphere bacterium involved in the formation of root nodules by getting associated with the roots of the leguminous plants.

Topic: Biology, Microbes in Human Welfare, NEET
An example of syncytium is

Options:
A pollen grain formation after meiosis
B Any normal cell division
C liquid endosperm in coconut
D Human zygote

The correct answer is:
liquid endosperm in coconut

Solution:
Syncytium means 'together' or 'fused'. It is multinucleate cell formed by formation of uninucleate cells. This is seen in plants too. Coconut contains liquid known as liquid endosperm. It is formed by free nuclear divisions without undergoing cytoplasmic division. Hence it is multinucleate or syncytium.

Topic: Biology, Cell Cycle and Cell Division, NEET
An excess of AgNO3 is added to 100 mL of a 0.01 M solution of dichlorotetraaquachromium (III) chloride. The number of moles of AgCl precipitated would be

Options:
A 0.002
B 0.003
C 0.01
D 0.001

The correct answer is:
0.001

Solution:
AgNO3excess+[Cr(H2O)4Cl2]Clmole→AgCl
n=MV=0.01×1001000=0.001

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
An excess of $\mathrm{AgNO}_3$ is added to $100 \mathrm{~mL}$ of a $0.01 \mathrm{M}$ solution of dichlorotetraaquachromium (III) chloride. The number of moles of AgCl precipitate would be

Options:
A 0.001
B 0.002
C 0.003
D 0.01

The correct answer is:
0.001

Solution:
The formula of dichlorotetraqua chromium
(III) chloride is $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_4 \mathrm{Cl}_2\right] \mathrm{Cl}$.
On ionisation it generates only one $\mathrm{Cl}^{-}$ion.
$$
\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_4 \mathrm{Cl}_2\right] \mathrm{Cl} \longrightarrow\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_4 \mathrm{Cl}_2\right]^{+}+\mathrm{Cl}^{-}
$$
Initial $100 \times 0.01$
0
$\mathrm{mmol}=1 \mathrm{mmol}$
After ionisation 0
$1 \mathrm{mmol} \quad 1 \mathrm{mmol}$
One mole of $\mathrm{Cl}^{-}$ions react with only 1 mole of $\mathrm{AgNO}_3$ molecule to produce 1 mole of $\mathrm{AgCl}$.
$\therefore 1$ mmol or $1 \times 10^{-3}$ mole reacts with $\mathrm{AgNO}_3$ to give $\mathrm{AgCl}$
$$
=\frac{1 \times 1 \times 10^{-3}}{1}=10^{-3} \text { or } 0.001 \mathrm{~mol} \mathrm{AgCl}
$$

Topic: Chemistry, Solutions, NEET, NEET 2013 (All India)
An excess of Na2S2O3 react with aqueous CuSO4 to give

Options:
A CuS2O3
B Cu2S2O3
C Na2CuS2O32
D Na4Cu6S2O35

The correct answer is:
Na4Cu6S2O35

Solution:
CuSO4+Na2S2O3⟶CuS2O3+Na2SO4
2 CuS 2 O 3 + Na 2 S 2 O 3 → Cu 2 S 2 O 3 + Na 2 S 4 O 6
3 Cu 2 S 2 O 3 +2 Na 2 S 2 O 3 → Na 4 Cu 6 S 2 O 3 5

Topic: Chemistry, s Block Elements, NEET
An excited He+ ion emits two photons in succession, with wavelengths 108.5 nm and 30.4 nm in making a transition to the ground state. The quantum number n, corresponding to its initial excited state is

(for a photon of wavelength Î», energy E=1240 eVÎ»(in nm))

Options:
A n=6
B n=5
C n=7
D n=4

The correct answer is:
n=5

Solution:
E1=1240λ1=1240108.5≈11.43 eV
E2=1240λ2=124030.4≈40.79 eV
ETotal=E1+E1=52.22 eV
52.22=13.6221-1n2
52.22=54.41-1n2
0.96=1-1n2
1n2=0.04
n2=10.04=1004=25
n=5

Topic: Physics, Atomic Physics, NEET
An experiment is performed to measure the specific heat of copper. A lump of copper is heated in an oven, then dropped into a beaker of water. To calculate the specific heat of copper, the experimenter must know or measure the value of all quantities below except the

Options:
A heat capacity of water and breaker
B original temperature of the copper and the water
C final (equilibrium) temperature of the copper and the water
D time taken to achieve equilibrium after the copper is dropped into the water

The correct answer is:
time taken to achieve equilibrium after the copper is dropped into the water

Solution:
Loss (copper) = gain (water + beaker)
mCusCuTCu-T=mwsw T-Tw+mbsbT-Tw
Hence, final temperature can be calculated.

Topic: Physics, Thermal Properties of Matter, NEET
An explant is

Options:
A bud of a plant.
B virus infected part of plant.
C the part of the plant used in tissue culture.
D the part of the plant that expresses a specific gene.

The correct answer is:
the part of the plant used in tissue culture.

Solution:
Any part of the plant taken out and grown out in a test tube, under sterile conditions in special nutrient media is called explants.

Topic: Biology, Strategies for Enhancement in Food Production, NEET
An explant is

Options:
A dead plant
B part of the plant that is not totipotent.
C part of the plant used in tissue culture
D part of the plant that expresses a specific gene.

The correct answer is:
part of the plant used in tissue culture

Solution:
Any part of the plant taken out and grown out in a test tube, under sterile conditions in special nutrient media, is called an explant.

Topic: Biology, Strategies for Enhancement in Food Production, NEET
An explosion blows a rock into three parts. Two parts go off at right angles to each other. These two are, $1 \mathrm{~kg}$ first part moving with a velocity of $12 \mathrm{~ms}^{-1}$ and $2 \mathrm{~kg}$ second part moving with a velocity of $8 \mathrm{~ms}^{-1}$. If the third part flies off with a velocity of $4 \mathrm{~ms}^{-1}$, its mass would be

Options:
A $5 \mathrm{~kg}$
B $7 \mathrm{~kg}$
C $17 \mathrm{~kg}$
D $3 \mathrm{~kg}$

The correct answer is:
$5 \mathrm{~kg}$

Solution:
Key Idea Apply law of conservation of linear momentum.
Momentum of first part $=1 \times 12=12 \mathrm{~kg} \mathrm{~ms}^{-1}$
Momentum of the second part
$$
=2 \times 8=16 \mathrm{~kg} \mathrm{~ms}^{-1}
$$
$\therefore$ Resultant momentum
$$
=\sqrt{(12)^2+(16)^2}=20 \mathrm{~kg} \mathrm{~ms}^{-1}
$$
The third part should also have the same momentum.
Let the mass of the third part be, then
$$
\begin{aligned}
4 \times M & =20 \\
M & =5 \mathrm{~kg}
\end{aligned}
$$
$$
\text { Alternative : }
$$
$$
\begin{aligned}
\mathrm{Mv} \cos \theta & =12 \\
\mathrm{Mv} \sin \theta & =16 \\
\tan \theta & =\frac{16}{12}=\frac{4}{3} \\
M & =\frac{12 \times 5}{4 \times 3}=\frac{60}{12}=5 \mathrm{~kg}
\end{aligned}
$$

Topic: Physics, Center of Mass Momentum and Collision, NEET, NEET 2009 (Screening)
An explosion breaks a rock into three parts in a horizontal plane. Two of them go off at right angles to each other. The first part of mass $1 \mathrm{~kg}$ moves with a speed of $12 \mathrm{~ms}^{-1}$ and the second part of mass $2 \mathrm{~kg}$ moves with $8 \mathrm{~ms}^{-1}$ speed. If the third part flies off with $4 \mathrm{~ms}^{-1}$ speed, then its mass is

Options:
A $3 \mathrm{~kg}$
B $5 \mathrm{~kg}$
C $7 \mathrm{~kg}$
D $17 \mathrm{~kg}$

The correct answer is:
$5 \mathrm{~kg}$

Solution:
We have $p_1+p_2+p_3=0 [\because p=m v]$
$\begin{aligned}
& \therefore 1 \times 12 i+2 \times 8 j+p_3=0 \\
& \Rightarrow 12 \mathrm{i}+16 \mathrm{j}+\mathrm{p}_3=0 \\
& \Rightarrow \mathrm{P}_3=-(12 \mathrm{i}+16 \mathrm{j}) \\
& \therefore \mathrm{p}_3=\sqrt{(12)^2+(16)^2} \\
& =\sqrt{144+256} \\
& =20 \mathrm{~kg}-\mathrm{m} / \mathrm{s}
\end{aligned}$
Now, $p_3=m_3 v_3$
$\Rightarrow m_3=\frac{p_3}{V_3}=\frac{20}{4}=5 \mathrm{~kg}$

Topic: Physics, Center of Mass Momentum and Collision, NEET, NEET 2013 (All India)
An explosion reaction takes place when concentrated H2SO4 is added to KMnO4 . Which of the following is formed

Options:
A Mn2O7
B Mn2O3
C MnSO4
D MnO2

The correct answer is:
Mn2O7

Solution:
H2SO4+2KMnO4→K2SO4+Mn2O7explosive+H2O
Reaction produces a highly stable magnese complex that can explode violently.

Topic: Chemistry, d and f Block Elements, NEET
An extra-chromosomal, self-replicating part of the cell that has proven to be a boon to biotechnology is:

Options:
A

Virus

B

Mitochondria

C

Nucleus

D

Plasmid

The correct answer is:

Plasmid

Solution:
Plasmids are extra-chromosomal, self-replicating. Usually, circular, double-stranded DNA molecule found naturally in many bacteria. Scientists have taken advantage of plasmids to use them as tools to clone, transfer, and manipulate genes. Plasmids that are used experimentally for these purposes are called vectors. Researchers can insert DNA fragments or genes into a plasmid vector, creating a so-called recombinant plasmid. This plasmid can be introduced into a bacterium by way of the process called transformation. Then, because bacteria divide rapidly, they can be used as factories to copy DNA fragments in large quantities.

In1993, Eily Lilly, an American company prepared two-DNA sequences corresponding to A and B, chain of human Insulin and introduced them in a plasmid of E. coil to produce Insulin chains, chain A and B. Each chain was produced separately, then extracted and combined by creating disulfide bonds to form human Insulin (humulin).

Topic: Biology, Biotechnology and Its Applications, NEET
An α-globin chain is coded by the gene present on _______ chromosome.

Options:
A 10th
B 12th
C 14th
D 16th

The correct answer is:
16th

Solution:
Haemoglobin is made up of alpha protein and beta protein. The alpha protein is synthesized by the gene HBA1 gene which is located in the 16th chromosome. The locus on the chromosome is known as alpha-globin locus.

Topic: Biology, Principles of Inheritance and Variation, NEET
An H-atom in the ground state is excited by monochromatic radiation of photon energy 13.056 eV. The number of emission lines will be (given its ionisation energy is 13.6 eV)

Options:
A one
B two
C four
D ten

The correct answer is:
ten

Solution:
If n, is the value of excited state, then
number of spectral lines = n n - 1 2
Energy of electron in ground state= – (ionisation energy) =-13.6 eV
Final energy -13.6+13.056=-0.544 eV
If n be the value of excited state, then
-13.6n2=-0.544
⇒ n2=25
⇒ n=5
Now, the number of spectral lines are
=nn-12=5×42=10 lines

Topic: Physics, Atomic Physics, NEET
An HIV infected individual will show a negative result for an antibody-antigen based test if the body has not produced antibodies. For such a situation, which of the following test could be used to confirm the presence of HIV?

Options:
A MRI
B PCR
C ELISA
D Widal test

The correct answer is:
PCR

Solution:
Polymerase chain reaction (PCR) is a process to amplify nucleic acid sequences. PCR technique can be used to detect the presence of HIV genetic material (RNA) in the blood sample even before the onset of antibodies development in the infected person. PCR is like a confirmatory test for HIV diagnosis.

Topic: Biology, Biotechnology and Its Applications, NEET
An ice skater spins at 3 Ï rad s-1 with her arms extended. If her moment of inertia with arms folded is 75% of that with arms extended, her angular velocity when she folds her arms is

Options:
A π rad s -1
B 2 π rad s -1
C 3 π rad s -1
D 4 π rad s -1

The correct answer is:
4 π rad s -1

Solution:
Here, ω 1 = 3 π rad s -1 , I 1 = I
I 2 = 7 5 1 0 0 I 1 = 3 4 I, ω 2 = ?
As I 2 ω 2 = I 1 ω 1
&therefore; ω 2 = I 1 I 2 × ω 1 = 4 3 × 3 π = 4 π rad s -1

Topic: Physics, Rotational Motion, NEET
An ideal battery of emf 2 V and a series resistance R are connected in the primary circuit of a potentiometer of length 1 m and resistance 5 Î©. The value of R, to give a potential difference of 5 mV across 10 cm of potentiometer wire is

Options:
A 1 8 0 Ω
B 1 9 0 Ω
C 1 9 5 Î©
D 2 0 0 Ω

The correct answer is:
1 9 5 Î©

Solution:
I=ER+r=2R+5 A
Therefore, the potential difference across the potentiometer wire of lengths L = 100 cm is
V=Ir=2R+5Ã5=10R+5 V
Given, V=5 mV=5Ã10-3 V for 10 cm of wire. Hence, we have
5Ã10-3=1R+5âR=195 Î©

Topic: Physics, Current Electricity, NEET
An ideal coil of 10 H is connected in series with a resistance of 5 Î© and a battery of 5 V . 2 seconds after the connection is made, the current flowing in ampere in the circuit is

Options:
A 1 - e -1
B 1 - e
C e
D e -1

The correct answer is:
1 - e -1

Solution:
During the growth of current in LR circuit is given by I = I 0 1 - e - R L t
or I = E R 1 - e - R L t = 5 5 1 - e - 5 10 Ã 2
I = 1 - e -1 .

Topic: Physics, Alternating Current, NEET
An ideal contraceptive should be

Options:
A easily unavailable.
B effective and irreversible
C with no or least side effects
D More than one option is correct.

The correct answer is:
with no or least side effects

Solution:
Contraceptives are birth control methods and thus couples can avoid unwanted pregnancies. An ideal contraceptives are cheap, easy to use, causes almost nil side effects, are reversible and causes no interference with sexual pleasure. The contraceptives are designed keeping these features in mind.

Topic: Biology, Reproductive Health, NEET
An ideal diatomic gas occupies a volume V1 at a pressure P1 . The gas undergoes a process in which the pressure is proportional to the volume. At the end of process the root mean square speed of the gas molecules has doubled from its initial value then the heat supplied to the gas in the given process is

Options:
A 7P1V1
B 8P1V1
C 9P1V1
D 10P1V1

The correct answer is:
9P1V1

Solution:
As P∝V
∴ PV-1= constant
Also, C=Cv-Rx-1
C=52R-R-1-1
C=3R
root mean square speed is, v=3RTM
But as root mean square speed is doubled therefore temperature becomes four times.
Hence, ΔQ=nCΔT=n×3R×4T1-T1=9nRT1
ΔQ=9P1V1

Topic: Physics, Kinetic Theory of Gases, NEET
An ideal fluid flows through a pipe of circular cross-section with diameters 5 cm and 10 cm as shown in the figure. The ratio of velocities of fluid at A and B is

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

The correct answer is:
4:1

Solution:
From continuity equation,
a1v1=a2v2
v1v2=a2a1
a1=Ïd122
â a2=Ïd222
Where d1 and d2 are the diameters of the openings.
â v1v2=Ïd224Ïd124=d22d12=10 cm5 cm2
=10Ã105Ã5=41 or 4:1

Topic: Physics, Mechanical Properties of Fluids, NEET
An ideal fluid is flowing in a non-uniform cross-sectional tube $X Y$ (as shown in the figure) from end $X$ to end $Y$. If $K_1$ and $K_2$ are the kinetic energy per unit volume of the fluid at $X$ and $Y$ respectively, then the correct option is :

Options:
A $K_1=K_2$
B $2 K_1=K_2$
C $K_1\gtK_2$
D $K_1 \lt K_2$

The correct answer is:
$K_1\gtK_2$

Solution:
According to Bernoulli's principle,
Kinetic energy per unit volume + Potential energy per unit volume + Pressure $=$ Constant
$\frac{1}{2} \rho V^2+\rho g h+P=$ constant
Apply Bernoulli's principle at point $X$ and $Y$,
$P+K_1+\rho g(0)=P+K_2+\rho g(\mathrm{~h})$
$K_1=K_2+\rho g h$
$K_1\gtK_2$

Topic: Physics, Mechanical Properties of Fluids, NEET, NEET 2024 (Re-NEET)
An ideal gas (1 mole, monatomic) is in the initial state P (see diagram) on an isothermal curve A at a temperature T0. It is brought under a constant volume 2V0 process to Q which lies on an adiabatic curve B intersecting the isothermal curve A at P0, V0, T0. The change in the internal energy of the gas (in terms of T0) during the process is 223=1.587

Options:
A 2.3T0
B -4.6T0
C -2.3T0
D 4.6T0

The correct answer is:
-4.6T0

Solution:
Temperature at state P=T0, since P lies on the isotherm of temperature T0. If T be the temperature at Q, then for the adiabatic process B, we have T0V0γ-1=T2V0γ-1
T=T2γ-1=T022/3
Change in the internal energy of the gas is
ΔU=CVT-T0=R-1T022/3-T0
=3RT01-22/32×22/3=-4.6T0

Topic: Physics, Thermodynamics, NEET
An ideal gas (γ=1.5) is expanded adiabatically. How many times has the gas to be expanded to reduce the root mean square velocity of molecules 2 times?

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

The correct answer is:
16  times

Solution:
vrms=3RTM or vrmsâT
vrms is to reduce two times, ie, the temperature of the gas will have to reduce four times or
Tâ²T=14
During adiabatic process,
TVÎ³-1=Tâ²Vâ²Î³-1
or Vâ²V=TTâ²1Î³-1
=411.5-1=42=16
â´ Vâ²=16 V

Topic: Physics, Kinetic Theory of Gases, NEET
An ideal gas Î³ = 1.5 is expanded adiabatically. How many times has the gas to be expanded to reduce the root-mean-square velocity of molecules becomes half?

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

The correct answer is:
16 times

Solution:
v rms = 3 RT M
Or v rms â T
Since vrms becomes half, so temperature is reduced to one-fourth of its previous value
T â² T = 1 4
During adiabatic process,
TV Î³ - 1 = T â² V Î³ - 1
or V â² V = T T â² 1 Î³ - 1
= 4 1 1.5 - 1 = 4 2 = 1 6
â´ V â² = 1 6 V

Topic: Physics, Thermodynamics, NEET
An ideal gas (Î³=1.5) is expanded adiabatically. How many times its initial volume should the gas be expanded to reduce the root mean square velocity of molecules by a factor of 2?

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

The correct answer is:
16 times

Solution:
vrms=3RTM or vrmsâT
vrms is to reduce two times, i.e. the temperature of the gas will have to reduce four times or
T'T=14
During the adiabatic process,
Tâ¡Vâ¡Î³-1=Tâ¡'Vâ¡'Î³-1
or Vâ¡'V=TT'1Î³-1
=411.5-1=42=16
â´ V'=16 V

Topic: Physics, Thermodynamics, NEET
An ideal gas A and a real gas B have their volumes increased from V to 2V under isothermal conditions. The increase in internal energy

Options:
A of A will be more than B
B of A will be less than B
C will be the same in both cases
D will be zero in both cases

The correct answer is:
will be zero in both cases

Solution:
Under isothermal conditions, T=constant
∴ Internal energy =constant or change in internal energy is zero

Topic: Physics, Thermodynamics, NEET
An ideal gas at 27 Â°C is compressed adiabatically to 827 of its original volume. If Î³=53, then the rise in temperature is

Options:
A 450 K
B 375 K
C 225 K
D 405 K

The correct answer is:
375 K

Solution:
For an adiabatic process TV Î³ - 1 = constant . Therefore,
T 1 T 2 = V 2 V 1 Î³ - 1
â T 2 = T 1 V 1 V 2 Î³ - 1
= 3 0 0 2 7 8 5 / 3 - 1 = 3 0 0 2 7 8 2 / 3 = 6 7 5 K
â Î T = 6 7 5 - 3 0 0 = 3 7 5 K

Topic: Physics, Thermodynamics, NEET
An ideal gas at 27°C is compressed adiabatically to 827 of its original volume. If γ=53, then the rise in temperature is

Options:
A 450 K
B 375 K
C 675 K
D 405 K

The correct answer is:
375 K

Solution:
T V Î³â1 = constant
300 V 5 3 â1 = T 2 8V 27 5 3 â1
300= T 2 8 27 2 3
300= T 2 4 9
â T 2 =675 K
Rise in temperature =(675â300)âK
=375 K

Topic: Physics, Thermodynamics, NEET
An ideal gas at a pressure 1 atm and temperature of 27 Â°C is compressed adiabatically until its pressure becomes 8 times the initial pressures. Then the final temperature is Î³=32

Options:
A 627 °C
B 527 °C
C 427 °C
D 327 °C

The correct answer is:
327 °C

Solution:
Here, p=1 atm, T1=27 °C
=27+273=300 K
p2=8 atm,
As changes are adiabatic,
∴ p 1 γ−1 T 1 −γ = p 2 γ−1 T 2 −γ
T2T1-γ=p1p2γ-1
T=600 K=600-273 °C=327 °C

Topic: Physics, Thermodynamics, NEET
An ideal gas equation can be written as, P=ρRTM0 where ρ and M0 are respectively,

Options:
A mass density, mass of the gas
B number density, molar mass
C mass density, molar mass
D number density, mass of the gas

The correct answer is:
mass density, molar mass

Solution:
Recall the ideal gas equation, in terms of pressure, volume and temperature, PV=nRT⇒P=1V·mM0·RT, here R, is the universal gas constant,

⇒P=mVRTM0⇒P=ρRTM0, the density is defined as mass per unit volume, ρ=mv,and, M0=the molar mass.

Topic: Physics, Kinetic Theory of Gases, NEET, NEET 2020 (Phase 2)
An ideal gas expands from 10-3m3 to 10-2m3 at 300 K, against a constant pressure of 105Nm-2 . The work done on it is

Options:
A -0.9 kJ
B -9 kJ
C -900 kJ
D -90 kJ

The correct answer is:
-0.9 kJ

Solution:
w=− P ext V 2 − V 1
=− 10 5 Nm −2 × 10 −2 m 3 − 10 −3 m 3
=− 10 5 ×9× 10 −3 Nm or J
=−9× 10 2  J= −0.9 kJ

Topic: Chemistry, Thermodynamics (C), NEET
An ideal gas expands in volume from 1Ã10-3 m3 to 1Ã10-2 m3 at 300 K against a constant pressure of 1Ã105 Nm-2 . The work done is

Options:
A –900 kJ
B –900 J
C 270 kJ
D 940 kJ

The correct answer is:
–900 J

Solution:
Work done due to change in volume against constant pressure is given as follows
W=-PV2-V1=-1×105 Nm-21×10-2-1×10-3 m3
= –900 Nm
= –900 J

Topic: Chemistry, States of Matter, NEET
An ideal gas expands isothermally from $10^{-3} \mathrm{~m}^3$ to $10^{-2} \mathrm{~m}^3$ at $300 \mathrm{~K}$ against a constant pressure of $10^5 \mathrm{Nm}^{-2}$. The work done on the gas is

Options:
A $+270 \mathrm{~kJ}$
B $-900 \mathrm{~J}$
C $+900 k J$
D $-900 k J$

The correct answer is:
$-900 \mathrm{~J}$

Solution:
For an isothermal irreversible expansion, Work done $(\mathrm{W})=-\mathrm{p}_{\mathrm{ext}}\left(\mathrm{V}_2-\mathrm{V}_1\right)$ where, $\mathrm{V}_1=$ initial volume
$$
\mathrm{V}_2=\text { final volume }
$$
Given, $\mathrm{p}_{\mathrm{ext}}=10^5 \mathrm{Nm}^{-2}, \mathrm{~V}_1=10^{-3} \mathrm{~m}^3, \mathrm{~V}_2=10^{-2} \mathrm{~m}^3$
On substituting the given values in Eq, (i),
We get,
$$
\begin{aligned}
W & =-10^5 \mathrm{Nm}^{-2}\left(10^{-2} \mathrm{~m}^3-10^{-3} \mathrm{~m}^3\right) \\
& =-10^5 \mathrm{Nm}^{-2} \times 10^{-3}(10-1) \mathrm{m}^3 \\
& =-900 \mathrm{Nm}=-900 \mathrm{~J}
\end{aligned}
$$

Topic: Chemistry, Thermodynamics (C), NEET, NEET 2019 (Odisha)
An ideal gas follows a process described by the equation $\mathrm{PV}^2=\mathrm{C}$ from the initial $\left(\mathrm{P}_1, \mathrm{~V}_1, \mathrm{~T}_1\right)$ to final ( $\left.\mathrm{P}_2, \mathrm{~V}_2, \mathrm{~T}_2\right)$ thermodynamic states, where $\mathrm{C}$ is a constant. Then:

Options:
A If $\mathrm{P}_1>{ }_2$ then $\mathrm{T}_1 < \mathrm{T}_2$
B If $\mathrm{V}_2>\mathrm{V}_1$ then $\mathrm{T}_2>\mathrm{T}_1$
C If $\mathrm{V}_2>\mathrm{V}_1$ then $\mathrm{T}_2 < \mathrm{T}_1$
D If $\mathrm{P}_1>\mathrm{P}_2$ then $\mathrm{V}_1>\mathrm{V}_2$

The correct answer is:
If $\mathrm{V}_2>\mathrm{V}_1$ then $\mathrm{T}_2 < \mathrm{T}_1$

Solution:
As per question,
$\mathrm{PV}^2=\mathrm{C}$
And by ideal gas equation, we have;
$\mathrm{PV}=n \mathrm{RT}$
Hence, $\left(\frac{n \mathrm{RT}}{\mathrm{V}}\right) \mathrm{V}^2=\mathrm{C}$
Hence, TV $=$ constant
$\mathrm{T} \propto \frac{1}{\mathrm{~V}} \therefore \mathrm{V}_2 > \mathrm{V}_1 \text { and } \mathrm{T}_1 > \mathrm{T}_2$

Topic: Physics, Thermodynamics, NEET, NEET 2022 (Phase 2)
An ideal gas heat engine is operating between 227 Â°C and 127 Â°C. It absorbs 104 J of heat at the higher temperature. The amount of heat converted into work is

Options:
A 2000 J

B 4000 J

C 8000 J

D 5600 J

The correct answer is:
2000 J

Solution:
Î·=1-T2T1=1-127+273227+273=15
W=Î·Q1=15Ã104 J=2000 J

Topic: Physics, Thermodynamics, NEET
An ideal gas heat engine operates in a Carnot cycle between $227^{\circ} \mathrm{C}$ and $127^{\circ} \mathrm{C}$. It absorbs $6 \mathrm{kcal}$ at the higher temperature. The amount of heat (in kcal) converted into work is equal to:

Options:
A 4.8
B 3.5
C 1.6
D 1.2

The correct answer is:
1.2

Solution:
$$
\begin{aligned}
& \text { Efficiency of carnot engine }=\frac{W}{Q_1} \\
& =1-\frac{T_2}{T_1} \\
& \frac{W}{6}=1-\frac{400}{500} \\
& \Rightarrow \quad W=\frac{6}{5}=1.2 \mathrm{Kcal}
\end{aligned}
$$

Topic: Physics, Thermodynamics, NEET, NEET 2003
An ideal gas heat engine operates in a Carnot cycle between 227 Â°C and 127 Â°C. It absorbs 6 kcal of heat at a higher temperature. The amount of heat in kcal rejected to sink is

Options:
A 4.8
B 2.4
C 1.2
D 6.0

The correct answer is:
4.8

Solution:
Here, T1=227 °C=227+273=500 K
T2=127 °C=127+273=400 K , Q 1 = 6 kcal ,
Q 2 Q 1 = T 2 T 1 = 400 500 = 4 5
Q 2 = 4 5 Q 1 = 4 5 × 6 kcal = 4.8 kcal.

Topic: Physics, Thermodynamics, NEET
An ideal gas heat engine operates in a Carnot's cycle between 227 Â°C and 127 Â°C. It absorbs 6Ã104 J at the higher temperature. The amount of heat converted into work is

Options:
A 1.6Ã104 J
B 1.2Ã104 J
C 4.8Ã104 J
D 3.5Ã104 J

The correct answer is:
1.2Ã104 J

Solution:
Using the relation,
WQ1=Q1-Q2 Q1
WQ1=1-Q2Q1
WQ1=1-T2T1 âµQ1Q2=T1T2
W=Q11-T2T1
â´ W=6Ã1041-(127+273)(227+273)
W=6Ã1041-400500
=6Ã104Ã100500
=1.2Ã104 J

Topic: Physics, Thermodynamics, NEET