datavorous/entrance-exam-dataset huggingface dataset:
Potatoes are

Options:
A Collenchymatous tubers
B Parenchymatous tubers
C Sclerenchymatous tubers
D None of these

The correct answer is:
Parenchymatous tubers

Solution:
Parenchyma is the most simple and unspecialized tissue which is concerned mainly with the vegetative activities of the plant. Parenchyma cells have a different shape and function, depending on its location and in which tissue it is present. However, all parenchyma cells always have a large central vacuole and only chlorenchyma cells contain high chloroplast content. chlorenchyma cells contain chloroplasts. Chloroplasts are organelles that carry out photosynthesis. The sugars are converted into other starch, fats, and oils, and stored in other parenchyma cells in the stems and roots. Potatoes, for example, are the tubers that are starch storing parenchyma cells.

Topic: Biology, Anatomy of Flowering Plants, NEET
Potential energy between a proton and an electron is given by U=Ke23R3, then radius of Bohr's orbit can be given by

Options:
A Ke2mh2
B 6π3Ke2mn3h2
C 2πnKe2mh2
D 4π2Ke2mn3h2

The correct answer is:
4π2Ke2mn3h2

Solution:
Since electrostatic force is a conservative force, therefore, F=-dUdx

⇒F=-dUdR=Ke2R4

This force provides the necessary centripetal force, FC=mv2R

⇒Ke2R4=mv2R⇒v2=Ke2mR3

Using Bohr quantization of angular momentum theory,

⇒mvr=nh2π⇒r=4π2Ke2mn2h2

Thus, Bohr's orbit is, r=4π2Ke2mn2h2

Topic: Physics, Atomic Physics, JEE Main
Potentiometer wire is $4 \mathrm{~m}$ long and potential difference of $3 \mathrm{~V}$ is maintained between the ends. The e.m.f. of the cell which balances against a length of $100 \mathrm{~cm}$ of the potentiometer wire is

Options:
A 0.60 V
B 0.20 V
C 0.45 V
D 0.75 V

The correct answer is:
0.75 V

Solution:
Potential gradient $=\frac{3}{4} \frac{\mathrm{V}}{\mathrm{m}}$
So, emf against $100 \mathrm{~cm}=\frac{3}{4} \times 1 \mathrm{~V}=0.75 \mathrm{~V}$

Topic: Physics, Current Electricity, MHT CET, MHT CET 2022 (08 Aug Shift 1)
Power applied to a particle varies with time as P=(3t2-2t+1) watt, where t is in second. What is the change in its kinetic energy (in joule) between t=2 s and t=4 s?

Options:

The correct answer is:
46

Solution:
P=3t2-2t+1=dEdt
â´ dE=(3t2-2t+1)dt
E=â«t=2st=4s(3t2-2t+1)dt
=3t33-2t22+tt=2st=4s
=[43-23-42-22+(4-2)]
E=56-12+2=46 J

Topic: Physics, Work Power Energy, JEE Main
Power dissipated across the $8 \Omega$ resistor in the circuit shown here is 2 watt. The power dissipated in watt units across the $3 \Omega$ resistor is:

Options:
A 3.0
B 2.0
C 1.0
D 0.5

The correct answer is:
3.0

Solution:
$1 \Omega$ and $3 \Omega$ are in series so,
$egin{aligned}
& R_s=R_1+R_2 \\
& R_s=3+1=4 \Omega
\end{aligned}$
and $R_2=8 \Omega$
net $i=$ current in the circuit
Current through $R_1$ is $i_1=\frac{i \times R_2}{R_1+R_2}$
$=\frac{i \times 8}{12}=\frac{21}{3}$
Current $R_2$ is $i_2=\frac{1 \times R_1}{R_1+R_2}=\frac{i \times 4}{12}$
$=\frac{i \times 8}{12}$
Power dissipated in $3 \Omega$ resistance is: $P_1=l_1^2 \times 3$
Power dissipated in $8 \Omega$ resistance is: $P_2=l_2^2 \times 8$
For (i) and (ii)
$\begin{aligned}
& \frac{P_1}{P_2}=\frac{i_1^2 \times 3}{i_2^2 \times 8} \\
& \Rightarrow \frac{P_1}{P_2}
\end{aligned}$
$\begin{aligned}
& =\frac{\left(\frac{2 i}{3}\right)^2 \times 3}{\left(\frac{i}{3}\right)^2 \times 8}=\frac{3}{2} \\
\therefore \quad P_1 & =\frac{3}{2} \times P_2=\frac{3}{2} \times 2 =3 \mathrm{~W}
\end{aligned}$
$\begin{aligned}
\because & P & =100 \mathrm{~W}, V=125 V \\
\therefore & P & =V I \\
\Rightarrow & I & =\frac{P}{V}=\frac{100}{125} A
\end{aligned}$

Topic: Physics, Current Electricity, JEE Main
Power dissipated in an LCR series circuit connected to an a.c. source of emf $\varepsilon$ is :

Options:
A $\varepsilon^2 R / \sqrt{R^2+\left(L \omega-\frac{1}{C \omega}\right)^2}$
B $\varepsilon^2 \mathrm{R} /\left[\mathrm{R}^2+\left(\mathrm{L} \omega-\frac{1}{\mathrm{C} \omega}\right)^2\right]$
C $\varepsilon^2 \sqrt{\left[\mathrm{R}^2+\left(\mathrm{L} \omega-\frac{1}{\mathrm{C} \omega}\right)^2\right]} / \mathrm{R}$
D $\frac{\varepsilon^2\left[\mathrm{R}^2+\left(\mathrm{L} \omega-\frac{1}{\mathrm{C} \omega}\right)^2\right]}{\mathrm{R}}$

The correct answer is:
$\varepsilon^2 \mathrm{R} /\left[\mathrm{R}^2+\left(\mathrm{L} \omega-\frac{1}{\mathrm{C} \omega}\right)^2\right]$

Solution:
$$
\begin{aligned}
\mathrm{P}_{\mathrm{av}} & =\mathrm{E}_{\mathrm{rms}} \cdot \mathrm{I}_{\mathrm{rms}} \cos \phi \\
& =\varepsilon \cdot \frac{\varepsilon}{\mathrm{z}} \cdot \frac{\mathrm{R}}{\mathrm{Z}}=\frac{\varepsilon^2 \mathrm{R}}{\mathrm{z}^2} \\
& =\frac{\varepsilon^2 \mathrm{R}}{\mathrm{R}^2+\left(\omega \mathrm{L}-\frac{1}{\omega \mathrm{C}}\right)^2}
\end{aligned}
$$

Topic: Physics, Alternating Current, NEET, NEET 2009 (Mains)
Power dissipated in an LCR series circuit connected to an a.c. source of emf $\varepsilon$ is :

Options:
A $\varepsilon^2 R / \sqrt{R^2+\left(L \omega-\frac{1}{C \omega}\right)^2}$
B $\varepsilon^2 \mathrm{R} /\left[\mathrm{R}^2+\left(\mathrm{L} \omega-\frac{1}{\mathrm{C} \omega}\right)^2\right]$
C $\varepsilon^2 \sqrt{\left[\mathrm{R}^2+\left(\mathrm{L} \omega-\frac{1}{\mathrm{C} \omega}\right)^2\right]} / \mathrm{R}$
D $\frac{\varepsilon^2\left[\mathrm{R}^2+\left(\mathrm{L} \omega-\frac{1}{\mathrm{C} \omega}\right)^2\right]}{\mathrm{R}}$

The correct answer is:
$\varepsilon^2 \mathrm{R} /\left[\mathrm{R}^2+\left(\mathrm{L} \omega-\frac{1}{\mathrm{C} \omega}\right)^2\right]$

Solution:
$$
\begin{aligned}
\mathrm{P}_{\mathrm{av}} & =\mathrm{E}_{\mathrm{rms}} \cdot \mathrm{I}_{\mathrm{rms}} \cos \phi \\
& =\varepsilon \cdot \frac{\varepsilon}{\mathrm{z}} \cdot \frac{\mathrm{R}}{\mathrm{Z}}=\frac{\varepsilon^2 \mathrm{R}}{\mathrm{z}^2} \\
& =\frac{\varepsilon^2 \mathrm{R}}{\mathrm{R}^2+\left(\omega \mathrm{L}-\frac{1}{\omega \mathrm{C}}\right)^2}
\end{aligned}
$$

Topic: Physics, Alternating Current, JEE Main
Power gain for common base amplifier is 400 and voltage gain is 420. Find the collector current (in mA) when base current is 100 Î¼A.

Options:

The correct answer is:
02.00

Solution:
Ap=αAv
⇒400=α×420
⇒α=4042=2021
⇒β=20
And, lc=β×lb=2 mA

Topic: Physics, Semiconductors, JEE Main
Power of the only force acting on a particle of mass m=1 kg moving in a straight line depends on its velocity as P=v2 where v is in m s-1 and P is in watt. If the initial velocity of the particle is 1 m s-1, then the displacement of the particle in ln(2) seconds will be

Options:
A ln2-1 m
B ln22 m
C 1 m
D 2 m

The correct answer is:
1 m

Solution:
P=Fv
v2=Fv ( since P=v2 )
therefore F=v
ma=v
1Ãvdvdx=v
â«1vdv=â«0xdx
v-1=x
v=x+1
dxdt=x+1
â«0xdxx+1= â«0tdt
lnx+1-ln0+1=t
x+1=et
x=et-1
x=et-1
At t=ln2
x=2-1=1 m

Topic: Physics, Work Power Energy, JEE Main
Power supplied to a particle of mass 2 kg varies with time as Pâ¡=3t22 W, where t is in seconds. If the velocity of the particle at t=0 is v=0, then what is the velocity (in m s-1) of the particle at t=2 s?

Options:

The correct answer is:
2

Solution:
Kf−Ki=∫P⋅dt12mv2−0=∫02(32t2)dt12(2)v2=32[t33]02=4v=2 ms−1

Topic: Physics, Work Power Energy, JEE Main
$P Q$ is a double ordinate of the parabola $y^2=4 a x$. The locus of the points of trisection of $P Q$ is

Options:
A $9 y^2=4 a x$
B $9 x^2=4 a y$
C $9 y^2+4 a x=0$
D $9 x^2+4 a y=0$

The correct answer is:
$9 y^2=4 a x$

Solution:
Required locus is $(3 y)^2=4 a x$
$\Rightarrow 9 y^2=4 a x$

Topic: Mathematics, Parabola, JEE Main
Predict $A$ and $B$ in the following reaction sequence :

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
Predict the age of a bottle of wine if the measured tritium $\left({ }^3 \mathrm{H}\right)$ content for the old bottle is $25 \%$ of that of the new wine? The half life of ${ }^3 \mathrm{H}$ is 12.5 years.

Options:
A 50 years
B 25 years
C 100 years
D 37.5 years

The correct answer is:
25 years

Solution:
To get reduced to $25 \%$ of its original amount, tritium would undergo two half-lives.
${ }^3 H \xrightarrow[\text { Half-life }]{\text { First }} 50$ percent ${ }^3 H$
50 percent ${ }^3 H \xrightarrow[\text { Half-life }]{\text { Second }} 25$ percent ${ }^3 H$
Hence, first half-life + second half-life $=12.5+12.5=25$ years

Topic: Physics, Nuclear Physics, JEE Main
Predict the direction of induced current in the situations described by the following Figure (a) to (f).

Options:

None

Solution:
(a) Apply Lenz's law, induced current will flow to oppose the magnet. So South-pole is developed at end $\mathrm{q}$ and current flows in qrpq direction.
(b) Same as above, South-pole developed at $\mathrm{q}$ and current is in prqp direction. For coil 2, North- is going away so attractive i.e., South pole is developed at end $x$ and direction is yzxy.
(c) As key is closed, magnetic flux rises in the first coil, in other coil, induced current would be such as to oppose this increasing magnetic flux. This happenes when magnetic field produced in this coil is from right to left hence current is in yzx direction.
(d) As rheostat is changed to decrease resistance, current will increase and magnetic flux linked with it will rise. Applying Lenz's law, the induced current would oppose this increase, hence direction of magnetic field should be right to left and current is in direction $z y x$.
(e) When current is flowing before release of tapping key magnetic flux is from left to right (right end is $\mathrm{N}$ pole). As key is released, current decreases magnetic flux decreases, hence induced flux would oppose this decrease and favour an increase in flux. Hence direction of induced current in adjoining coil is from left to right in direction xry.
(f) As current decreases in straight conductor, magnetic field in the plane of the coil remains to be in the same plane. The circular coil does not see the change in magnetic field (same plane) hence no current (or flux) is induced.

Topic: Physics, Electromagnetic Induction
Predict the major product P in the following reaction.

Options:
A
B
C
D

The correct answer is:

Solution:
This reaction occurs as follows

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Predict the order of Îâ for the following compounds

I MnH2O62+

II MnCN2H2O4

III MnCN4H2O22-

Options:
A Δ∘ I<Δ∘ II<Δ∘III
B Δ∘II<Δ∘I<Δ∘III
C Δ∘III<Δ∘II<Δ∘I
D Δ∘I<Δ∘III<Δ∘II

The correct answer is:
Δ∘ I<Δ∘ II<Δ∘III

Solution:
CN- is a strong field ligand so greater the number of CN- , more will be the strength of ligands and more will be the value of Δo (crystal field splitting energy).

Topic: Chemistry, Coordination Compounds, JEE Main
$$
\text { Predict the product : }
$$

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
Predict the product,

Options:
A

B

C

D

The correct answer is:

Solution:
Key Idea With nitric acid $\left(\mathrm{NaNO}_2+\mathrm{HCl}\right)$ primary aromatic amines give azo compounds and secondary aromatic amines give nitroso compound

Topic: Chemistry, Amines, JEE Main
Predict the product ' $B$ ' in the sequence of reaction
$\mathrm{HC} \equiv \mathrm{CH} \xrightarrow[\mathrm{HgSO}_4]{30 \% \mathrm{H}_2 \mathrm{SO}_4} A \xrightarrow{\mathrm{NaOH}} B$

Options:
A $\mathrm{CH}_3 \mathrm{COONa}$
B $\mathrm{CH}_3 \mathrm{COOH}$
C $\mathrm{CH}_3 \mathrm{CHO}$
D

The correct answer is:

Solution:

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Predict the product of the following reaction :

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Amines, JEE Main
$$
\text { Predict the product : }
$$

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, NEET, NEET 2009 (Mains)
Predict the products of electrolysis in each of the following.
(i) An aqueous solution of \(\mathrm{AgNO}_3\) with silver electrodes.
(ii) An aqueous solution of \(\mathrm{AgNO}_3\) with platinum electrodes.
(iii) A dilute solution of \(\mathrm{H}_2 \mathrm{SO}_4\) with platinum electrodes.
(iv) An aqueous solution of \(\mathrm{CuCl}_2\) with platinum electrodes.

Options:

None

Solution:
(i) \(\mathrm{AgNO}_3(s) \longrightarrow \mathrm{Ag}^{+}(a q)+\mathrm{NO}_3^{-}(a q)\)
At cathode: \(\mathrm{Ag}^{+}\)ions have lower discharge potential than \(\mathrm{H}^{+}\)ions. Hence, \(\mathrm{Ag}^{+}\)ions will be deposited as Ag in preference to \(\mathrm{H}^{+}\)ions.
\(\mathrm{Ag}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s)\)
At anode: \(\quad\) As Ag anode is attacked by \(\mathrm{NO}_3{ }^{-}\)ions, \(\mathrm{Ag}\) of the anode will dissolve to form \(\mathrm{Ag}^{+}\)ions in the solution.
\(\mathrm{Ag}(s) \longrightarrow \mathrm{Ag}^{+}(a q)+e^{-}\)
(ii) At cathode: \(\mathrm{Ag}^{+}\)ions have lower discharge potential than \(\mathrm{H}^{+}\)ions. Hence, \(\mathrm{Ag}^{+}\)ions will be deposited as Ag in preference to \(\mathrm{H}^{+}\)ions.
At anode: Since Pt electrode are inert, the anode is not attacked by \(\mathrm{NO}_3^{-}\)ions. Out of \(\mathrm{OH}^{-}\)and \(\mathrm{NO}_3^{-}\) ions, \(\mathrm{OH}^{-}\)ions have lower discharge potential. Hence, \(\mathrm{OH}^{-}\)ions will be discharged in preference to \(\mathrm{NO}_3^{-}\)ions, which then decompose to give out \(\mathrm{O}_2\).
\(\begin{aligned}
&\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{OH}+\mathrm{e}^{-} \\
&4 \mathrm{OH} \longrightarrow 2 \mathrm{H}_2 \mathrm{O}(\ell)+\mathrm{O}_2(g)
\end{aligned}\)
\(\begin{aligned}
\text { (iii) } \mathrm{H}_2 \mathrm{SO}_4(a q) \longrightarrow 2 \mathrm{H}^{+}(a q)+\mathrm{SO}_4^{2-(a q)} \\
& \mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}^{+}+\mathrm{OH}^{-} \\
\text {At cathode: } & \mathrm{H}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{H} \\
& \mathrm{H}+\mathrm{H} \longrightarrow \mathrm{H}_2(g) \\
\text { At Anode: } & \mathrm{OH}^{-} \longrightarrow \mathrm{OH}^{-} \\
& 4 \mathrm{OH} \longrightarrow 2 \mathrm{e}_2^{-} \\
& 2 \mathrm{O}_2+\mathrm{O}_2(g)
\end{aligned}\)
Thus, \(\mathrm{H}_2\) gas is liberated at the cathode and \(\mathrm{O}_2\) gas at the anode.
(iv) \(\mathrm{CuCl}_2(s) \longrightarrow \mathrm{Cu}^{2+}(a q)+2 \mathrm{Cl}^{-}(a q)\)
At cathode: \(\mathrm{Cu}^{2+}\) ions will be reduced in preference to \(\mathrm{H}^{+}\)ions and copper will be deposited at cathode.
\(\mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}\)
At anode: \(\mathrm{Cl}^{-}\)ions will be discharged in preference to \(\mathrm{OH}^{-}\)ions which remains in solution.
Thus, \(\mathrm{Cu}\) will be deposited on the cathode and \(\mathrm{Cl}_2\) gas will be liberated at the anode.

Topic: Chemistry, Electrochemistry
Preeti reached the metro station and found that the escalator was not working. She walked up the stationary escalator in time t1. On other days, if she remains stationary on the moving escalator, then the escalator takes her up in time t2. The time taken by her to walk up on the moving escalator will be

Options:
A t1+t22
B t1t2t2-t1
C t1t2t2+t1
D t1-t2

The correct answer is:
t1t2t2+t1

Solution:
Let d be the displacement of the escalator.

Velocity of girl w.r.t. elevator =dt1=vge
Velocity of elevator w.r.t ground veG=dt2

Then velocity of girl w.r.t ground
v→gG= v→ge+v→eG
i.e., VgG=Vge+VeG
dt=dt1+dt2
1t=1t1+1t2
t=t1 t2t1+t2

Topic: Physics, Motion In One Dimension, NEET, NEET 2017
Preparation of Bakelite proceeds via reactions:

Options:
A Electrophilic addition and dehydration
B Condensation and elimination
C Electrophilic substitution and dehydration
D Nucleophilic addition and dehydration

The correct answer is:
Electrophilic substitution and dehydration

Solution:
Formation of Bakelite follows electrophilic substitution reaction of phenol with formaldehyde followed by condensation.

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2020 (08 Jan Shift 2)
Preparation of Bakelite proceeds via reactions

Options:
A Electrophilic substitution and dehydration
B Nucleophilic addition and dehydration
C Electrophilic addition and dehydration
D Condensation and elimination

The correct answer is:
Electrophilic substitution and dehydration

Solution:
Formation of Bakelite follows electrophillic substitution of phenol and formaldehyde followed by dehydration.

Topic: Chemistry, Polymers, JEE Main
Presence of a nitro group in a benzene ring

Options:
A
activates the ring towards electrophilic substitution

B
renders the ring basic

C
deactivates the ring towards nucleophilic substitution

D
deactivates the ring towards electrophilic substitution

The correct answer is:
deactivates the ring towards electrophilic substitution

Solution:
$-\mathrm{NO}_2$ group shows $-\mathrm{M}$ effect, so withdraws the electron density from the ring and hence deactivate the ring towards electrophilic aromatic substitution.
Hence, (D) is correct.

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2007
Presence of which reagent will affect the reversibility of the following reaction, and change it to a irreversible reaction :

CH4+I2⇌ReversiblehvCH3-I + HI

Options:
A HOCl
B dilute HNO2
C Liquid NH3
D Concentrated HIO3

The correct answer is:
Concentrated HIO3

Solution:
Iodination of alkane is reversible reaction.
It can be irreversible in the presence of strong oxidising agent like conc. HNO3 or conc. HIO3

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2021 (27 Jul Shift 1)
Pressure exerted by 1 mole of methane in a 0.25 litre container at $300 \mathrm{~K}$ using vander Waal's equation (given $\left.1=2.253 \mathrm{atml}^2 \mathrm{~mol}^{-2}, b=0.0428 / \mathrm{litmol}^{-1}\right)$ is

Options:
A $82.82 \mathrm{~atm}$
B $152.51 \mathrm{~atm}$
C $190.52 \mathrm{~atm}$
D $70.52 \mathrm{~atm}$

The correct answer is:
$82.82 \mathrm{~atm}$

Solution:
$\left(P+\frac{n^2 a}{V^2}\right)(V-n b)=n R T$
$\left(P+\frac{2.253}{0.25 \times 0.25}\right)(0.25-0.0428)=0.0821 \times 300$
or $(P+36.048)(0.2072)=24.63$
$\Rightarrow P+36.048=118.87 \Rightarrow P=82.82 \mathrm{~atm}$.

Topic: Chemistry, States of Matter, JEE Main
Pressure inside a soap bubble is greater than the pressure outside by an amount :
(given : $\mathrm{R}=$ Radius of bubble $\mathrm{S}=$ Surface tension of bubble)

Options:
A $\frac{2 S}{R}$
B $\frac{4 R}{S}$
C $\frac{S}{R}$
D $\frac{4 \mathrm{~S}}{\mathrm{R}}$

The correct answer is:
$\frac{4 \mathrm{~S}}{\mathrm{R}}$

Solution:
There are two liquid-air surfaces in bubble so
$\Delta \mathrm{P}=2\left(\frac{2 \mathrm{~S}}{\mathrm{R}}\right)=\frac{4 \mathrm{~S}}{\mathrm{R}}$

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2024 (06 Apr Shift 2)
Pressure inside two soap bubbles are 1.01 and 1.02 atmosphere, respectively. The ratio of their volumes is :

Options:
A 4:1
B 0.8:1
C 8:1
D 2:1

The correct answer is:
8:1

Solution:
Excess pressure inside a soap bubble is 4TR. Therefore,

P1-P0=4TR1=0.01 atm

P2-P0=4TR2=0.02 atm

Dividing,

12=R2R1

⇒R1=2R2

⇒V1V2=R13R23=8R23R23=81

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2020 (03 Sep Shift 1)
Pressure of an ideal gas is increased by keeping temperature constant. The kinetic energy of
molecules

Options:
A Decreases
B Increases
C Remains same
D Increses or decreases depending on the nature of gas

The correct answer is:
Remains same

Solution:
Kinetic energy of molecules depends only on the temperature, that is,
\( \mathrm{K} \propto \mathrm{T} \)
Therefore, if the pressure of an ideal gas is increased by keeping temperature constant, then kinetic energy of molecules
remains same.

Topic: Physics, Kinetic Theory of Gases, KCET, KCET 2018
Pressure of an ideal gas is increased by keeping temperature constant. What is effect on kinetic energy of molecules?

Options:
A Increase
B Decrease
C No change
D Cannot be determined

The correct answer is:
No change

Solution:
Kinetic energy of an ideal gas depends only on its temperature. Hence, it remains constant whether its pressure is increased or decreased.

Topic: Physics, Kinetic Theory of Gases, JIPMER, JIPMER 2011
Pressure of an ideal gas is increased by keeping temperature constant. What is the effect on kinetic energy of molecules?

Options:
A Increases
B Decrease
C No change
D Can't be determined

The correct answer is:
No change

Solution:
Kinetic energy of ideal gas depends only on its temperature. Hence, it remains constant whether its pressure is increased or decreased.

Topic: Physics, Kinetic Theory of Gases, JEE Main
Pressure of an ideal gas is increased by keeping temperature constant. What is the effect on kinetic energy of molecules?

Options:
A Increase
B Decrease
C No change
D Can't be determined

The correct answer is:
No change

Solution:
Kinetic energy of ideal gas depends only on its temperature. Hence, it remains constant whether its pressure is increased on decreased.

Topic: Physics, Kinetic Theory of Gases, JIPMER, JIPMER 2008
Pressure of an ideal gas is increased by keeping temperature constant. What is the effect on kinetic energy of molecules?

Options:
A Increases
B Decrease
C No change
D Can't be determined

The correct answer is:
No change

Solution:
Kinetic energy of ideal gas depends only on its temperature. Hence, it remains constant whether its pressure is increased or decreased.

Topic: Physics, Kinetic Theory of Gases, NEET
Pressure of ideal gas at constant volume is proportional to

Options:
A force between the molecules
B average potential energy of the molecules
C total energy of the gas
D average kinetic energy of the molecules

The correct answer is:
average kinetic energy of the molecules

Solution:
Gay Lussac's law states that the pressure of a
fixed mass of a gas at constant volume is directly proportional to the temperature.
i.e. $p \propto T$
or
$p \propto \frac{1}{2} K_B T$
Since kinetic energy of the molecules is directly proportional to the temperature.

Topic: Physics, Kinetic Theory of Gases, JEE Main
Pressure $(\mathrm{P})$ versus density (d) graph of an ideal gas is shown in the figure. Choose the correct statement.

Options:
A During the process $\mathrm{AB}$, workdone by the gas is positive
B During the process $\mathrm{AB}$, workdone by the gas is negative
C During the process $B C$, internal energy of the gas increases
D During the process DA, internal energy of the gas remains constant

The correct answer is:
During the process DA, internal energy of the gas remains constant

Solution:
No solution. Refer to answer key.

Topic: Physics, Thermodynamics, AP EAMCET, AP EAMCET 2017 (25 Apr Shift 1)
Pressure P, volume V and temperature T of a certain material are related by P= α T 2 V , where α is constant. Work done by the material when temperature changes from T0 to 2T0 and pressure remains constant is

Options:
A 3αT02
B 5αT02
C 32αT02
D 7αT02

The correct answer is:
3αT02

Solution:
The given relation is p=αT2V ∴ V=αT2P
As pressure is kept constant, dV=2αTPdT
W=∫p dV=∫T02T0p2αTpdT
=2αT22T02T0=3αT02

Topic: Physics, Thermodynamics, NEET
Pressure remaining the same, the volume of a given mass of an ideal gas increases for every degree centigrade rise in temperature by definite fraction of its volume at

Options:
A $0^{\circ} \mathrm{C}$
B Its critical temperature
C Absolute zero
D Its Boyle temperature

The correct answer is:
$0^{\circ} \mathrm{C}$

Solution:
$V_t=V_o\left(1+\alpha_v t\right)$
$\because\left(V_2-V_1\right)=\Delta V=V_o \alpha\left(t_2-t_1\right)$
if $t_2-t_1=1^\rho$ then $\Delta V=\alpha V_o$
For every $1^{\circ} \mathrm{C}$ increase in temperature, the volume of a given mass of an ideal gas increases by a definite fraction
$\frac{1}{273.15}$ of $V_o$. Here $V_o$ is volume at $0^{\circ} \mathrm{C}$ temperature.

Topic: Chemistry, States of Matter, JEE Main
Primary amines can be distinguished from secondary and tertiary amines by reacting with

Options:
A
Chloroform and alcoholic $\mathrm{KOH}$
B
Methyl iodide
C
Chloroform alone
D
Zinc Dust

The correct answer is:
Chloroform and alcoholic $\mathrm{KOH}$

Solution:
$1^{\circ}$ amines (aliphatic and aromatic) react with $\mathrm{CHCl}_3 / \mathrm{KOH}$ to yield isocyanide (foul smelling) This is known as carbylamine test which is not given by $2^{\circ}$ and $3^{\circ}$ amines.

Topic: Chemistry, Amines, JEE Main