datavorous/entrance-exam-dataset huggingface dataset:
At STP, $0.50 \mathrm{~mol} \mathrm{H}_2$ gas and $1.0 \mathrm{~mol} \mathrm{He}$ gas

Options:
A Have equal average kinetic energies
B Have equal molecular speeds
C Occupy equal volumes
D Have equal effusion rates

The correct answer is:
Have equal average kinetic energies

Solution:
(a) For same temperature kinetic energies of $\mathrm{H}_2 \& \mathrm{He}$ molecules will be same because kinetic energy depends only on temperature.

Topic: Chemistry, States of Matter, JEE Main
At STP, 5.6 litre of a gas weighs 60 g. The vapour density of gas is -

Options:

The correct answer is:
120

Solution:
5.6 lit gas weight = 60 gm
22.4 lit gas weight = 605.6Ã22.4 gm
= 240 gm ( molecular weight of gas )
Vapour density= Molecular weight 2 = 120

Topic: Chemistry, States of Matter, JEE Main
At STP, a container has 1 mole of Ar, 2 moles of CO2 , 3 moles of O2 and 4 moles of N2 without changing the total pressure if one mole of O2 is removed, the partial pressure of O2

Options:
A is changed by about 26%
B is halved
C is unchanged
D changes by 30%

The correct answer is:
is changed by about 26%

Solution:
In the first case,
nAr=1,nCO2=2,nO2=3,nNA=4
So total number of moles =1+2+3+4=10
If P in the total pressure, PO2=3P10
In the second case,
nO2=2,ntotal=9
As P remains the same, PO2=2P9
So decrease in partial pressure of O2
=3P10-2P9=7P90
So % decrease in partial pressure of O2
=7P903P10×100=70027=26%

Topic: Chemistry, States of Matter, JEE Main
At t=0, a transverse wave pulse in a wire is described by the function where x and y are in metre. The function y(x, t) that describes this wave equation, if it is travelling in the positive x-direction with a speed of 4.5 m s-1 is

Options:
A y = 6 x + 4 . 5 t 3 - 3
B y = 6 x - 4 . 5 t 2 + 3
C y = 6 x + 4 . 5 t 2 - 3
D y = 6 x - 4 . 5 t 2 + 3

The correct answer is:
y = 6 x - 4 . 5 t 2 + 3

Solution:
y x , t = a x Â± vâ¡ t 2 + b is another form of progressive wave equation propagating with a speed v. Negative sign to be taken for propagation along + x-axis and positive sign to be taken for propagation along - x-axis.

Topic: Physics, Waves and Sound, JEE Main
At t=0 the positions and velocities of two particles are as shown in the figure. They are kept on a smooth surface and being mutually attracted by gravitational force. Then the position of the centre of mass at t=2 s is

Options:
A X=5 m
B X=7 m
C X=3 m
D X=2 m

The correct answer is:
X=7 m

Solution:
Initial position of center of mass,
xi=10+1101+1=5 m
Initial velocity of center of mass,
v=15+1-31+1=1 m
This velocity remains constant as the internal forces during collision do not change the velocity of the center of mass. Therefore distance travelled by the center of mass in 2s ,
Δx=vΔt
Δx=12=2 m
Final position of center of mass,
xf=xi+Δx
xf=5+2=7 m

Topic: Physics, Center of Mass Momentum and Collision, NEET
At $T(\mathrm{~K})$, the following data were obtained for a general reaction, $A+B+C \longrightarrow$ products
\begin{array}{ccccc}
\hline Expt. & \begin{array}{c}
Initial \\
{[A]}
\end{array} & \begin{array}{c}
Initial \\
{[B]}
\end{array} & \begin{array}{c}
Initial \\
{[C]}
\end{array} & Initial rate \\
\hline 1. & 0.02 \mathrm{M} & 0.1 \mathrm{M} & 0.03 \mathrm{M} & 2.4 \times 10^{-6} \mathrm{~ms}^{-1} \\
\hline 2. & 0.02 \mathrm{M} & 0.2 \mathrm{M} & 0.03 \mathrm{M} & 4.8 \times 10^{-6} \mathrm{~ms}^{-1} \\
\hline 3. & 0.02 \mathrm{M} & 0.2 \mathrm{M} & 0.06 \mathrm{M} & 9.6 \times 10^{-6} \mathrm{~ms}^{-1} \\
\hline 4. & 0.04 \mathrm{M} & 0.2 \mathrm{M} & 0.06 \mathrm{M} & 9.6 \times 10^{-6} \mathrm{~ms}^{-1} \\
\hline
\end{array}
The rate constant for the above reaction is

Options:
A $8.0 \times 10^{-4} \mathrm{~s}^{-1}$
B $8.0 \times 10^{-4} \mathrm{~L} \mathrm{~mol}^{-1} \mathrm{~s}^{-1}$
C $8.0 \times 10^4 \mathrm{~L} \mathrm{~mol}^{-1} \mathrm{~s}^{-1}$
D $8.0 \times 10^{-4} \mathrm{~L}^2 \mathrm{~mol}^{-2} \mathrm{~s}^{-1}$

The correct answer is:
$8.0 \times 10^{-4} \mathrm{~L} \mathrm{~mol}^{-1} \mathrm{~s}^{-1}$

Solution:
Comparing expt. (1) and expt. (2).
When $[B]$ is doubled rate of reaction doubles.
$\therefore$ Reaction is first order is $B$.
Comparing expt. (2) and expt (3).
When $[C]$ is doubled rate of reaction doubles.
$\therefore$ Reaction is first order in $C$.
Comparing expt. [3] and expt [4]. When $[A]$ is doubled rate of reaction remains unchanged.
$\therefore$ Reaction is zero order in $A$.
Rate law $=K[A]^0[B]^{\prime}[C]^{\prime}$
Using data of expt. (1) we get
$$
\begin{aligned}
& 2.4 \times 10^{-6}=K[0.02]^0[0.1]^1[0.03]^1 \\
& 2.4 \times 10^{-6}=K \times 1 \times 0.1 \times 0.03 \\
& K=\frac{2.4 \times 10^{-6}}{3 \times 10^{-3}}=0.8 \times 10^{-3}=8.0 \times 10^{-4} \mathrm{~L} \mathrm{~mol}^{-1} \mathrm{~s}^{-1}
\end{aligned}
$$

Topic: Chemistry, Chemical Kinetics, AP EAMCET, AP EAMCET 2018 (23 Apr Shift 2)
At \( t=0 \), the function \( f(t)=\frac{\sin t}{t} \) has

Options:
A A minimum
B A discontinuity
C A point of inflexion
D A maximum

The correct answer is:
A maximum

Solution:
Given, ft=sinâ¡tt
At t = 0, first we will check continuity of the function.
Now, LHL=f0-h
=limhâ0â¡sinâ¡(0-h)0-h
=limhâ0â¡-sinâ¡h-h
=1
RHL=f0+h
=limhâ0â¡sinâ¡(0+h)0+h
=limhâ0â¡sinâ¡hh=1
And f0=1
Since, LHL=RHL=f0
So, the function is continuous at t = 0
Now, we check the function is maximum or minimum
f′t=1tcosâ¡t-1t2sinâ¡t
and f′′t=-1tsinâ¡t-1t2cosâ¡t-1t2cosâ¡t+2t3sinâ¡t
=-sinâ¡tt-2cosâ¡tt2+2sinâ¡tt2
For maximum or minimum value of f(x) , put
f′x=0
âÂ Â cosâ¡tt-sinâ¡tt2=0
âÂ tanâ¡tt=1
Now, limtâ0â¡f"(t)Â
=-limtâ0â¡sinâ¡tt-2Â limtâ0â¡tcosâ¡t-sinâ¡tt3Â Â Â Â Â 00Â from
=-1-2Â limtâ0â¡cosâ¡t-sinâ¡t-cosâ¡t3t2Â [using L' Hospital rule]
=-1+23Â limtâ0â¡sinâ¡tt
=-1+23Ã1=-13<0
So, function f(t) is maximum at t = 0

Topic: Mathematics, Limits, JEE Main
At temperature T, the average kinetic energy of any particle is 32kT. The de Broglie wavelength follows the order:

Options:
A Visible photon>thermal electron>thermal neutron.
B Thermal neutron>visible photon>thermal electron.
C Thermal neutron>thermal electron>visible photon.
D Visible photon>thermal neutron>thermal electron.

The correct answer is:
Visible photon>thermal electron>thermal neutron.

Solution:
λn=h2mE
(E=kT for thermal neutron)
λn=h2mnkT=30.835T=30.83517.32=1.78 Å
λe=150V
Kinetic energy for electron in hydrogen atom =13.60 eV V=13.6 V
λe=15013.6=3.32 Å
λphotonvisible range=6000 Å
λphoton>λelectron>λneutron

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2015 (11 Apr Online)
At the centre of a current-carrying circular coil of radius 5 cm, magnetic field due to earth is 0.5Ã10-5 Wb m-2. The current flowing through the coil, so that it equals the Earth's magnetic field, is

Options:
A 40 A
B 4 A
C 0.4 A
D 0.2 A

The correct answer is:
0.4 A

Solution:
Magnetic field (B) at the centre of a coil of radius (r) carrying current (i) is
B=Î¼0i2r
â i=2rBÎ¼0
Given, B=0.5Ã10-5 T, r=0.05 m
i=2Ã0.5Ã10-5Ã0.054ÏÃ10-7=0.398 A
âiâ0.4 A

Topic: Physics, Magnetic Effects of Current, JEE Main
At the centre of a fixed large circular coil of radius $\mathrm{R}$, a much smaller circular coil of radius $r$ is placed. The two coils are concentric and are in the same plane. The larger coil carries a current I. The smaller coil is set to rotate with a constant angular velocity $\omega$ about an axis along their common diameter. Calculate the emf induced in the smaller coil after a time $t$ of its start of rotation.

Options:
A
$\frac{\mu_0 \mathrm{I}}{2 \mathrm{R}} \omega \mathrm{r}^2 \sin \omega \mathrm{t}$

B
$\frac{\mu_0 I}{4 R} \omega \pi r^2 \sin \omega t$

C
$\frac{\mu_0 I}{2 R} \omega \pi r^2 \sin \omega t$

D
$\frac{\mu_0 \mathrm{I}}{4 \mathrm{R}} \omega \mathrm{r}^2 \sin \omega \mathrm{t}$

The correct answer is:
$\frac{\mu_0 I}{2 R} \omega \pi r^2 \sin \omega t$

Solution:
According to Faraday's law of electromagnetic induction,
$$
\begin{array}{r}
e=-\frac{d \phi}{d t} \text { and } \phi=B A \cos \omega t=B \pi r^2 \cos \omega t \\
\Rightarrow \quad e=-\frac{d}{d t}\left(\pi r^2 B \cos \omega t\right)=\pi r^2 B \sin \omega t(\omega) \\
\therefore e=\frac{\mu_0 I}{2 R} \pi \omega r^2 \sin \omega t\left(\because B=\frac{\mu_0 I}{2 R}\right)
\end{array}
$$

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
At the centre of a half ring of radius $\mathrm{R}=10 \mathrm{~cm}$ and linear charge density $4 \mathrm{n} \mathrm{C} \mathrm{m}^{-1}$, the potential is $x \pi \mathrm{V}$. The value of $x$ is _______

Options:

The correct answer is:
36

Solution:
Potential at centre of half ring
$\begin{aligned} & V=\frac{K Q}{R} \\ & V=\frac{K \lambda \pi R}{R} \\ & V=K \lambda \pi \Rightarrow V=9 \times 10^9 \times 4 \times 10^{-9} \pi \\ & V=36 \pi\end{aligned}$

Topic: Physics, Electrostatics, JEE Main, JEE Main 2024 (09 Apr Shift 1)
At the critical micelle concentration, the surfactant molecules

Options:
A Decompose
B Dissociate
C Associate
D Become completely soluble

The correct answer is:
Dissociate

Solution:
The micelles are formed by the association of dispersed particles above a certain concentration which is known as critical micelle concentration. Above CMC, the surfactant molecules associate to form a larger aggregate.

Topic: Chemistry, Surface Chemistry, JEE Main
At the end of the G2 phase, the number of chromosomes and the amount of DNA in a diploid cell, respectively, is

Options:
A n and 2C
B 2n and C
C n and 4C
D 2n and 4C

The correct answer is:
2n and 4C

Solution:
If a cell at G1 phase has 2n chromosome and 2CââDNA, then during S phase, DNA replication occurs but chromosome number remains unchanged. As a result, chromosome number remains 2n but the quantity of DNA becomes 4C.

Topic: Biology, Cell Cycle and Cell Division, NEET
At the end of the G2 phase, the amount of DNA in a diploid cell having the amount of chromosome 2C will be

Options:
A C.
B 2C.
C 4C.
D C/2.

The correct answer is:
4C.

Solution:
If a cell at G1 phase has 2n chromaosome and 2C DNA than during S phase DNA replication occurs but chromosome number remains unchanged as a result chromosome number remains 2n but quantity of DNA becomes 4C.

Topic: Biology, Cell Cycle and Cell Division, NEET
At the first minimum adjacent to the central maximum of a single-slit diffraction pattern, the phase difference between the Huygen's wavelet from the edge of the slit and the wavelet from the midpoint of the slit is:

Options:
A Ï2 radian
B π radian
C Ï8 radian
D Ï4 radian

The correct answer is:
π radian

Solution:
For first minima
AP-BP=λ
AP-MP=λ2
So phase difference =2πλ×λ2=π

Topic: Physics, Wave Optics, JEE Main
At the foot of the mountain, the angle of elevation of its summit is 45o, after ascending 1000 m towards the mountain up a slope of 30o inclination, the angle of elevation is found to be 60o. The height (in km) of the mountain is (take 3=1.7)

Options:

The correct answer is:
1.35

Solution:
Let P be the summit of the mountain and Q be the foot.

Let A be the first position and B the second position of observation.

BN and BM are ⊥s from B to PQ and AQ respectively.
Then, AB=1000m=1km ,
∠MAB=30o,
∠MAP=45o,

∠NBP=60o
Now, ∠BAP=∠MAP-∠MAB=45o-30o=15o
∠APB=∠APN-∠BPN=45o-30o=15o
∴∆ABP is isosceles

∴AB=BP
But, AB=1 kilometre

∴BP=1 kilometre.
Now, PQ=PN+NQ=PN+BM
=BPsin⁡60o+ABsin⁡30o
=1⋅32+1⋅12=3+12=1.35 km

Topic: Mathematics, Heights and Distances, JEE Main
At the instant t=0, a force F=kt ( k is a constant) acts on a small body of mass m resting on a smooth horizontal surface. The time, when the body leaves the surface is

Options:
A mgksin⁡α
B k sin αmg
C mg sin⁡αk
D mgk sinâ¡Î±

The correct answer is:
mgk sinâ¡Î±

Solution:
N+F sinâ¡Î±=mg
The body leaves the surface when the normal reaction on it becomes zero.
N=0
âkt sinâ¡Î±=mg
t=mgksinâ¡Î±

Topic: Physics, Laws of Motion, JEE Main
At the magnetic north pole of the earth, the value of horizontal component of earth's magnetic field and angle of dip are, respectively

Options:
A zero, maximum
B maximum, minimum
C maximum, maximum
D minimum, minimum

The correct answer is:
zero, maximum

Solution:
At the magnetic north pole, the magnetic needle will point vertically. There is no component of earth's magnetic field in the horizontal direction and the angle of dip (the angle that the resultant magnetic field at the place makes with the horizontal) is $90^{\circ}$.
$$
\mathrm{H}=0, \delta=90^{\circ}(\text { maximum })
$$

Topic: Physics, Magnetic Properties of Matter, JEE Main
At the moment t=0, when the charge on the capacitor C1 is zero, the switch is closed. If I0 be the current through inductor at t=0, then for t>0 (initially C2 is uncharged)

Options:
A maximum current through inductor equals I02
B maximum current through inductor equals C1I0C1+C2
C maximum charge on C1=C1I0LC2C1+C2
D maximum charge on C1=C1I0LC1+C2

The correct answer is:
maximum charge on C1=C1I0LC1+C2

Solution:
The maximum current through the inductor is I0.
By conservation of energy,
12LI02=12C1+C2V2
V=LC1+C2I0
Q1=C1V=C1I0LC1+C2

Topic: Physics, Alternating Current, JEE Main
At the point of intersection of the two curves shown, the conc. of B is given by………for, A→nB

Options:
A nA02
B A0n-1
C nA0n+1
D n-1n+1A0

The correct answer is:
nA0n+1

Solution:
Aleft=Bformed=n×Adecayed
A0e-λt=n×A01-e-λt
So e-λt=nn+1
Hence Bformed=n×A0×1-e-λt
=n×A0×1-nn+1=nA0n+1

Topic: Chemistry, Chemical Kinetics, JEE Main
At the point of intersection of the two curves shown, the concentration of B is given by________, for the first reaction A â nB.

Options:
A n A 0 2
B A 0 n - 1
C n A 0 n + 1
D n - 1 n + 1 A 0

The correct answer is:
n A 0 n + 1

Solution:
According to question, at presentation point ⇒ A = B
⇒ A 0 −x=nx
⇒ A 0 = n+1 x
⇒x= A 0 n+1
B =nx
= n× A 0 n+1

Topic: Chemistry, Chemical Kinetics, JEE Main
At the point $x=1$, the given function $f(x)=\left\{\begin{array}{l}
x^3-1 ; 1 \lt x \lt \infty \\
x-1 ;-\infty \lt x \leq 1
\end{array}\right.$ is

Options:
A Continuous and differentiable
B Continuous and not differentiable
C Discontinuous and differentiable
D Discontinuous and not differentiable

The correct answer is:
Continuous and not differentiable

Solution:
We have $R f^{\prime}(1)=\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h}$
$=\lim _{h \rightarrow 0} \frac{\left\{(1+h)^3-1\right\}-0}{h}=3$
$L f^{\prime}(1)=\lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{-h}=\lim _{h \rightarrow 0} \frac{\{(1-h)-1\}-0}{-h}=1$
$\therefore R f^{\prime}(1) \neq L f^{\prime}(1) \Rightarrow f(x)$ is not differentiable at $x=1$.
Now,$f(1+0)=\lim _{h \rightarrow 0} f(1+h)=0$
and $f(1-0)=\lim _{h \rightarrow 0} f(1-h)=0$
$\therefore f(1+0)=f(1-0)=f(0) \Rightarrow f(x)$ is continuous at $x=1$. Hence at $x=1, f(x)$ is continuous and not differentiable.

Topic: Mathematics, Continuity and Differentiability, JEE Main
At the stage of generation of action potential, the axolemma is:

Options:
A Positive outside
B Negative outside
C Neither positive nor negative
D Initially positive than negative

The correct answer is:
Positive outside

Solution:
Resting state of neuron
- High concentration of K+ and negatively charged proteins and low concentration of Na+ in axoplasm.
- In contrast, the fluid outside the axon contains a low concentration of K+, a high concentration of Na+
and thus form a concentration gradient.
- These ionic gradients across the resting membrane are maintained by the Na-K pump which actively
transports 3Na+ outwards for 2K+ into the cell.
- Outer surface of sarcolemma: Positively charged.
Inner surface: Negatively charged.
- Membrane is polarised.

Topic: Biology, Neural Control and Coordination, NEET
At the time t=0, some radioactive gas is injected into a sealed vessel. At the time T, some more of the same gas is injected into the same vessel. Which one of the following graphs, best represents the variation of the logarithm of the activity A of the gas with time t?

Options:
A
B
C
D None of these

The correct answer is:

Solution:
A=A0e-λt
Take log both side
ℓn  A  =  ℓn   A 0 − λ t

Topic: Physics, Nuclear Physics, JEE Main
At the top of the trajectory of a projectile, the directions of its velocity and acceleration are

Options:
A Perpendicular to each other
B Parallel to each other
C Inclined to each other at an angle of $45^{\circ}$
D Antiparallel to each other

The correct answer is:
Perpendicular to each other

Solution:
Direction of velocity is always tangent to the path so at the top of trajectory, it is in horizontal direction and acceleration due to gravity is always in vertically downward direction. It means angle between $\vec{v}$ and $\vec{g}$ are perpendicular to each other.

Topic: Physics, Motion In Two Dimensions, JEE Main
At the uppermost point of a projectile, its velocity and acceleration are at an angle of

Options:
A $0^{\circ}$
B $90^{\circ}$
C $45^{\circ}$
D $180^{\circ}$.

The correct answer is:
$90^{\circ}$

Solution:
At the uppermost point of a projectile, the vertical component of the velocity of projection becomes zero, while the horizontal component remains constant. And the acceleration (due to gravity) is always vertically downwards. Therefore at the uppermost point of a projectile, its velocity and acceleration are at an angle of $90^{\circ}$.

Topic: Physics, Motion In Two Dimensions, AIIMS, AIIMS 2002
At time t=0, a material is composed of two radioactive atoms A and B, where NA(0)=2NB(0) The decay constant of both kind of radioactive atoms is λ. However, A disintegrates to B and B disintegrates to C. Which of the following figures represents the evolution of NBtNB0 with respect to time t?

Options:
A
B
C
D

The correct answer is:

Solution:
The rate at which nuclei of B is forming,

dNBdt=λNA-λNB

dNBdt=2λNB0e-λt-λNB

e-λtdNBdt+λNB=2λNB0e-λteλt

ddtNBe-λt=2λNB0

on integrating above the equation,
NBeλt=2λtNB+NB0
NBeλt=2λtNB0+NB0

NB=NB1+2λte-λtddtNB=0

-λ1+2λte-λt+2λe-λt=0NBmax at t=12λ

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2021 (26 Aug Shift 2)
At time t=0 a particle starts travelling from a height 7z^ cm in a plane keeping z coordinate constant. At any instant of time, it's position along the x and y directions are defined as 3t and 5t3 respectively. At t=1 s acceleration of the particle will be

Options:
A -30y
B 30y
C 3x+15y
D 3x+15y+7z^

The correct answer is:
30y

Solution:
The position vector of the particle can be written as,

r→=3ti^+5t3j^+7k^

The velocity of the particle will be,

v→=dr→dt=3i^+15t2j^

Now the acceleration of the particle will be,

a→=dv→dt=d2r→dt2=30tj^

At t=1 s⇒d2rdt2=30j^

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2022 (28 Jul Shift 2)
At two different places, the angles of dip are respectively $30^{\circ}$ and $45^{\circ} .$ At these two places the ratio of horizontal component of earth's magnetic field is

Options:
A $\sqrt{3}: \sqrt{2}$
B $1: \sqrt{2}$
C 1: 2
D $1: \sqrt{3}$

The correct answer is:
$\sqrt{3}: \sqrt{2}$

Solution:
The horizontal component of earth's magnetic field is given by, $H=R \cos \delta$
where, $\delta$ is the angle of dip so $H_{1}=R \cos 30^{\circ}$ and $\mathrm{H}_{2}=\mathrm{Rcos} 45^{\circ}$
$\therefore \quad \frac{H_{1}}{H_{2}}=\frac{R \cos 30^{\circ}}{R \cos 45^{\circ}}=\frac{\sqrt{3} / 2}{1 / \sqrt{2}}=\frac{\sqrt{3}}{\sqrt{2}}$

Topic: Physics, Magnetic Properties of Matter, JEE Main
At very high frequencies, the effective impedance of the given circuit will be _________Ω.

Options:

The correct answer is:
2

Solution:
For ω→∞
1ωC→0⇒ capacitance acts as short circuit
& ωL→∞ Inductance acts as open circuit
Z=1+2×22+2=2Ω

Topic: Physics, Alternating Current, JEE Main, JEE Main 2021 (31 Aug Shift 2)
At very high pressures, the compressibility factor of one mole of a gas is given by:

Options:
A 1+PbRT
B PbRT
C 1-PbRT
D 1-bVRT

The correct answer is:
1+PbRT

Solution:
P+aV2V-b=RT
at very high pressure P≫aV2
PV-b=RT
PV-Pb=RT
∴Z=1+PbRT compressibility factor

Topic: Chemistry, States of Matter, JEE Main, JEE Main 2016 (09 Apr Online)
At very high pressures, the compressibility factor of one mole of a gas is given by:

Options:
A 1+PbRT
B PbRT
C 1-PbRT
D 1-bVRT

The correct answer is:
1+PbRT

Solution:
P+aV2V-b=RT
at very high pressure P≫aV2
PV-b=RT
PV-Pb=RT
∴Z=1+PbRT compremibility factor

Topic: Chemistry, States of Matter, JEE Main
At what distance (in m) from a convex mirror of focal length 2.5 m should a boy stand so that his image has a height equal to half of his height? The principal axis of the mirror is perpendicular to the height of the boy.

Options:

The correct answer is:
2.5

Solution:
We have,
m=-vu=12
v=-u2
1u⁡+1v⁡=1f⁡
1u⁡+1-u⁡/2=12.5 m
-1u⁡=12.5 m
u⁡=-2.5 m
Thus, he should stand at a distance of 2.5 m from the mirror.

Topic: Physics, Ray Optics, JEE Main
At what point on the parabola $y^2=4 x$, the normal makes equal angles with the co-ordinate axes

Options:
A $(4,4)$
B $(9,6)$
C $(4,-4)$
D $(1,-2)$

The correct answer is:
$(1,-2)$

Solution:
The equation of a normal to $y^2=4 x$ at $\left(m^2,-2 m\right)$ is $y=m x-2 m-m^3$. If the normal makes equal angles with the coordinates axes, then $m=\tan \frac{\pi}{4}=1$ Thus, the required point is $(1,-2)$.

Topic: Mathematics, Parabola, JEE Main
At what temperature does carbon dioxide remain as a gas only up to a certain point?

Options:
A 0°C
B 30.98°C
C 73°C
D 21.5°C

The correct answer is:
21.5°C

Solution:
At 21.5°C, carbon dioxide remains as a gas only up to point B.

Topic: Chemistry, States of Matter, NEET
At what temperature does carbon dioxide remain in the gaseous state up to a pressure of 73 atmospheric pressure?

Options:
A 0°C
B 30.98°C
C 73°C
D 21.5°C

The correct answer is:
30.98°C

Solution:
At 30.98°C, carbon dioxide remains a gas up to 73 atmospheric pressure.

Topic: Chemistry, States of Matter, NEET
At what temperature (in kelvin) is the rms speed of hydrogen molecules the same as oxygen molecules at 1327℃?

Options:

The correct answer is:
100

Solution:
vH2=vO2
3RTH2MH2=3RTO2MO2
So,MO2TH2=MH2TO2
32×TH2=2×1600
T H 2 =2× 1600 32 =100

Topic: Chemistry, States of Matter, JEE Main
At what temperature (in kelvin) is the rms speed of hydrogen molecules the same as oxygen molecules at \( 1327^{\circ} \mathrm{C} ? \)

Options:

The correct answer is:
100

Solution:
vH2=vO2
3RTH2MH2=3RTO2MO2
So,MO2TH2=MH2TO2
32×TH2=2×1600
TH2=2×160032=100K

Topic: Chemistry, States of Matter, JEE Main
At what temperature is ethoxyethane the main product of ethanol dehydration in the presence of sulfuric acid?

Options:
A 273 K
B 343 K
C 413 K
D 443 K

The correct answer is:
413 K

Solution:
Ethoxyethane is the main product of ethanol dehydration at 413 K.

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
At what temperature is the r.m.s. velocity of a hydrogen molecule equal to that of an oxygen molecule at $47^{\circ} \mathrm{C}$ ?

Options:
A
80 K

B
73 K

C
3 K

D
20 K

The correct answer is:
20 K

Solution:
$\mathrm{v}_{\mathrm{ms}} \alpha \sqrt{\mathrm{T} / \mathrm{m}} ; \quad \sqrt{\frac{273+47}{32}}=\sqrt{\frac{T}{2}}$ or $\mathrm{T}=20 \mathrm{~K}$

Topic: Physics, Kinetic Theory of Gases, JEE Main
At what temperature is the r.m.s. velocity of a hydrogen molecule equal to that of an oxygen molecule at $47^{\circ} \mathrm{C}$ ?

Options:
A
80 K

B
73 K

C
3 K

D
20 K

The correct answer is:
20 K

Solution:
$\mathrm{v}_{\mathrm{ms}} \alpha \sqrt{\mathrm{T} / \mathrm{m}} ; \quad \sqrt{\frac{273+47}{32}}=\sqrt{\frac{T}{2}}$ or $\mathrm{T}=20 \mathrm{~K}$

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2002
At what temperature rms speed of air molecules is doubled of that at NIP ?

Options:
A $819^{\circ} \mathrm{C}$
B $719^{\circ} \mathrm{C}$
C $909^{\circ} \mathrm{C}$
D None of these

The correct answer is:
$819^{\circ} \mathrm{C}$

Solution:
$$
v_{\mathrm{rms}} \propto \sqrt{T}
$$
$$
\begin{aligned}
\Rightarrow & & \frac{v_{1}^{2}}{v_{2}^{2}} &=\frac{T_{1}}{T_{2}} \\
\Rightarrow & & \frac{v^{2}}{4 v^{2}} &=\frac{273}{T_{2}} \\
\Rightarrow & & T_{2} &=1092 \mathrm{~K} \\
& &=819{ }^{\circ} \mathrm{C}
\end{aligned}
$$

Topic: Physics, Kinetic Theory of Gases, JEE Main