datavorous/entrance-exam-dataset huggingface dataset:
The special arrangement of lipids in the cell membrane ensures that the

Options:
A polar tail of saturated hydrocarbons is protected from the aqueous environment.
B non-polar tail of unsaturated hydrocarbons is protected from the aqueous environment.
C polar tail of unsaturated hydrocarbons is protected from the aqueous environment.
D non-polar tail of saturated hydrocarbons is protected from the aqueous environment.

The correct answer is:
non-polar tail of saturated hydrocarbons is protected from the aqueous environment.

Solution:
The detailed structure of the membrane was studied only after the advent of the electron microscope in the 1950s. Meanwhile, chemical studies on the cell membrane, especially in human red blood cells (RBCs), enabled the scientists to deduce the possible structure of plasma membrane. These studies showed that the cell membrane is composed of lipids that are arranged in a bilayer. Also, the lipids are arranged within the membrane with the polar head towards the outer sides and the hydrophobic tails towards the inner part.This ensures that the nonpolar tail of saturated hydrocarbons is protected from the aqueous environment. The lipid component of the membrane mainly consists of phosphoglycerides.

Topic: Biology, Cell - The Unit of Life, NEET
The species ${ }^{19} \mathrm{Ne}$ and ${ }^{14} \mathrm{C}$ emit a positron and $\beta$-particle respectively.
The resulting species formed are respectively

Options:
A ${ }^{19} \mathrm{Na}$ and ${ }^{14} \mathrm{~B}$
B ${ }^{19} \mathrm{~F}$ and ${ }^{14} \mathrm{~N}$
C ${ }^{19} \mathrm{Na}$ and ${ }^{14} \mathrm{~N}$
D ${ }^{19} \mathrm{~F}$ and ${ }^{14} \mathrm{~B}$

The correct answer is:
${ }^{19} \mathrm{~F}$ and ${ }^{14} \mathrm{~N}$

Solution:
When ${ }^{19} \mathrm{Ne}$ emits a positron, it undergoes beta plus decay. In this process, a proton in the nucleus is transformed into a neutron, and a positron (a positively charged electron) is emitted, resulting in a change of the atomic number. So, ${ }^{19} \mathrm{Ne}$ becomes ${ }^{19} \mathrm{~F}$. The reaction can be represented as follows:
${ }_{10}^{19} \mathrm{Ne} \xrightarrow{-\mathrm{e}_{+1} \text { (positron) }}{ }_9^{19} \mathrm{~F}$
When ${ }^{14} \mathrm{C}$ emits a beta particle ( $\beta^{-}$particle), it undergoes beta minus decay, which is also known as electron emission. In this process, a neutron in the nucleus is transformed into a proton, and a beta minus particle or electron (denoted as $\beta^{-}$) is emitted, which increases its atomic number. So, ${ }^{14} \mathrm{C}$ becomes ${ }^{14} \mathrm{~N}$. The reaction can be represented as follows:
${ }_6^{14} \mathrm{C} \xrightarrow{-\mathrm{e}-1 \text { (beta particle) }}{ }_7^{14} \mathrm{~N}$

Topic: Physics, Nuclear Physics, JEE Main
The species given below that does NOT show disproportionation reaction is:

Options:
A BrO4-
B BrO-
C BrO2-
D BrO3-

The correct answer is:
BrO4-

Solution:
In BrO4⊖,Br is in highest oxidation state (+7), So it cannot oxidise further hence, it cannot show disproportionation reaction.

Topic: Chemistry, Redox Reactions, JEE Main, JEE Main 2021 (20 Jul Shift 1)
The species in which the N atom is a state of sp hybridization is:

Options:
A NO3-
B NO2
C NO2+
D NO2-

The correct answer is:
NO2+

Solution:

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2016 (03 Apr)
The species in which the N atom is a state of sp hybridization is:

Options:
A NO3-
B NO2
C NO2+
D NO2-

The correct answer is:
NO2+

Solution:

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The species in which the N-atom is in a state of sp hybridization is

Options:
A NO2
B NO2+
C NO2-
D NO3-

The correct answer is:
NO2+

Solution:
Option A:
Option B:
Option C:
Option D:

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The species that can have a trans-isomer is:
(en = ethane-1, 2-diamine, ox=oxalate)

Options:
A Cren2ox+
B PtenCl2
C ZnenCl2
D Pten2Cl22+

The correct answer is:
Pten2Cl22+

Solution:
Pten2Cl22+ Dichlorido-bis (ethane – 1, 2 – diamine) Platinum (IV) ion
Show geometrical isomers

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2019 (10 Apr Shift 1)
The species that cannot exist is

Options:
A SiF62-
B BF63-
C SF6
D AlF63-

The correct answer is:
BF63-

Solution:
As Boron does not have d orbital in its valence orbit. Hence, maximum covalency of boron cannot exceed more than 4 so BF63- does not exist. Rest all exist as Si, S,Al have d- orbitals to exceed covalency more than 4.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The species that has a spin-only magnetic moment of 5.9BM, is :
(Td=tetrahedral)

Options:
A Ni(CN)42- (square planar)
B NiCl42-Td
C Ni(CO)4Td
D MnBr42-Td

The correct answer is:
MnBr42-Td

Solution:
Ni(CO)43d10 n=0

Ni(CN)42- 3d8 n=0

NiCl42- 3d8 n=2

MnBr42- 3d5 n=5

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2020 (06 Sep Shift 1)
The species which can best serve as an initiator for the cationic polymerization is :

Options:
A
$\mathrm{LiAlH}_4$

B
$\mathrm{HNO}_3$

C
$\mathrm{AlCl}_3$

D
$\mathrm{BuLi}$

The correct answer is:
$\mathrm{AlCl}_3$

Solution:
Lewis acids can initiate the cationic polymerization.

Topic: Chemistry, Polymers, JEE Main, JEE Main 2012 (Offline)
The specific conductance at \( 25^{\circ} \mathrm{C} \) of a saturated solution of \( \mathrm{AgCl} \) in water is \( 1.826 \times 10^{-6} \mathrm{mho} \mathrm{cm}^{-1} \). If \( \Lambda_{\mathrm{AgCl}}^{\circ} \) is equal to \( 138.26 \mathrm{mho} \mathrm{cm}^{2} \mathrm{~mol}^{-1} \), find out the solubility of \( \mathrm{AgCl} \) in water in milligrams per litre.

Options:

The correct answer is:
1.89

Solution:
Given,

Specific conductance, k= 1.826×10-6 mho cm-1

Limiting molar conductivity, ΛAgCl°=138.26 mho cm2 mol-1

And molar mass of AgCl=143.5 g

We know that,

Solubility, S=κ×1000ΛAgCl°

Substituting the values,

⇒S=1.826×10-6×1000138.26=0.0132×10-3 mol/L

Solubility of AgCl=0.0000132×143.5=0.00189 g/L=1.89×10-3 g/L

Topic: Chemistry, Electrochemistry, JEE Main
The specific conductance of 0.0025M acetic acid is 5×10-5 S cm-1 at a certain temperature. The dissociation constant of acetic acid is ___________ ×10-7.(Nearest integer)

Consider limiting molar conductivity of CH3COOH as 400 S cm2 mol-1

Options:

The correct answer is:
66

Solution:
The relation between Molar conductance λm and specific conductance κ is

λm=k×1000M

M = molarity

λm=5×10-5×1032.5×10-3=20 S cm2 mol-1

Now, the degree of dissociation, α=λmλ∞

α=20400=120

The acid dissociation constant, Ka=Cα2(1-α)=2.5×10-3120×1201920

= 65.789 × 10–7

≈ 66 × 10–7

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2023 (10 Apr Shift 2)
The specific conductance of a solution is $0.2 \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}$ and conductivity is $0.04 \mathrm{ohm}^{-1}$. The cell constant would be

Options:
A $1 \mathrm{~cm}^{-1}$
B $0 \mathrm{~cm}^{-1}$
C $5 \mathrm{~cm}^{-1}$
D $0.2 \mathrm{~cm}^{-1}$

The correct answer is:
$5 \mathrm{~cm}^{-1}$

Solution:
$K=C \times$ Cell constant $=\frac{K}{C}=\frac{0.2}{0.04}=5 \mathrm{~cm}^{-1}$

Topic: Chemistry, Electrochemistry, JEE Main
The specific heat at constant pressure and at constant volume for an ideal gas are ${ }^{C_p}$ and $C_v$ and its adiabatic and isothermal elasticities are $E_\phi$ and $E_\theta$ respectively. The ratio of $E_\phi$ to $E_\theta$ is

Options:
A $C_v / C_p$
B $C_p / C_v$
C $C_p C_v$
D $1 / C_p C_v$

The correct answer is:
$C_p / C_v$

Solution:
Ratio of adiabatic and isothermal elasticities
$\frac{E \phi}{E \theta}=\frac{\gamma P}{P}=\gamma=\frac{C_p}{C_\gamma}$

Topic: Physics, Mechanical Properties of Solids, JEE Main
The specific heat capacity of a copper block of mass $\mathrm{M}$ is 's'. If the mass of the copper block is doubled, the specific heat capacity will be

Options:
A $2 \mathrm{~s}$
B $\frac{s}{2}$
C $\mathrm{s}$
D $\sqrt{\frac{3}{2}} \mathrm{~s}$

The correct answer is:
$\mathrm{s}$

Solution:
Specific heat capacity of copper is $386.4 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}$ therefore specific heat capacity will remain same. It is independent of mass of the substance.

Topic: Physics, Thermal Properties of Matter, AP EAMCET, AP EAMCET 2023 (15 May Shift 1)
The specific heat of helium at constant volume is 12.6 $\mathrm{Jmol}^{-1} \mathrm{~K}^{-1}$. The specific heat of helium at constant pressure in $\mathrm{Jmol}^{-1} \mathrm{~K}^{-1}$ is about (Assume the temperature of the gas is moderate, universal gas constant, $R=8.314 \mathrm{Jmol}^{-1} \mathrm{~K}^{-1}$ )

Options:
A $12.6$
B $16.8$
C $18.9$
D $21$

The correct answer is:
$21$

Solution:
As $C_V=12.6 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$ is given in SI unit
So, $R=8314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$
Then, $C_p=C_V+R=12.6+8.314=21$

Topic: Physics, Kinetic Theory of Gases, JEE Main
The specific heat of hydrogen gas at constant pressure is $C_P=3.4 \times 10^3 \mathrm{cal} / \mathrm{kg}{ }^{\circ} \mathrm{C}$ and at constant volume is $C_v=2.4 \times 10^3 \mathrm{cal} / \mathrm{kg}^{\circ} \mathrm{C}$. If one kilogram hydrogen gas is heated from $10^{\circ} \mathrm{C}$ to $20^{\circ} \mathrm{C}$ at constant pressure, the external work done on the gas to maintain it at constant pressure is

Options:
A $10^5 \mathrm{cal}$
B $10^4 \mathrm{cal}$
C $10^3 \mathrm{cal}$
D $5 \times 10^3 \mathrm{cal}$

The correct answer is:
$10^4 \mathrm{cal}$

Solution:
From FLOT $\Delta Q=\Delta U+\Delta W$
Work done at constant pressure $(\Delta W)_P=(\Delta Q)_P-\Delta U$
$(\Delta Q)_P-(\Delta Q)_V$
(As we know $(\Delta Q)_V=\Delta U$ )
$\begin{gathered}\text { Also }(\Delta Q)_P=m c_P \Delta T \text { and }(\Delta Q)_V=m c_V \Delta T \\ \Rightarrow(\Delta W)_P=m\left(c_P-c_V\right) \Delta T \\ \Rightarrow(\Delta W)_P=1 \times\left(3.4 \times 10^3-2.4 \times 10^3\right) \times 10=10^4 \mathrm{cal}\end{gathered}$

Topic: Physics, Thermodynamics, JEE Main
The specific heat of water =4200 J kg–1 K–1 and the latent heat of ice =3.4×105 J kg–1. 100 grams of ice at 0 oC is placed in 200 g of water at 25 oC. The amount of ice that will melt as the temperature of water reaches 0 oC is close to (in grams)

Options:
A 61.7
B 63.8
C 69.3
D 64.6

The correct answer is:
61.7

Solution:
MSΔT=Ml

2001000×4200×25=m×340×103

m=61.7

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2020 (04 Sep Shift 1)
The specific heats, \( \mathrm{C}_{p} \) and \( \mathrm{C}_{v} \) of a gas of diatomic molecules, \( \mathrm{A} \), are given (in units of \( \mathrm{J} \mathrm{mol}^{-1} \mathrm{~K}^{-1} \) ) by \( 29 \) and \( 22 \), respectively. Another gas of diatomic molecules, \( \mathrm{B} \), has the corresponding values \( 30 \) and \( 21 . \) If they are treated as ideal gases, then:

Options:
A A has one vibrational mode and B has two
B A has a vibrational mode but B has none.
C Both A and B have a vibrational mode each.
D A is rigid but B has a vibrational mode.

The correct answer is:
A has a vibrational mode but B has none.

Solution:
For A :
R=Cp-Cv=29-22=7
Cv=fR2
⇒22=f×72
⇒f=447=6.3
⇒f≈6

(5 rotational + translational, 1 vibrational)

For B :
R=Cp-CV=30-21=9
Cv=fR2=21
⇒f=21×29=429=4.66≃5
∴ f≃5

(5 rotational + transitional, 0 vibrational)

Topic: Physics, Kinetic Theory of Gases, JEE Main
The specific heats of an ideal gas at constant pressure and constant volume are 525 J kgÂ°C-1 and 315 J kgÂ° C-1 respectively. Its density at NTP is

Options:
A 0.64 kg m-3
B 1.20 kg m-3
C 1.75 kg m-3
D 2.62 kg m-3

The correct answer is:
1.75 kg m-3

Solution:
If M is molecular mass of the gas, then from
MCp-Cv=R
M=8.31210=0.0392
If Ï is density of the gas at NTP, then mass of 1 m3of gas at NTP=Ï kg
â´ Mass of 22.4 L (=22.4Ã10-3m3) of gas at NTP =ÏÃ22.4Ã10-3 kg, which is the molecular mass of the gas
â´ ÏÃ22.4Ã10-3=0.0392
Ï=0.039222.4Ã10-3=1.75 kg m-3

Topic: Physics, Thermodynamics, JEE Main
The specific rate constant of decomposition of a compound is given by $\ln k=5.0-\frac{12000}{T}$. The activation energy of decomposition for this compound at $300 \mathrm{~K}$ is

Options:
A $24 \mathrm{ kcal} \mathrm{mol}^{-1}$
B $12 \mathrm{kcal} \mathrm{mol}^{-1}$
C $24 \mathrm{cal} \mathrm{mol}^{-1}$
D $12 \mathrm{cal} \mathrm{mol}^{-1}$

The correct answer is:
$24 \mathrm{ kcal} \mathrm{mol}^{-1}$

Solution:
Given,
$\text { In } k=5.0-\frac{12000}{T}$
Also,
$\because \quad \text { In } k=\log A-\frac{E_a}{R T} \text { and } k=A e^{-E_a / R T}$
or, $\quad \log _e K=\log _e A=\frac{E_a}{R T}$
where, $A=$ pre-exponential constant i.e. frequency factor
$E_a=$ activation energy
$R=$ gas constant $=2 \mathrm{kcal} \mathrm{mol}^{-1}$
$T_k=$ temperature
$k=$ rate constant
Thus,
$\frac{E_a}{R T}=\frac{12000}{T} \text { or } E_a=12000 \times R$
$\therefore \quad E_a=\frac{12000 \times 2}{1000} \quad\left(\right.$ in kcal mol$\left.{ }^{-1}\right)$
i.e. $\quad E_a=24 \mathrm{kcal} \mathrm{mol}^{-1}$

Topic: Chemistry, Chemical Kinetics, JEE Main
The specific resistance of a wire is $\rho$, its volume is $3 \mathrm{~m}^3$ and its resistance is $3 \mathrm{ohms}$, then its length will be

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

The correct answer is:
$\frac{3}{\sqrt{\rho}}$

Solution:
$\text { Volume }=A l=3 \Rightarrow A=\frac{3}{l}$
Now $R=\rho \frac{l}{A} \Rightarrow 3=\frac{\rho \times l}{3 / l}=\frac{\rho l^2}{3} \Rightarrow l^2=\frac{9}{\rho}=\frac{3}{\sqrt{\rho}}$

Topic: Physics, Current Electricity, JEE Main
The specific rotation of a pure enantiomer is +10° . The observed rotation, if it is isolated from a reaction with 30% racemisation and 70% inversion is

Options:
A +10°
B -10°
C 3°
D -7°

The correct answer is:
-7°

Solution:
Specific rotation of inverted isomer =-10°
Angle of rotation =(Specific rotation of inverted isomer)×(Fraction of inversion)
=-10×70100=-7°

Topic: Chemistry, General Organic Chemistry, JEE Main
The spectral emissive power E Î» for a body at temperature T1 is plotted against the wavelength (see figure) and area under the curve is found to be A. At a different temperature T2 the area is found to be 9A. Then Î» 1 / Î» 2 =

Options:
A 3
B 1 3
C 1 3
D 3

The correct answer is:
3

Solution:
E1E2=19=T1T24
λ1T1=λ2T2⇒T1T2=λ2λ1
19=λ2λ14⇒λ2λ1=13
λ1λ2=3

Topic: Physics, Thermal Properties of Matter, JEE Main
The speed of a projectile at its maximum height is half of its initial speed. The angle of projection is

Options:
A $60^{\circ}$
B $15^{\circ}$
C $30^{\circ}$
D $45^{\circ}$

The correct answer is:
$60^{\circ}$

Solution:
The speed of a projectile at its maximum height
$$
\begin{aligned}
\mathrm{v}^{\prime} & =\mathrm{v}_0 \cos \theta \\
\frac{\mathrm{v}_0}{2} & =\mathrm{v}_0 \cos \theta \\
\cos \theta & =\frac{1}{2} \\
\theta & =60^{\circ}
\end{aligned}
$$

Topic: Physics, Motion In Two Dimensions, NEET, NEET 2010 (Mains)
The speed of a swimmer is 4 km h-1 in still water. If the swimmer makes his strokes normal to the flow of river of width 1 km, he reaches a point 750 m down the stream on the opposite bank. The speed of the river water is ______ km h-1.

Options:

The correct answer is:
3

Solution:
Let vr be velocity of river and vm be velocity of man.

Time taken by man to cross the river of width =1000 m is t=14 h.

Now, drift x=vr×t

⇒0.750=vr×14⇒vr=3 km h-1

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2023 (31 Jan Shift 1)
The speed of a transverse wave passing through a string of length 50 cm and mass 10 g is 60 m s-1. The area of cross-section of the wire is 2.0 mm2 and its Young's modulus is 1.2×1011 N m-2. The extension of the wire over its natural length due to its tension will be x×10-5 m. The value of x is _____ .

Options:

The correct answer is:
15

Solution:
The wave speed in string is given by vw=Tμ, where, T is tension and μ=ml is linear mass density.

Putting the values, we have

⇒60=T10×10-3×0.5

⇒T=602×10-20.5=72 N

The Young's modulus is given by Y=FlA∆l, here, A is area.

From above the extension of the wire is Δl=TlAY=72×0.52×10-6×1.2×1011=72×524×10-5=15×10-5 m.

Hence, the value of x=15.

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2022 (29 Jul Shift 2)
The speed of a wave produced in water is given by ν=λagbρc. Where λ, g and ρ are wavelength of wave, acceleration due to gravity and density of water respectively. The values of a, b and c respectively, are

Options:
A 1, -1, 0
B 12, 0, 12
C 1, 1, 0
D 12, 12, 0

The correct answer is:
12, 12, 0

Solution:
The dimensional form of velocity can be written as

v=λagbρc

⇒M0LT-1=LaLT-2bML-3c

Comparing the exponents for M,L,T,

For M, Mc=M0For L,La+b-3c=L1For T,T-2b=T-1

So,

c=0 ...(i)a+b-3c=1 ...(ii)-2b=-1 ...(iii)

From equations (i),(ii),(iii),

a=12, b=12,c=0

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2023 (15 Apr Shift 1)
The speed of electrons in a scanning electron microscope is 1×107 ms-1. If the protons having the same speed are used instead of electrons, then the resolving power of scanning proton microscope will be changed by a factor of:

Options:
A 1837
B 11837
C 1837
D 11837

The correct answer is:
1837

Solution:
Resolving power RP∝1λ

λ=hP=hmv

So, RP∝mvh
RP∝P

RP∝mv

RP∝m

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2021 (18 Mar Shift 2)
The speed of light c, gravitational constant G and Planck’s constant h are taken as fundamental units in a system. The dimensions of time in this new system should be:

Options:
A

G 1 2 â h 1 2 â c â5 2

B

G −1 2   h 1 2   c 1 2

C

G 1 2   h 1 2   c −3 2

D

G 1 2   h 1 2   c 1 2

The correct answer is:

G 1 2 â h 1 2 â c â5 2

Solution:
Let time, Tâ c X G Y h Z

âT=k c X G Y h Z

Taking dimensions on both sides

[ M 0 L 0 T 1 ]= [L T â1 ] X [ M â1 L 3 T â2 ] Y [M L 2 T â1 ] Z

i.e.,

[ M 0 L 0 T 1 ]=[ M âY+Z L X+3Y+2Z T âXâ2YâZ ]

Equating powers of M, L, and T, on both sides, we get â

âY+Z=0âââââ...(1)

X+3Y+2Z=0ââââââ...(2)

âXâ2YâZ=1âââââââ...(3)

From equation (i)

Adding (2) and (3)

Y+Z=1

or 2Y=1ââââ( âµââZ=Y )

â´Z=Y= 1 2 âââââ

Putting these values in (2), we get â

X+ 3 2 +1=0ââorââX= â5 2

Hence, [T]=[ G 1 2 â h 1 2 â c â5 2 ]

Topic: Physics, Units and Dimensions, JEE Main
The speed of light in media $\mathbf{M}_1$ and $\mathrm{M}_2$ is $1.5 \times 10^8 \mathrm{~m} / \mathrm{s}$ and $2.0 \times 10^8 \mathrm{~m} / \mathrm{s}$ respectively. A ray of light enters from medium $\mathrm{M}_1$ to $\mathrm{M}_2$ at an incidence angle $i$. If the ray suffers total internal reflection, the value of $i$ is

Options:
A equal to $\sin ^{-1}\left(\frac{2}{3}\right)$
B equal to or less than $\sin ^{-1}\left(\frac{3}{5}\right)$
C equal to or greater than $\sin ^{-1}\left(\frac{3}{4}\right)$
D less than $\sin ^{-1}\left(\frac{2}{3}\right)$

The correct answer is:
equal to or greater than $\sin ^{-1}\left(\frac{3}{4}\right)$

Solution:
In total internal reflection, the angle of incidence (i) must be greater than critical angle $C$
$$
\because \quad \begin{aligned}
2 \mu_1 & =2 \text { and } \mu_2=\frac{3}{2} \\
2 \sin \mathrm{i} & \geq \frac{3}{2} \sin 90^{\circ} \\
\sin \mathrm{i} & \geq \frac{3}{4} \\
\mathrm{i} & \geq \sin ^{-1}\left(\frac{3}{4}\right)
\end{aligned}
$$

Topic: Physics, Ray Optics, JEE Main
The speed of light in media $\mathbf{M}_1$ and $\mathrm{M}_2$ is $1.5 \times 10^8 \mathrm{~m} / \mathrm{s}$ and $2.0 \times 10^8 \mathrm{~m} / \mathrm{s}$ respectively. A ray of light enters from medium $\mathrm{M}_1$ to $\mathrm{M}_2$ at an incidence angle $i$. If the ray suffers total internal reflection, the value of $i$ is

Options:
A equal to $\sin ^{-1}\left(\frac{2}{3}\right)$
B equal to or less than $\sin ^{-1}\left(\frac{3}{5}\right)$
C equal to or greater than $\sin ^{-1}\left(\frac{3}{4}\right)$
D less than $\sin ^{-1}\left(\frac{2}{3}\right)$

The correct answer is:
equal to or greater than $\sin ^{-1}\left(\frac{3}{4}\right)$

Solution:
In total internal reflection, the angle of incidence (i) must be greater than critical angle $C$
$$
\because \quad \begin{aligned}
2 \mu_1 & =2 \text { and } \mu_2=\frac{3}{2} \\
2 \sin \mathrm{i} & \geq \frac{3}{2} \sin 90^{\circ} \\
\sin \mathrm{i} & \geq \frac{3}{4} \\
\mathrm{i} & \geq \sin ^{-1}\left(\frac{3}{4}\right)
\end{aligned}
$$

Topic: Physics, Ray Optics, NEET, NEET 2010 (Mains)
The speed of ripples (v) on water surface depends on surface tension (σ), density (ρ), and wavelength (λ). Then the square of speed (v) is proportional to

Options:
A σρλ
B ρσλ
C λσρ
D ρλσ

The correct answer is:
σρλ

Solution:
Speed of ripples, v depends on σ,ρ,λ

⇒v∝σaρbλc

⇒M0LT-1∝MT-2aML-3bLc

⇒M0LT-1∝Ma+bL-3b+cT-2a

By equating the powers, ⇒a+b=0, -3b+c=1, -2a=-1

⇒a=12, b=-12, c=-12

⇒v∝σ12ρ-12λ-12

⇒v∝σρλ⇒v2∝σρλ

Topic: Physics, Units and Dimensions, JEE Main
The speed of sound in air at temperature $T$ and pressure $p$ is $v$. When the temperature is increased to $2 T$ and the pressure is reduced to $\frac{p}{2}$, then the speed is changed to

Options:
A $2 \mathrm{v}$
B $v$
C $\sqrt{2} v$
D $\frac{v}{\sqrt{2}}$

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

Solution:
Speed of sound in air does not depend on pressure, but directly proportional to the root of temperature, i.e.
$$
v \propto \sqrt{T} \Rightarrow \frac{v_2}{v_1}=\sqrt{\frac{T_2}{T_1}}
$$
Given, $T_1=T$ and $T_2=2 T$,
$$
\begin{aligned}
& v_1=v_1 v_2=? \\
& \frac{v_2}{v}=\sqrt{\frac{2 T}{T}} \Rightarrow v_2=\sqrt{2} v
\end{aligned}
$$

Topic: Physics, Waves and Sound, JEE Main
The speed of sound in an ideal gas at a given temperature $T$ is $v$. The rms speed of gas molecules at that temperature is $v_{\mathrm{rms}}$. The ratio of the velocities $v$ and $v_{\text {rms }}$ for helium and oxygen gases are $X$ and $X^{\prime}$. respectively. Then, $\frac{X}{X^{\prime}}$ is equal to

Options:
A $\frac{21}{\sqrt{5}}$
B $\frac{5}{\sqrt{21}}$
C $\sqrt{\frac{5}{21}}$
D $\frac{21}{5}$

The correct answer is:
$\frac{5}{\sqrt{21}}$

Solution:
The speed of sound at a given temperature is
given by
$v_{\text {sound }}=\sqrt{\frac{r R T}{M}} \Rightarrow v_{\mathrm{rms}}=\sqrt{\frac{3 R T}{M}}$
For oxygen molecules; $\frac{v_0}{v_{\text {rms }}}=\frac{\sqrt{\frac{r_0 R T}{M}}}{\sqrt{\frac{3 R T}{M}}}=\sqrt{\frac{r_0}{3}}$
For helium molecules; $\frac{v_{\mathrm{He}}}{v_{\mathrm{rms}}}=\frac{\sqrt{r_{\mathrm{He}} \frac{R T}{M}}}{\sqrt{\frac{3 R T}{M}}}=\sqrt{\frac{r_{\mathrm{He}}}{3}}$ $\left(r_{\mathrm{Be}}=\frac{5}{3}, r_{\mathrm{O}}=\frac{7}{5}\right)$
$\frac{x^{\prime}}{x}=\frac{\sqrt{\frac{5}{9}}}{\sqrt{\frac{7}{15}}}=\sqrt{\frac{25}{21}}=\frac{5}{\sqrt{21}}$

Topic: Physics, Kinetic Theory of Gases, JEE Main
The speed of sound in hydrogen gas at N.T.P is 1328 m s-1. If the density of hydrogen is 116th of that of air, then the speed of sound in air at N.T.P is

Options:
A 340 m s-1
B 332 m s-1
C 320 m s-1
D 280 m s-1

The correct answer is:
332 m s-1

Solution:
Here, vH2=1328 m s−1 ;  ρH2=ρair16 ;  v=γP8Let vair be the velocity of sound in air at N.T.PNow,  vairvH2=ρH2ρair                {Since          γH2=γair}or     vair=vH2ρH2ρairor   vair=1328ρair16ρair=1328116=13284 vair           =332 m s−1

Topic: Physics, Waves and Sound, JEE Main
The speed of sound in hydrogen gas at N.T.P is 1328 ms-1. If the density of hydrogen is 116th of that of air, then the speed of sound in air at N.T.P is

Options:
A 340 ms-1
B 332 ms-1
C 320 ms-1
D 280 ms-1

The correct answer is:
332 ms-1

Solution:
Here, v H 2 =1328âmâ s â1 â;ââ Ï H 2 = Ï air 16 â;ââv= Î³P 8 Let v air âbeâtheâvelocityâofâsoundâinâairâatâN.T.P Now,ââ v air v H 2 = Ï H 2 Ï air ââââââââââââââââ{ Since âââââââââ Î³ H 2 = Î³ air } orâââââ v air = v H 2 Ï H 2 Ï air orâââ v air =1328 Ï air 16 Ï air =1328 1 16 = 1328 4 âââââââââââââââ=332âmâ s â1

Topic: Physics, Waves and Sound, JEE Main
The speed of sound in oxygen at S.T.P. will be approximately:

(Given, R=8.3 J K-1, γ=1.4)

Options:
A 310 m s-1
B 333 m s-1
C 341 m s-1
D 325 m s-1

The correct answer is:
310 m s-1

Solution:
The speed of sound is given by,

v=γRTM=1.4×8.3×27332×10-3

=314.8541≃315 m s-1

Nearest answer given in the option is 310 m s-1.

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2024 (31 Jan Shift 2)