datavorous/entrance-exam-dataset huggingface dataset:
The pressure applied from all directions on a cube is $P$. How much its temperature should be raised to maintain the original volume? The volume elasticity of the cube is $\beta$ and the coefficient of volume expansion is $\alpha$

Options:
A $\frac{P}{\alpha \beta}$
B $\frac{P \alpha}{\beta}$
C $\frac{P \beta}{\alpha}$
D $\frac{\alpha \beta}{P}$

The correct answer is:
$\frac{P}{\alpha \beta}$

Solution:
If coefficient of volume expansion is $\alpha$ and rise in temperature is $\Delta \theta$ then $\Delta V=V \alpha \Delta \theta \Rightarrow \frac{\Delta V}{V}=\alpha \Delta \theta$
Volume elasticity $\beta=\frac{P}{\Delta V / V}=\frac{P}{\alpha \Delta \theta} \Rightarrow \Delta \theta=\frac{P}{\alpha \beta}$

Topic: Physics, Mechanical Properties of Solids, JEE Main
The pressure applied from all directions on a cube is p. How much its temperature should be raised to maintain the original volume? The volume elasticity of the cube is β and the coefficient of volume expansion is α.

Options:
A pαβ
B pαβ
C pβα
D αβp

The correct answer is:
pαβ

Solution:
If coefficient of volume expansion is α and rise in temperature is ∆θ, then ∆V=V α∆θ
⇒ ∆VV=α∆θ
∴ Volume elasticity, β=p∆V/V
=pα∆θ
⇒ ∆θ=pαβ

Topic: Physics, Mechanical Properties of Solids, JEE Main
The pressure at the bottom of a tank of water is 3P where P is the atmospheric pressure. If the water is drawn out till the level of water is lowered by one-fifth, the pressure at the bottom of the tank will now be

Options:
A 2P
B 135P
C 85P
D 45P

The correct answer is:
135P

Solution:
P + h Ï g = 3 P
h Ï g = 2 P ...(1)
P â² = P + 4 5 h Ï g ...(2)
From (1) and (2)
P â² = P + 4 5 Ã 2 P â P â² = 1 3 P 5

Topic: Physics, Mechanical Properties of Fluids, JEE Main
The pressure exerted by $6.0 \mathrm{~g}$ of methane gas in a $0.03 \mathrm{~m}^3$ vessel at $129^{\circ} \mathrm{C}$ is (Atomic masses : $\mathrm{C}=12.01, \mathrm{H}=1.01$ and $\mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$ )

Options:
A $215216 \mathrm{~Pa}$
B $13409 \mathrm{~Pa}$
C $41648 \mathrm{~Pa}$
D $31684 \mathrm{~Pa}$

The correct answer is:
$41648 \mathrm{~Pa}$

Solution:
Given,
volume, $\mathrm{V}=0.03 \mathrm{~m}^3$
temperature, $\mathrm{T}=129+273=402 \mathrm{~K}$
mass of methane, $w=6.0 \mathrm{~g}$
mol. mass of methane, $M=12.01+4 \times 1.01$
$=16.05$
From, ideal gas equation,
$$
\begin{aligned}
& \mathrm{pV}=\mathrm{nRT} \\
& \mathrm{p}=\frac{6}{16.05} \times \frac{8.314 \times 402}{0.03}=41648 \mathrm{~Pa}
\end{aligned}
$$

Topic: Chemistry, States of Matter, NEET, NEET 2010 (Mains)
The pressure exerted by a non-reactive gaseous mixture of 6.4 g of methane and 8.8 g of carbon dioxide in a 10 L vessel at 27°C is _____kPa. (Round off to the Nearest Integer).

[Assume gases are ideal, R=8.314 J mol-1 K-1 Atomic masses: C :12.0 u, H :1.0 u, O :16.0 u]

Options:

The correct answer is:
150

Solution:
Total moles of gases, n=nCH4+nCO2

=6.416+8.844=0.6

Now, P=nRTV=0.6×8.314×30010×10-3

=1.49652×105 Pa=149.652 kPa

≈150 kPa

Topic: Chemistry, States of Matter, JEE Main, JEE Main 2021 (17 Mar Shift 1)
The pressure exerted by an ideal gas on the walls of a container is

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

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

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

Topic: Physics, Kinetic Theory of Gases, JEE Main
The pressure exerted on the walls of the container by a gas is due to the fact that the gas molecules ______.

Options:
A stick to the walls
B change their momentum due to collision with the walls
C lose their kinetic energy
D are accelerated towards the walls

The correct answer is:
change their momentum due to collision with the walls

Solution:
At microscopic level, pressure is the force exerted by the gas molecules on the wall by virtue of momentum change during their elastic collision with the wall atoms.

Topic: Physics, Kinetic Theory of Gases, JEE Main
The pressure in the tyre of a car is four times the atmospheric pressure at $300 \mathrm{~K}$. If this tyre suddenly bursts, its new temperature will be $(\gamma=1.4)$

Options:
A $300(4)^{1.4 / 0.4}$
B $300\left(\frac{1}{4}\right)^{-0.4 / 1.4}$
C $300(2)^{-0.4 / 1.4}$
D $300(4)^{-0.4 / 1.4}$

The correct answer is:
$300(4)^{-0.4 / 1.4}$

Solution:
For adiabatic process $\frac{T^\gamma}{P^{\gamma-1}}=$ constant
$\Rightarrow \frac{T_2}{T_1}=\left(\frac{P_1}{P_2}\right)^{\frac{1-\gamma}{\gamma}} \Rightarrow \frac{T_2}{300}=\left(\frac{4}{1}\right)^{\frac{(1-1.4)}{1.4}} \Rightarrow T_2=300(4)^{-\frac{0.4}{1.4}}$

Topic: Physics, Thermodynamics, JEE Main
The pressure of a gas changes linearly with volume from A to B as shown in figure. If no heat is supplied to or extracted from the gas then change in the internal energy of the gas will be

Options:
A 6 J
B Zero
C -4.5 J
D 4.5 J

The correct answer is:
4.5 J

Solution:
Given here, ∆Q=0

From first law of thermodynamics, ∆Q=∆U+W, where, ∆U is change in internal energy and W is work done.

We have, ∆U=-W

W=Area under P-V curve=∫PdV

Here, volume decreases, hence work done by system will be negative.

∆U=-W=30×103×150×10-6

=4500×10-3

=4.5 J

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2023 (31 Jan Shift 1)
The pressure of an ideal gas is increased by keeping the temperature constant. The kinetic energy of molecules

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

The correct answer is:
Increases

Solution:
p=23E ⇒p∝E
So, as p increases, E also increases.

Topic: Physics, Kinetic Theory of Gases, NEET
The pressure of an ideal gas varies with volume as $P=\alpha V$, where $\alpha$ is a constant. One mole of the gas is allowed to undergo expansion such that its volume becomes ' $m$ ' times its initial volume. The work done by the gas in the process is

Options:
A
$\frac{\alpha V}{2}\left(m^2-1\right)$

B
$\frac{\alpha^2 V^2}{2}\left(m^2-1\right)$

C
$\frac{\alpha}{2}\left(m^2-1\right)$

D
$\frac{\alpha V^2}{2}\left(m^2-1\right)$

The correct answer is:
$\frac{\alpha V^2}{2}\left(m^2-1\right)$

Solution:
Given $P=\alpha V$
Work done, $w=\int_V^{m V} P d V$
$=\int_V^{m V} \alpha V d V=\frac{\alpha V^2}{2}\left(m^2-1\right)$.

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2012 (19 May Online)
The pressure of H2 required to make the potential of hydrogen electrode zero in pure water at 298 K is:

Options:
A 10-14 atm
B 10-12Â atm
C 10-10Â atm
D 10-4Â atm

The correct answer is:
10-14 atm

Solution:
Pt,H2gH+ Hydrogen electrode EH2/H+o=0.0 Volt
H2gâ2Haq++2e-
H+=10-7 M at 25oC (for Pure water)
E=-0.05912logâ¡H+2PH2
E=0=logâ¡H+PH2=0
â´ H + 2 = P H 2 = 10 â7 2
â´PH2=10-14 atm

Topic: Chemistry, Electrochemistry, JEE Main
The pressure on a square plate is measured by measuring the force on the plate and the length of the sides of the plate. If the maximum errors in the measurement of force and length are respectively, 4% and 2%, then the maximum error in the measurement of pressure is

Options:
A 1%
B 2%
C 6%
D 8%

The correct answer is:
8%

Solution:
∴p=FA=Fl2, so maximum error in pressure p is
Δpp×100max=ΔFF×100+2Δll×100
=4%+2×2%
=8%

Topic: Physics, Units and Dimensions, JEE Main
The pressure P1 and density d1 of diatomic gas γ=75 changes suddenly to P2>P1 and d2 respectively during an adiabatic process. The temperature of the gas increases and becomes____times of its initial temperature.

(given d2 d1=32)

Options:

The correct answer is:
4

Solution:
For adiabatic process, PVγ= constant

We know, density d=massVolume.

Then, we have Pdγ= constant

Given here, d2 d1=32 and γ=75.

The ratio P1P2=d1 d2γ=13275=1128

The ratio of temperature using ideal gas equation is T1 T2=P1 V1P2 V2=1128×32=14

Thus, T2=4T1.

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2022 (29 Jul Shift 1)
The pressure (P) and temperature (T) relationship of an ideal gas obeys the equation PT2= constant. The volume expansion coefficient of the gas will be :

Options:
A 3T2
B 3T2
C 3T3
D 3T

The correct answer is:
3T

Solution:
Given here,

PT2=constant ...1

Using ideal gas equation,

PV=nRT

We have, P=nRTV

Putting in equation 1,

PT2=nRTV×T2=constant

⇒T3=KV ...2

Differentiating the equation with respect to T,

So, ddTKV=3T2

⇒KdVdT=3T2

⇒dV=3T2KdT

⇒γV=3T2Kas dV=γVdT

⇒γ=3T2KV=3T2T3=3T

Topic: Physics, Kinetic Theory of Gases, JEE Main
The pressure (P) and temperature (T) relationship of an ideal gas obeys the equation PT2= constant. The volume expansion coefficient of the gas will be :

Options:
A 3T2
B 3T2
C 3T3
D 3T

The correct answer is:
3T

Solution:
Given here,

PT2=constant ...1

Using ideal gas equation,

PV=nRT

We have, P=nRTV

Putting in equation 1,

PT2=nRTV×T2=constant

⇒T3=KV ...2

Differentiating the equation with respect to T,

So, ddTKV=3T2

⇒KdVdT=3T2

⇒dV=3T2KdT

⇒γV=3T2Kas dV=γVdT

⇒γ=3T2KV=3T2T3=3T

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2023 (30 Jan Shift 1)
The pressure $p$ of a gas is plotted against its absolute temperature $T$ for two different constant volumes, $V_1$ and $V_2$. When $V_1 \gt V_2$, the

Options:
A Curves have the same slope and do not intersect
B Curves must intersect at some point other than $T=0$
C Curve for $V_2$ has a greater slope than that for $V_1$
D Curve for $V_1$ has a greater slope than that for $V_2$

The correct answer is:
Curve for $V_2$ has a greater slope than that for $V_1$

Solution:
At constant volumes $P \propto T$
$P=$ constant $T ; \quad P V=n R T \therefore P=\frac{n R}{V} T$
slope $=m=\frac{\eta R}{V} \because V_2 \lt V_1$
$\frac{m_1}{m_2}=\frac{V_2}{V_1} \therefore m_1 \lt m_2$ is curve for $V_2$ has a greater slope than for $V_1$

Topic: Chemistry, States of Matter, JEE Main
The pressure $p$, volume $V$ and temperature $T$ for a certain gas are related by $p=\frac{A T-B T^{2}}{V}$ where $A$ and $B$ are constants. The work done by the gas when the temperature changes from $T_{1}$ to $T_{2}$ while the pressure remains constant, is given by

Options:
A $A\left(T_{2}-T_{1}\right)+B\left(T_{2}^{2}-T_{1}^{2}\right)$
B $\frac{A\left(T_{2}-T_{1}\right)}{V_{2}-V_{1}}-\frac{B\left(T_{2}^{2}-T_{1}^{2}\right)}{V_{2}-V_{1}}$
C $ A\left(\tau_{2}-T_{1}\right)-\frac{B}{2}\left(T_{2}^{2}-T_{1}^{2}\right)$
D $\frac{A\left(T_{2}-T_{1}^{2}\right)}{V_{2}-V_{1}}$

The correct answer is:
$ A\left(\tau_{2}-T_{1}\right)-\frac{B}{2}\left(T_{2}^{2}-T_{1}^{2}\right)$

Solution:
Given $P=\frac{A T-B T^{2}}{V}$
$\begin{array}{lc}\Rightarrow & P V=A T-B T^{2} \\ \Rightarrow & P \Delta V=A \Delta T-B T \Delta T\end{array}$
On integrating, we get Work $=\int P d V=A \int_{T_{1}}^{T_{2}} d T-B \int_{T_{1}}^{T_{2}} T d T$
$=A\left(T_{2}-T_{1}\right)-\frac{B}{2}\left[\left(T_{2}\right)^{2}-\left(T_{1}\right)^{2}\right]$

Topic: Physics, Thermodynamics, JEE Main
The pressure $p$, volume $V$ and temperature $T$ for a certain gas are related by $p=\frac{A T-B T^{2}}{V}$ where $A$ and $B$ are constants. The work done by the gas when the temperature changes from $T_{1}$ to $T_{2}$ while the pressure remains constant, is given by

Options:
A $A\left(T_{2}-T_{1}\right)+B\left(T_{2}^{2}-T_{1}^{2}\right)$
B $\frac{A\left(T_{2}-T_{1}\right)}{V_{2}-V_{1}}-\frac{B\left(T_{2}^{2}-T_{1}^{2}\right)}{V_{2}-V_{1}}$
C $ A\left(\tau_{2}-T_{1}\right)-\frac{B}{2}\left(T_{2}^{2}-T_{1}^{2}\right)$
D $\frac{A\left(T_{2}-T_{1}^{2}\right)}{V_{2}-V_{1}}$

The correct answer is:
$ A\left(\tau_{2}-T_{1}\right)-\frac{B}{2}\left(T_{2}^{2}-T_{1}^{2}\right)$

Solution:
Given $P=\frac{A T-B T^{2}}{V}$
$\begin{array}{lc}\Rightarrow & P V=A T-B T^{2} \\ \Rightarrow & P \Delta V=A \Delta T-B T \Delta T\end{array}$
On integrating, we get Work $=\int P d V=A \int_{T_{1}}^{T_{2}} d T-B \int_{T_{1}}^{T_{2}} T d T$
$=A\left(T_{2}-T_{1}\right)-\frac{B}{2}\left[\left(T_{2}\right)^{2}-\left(T_{1}\right)^{2}\right]$

Topic: Physics, Thermodynamics, WBJEE, WBJEE 2015
The pressure P, volume V and temperature T of a certain material are related by the equation P=Î±T2/V. The work done by the material at constant pressure, when the temperature changes from T0 to 2T0, is

Options:
A 3αT02
B 5αT02
C 23αT02
D 7αT02

The correct answer is:
3αT02

Solution:
The given relation is
P = Î± T 2 V â´ V = Î± T 2 P
As pressure is kept constant, dV = 2 Î± T P dT
W = â« P Â· dV = â« T 0 2T 0 P 2 Î± T P dT = 2 Î± T 2 2 T 0 2T 0 = 3 Î± T 0 2

Topic: Physics, Thermodynamics, JEE Main
The pressure that has to be applied to the ends of a steel wire of length 10 cm to keep its length constant when its temperature is raised by 100 °C is :
(For steel, Young's modulus is 2×1011 N m-2 and coefficient of thermal expansion is 1.1×10-5 K-1)

Options:
A 2.2×108 Pa
B 2.2×109 Pa
C 2.2×107 Pa
D 2.2×106 Pa

The correct answer is:
2.2×108 Pa

Solution:
Young's modulus, Y=FALΔL
∴FA=YΔLL=YαΔT
∴P=2×​ 1011×1.1×10-5×102
∴ P=2.2×108 Pa

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2014 (06 Apr)
The pressures inside two soap bubbles are 1.02 atm and 1.03 atm respectively. The ratio of their volumes is

Options:
A 102Â :Â 103
B 8 : 27
C 1033Â :Â 1023
D 27 : 8

The correct answer is:
27 : 8

Solution:
Pin- Patm= excess pressure=4TR
Patm=1 atm
∴ R1R2=ΔP2ΔP1= 32
∴ V1V2= R13R23=278

Topic: Physics, Mechanical Properties of Fluids, JEE Main
The primary and secondary coils of a transformer have 50 and 1500 turns respectively. If the magnetic flux $\phi$ linked with the primary coil is given by $\phi=\phi_v+4 t$, where $\phi$ is in webers, $t$ is time in seconds and $\phi_v$ is a constant, the output voltage across the output voltage across the secondary coil is:

Options:
A 120 volts
B 220 volts
C 30 volts
D 90 volts.

The correct answer is:
120 volts

Solution:
The magnetic flux linked with the primary coil is given by
\(\phi=\phi_0+4 t\)
So, voltage across primary
\(\begin{aligned}
& V_p=\frac{d \phi}{d t}=\frac{d}{d t}\left(\phi_0+4 t\right) \\
& \left.=4 \text { volt (as } \phi_0=\text { constant }\right)
\end{aligned}\)
Also, we have
\(N_p=50 \text { and } N_s=1500\)
From relation,
\(\begin{aligned}
& \frac{V_s}{V_p}=\frac{N_s}{N_p} \\
& \text {or } V_s=V_p \frac{N_s}{N_p} \\
& =4\left(\frac{1500}{50}\right) \\
& =120 \mathrm{~V}
\end{aligned}\)

Topic: Physics, Alternating Current, JEE Main
The primary and secondary valencies of cobalt respectively in Co(NH3)5ClCl2 are:

Options:
A 3 and 5
B 2 and 6
C 2 and 8
D 3 and 6

The correct answer is:
3 and 6

Solution:
Co(NH3)5ClCl2

The primary valency is the ionisable valency. Oxidation number of Co is +3. So, primary valency is 3. Secondary valency refers the coordination number. It is an octahedral complex so, secondary valency is 6.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2023 (24 Jan Shift 1)
The primary pollutant that leads to photochemical smog is:

Options:
A nitrogen oxides
B ozone
C acrolein
D Sulphur dioxide

The correct answer is:
nitrogen oxides

Solution:
Nitrogen oxides and hydrocarbons (unburnt fuel) are primary pollutant that leads to photochemical smog.

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2019 (12 Apr Shift 2)
The primary valence of the metal ion in the co-ordination compound $\mathrm{K}_2\left[\mathrm{Ni}(\mathrm{CN})_4\right]$ is

Options:
A Four
B Zero
C Two
D Six

The correct answer is:
Two

Solution:
Primary valencies are also known as the oxidation state of the central metal atom.
\(K_2\left[N i(C N)_4\right]\). Let \(x\) be the oxidation state of \(\mathrm{Ni}\).
Now,
\(2 \times 1+x+4 \times(-1)=0 ; \Rightarrow x=+2\)

Topic: Chemistry, Coordination Compounds, JEE Main
The primary winding of a transformer has 100 turns and its secondary winding has 200 turns. The primary is connected to an AC supply of 120 V and the current flowing in it is 10 A. The voltage and the current in the secondary are

Options:
A 240 V, 5 A
B 240 V, 10 A
C 60 V, 20 A
D 120 V, 20 A

The correct answer is:
240 V, 5 A

Solution:
EsEp=nsnp or Es=EpÃnsnp
â´ Es=120 Ã200100=240 V
IpIs=nsnp or Is=Ipnpns
â´ Is=10100200=5 A

Topic: Physics, Alternating Current, JEE Main
The primary winding of transformers has \( 500 \) turns whereas its secondary has \( 5000 \) turns. The primary is connected to an \( A . C \). supply of \( 20 \mathrm{~V}, 50 \mathrm{~Hz} \). The secondary will have an output of

Options:
A \( 2 \mathrm{~V}, 5 \mathrm{~Hz} \)
B \( 200 \mathrm{~V}, 500 \mathrm{~Hz} \)
C \( 2 \mathrm{~V}, 50 \mathrm{~Hz} \)
D \( 200 \mathrm{~V}, 50 \mathrm{~Hz} \)

The correct answer is:
\( 200 \mathrm{~V}, 50 \mathrm{~Hz} \)

Solution:
Given,

Vp=20 V,

Np=500,

Ns=5000,

f=50 Hz,

We know that for a transformer,

NpNs=VpVs

Putting in the given values,

110=20Vs

⇒Vs=200 V

Also, we know that transformer doesn't affect the frequency of the source.

Topic: Physics, Alternating Current, JEE Main
The principal organ in humans involved in the absorption of nutrients like glucose, fructose, fatty acids, glycerol and amino acids is

Options:
A stomach
B small intestine
C mouth
D large intestine

The correct answer is:
small intestine

Solution:
The small intestine is the main organ for absorption of nutrients. It has a large number of villi and microvilli which increase the surface area of intestine for the absorption of food. Microbial eye forms as a result of holding of plasma membrane of cells, giving it a brush-bordered appearance.

Topic: Biology, Digestion and Absorption, NEET
The principal organic product formed in the following reaction is
$\mathrm{CH}_2=\mathrm{CH}\left(\mathrm{CH}_2\right)_8 \mathrm{COOH}+\mathrm{HBr} \xrightarrow{\text { peroxide }}$

Options:
A $\mathrm{CH}_3 \mathrm{CHBr}\left(\mathrm{CH}_2\right)_8 \mathrm{COOH}$
B $\mathrm{CH}_2=\mathrm{CH}\left(\mathrm{CH}_2\right)_8 \mathrm{COBr}$
C $\mathrm{CH}_2 \mathrm{Br} \mathrm{CH}_2\left(\mathrm{CH}_2\right)_8 \mathrm{COOH}$
D $\mathrm{CH}_2=\mathrm{CH}\left(\mathrm{CH}_2\right), \mathrm{CHBr} \mathrm{COOH}$

The correct answer is:
$\mathrm{CH}_2 \mathrm{Br} \mathrm{CH}_2\left(\mathrm{CH}_2\right)_8 \mathrm{COOH}$

Solution:
$\mathrm{CH}_2=\mathrm{CH}\left(\mathrm{CH}_2\right)_8 \mathrm{COOH}+\mathrm{HBr} \xrightarrow{\text { peroxide }} \mathrm{CH}_2 \mathrm{BrCH}_2\left(\mathrm{CH}_2\right)_8 \mathrm{COOH}$

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
The principal section of a glass prism is an isosceles triangle \( A B C \) with \( A B=A C . \) The face \( A C \) is silvered. A ray of light is incident normally on the face \( A B \) and after tworeflections, it emerges from the base \( B C \) perpendicular to the base. Angle \( B A C \) of the prism is

Options:
A \( 30^{\circ} \)
B \( 36^{\circ} \)
C \( 60^{\circ} \)
D \( 72^{\circ} \)

The correct answer is:
\( 36^{\circ} \)

Solution:
From the figure
i1=90°-90°-A=A
and ∝ =90°-2i1=90°-2A
∴ i2=90°-α=90°-90°-2A=2A
∴ β=90°-i2=90°-2A
From the geometry of the figure
A+2A+2A=180°
∴ A=36°

Topic: Physics, Ray Optics, JEE Main
The principal value of
$\sin ^{-1}\left(-\frac{1}{2}\right)$ is

Options:
A $\frac{\pi}{3}$
B $\frac{\pi}{6}$
C $-\frac{\pi}{3}$
D $-\frac{\pi}{6}$

The correct answer is:
$-\frac{\pi}{6}$

Solution:
The principal value of
$\sin ^{-1}\left(-\frac{1}{2}\right)=\sin ^{-1} \sin \left(-30^{\circ}\right)=-\frac{\pi}{6}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The principal value of $\sin ^{-1}\left(-\frac{1}{2}\right)$ is

Options:
A $\frac{\pi}{3}$
B $\frac{\pi}{6}$
C $-\frac{\pi}{3}$
D $-\frac{\pi}{6}$

The correct answer is:
$-\frac{\pi}{6}$

Solution:
$\sin ^{-1}\left(-\frac{1}{2}\right)=-\sin ^{-1} \frac{1}{2}=\frac{-\pi}{6}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The principal value of
$\sin ^{-1}\left(-\frac{\sqrt{3}}{2}\right)$

Options:
A $\frac{-2 \pi}{3}$
B $\frac{-\pi}{3}$
C $\frac{4 \pi}{3}$
D $\frac{5 \pi}{3}$

The correct answer is:
$\frac{-\pi}{3}$

Solution:
$\sin ^{-1}\left(\frac{-\sqrt{3}}{2}\right)=-\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)=-\frac{\pi}{3}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The principal value of $\sin ^{-1}\left[\sin \left(\frac{2 \pi}{3}\right)\right]$

Options:
A $-\frac{2 \pi}{3}$
B $\frac{2 \pi}{3}$
C $\frac{4 \pi}{3}$
D None of these

The correct answer is:
None of these

Solution:
$\begin{aligned}
& \text {The principal value of } \sin ^{-1}\left[\sin \left(\pi-\frac{2 \pi}{3}\right)\right] \\
& =\sin ^{-1} \sin \left(\frac{\pi}{3}\right)=\frac{\pi}{3}
\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The principal value of
$\sin ^{-1}\left(\sin \frac{5 \pi}{3}\right)$ is

Options:
A $\frac{5 \pi}{3}$
B $-\frac{5 \pi}{3}$
C $-\frac{\pi}{3}$
D $\frac{4 \pi}{3}$

The correct answer is:
$-\frac{\pi}{3}$

Solution:
$\sin ^{-1}\left(\sin \frac{5 \pi}{3}\right)=\sin ^{-1}\left(-\frac{\sqrt{3}}{2}\right)=-\frac{\pi}{3}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The principal value of $\sin ^{-1}\left(\sin \frac{5 \pi}{3}\right)$ is

Options:
A $\frac{5 \pi}{3}$
B $-\frac{5 \pi}{3}$
C $-\frac{\pi}{3}$
D $\frac{4 \pi}{3}$

The correct answer is:
$-\frac{\pi}{3}$

Solution:
$\sin ^{-1}\left(\sin \frac{5 \pi}{3}\right)=\sin ^{-1}\left(-\frac{\sqrt{3}}{2}\right)=-\frac{\pi}{3}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The principal value of tan-1cot43π4 is

Options:
A π4
B -π4
C 3π4
D - 3 π 4

The correct answer is:
-π4

Solution:
-π2<tan-1x<π2

cot43π4=cot10π+3π4

=cot3π4

=-1

tan-1cot43π4=tan-1-1

=-π4

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main, JEE Main 2014 (19 Apr Online)
The principle of column chromatography is:

Options:
A Differential adsorption of the substances on the solid phase.
B Differential absorption of the substances on the solid phase.
C Gravitational force.
D Capillary action

The correct answer is:
Differential adsorption of the substances on the solid phase.

Solution:
Principal of chromatography is based on adsorption of the substance on the solid phase.

A Chromatography column is a device used in chromatography for the separation of chemical compounds. A chromatography column contains the stationary phase, allowing the mobile phase to pass through it. Chromatography columns of different types are used in both gas and liquid chromatography.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2019 (10 Apr Shift 1)
The principle of conservation of linear momentum can be strictly applied during a collision between two particles provided the time of impact

Options:
A is extremely small
B is moderately small
C is extremely large
D depends on particular case

The correct answer is:
is extremely small

Solution:
In physics, collision does not means that one particle strike another particle. Infact, two particles may not even touch each other \& may still said to be colliding.
The necessary requirements of collision are
(i) A large force for a relatively short time (i.e., an impulse) acts on each colliding particle.
(ii) The motion of the particles (at least one of the particle) is changed abruptly.
(iii) The total momentum (as also the total energy) of particles remains conserved.

Topic: Physics, Center of Mass Momentum and Collision, VITEEE, VITEEE 2018