datavorous/entrance-exam-dataset huggingface dataset:
The real number $x$ when added to its inverse gives the minimum value of the sum at $x$ equal to

Options:
A
$-2$

B
2

C
1

D
$-1$

The correct answer is:
1

Solution:
$y=x+\frac{1}{x} \text { or } \frac{d y}{d x}=1-\frac{1}{x^2}$
For $\max$. or $\min ., 1-\frac{1}{x^2}=0 \Rightarrow x=\pm 1$
$\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}=\frac{2}{\mathrm{x}^3} \Rightarrow\left(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\right)_{\mathrm{x}=2}=2(+$ ve minima)
Therefore $x=1$

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2003
The real part of $(1-\cos \theta+2 i \sin \theta)^{-1}$ is

Options:
A $\frac{1}{3+5 \cos \theta}$
B $\frac{1}{5-3 \cos \theta}$
C $\frac{1}{3-5 \cos \theta}$
D $\frac{1}{5+3 \cos \theta}$

The correct answer is:
$\frac{1}{5+3 \cos \theta}$

Solution:
$\begin{aligned} & \{(1-\cos \theta)+i .2 \sin \theta\}^{-1}=\left\{2 \sin ^2 \frac{\theta}{2}+i .4 \sin \frac{\theta}{2} \cos \frac{\theta}{2}\right\}^{-1} \\ = & \left(2 \sin \frac{\theta}{2}\right)^{-1}\left\{\sin \frac{\theta}{2}+i .2 \cos \frac{\theta}{2}\right\}^{-1} \\ = & \left(2 \sin \frac{\theta}{2}\right)^{-1} \frac{1}{\sin \frac{\theta}{2}+i .2 \cos \frac{\theta}{2}} \times \frac{\sin \frac{\theta}{2}-i .2 \cos \frac{\theta}{2}}{\sin \frac{\theta}{2}-i .2 \cos \frac{\theta}{2}} \\ = & \frac{\sin \frac{\theta}{2}-i .2 \cos \frac{\theta}{2}}{2 \sin \frac{\theta}{2}\left(\sin ^2 \frac{\theta}{2}+4 \cos ^2 \frac{\theta}{2}\right)} .\end{aligned}$
it's real part
$=\frac{\sin \frac{\theta}{2}}{2 \sin \frac{\theta}{2}\left(1+3 \cos ^2 \frac{\theta}{2}\right)}=\frac{1}{2\left(1+3 \cos ^2 \frac{\theta}{2}\right)}=\frac{1}{5+3 \cos \theta}$

Topic: Mathematics, Complex Number, JEE Main
The real part of the complex number z satisfying z-1-2i≤1 and having the least positive argument, is

Options:
A 45
B 85
C 65
D 75

The correct answer is:
85

Solution:
Here, z-1-2i=1 represents a circle with centre 1,2 and radius 1 unit.

The complex number z=x+iy satisfying the given inequality and having the least positive argument is the point of contact of the tangent from the origin to the circle with the least positive slope.

From the diagram,

tan⁡ϕ=12
∴tan⁡2ϕ=2tanϕ1-tan2ϕ=43

For the least positive argument, argz=θ (let)
tan⁡θ=tanπ2-2ϕ=cot2ϕ=34

Also, from the diagram,

x2+y2=4 and tanθ=yx=34

i.e. y=34x

⇒x2+9x216=4⇒x=85

Hence, for the least positive argument, the real part of z is equal to 85

Topic: Mathematics, Complex Number, JEE Main
The real part of the principle value of $2^{-i}$ is

Options:
A $\sin (\log 2)$
B $\cos \left(\frac{1}{\log 2}\right)$
C $\cos \left[\log \left(\frac{1}{2}\right)\right]$
D $\cos (\log 2)$

The correct answer is:
$\cos \left[\log \left(\frac{1}{2}\right)\right]$

Solution:
Let $z=2^{-i}$
Taking $\log$ on both sides, we get $\begin{aligned} & \log z=\log \left(2^{-i}\right) \\ \Rightarrow & \log z=-i \log 2 \\ \Rightarrow & \log z=i \log \left(\frac{1}{2}\right) \\ \Rightarrow & z=e^{i \log (1 / 2)} \quad\left(\because e^{i \theta}=\cos \theta+i \sin \theta\right) \\ \Rightarrow & z=\cos \left(\log \frac{1}{2}\right)+i \sin \left(\log \frac{1}{2}\right) \\ \therefore \text { The real part of } z=2^{-i} \text { is } \cos \left(\log \frac{1}{2}\right) \end{aligned}$

Topic: Mathematics, Complex Number, JEE Main
The real solutions of the equation tan -1 x x + 1 + sin -1 x 2 + x + 1 = Ï 2 are

Options:
A -1, 0
B 0, 1
C -1, 1
D -1, 2

The correct answer is:
-1, 0

Solution:
tan -1 x x + 1 + sin -1 x 2 + x + 1 = π 2
⇒ tan -1 x 2 + x + sin -1 x 2 + x + 1 = π 2
This equation holds true, if
x 2 + x ≥ 0 and 0 ≤ x 2 + x + 1 ≤ 1
Now, x 2 + x ≥ 0 and 0 ≤ x 2 + x + 1 ≤ 1
⇒ x 2 + x ≥ 0 and x 2 + x + 1 ≤ 1 ∵ x 2 + x + 1 > 0 for all x
⇒ x 2 + x ≥ 0 and x 2 + x ≤ 0
⇒ x 2 + x = 0 ⇒ x = 0 , - 1 .
Clearly, these two values satisfy the given equation. Hence, x=-1, 0 are the solutions of the given equation.

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The real valued function fx=cosec-1xx-[x], where x denotes the greatest integer less than or equal to x, is defined for all x belonging to:

Options:
A all reals except integers
B all non-integers except the interval -1,1
C all integers except 0,-1,1
D all reals except the Interval -1,1

The correct answer is:
all non-integers except the interval -1,1

Solution:
Given fx=cosec-1x{x}

Domain of cosec-1x∈(-∞,-1]∪[1,∞)

We know, {x}≠0, so x≠integers.

So, x belongs to all non-integers except the interval -1,1.

Topic: Mathematics, Functions, JEE Main, JEE Main 2021 (18 Mar Shift 1)
The real values of x and y for which the equation ( x 4 +2xi)−(3 x 2 +yi)= (3−5i)+(1+2yi) is satisfied, are

Options:
A x=2,y=3
B x=−2,y= 1 3
C

x=±2 and y=3, 1 3

D None of these

The correct answer is:

x=±2 and y=3, 1 3

Solution:
Given equation ( x 4 +2xi)−(3 x 2 +yi)=(3−5i)+(1+2yi)

⇒ ( x 4 −3 x 2 )+i(2x−3y)=4−5i

Equating real and imaginary parts, we get x 4 −3 x 2 =4 ......(i)

and 2x−3y=−5 .....(ii)

From (i) and (ii), we get

x=±2 and y=3, 1 3

Topic: Mathematics, Complex Number, JEE Main
The real values of x and y satisfying the equation 1+ix-2i3+i+2-3iy+i3-i=i are

Options:
A x=-1,Â y=3
B x=3, y=-1
C x=0,Â y=1
D x=1,Â y=0

The correct answer is:
x=3, y=-1

Solution:
Given, 1+ix-2i3+i+2-3iy+i3-i=i
â ( 1+i )( 3âi )xâ2i( 3âi )+( 2â3i )( 3+i )y+i( 3+i ) ( 9+1 ) =i
â 4+2ix+9-7iy-3i-3=10i
â( 4x+9yâ3 )+i( 2xâ7y-3 )=10i
Equating the real and imaginary parts, we get
2x-7y=13 and 4x+9y=3, hence x=3 and y=-1

Topic: Mathematics, Complex Number, JEE Main
The rear side of a truck is open and a box of mass 2 kg is placed on the truck 8 m away from the open end. The truck starts from rest with an acceleration of 2 m s-2 on a straight road. The box will fall off the truck when it is at distance from the starting point equal to (in meter)

(Given, Î¼=0.1 and g=10 m s-2 ).

Options:

The correct answer is:
16

Solution:
Let a be the acceleration of truck,
For block,
ma-Î¼mg=ma0
a0=a-Î¼g
=2-0.1Ã10=1 m/s2
8=12a0t2
=12Ã1Ãt2
â t=4 s
The block will fall off the truck after 4 s.
For truck,
Distance travelled by the truck in this time.
s=12at2=12Ã2Ã42=16 m

Topic: Physics, Laws of Motion, JEE Main
The reason for almost doubling the rate of reaction on increasing the temperature of the reaction system by \( 10^{\circ} \mathrm{C} \) is:

Options:
A The value of threshold energy increases
B Collision frequency increases
C The fraction of the molecules having energy equal to threshold energy increases
D Activation energy decreases

The correct answer is:
Collision frequency increases

Solution:
On increasing the temperature the average kinetic energy of the reacting molecules increases and hence, number of collisions increases. So, the rate of reaction will also be increased.

According to Arrhenius equation:

K = Ae-Ea/RTlog K = log A -Ea2.303RT Remember log1010 = 1

If you increases temperature by 10°C at same A and Ea Rate increases approximately double.

Topic: Chemistry, Chemical Kinetics, JEE Main
The reason for double helical structure of DNA is operation of

Options:
A
dipole-dipole interaction

B
hydrogen bonding

C
electrostatic attractions

D
van der Waals' forces

The correct answer is:
hydrogen bonding

Solution:
No Solution Available

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2003
The reason for double helical structure of DNA is the operation of :

Options:
A Van der Waals forces
B Electrostatic attractions
C Hydrogen bonding
D Dipole - Dipole interactions

The correct answer is:
Hydrogen bonding

Solution:
DNA is a chain of two strands of nucleotides that are linked to one another through hydrogen bonds and arranged into a structure known as a double-helix.

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2014 (19 Apr Online)
The reason for "drug induced poisoning" is:

Options:
A binding irreversibly to the active site of the enzyme.
B binding at the allosteric sites of the enzyme.
C binding reversibly at the active site of the enzyme.
D brining conformational change in the binding site of enzyme.

The correct answer is:
binding at the allosteric sites of the enzyme.

Solution:
The reason for drug induced poisoning is the binding of drug, irreversibly, to the active site of the enzyme. If this happens, then body degenerates the enzyme and releases a new enzyme.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main, JEE Main 2017 (08 Apr Online)
The reason for shining of air bubble in water is

Options:
A Diffraction of light
B Dispersion of light
C Scattering of light
D Total internal reflection of light

The correct answer is:
Total internal reflection of light

Solution:
CONCEPT:
- Optical fibers are transparent fibers and act as a light pipe to transmit light between its two ends. They are made up of silicon dioxide.
- The total internal reflection occurs when the angle of incidence is greater than the critical angle.
- In optical fiber glasses of a high and lower index are assembled in the precise order.
- The light comes in from one end of the fiber and after thousands of successive internal reflections, the light reaches the opposite end of the fiber with almost zero loss.
EXPLANATION:
- The refractive index of the material of the core $\left(\mu_1\right)$ is higher than that of the cladding $\left(\mu_2\right)$. Therefore option 1 is correct.
- When the light is incident on one end of the fiber at a small angle, the light passes inside, undergoes repeated total internal reflections along with the fiber, and finally comes out.
- Hence, the function of the cladding is to provide a lower refractive index at the core interface in order to cause reflection within the core so that light waves are transmitted through the fiber.
- Thus, cladding is used in the optical fiber in order to cause Total internal reflection.

Topic: Physics, Ray Optics, JEE Main
The reason of bleaching properties of \( \mathrm{H}_{2} \mathrm{O}_{2} \) is

Options:
A Unstable nature
B Acidic nature
C Reducing nature
D Oxidising nature

The correct answer is:
Oxidising nature

Solution:
Oxidising nature

Topic: Chemistry, Hydrogen, JEE Main
The recent record of __________ years old viable seed is of the date palm, Phoenix dactylifera, discovered during archaeological excavation at King Herod's palace near the dead sea.

Options:
A 1000
B 2000
C 3000
D 500

The correct answer is:
2000

Solution:
In a few species, seeds lose viability within a few months. Seeds of a large number of species remain alive for several years. Some seeds remain alive for hundreds of years. A recent record of 2000-year-old viable seed is of the date palm, Phoenix dactylifera, discovered during the archeological excavation at King Herodâs palace near the Dead Sea.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
The reciprocal of the eccentricity of rectangular hyperbola, is

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

The correct answer is:
$\frac{1}{\sqrt{2}}$

Solution:
Since eccentricity of rectangular hyperbola is $\sqrt{2}$, So reciprocal \(=\frac{1}{\sqrt{2}}\).

Topic: Mathematics, Hyperbola, JEE Main
The recoil speed of a hydrogen atom after it emits a photon in going from n=5 state to n=5 state will be

Options:
A 4.17 m s-1
B 2.19 m s-1
C 4.34 m s-1
D 3.25 m s-1

The correct answer is:
4.17 m s-1

Solution:
ΔE Releases when photon going from n=5 to n=ΔE

=13.6-0.54 eV=13.06 eV

Pi=Pf (By linear momentum conservation)

0=hλ-Mv=VRecoil=hλM ...1

& ΔE=hcλ=hcλM×M⇒McVRecoil

VRecoil =ΔEMc=13.06×1.6×10-191.67×10-27×3×108=4.17 m s-1

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2021 (26 Feb Shift 2)
The recommended concentration of fluoride ion in drinking water is up to 1 ppm as fluoride ion is required to make teeth enamel harder by converting 3Ca3PO42.CaOH2 to:

Options:
A 3CaOH2.CaF2
B CaF2
C 3CaF2.CaOH2
D 3Ca3PO42.CaF2

The correct answer is:
3Ca3PO42.CaF2

Solution:
The recommended concentration of fluoride ion in drinking water is up to 1 ppm as fluoride ion is required to make teeth enamel harder by converting

[ 3 Ca3(PO4)2.Ca(OH)2 ] to [ 3 Ca3(PO4)2.CaF2 ].

[ 3 Ca3(PO4)2.Ca(OH)2 ] + 2 F- → [ 3 Ca3(PO4)2.CaF2 ]

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2018 (08 Apr)
The recommended concentration of fluoride ion in drinking water is up to 1 ppm as fluoride ion is required to make teeth enamel harder by converting [3Ca3PO42.CaOH2] to

Options:
A [3CaOH)2.CaF2
B CaF2
C 3CaF2.CaOH2
D [3Ca3PO4)2.CaF2

The correct answer is:
[3Ca3PO4)2.CaF2

Solution:
[3Ca3(PO4)2.CaOH)2+2F-→3Ca3PO42.CaF2+2OH-

Topic: Chemistry, Environmental Chemistry, JEE Main
The red colour of oxyhaemoglobin is mainly due to

Options:
A d-d transition
B metal to ligand charge transfer transition
C ligand to metal charge transfer transition
D intraligand π-π* transition

The correct answer is:
d-d transition

Solution:
When oxygen is bound to Fe2+ in heme, all the d-electrons of the Fe atom are forced into the three lower energy t2g orbitals and it produces low spin oxyhaemoglobin.
In oxyhaemoglobin, Fe2+=Ar3d6
The sharp high energy transition between t2g and eg levels of the d-electrons is responsible for the bright red colour of oxyhaemoglobin.

Topic: Chemistry, Coordination Compounds, JEE Main
The red list contains data or information on

Options:
A All economically important plants.
B Plants whose products are in international trade.
C Threatened species.
D Marine vertebrates only.

The correct answer is:
Threatened species.

Solution:
IUCN [International Union Conservation of Nature and Natural Resources] (or) WCU [World Conservation Union] maintains a red list which is a catalogue for threatened species.

Topic: Biology, Biodiversity and its Conservation, NEET, NEET 2016 (Phase 2)
The reddish brown coloured gas formed when nitric oxide is oxidised by air is

Options:
A \( \mathrm{N}_{2} \mathrm{O}_{5} \)
B \( \mathrm{N}_{2} \mathrm{O}_{4} \)
C \( \mathrm{NO}_{2} \)
D \( \mathrm{N}_{2} \mathrm{O}_{3} \)

The correct answer is:
\( \mathrm{NO}_{2} \)

Solution:
The reddish brown coloured gas formed when nitric oxide is oxidised by air is NO2.

2NO+O2→2NO2.

Nitric oxide is colourless, flammable gas with a slight odour. Nitrogen dioxide is a deep red-orange gas that is poisonous but not flammable.

NO2 is an intermediate in the industrial synthesis of nitric acid.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
The redox reaction among the following is

Options:
A formation of ozone from atmospheric oxygen in the presence of sunlight
B reaction of CoH2O6Cl3 with AgNO3
C reaction of H2SO4 with NaOH
D combination of dinitrogen with dioxygen at 2000K

The correct answer is:
combination of dinitrogen with dioxygen at 2000K

Solution:
(i) N 2 + O 2 → 2000 K 2NO  (Re dox reaction)
during the reaction, oxidation of nitrogen take place from 0 to 2 and reduction of oxygen take place from 0 to –2. It means this reaction is redox reaction.

Topic: Chemistry, Redox Reactions, JEE Main, JEE Main 2020 (07 Jan Shift 2)
The redox reaction among the following is

Options:
A 2 NaOH +  Cl 2  → NaCl + NaOCl +  H 2 O
B formation of ozone from atmospheric oxygen in the presence of sunlight
C reaction of [Co(H2O)6]Cl3 with AgNO3
D reaction of H2SO4 with NaOH

The correct answer is:
2 NaOH +  Cl 2  → NaCl + NaOCl +  H 2 O

Solution:
2 NaOH +  Cl 2  → NaCl + NaOCl +  H 2 O
Here Cl 2 is undergoing oxidation as well as reduction.In rest of these reactions oxidation states are not changing so these are not redox reactions.
3O2→2O3
2NaOH+H2SO4→Na2SO4+2H2O
AgNO3+NaCl→NaNO3+AgCl

Topic: Chemistry, Redox Reactions, JEE Main
The reduced temperature =Î¸=TTC

The reduced pressure =Ï=PPC

The reduced volume =Ï=VVC

Hence, it can be said that the reduced equation of state may be given as

Options:
A π3+1ϕ23ϕ-1=83θ
B π3+1ϕϕ-1=83θ
C π4+1ϕ3θ-1=83ϕ
D π3+1ϕ3ϕ-1=83θ

The correct answer is:
π3+1ϕ23ϕ-1=83θ

Solution:
P=Pcπ=a27b2π
V=Vcϕ=3bϕ
T=Tcθ=8a27Rbθ
Hence substituting in the van der Waal's equation
P+aV2V-b=RT
⇒π+3ϕ23ϕ-1=8θ
This is reduced state equation
π+3ϕ23ϕ8θ-18θ=1
π3+1ϕ23ϕ-1=83θ

Topic: Chemistry, States of Matter, JEE Main
the reducing power of a metal depends on various factors. Suggest the factor which makes Li metal the strongest reducing agent in aqueous solution

Options:
A Sublimation enthalpy
B Ionisation enthalpy
C Hydration enthalpy
D Electron-gain enthalpy

The correct answer is:
Hydration enthalpy

Solution:
Due to high hydration energy of Li+ cation, the standard reduction potential Li+ is more negative among all alkali metal cations hence Li acts as strong reducing agent in water.

Topic: Chemistry, s Block Elements, JEE Main
The reduction of nitrobenzene with \( \mathrm{LiAlH}_{4} \) gives:

Options:
A Nitrosobenzene
B Aniline
C Azobenzene
D Phenylhydroxylamine

The correct answer is:
Azobenzene

Solution:
This is a reduction reaction of the nitro compound. The nitro compound can get reduced in an acidic, a basic, or neutral medium.

Nitrobenzene on reduction with lithium aluminium hydride (LiAlH4) gives azobenzene.

Ph-NO2→LiAlH4Ph-N=N-Ph

Topic: Chemistry, Amines, JEE Main
The reduction of peroxydisulphate ion by I ion

is expressed by $\mathrm{S}_{2} \mathrm{O}_{8}^{2-}+3 \mathrm{I}^{-} \rightarrow 2 \mathrm{SO}_{4}^{2-}+\mathrm{I}_{3}^{-}$. If

rate of disappearance of $\mathrm{I}^{-}$ is $9 / 2 \times 10^{-3} \mathrm{~mol}$ lit $^{-1}$

$\mathrm{s}^{-1},$ what is the rate of formation of $2 \mathrm{SO}_{4}^{2-}$ during same time?

Options:
A $3 \times 10^{-3} \mathrm{~mol} \mathrm{Lit}^{-1} \mathrm{~s}^{-1}$
B $2 \times 10^{-3} \mathrm{~mol} \mathrm{Lit}^{-1} \mathrm{~s}^{-1}$
C $10^{-3} \mathrm{~mol} \mathrm{Lit}^{-1} \mathrm{~s}^{-1}$
D $4 \times 10^{-3} \mathrm{~mol} \mathrm{Lit}^{-1} \mathrm{~s}^{-1}$

The correct answer is:
$3 \times 10^{-3} \mathrm{~mol} \mathrm{Lit}^{-1} \mathrm{~s}^{-1}$

Solution:
$\frac{1 d\left[\mathrm{SO}_{4}^{2-}\right]}{2}=\frac{1}{3}\left(-\frac{\mathrm{d}\left(\mathrm{I}^{-}\right)}{d t}\right)$

$\frac{1}{2} \times \frac{d\left[\mathrm{SO}_{4}^{2-}\right]}{d t}=\frac{1}{3} \times \frac{9}{2} \times 10^{-3}$

$\therefore \quad \frac{d\left[\mathrm{SO}_{4}^{2-}\right]}{d t}=3 \times 10^{-3} \mathrm{~mol} \mathrm{Lit}^{-1} \mathrm{~s}^{-1}$

Topic: Chemistry, Chemical Kinetics, JEE Main
The reduction potential of hydrogen half cell will be negative if :

Options:
A
$\mathrm{p}\left(\mathrm{H}_2\right)=1 \mathrm{~atm}$ and $\left[\mathrm{H}^{+}\right]=1.0 \mathrm{M}$

B
$\mathrm{p}\left(\mathrm{H}_2\right)=2 \mathrm{~atm}$ and $\left[\mathrm{H}^{+}\right]=1.0 \mathrm{M}$

C
$\mathrm{p}\left(\mathrm{H}_2\right)=2 \mathrm{~atm}$ and $\left[\mathrm{H}^{+}\right]=2.0 \mathrm{M}$

D
$\mathrm{p}\left(\mathrm{H}_2\right)=1 \mathrm{~atm}$ and $\left[\mathrm{H}^{+}\right]=2.0 \mathrm{M}$

The correct answer is:
$\mathrm{p}\left(\mathrm{H}_2\right)=2 \mathrm{~atm}$ and $\left[\mathrm{H}^{+}\right]=1.0 \mathrm{M}$

Solution:
$2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_2(\mathrm{~g})$
$\mathrm{E}=\mathrm{E}^{\circ}-0.059 \log \left(\frac{\mathrm{P}_{\mathrm{H}_2}}{\left[\mathrm{H}^{+}\right]^2}\right)$ (here $\mathrm{E}$ is -ve when $\mathrm{P}_{\mathrm{H}_2}>\left[\mathrm{H}^{+}\right]^2$ )
$=\frac{-0.0591}{2} \log _{10}\left(\frac{2}{1}\right)=\frac{-.0591}{2} \times .3010=$ negative value

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2011
The reductions of an oxide by aluminium is called.

Options:
A Beeyer's process
B Goldschmidt's aluminothermite process
C Hall's process
D van Arkel process

The correct answer is:
Goldschmidt's aluminothermite process

Solution:
The reduction of metal oxides by aluminium powder is called Gold schmidt's aluminothermite process.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
The refining method used when the metal and the impurities have low and high melting temperatures, respectively, is

Options:
A liquation.
B vapour phase refining.
C zone refining.
D distillation.

The correct answer is:
liquation.

Solution:
Liquation method is used when the melting point of metal is less compare to the melting point of the associated impurity.

In liquation process, the impure metal is heated on slopy hearth of furnace, due to less melting point, metal melts first and flow down leaving impurities behind.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2020 (07 Jan Shift 2)
The reflection of the point \((4,-13)\) in the line \(5 x+y+6=0\), is

Options:
A \((-1,-14)\)
B \((3,4)\)
C \((1,2)\)
D \((-4,13)\)

The correct answer is:
\((-1,-14)\)

Solution:
Let \(\mathrm{Q}(a, b)\) be the reflection of \(\mathrm{P}(4,-13)\) in the line \(5 x+y+6=0\)
Then the mid-point \(R\left(\frac{a+4}{2}, \frac{b-13}{2}\right)\) lies on
\(\begin{aligned}
& 5 x+y+6=0 \\
& \therefore 5\left(\frac{a+4}{2}\right)+\frac{b-13}{2}+6=0 \\
& \Rightarrow 5 a+b+19=0 \quad ...(i)
\end{aligned}\)
Also \(\mathrm{PQ}\) is perpendicular to \(5 x+y+6=0\)
Therefore \(\frac{b+13}{a-4} \times\left(-\frac{5}{1}\right)=-1\)
\(\Rightarrow a-5 b-69=0 \quad ...(ii)\)
Solving (i) and (ii), we get \(a=-1, b=-14\)

Topic: Mathematics, Straight Lines, JEE Main
The refracting angle of a prism is A and refractive index of the material of the prism is cotA2. Then the angle of minimum deviation will be

Options:
A 180°-2A
B 90°-A
C 180°+2A
D 180°-3A

The correct answer is:
180°-2A

Solution:
As we know, angle of minimum deviation is given by, ⇒μ=sinδm+A2sinA2

⇒cotA2=sinδm+A2sinA2

⇒cosA2sinA2=sinδm+A2sinA2

⇒sin90°-A2=sinδm+A2

Therefore, 90°-A2=δm+A2 ⇒δm=180°-2A

Topic: Physics, Ray Optics, JEE Main, JEE Main 2022 (28 Jun Shift 1)
The refractive index and the permeability of a medium are respectively $1.5$ and $5 \times 10^{-7} \mathrm{Hm}^{-1}$. The relative permittivity of the medium is nearly

Options:
A $25$
B $15$
C $6$
D $10$

The correct answer is:
$6$

Solution:
Given, refractive index of the medium, $n=15$
Permeability of the medium,
$$
\mu=5 \times 10^{-7} \mathrm{Hm}^{-1}
$$
Relative permeability of the medium,
$$
\mu_{r}=\frac{\mu}{\mu_{0}}=\frac{5 \times 10^{-7}}{4 \pi \times 10^{-7}}=\frac{5}{4 \pi}
$$
We know that, refractive index,
$$
\begin{aligned}
n &=\sqrt{\varepsilon_{r} \mu_{r}} \\
n^{2} &=\varepsilon_{r} \mu_{r} \\
\Rightarrow \quad \varepsilon_{r} &=\frac{n^{2}}{\mu_{r}}=\frac{(1.5)^{2}}{5 / 4 \pi} \\
&=\frac{(1.5)^{2} \times 4 \pi}{5}=5.7 \simeq 6
\end{aligned}
$$

Topic: Physics, Electromagnetic Waves, JEE Main
The refractive index of a certain glass is 1.5 for light whose wavelength in vacuum is $6000 \hat{A}$. The wavelength of this light when it passes through glass is

Options:
A $4000 Å$
B $6000 Å$
C $9000 Å$
D $15000 Å$

The correct answer is:
$4000 Å$

Solution:
The correct option is $\mathbf{1}$
$4000 Å$
$\lambda_{\text {madtm }}=\frac{\lambda_{\text {as }}}{\mu}=\frac{6000}{1.5}=4000 \mathrm{~A}$

Topic: Physics, Ray Optics, JEE Main
The refractive index of a converging lens is 1.4. What will be the focal length of this lens if it is placed in a medium of same refractive index ? (Assume the radii of curvature of the faces of lens are R1 and R2 respectively)

Options:
A 1
B Infinite
C R1R2R1-R2
D Zero

The correct answer is:
Infinite

Solution:
1 F=μLμS-11R1-1R2

If μL=μS⇒1 F=0⇒F=∞

Topic: Physics, Ray Optics, JEE Main, JEE Main 2021 (16 Mar Shift 2)
The refractive index of a material is given by the equation \( n=\frac{A+B}{\lambda^{2}} \), where \( A \) and \( B \) are constant. The dimensional formula for \( B \) is

Options:
A \( \left[\mathrm{M}^{0} \mathrm{~L}^{2} \mathrm{~T}^{-1}\right] \)
B \( \left[\mathrm{M}^{0} \mathrm{~L}^{-2} \mathrm{~T}^{0}\right] \)
C \( \left[\mathrm{M}^{0} \mathrm{~L}^{2} \mathrm{~T}^{-2}\right] \)
D \( \left[\mathrm{M}^{0} \mathrm{~L}^{2} \mathrm{~T}^{0}\right] \)

The correct answer is:
\( \left[\mathrm{M}^{0} \mathrm{~L}^{2} \mathrm{~T}^{0}\right] \)

Solution:
The given equation is
n=A+Bλ2
Where A and B are constants.
By homogeneity principle the dimensions of all the terms on both sides should be same
ie, B=A=nλ2
∴ B=M0L0T0L2
=M0L2T0

Topic: Physics, Units and Dimensions, JEE Main