datavorous/entrance-exam-dataset huggingface dataset:
The refractive index of a material of a prism of angles $45^{\circ}-45^{\circ}-90^{\circ}$ is 1.5. The path of the ray of light incident normally on the hypotenuse side is shown in

Options:
A

B

C

D

The correct answer is:

Solution:
According to given conditions TIR must take place at both the surfaces $A B$ and $A C$. Hence only option $(1)$ is correct.

Topic: Physics, Ray Optics, JEE Main
The refractive index of a piece of transparent quartz is the greatest for

Options:
A Red light
B Violet light
C Green light
D Yellow light

The correct answer is:
Violet light

Solution:
The correct option is 2 Violet light
$\mu \infty \frac{1}{\lambda}, \lambda,\gt\lambda_v$

Topic: Physics, Ray Optics, JEE Main
The refractive index of a prism for a monochromatic wave is $\sqrt{2}$ and its refracting angle is $60^{\circ}$. For minimum deviation, the angle of incidence will be

Options:
A $30^{\circ}$
B $45^{\circ}$
C $60^{\circ}$
D $75^{\circ}$

The correct answer is:
$45^{\circ}$

Solution:
$\mu=\frac{\sin i}{\sin A / 2} \Rightarrow \sqrt{2}=\frac{\sin i}{\sin \left(\frac{60}{2}\right)}$
$\Rightarrow \sqrt{2} \times \sin 30=\sin i \Rightarrow i=45^{\circ}$

Topic: Physics, Ray Optics, JEE Main
The refractive index of a prism for a monochromatic wave is \( \sqrt{2} \) and its refracting angle is \( 60^{\circ} \). For minimum deviation, the angle of incidence will be

Options:
A \( 30^{\circ} \)
B \( 45^{\circ} \)
C \( 60^{\circ} \)
D \( 75^{\circ} \)

The correct answer is:
\( 45^{\circ} \)

Solution:
Using the prism formula and snell's law, under the condition of minimum deviation, we have,

μ=sin⁡A+δm2sin⁡A2
2=sin⁡60°+δm2sin⁡60°2
12=sin⁡60°+δm2
Or sin⁡45°=sin⁡60°+δm2

Therefore, the minimum deviation is,

δm=30°

And, the required angle of incidence for it,

Or i=A+δm2
=60+302=902=45°

Topic: Physics, Ray Optics, JEE Main
The refractive index of a prism with apex angle A is cotA2. The angle of minimum deviation is :

Options:
A δm=180°-A
B δm=180°-3A
C δm=180°-4A
D δm=180°-2A

The correct answer is:
δm=180°-2A

Solution:
As we know, μ=sinA+δm2sinA2

⇒cosA2sinA2=sinA+δm2sinA2

⇒sinπ2-A2=sinA+δm2

⇒π2-A2=A2+δm2

⇒δm=π-2A=180°-2A

Topic: Physics, Ray Optics, JEE Main, JEE Main 2024 (31 Jan Shift 1)
The refractive index of a transparent liquid filled in an equilateral hollow prism is 2. The angle of minimum deviation for the liquid will be _______°.

Options:

The correct answer is:
30

Solution:
For minimum deviation in a prism,

μ=sinDmin+A2sinA2

The value of minimum deviation is

⇒2=sinDmin+60°2sin30°

⇒12=sinDmin+60°2⇒45°=Dmin+60°2⇒Dmin=30°

Topic: Physics, Ray Optics, JEE Main, JEE Main 2023 (15 Apr Shift 1)
The refractive index of glass is $1.520$ for red light and $1.525$ for blue light. Let $D_1$ and $D_2$ be an of minimum deviation for red and blue light respectively in a prism of this glass. Then

Options:
A
$D_1>D_2$

B
$\mathrm{D}_1 < \mathrm{D}_2$
C
$D_1=D_2$

D
$D_1$ can be less than or greater than depending upon the angle of prism

The correct answer is:
$\mathrm{D}_1 < \mathrm{D}_2$

Solution:
$D=(\mu-1) A$
$D_2>D_1$

Topic: Physics, Ray Optics, JEE Main, JEE Main 2006
The refractive index of the material of a concave lens is $\mu$. It is immersed in a medium of refractive index $\mu_1$. A parallel beam of light is incident on the lens. The path of the emergent rays when $\mu_1>\mu$ is:

Options:
A

B

C

D

The correct answer is:

Solution:
If a lens of refractive index $\mu$ is immersed in a medium of refractive index $\mu_1$, then its focal length in medium is given by
$$
\frac{1}{\mathrm{f}_{\mathrm{m}}}=\left({ }_{\mathrm{m}} \mu_l-1\left(\frac{1}{\mathrm{R}_1}\right)-\frac{1}{\mathrm{R}_2}\right)
$$
If $\mathrm{f}_{\mathrm{a}}$ is the focal length of lens in air, then
$$
\begin{aligned}
&\frac{1}{\mathrm{f}_{\mathrm{a}}}=\left({ }_{\mathrm{a}} \mu_l-1\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)\right. \\
&\Rightarrow \frac{\mathrm{f}_{\mathrm{m}}}{\mathrm{f}_{\mathrm{a}}}=\frac{\left({ }_{\mathrm{a}} \mu_l-1\right)}{\left({ }_{\mathrm{m}} \mu_l-1\right)}
\end{aligned}
$$
If $\mu_1>\mu$, then $\mathrm{f}_{\mathrm{m}}$ and $\mathrm{f}_{\mathrm{a}}$ have opposite signs and the nature of lens changes i.e. a convex lens diverges the light rays and concave lens converges the light rays. Thus given option (a) is correct.

Topic: Physics, Ray Optics, JEE Main, JEE Main 2014 (12 Apr Online)
The refractive index of the material of a prism is $\sqrt{2}$ and its refracting angle is $30^{\circ}$. One of the refracting surface of the prism is made a mirror inward. A beam of monochromatic light entering the prism from the other face will retrace its path after reflection from the mirrored surface if its angle of incidence on the prism is:

Options:
A $45^{\circ}$
B $60^{\circ}$
C 0
D $30^{\circ}$

The correct answer is:
$45^{\circ}$

Solution:
Here $\angle r=30^{\circ}$
(Using law of triangle)
$$
\begin{aligned}
& \Rightarrow \quad \mu=\frac{\sin i}{\sin r} \\
& \Rightarrow \sqrt{2} \cdot \sin 30^{\circ}=\sin i \\
& \Rightarrow \quad \sin i=\frac{1}{\sqrt{2}} \\
& \Rightarrow \quad i=45^{\circ}
\end{aligned}
$$

Topic: Physics, Ray Optics, JEE Main
The refractive index of water is $4 / 3$ and that of glass is $5 / 3$. What will be the critical angle for the ray of light entering water from the glass

Options:
A $\sin ^{-1} \frac{4}{5}$
B $\sin ^{-1} \frac{5}{4}$
C $\sin ^{-1} \frac{1}{2}$
D $\sin ^{-1} \frac{2}{1}$

The correct answer is:
$\sin ^{-1} \frac{4}{5}$

Solution:
$\begin{aligned} & { }_w \mu_g=\frac{1}{\sin C} \Rightarrow \frac{\mu_g}{\mu_w}=\frac{5 / 3}{4 / 3}=\frac{1}{\sin C} \\ & \Rightarrow \sin C=\frac{4}{5} \Rightarrow C=\sin ^{-1}\left(\frac{4}{5}\right)\end{aligned}$

Topic: Physics, Ray Optics, JEE Main
The region between two concentric spheres of radii 'a' and 'b', respectively (see figure), has volume charge density ρ=Ar , where A is a constant and r is the distance from the centre. At the centre of the spheres is a point charge Q. The value of A such that the electric field in the region between the spheres will be constant, is:

Options:
A 2Qπa2-b2
B 2Qπa2
C Q2πa2
D Q2πb2-a2

The correct answer is:
Q2πa2

Solution:
E=E1+E2

⇒E=KQ+∫ar4πx2dxAxr2
E=KQr2+4πAr22r2-a22r2
dEdr=0
⇒-2Qr3+8πAa22r3=0
A=Q2πa2

Topic: Physics, Electrostatics, JEE Main, JEE Main 2016 (03 Apr)
The region between $y=0$ and $y=\mathrm{d}$ contains a magnetic field $\overrightarrow{\mathrm{B}}=\mathrm{B} \hat{\mathrm{z}}$. A particle of mass $\mathrm{m}$ and charge $\mathrm{q}$ enters

the region with a velocity $\vec{v}=v \hat{i} .$ if $\mathrm{d}=\frac{\mathrm{m} v}{2 \mathrm{qB}},$ the acceleration of the charged particle at the point of its emergence at the other side is :

Options:
A $\frac{\mathrm{q} v \mathrm{~B}}{\mathrm{~m}}\left(\frac{1}{2} \hat{i}-\frac{\sqrt{3}}{2} \hat{j}\right)$
B $\frac{\mathrm{q} v \mathrm{~B}}{\mathrm{~m}}\left(\frac{\sqrt{3}}{2} \hat{i}+\frac{1}{2} \hat{j}\right)$
C $\frac{\mathrm{q} v \mathrm{~B}}{\mathrm{~m}}\left(\frac{-\hat{j}+\hat{i}}{\sqrt{2}}\right)$
D None of the above

The correct answer is:
None of the above

Solution:
Assuming particle enters from $(0, \mathrm{~d})$

$r=\frac{m v}{q B}, \quad d=\frac{r}{2}$

$a=\frac{q V B}{m}\left[\frac{-\sqrt{3} i-j}{2}\right]$

this option is not given in the all above four choices.

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2019 (11 Jan Shift 2)
The region in the electromagnetic spectrum where the Balmer series lines appear is:

Options:
A Visible
B Microwave
C Infrared
D Ultraviolet

The correct answer is:
Visible

Solution:
In hydrogen spectrum maximum lines of Balmer series lies in visible region.

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2020 (04 Sep Shift 1)
The region of chromatin that is stained lightly is

Options:
A Heterochromatin
B Transcriptionally inactive region
C Centromere
D Transcriptionally active region

The correct answer is:
Transcriptionally active region

Solution:
In a typical nucleus, some regions of chromatin are loosely packed (and stains light) and are referred to as euchromatin. The chromatin that is more densely packed and stains dark is called heterochromatin. Euchromatin is said to be transcriptionally active chromatin, whereas heterochromatin is inactive. The centromere is mainly composed of heterochromatin.

Topic: Biology, Cell - The Unit of Life, NEET, NEET 2016 (Phase 1)
The region of chromatin that is stained lightly is

Options:
A Heterochromatin
B Transcriptionally inactive region
C Centromere
D Transcriptionally active region

The correct answer is:
Transcriptionally active region

Solution:
In a typical nucleus, some regions of chromatin are loosely packed (and stains light) and are referred to as euchromatin. The chromatin that is more densely packed and stains dark is called heterochromatin. Euchromatin is said to be transcriptionally active chromatin, whereas heterochromatin is inactive. The centromere is mainly composed of heterochromatin.

Topic: Biology, Molecular Basis of Inheritance, NEET
The region outside the seminiferous tubules is called interstitial spaces which contain all, except

Options:
A Blood vessels
B Immunologically competent cells
C Leydig cell
D Sertoli cells

The correct answer is:
Sertoli cells

Solution:
Testes consists of structural units known as Seminiferous tubules. These are present inside the lobules of testes. Each lobule consists of one to three such tubules in highly coiled manner. Seminiferous tubules consist of two cells. The main Spermatogonia i.e. mother cells that form the sperms and Sertoli cells or nurse cells that help to mature the sperms. These cells are present in between Spermatogonia. In between the seminiferous tubules is interstitial tissue that contains Leydig cells, Blood vessels, immune cells etc.

Topic: Biology, Human Reproduction, NEET
The region represented by x-y≤2 and x+y≤2 is bounded by a

Options:
A rhombus of area 82 sq. units.
B rhombus of side length 2 units.
C square of area 16 sq. units.
D square of side length 22 units.

The correct answer is:
square of side length 22 units.

Solution:
x+y=2⇒x+y=2 and x+y=–2

x–y=2 ⇒x –y=2 and x–y=–2

So, the region bounded by the given curves, i.e., pairs of straight lines, is a square.

Now, length of side of square=distance between parallel lines x+y=2 and x+y=-2

=2--21+1=42=22

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2019 (10 Apr Shift 1)
The region represented by {z=x+iy∈C:|z|-Re(z)≤1} is also given by the inequality

Options:
A y2≥2(x+1)
B y2≤2x+12
C y2≤x+12
D y2≥x+1

The correct answer is:
y2≤2x+12

Solution:
x2+y2-x≤1
⇒ x2+y2≤x+1
⇒ x2+y2≤x2+2x+1
⇒ y2≤2x+1

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2020 (06 Sep Shift 1)
The regions of the atmosphere, where clouds form and where we live, respectively, are:

Options:
A Troposphere and Troposphere
B Stratosphere and Stratosphere
C Stratosphere and Troposphere
D Troposphere and Stratosphere

The correct answer is:
Troposphere and Troposphere

Solution:
Troposphere contains about 75% of air in atmosphere and all amount is the water vapour.

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2019 (10 Apr Shift 1)
The relation between $\Delta E$ and $\Delta H$ is

Options:
A $\Delta H=\Delta E-P \Delta V$
B $\Delta H=\Delta E+P \Delta V$
C $\Delta E=\Delta V+\Delta H$
D $\Delta E=\Delta H+P \Delta V$

The correct answer is:
$\Delta H=\Delta E+P \Delta V$

Solution:
$\Delta H=\Delta E+P \Delta V$.

Topic: Chemistry, Thermodynamics (C), JEE Main
The relation between equilibrium constant $K_p$ and $K_c$ is

Options:
A $K_c=K_p(R T)^{\Delta n}$
B $K_p=K_c(R T)^{\Delta n}$
C $K_p=\left(\frac{K_c}{R T}\right)^{s n}$
D $K_p-K_c=(R T)^{\Delta n}$

The correct answer is:
$K_p=K_c(R T)^{\Delta n}$

Solution:
$K_p=K_c(R T)^{\Delta n}$

Topic: Chemistry, Chemical Equilibrium, JEE Main
The relation between KP and KC for the reaction A g +B g ⇌ C g +2D g is-

Options:
A KP=KCRT–1
B KP.KC–1=RT
C KCKP–1=RT
D KP=KCRT3

The correct answer is:
KP.KC–1=RT

Solution:
Δn g = (No. Moles of products- No. of moles of reactant) =1
Δn g  = 1
Kp =  K c Δn g RT

Topic: Chemistry, Chemical Equilibrium, JEE Main
The relation between pressure P and volume V is given by PV14 = constant. If the percentage decrease in volume is 14 , then the approximate percentage increase in pressure is

Options:
A 116
B 14
C 18
D 12

The correct answer is:
116

Solution:
P V14=Constant
% decreases in volume =14
dvv×100=14
Now, PV14=a
log⁡P+14log⁡v=log⁡a
dpp+14dvv=0
100×dPP+14dVV×100=0
dPP×100=-14dVV×100
=14×14=116

Topic: Chemistry, Thermodynamics (C), JEE Main
The relation between root mean square speed vrms and most probable speed vp for the molar mass M of oxygen gas molecule at the temperature of 300 K will be

Options:
A vrms=23vp
B vrms=13vp
C vrms=32vp
D vrms=vp

The correct answer is:
vrms=32vp

Solution:
RMS speed of O2 gas is vrms=3RTM=3RT32 and most probable speed of O2 gas is vmost probable=2RTM=2RT32.

Here temperature is same.

Thus, vrmsvmost probable=3RT322RT32=32

Hence, vrms=32vp

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2022 (25 Jun Shift 1)
The relation between the gas pressure $P$ and average kinetic energy per unit volume $E$ is

Options:
A $P=\frac{1}{2} E$
B $P=E$
C $P=\frac{3}{2} E$
D $P=\frac{2}{3} E$

The correct answer is:
$P=\frac{2}{3} E$

Solution:
$\begin{aligned} & E=\text { Average Kinetic energy per unit volume }= \\ & {\left[\left\{(1 / 2) M V^2 r m s\right\} / V\right]} \\ & =(1 / 2) \times(M / V) \times V_{r m s}^2 \\ & \therefore E=(1 / 2) \rho V_{r m s}{ }^2(1) \\ & \text { Pressure }=P=(1 / 3) \rho V_{r m s}{ }^2 \text { hence } \rho \cdot V_{r m s}{ }^2=3 \\ & \therefore \text { from }(1), E=(1 / 2) \times 3 P=(3 / 2) P \\ & \therefore P=(2 / 3) E\end{aligned}$

Topic: Physics, Kinetic Theory of Gases, JEE Main
The relation between the linear magnification $m$, the object distance ${ }^u$ and the focal length $f$ is

Options:
A $m=\frac{f-u}{f}$
B $m=\frac{f}{f-u}$
C $m=\frac{f+u}{f}$
D $m=\frac{f}{f+u}$

The correct answer is:
$m=\frac{f}{f-u}$

Solution:
$\because m=-\frac{v}{u}$ also $\frac{1}{f}=\frac{1}{v}+\frac{1}{u} \Rightarrow \frac{u}{f}=\frac{u}{v}+1$
$\Rightarrow-\frac{u}{v}=1-\frac{u}{f} \Rightarrow \frac{-v}{u}=\frac{f}{f-u}$ so $m=\frac{f}{f-u}$.

Topic: Physics, Ray Optics, JEE Main
The relation between the stopping potential V0 and frequency v is correctly represented in [ϕ= Work function]

Options:
A V0=ϕe-hv2e
B V0=hev+ϕe
C V0=hve-ϕe
D V0=hve2

The correct answer is:
V0=hve-ϕe

Solution:
According to the question, Stopping potential or cut-off potential is defined as the required potential for stopping the removal of an electron from a metal surface when the incident light energy is greater than the work potential of the metal on which the incident light is focused.

(a) Stopping potential does not depend on the intensity of incident radiation. On increasing intensity, the value of saturated current increases, whereas the stopping potential remains unchanged.

(b) For a given intensity of radiation, the stopping potential depends on the frequency. Higher the frequency of incident light higher the value of stopping potential.

Frequency means cycles per second, or the number of complete waves that pass through a given point per second. Frequency depends on wavelength and is given by the equation λv=c, where λ is wavelength, ν is frequency and c is the speed of light which is a constant.

By einsteins photoelectric equation-

K.E max=hv-ϕ =hcλ-ϕK.E max = eVseVs=hcλ-ϕ

Topic: Chemistry, Structure of Atom, TS EAMCET, TS EAMCET 2021 (05 Aug Shift 2)
The relation between time and displacement for two particles is given by
$y_1=0.06 \sin 2 \pi\left(0.04 t+\phi_1\right), y_2=0.03 \sin 2 \pi\left(1.04 t+\phi_2\right)$
The ratio of the intensity of the waves produced by the vibrations of the two particles will be

Options:
A $2: 1$
B $1: 2$
C $4: 1$
D $1: 4$

The correct answer is:
$4: 1$

Solution:
$\frac{I_1}{I_2}=\frac{a_1^2}{a_2^2}=\left(\frac{0.06}{0.03}\right)^2=\frac{4}{1}$

Topic: Physics, Waves and Sound, JEE Main
The relation between time $t$ and distance $\mathrm{x}$ is $\mathrm{t}=a \mathrm{x}^2+\mathrm{bx}$ where $\mathrm{a}$ and $\mathrm{b}$ are constants. The acceleration is

Options:
A
$-2 a b v^2$

B
$2 b v^3$

C
$-2 a v^3$

D
$2 a v^2$

The correct answer is:
$-2 a v^3$

Solution:
$$
t=a x^2+b x
$$
by differentiating acceleration $=-2 a v^3$

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2005
The relation between two specific heats of a gas is

Options:
A $C_P-C_V=\frac{R}{J}$
B $C_V-C_P=\frac{R}{J}$
C $C_P-C_V=J$
D $C_V-C_P=J$

The correct answer is:
$C_P-C_V=\frac{R}{J}$

Solution:
When $C_P$ and $C_V$ are given with calorie and $R$ with Joule then $C_P-C_V=R / J$

Topic: Physics, Kinetic Theory of Gases, JEE Main
The relation between volume $(\mathrm{V})$ and absolute temperature ( $\mathrm{T}$ ) of a gas in an adiabatic process is

Options:
A $\mathrm{TV}^{\mathrm{g}}=$ constant
B $\mathrm{VTg}=$ constant
C $\mathrm{TV}^{1-\mathrm{g}}=$ constant
D $\mathrm{TV}^{\mathrm{g}-1}=$ constant

The correct answer is:
$\mathrm{TV}^{\mathrm{g}-1}=$ constant

Solution:
The process in which heat remains constant is called adiabatic process.
In adiabatic process relation between temperature and volume is given as:
$$
\mathrm{TV}^{\gamma-1}=\text { constant }
$$

Topic: Physics, Thermodynamics, AP EAMCET, AP EAMCET 2023 (18 May Shift 2)
The relation "less than" in the set of natural numbers is

Options:
A only symmetric
B only transitive
C only reflexive
D an equivalence relation

The correct answer is:
only transitive

Solution:
Since, x<y, y<zâx<zâââx, y, zâN
â´ x R y, yR zâx R z,
â´ Relation is transitive,
âµ x<y does not give y<x,
â´ Relation is not symmetric.
Since, x<x does not hold, hence the relation is not reflexive.

Topic: Mathematics, Sets and Relations, JEE Main
The relation R=a,b:gcda,b=1, 2a≠b, a, b∈ℤ is:

Options:
A transitive but not reflexive
B symmetric but not transitive
C reflexive but not symmetric
D neither symmetric nor transitive

The correct answer is:
neither symmetric nor transitive

Solution:
Given relation is

R=a,b:gcda,b=1, 2a≠b, a, b∈ℤ

Now,

Reflexive:

gcda,a≠1 ∀a∈Z, hence R is not reflexive.

Symmetric:

a=2, b=1⇒gcd2,1=1

Also, 2a=4≠b

Now when a=1, b=2⇒gcd(1, 2)=1

But, 2a=2=b

Hence, a=2b

⇒R is not Symmetric

Transitive:

Let a=14, b=19, c=21

gcd(a, b)=1

gcd(b, c)=1

gcd(a, c)=7

Hence, not transitive

So, R is neither symmetric nor transitive.

Topic: Mathematics, Sets and Relations, JEE Main, JEE Main 2023 (24 Jan Shift 1)
The relation R given by x,y: x2-3xy+2y2=0, ∀x,y∈R is

Options:
A reflexive but not symmetric
B symmetric but not transitive
C symmetric and transitive
D an equivalence relation

The correct answer is:
reflexive but not symmetric

Solution:
∵ x2-3xy+2y2=0⇒x2-xy-2xy+2y2=0
⇒xx-y-2yx-y=0
⇒x-2yx-y=0
⇒x=y or x=2y
Now, ∵ in R all ordered pairs x,x are present
∴ It is reflexive
Now, 4,2∈R as 4=22
but 2,4∉R as 2≠24
∴ It is not symmetric
Also 4,2 & 2,1∈R but 4,1∉R
∴ It is not transitive

Topic: Mathematics, Sets and Relations, JEE Main
The relation R is defined in the set 1, 2, 3, 4, 5, 6 as R=(a,b):b=a+1, then

Options:
A R is neither reflexive nor symmetric nor transitive
B R is neither reflexive nor symmetric but transitive
C R is not reflexive but symmetric and transitive
D R is reflexive, symmetric and transitive

The correct answer is:
R is neither reflexive nor symmetric nor transitive

Solution:
Let, A=1, 2, 3, 4, 5, 6
The relation R is defined on set A is
R=a, b : b=a+1. Therefore, R=(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)

Now, 6 â A but (6, 6) â R.
Therefore, R is not reflexive.
It can be observed that (1, 2) â R but (2,1) âR. Therefore, R is not symmetric.

Now, (1, 2), (2, 3) â R but (1, 3)â R. Therefore, R is not transitive.
Hence, R is neither reflexive nor symmetric nor transitive

Topic: Mathematics, Sets and Relations, JEE Main
The relation tan-1⁡1+x1-x=π4+tan-1⁡x holds true for all

Options:
A xâR
B xâ(-â,1)
C x∈(-1,∞)
D xâ(-â,2)

The correct answer is:
x∈(-∞,1)

Solution:
Given, tan-1â¡1+x1-x=Ï4+tan-1â¡x
RHS =Ï4+tan-1â¡x=tan-1â¡1+tan-1â¡x
=tan-1â¡1+x1-x, if x<1
â´ xâ(-â, 1)

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The relationship among most probable velocity, average velocity and root mean square velocity is respectively

Options:
A
$\sqrt{2}: \sqrt{3}: \sqrt{8 / \pi}$

B
$\sqrt{2}: \sqrt{8 / \pi}: \sqrt{3}$

C
$\sqrt{8 / \pi}: \sqrt{3}: \sqrt{2}$

D
$\sqrt{3}: \sqrt{8 / \pi}: \sqrt{2}$

The correct answer is:
$\sqrt{2}: \sqrt{8 / \pi}: \sqrt{3}$

Solution:
Most probable speed $\left(C^*\right)=\sqrt{\frac{2 R T}{M}}$ Average speed $(\bar{C})=\sqrt{\frac{8 R T}{\pi M}}$ Root mean square velocity $(C)=\sqrt{\frac{3 R T}{M}}$ $C^*: \bar{C}: C=\sqrt{2}: \sqrt{\frac{8}{\pi}}: \sqrt{3}$

Topic: Chemistry, States of Matter, JEE Main, JEE Main 2012 (26 May Online)
The relationship between rate constant and half life period of zero order reaction is given by

Options:
A \( \mathrm{t}_{1 / 2}=[\mathrm{A}]_{0} 2 \mathrm{~K} \)
B \( \mathrm{t}_{1 / 2}=\frac{0.693}{\mathrm{~K}} \)
C \( \mathrm{t}_{1 / 2}=\frac{[\mathrm{A}]_{0}}{2 \mathrm{~K}} \)
D \( \mathrm{t}_{1 / 2}=\frac{2[\mathrm{~A}]_{0}}{\mathrm{~K}} \)

The correct answer is:
\( \mathrm{t}_{1 / 2}=\frac{[\mathrm{A}]_{0}}{2 \mathrm{~K}} \)

Solution:
For zeroth order reaction,

At=A0-Kt

where, At=concentration after time t

A0=initial concentration

K=rate constant

t=time

At half life, At =A02

A02= A0 - Kt12Kt12 = A02t12 = A02 K

Topic: Chemistry, Chemical Kinetics, JEE Main
The relationship between relative magnetic permeability ' $\mu_r$ ' of a substance and its magnetic susceptibility ' $x$ ' is given by $\qquad$

Options:
A $\mu_r=1-\chi$
B $\mu_r=1+\chi^2$
C $\mu_r=1+\chi$
D $\mu_r=1-\chi^2$

The correct answer is:
$\mu_r=1+\chi$

Solution:
The above derivation can be expressed as
$\begin{aligned}
B & =B_0+B_m \Rightarrow B=\mu_0 H+\mu_0 M=\mu_0 H(1+\chi) \\
\mu_r & =1+\chi\left(\because \mu_r=B / H\right)
\end{aligned}$

Topic: Physics, Magnetic Properties of Matter, JEE Main
The relationship between standard reduction potential of cell and equilibrium constant is shown by

Options:
A $E_{\text {cell }}^0=\frac{n}{0.059} \log K_c$
B $E_{\text {cell }}^0=\frac{0.059}{n} \log K_c$
C $E_{\text {cell }}^0=0.059 n \log K_c$
D $E_{\text {cell }}^0=\frac{\log K_c}{n}$

The correct answer is:
$E_{\text {cell }}^0=\frac{0.059}{n} \log K_c$

Solution:
$E_{\text {cell }}^o=\frac{2.303 R T}{n F} \log K=\frac{0.0591}{n} \log K_c$ at $298 K$.

Topic: Chemistry, Electrochemistry, JEE Main
The relationship between the force \( F \) and position \( x \) of a body is as shown in figure. The work done in displacing the body form \( x=1 \mathrm{~m} \) to \( x=5 \mathrm{~m} \) will be

Options:
A \( 30 \mathrm{~J} \)
B \( 15 \mathrm{~J} \)
C \( 25 \mathrm{~J} \)
D \( 20 \mathrm{~J} \)

The correct answer is:
\( 15 \mathrm{~J} \)

Solution:
Work done = area under F-x graph

= area of rectangle ABDC+ area of rectangle EFHD+ area of rectangle GJHI+ area of triangle JKL.

=2-1(10-0)+3-25-0+4-3-5-0+125-410-0=15 J.

Topic: Physics, Work Power Energy, JEE Main
The relationship between the force $F$ and position $x$ of a body is as shown in the figure. The work done in displacing the body from $x=1 \mathrm{~m}$ to $x=5 \mathrm{~m}$ will be

Options:
A $30 J$
B $15 \mathrm{~J}$
C $25 \mathrm{~J}$
D $20 \mathrm{~J}$

The correct answer is:
$15 \mathrm{~J}$

Solution:
The work done in a $F-x$ graph is calculated by the area under the graph. The given graph can be shown as
$\therefore$ Work done in displacing the object from $x=1 \mathrm{~m}$ to $x=5 \mathrm{~m}$ will be
$\begin{aligned}
W &=\text { area under } F-x \text { graph } \\
=& \text { area } A B J K+\text { areaCLDJ } \\
& \quad+\text { area } D G F E+\text { area } G H I \\
=&(10 \times 1)+(5 \times 1)+(-5 \times 1)+\left(\frac{1}{2} \times 1 \times 10\right) \\
=& 10+5-5+5=15 \mathrm{~J}
\end{aligned}$

Topic: Physics, Work Power Energy, JEE Main