datavorous/entrance-exam-dataset huggingface dataset:
The diameter of a flywheel is $1 \mathrm{~m}$. It has a mass of $20 \mathrm{~kg}$. It is rotating about its axis with a speed of 120 rotations in one minute. Its angular momentum in $\mathrm{kg}-\mathrm{m}^2 / \mathrm{s}$ is

Options:
A $13.4$
B $31.4$
C $41.4$
D $43.4$

The correct answer is:
$31.4$

Solution:
$\begin{aligned} & r=\frac{1}{2} \mathrm{~m}=0.5 \mathrm{~m}, m=20 \mathrm{~kg}, \\ & n=120 \text { rotation per minute, } \\ &=2 \text { rotation per sec. } \\ & \text { Angular momentum } L=I \omega \\ &=\frac{M r^2}{2} 2 \pi n=\frac{20 \times(0.5)^2}{2} \times 2 \times 3.14 \times 2 \\ &=5 \times 3.14 \times 2=31.4 \mathrm{~kg}-\mathrm{m}^2 / \mathrm{s}\end{aligned}$

Topic: Physics, Rotational Motion, JEE Main
The diameter of a solid disc is $0.5 \mathrm{~m}$ and its mass is $16 \mathrm{~kg}$. What torque is required to increase its angular velocity about an axis perpendicular to its plane from zero to 120 rotations/minute in 8 seconds?

Options:
A $\pi N / m$
B $2 \pi N / m$
C $\frac{\pi}{2} \mathrm{~N} / \mathrm{m}$
D $\frac{\pi}{4} \mathrm{~N} / \mathrm{m}$

The correct answer is:
$\frac{\pi}{4} \mathrm{~N} / \mathrm{m}$

Solution:
We have torque as $\tau=I \alpha$.
Also the final angular velocity $\omega$ is given as 120 rotations $/ \mathrm{min}$ or 2 rotations/second or $4 \pi$ radians/sec.
Thus we have $\alpha$ as $\frac{4 \pi-0}{8}=\frac{\pi}{2} \mathrm{rad} / \mathrm{s}^2$
And moment of inertia I for solid disc is $\frac{\mathrm{mr}^2}{2}$
Thus we get torque as
$\tau=\left(\frac{\mathrm{mr}^2}{2} \times \frac{\pi}{2}\right)$
Substituting the values we get-
$\tau=\frac{\pi}{4} \mathrm{~N} / \mathrm{m}$

Topic: Physics, Rotational Motion, JEE Main
The diameter of a spherical bob is measured using a vernier callipers. 9 divisions of the main scale, in the vernier callipers, are equal to 10 divisions of vernier scale. One main scale division is 1 mm. The main scale reading is 10 mm and 8th division of vernier scale was found to coincide exactly with one of the main scale division. If the given vernier callipers has positive zero error of 0.04 cm, then the radius of the bob is _____________ ×10-2 cm.

Options:

The correct answer is:
52

Solution:
9 MSD=10 VSD
1 VSD=0.9 mm
1 MSD=1 mm
∴ L.C.=MSD-VSDMSD=1-0.91=0.1
Reading = Main scale reading + L.C. × vernier scale reading
=10+0.1×8
=10.8 mm

Reading of diameter=10.8 mm-0.4 mm=10.4 mm
radius of the bob=5.2 mm
=0.52 cm=52×10-2 cm

Topic: Physics, Experimental Physics, JEE Main, JEE Main 2021 (31 Aug Shift 2)
The diameter of an air bubble which was initially 2 mm, rises steadily through a solution of density $1750 \mathrm{~kg} / \mathrm{m}^3$ at the rate of $0.35 \mathrm{~cm} / \mathrm{s}$. Coefficient of viscosity of the solution is ______ (Assume mass of the bubble to be negligible) (Answer in Poise to the nearest integer)

Options:

The correct answer is:
11

Solution:
As the bubble is rising steadily the net force acting on it will be zero.

(Because of density of air is negligible, so the value of mg can be neglected)

So, the force of buoyancy B is equal to the weight of the displaced liquid acting upward is equal to the viscous force acting downward, i.e., B=F.

⇒4π3R3ρg=6πηRv

Putting R=1 mm=10-3 m, ρ=1.75×103 kg m-3, g=10 m s-2 and v=0.35×10-2 m s-1

η=109=1·11

In SI unit, η=11 poise CGS.

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2022 (28 Jul Shift 1)
The diameter of one of the bases of a truncated cone is $100 \mathrm{~mm}$. If the diameter of this base is increased by $21 \%$ such that it still remains a truncated cone with the height and the other base unchanged, the volume also increases by $21 \%$. The radius of the other base (in $\mathrm{mm}$ ) is-

Options:
A 65
B 55
C 45
D 35

The correct answer is:
55

Solution:
Let initially 2 bases have radii $5 \mathrm{~cm}$ and $\mathrm{r} \mathrm{cm}$. Finally base have radii $(1.21 \times 5)$ and $\mathrm{r}$.
Ratios of volumes $=\frac{V_{2}}{V_{1}}=1.21$
$\begin{array}{l}
\mathrm{V}_{2}=\frac{\pi \mathrm{h}}{3}\left[(6.05)^{2}+(6.05) \mathrm{r}+\mathrm{r}^{2}\right] \\
\mathrm{V}_{1}=\frac{\pi \mathrm{h}}{3}\left[5^{2}+5 \mathrm{r}+\mathrm{r}^{2}\right] \\
\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}=1.21 \Rightarrow \frac{(6.05)^{2}+(6.05) \mathrm{r}+\mathrm{r}^{2}}{5^{2}+5 \mathrm{r}+\mathrm{r}^{2}}=1.21 \\
\Rightarrow \mathrm{r}^{2}=\frac{6.3525}{21} \\
\Rightarrow \mathrm{r}=\frac{11}{2} \mathrm{~cm}=55 \mathrm{~mm}
\end{array}$

Topic: Mathematics, Application of Derivatives, KVPY, KVPY 2011 (SA)
The diameter of the circle, whose Centre lies on the line x+y=2 in the first quadrant and which touches both the lines x=3 and y=2 is

Options:

The correct answer is:
3

Solution:
Centre of circle lies on the line x+y=2.

Any point on the line can be assumed as α, 2-α.

So let Centre of circle is α, 2-α.

Since both lines x=3 and y=2 touches the circle means distance from the Centre of circle to the line is equal to radius.

Let r is radius of circle.

For the line x=3 ⇒x-3=0

r=α-31=α-3 ...1

For the line y=2 ⇒y-2=0

r=2-α-21=α ...2

From 1 and 2

α=α-3

Squaring both side

⇒α2=α-32 ⇒α2=α2-6α+9

⇒6α=9 ⇒α=32

We know r=α ⇒r=32.

Diameter=2r=232=3.

Topic: Mathematics, Circle, JEE Main, JEE Main 2020 (03 Sep Shift 1)
The diameter of the circumcircle of a triangle with sides $5 \mathrm{~cm}, 6 \mathrm{~cm}$ and $7 \mathrm{~cm}$ is

Options:
A $\frac{3 \sqrt{6}}{2} \mathrm{~cm}$
B $2 \sqrt{6} \mathrm{~cm}$
C $\frac{35}{48} \mathrm{~cm}$
D None of these

The correct answer is:
None of these

Solution:
Radius of circumcircle is given by $R=\frac{a b c}{4 \Delta}$ and $\Delta=\sqrt{s(s-a)(s-b)(s-c)}$ where $s=\frac{a+b+c}{2}$ Here $a=5 \mathrm{~cm}, b=6 \mathrm{~cm}$, and $\mathrm{c}=7 \mathrm{~cm}$
$\therefore \quad s=\frac{5+6+7}{2}=9$
$\Delta=\sqrt{9(9-5)(9-6)(9-7)}=\sqrt{216}=6 \sqrt{6}$
$\Rightarrow R=\frac{5.6 .7}{4.6 \cdot \sqrt{6}}=\frac{35}{4 \sqrt{6}}$
Diameter $=2 R=\frac{35}{2 \sqrt{ } 6}$

Topic: Mathematics, Properties of Triangles, JEE Main
The diameter of the objective lens of the microscope makes an angle β at the focus of the microscope. Further, the medium between the object and the lens is the oil of the refractive index n. Then the resolving power of the microscope.

Options:
A Increases with decreasing value of β
B Increases with increasing value of nsin2β
C Increases with increasing value of 1nsin2β
D Increases with decreasing value of n

The correct answer is:
Increases with increasing value of nsin2β

Solution:
R.P.microscope=2nsin2βλ

⇒R.P.∝nsin2β

Topic: Physics, Ray Optics, JEE Main, JEE Main 2014 (19 Apr Online)
The diameter of the objective of a telescope is a, its magnifying power is $m$ and wavelength of light is $\lambda$. The resolving power of the telescope is

Options:
A $(1.22 \lambda) / a$
B $(1.22 a) / \lambda$
C $\lambda m /(1.22 a)$
D $a /(1.22 \lambda)$

The correct answer is:
$a /(1.22 \lambda)$

Solution:
Resolving power of telescope:
$\mathrm{R}=\frac{1}{\theta}=\frac{\mathrm{a}}{1.22 \lambda}$
where $\theta$ is angular resolution, $a$ is diameter of the objective and $\lambda$ is wavelength of light.

Topic: Physics, Ray Optics, JEE Main
The diameters of a circle are along $2 x+y-7=0$ and $x+3 y-11=0$. Then, the equation of this circle, which also passes through $(5,7)$, is

Options:
A $x^2+y^2-4 x-6 y-16=0$
B $x^2+y^2-4 x-6 y-20=0$
C $x^2+y^2-4 x-6 y-12=0$
D $x^2+y^2+4 x+6 y-12=0$

The correct answer is:
$x^2+y^2-4 x-6 y-12=0$

Solution:
The intersection point of diameter lines is $(2,3)$ which is the centre of circle.
Now, radius
$\begin{aligned}
& =\sqrt{(5-2)^2+(7-3)^2} \\
& =\sqrt{9+16}=5
\end{aligned}$
$\therefore$ Required equation of circle is
$\begin{aligned}
(x-2)^2+(y-3)^2 & =5^2 \\
\Rightarrow \quad x^2+y^2-4 x-6 y-12 & =0
\end{aligned}$

Topic: Mathematics, Circle, JEE Main
The diameters range of colloidal particles is approximately.

Options:
A 1 to 1000 nm
B 1000 to 2000 nm
C 2000 to 3000 nm
D 3000 to 4000 nm

The correct answer is:
1 to 1000 nm

Solution:
A colloid is a heterogeneous system in which one substance is dispersed (dispersed phase) as very fine particles in another substance called dispersion medium. The essential difference between a solution and a colloid is that of particle size. While in a solution, the constituent particles are ions or small molecules, in a colloid, the dispersed phase may consist of particles of a single macromolecule (such as protein or synthetic polymer) or an aggregate of many atoms, ions or molecules. Colloidal particles are larger than simple molecules but small enough to remain suspended. Their range of diameters is between 1 and 1000 nm (10–9 to 10–6 m)

Topic: Chemistry, Surface Chemistry, AP EAMCET, AP EAMCET 2022 (04 Jul Shift 1)
The diazonium salt of which of the following compounds will form a coloured dye on reaction with β-Naphthol in NaOH?

Options:
A
B
C
D

The correct answer is:

Solution:
The colour obtained is the colour of the compound formed in the reaction of aniline with NaNO2​/HCl at 0oC followed by coupling with β-naphthol.

Aniline on reaction with NaNO2/HCl forms a diazonium salt. This diazonium salt is then coupled with β-naphthol.

Topic: Chemistry, Amines, JEE Main, JEE Main 2021 (24 Feb Shift 2)
The dielectric constant $k$ of an insulator cannot be

Options:
A 3
B 6
C 8
D $\infty$

The correct answer is:
$\infty$

Solution:
Because for metal $K=$.infinite

Topic: Physics, Capacitance, JEE Main
The dielectric constant of H 2 O is 80. The electrostatic force of attraction between N a + and C l â will be

Options:
A Reduced to 1 80 in water than in air
B Reduced to 1 40 in water than in air
C Will remain unchanged
D Will be increased to 80 in water than in air

The correct answer is:
Reduced to 1 80 in water than in air

Solution:
Electrostatic forces of attraction is inversely proportional to dielectric constant of medium ( H 2 O ) i.e. Fâ 1 Îµ therefore, electrostatic forces of attraction are reduced to 1 80âth in water.

Topic: Chemistry, s Block Elements, NEET
The dielectric slab is introduced between the plates of a parallel plate charged capacitor. Which one of the following quantities will not change?

Options:
A The charge on the capacitor.
B The potential difference between the plates.
C The energy stored in the capacitor.
D The electric field between the plates of a capacitor.

The correct answer is:
The charge on the capacitor.

Solution:
Only the amount of charge stored in the capacitor will remain constant when the dielectric is inserted between the plates (isolated system).
Charge will redistribute, such on the plate $Q\left(1-\frac{1}{K}\right)$ and $\frac{Q}{K}$ on the dielectric interface.
The electric field present between the plates of capacitor changes because it varies with the dielectric constant:
$E^{\prime}=\frac{Q}{4 \pi \varepsilon_0 K}=\frac{E}{K}$
The capacitance increases as:
$C^{\prime}=\frac{K \varepsilon_0 A}{d}=K C$
The dielectric constant also affects the potential difference as:
$\begin{aligned} & Q=C^{\prime} V^{\prime}=C K V^{\prime}=C V \\ & \therefore V^{\prime}=\frac{V}{K}\end{aligned}$
and the energy stored in the capacitor:
$U^{\prime}=\frac{1}{2} \frac{Q^2}{C^{\prime}}=\frac{U}{K}$

Topic: Physics, Capacitance, MHT CET, MHT CET 2022 (10 Aug Shift 1)
The difference between any two consecutive interior angles of a polygon is \( 5^{\circ} \). If the smallest angle is \( 120^{\circ} \), then the number of the sides of the polygon is

Options:

The correct answer is:
9.00

Solution:
Here the angles of a polygon of n sides form an A.P. whose first term is 120o and common difference is 5o.
Now, the sum of interior angles
Sn=n2[2a+(n−1)d]=n2[2×120+(n−1)5]
Sn=n2[240+5n−5]=n2(235+5n) ...(1)
Also the sum of interior angles =180( n−2 )
From equation (1),
∴n2235+5n=180n-360
⇒5n2(47+n)=180(n−2)
⇒n(47+n)=72(n−2)
⇒n2+47n=72n−144
⇒n2-25n+144=0
⇒(n-16)(n-9)=0

⇒n≠16∴ n=9.
∴ n≠16, because for n=16, one of the interior angle will become 180 o , which is not possible.
For n=16⇒ 120 o ,  125 o ,......., 180 o ,....., 195 o and for n=9⇒ 120 o ,  125 o , ......, 160 o .

Topic: Mathematics, Sequences and Series, JEE Main
The difference between bond orders of CO and NO⊕ is x2 where x=
(Round off to the Nearest Integer)

Options:

The correct answer is:
0

Solution:
Bond order of CO=3
Bond order of NO+=3
Difference =0=x2
x=0

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2021 (27 Jul Shift 1)
The difference between bond orders of \( \mathrm{CO} \) and \( \mathrm{NO}^{\oplus} \) is \( \frac{\mathrm{x}}{2} \) where \( \mathrm{x}= \)
(Round off to the Nearest Integer)

Options:

The correct answer is:
0

Solution:
Bond order of CO=3
Bond order of NO+=3
Difference =0=x2
x=0

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The difference between degree and order of a differential equation that represents the family of curves given by y2=ax+a2,a>0 is _______.

Options:

The correct answer is:
2

Solution:
y2=ax+a2=ax+a3/22 ...1

⇒2yy'=a

Put in equation 1 we get,

y2=2yy'x+2yy'3/22

y2-2xyy'=2yy'3/22

Squaring both the sides we get,

y2-2xyy'2=2y3y'3

∴ Order =1

Degree =3

Degree - order =3-1=2

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2021 (26 Feb Shift 1)
The difference between electron gain enthalpies will be maximum between :

Options:
A Ne and F
B Ar and F
C Ne and Cl
D Ar and Cl

The correct answer is:
Ne and Cl

Solution:
Electron gain enthalpy is defined as the amount of energy released when an electron is added to an isolated gaseous atom. During the addition of an electron, energy can either be released or absorbed. Electron gain enthalpy decreases down the group because as we move down the group, the size of an element increases, and the effective nuclear charge decreases. Due to this, the incoming electron will experience less attraction. Thus, electron gain enthalpy decreases as we move down the group.

ΔHeg for chlorine=-349 kJ mole-1

ΔHeg for Neon=+116 kJ mole-1

ΔHeg for Argon =+96 kJmol-1

ΔHeg for fluorine =-333 kJ mol-1

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2023 (06 Apr Shift 1)
The difference between kinetic energies of photoelectrons emitted from a surface by the light of wavelength 2500 Aâ and 5000 Aâ will be

Options:
A 1.61 eV

B 2.47 eV
C 3.96 eV
D 3.96Ã10-19eV

The correct answer is:
2.47 eV

Solution:
ÎE=hcÎ»1-hcÎ»2=hc(Î»2-Î»1)Î»1Î»2 (in eV)
=6.62Ã10-34Ã3Ã108Ã(5000-2500)Ã10-102500Ã5000Ã10-20Ã1.6Ã10-19
=2.47 eV

Topic: Physics, Dual Nature of Matter, JEE Main
The difference between lectin and lecithin is:

Options:
A Lectin is a secondary metabolite while lecithin is a phospholipid.
B Lectin is a fat binding protein while lecithin is a phospholipid.
C Lectin is a phospholipid with choline while lecithin is a carbohydrate-binding protein.
D Lectin is a phospholipid while lecithin is a secondary metabolite.

The correct answer is:
Lectin is a secondary metabolite while lecithin is a phospholipid.

Solution:
The main differentiating factor between lectin and lecithin is that lectin is a secondary metabolite. Concanavalin A is a plant protein which binds with carbohydrate whereas lecithin is a fatty substance which is found in plant and animal tissues. Lectin, due to its ability to bind with carbohydrate reduces nutrient absorption. Lecithin itself is an essential fatty nutrient, a major component of phosphatidylcholine with the ability to lower the bad cholesterol in the body.

Topic: Chemistry, Biomolecules, NEET
The difference between oogenesis and spermatogenesis is

Options:
A oogenesis begins at puberty and spermatogenesis during embryonic development.
B oogenesis results in formation of four haploid ova from a primary oocyte, while during spermatogenesis only one sperm is formed from a primary spermatocyte.
C the cytoplasmic division is unequal in oogenesis and equal in spermatogenesis.
D spermatogenesis is not completed until after fertilization occurs, but oogenesis is completed at the time of birth.

The correct answer is:
the cytoplasmic division is unequal in oogenesis and equal in spermatogenesis.

Solution:
Oogenesis is formation of female gametes while spermatogenesis is formation of male gametes. Both begins with primary mother cells in the gonads. Primary oocyte undergoes meiosis 1 and forms secondary oocyte and first polar body while Primary spermatocyte undergoes meiosis1 and forms two secondary spermatocyte. The process is the same in both but the daughter cells formed are different. In oogenesis, the daughter cells are of unequal size i.e. secondary oocyte is larger than the first polar body because secondary oocyte has maximum amount of cytoplasm retained in it. This unequal division is important because incase sperms approach these gametes, there is a chance of fertilization and zygote formation. Hence cytoplasm can act as stored food for the zygote. While in spermatogenesis the daughter cells formed are of same size as seen in Meiosis of other cells. This is the main difference.

Topic: Biology, Human Reproduction, NEET
The difference between spin only magnetic moment values of CoH2O6Cl2 and CrH2O6Cl3 is

Options:

The correct answer is:
0

Solution:
CoH2O62+

Number of unpaired e-=3

μ=nn+2BMn=number of unpaired electrons

μ=15BM

CrH2O63+

number of unpaired e-=3

μ=15BM

Difference in spin only magnetic moment =0

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2022 (26 Jul Shift 1)
The difference between the fourth term and the first term of a Geometrical Progresssion is 52. If the sum of its first three terms is 26 , then the sum of the first six terms of the progression is

Options:
A
63

B
189

C
728

D
364

The correct answer is:
728

Solution:
Let $a, a r, a r^2, a r^3, a r^4, a r^5$ be six terms of a G.P. where ' $a$ ' is first term and $\mathrm{r}$ is common ratio.
According to given conditions, we have $a r^3-a=5 \Rightarrow a\left(r^3-1\right)=52$ and $a+a r+a r^2=26$ $\Rightarrow a\left(1+r+r^2\right)=26$
To find: $a\left(1+r+r^2+r^3+r^4+r^5\right)$
Consider
$$
\begin{aligned}
& a\left[1+r+r^2+r^3+r^4+r^5\right] \\
& =a\left[1+r+r^2+r^3\left(1+r+r^2\right)\right] \\
& =a\left[1+r+r^2\right]\left[1+r^3\right]
\end{aligned}
$$
Divide (1) by (2), we get
$$
\frac{r^3-1}{1+r+r^2}=2
$$
we know $r^3-1=(r-1)\left(1+r+r^2\right)$
$$
\begin{aligned}
& \therefore r-1=2 \Rightarrow r=3 \text { and } a=2 \\
& \therefore \quad a\left(1+r+r^2+r^3+r^4+r^5\right) \\
& \quad=a\left(1+r+r^2\right)\left(1+r^3\right) \\
& \quad=2(1+3+9)(1+27) \\
& \quad=26 \times 28=728
\end{aligned}
$$

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2012 (07 May Online)
The difference between the greatest and the least possible value of the expression 3-cosâ¡x+sin2â¡x is

Options:
A 134
B 174
C 94
D 14

The correct answer is:
94

Solution:
3-cos⁡x+1-cos2⁡x
=-cos2⁡x+cos⁡x+4
=-cos2⁡x+cos⁡x+14+4+14
=-cos⁡x+122+174
=174-cos⁡x+122
Maximum value at cos⁡x=-12=174
Minimum value at cos⁡x=1=174-94=2
Difference =174-2=94

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
The difference between the maximum and minimum values of the function fx=sin3â¡x-3sinâ¡x, âxâ0,Ï6 is

Options:
A 2
B 12
C 118
D 76

The correct answer is:
118

Solution:
Let, sin⁡x=t
∴ If x∈0,π6 ⇒t∈0,12
Now, ft=t3-3t
⇒f't=3t2-3=3t-1t+1
Thus, ft is decreasing ∀t∈0,12
∴ftmax=f0=0
ftmin=f12=18-32=-118
∴fmax-fmin=118

Topic: Mathematics, Application of Derivatives, JEE Main
The difference between the maximum and minimum values of the function fx=x3-3x+4, ∀x∈0,1 is

Options:

The correct answer is:
2

Solution:
As, f ′ x =3 x 2 −3 =3x-1x+1
Hence, fx is decreasing ∀x∈0,1
Thus, f(x)max=f0=4 & f(x)min=f1=2
∴ the required difference =2

Topic: Mathematics, Application of Derivatives, JEE Main
The difference between the radii of 3rd and 4th orbits of Li2+ is ΔR1. The difference between the radii of 3rd and 4th orbits of He+ ΔR2. Ratio ΔR1:ΔR2 is :

Options:
A 8:3
B 3:8
C 2: 3
D 3: 2

The correct answer is:
2: 3

Solution:
r=0.529n27A

For Li2+

rL2+n=4-rL2+n=3=0.5293[16-9]=ΔR1

For He+

rHe+n=4-rHe+n=3=0.5292[16-9]=ΔR2

ΔR1ΔR2=23

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2020 (05 Sep Shift 1)
The difference between the reaction enthalpy change $\left(\Delta_{\mathrm{r}} \mathrm{H}\right)$ and reaction internal energy change $\left(\Delta_{\mathrm{r}} \mathrm{U}\right)$ for the reaction:
$$
2 \mathrm{C}_6 \mathrm{H}_6(\mathrm{l})+15 \mathrm{O}_2(\mathrm{~g}) \longrightarrow
$$
at $300 \mathrm{~K}$ is $\left(\mathrm{R}=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right)$

Options:
A
$0 \mathrm{~J} \mathrm{~mol}^{-1}$

B
$2490 \mathrm{~J} \mathrm{~mol}^{-1}$

C
$-2490 \mathrm{~J} \mathrm{~mol}^{-1}$

D
$-7482 \mathrm{~J} \mathrm{~mol}^{-1}$

The correct answer is:
$-7482 \mathrm{~J} \mathrm{~mol}^{-1}$

Solution:
$\Delta H=\Delta U+\Delta n_{\mathrm{g}} R T$
For the reaction $\Delta n_{\mathrm{g}}=12-15=-3$
$$
\begin{aligned}
\Delta H-\Delta U & =-3 \times 8.314 \times 300 \\
& =-7482 \mathrm{~J} \mathrm{~mol}^{-1}
\end{aligned}
$$

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2012 (12 May Online)
The difference between the wavelengths of the stokes line and anti-stokes lines in the Raman spectrum of H-Br molecule is $100 Å$. If the wavelength of the anti-stokes line is $5000 Å$, then the wavelength of the incident radiation is

Options:
A $5050 Å$
B $4950 Å$
C $5100 Å$
D $4900 Å$

The correct answer is:
$5050 Å$

Solution:
Stokes lines are the radiation of certain wavelengths present in the line spectra, that is associated with the phenomena of fluorescence and the Raman effect. Anti-stokes lines are produced due to already excited atoms or molecules. These lines are symmetrical about the main lines and the wavelength of anti-stokes lines is less than the main line. So, if the difference between wavelengths of stokes line and corresponding anti-stokes line is $100 Å$, then the difference between anti-stokes line and main line is $50 Å$. Therefore, wavelength of incident radiation is $5050 Å$.

Topic: Physics, Atomic Physics, COMEDK, COMEDK 2012
The difference between threshold wavelengths for two metal surfaces A and B having work function ϕA=9eV and ϕB=4.5eV in nm is:

{Given, hc=1242 eV nm}

Options:
A 540
B 276
C 264
D 138

The correct answer is:
138

Solution:
The work function is given by

hcλ=ϕ

⇒λ=hcϕ

By the given data,

λA=12429 nm, λB=12424.5 nm,

The difference between the wavelengths,

λB-λA=12424.5-12429 nm=12429 nm

=138 nm

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2023 (13 Apr Shift 1)
The difference between two acute angle of a right angle triangle is $\left(\frac{\pi}{9}\right)$. Then the angles in degree are -

Options:
A $50^{\circ}, 30^{\circ}$
B $25^{\circ}, 45^{\circ}$
C $20^{\circ}, 40^{\circ}$
D $35^{\circ}, 55^{\circ}$

The correct answer is:
$35^{\circ}, 55^{\circ}$

Solution:
In triangle $\mathrm{ABC}$ let $\angle \mathrm{C}=90^{\circ}$
So $\angle A-\angle B=\left(\frac{\pi}{9}\right)^c=20^{\circ}$
and sum of all the angles in $\triangle \mathrm{ABC}$
$\begin{aligned}
& \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180^{\circ} \\
& \because \angle \mathrm{C}=90^{\circ} \\
& \angle \mathrm{A}+\angle \mathrm{B}=90^{\circ}
\end{aligned}$
Solving (i) & (ii)
$\angle A=55^{\circ}, \angle B=35^{\circ}$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
The difference in energy between the actual structure and the lowest energy resonance structure for the given compound is:

Options:
A electromeric energy
B resonance energy
C ionization energy
D hyperconjugation energy

The correct answer is:
resonance energy

Solution:
The difference in energy between the actual structure and the lowest energy resonance structure for the given compound is known as resonance energy.

The resonance energy of a compound is a measure of the extra stability of the conjugated system compared to the corresponding number of isolated double bonds.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2024 (29 Jan Shift 1)