datavorous/entrance-exam-dataset huggingface dataset:
In a Young's double slit experiment, the intensity at a point is $\left(\frac{1}{4}\right)^{\text {th }}$ of the maximum intensity, the minimum distance of the point from the central maximum is ________ $\mu \mathrm{m}$.
(Given : $\lambda=600 \mathrm{~nm}, \mathrm{~d}=1.0 \mathrm{~mm}, \mathrm{D}=1.0 \mathrm{~m}$ )

Options:

The correct answer is:
200

Solution:
$\begin{aligned} & \mathrm{I}=\mathrm{I}_0 \cos ^2\left(\frac{\Delta \phi}{2}\right) \\ & \frac{\mathrm{I}_0}{4}=\cos ^2\left(\frac{\Delta \phi}{2}\right) \\ & \Delta \phi=\frac{2 \pi}{3} \\ & \frac{2 \pi}{\lambda}\left(\frac{\mathrm{yd}}{\mathrm{D}}\right)=\frac{2 \pi}{3} \\ & \mathrm{y}=\frac{\lambda \mathrm{D}}{3 \mathrm{~d}}=\frac{600 \times 10^{-9} \times 1}{3 \times 10^{-3}}=2 \times 10^{-4} \mathrm{~m}\end{aligned}$

Topic: Physics, Wave Optics, JEE Main, JEE Main 2024 (09 Apr Shift 1)
In a Young's double slit experiment the intensity at a point where the path difference is $\frac{\lambda}{6}$ ($\lambda$ being the wavelength of the light used) is $\mathrm{I}$. If $\mathrm{I}_0$ denotes the maximum intensity, $\frac{\mathrm{I}}{\mathrm{I}_0}$ is equal to

Options:
A
$\frac{1}{\sqrt{2}}$

B
$\frac{\sqrt{3}}{2}$

C
$1 / 2$

D
$3 / 4$

The correct answer is:
$3 / 4$

Solution:
$\frac{\mathrm{I}}{\mathrm{I}_{\max }}=\cos ^2\left(\frac{\phi}{2}\right)$

Topic: Physics, Wave Optics, JEE Main, JEE Main 2007
In a Young's double slit experiment, the path difference, at a certain point on the screen, betwen two interfering waves is $\frac{1}{8}$ th of wavelength. The ratio of the intensity at this point to that at the centre of a bright fringe is close to:

Options:
A 0.74
B 0.85
C 0.94
D 0.8

The correct answer is:
0.85

Solution:
Given, path difference, $\Delta x=\frac{\lambda}{8}$

Phase difference $(\Delta \phi)$ is given by

$\Delta \phi=\frac{2 \pi}{\lambda}(\Delta \mathrm{x})$

$\Delta \phi=\frac{(2 \pi)}{\lambda} \frac{\lambda}{8}=\frac{\pi}{4}$

For two sources in different phases,$$

\begin{array}{l}

I=I_{0} \cos ^{2}\left(\frac{\pi}{8}\right) \\

\frac{1}{I_{0}}=\cos ^{2}\left(\frac{\pi}{8}\right) \\

=\frac{1+\cos \frac{\pi}{4}}{2}=\frac{1+\frac{1}{\sqrt{2}}}{2}=0.85

\end{array}

$$

Topic: Physics, Wave Optics, JEE Main, JEE Main 2019 (11 Jan Shift 1)
In a Young’s double slit experiment, the ratio of amplitude of light coming from slits is 2 : 1. The ratio of the maximum to minimum intensity in the interference pattern is

Options:
A 9 : 4
B 25 : 9
C 2 : 1
D 9 : 1

The correct answer is:
9 : 1

Solution:
The ratio of the maximum to minimum intensity is given by

ImaxImin=I1+I22I1-I22

It is given that A1A2=21

ImaxImin=A1+A2A1-A22=A222+12A22(2-1)2=2+12-12=91

Topic: Physics, Wave Optics, JEE Main, JEE Main 2023 (13 Apr Shift 2)
In a Young’s double slit experiment, the separation between the slits is 0.15mm . In the experiment, a source of light of wavelength 589nm is used and the interference pattern is observed on a screen kept 1.5m away. The separation between the successive bright fringes on the screen is:

Options:
A 6.9mm
B 3.9mm
C 5.9mm
D 4.9mm

The correct answer is:
5.9mm

Solution:
β=λDd=589×10-9×1.50.15×10-3=5.9mm

Topic: Physics, Wave Optics, JEE Main, JEE Main 2020 (07 Jan Shift 2)
In a Young's double slit experiment, the separation of the two slits is doubled. To keep the same spacing of fringes, the distance $D$ of the screen from the slits should be made

Options:
A $\frac{D}{2}$
B $\frac{D}{\sqrt{2}}$
C $2 D$
D $4 D$

The correct answer is:
$2 D$

Solution:
$\beta=\frac{\lambda D}{d}$

Topic: Physics, Wave Optics, JEE Main
In a young's double slit experiment, the slits are placed 0.320 mm apart. Light of wavelength λ=500 nm is incident on the slits. The total number of bright fringes that are observed in the angular range -30o≤θ≤30o is:

Options:
A 321
B 641
C 320
D 640

The correct answer is:
641

Solution:
dsinθ=nλ
n=0.320×10-3×12500×10-9
=320
Hence number of maxima =320×2+1=641

Topic: Physics, Wave Optics, JEE Main, JEE Main 2019 (09 Jan Shift 2)
In a Young's double slit experiment, the slits are separated by 0.3 mm and the screen is 1.5 m away from the plane of slits. Distance between fourth bright fringes on both sides of central bright fringe is 2.4 cm. The frequency of light used is x×1014 Hz.

Options:

The correct answer is:
5

Solution:
The distance between nth bright fringe on both side of central maxima is 2.4 cm. The f=?

Δy=Dλd×8
λ=dΔy8D=0.3×10-3×2.4×10-28×1.5=0.12×10-52=0.06×10-5
λ=6×10-7 m
f=C6×10-7=0.5×10+15
f=5×1014 Hz
So, x=5

Topic: Physics, Wave Optics, JEE Main, JEE Main 2021 (31 Aug Shift 2)
In a Young's double slit experiment, the width of the one of the slit is three times the other slit. The amplitude of the light coming from a slit is proportional to the slit-width. Find the ratio of the maximum to the minimum intensity in the interference pattern.

Options:
A 1:4
B 2:1
C 3:1
D 4:1

The correct answer is:
4:1

Solution:
⇒A2=3 A1

ImaxImin=I1+I2I1-I22

∵ Intensity I∝A2

⇒ImaxImin=A1+A2A1−A22

=A1+3A1A1−3A12

=4A12A12=4:1

Topic: Physics, Wave Optics, JEE Main, JEE Main 2021 (24 Feb Shift 1)
In a Young’s double slit experiment, two slits are illuminated with a light of wavelength 800 nm. The line joining A1P is perpendicular to A1A2 as shown in the figure. If the first minimum is detected at P, the value of slits separation a will be :

The distance of screen from slits D=5 cm.

Options:
A 0.4 mm
B 0.5 mm
C 0.2 mm
D 0.1 mm

The correct answer is:
0.2 mm

Solution:
The position of first minima is

y=λD2×a(y is height of the point from the axis passing through the mid-point of A1 and A2)

Since the first minima is in front of the slit as shown in the figure,

λD2×a=a2

⇒a=λD=800×10-9×5×10-2=2×10-4 m=0.2 mm

Topic: Physics, Wave Optics, JEE Main, JEE Main 2023 (29 Jan Shift 1)
In a Young's double slit experiment two slits are separated by 2 mm and the screen is placed one meter away. When a light of wavelength 500 nm is used, the fringe separation will be :

Options:
A 1 mm
B 0.50 mm
C 0.75 mm
D 0.25 mm

The correct answer is:
0.25 mm

Solution:
β=λDd=500×10-9×12×10-3

β=52×10-4 m=2.5×10-1 mm

b=0.25 mm

Topic: Physics, Wave Optics, JEE Main, JEE Main 2021 (26 Feb Shift 1)
In a Young's double slit experiment using monochromatic light of wavelengths λ, the intensity of light at a point on the screen with a path difference λ is M units and the intensity of light at a point where the path difference is λ3 is nM units. What is the value of n?

Options:

The correct answer is:
0.25

Solution:
Resultant intensity IR=I1+I2+2I1I2cosθ
If path difference, =λ then phase difference =2π ….∵ϕ=2πλ×Δx
∴ IR=I+I+2I×Icos2π …..(∵I1=I2=I)
=4I=M
If path difference =λ3
then phase difference =2πλ×λ3=2π3rad
IR'=I+I+2I×Icos2π3
=2I+2I-12=I=M4=0.25M

Topic: Physics, Wave Optics, JEE Main
In a Young's double slit experiment with light of wavelength $\lambda$, fringe pattern on the screen has fringe width $\beta$. When two thin transparent glass (refractive index $\mu$ ) plates of thickness $t_1$ and $t_2$ $\left(t_1>t_2\right)$ are placed in the path of the two beams respectively, the fringe pattern will shift by a distance

Options:
A
$\frac{\beta(\mu-1}{\lambda}\left(\frac{t_1}{t_2}\right)$

B
$\frac{\mu \beta}{\lambda} \frac{t_1}{t_2}$

C
$\frac{\beta(\mu-1}{\lambda}\left(t_1-t_2\right.$

D
$\left(\mu-1 \frac{\lambda}{\beta}\right)\left(t_1+t_2 \quad\right)$

The correct answer is:
$\frac{\beta(\mu-1}{\lambda}\left(t_1-t_2\right.$

Solution:
$$
\text { } \begin{aligned}
\text { Shift } & =\frac{\beta(\mu-1)}{\lambda} t_1-\frac{\beta(\mu-1)}{\lambda} t_2 \\
& =\frac{\beta(\mu-1)}{\lambda}\left(t_1-t_2\right)
\end{aligned}
$$

Topic: Physics, Wave Optics, JEE Main, JEE Main 2012 (07 May Online)
In a Young's double slit experiment with light of wavelength λ, the separation of slits is d and distance of screen is D such that D≫d≫λ . If the Fringe width is β , the distance from point of maximum intensity to the point where intensity falls to half of the maximum intensity on either side is:

Options:
A β4
B β3
C β6
D β2

The correct answer is:
β4

Solution:
As Imax=4I
Then Inet=12 Imax=2I
⇒2I=2I1+cosϕ
⇒1+cosϕ=1
⇒cosϕ=0
⇒ϕ=π2,3π2,5π2, …
ϕ=2πλΔx
⇒π2=2πλΔx
⇒Δx=λ4,3λ4,5λ4, …
Also path difference Δx=dsin⁡θ =λ4
At desired location
⇒d ynD=λ4
⇒ yn=λD4d=β4 ∵β=λDd
⇒ Separation between central maxima and desired y=β4

Topic: Physics, Wave Optics, JEE Main, JEE Main 2015 (11 Apr Online)
In a Young's double slit experiment with light of wavelength λ, the separation of slits is d and distance of screen is D such that D≫d≫λ . If the Fringe width is β , the distance from point of maximum intensity to the point where intensity falls to half of the maximum intensity on either side is:

Options:
A β4
B β3
C β6
D β2

The correct answer is:
β4

Solution:
As Imax=4I
Then Inet=12 Imax=2I
â2I=2I1+cosÏ
â1+cosÏ=1
âcosÏ=0
âÏ=Ï2,3Ï2,5Ï2, â¦
Ï=2ÏÎ»Îx
âÏ2=2ÏÎ»Îx
âÎx=Î»4,3Î»4,5Î»4, â¦
Also path difference Îx=dsinâ¡Î¸ =Î»4
At desired location
âd ynD=Î»4
â yn=Î»D4d=Î²4 âµÎ²=Î»Dd
â Separation between central maxima and desired y=Î²4

Topic: Physics, Wave Optics, JEE Main
In a Young's double-slit experiment, the ratio of the slit's width is 4 :1 . The ratio of the intensity of maxima to minima, close to the central fringe on the screen, will be

Options:
A 25 :9
B 9 :1
C 3+14 :16
D 4 :1

The correct answer is:
9 :1

Solution:
Intensity ∝ width of slit
I1I2=41.
ImaxImin=I1+I2I1-I22=I1I2+1I1I2-12
=4+14-12=312=91
Imax :Imin=9:1

Topic: Physics, Wave Optics, JEE Main, JEE Main 2019 (10 Apr Shift 2)
In a zener diode, the reverse bias voltage is \( 3 \mathrm{~V} \) and the width of the depletion region is \( 300 \mathrm{~A} \). The electric field intensity will be \( _{-} V \mathrm{~cm}^{-1} \).

Options:
A \( 10^{4} \)
B \( 10^{6} \)
C \( 10^{8} \)
D \( 10^{-2} \)

The correct answer is:
\( 10^{6} \)

Solution:
Given, reverse bias voltage of zener diode, V=3 V and the width of the depletion region, d=300 Ao.

The electric field intensity at depletion region is given by,

E=Vd

E=3300×10-10 V m-1

E=106 V cm-1 will be electric field at depletion region.

Topic: Physics, Semiconductors, JEE Main
In a Zener regulated power supply a Zener diode with $V_z=6 \mathrm{~V}$ is used for regulation. The load current is $4 \mathrm{~mA}$ and the unregulated input voltage is $10 \mathrm{~V}$. To get a Zener current
five times the load current the value of series resistor $R_S$ is nearly

Options:
A $150 \Omega$
B $167 \Omega$
C $175 \Omega$
D $159 \Omega$

The correct answer is:
$167 \Omega$

Solution:
Given, $I_Z=5 I_L=20 \mathrm{~mA}$
Total current $I_s$ through $R_s$ is $I_s=I_Z+I_L$
$=20+4=24 \mathrm{~mA}$
As a potential drop of $6 \mathrm{~V}$ occurs across Zener diode and supply voltage is $10 \mathrm{~V}$.
So, potential drop across $R_s, V_s=10-6=4 \mathrm{~V}$ $\therefore$ Series resistor,
$R_s=\frac{V_s}{I_s}=\frac{4}{24 \times 10^{-3}} \approx 167 \Omega$

Topic: Physics, Semiconductors, JEE Main
In a zero-order reaction, what happens to the half-life if the initial concentration of the reactants increases?

Options:
A The half-life increases.
B The half-life decreases.
C The half-life remains constant.
D The half-life becomes zero.

The correct answer is:
The half-life increases.

Solution:
In a zero-order reaction, the half-life is directly proportional to the initial concentration of the reactants. Therefore, if the initial concentration increases, the half-life also increases.

Topic: Chemistry, Chemical Kinetics, NEET
In $\triangle \mathrm{ABC}$, if a, b, c are in arithmetic progression and $\mathrm{C}=2 \mathrm{~A}$, then $\mathrm{a}: \mathrm{c}=$

Options:
A $4: 5$
B $2: 3$
C $5: 6$
D $\sqrt{3}: 2$

The correct answer is:
$2: 3$

Solution:
Given that $a, b, c$ are in A.P
$\begin{aligned}
& \Rightarrow \mathrm{b}=\frac{\mathrm{a}+\mathrm{c}}{2} \ldots \text { (i) } \\
& \text { And } \mathrm{C}=2 \mathrm{~A} \Rightarrow \sin \mathrm{C}=\sin 2 \mathrm{~A}=2 \sin \mathrm{A} \cdot \cos \mathrm{A} \\
& \Rightarrow \mathrm{C}=2 \mathrm{a} \cdot \frac{\left(\mathrm{b}^2+\mathrm{c}^2-\mathrm{a}^2\right)}{2 \mathrm{bc}} \\
& \Rightarrow 2 \mathrm{c}^3+3 \mathrm{a}^3-3 \mathrm{ac}^2-2 \mathrm{a}^2 \mathrm{c}=0 \\
& \Rightarrow 3\left(\frac{\mathrm{a}}{\mathrm{c}}\right)^3-2\left(\frac{\mathrm{a}}{\mathrm{c}}\right)^2-3\left(\frac{\mathrm{a}}{\mathrm{c}}\right)+2=0 \\
& \Rightarrow\left(3\left(\frac{\mathrm{a}}{\mathrm{c}}\right)-2\right)\left[\left(\frac{\mathrm{a}}{\mathrm{c}}\right)^2-1\right]=0 \Rightarrow \frac{\mathrm{a}}{\mathrm{c}}=\frac{2}{3}=2: 3
\end{aligned}$

Topic: Mathematics, Sequences and Series, JEE Main
In $\triangle \mathrm{ABC}$, with tustal notations, $\mathrm{m} \angle \mathrm{C}=\frac{\pi}{2}$, if $\tan \left(\frac{\mathrm{A}}{2}\right)$ and $\tan \left(\frac{\mathrm{B}}{2}\right)$ are the roots of the equation. $\mathrm{a}_1 x^2+\mathrm{b}_1 x+\mathrm{c}_1=0\left(\mathrm{a}_1 \neq 0\right)$, then

Options:
A $a_1+b_1=c_1$
B $b_1+c_1=a_1$
C $\mathrm{a}_1+\mathrm{c}_1-\mathrm{b}_1$
D $b_1=c_1$

The correct answer is:
$a_1+b_1=c_1$

Solution:
$\begin{array}{ll}
& \text { In } \triangle \mathrm{ABC}, \\
& \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180^{\circ} \\
\therefore \quad & \angle \mathrm{A}+\frac{\pi}{2}+\angle \mathrm{B}=180^{\circ} \\
\therefore \quad & \angle \mathrm{A}+\angle \mathrm{B}=\frac{\pi}{2} \\
\therefore \quad & \frac{\angle \mathrm{A}}{2}+\frac{\angle \mathrm{B}}{2}=\frac{\pi}{4}
\end{array}$
$\tan \left(\frac{A}{2}\right)$ and $\tan \left(\frac{B}{2}\right)$ are roots of equation $\mathrm{a}_1 x^2+\mathrm{b}_1 x+\mathrm{c}_1=0 ... [Given]$
$\begin{aligned} & \therefore \quad \text { Sum of roots }=\frac{-b_1}{a_1} \\ & \tan \left(\frac{A}{2}\right)+\tan \left(\frac{B}{2}\right)=\frac{-b_1}{a_1}\end{aligned}$
Also, $\tan \left(\frac{\mathrm{A}}{2}\right) \cdot \tan \left(\frac{\mathrm{B}}{2}\right)=\frac{\mathrm{c}_1}{\mathrm{a}_1}$
Using $\tan \left(\frac{\mathrm{A}}{2}+\frac{\mathrm{B}}{2}\right)=\frac{\tan \frac{\mathrm{A}}{2}+\tan \frac{\mathrm{B}}{2}}{1-\tan \frac{\mathrm{A}}{2} \tan \frac{\mathrm{B}}{2}}$, we get
$\begin{aligned}
& \tan \left(\frac{\pi}{4}\right)=\frac{\frac{-b_1}{a_1}}{1-\frac{c_1}{a_1}} \\
& 1=\frac{-b_1}{a_1-c_1} \\
& a_1-c_1=-b_1 \\
& a_1+b_1=c_1
\end{aligned}$

Topic: Mathematics, Quadratic Equation, JEE Main
In absence of formation of complex ions by the addition of a common ion, the solubility of a given salt is

Options:
A Increased
B Decreased
C Unaffected
D First increased and then decreased

The correct answer is:
Decreased

Solution:
Due to the common ion effect.

Topic: Chemistry, Ionic Equilibrium, JEE Main
In acetylene molecule, the two carbon atoms are linked by

Options:
A three $\sigma$-bonds
B three $\pi$-bonds
C one $\sigma$-bond and two $\pi$-bonds
D two $\sigma$-bonds and one $\pi$-bond

The correct answer is:
one $\sigma$-bond and two $\pi$-bonds

Solution:
In acetylene or ethyne molecule, the two carbon atoms are linked by one $\sigma$ and two $\pi$-bonds.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
In acidic medium, K2Cr2O7 shows oxidising action as represented in the half reaction

Cr2O72-+XH++Ye-→2 A+ZH2O X, Y, Z and A are respectively are:

Options:
A 8, 6, 4 and Cr2O3
B 14, 7, 6 and Cr3+
C 8, 4, 6 and Cr2O3
D 14, 6, 7 and Cr3+

The correct answer is:
14, 6, 7 and Cr3+

Solution:
Potassium dichromate, K2Cr2O7, is a common inorganic chemical reagent, most commonly used as an oxidizing agent in various laboratory and industrial applications.

The balanced reaction is,

Cr2O72-+14H++6e-→2Cr3++7H2O

X=14

Y=6

A=7

Topic: Chemistry, Redox Reactions, JEE Main, JEE Main 2024 (01 Feb Shift 1)
In acidic medium, potassium permanganate oxidises oxalic acid to

Options:
A oxalate
B carbon dioxide
C acetate
D acetic acid

The correct answer is:
carbon dioxide

Solution:
In acidic medium, oxalic acid is oxidised to $\mathrm{CO}_{2}$ (carbon dioxide) by $\mathrm{KMnO}_{4}$.
$$
\begin{aligned}
5(\mathrm{COOH})_{2}+2 \mathrm{KMnO}_{4}+3 \mathrm{H}_{2} \mathrm{SO}_{4} & \longrightarrow \\
\mathrm{K}_{2} \mathrm{SO}_{4}+2 \mathrm{MnSO}_{4}+10 \mathrm{CO}_{2}+8 \mathrm{H}_{2} \mathrm{O}
\end{aligned}
$$

Topic: Chemistry, d and f Block Elements, JEE Main
In acidiic medium Zn reduces nitrate ion to NH4+ ion according to the reaction

Zn +  NO 3 − → Zn 2+  +  NH 4 +  +  H 2 O (unbalanced)

How many moles of HCl are required to reduce half a mole of NaNO3 completely? Assume the availability of sufficient Zn.

Options:
A 5
B 4
C 3
D 2

The correct answer is:
5

Solution:
â´4Zn+NO3-+10H+â¶4Zn2++NH4++3H2O(Net equation)
Here H + will be obtained from HCl
âµ1 mole of NO3- (0r NaNO3) is reduced by
=10 moles of HCl
â´12 mole of NO3- will be reduced by
= 10Ã12 moles of HCl
= 5 moles of HCl

Topic: Chemistry, Redox Reactions, JEE Main
In aerosol, the dispersion medium is

Options:
A Solid.
B Liquid.
C Gas.
D Any of these.

The correct answer is:
Gas.

Solution:
Aerosol is a colloidal system of solid in gas, e.g, smoke. So, the dispersion medium in aerosol is gas.

Topic: Chemistry, Surface Chemistry, JEE Main
In alanylglycylleucylalanylvaline the number of peptide linkages is:

Options:

The correct answer is:
4

Solution:
A peptide bond is a chemical bond formed between two molecules when the carboxyl group of one molecule reacts with the amino group of the other molecule, releasing a molecule of water. This linkage is found along a peptide or protein chain.

There are five amino acids in the given sequence, so they will have four peptide bonds in between them.

Topic: Chemistry, Biomolecules, JEE Main
In alanylglycylleucylalanylvaline the number of peptide linkages is:

Options:

The correct answer is:
4

Solution:
A peptide bond is a chemical bond formed between two molecules when the carboxyl group of one molecule reacts with the amino group of the other molecule, releasing a molecule of water. This linkage is found along a peptide or protein chain.

There are five amino acids in the given sequence, so they will have four peptide bonds in between them.

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2022 (24 Jun Shift 2)
In aldehydes and ketones, carbon of carbonyl group is

Options:
A $s p^3$ hybridised
B $s p^2$ hybridised
C sp hxbridised
D Unhybridized

The correct answer is:
$s p^2$ hybridised

Solution:
$\begin{gathered} \gt C=O \\ s p^2 \text { hytridised }\end{gathered}$

Topic: Chemistry, Aldehydes and Ketones, JEE Main
In alkaline medium $\mathrm{ClO}_2$ oxidize $\mathrm{H}_2 \mathrm{O}_2$ in $\mathrm{O}_2$ and reduced itself in $\mathrm{Cl}^{-}$then how many mole of $\mathrm{H}_2 \mathrm{O}_2$ will oxidize by one mole of $\mathrm{ClO}_2$

Options:
A $1.0$
B $1.5$
C $2.5$
D $3.5$

The correct answer is:
$2.5$

Solution:

Topic: Chemistry, Redox Reactions, JEE Main
In alkaline medium, the reduction of permanganate anion involves a gain of ____________ electrons.

Options:

The correct answer is:
3

Solution:
In Faintly alkaline medium potassium permanganate changes to manganese dioxide in its oxidising action.

MnO4-+2H2O+3e-→MnO2+4OH-

The oxidation state of manganese in MnO4- is +7 and in MnO2 is +4.

Hence, it transfers 3 electrons during the oxidising action in alkaline medium.

Topic: Chemistry, Redox Reactions, JEE Main, JEE Main 2023 (10 Apr Shift 2)
In all its properties, hydrogen resembles

Options:
A Alkali metals only
B Halogen only
C Both alkali metals and halogens
D Neither alkali metals nor halogens

The correct answer is:
Both alkali metals and halogens

Solution:
$\mathrm{H}_2$ like halogens gains one electron to form stable duplet.
Also, it looses one electron like alkali metals to form $H^{+}$ion

Topic: Chemistry, Hydrogen, JEE Main
In allene $\left(\mathrm{C}_3 \mathrm{H}_4\right)$, the type(s) of hybridization of the carbon atoms is (are) :

Options:
A
$\mathrm{sp}$ and $\mathrm{sp}^3$

B
$\mathrm{sp}^2$ and $\mathrm{sp}$

C
only $\mathrm{sp}^2$

D
$\mathrm{sp}^2$ and $\mathrm{sp}^3$

The correct answer is:
$\mathrm{sp}^2$ and $\mathrm{sp}$

Solution:

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2014 (11 Apr Online)
In AM modulation, a signal is modulated on a carrier wave such that maximum and minimum amplitude are found to be 6 V and 2 V respectively. The modulation index is

Options:
A 100%
B 80%
C 60%
D 50%

The correct answer is:
50%

Solution:
Given here, Vmax=6 V and Vmin=2 V

Now, modulation index =Vmax-VminVmax+Vmin.×100%

=6-26+2×100%=50%

Topic: Physics, Communication System, JEE Main, JEE Main 2022 (25 Jul Shift 2)
In ammonium-phosphomolybdate, the oxidation state of Mo is +_______

Options:

The correct answer is:
6

Solution:
Phosphomolybdate is an anion containing both phosphorus and molybdenum, with the chemical formula NH43PO4·12MoO3.

The ammonium ion has a total charge of +3, and the phosphate ion has a total charge of -3. Therefore, the 12 MoO3 units in the compound must have a total charge of +3 - (-3) = +6.

Each MoO3 unit contains one Mo atom and three O atoms. The total oxidation state of the three O atoms is -6, since each O atom has an oxidation state of -2.

So,

Mo + (-6) = 0 Mo = +6

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2023 (06 Apr Shift 1)
In amplitude modulation

Options:
A Amplitude remains constant but frequency change
B Both amplitude and frequency do not change
C Both amplitude and frequency change
D Amplitude of the carrier wave changes according to information signal

The correct answer is:
Amplitude of the carrier wave changes according to information signal

Solution:
In amplitude modulation amplitude of the carrier wave changes according to information signal.

Topic: Physics, Communication System, MHT CET, MHT CET 2018
In amplitude modulation,

Options:
A both amplitude and frequency do not change according to information signal.
B amplitude remains constant but frequency changes according to information
signal.
C both amplitude and frequency change according to information signal.
D amplitude of carrier wave changes according to information signal.

The correct answer is:
amplitude of carrier wave changes according to information signal.

Solution:
amplitude of carrier wave changes according to information signal

Topic: Physics, Communication System, MHT CET, MHT CET 2020 (16 Oct Shift 1)
In amplitude modulation, the message signal \( V_{\mathrm{m}}(t)=10 \sin \left(2 \pi \times 10^{5} t\right) \) volts and carrier signal \( V_{\mathrm{C}}(t)=20 \sin \left(2 \pi \times 10^{7} t\right) \) volts. The modulated signal now contains the message signal with lower side band and upper side band frequency, therefore the bandwidth of modulated signal is \( \alpha \mathrm{kHz} \). The value of \( \alpha \) is:

Options:
A \( 200 \mathrm{kHz} \)
B \( 500 \mathrm{kHz} \)
C \( 100 \mathrm{kHz} \)
D \( 0 \)

The correct answer is:
\( 200 \mathrm{kHz} \)

Solution:
Bandwidth =2×fm

=2×105 Hz=200 kHz

Topic: Physics, Communication System, JEE Main
In amplitude modulation, the sinusoidal carrier frequency used is denoted by ωc and the signal frequency is denoted by ωm . The bandwidth Δωm of the signal is such that Δωm≪ωc. Which of the following frequencies is not contained in the modulated wave?

Options:
A ωc-ωm
B ωm

C ωc
D ωm+ωc

The correct answer is:
ωm

Solution:
∵ Δωm≪ωc
ωm will not be there in the frequencies.
In modulated wave

Topic: Physics, Communication System, JEE Main, JEE Main 2017 (02 Apr)
In amplitude modulation, the sinusoidal carrier frequency used is denoted by \( \omega_{c} \) and the signal frequency is denoted by \( \omega_{m} \). The bandwidth \( \left(\Delta \omega_{m}\right) \) of the signal is such that \( \Delta \omega_{m} \ll \omega_{c} \). Which of the following frequencies is not contained in the modulated wave?

Options:
A \( \omega_{c}-\omega_{m} \)
B \( \omega_{m} \)
C \( \omega_{c} \)
D \( \omega_{m}+\omega_{c} \)

The correct answer is:
\( \omega_{m} \)

Solution:
∵ Δωm≪ωc
ωm will not be there in the frequencies.
In modulated wave

Topic: Physics, Communication System, JEE Main