datavorous/entrance-exam-dataset huggingface dataset:
The walls of a closed cubical box of edge $60 \mathrm{~cm}$ are made of material of thickness $1 \mathrm{~mm}$ and thermal conductivity, $4 \times 10^{-4} \mathrm{cal} \mathrm{s}^{-1} \mathrm{~cm}^{-1}{ }^{\circ} \mathrm{C}^{-1}$. The interior of the box is maintained $1000^{\circ} \mathrm{C}$ above the outside temperature by a heater placed inside the box and connected across $400 \mathrm{~V}$ DC supply. The resistance of the heater is

Options:
A $4.41 \Omega$
B $44.1 \Omega$
C $0.441 \Omega$
D $441 \Omega$

The correct answer is:
$0.441 \Omega$

Solution:
Key Idea Heat transfer through conduction wall is given by mathematical expression,
$$
\frac{d Q}{d t}=k A \frac{\left(T_1-T_0\right)}{x}
$$
where, $\frac{d Q}{d t}=$ power transferred through the wall and $x=$ thickness of the wall
Here, side of cube, $a=60 \mathrm{~cm}$. Hence, area $A=6 a$,
$\because$ Total area $=6$ (area of one side of the cube)
thickness $x=0.1 \mathrm{~cm}$, thermal conductivity, $k=4 \times 10^{-4} \mathrm{cal} \mathrm{s}^{-1} \mathrm{~cm}^{-1}{ }^{\circ} \mathrm{C}^{-1}$, temperature difference, $\left(T_1-T_0\right)=1000^{\circ} \mathrm{C}$ and potential of DC source $V=400 \mathrm{~V}$
Hence, the power
$$
\begin{aligned}
P & =\frac{k A \times 4.184 \times\left(T_1-T_0\right)}{x} \\
P & =\frac{k 6 a^2 \times 4.184 \times 10^3}{x} \\
P & =\frac{4 \times 10^{-4} \times 6 \times(60)^2 \times 10^3 \times 4.184}{0.1} \mathrm{~J} \\
P & =361.49 \mathrm{~W}
\end{aligned}
$$
Given, voltage supply of $\mathrm{DC}=400 \mathrm{~V}$ Hence, power generated in a resistance,
$$
\begin{aligned}
& P=\frac{V^2}{R}=361.49 \Rightarrow R=\frac{V^2}{P} \\
& R=\frac{400 \times 400}{361.49 \times 10^3}=0.4426 \Omega
\end{aligned}
$$
Hence, the correct option is (c).

Topic: Physics, Thermal Properties of Matter, AP EAMCET, AP EAMCET 2019 (21 Apr Shift 1)
The walls of sieve tubes develop into sieve plates together getting associated with _________.

Options:
A Chiasmata
B Plasmodesmata
C Tractile fibres
D Ultra-cellular strands

The correct answer is:
Plasmodesmata

Solution:
The plasmodesmata act as channels between the cell walls of adjacent plant cells. They are mainly involved in transport and cell to cell recognition. Structurally, the sieve tubes wall is dispersed with plasmodesmata and are grouped together. Later, these areas of the tube walls and plasmodesmata develop into sieve plates.

Topic: Biology, Cell - The Unit of Life, NEET
The walls of the halls built for music concerts should

Options:
A Amplify sound
B Transmit sound
C Reflect sound
D Absorb sound

The correct answer is:
Absorb sound

Solution:
The wall of the hall build for the music concerts should absorb the sound completely and it should not reflect the sound back in the hall. The reflected sound results in the echo.

Topic: Physics, Waves and Sound, JEE Main
The water flows from a tap of diameter \( 1.25 \mathrm{~cm} \) with a rate of \( 5 \times 10^{-5} \mathrm{~m}^{3} \mathrm{~s}^{-1} \). The density and coefficient of viscosity of water are \( 10^{3} \mathrm{k} \mathrm{gm}^{-3} \) and \( 10^{-3} \mathrm{~Pa} . \mathrm{s} \) respectively. The flow of water is

Options:
A steady with Reynold's number \( 5100 \).
B turbulent with Reynold's number \( 5100 \).
C steady with Reynold's number \( 3900 \)
D turbulent with Reynold's number \( 3900 \)

The correct answer is:
turbulent with Reynold's number \( 5100 \).

Solution:
Given, diameter of pipe D=1.25 cm=1.25×10-2 m

Rate of flow of water Q=5×10-5 m3 s-1

Density of water ρ=1000 kg m-3

Viscosity of water η=10-3 Pa s

Now, we know that,

Reynold's number NR=ρvDη ...1; where v is the velocity of water.

Now, rate of flow of water Q=Area of cross section×cross timesspeed of flow.

⇒Q=πr2×v=πD24×v

⇒v=4QπD2

On putting this value in equation 1

NR=ρD4QηπD2=4ρQηπD

On substituting the values,

NR=4×1000×5×10-510-3×227×1.25×10-2

NR=5100

For NR>3000, the flow is turbulent. Hence, the flow of water is turbulent with Reynold’s number 5100.

Topic: Physics, Mechanical Properties of Fluids, JEE Main
The water gas on reacting with cobalt as a catalyst forms

Options:
A Methanal
B Methanoic acid
C Ethanol
D Methanol

The correct answer is:
Methanol

Solution:
Methanol can be produced from syn gas (a mixture of carbon monoxide and hydrogen) through a different process, mainly through synthesis reaction. In this process, syn gas is passed over a heterogeneous catalyst, at high temperatures and pressures.

COg+2H2g⏟Water gas→catalystCobaltCH3OHlMethanol

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2023 (08 Apr Shift 1)
The water having more dissolved O2 is :

Options:
A boiling water
B water at 80°C
C polluted water
D water at 4°C

The correct answer is:
water at 4°C

Solution:
On heating concentration of O2 in water decreases. So boiling water and water at 80°C having less O2 concentration. Polluted water also having less O2 concentration. So water at 4°C is having maximum O2 concentration.

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2021 (22 Jul Shift 1)
The water is filled up to a height of 12 m in a tank having vertical sidewalls. A hole is made in one of the walls at a depth h below the water level. The value of h for which the emerging stream of water strikes the ground at the maximum range is _____ m.

Options:

The correct answer is:
6

Solution:
R=2gh×(12-h)×2 g

4 h(12-h)=R

For maximum R

dRdh=0

⇒h=6 m

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2021 (27 Jul Shift 2)
The water quality of a pond was analysed and its BOD was found to be 4. The pond has

Options:
A Highly polluted water
B Water has high amount of fluoride compounds
C Very clean water
D Slightly polluted water

The correct answer is:
Very clean water

Solution:
The amount of BOD in the water is a measure of the amount of organic material in the water, in terms of how much oxygen will be required to break it down biologically.

Clean water has BOD value of less than 5 ppm whereas polluted water has a BOD value of 17 ppm or more.

So, the pond has very clean water.

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2023 (30 Jan Shift 2)
The water soluble protein is:

Options:
A Fibrin
B Albumin
C Myosin
D Collagen

The correct answer is:
Albumin

Solution:
Albumin is water soluble.

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2021 (25 Jul Shift 1)
The wave described by $y=0.25 \sin (10 \pi x-2 \pi t)$, where $\mathrm{x}$ and $\mathrm{y}$ are in meters and $\mathrm{t}$ in seconds, is a wave travelling along the

Options:
A -ve $\mathrm{x}$ direction with amplitude $0.25 \mathrm{~m}$ and wavelength $\lambda=0.2 \mathrm{~m}$
B -ve $x$ direction with frequency $1 \mathrm{~Hz}$
C +ve $\mathrm{x}$ direction with frequency $\pi \mathrm{Hz}$ and wavelength $\lambda=0.2 \mathrm{~m}$
D +ve $\mathrm{x}$ direction with frequency $1 \mathrm{~Hz}$ and wavelength $\lambda=0.2 \mathrm{~m}$

The correct answer is:
+ve $\mathrm{x}$ direction with frequency $1 \mathrm{~Hz}$ and wavelength $\lambda=0.2 \mathrm{~m}$

Solution:
Give, $y=0.25 \sin (10 \pi x-2 \pi t)$
Comparing with $y=A \sin \left(\frac{2 \pi}{\lambda} \cdot x-2 \pi n t\right)$, we get,
$$
\begin{aligned}
& \lambda=0.2 \mathrm{~m} \\
& n=1 \mathrm{~Hz}
\end{aligned}
$$
-ve sign indicates, the $x$ direction.

Topic: Physics, Waves and Sound, NEET, NEET 2008 (Mains)
The wave described by $y=0.25 \sin (10 \pi x-2 \pi t)$, where $x$ and $y$ are in metre and $t$ in second, is a wave travelling along the

Options:
A - ve $x$ direction with frequency $1 \mathrm{~Hz}$
B +ve $x$ direction with frequency $\pi \mathrm{Hz}$ and wavelength $\lambda=0.2 \mathrm{~m}$
C + ve $x$ direction with frequency $1 \mathrm{~Hz}$ and wavelength $\lambda=0.2 \mathrm{~m}$
D -ve $x$ direction with amplitude $0.25 \mathrm{~m}$ and wavelength $\lambda=0.2 \mathrm{~m}$

The correct answer is:
+ ve $x$ direction with frequency $1 \mathrm{~Hz}$ and wavelength $\lambda=0.2 \mathrm{~m}$

Solution:
Key Idea : The sign between two terms in argument of sine will define its direction.
Writing the given wave equation
$y=0.25 \sin (10 \pi x-2 \pi t)$
The minus $(-)$ between $(10 \pi x)$ and $(2 \pi t)$ implies that the wave is travelling along positive $x$ direction.
Now comparing Eq. (i) with standard wave equation
$y=a \sin (k x-\omega t)$
We have
$\begin{array}{l}
a=0.25 \mathrm{~m}, \omega=2 \pi, k=10 \pi \mathrm{m} \\
\therefore \frac{2 \pi}{T}=2 \pi \\
\Rightarrow f=1 \mathrm{~Hz}
\end{array}$
Also, $\lambda=\frac{2 \pi}{k}=\frac{2 \pi}{10 \pi}=0.2 \mathrm{~m}$
Therefore, the wave is travelling along +ve $x$ direction with frequency $1 \mathrm{~Hz}$ and wavelength $0.2 \mathrm{~m}$

Topic: Physics, Magnetic Properties of Matter, JEE Main
The wave function Ψ of 2 s is given by

Ψ2s=122π1a01/22-ra0e-r/2a0

At r=r0, radial node is formed. Thus, r0 in terms of a0

Options:
A r0=a0
B r0=4a0
C r0=a02
D r0=2a0

The correct answer is:
r0=2a0

Solution:
The wave function (Ψ) of 2s is given by,

Ψ2s=122π1a01/22-ra0e-r/2a0

At r = r0, radial node is formed.

Ψ2s2 = Probability density of finding an electron

At node, Ψ2s2 = 0 (∵Probability of finding an electron is zero at the node)

Ψ2s2 =122π1a01/22-ra0e-r/2a02

⇒ 0 =18π1a0 2-r0a02e-2r0/a0

⇒ 2 - r0a0 = 0

⇒ 2 = r0a0

⇒ 2a0 = r0

So, the correct option is D.

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2023 (30 Jan Shift 2)
The wave function of atomic orbital of H like species is given as
ψ2s=142πZ322-Zre-Zr2
The radius for nodal surface for He+ ion in Å .

Options:

The correct answer is:
1

Solution:
As at nodal surface Ï=0
So 2-Zr=0
r= 2 Z = 2 2 =1 Å

Topic: Chemistry, Structure of Atom, JEE Main
The wave function of atomic orbital of H-like species is given as

ψ2s=142πZ3/2(2-Zr)e-Zr/2

The radius for nodal surface for He+ ion in Å is

Options:

The correct answer is:
1

Solution:
At nodal surface Ï2=0
So 2-Zr=0
r=2Z=22
=1 Ã.

Topic: Chemistry, Structure of Atom, JEE Main
The wave length of light in visible part $\left(\lambda_V\right)$ and for sound $\left(\lambda_S\right)$ are related as

Options:
A $\lambda_V\gt\lambda_S$
B $\lambda_S\gt\lambda_V$
C $\lambda_S=\lambda_V$
D None of these

The correct answer is:
$\lambda_S\gt\lambda_V$

Solution:
We know that the speed of the sound is comparatively very small as compared to the speed of the light. So, the wavelength of the sound wave will be very large as compared to the wavelength of the visible part.
$\therefore \lambda_S\gt\lambda_V$

Topic: Physics, Waves and Sound, JEE Main
The wave nature of electrons was first proved by whom?

Options:
A Photoelectric effect
B Double slit experiment
C Davison and Germer experiment
D Compton effect

The correct answer is:
Davison and Germer experiment

Solution:
The correct answer is Davison and Germer experiment.
The Davisson and Germer experiment demonstrated the wave nature of the electrons, confirming the earlier hypothesis of de Broglie.

Topic: Physics, Dual Nature of Matter, JEE Main
The wave number of first emission line in the atomic spectrum of hydrogen in the Balmer series is

Options:
A 5R36cm-1
B 3R4cm-1
C 7R144cm-1
D 9R400cm-1

The correct answer is:
5R36cm-1

Solution:
Balmer series starts from n=2 . So first emission line in atomic spectrum of H is 2→3
1 λ  =  RZ 2 1 n 1 2 −1 n 2 2
ν-=R122-132cm-1=5R36cm-1

Topic: Chemistry, Structure of Atom, JEE Main
The wave number of first line of Balmer series of hydrogen is 1520 m-1 . The wave number of first Balmer line of Li2+ ion in m-1 is x . Find the value of x 100 .

Options:

The correct answer is:
136.8

Solution:
For a given line of hydrogen-like species, frequency, vÌâZ2
So vÌ1vÌ2=Z12Z22
or vÌ2=vÌ2ÃZ22Z12
=1520 m-1Ã(3)2(1)2
=13680 m-1

Topic: Chemistry, Structure of Atom, JEE Main
The wave number of the energy emitted when electron comes from fourth orbit to second orbit in hydrogen is \( 20,397 \mathrm{~cm}^{-1} \),
The wave number of the energy for the same transition in \( \mathrm{He}^{+} \)is

Options:
A \( 5,099 \mathrm{~cm}^{-1} \)
B \( 20,497 \mathrm{~cm}^{-1} \)
C \( 14,400 Å \)
D \( 81,588 \mathrm{~cm}^{-1} \)

The correct answer is:
\( 81,588 \mathrm{~cm}^{-1} \)

Solution:
v-=R122-142=3R4=20397cm-1
For the same transaction in He atom (Z=2)
v-=RZ2122-142=3R×224
=20397×4=81588 cm-1

Topic: Physics, Atomic Physics, JEE Main
The wave number of the first emission line in the Balmer series of $\mathrm{H}$-Spectrum is : $(\mathrm{R}=$ Rydberg constant $)$ :

Options:
A
$\frac{5}{36} R$

B
$\frac{9}{400} R$

C
$\frac{7}{6} R$

D
$\frac{3}{4} R$

The correct answer is:
$\frac{5}{36} R$

Solution:
$\bar{v}=\mathrm{RZ}^2\left(\frac{1}{2^2}-\frac{1}{3^2}\right)$
$$
=\mathrm{R}\left(\frac{1}{4}-\frac{1}{9}\right)=\frac{5 \mathrm{R}}{36}
$$

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2013 (22 Apr Online)
The wave number of the first emission line in the Balmer series of H-Spectrum is n36R . The value of 'n' is. (R = Rydberg constant):

Options:

The correct answer is:
5

Solution:
v¯=RZ2122-132
=R14-19=5R36=n36R
Hence n = 5.

Topic: Chemistry, Structure of Atom, JEE Main
The wavelength of a photon and de-Broglie wavelength an electron have the same value. Given that v is the speed of electron and c is the velocity of light. Ee, Ep is the kinetic energy of electron and energy of photon respectively while pe, ph is the momentum of electron and photon respectively. Then which of the following relation is correct?

Options:
A EeEp=v2c
B EeEp=2cv
C peph=c2v
D peph=2cv

The correct answer is:
EeEp=v2c

Solution:
Ee=12mv2=12mvv=12hÎ»v
and Ep=hcÎ»;
â´ EeEp=v2c
and according to De Broglie's equation
ph=hÎ»
â´ peph=1

Topic: Physics, Atomic Physics, JEE Main
The wavelength of an electron and a neutron will become equal when the velocity of the electron is x times the velocity of neutron. The value of x is____(the nearest integer)(Mass of electron is 9.1×10-31 kg and mass of neutron is 1.6×10-27 kg)

Options:

The correct answer is:
1758

Solution:
λe=λn

hme.ve=hmn.vn

λe=Wavelength of electron

λn=Wave length of neutron

me and Mn=Mass of electron and neutron respectively

ve and vn=Velocity of electron and neutron respectively

h=6.626×10-34Js=Planck's constant

me.x.vn=mn.vn

x=mnme=1.6×10-279.1×10-31

⇒x=169×103

=1.758×103

=1758

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2022 (26 Jul Shift 1)
The wavelength of an electron of kinetic energy 4.50×10-29 J is ________×10–5 m. (Nearest integer) Given: mass of electron is9×10–31 kg, h=6.6×10–34 Js

Options:

The correct answer is:
7

Solution:
Given:
Kinetic energy(K.E.)=4.55×10−29J
Mass of electron(me)=9.1×10−31kgh=6.6×10−34J−sec

debroglie wavelength λd=hmv=h2mKE

Here λ is the de Broglie wavelength, h is Planck's constant, m is the mass of the particle, v is its velocity, and p is the momentum of the particle, which is equal to mv.

λ=h2mKE

=6.6×10-342×9×10-31×4.5×10-29=6.6×10-349×10-30=6.69×10-4

=669×10-5

=7.33×10-5

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2023 (06 Apr Shift 1)
The wavelength of second Balmer line in hydrogen &spectrum is 600 nm The wavelength for its third line in Lyman series is

Options:
A $800 \mathrm{nm}$
B $600 \mathrm{nm}$
C $400 \mathrm{nm}$
D None of the above

The correct answer is:
None of the above

Solution:
$\frac{1}{\lambda}=R\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right]$
For Balmer series,
$\begin{array}{l}
\therefore n_{1}=2, n_{2}=3,4,5 \ldots \\
\frac{1}{\lambda_{B}}=R\left[\frac{1}{2^{2}}-\frac{1}{4^{2}}\right]
\end{array}$
$\begin{array}{l}
=R\left[\frac{1}{4}-\frac{1}{16}\right] \\
\frac{1}{\lambda_{B}}=R \cdot \frac{3}{16}
\end{array}$
Lyman series is obtained when $n_{1}=1, n_{2}=2,3,4, \ldots$ For third line in Lyman series,
$\begin{aligned}
\frac{1}{\lambda_{L}} &=R\left[\frac{1}{1^{2}}-\frac{1}{4^{2}}\right]=R\left[1-\frac{1}{16}\right] \\
\frac{1}{\lambda_{L}} &=R \cdot \frac{15}{16} \\
\frac{1 / \lambda_{B}}{\frac{1}{\lambda_{L}}} &=\frac{R \cdot \frac{3}{16}}{R \cdot \frac{15}{16}} \\
\frac{\lambda_{L}}{\lambda_{B}} &=\frac{3}{15} \\
\lambda_{L} &=\frac{3}{15} \times \lambda_{\theta} \\
\lambda_{B} &=600 \mathrm{nm}=600 \times 10^{-9} \mathrm{m} \\
\lambda_{L} &=\frac{3}{15} \times 600 \times 10^{-9} \\
&=120 \times 10^{-9} \mathrm{m}=120 \mathrm{nm}
\end{aligned}$
Hence, none of the given options is correct.

Topic: Physics, Atomic Physics, JEE Main
The wavelength of the carrier waves in a modern optical fiber communication network is close to:

Options:
A 2400 nm
B 900 nm
C 600 nm
D 1500 nm

The correct answer is:
1500 nm

Solution:
For minimum attenuation, the wavelength of carrier waves is close to 1500 nm

Topic: Physics, Communication System, JEE Main, JEE Main 2019 (08 Apr Shift 1)
The wavelength of the carrier waves in a modern optical fibre communication network is close to

Options:
A $600 \mathrm{~nm}$
B $2400 \mathrm{~nm}$
C $1500 \mathrm{~nm}$
D $900 \mathrm{~nm}$

The correct answer is:
$1500 \mathrm{~nm}$

Solution:
Fibre optics communication is mainly conducted in a wavelength range from 1260 $\mathrm{nm}$ to $1625 \mathrm{~nm}$

Topic: Physics, Communication System, JEE Main
The wavelength of the first line of Lyman series for $\mathrm{H}$ - atom is equal to that of the second line of Balmer series for a $\mathrm{H}$-like ion. The atomic number $\mathrm{Z}$ of $\mathrm{H}$-like ion is

Options:
A 4
B 1
C 2
D 3

The correct answer is:
2

Solution:
Lyman series of $\mathrm{H}$-atom, we can write
$\frac{h c}{\lambda}=R h c\left(\frac{1}{1^2}-\frac{1}{2^2}\right)$
where, symbols have their usual meaning and for second line of Balmer series of $\mathrm{H}$-like ion
$\frac{h c}{\lambda}=Z^2 R h c\left(\frac{1}{2^2}-\frac{1}{4^2}\right) \text {. }$
Therefore, $\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=Z^2\left(\frac{1}{4}-\frac{1}{16}\right)$
$\begin{aligned}
\left(1-\frac{1}{4}\right) & =Z^2\left(\frac{1}{4}-\frac{1}{16}\right) \\
Z & =2.19 \simeq 2
\end{aligned}$
The approximately value of $Z$ will be 2 .

Topic: Physics, Atomic Physics, JEE Main
The wavelength of the first spectral line in the Balmer series of the hydrogen atom is 6561 Aâ. The wavelength of the second spectral line in the Balmer series of singly ionized helium atom is

Options:
A 1215 A∘
B 1640 A∘
C 2430 A∘
D 4687 A∘

The correct answer is:
1215 A∘

Solution:
For hydrogen or hydrogen type atoms
1λ =RZ21nf2-1ni2
In the transition from ni→ nf
∴ λ ∝ 1Z21nf2-1ni2
∴ λ2λ1= Z121nf2-1ni21Z221nf2-1ni22
λ2= λ1Z121nf2-1ni21Z221nf2-1ni22
Substituting the values, we have
=(6561)12122-13222122 - 142 =1215 A∘

Topic: Physics, Atomic Physics, JEE Main
The wavelength of the photon emitted by a hydrogen atom when an electron makes a transition from n=2 to n=1 state is:

Options:
A 121.8 nm
B 194.8 nm
C 490.7 nm
D 913.3 nm

The correct answer is:
121.8 nm

Solution:
1λ=R112-122

λ=121.8 nm

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2021 (25 Feb Shift 2)
The wavelength of the radiation emitted is λ0 when an electron jumps from the second excited state to the first excited state of hydrogen atom. If the electron jumps from the third excited state to the second orbit of the hydrogen atom, the wavelength of the radiation emitted will be 20xλ0. The value of x is________.

Options:

The correct answer is:
27

Solution:
_____ n=4

_____ n=3

_____ n=2

_____ n=1

The wavelength of the light released when an electron jumps from higher energy state n2 to lower energy state n1 is 1λ=R1n12-1n22, here, λ is wavelength, R is Rydberg Constant.

When transition is from second excited state to first excited state,
n=3→n=2

1λ0=R122-132 ...1

When transition is from third excited state to second orbit

n=4→n=2

120xλ0=R122-142 ...2

Dividing 2 by 1

⇒x20=122-142122-132=14-11614-19=316536=2720

⇒x=27

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2023 (25 Jan Shift 1)
The wavelength of the radiation emitted, when in a hydrogen atom electron falls from infinity to stationary state 1 , would be (Rydberg constant $=1.097 \times 10^7 \mathrm{~m}^{-1}$ )

Options:
A $406 \mathrm{~nm}$
B $192 \mathrm{~nm}$
C $91 \mathrm{~nm}$
D $9.1 \times 10^{-8} \mathrm{~nm}$

The correct answer is:
$91 \mathrm{~nm}$

Solution:
$\frac{1}{\lambda}=R\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]$
$\frac{1}{\lambda}=1.097 \times 10^7 \mathrm{~m}^{-1}\left[\frac{1}{1^2}-\frac{1}{\infty^2}\right]$
$\therefore \lambda=91 \times 10^{-9} \mathrm{~m}$
We know $10^{-9}=1 \mathrm{~nm}$ So $\lambda=91 \mathrm{~nm}$

Topic: Chemistry, Structure of Atom, JEE Main
The wavelength of the radiation emitted, when in hydrogen atom electron falls from infinity to stationary state 1 , would be (Rydberg constant $\left.=1.097 \times 10^7 \mathrm{~m}^{-1}\right)$

Options:
A
$91 \mathrm{~nm}$

B
$9.1 \times 10^{-8} \mathrm{~nm}$

C
$406 \mathrm{~nm}$

D
$192 \mathrm{~nm}$

The correct answer is:
$91 \mathrm{~nm}$

Solution:
$\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right]$
$=1.097 \times 10^7\left(\frac{1}{1}\right)$
$\lambda=\frac{1}{1.097} \times 10^{-7} \mathrm{~m}$

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2004
The wavelength of the third line of the Balmer series for a hydrogen atom is

Options:
A 21100RH
B 10021RH
C 21RH100R
D 100 R H 21

The correct answer is:
10021RH

Solution:
1λ=RH1n12-1n22 for the H-atom
For the Balmer series,
n1=2 and n2=5 (for the third line).
1λ=RH122-152=21100RH
So λ =  100 21R H

Topic: Chemistry, Structure of Atom, JEE Main
The wavelengths involved in the spectrum of deuterium $\left({ }_1^2 \mathrm{D}\right)$ are slightly different from that of hydrogen spectrum, because

Options:
A
the size of the two nuclei are different

B
the nuclear forces are different in the two cases

C
the masses of the two nuclei are different

D
the attraction between the electron and the nucleus is different in the two cases

The correct answer is:
the masses of the two nuclei are different

Solution:
No Solution Available

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2003
The wavelengths of two notes in the air are 36195 m and 36193 m. each note produces 10 beats per second separately with a third note of fixed frequency. The velocity of sound in air in m s-1 is

Options:
A 330
B 340
C 350
D 360

The correct answer is:
360

Solution:
Beat frequency=v1~v2
Let the frequency of third note be n.
Then,
195v36-n=10 …i
And
n-193v36=10 …ii
Adding Eqs. (i) and (ii)
v18=20
⟹ v=360 m s-1

Topic: Physics, Waves and Sound, JEE Main
The waves emitted when a metal target is bombarded with high energy electrons are

Options:
A Microwaves
B Infrared rays
C X-rays
D Radio Waves

The correct answer is:
X-rays

Solution:
The bombarding electrons can eject electrons from the inner shells of the atoms of the metal target. Those vacancies will be quickly filled by electrons dropping from higher levels, emitting x-rays with sharply defined frequencies associated with the difference between the atomic energy levels of the target atoms. Hence, X-rays are emitted when a metal target is bombarded with high energy electrons.

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2023 (08 Apr Shift 2)
The waves in which the particles of the medium vibrate in a direction perpendicular to the direction of wave motion is known as

Options:
A Transverse waves
B Longitudinal waves
C Propagated waves
D None of these

The correct answer is:
Transverse waves

Solution:
In transverse waves medium particles vibrate perpendicular to the direction of propagation of wave.

Topic: Physics, Waves and Sound, JEE Main
The way the fringe pattern changes when the screen is moved away from the slits is

Options:
A width of fringe increases
B width of fringe decreases
C no changes observed
D change will vary

The correct answer is:
width of fringe increases

Solution:
Since fringe width Î²= Î»D d
When screen is moved away, D increases. Therefore width of the fringes increases but the angular separation ( Î»/d ) remains the same.

Topic: Physics, Wave Optics, NEET