datavorous/entrance-exam-dataset huggingface dataset:
The effective capacitances of two capacitors are 3 $\mu \mathrm{F}$ and $16 \mu \mathrm{F}$, when they are connected in series and parallel respectively. The capacitance of two capacitors are:

Options:
A $10 \mu \mathrm{F}, 6 \mu \mathrm{F}$
B $8 \mu \mathrm{F}, 8 \mu \mathrm{F}$
C $12 \mu \mathrm{F}, 4 \mu \mathrm{F}$
D $1.2 \mu \mathrm{F}, 1.8 \mu \mathrm{F}$

The correct answer is:
$12 \mu \mathrm{F}, 4 \mu \mathrm{F}$

Solution:
given: \(C_S=3 M f\)
\(\begin{aligned}
& C_P=16 M f \\
& \text {as } C_S=\left[\left(C_1 C_2\right) /\left(C_1+C_2\right)\right]=3 \\
& C_P=C_1+C_2=16(2) \\
& \therefore \text {from (1), }\left[\left(C_1 C_2\right) /(16)\right]=3 \\
& C_1 C_2=48 \\
& \text {from (2), } C_2=16-C_1 \\
& C_2=16-\left(48 / C_2\right) \\
& \therefore C_2+\left(48 / C_2\right)=16 \\
& \therefore C_2^2+48-16 C_2=0 \\
& \therefore C_2=12 \mu F, C_2=4 \mu F
\end{aligned}\)
i.e. capacitance are \(4 \mu \mathrm{F} ~\&~ 12 \mu \mathrm{F}\)

Topic: Physics, Capacitance, JEE Main
The effective length of a magnet is $31.4 \mathrm{~cm}$ and its pole strength is $0.5 \mathrm{Am}$. The magnetic moment, if it is bent in the form of a semicircle will be

Options:
A $0.01 \mathrm{Am}^2$
B $0.2 \mathrm{Am}^2$
C $0.1 \mathrm{Am}^2$
D $1.2 \mathrm{Am}^2$

The correct answer is:
$0.1 \mathrm{Am}^2$

Solution:
New magnetic moment
$\mathrm{M}^{\prime}=\frac{2 \mathrm{M}}{\pi}=\frac{2 \mathrm{~mL}}{\pi}=\frac{2 \times 0.5 \times 31.4 \times 10^{-2}}{3.14}=0.1 \mathrm{amp} \times \mathrm{m}^2$

Topic: Physics, Magnetic Properties of Matter, JEE Main
The effective nuclear charge of an element with three valence electrons is 2.60. What is the minimum atomic number of the element?

Options:
A 5
B 4
C 3
D 2

The correct answer is:
5

Solution:
As effective nuclear charge is 2.60 and valence shell contain 3 electrons so the minimum number of main shells for the given element are two i.e., (n = 2) and its configuration will be 1s2,2s2,2p1 .
Hence the given element has 5 electrons in all.
Also, for a neutral atom.
As number of electrons = atomic number = 5.
Thus, the atomic number of the element is 5.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
The effective resistance between $A$ and $B$, if resistance of each resistor is $R$, will be

Options:
A $\frac{8 R}{3}$
B $\frac{5 R}{3}$
C $\frac{4 R}{3}$
D $\frac{2}{3} R$

The correct answer is:
$\frac{8 R}{3}$

Solution:
From symmetry we can remove two middle resistance.
New circuit is

Topic: Physics, Current Electricity, JEE Main, JEE Main 2024 (09 Apr Shift 2)
The effective resistance between $A$ and $B$ in the figure is $\frac{7}{12} \Omega$ if each side of the cube has $1 \Omega$ resistance. The effective resistance between the same two points, when the link $A B$ is removed, is

Options:
A $\frac{7}{12} \Omega$
B $\frac{5}{12} \Omega$
C $\frac{7}{5} \Omega$
D $\frac{5}{7} \Omega$

The correct answer is:
$\frac{7}{5} \Omega$

Solution:
Let when link AB is removed, $x$ is an equivalent of remaining part of cube without link.

AB and rest part will be in paralel, so according to question
$$
\begin{array}{c}
\frac{7}{12}=\frac{R_{1} R_{2}}{R_{1}+R}=\frac{1(x)}{1+x}=\frac{x}{1+x} \\
\frac{7}{12}=\frac{x}{1+x} \\
12 x=7+7 x \\
12 x-7 x=7
\end{array}
5 x=7
$$
$$
x=\frac{7}{5} \Omega
$$

Topic: Physics, Current Electricity, WBJEE, WBJEE 2016
The effective resistance between points P and Q of the electrical circuit shown in the figure is

Options:
A R2
B R
C 3R2
D 2R

The correct answer is:
R2

Solution:
According balanced wheat stone bridge
1Req=12R+12R+1R
=1+1+22R=2R
Req=R2

Topic: Physics, Current Electricity, JEE Main
The effective resistance between the points $A$ and $B$ in the figure is

Options:
A $5 \Omega$
B $2 \Omega$
C $3 \Omega$
D $4 \Omega$

The correct answer is:
$2 \Omega$

Solution:
Given circuit is equivalent to
So the equivalent resistance between points $A$ and $B$ is equal to $R=\frac{6 \times 3}{6+3}=2 \Omega$

Topic: Physics, Current Electricity, JEE Main
The efficiency of a Carnot's engine, working between steam point and ice point, will be

Options:
A 26.81%
B 37.81%
C 47.81%
D 57.81%

The correct answer is:
26.81%

Solution:
Steam point =100 °C=100+273 K=373 K

Ice point =0 °C=0+273 K=273 K

Now,

η=1-TsinkTsource =1-273373

=100373×100% =26.81%

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2022 (26 Jun Shift 1)
The efficiency of an engine is given by $\eta=\frac{\alpha \beta}{\sin \theta} \cdot \log _{\mathrm{e}} \frac{\beta \mathrm{x}}{\mathrm{kT}}$, where $\alpha$ and $\beta$ are constants. If $T$ is the absolute temperature, $k$ - Boltzmann constant, $\theta$-angular displacement and $\mathbf{x}$ is distance, then the incorrect statement is

Options:
A Dimensions of $\beta$ are same as that of force
B Dimensions of $\alpha^{-1} \mathrm{x}$ are same as that of energy
C Dimensions of $\eta^{-1} \sin \theta$ are same as that of $\alpha \beta$
D Dimensions of $\alpha$ are same as that of $\beta$

The correct answer is:
Dimensions of $\alpha$ are same as that of $\beta$

Solution:
The efficiency of an engine is given by
$\begin{aligned}
& \eta=\frac{\alpha \beta}{\sin \theta} \log _{\mathrm{e}} \frac{\beta \mathrm{x}}{\mathrm{kT}} \\
& {[\alpha \beta]=[\eta]=[\sin \theta]=\text { Dimension less }} \\
& {\left[\eta^{-1} \sin \theta\right]=[\alpha \cdot \beta]=\text { Dimension less }} \\
& \frac{\beta \mathrm{x}}{\mathrm{kt}}=1 \\
& \beta=\frac{\mathrm{kT}}{\mathrm{x}}
\end{aligned}$
Here, Boltzmann constant, $\mathrm{K}=\left[\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~K}^{-1}\right]$
$\begin{aligned}
& \beta=\frac{\left[\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~K}^{-1}\right][\mathrm{K}]}{[\mathrm{L}]} \\
& =\left[\mathrm{ML} \mathrm{T}^{-2}\right]
\end{aligned}$
Force, $\mathrm{F}=\mathrm{ma}=\mathrm{m} \frac{\mathrm{dv}}{\mathrm{dt}}$
Force, $\mathrm{F}=\left[\mathrm{ML} \mathrm{T}^{-2}\right]$
Hence, Dimension of $\beta$ are same as that of force.
$\begin{aligned}
& \beta=\alpha^{-1}=\left[\mathrm{ML} \mathrm{T}^{-2}\right] \\
& \alpha^{-1} \mathrm{x}=\left[\mathrm{ML} \mathrm{T}^{-2}\right][\mathrm{L}]=\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]
\end{aligned}$
Energy, $\mathrm{E}=\mathrm{mc}^2=\mathrm{ML}^2 \mathrm{~T}^{-2}$
Hence, dimension of $\alpha^{-1} \mathrm{x}$ are same as that of energy.

Topic: Physics, Units and Dimensions, JEE Main
The efficiency of carnot engine is η when its hot and cold reservoirs are maintained at temperature T1 and T2, respectively. To increase the efficiency to 1.5 η, the increase in temperature ΔT of the hot reservoir by keeping the cold one constant at T2 is

Options:
A T1T21-η1-1.5η
B 0.5T2η1-1.5η1-η
C T11-η-T21-1.5η
D 1-η1-1.5ηT1T2

The correct answer is:
0.5T2η1-1.5η1-η

Solution:
The efficiency of Carnot engine is

η=1-T2T1 ⇒T1T2=11-η

The efficiency has to be increased to 1.5 η by maintaining T2 constant.

∴ 1.5η=1-T2T1+∆T⇒1.5η=1-T2T1+∆T

∴ T2T1+∆T=1-1.5η

∴ T1+∆TT2=11-1.5η

∴ T1T2+∆TT2=11-1.5η⇒11-η+∆TT2=11-1.5η

∴ ∆TT2=11-1.5η-11-η

∴ ∆T=0.5T2η1-1.5η1-η

Topic: Physics, Thermodynamics, TS EAMCET, TS EAMCET 2018 (04 May Shift 1)
The efficiency of Carnot's heat engine is $0.5$ when the temperature of the source is $\mathrm{T}_{1}$ and that of $\operatorname{sink}$ is $\mathrm{T}_{2}$. The efficiency of another Carnot's heat engine is also 0.5. The temperatures of source and sink of the second engine are respectively

Options:
A $2 \mathrm{~T}_{1}, 2 \mathrm{~T}_{2}$
B $2 \mathrm{~T}_{1}, \frac{\mathrm{T}_{2}}{2}$
C $\mathrm{T}_{1}+5, \mathrm{~T}_{2}-5$
D $\mathrm{T}_{1}+10, \mathrm{~T}_{2}-10$

The correct answer is:
$2 \mathrm{~T}_{1}, 2 \mathrm{~T}_{2}$

Solution:
Efficiency of Carnot's heat engine, $\eta=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}$
Efficiency remains same when both $\mathrm{T}_{1}$ and $\mathrm{T}_{2}$ are increased by same factor.

Topic: Physics, Thermodynamics, KCET, KCET 2010
The efficiency of fuel cell is given by the expression, $\eta$ is

Options:
A $\eta=-\frac{\mathrm{nFE}_{\text {cell }}}{\Delta \mathrm{H}} \times 100$
B $\eta=-\frac{\mathrm{nFE}_{\text {cell }}}{\Delta \mathrm{S}} \times 100$
C $\eta=-\frac{\mathrm{nFE}_{\text {cell }}}{\Delta \mathrm{A}} \times 100$
D None of the above

The correct answer is:
$\eta=-\frac{\mathrm{nFE}_{\text {cell }}}{\Delta \mathrm{H}} \times 100$

Solution:
Efficiency of fuel cell is:

$\eta=\frac{-n F E_{\text {cell }}}{\Delta H} \times 100$

Topic: Chemistry, Electrochemistry, JEE Main
The efficiency of packing in the case of a metal crystal for face-centered cubic is

Options:

The correct answer is:
74

Solution:
In face centred cubic, packing efficiency
= Volume occupied by 4 atoms in the unit cell × 1 0 0 Volume of the unit cell
=4×4/3πr⁡3×10022r⁡3=16/3πr⁡3×100162r⁡3=74%

Topic: Chemistry, Solid State, JEE Main
The efficiency of the reversible cycle shown in the figure will be

Options:
A 33.33%
B 56%
C 66%
D 25%

The correct answer is:
25%

Solution:
Efficiency of cycle =Area of closed cycleArea of under the curve×100
=12×(1500-1000)×(250-150)12×(1500-1000)×(250-150)+(1500-1000)×(150-0)×100
=12×500×10012×500×100+500×150
=500×50×100500×50+500×150
=25 .

Topic: Chemistry, Thermodynamics (C), JEE Main
The effluent from the ______ treatment plant is generally released into natural water bodies like rivers and streams.

Options:
A primary
B secondary
C primordial
D tertiary

The correct answer is:
secondary

Solution:
The primary effluent is passed to the aeration tanks in the secondary treatment where the aerobic microbes grow and form flocs. These microbes help in consuming the organic matter and in doing so reduce the BOD of the effluent significantly.
After reducing the BOD, the effluent is passed in to settling tanks where the settled flocs formed the activated sludge. Some part of it is used as inoculum and remaining is passed into large tanks called anaerobic sludge digesters. In the digestor, produce a mixture of gases which can be used as source of energy.
The effluent from the secondary treatment plant is generally released into natural water bodies like rivers and streams.

Topic: Biology, Microbes in Human Welfare, NEET
The Einstein's photoelectric equation is $h \nu=\phi+k$. Here $\mathrm{k}$ represents(h is planck's constant, $c$ is speed of light, $\lambda$ is wavelength, and $\phi$ is work function)

Options:
A Minimum kinetic energy of electrons
B Maximum kinetic energy of electrons
C Mean kinetic energy of electrons
D None of the above

The correct answer is:
Maximum kinetic energy of electrons

Solution:
- According to Einstein's photoelectric equation:
$\begin{aligned}& \Rightarrow E=\varphi+K E_{\max } \\& \Rightarrow E=h v \\& \Rightarrow h v=\varphi+K E_{\max }...(2)\end{aligned}$
On comparing equation 1 and 2 , we get to know that,
$\Rightarrow k=K E_{\max }$
- Therefore $\mathrm{k}$ represents the maximum kinetic energy of electrons. Hence option 2 is correct.

Topic: Physics, Dual Nature of Matter, JEE Main
The elastic limit of brass is 379 MPa. The minimum diameter of a brass rod if it is to support a 400 N load without exceeding its elastic limit will be

Options:
A 1.00 mm
B 1.36 mm
C 1.16 mm
D 0.90 mm

The correct answer is:
1.16 mm

Solution:
Stress =FA
Elastic limit is 379 MPa
⇒379×106=400A=400πd2/4
d=400×4π×379×106
=1.16×10-3 m
=1.16 mm

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2019 (10 Apr Shift 2)
The elastic limit of brass is 379 MPa. The minimum diameter of a brass rod if it is to support a 400 N load without exceeding its elastic limit will be

Options:
A 1.00 mm
B 1.36 mm
C 1.16 mm
D 0.90 mm

The correct answer is:
1.16 mm

Solution:
Stress =FA
Elastic limit is 379 MPa
⇒379×106=400A=400πd2/4
d=400×4π×379×106
=1.16×10-3 m
=1.16 mm

Topic: Physics, Mechanical Properties of Solids, JEE Main
The elastic potential energy stored in a steel wire of length 20 m stretched through 2 cm is 80 J. The cross sectional area of the wire is _____ mm2. (Given, Y=2.0×1011 N m–2)

Options:

The correct answer is:
40

Solution:
The formula to calculate the Young's modulus of the material is given by

Y=stressstrain=stress∆lL ...1

The formula to calculate the potential energy U stored into the wire is given by

U=12×stress×strain×volume=12×stress×∆lL×A×L= 12×stress×A×∆l ...2

From equation (1) and equation (2), it can be written that

U=12×Y×∆l2L×A ...3

Substitute the values of the known parameters into equation (3) and solve to calculate the required cross-sectional area of the wire.

80 J=12×2.0×1011 Nm-2×0.02 m220 m×A⇒A= 2×80 J×202.0×1011×4×10-4 Nm-1=4×10-5 m2×106 mm21 m2= 40 mm2

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2023 (13 Apr Shift 1)
The electric and magnetic field of an electromagnetic wave are:

Options:
A in opposite phase and perpendicular to each other
B in opposite phase and parallel to each other
C in phase and perpendicular to each other
D in phase and parallel to each other

The correct answer is:
in phase and perpendicular to each other

Solution:
In EM waves $\vec{E}$ and $\vec{B}$ are in same phase and perpendicular to each other.

Topic: Physics, Electromagnetic Waves, JEE Main
The electric charge in uniform motion produces

Options:
A An electric field only
B A magnetic field only
C Both electric and magnetic field
D Neither electric nor magnetic field

The correct answer is:
Both electric and magnetic field

Solution:
The presence of charge produces electric field and a uniformly moving charge generates current which produces magnetic field. Thus a uniformly moving charge produces both electric and magnetic field.

Topic: Physics, Electrostatics, JEE Main
The electric current in a circular coil of 2 turns produces a magnetic induction B1 at its centre. The coil is unwound and is rewound into a circular coil of 5 turns and the same current produces a magnetic induction B2 at its centre.
The ratio of B2B1 is:

Options:
A 52
B 254
C 54
D 252

The correct answer is:
254

Solution:
Magnetic field on the centre of a circular current coil is given by B=Nμ0i2R, here, N is number of turns, i is current and R is radius of loop.

Now, for, N=2, magnetic field, B1=N1μ0i2R1

For N2=5

Radius of coil =R2=N1×R1 N2

Magnetic field, B2=N2μ0i2R2

Thus, the ratio is B2 B1=N2 N1R1R2=N2 N1×N2 N1

B2 B1=254

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2022 (25 Jul Shift 2)
The electric current in a circular coil of four turns produces a magnetic induction 32 T at its centre. The coil is unwound and is rewound into a circular coil of single turn, the magnetic induction at the centre of the coil by the same current will be :

Options:
A 8 T
B 4 T
C 2 T
D 16 T

The correct answer is:
2 T

Solution:
Magnetic field at the centre of the coil is given by, B=μ0i2R.

Now in the first case:

B=4μ0i2R

and for the second case:

B'=μ0i2R'.

As the length of the wire will remain same, therefore

2πR'=42πR⇒R'=4R

Hence, required ratio will be

B'B=R4R'=116

⇒B'=2 T

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2023 (29 Jan Shift 2)
The electric current through a wire varies with time as I=I0+βt, where I0=20 A and β=3 A s-1. The amount of electric charge crossed through a section of the wire in 20 s is:

Options:
A 80 C
B 1000 C
C 800 C
D 1600 C

The correct answer is:
1000 C

Solution:
Given that
Current I=I0+βt, I0=20 A and β=3 A s-1

The instantaneous current is given by

I=20+3t ...1

Integrating equation (1), we have

dqdt=20+3t⇒∫0qdq=∫020(20+3t)dt⇒q=∫02020dt+∫0203tdt⇒q=20t+3t22020=1000 C

Topic: Physics, Current Electricity, JEE Main, JEE Main 2024 (29 Jan Shift 1)
The electric field and electric potential at a point due to a point charge are $500 \mathrm{NC}^{-1}$ and $30 \mathrm{~V}$ respectively, then the magnitude of the charge is

Options:
A $1.3 \times 10^{-9} \mathrm{C}$
B $3 \times 10^{-12} \mathrm{C}$
C $2 \times 10^{-10} \mathrm{C}$
D $1.6 \times 10^{-20} \mathrm{C}$

The correct answer is:
$2 \times 10^{-10} \mathrm{C}$

Solution:
Electric field, $\mathrm{E}=500 \mathrm{~N} / \mathrm{C}$
Electric potential, $\mathrm{V}=30 \mathrm{~V}$.
$$
\begin{aligned}
& \mathrm{E}=\frac{\mathrm{V}}{\mathrm{r}} \\
& \mathrm{r}=\frac{\mathrm{V}}{\mathrm{E}}=\frac{30}{500}=\frac{3}{50} \mathrm{~m} .
\end{aligned}
$$
$$
\begin{aligned}
& \text { Now, } V=\frac{K q}{r} \Rightarrow q=\frac{V r}{K} \\
& \Rightarrow q=\frac{3 \times 30}{50 \times 9 \times 10^9} \\
& \Rightarrow q=\frac{1}{5} \times 10^{-9}=0.2 \times 10^{-9} \\
& q=2 \times 10^{-10} C
\end{aligned}
$$

Topic: Physics, Electrostatics, JEE Main
The electric field and magnetic field components of an electromagnetic wave going through vacuum is described by

Ex=E0sinkz-ωt

By=B0sinkz-ωt

Then the correct relation between E0 and B0 is given by

Options:
A E0B0=ωk
B E0=kB0
C kE0=ωB0
D ωE0=kB0

The correct answer is:
kE0=ωB0

Solution:
As we know, electric field and magnetic field are related as

E0=cB0, where, c is speed of light in vacuum

Also c=ωk

⇒E0=ωkB0

⇒ E0k=ωB0

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2023 (24 Jan Shift 2)
The electric field associated with a light wave is given by E=E0sinâ¡1.57Ã107x-ct (where x and t are in metre and second respectively). This light is used in an experiment on photoelectric effect with the emitter having work function Ï=1.9 eV. What is the maximum kinetic energy (in eV) of the ejected photoelectron? [Take Ï=3.14, hc=12400 eV Aâ]

Options:

The correct answer is:
1.2

Solution:
Given, E=E0sinâ¡1.57Ã107x-ct ...(i)
â´E=E0sinâ¡Ïcx-ct ...(ii)
On comparing equations (i) and (ii), we get
Ïc=1.57Ã107
=Ï2Ã107
Î»=4Ã10-7 m
=400 nm
As, Ï0=hcÎ»-EK
1.9=1240400-EK
EK=3.1-1.9 eV=1.2 eV

Topic: Physics, Dual Nature of Matter, JEE Main
The electric field associated with a light wave is given by E=E0sinâ¡1.57Ã107x-ct (where x and t are in metre and second respectively). The stopping potential when this light is used in an experiment on photoelectric effect with the emitter having work function Ï=1.9 eV is.

Options:
A 0.6 eV
B 1.2 eV
C 1.8 eV
D 2.4 eV

The correct answer is:
1.2 eV

Solution:
Given, E=E0sinâ¡1.57Ã107x-ct ...(i)
â´ E=E0sinâ¡Ïcx-ct ...(ii)
On comparing Equations (i) and (ii), we get
Ïc=1.57Ã107
=Ï2Ã107
Î»=4Ã10-7 m
=400 nm
As, Ï0=hcÎ»-EK
1.9=1242400-EK
EK=3.1-1.9 eV=1.2 eV

Topic: Physics, Dual Nature of Matter, JEE Main
The electric field associated with an electro magnetic wave in vacuum is given by $\mathbf{E}=\mathbf{i}$ $40 \cos \left(k z-6 \times 10^8 t\right)$, where $E$, $z$ and $t$ are in volt/m, metre and second respectively. The value of wave vector $k$ is

Options:
A $2 \mathrm{~m}^{-1}$
B $0.5 \mathrm{~m}^{-1}$
C $6 \mathrm{~m}^{-1}$
D $3 \mathrm{~m}^{-1}$

The correct answer is:
$2 \mathrm{~m}^{-1}$

Solution:
Electromagnetic wave equation
$E=E_0 \cos (k w-\omega t)$
Speed of electromagnetic wave $v=\frac{\omega}{k}$
Given, equation
$\mathbf{E}= 40 \cos \left(k z-6 \times 10^8 t\right) \mathbf{i}$
Comparing Eqs. (i) and (ii), we get
$\omega=6 \times 10^8$
and $E_0=40 \mathbf{i}$
Here, wave factor $k=\frac{\omega}{v}=\frac{6 \times 10^8}{3 \times 10^8}=2 \mathrm{~m}^{-1}$

Topic: Physics, Electromagnetic Waves, JEE Main
The electric field at a distance $\frac{3 R}{2}$ from the centre of a charged conducting spherical shell of radius $R$ is $E$. The electric field at a distance $\frac{R}{2}$ from the centre of the sphere is

Options:
A zero
B $E$
C $\frac{E}{2}$
D $\frac{E}{3}$

The correct answer is:
zero

Solution:
The radius of spherical shell is $R$. Because inside the conducting charge sphere, electric field is zero. So, the electric field at a distance $\mathrm{R} / 2$ from the centre of the sphere will be zero.

Topic: Physics, Electrostatics, NEET, NEET 2010 (Mains)
The electric field at point $\mathrm{p}$ due to an electric dipole is $\mathrm{E}$. The electric field at point $\mathrm{R}$ on equitorial line will be $\frac{\mathrm{E}}{x}$. The value of $x$ :

Options:

The correct answer is:
16

Solution:
$\begin{aligned} & E_P=\frac{2 K P}{r^3}=E \\ & E_R=\frac{K P}{(2 r)^3}=\frac{E}{16} \\ & x=16\end{aligned}$

Topic: Physics, Electrostatics, JEE Main, JEE Main 2024 (05 Apr Shift 2)
The electric field component of a monochromatic radiation is given by E→=2E0coskzcosωt i^, Its magnetic field B→ is then given by:

Options:
A 2E0csinkzsinωt j^
B 2E0ccoskzcosωt j^
C 2E0csinkzcosωt j^
D -2E0csinkzsinωt j^

The correct answer is:
2E0csinkzsinωt j^

Solution:
dEdz=-dBdt
dEdz=-2E0Ksin⁡kzcos⁡ωt=-dBdt
dB=+2E0Ksin⁡kzcos⁡ωt dt
B= +2E0Ksin⁡kz∫cos⁡ωt dt
=+2E0kωsin⁡kzsin⁡ωt
E0B0=ωk=C
B=2E0Csin⁡kzsin⁡ωt

Propagation of EM radiation, Electric field oscillation and Magnetic field oscillation are perpendicular to each other.

Since propagation of EM radiation is x-axis and electric field oscillation is along z-axis, the Magnetic field oscillation is along y-axis.

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2017 (09 Apr Online)
The electric field due to a charge at a distance of $3 \mathrm{~m}$ from it is $500 \mathrm{~N} /$ coulomb. The magnitude of the charge is $\left[\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \frac{N-\mathrm{m}^2}{\text { coulomb }^2}\right]$

Options:
A $2.5 micro-coulomb$
B $2.0 micro-coulomb$
C $1.0 micro-coulomb$
D $0.5 micro-coulomb$

The correct answer is:
$0.5 micro-coulomb$

Solution:
$E=9 \times 10^9 \times \frac{Q}{r^2} \Rightarrow 500=9 \times 10^9 \times \frac{Q}{(3)^2} \Rightarrow Q=0.5$ $\mu C$

Topic: Physics, Electrostatics, JEE Main
The electric field due to a short electric dipole at a large distance r from center of dipole on the equatorial plane varies with distance as :

Options:
A r
B 1r2
C 1r3
D 1r

The correct answer is:
1r3

Solution:
An electric dipole is a system of two charges of equal and opposite magnitude placed at a short distance.

The dipole moment is, p=2ql.

The electric field due to the dipole is,

E=2kpr3

Clearly, the electric field varies as 1r3.

Topic: Physics, Electrostatics, JEE Main, JEE Main 2023 (15 Apr Shift 1)
The electric field in a plane electromagnetic wave is given by, E=50sin500x-10×1010t V m-1. The velocity of an electromagnetic wave in this medium is: (Given c= the speed of light in vacuum).

Options:
A 32c
B c
C c2
D 23c

The correct answer is:
23c

Solution:
The equation of electric field is given in the question, E=50sin500x-10×1010t V m-1 now compare it the equation of electric field in an electromagnetic wave, E=E0sinkx-ωt V m-1

⇒ω=1010 rad s-1 and, k=50 m-1, so the velocity of electromagnetic waves is, v=ωk=101050=2×108=23c.

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2021 (27 Aug Shift 1)
The electric field in a plane electromagnetic wave is given by
E→=200cos0.5×103 m-1x-1.5×1011 rad s-1t V m-1 j^.
If this wave falls normally on a perfectly reflecting surface having an area of 100 cm2. If the radiation pressure exerted by the E.M. wave on the surface during a 10 min exposure is k109 N m-2. Find the value of k

Options:

The correct answer is:
354

Solution:
E0=200

I=12ε0E02c
Radiation pressure
P=2lc=2c12ε0E02C=ε0E02P=8.85×10-12×2002=8.85×10-8×4=354109So,k=354

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2021 (26 Aug Shift 1)
The electric field in a region is given by E→=25E0i^+35E0j^ with E0=4.0×103 N C-1. The flux of this field through a rectangular surface, area 0.4 m2 parallel to the Y-Z plane is _______ N m2 C-1.

Options:

The correct answer is:
640

Solution:
ϕ=ExA⇒25×4×103×0.4=640 N m2 C-1

Topic: Physics, Electrostatics, JEE Main, JEE Main 2021 (17 Mar Shift 2)
The electric field in a region is given by E→=Ax+B i^ , where E is in NC-1 and x is in metres. The values of constants are A=20 SI unit and B=10 SI unit. If the potential at x=1 is V1 and that at x=-5 is V2, then V1-V2 is

Options:
A 320 V
B -520 V
C 180 V
D -48 V

The correct answer is:
180 V

Solution:
V1-V2=-∫E.dx
=-∫-51ax+b dx
=-ax22+bx-51
= -a2+b-a×252-5b
= -202+10-250+50
=180 V

Topic: Physics, Electrostatics, JEE Main, JEE Main 2019 (08 Apr Shift 2)
The electric field in a region of space is given by, E→=E0i^+2E0j^ where E0=100 N C-1. The flux of this field through a circular surface of radius 0.02 m parallel to the Y‐Z plane is nearly

Options:
A 0.02 N m2 C-1
B 0.005 N m2 C-1
C 0.125 N m2 C-1
D 3.14 N m2 C-1

The correct answer is:
0.125 N m2 C-1

Solution:
ϕ=E→·A→=E0i^+2E0j^·πr2i^

=E0πr2=0.125 N m2 C-1

Topic: Physics, Electrostatics, JEE Main, JEE Main 2014 (19 Apr Online)