datavorous/entrance-exam-dataset huggingface dataset:
An electric dipole is formed by two equal and opposite charges q with separation d . The charges have same mass m. It is kept in a uniform electric field E. If it is slightly rotated from its equilibrium orientation, then its angular frequency ω is:

Options:
A 2qEmd
B 2qEmd
C qE2md
D qEmd

The correct answer is:
2qEmd

Solution:
τ=pEsin⁡θ
For small angle,
τ=pE θ
α=τI=pEI θ
∵ α =ω2θ Angular frequency =pEI
=qd Emd24+md24
=2qEmd

Topic: Physics, Electrostatics, JEE Main, JEE Main 2019 (08 Apr Shift 2)
An electric dipole is placed along the $x$ - axis at the origin $O$. A point $P$ is at a distance of $20 \mathrm{~cm}$ from this origin such that $O P$ makes an angle ${ }^{\frac{\pi}{3}}$ with the $x$-axis. If the electric field at $P$ makes an angle $\theta$ with the $x$-axis, the value of $\theta$ would be

Options:
A $\frac{\pi}{3}$
B $\frac{\pi}{3}+\tan ^{-1}\left(\frac{\sqrt{3}}{2}\right)$
C $\frac{2 \pi}{3}$
D $\tan ^{-1}\left(\frac{\sqrt{3}}{2}\right)$

The correct answer is:
$\frac{\pi}{3}+\tan ^{-1}\left(\frac{\sqrt{3}}{2}\right)$

Solution:
$\theta=\frac{\pi}{3}+\alpha$ where $\tan \alpha=\frac{1}{2} \tan \frac{\pi}{3}$
$\Rightarrow \alpha=\tan ^{-1} \sqrt{3} / 2$ so, $\theta=\frac{\pi}{3}+\tan ^{-1} \sqrt{3} / 2$

Topic: Physics, Electrostatics, JEE Main
An electric dipole is placed at the origin and is directed along the x-axis. At a point P, far away from the dipole, the electric field is parallel to the y-axis. OP makes an angle Î¸ with the x-axis, then

Options:
A tan θ = 3
B tan Î¸ = 2
C θ = 4 5 â
D tan θ = 1 / 2

The correct answer is:
tan θ = 2

Solution:
Direction of the electric field from the axis of the dipole is given by = Î¸ + tan - 1 tan Î¸ 2
Here given that E is along the y-axis.
Î¸ + tan - 1 tan Î¸ 2 = Ï 2
tan Î¸ 2 = cot Î¸
tan Î¸ = 2 cot Î¸ â tan 2 Î¸ = 2
â´ tan Î¸ = 2

Topic: Physics, Electrostatics, JEE Main
An electric dipole is situated in an electric field of uniform intensity E whose dipole moment is p and moment of inertia is I. If the dipole is displaced slightly from the equilibrium position, then the angular frequency of its oscillations is

Options:
A pEI1/2
B pEI3/2
C IpE1/2
D pIE1/2

The correct answer is:
pEI1/2

Solution:
τ = P × E = P E sin θ .... (i)
-PEsinθ=Iα
[ I is the moment of inertia and α the angular acceleration]
Since displacement is small sin θ = ∼ θ
∴Iα=-PEθ⇒α=-PEIθ
The angular frequency of oscillation is given by
ω=PEI
The potential energy is lowest when the dipole moment is aligned with the field and highest when the two are antialigned.

Topic: Physics, Electrostatics, JEE Main
An electric dipole of dipole moment is 6.0 × 10-6 C m placed in a uniform electric field of 1.5 × 103 N C-1 in such a way that dipole moment is along electric field. The work done in rotating dipole by 180° in this field will be_______mJ.

Options:

The correct answer is:
18

Solution:
The work done for rotating a dipole is given by the change in the potential energy. The potential energy is given by

U=-p→·E→

W =∆U = Uf−Ui

So, the work done is

W=-pEcos180°-(-pEcos0°)=2pE=2×6×10-6×1.5×103=18 mJ

Topic: Physics, Electrostatics, JEE Main, JEE Main 2023 (08 Apr Shift 1)
An electric dipole of moment p→=-i^-3j^+2k^×10-29 C m at the origin 0,0,0 . The electric field due to this dipole at r→=+i^+3j^+5k^ (note that r→∙p→=0 ) is parallel to:

Options:
A +i^-3j^-2k^
B -i^+3j^-2k^
C +i^+3j^-2k^
D -i^-3j^+2k^

The correct answer is:
+i^+3j^-2k^

Solution:
As r→·p→=0, this means that r→ is perpendicular to p→.

For an electric dipole, electric field at equatorial position is antiparallel to p→.

⇒E→=-λp→,

where, λ is an arbitrary positive constant.

⇒E→=-λ-i^-3j^+2k^×10-29 C m-2

⇒E→=λi^+3j^-2k^×10-29 C m-2

For λ=110-29

⇒E→=i^+3j^-2k^ C m-2

Topic: Physics, Electrostatics, JEE Main, JEE Main 2020 (09 Jan Shift 1)
An electric dipole when placed in a uniform electric field $\boldsymbol{E}$ will have minimum potential energy, if the positive direction of dipole moment makes the following angle with $\boldsymbol{E}$

Options:
A $\pi$
B $\pi / 2$
C Zero
D $3 \pi / 2$

The correct answer is:
Zero

Solution:
Potential energy $=-p E \cos \theta$
When $\theta=0$. Potential energy $=-p E$ (minimum)

Topic: Physics, Electrostatics, JEE Main
An electric field $\vec{E}=(2 x \hat{i}) N C^{-1}$ exists in space. A cube of side $2 \mathrm{~m}$ is placed in the space as per figure given below. The electric flux through the cube is ______ $\mathrm{Nm}^2 / \mathrm{C}$.

Options:

The correct answer is:
16

Solution:
$\begin{aligned} & \overrightarrow{\mathrm{E}}=2 x \hat{\mathrm{i}} \\ & \phi=\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{A}}\end{aligned}$
$\begin{aligned} & \phi_{\text {in }}=-4 \times 4=-16 \mathrm{Nm}^2 / \mathrm{c} \\ & \phi_{\text {out }}=8 \times 4=32 \mathrm{Nm}^2 / \mathrm{c} \\ & d_{\text {net }}=\phi_{\text {in }+} \phi_{\text {out }}=-16+32=16 \mathrm{Nm}^2 / \mathrm{c}\end{aligned}$

Topic: Physics, Electrostatics, JEE Main, JEE Main 2024 (09 Apr Shift 2)
An electric field E→=4xi^-y2+1j^N/C passes through the box shown in figure. The flux of the electric field through surfaces ABCD and BCGF are marked as ϕI and ϕII respectively. The difference between ϕI-ϕII is (in Nm2/C ) ____________.

Options:

The correct answer is:
48

Solution:
Flux via ABCD
ϕ1=∫E.dA=0
Flux via BCEF
ϕ2=∫E.dA
ϕ2=E.A=4xi^-y2+1j^.4i^
=16x where, x=3
ϕ2=48N.m2C ϕ1-ϕ2=-48N.m2C

Topic: Physics, Electrostatics, JEE Main
An electric field E→=4xi^-y2+1j^N/C passes through the box shown in figure. The flux of the electric field through surfaces ABCD and BCGF are marked as ϕI and ϕII respectively. The difference between ϕI-ϕII is (in Nm2/C ) ____________.

Options:

The correct answer is:
48

Solution:
Flux via ABCD
ϕ1=∫E.dA=0
Flux via BCEF
ϕ2=∫E.dA
ϕ2=E.A=4xi^-y2+1j^.4i^
=16x where, x=3
ϕ2=48N.m2C ϕ1-ϕ2=-48N.m2C

Topic: Physics, Electrostatics, JEE Main, JEE Main 2020 (09 Jan Shift 2)
An electric field, $\overrightarrow{\mathrm{E}}=\frac{2 \hat{i}+6 \hat{j}+8 \hat{k}}{\sqrt{6}}$ passes through the surface of $4 \mathrm{~m}^2$ area having unit vector $\hat{n}=\left(\frac{2 \hat{i}+\hat{j}+\hat{k}}{\sqrt{6}}\right)$. The electric flux for that surface is ______ $\mathrm{Vm}$.

Options:

The correct answer is:
12

Solution:
$\begin{aligned} \phi & =\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{A}} \\ & =\left(\frac{2 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}+8 \hat{\mathrm{k}}}{\sqrt{6}}\right) \cdot 4\left(\frac{2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{6}}\right) \\ & =\frac{4}{6} \times(4+6+8)=12 \mathrm{Vm}\end{aligned}$

Topic: Physics, Electrostatics, JEE Main, JEE Main 2024 (08 Apr Shift 1)
An electric field Eâ=10x i^ N C-1 exists in space where x is measured in metre. Considering the potential at 10 m, 20 m to be zero, the potential at the origin is

Options:
A 1000Â V
B 100Â V
C 375 V
D 500 V

The correct answer is:
500 V

Solution:
dV=– E→.dr→

If V0 is the potential at the origin, then

∫0V0dV=-∫100E dx

⇒ V0=101022=500 V

Topic: Physics, Electrostatics, JEE Main
An electric field E→=25 i^+30 j^ N C-1 exists in a region of space. If the potential at the origin is taken to be zero then the potential at x=2 m, y=2 m is:

Options:
A -110 J C-1
B -140 J C-1
C -130 J C-1
D -120 J C-1

The correct answer is:
-110 J C-1

Solution:
∵∆V=-E→.∆r→
V-0=-25i^+30j^. 2i^+2j^
=-50+60
V=-110 volt=-110 J C-1

Topic: Physics, Electrostatics, JEE Main, JEE Main 2015 (11 Apr Online)
An electric field is given by (6i^+5j^+3k^) N C-1. The electric flux through a surface area 30i^ m2 lying in YZ-plane(in SI unit) is:

Options:
A 90
B 150
C 180
D 60

The correct answer is:
180

Solution:
Given:

E→=(6i^+5j^+3k^) N C-1

A→=30i^ m2

ϕ=E→·A→

⇒ϕ=(6i^+5j^+3k^)·(30i^)=180 SI unit

Topic: Physics, Electrostatics, JEE Main, JEE Main 2024 (29 Jan Shift 2)
An electric field of $1000 \mathrm{~V} / \mathrm{m}$ is applied to an electric dipole at angle of $45^{\circ}$. The value of electric dipole moment is $10^{-29} \mathrm{C} \cdot \mathrm{m} .$ What is the potential energy of the electric dipole?

Options:
A $-20 \times 10^{-18} \mathrm{~J}$
B $-7 \times 10^{-27} \mathrm{~J}$
C $-10 \times 10^{-29} \mathrm{~J}$
D $-9 \times 10^{-20} \mathrm{~J}$

The correct answer is:
$-7 \times 10^{-27} \mathrm{~J}$

Solution:
Potential energy of a dipole is given by $\mathrm{U}=-\overrightarrow{\mathrm{P}} \overrightarrow{\mathrm{E}}$

$=-P E \cos \theta$

[Where $\theta=$ angle between dipole and perpendicular to the field $=-\left(10^{-29}\right)\left(10^{3}\right) \cos 45^{\circ}$

$=-0.707 \times 10^{-26} \mathrm{~J}=-7 \times 10^{-27} \mathrm{~J}$

Topic: Physics, Electrostatics, JEE Main, JEE Main 2019 (11 Jan Shift 2)
An electric heater is placed inside a room of total wall area 137 m2 and maintained at a temperature 20 Â°C inside, outside temperature â10 Â°C. The walls are made up of three composite materials. The innermost layer is made up of wood of thickness 2.5 cm, the middle layer is of cement of thickness 1 cm and the exterior layer is 25 cm thick. Assuming there is no loss of heat through any other way, the power of the electric heater is (the thermal conductivity of wood, cement and brick are 0.125 W m2 Â°C-1, 1.5 W m-2 Â°C-1 and 1 W m-2 Â°C-1 respectively)

Options:
A 9000 W

B 8500 W
C 8800 W
D 9400 W

The correct answer is:
9000 W

Solution:
(RTh)w = lkA=2.5×10-20.125×137=0.2137 °C W-1
(RTh)C = lkA=1×10-21.5×137=0.0067137 °C W-1
(RTh)B = lkA=25×10-21×137=0.25137 °C W-1
RTh = (RTh)w +(RTh)C+ (RTh)B
= 0.2+0.0067+0.25137 = 3.33×10-3 °C W-1
Rate of heat flow = θ1-θ2RTh = 20--103.33×10-3 = 9000 W

Topic: Physics, Thermal Properties of Matter, JEE Main
An electric heater rated $220 \mathrm{~V}$ and $550 \mathrm{~W}$ is connected to AC mains. The current drawn by it is

Options:
A $0.8 \mathrm{~A}$
B $2.5 \mathrm{~A}$
C $0.4 \mathrm{~A}$
D $1.25 \mathrm{~A}$

The correct answer is:
$2.5 \mathrm{~A}$

Solution:
$I=\frac{P}{V}=\frac{550}{220}=2.5 \mathrm{~A}$

Topic: Physics, Current Electricity, KCET, KCET 2009
An electric heater with constant heat supply rate is used to convert a certain amount of liquid ammonia to saturated vapour at high pressure. The heater takes 14 minutes to bring the liquid at $15^{\circ} \mathrm{C}$ to the boiling point of $50{ }^{\circ} \mathrm{C}$ and 92 minutes to convert the liquid at the boiling point wholly to vapour. If the specific heat capacity of liquid ammonia is $4.9 \mathrm{~kJ} / \mathrm{kg} \mathrm{K}$, the latent heat of vaporisation of ammonia in $\mathrm{kJ} / \mathrm{kg}$ is

Options:
A 557
B 981
C 1127
D 2250

The correct answer is:
1127

Solution:
Let mass of liquid ammonia taken is $m$ and rate of heat supplied is $r$.
Then, as per given condition, $r \times 14=m \times 4.9 \times(50-15)$ and $r \times 92=m L$
Dividing these, we get
$$
\begin{aligned}
\frac{92}{14} & =\frac{L}{4.9 \times(50-15)} \text { or } L=\frac{92 \times 4.9 \times 35}{14} \\
& =1127 \mathrm{~kJ} / \mathrm{kg}
\end{aligned}
$$

Topic: Physics, Thermal Properties of Matter, JEE Main
An electric instrument consists of two units. Each unit must function independently for the instrument to operate. The probability that the first unit functions is 0.9 and that of the second unit is 0.8. The instrument is switched on and it fails to operate. If the probability that only the first unit failed and second unit is functioning is p, then 98p is equal to

Options:

The correct answer is:
28

Solution:
Given that

An electric instrument consists of two units. Each unit must function independently for the instrument to operate. The probability that the first unit functions is 0.9 and that of the second unit is 0.8. The instrument is switched on and it fails to operate. If the probability that only the first unit failed and second unit is functioning is p.

Let E1= Event of first unit is working

E2= Event of second unit is working

PE1=0.9 PE2=0.8

We know that PE = 1 - PE

PE¯1=0.1 PE¯2=0.2

By using Baye's theorem

Required probability

p = PE2×PE1PE2×PE1+PE1×PE2+PE2×PE1

p=0.8×0.10.1×0.2+0.9×0.2+0.1×0.8=81002 +18 +8100= 828

So, 98P=98×828=28

Topic: Mathematics, Probability, JEE Main, JEE Main 2021 (31 Aug Shift 1)
An electric kettle has two heating coils. When one of the coils is connected to a.c. source, the water in the kettle boils in 10 minute. When the other coil is used the water boils in 40 minute. If both the coils are connected in parallel, the time taken by the same quantity of water to boil will be:

Options:
A 8 minute
B 4 minute
C 25 minute
D 15 minute

The correct answer is:
8 minute

Solution:
Here $Q=\frac{V^2}{R_1} \times t_1=\frac{V^2}{R_2} \times t_2$
$$
\begin{aligned}
& =\frac{V^2}{R} \times t \\
& \therefore \quad \frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2} \\
& \Rightarrow \quad \frac{Q}{V^2 t}=\frac{Q}{V^2 t_1}+\frac{Q}{V^2 t_2} \\
& \Rightarrow \quad \frac{1}{t}=\frac{1}{t_1}+\frac{1}{t_4} \\
& \Rightarrow \quad t=\frac{t_1 t_2}{t_1+t_2} \\
& =\frac{10 \times 40}{10+40}=8 \mathrm{~min} \\
&
\end{aligned}
$$

Topic: Physics, Alternating Current, JEE Main
An electric kettle takes $4 \mathrm{~A}$ current at $220 \mathrm{~V}$. How much time will it take to boil $1 \mathrm{~kg}$ of water from temperature $20^{\circ} \mathrm{C}$ ? The temperature of boiling water is $100^{\circ} \mathrm{C}$.

Options:
A $4.2 \mathrm{~min}$
B $6.3 \mathrm{~min}$
C $8.4 \mathrm{~min}$
D $12.6 \mathrm{~min}$

The correct answer is:
$6.3 \mathrm{~min}$

Solution:
$$
\begin{aligned}
& \text { Vit }=m c \Delta \theta \\
& t=\frac{1 \times 4200 \times 80}{220 \times 4}=381.8 \mathrm{~s}=6.36 \mathrm{~min}
\end{aligned}
$$

Topic: Physics, Thermal Properties of Matter, NEET, NEET 2008 (Mains)
An electric kettle takes $4 \mathrm{~A}$ current at $220 \mathrm{~V}$. How much time will it take to boil $1 \mathrm{~kg}$ of water from temperature $20^{\circ} \mathrm{C}$ ? The temperature of boiling water is $100^{\circ} \mathrm{C}$.

Options:
A $4.2 \mathrm{~min}$
B $6.3 \mathrm{~min}$
C $8.4 \mathrm{~min}$
D $12.6 \mathrm{~min}$

The correct answer is:
$6.3 \mathrm{~min}$

Solution:
$$
\begin{aligned}
& \text { Vit }=m c \Delta \theta \\
& t=\frac{1 \times 4200 \times 80}{220 \times 4}=381.8 \mathrm{~s}=6.36 \mathrm{~min}
\end{aligned}
$$

Topic: Physics, Thermal Properties of Matter, JEE Main
An electric motor exerts a force of $40 \mathrm{~N}$ on a cable and pulls it by a distance of $30 \mathrm{~m}$ in one minute. The power supplied by the motor (in Watts) is

Options:
A 20
B 200
C 2
D 10

The correct answer is:
20

Solution:
$P=F v=F \times \frac{s}{t}=40 \times \frac{30}{60}=20 \mathrm{~W}$

Topic: Physics, Work Power Energy, JEE Main
An electric toaster has resistance of 60 Ω at room temperature 27 °C. The toaster is connected to a 220 V supply. If the current flowing through it reaches 2.75 A, the temperature attained by toaster is around: (if α=2×10-4 C-1∘ )

Options:
A 694°C
B 1235°C
C 1694°C
D 1667°C

The correct answer is:
1694°C

Solution:
Resistance at 27 °C, RT=27=60 Ω.

Final resistance, RT=2202.75=80 Ω.

Now, we can write R=R0(1+αΔT)

⇒80=601+2×10-4( T-27)

⇒T≈1694°C

Topic: Physics, Current Electricity, JEE Main, JEE Main 2024 (30 Jan Shift 1)
An electric wire is connected across a cell of e.m.f. E. The current $I$ is measured by an ammeter of resistance $R$. According to ohm's law

Options:
A $E=I^2 R$
B $E=I R$
C $E=R / I$
D $E=I / R$

The correct answer is:
$E=I R$

Solution:
Correct option is B. \(E=I R\)
From ohm's law potential drop \(=\) current \(\mathrm{x}\) resistance mathmatically \(E=I R\)

Topic: Physics, Current Electricity, JEE Main
An electrical bulb rated 220 V,100 W, is connected in series with another bulb rated 220 V, 60 W. If the voltage across combination is 220 V, the power consumed by the 100 W bulb will be about _____ W.

Options:

The correct answer is:
14

Solution:
Electric power is given by P=V2R, here, V is voltage and R is resistance.

Now, for each bulb, R1=V2P=2202100=484 Ω and R2=V2P=220260=484106 Ω

Net current through the series combination is I=220484+484×106 A=0.171 A

Hence, the power consumed by the 100 W bulb is P1=I2R1=14.06 W

Topic: Physics, Current Electricity, JEE Main, JEE Main 2022 (28 Jul Shift 2)
An electrical cable having a resistance of $0.2 \Omega$ delivers $10 \mathrm{kw}$ at $200 \mathrm{~V}$ D.C. to a factory. What is the efficiency of transmission?

Options:
A $65 \%$
B $75 \%$
C $85 \%$
D $95 \%$

The correct answer is:
$95 \%$

Solution:
$P=V I \Rightarrow I=\frac{10 \times 10^3}{200}=50 A$, Power loss $=(50)^2(0.2)=500 \mathrm{~W}$
Efficiency $=\frac{10000 \times 100}{10000+500}=95.23 \%$

Topic: Physics, Current Electricity, JEE Main
An electrical device draws $2 \mathrm{~kW}$ power from $\mathrm{AC}$ mains (voltage $223 \mathrm{~V}(\mathrm{rms})=\sqrt{50000} \mathrm{~V}$ ). The current differs (lags) in phase by $\phi\left(\tan \phi=\frac{-3}{4}\right)$ as compared to voltage.
Find (a) $\mathrm{R}$, (b) $\mathrm{X}_{\mathrm{C}}-\mathrm{X}_{\mathrm{L}}$ and (c) $\mathrm{I}_{\mathrm{M}}$. Another device has twice the values for $\mathrm{R}, \mathrm{X}_{\mathrm{C}}$ and $\mathrm{X}_{\mathrm{L}}$. How are the answers affected?

Options:

None

Solution:
As given that, the power is
$$
\mathrm{P}=2 \mathrm{~kW}=2000 \mathrm{~W}
$$
The voltage when current lags to the voltage
$$
\begin{aligned}
\mathrm{V} &=223 \mathrm{~V}, \\
\tan \phi &=-\frac{3}{4} \\
\mathrm{I}_{\mathrm{M}} &=\mathrm{I}_0=? \\
\mathrm{X}_{\mathrm{C}}-\mathrm{X}_{\mathrm{L}} &=? \\
\mathrm{R} &=?
\end{aligned}
$$
As we know that,
$$
\text { Power }(\mathrm{P})=\frac{\mathrm{V}^2}{\mathrm{Z}}
$$
So, $\quad \mathrm{Z}=\frac{\mathrm{V}^2}{\mathrm{P}}=\frac{223 \times 223}{2 \times 10^3}=25$ Impedance $\mathrm{Z}=25 \Omega$
$$
\begin{array}{rlrlr}
\text { Impedance } Z & =\sqrt{R^2+\left(X_L-X_C\right)^2} \\
\Rightarrow & \text { or } & 25 & =\sqrt{R^2+\left(X_L-X_C\right)^2} \\
& \text { And, } & 625 & =R^2+\left(X_L-X_C\right)^2 \\
& \tan \phi & =\frac{X_L-X_C}{R}=\frac{3}{4}
\end{array}
$$
or $\quad X_L-X_C=\frac{3 R}{4}$
By putting the value of $\left(X_L-X_C\right)=\frac{3 R}{4}$ in Eq. (i), then,
$$
625=R^2+\left(\frac{3 R}{4}\right)^2=R^2+\frac{9 R^2}{16}
$$
or $625=\frac{25 \mathrm{R}^2}{16}$
(a) Resistance $\mathrm{R}=\sqrt{25 \times 16}=\sqrt{400}$ $=20 \Omega$
(b) $\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}=\frac{3 \mathrm{R}}{4}=\frac{3}{4} \times 20=15 \Omega$
(c) Main current
$$
\begin{aligned}
\mathrm{I}_{\mathrm{M}} &=\sqrt{ } 2 \mathrm{I}=\sqrt{2} \frac{\mathrm{V}}{\mathrm{Z}} \\
&=\frac{223}{25} \times \sqrt{2}=12.6 \mathrm{~A}
\end{aligned}
$$
So, $R, X_C, X_L$ are all doubled, $\tan \phi=\left(\frac{X_C-X_L}{R}\right)$ does not change. $Z$ will be doubled, then the current $\left(I=\frac{V}{Z}\right)$ become halved. As I become half $(\mathrm{P}=$ VI) power is also halved as voltage remain same.

Topic: Physics, Alternating Current
An electrical device draws 2kW power from AC mains (voltage 223V (rms) = $\sqrt[{}]{50,000}$ V). The current differs (lags) in phase by (tanφ= –¾) as compared to voltage. Find R

Options:
A 20Ω
B 40Ω
C 60Ω
D 80Ω

The correct answer is:
20Ω

Solution:
Correct Option is : (A)
20Ω

Topic: Chemistry, Alternating Current, NEET
An electrical power line, having a total resistance of 2Ω, delivers 1 kW at 220 V. The efficiency of the transmission line is approximately :

Options:
A 72%
B 91%
C 85%
D 96%

The correct answer is:
96%

Solution:
i=PV=1000220

PR=i2R

η=1000×1001000+41.32=96%

Topic: Physics, Current Electricity, JEE Main, JEE Main 2020 (05 Sep Shift 1)
An electrolyte of a polymer-salt complex of poly (ethylene oxide) - LiCF3SO3 is shaped into a free standing circular film of 20 mm diameter and a thickness of 20μm.When it is sandwiched between 2 stainless steel circular electrodes of the same diameter, this cell exhibits a conductance of 3145 S.What is the specific conductivity of the electrolyte?

Options:
A 4mS cm-1
B 0.4S cm-1
C 40mS cm-1
D 0.004 S cm-1

The correct answer is:
40mS cm-1

Solution:
Specific conductance is given by

κ=1R×lA

Here thickness is 20×10-4 cm and radius is 1 cm

So κ=3145×20×10-43.14×12 =0.04 Scm-1

κ=40 mS cm-1

Topic: Chemistry, Solutions, JEE Main
An electrolytic cell is constructed for preparing hydrogen. For an average current of 2 ampere in the circuit, the time in seconds required to produce 450 mL of hydrogen at NTP is approximately

Options:

The correct answer is:
1930

Solution:
22400 mL H2 at STP =1molH2
So 450 ml H2 at STP
=450mL22400mLÃ1molH2
=0.02 mol
Electrode reaction
2H++2e-2FâH21mol
The liberation of 1 mol H2 requires =2Ã96500C
So the liberation of 0.02 mol H2 requires =2Ã96500Ã0.02C
=3860 C
Time required to produce 0.02 mol H2 , t=Qi
=38602A=1930 sec

Topic: Chemistry, Electrochemistry, JEE Main
An electromagnetic wave in vacuum has the electric and magnetic fields $\vec{E}$ and $\vec{B}$, which are always perpendicular to each other. The direction of polarization is given by $\vec{X}$ and that of wave propagation by $\overrightarrow{\mathrm{k}}$. Then :

Options:
A
$\vec{X} \| \vec{B}$ and $\vec{k} \| \vec{B} \times \vec{E}$

B
$\vec{X} \| \vec{E}$ and $\vec{k} \| \vec{E} \times \vec{B}$

C
$\vec{X} \| \vec{B}$ and $\vec{k} \| \vec{E} \times \vec{B}$

D
$\vec{X} \| \vec{E}$ and $\vec{k} \| \vec{B} \times \vec{E}$

The correct answer is:
$\vec{X} \| \vec{B}$ and $\vec{k} \| \vec{E} \times \vec{B}$

Solution:
Direction of polarization is parallel to magnetic field,
$\therefore \quad \overrightarrow{\mathrm{X}} \| \overrightarrow{\mathrm{B}}$
and direction of wave propagation is parallel to $\overrightarrow{\mathrm{E}} \times \overrightarrow{\mathrm{B}}$
$\therefore \quad \overrightarrow{\mathrm{K}} \| \overrightarrow{\mathrm{E}} \times \overrightarrow{\mathrm{B}}$

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2012 (Offline)
An electromagnetic wave is propagating in vacuum along -j^ direction. The magnetic field of the wave is given by B→=2×10-8cosπ×1015t+yc k^ T. The electric field E→ of this wave is (c= speed of light)

Options:
A E→=4cosπ×1015t+yc j^ V m-1
B E→=6cosπ×1015t+yc i^ V m-1
C E→=6cosπ×1015t-yc j^ V m-1
D E→=4cosπ×1015t-yc i^ V m-1

The correct answer is:
E→=6cosπ×1015t+yc i^ V m-1

Solution:
As we know,

c=E0B0⇒E0=B0c=2×10-83×108=6 V m-1

Now, direction of propagation of wave

=E^×B^⇒-j^=E^×k^⇒E^=i^

Angular frequency will remain the same. Therefore electric field can be written as,

E→=6cosπ×1015t+yc i^ V m-1

Topic: Physics, Electromagnetic Waves, TS EAMCET, TS EAMCET 2022 (18 Jul Shift 1)
An electromagnetic wave is represented by the electric filed E→=E0n^sin⁡ωt+6y-8z. Taking unit vectors in x,y and z directions to be i^,j^,k^, the directions of propogations s^, is:

Options:
A s^=3i^-4j^5
B s^=-3i^+4j^5
C s^=4i^-3k^5
D s^=-4i^-3k^5

The correct answer is:
s^=-3i^+4j^5

Solution:
The standard equation of electromagnetic wave E→=E0n^sinωt-kx^

here x^ represents the direction of propagation wave.

On comparing the given equation with the standard equation, the direction of propagation is along line -6y+8z . Thus, unit vector along the direction of propagation is,
s^=-6j^+8k^10=-35j^+45k^

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2019 (12 Apr Shift 1)
An electromagnetic wave of frequency $1 \times 10^{14}$ hertz is propagating along $\mathrm{z}$-axis. The amplitude of electric field is $4 \mathrm{~V} / \mathrm{m}$. If $\varepsilon_0=8.8 \times 10^{-12} \mathrm{C}^2 / \mathrm{N}-$ $\mathrm{m}^2$, then average energy density of electric field will be:

Options:
A
$35.2 \times 10^{-10} \mathrm{~J} / \mathrm{m}^3$

B
$35.2 \times 10^{-11} \mathrm{~J} / \mathrm{m}^3$

C
$35.2 \times 10^{-12} \mathrm{~J} / \mathrm{m}^3$

D
$35.2 \times 10^{-13} \mathrm{~J} / \mathrm{m}^3$

The correct answer is:
$35.2 \times 10^{-12} \mathrm{~J} / \mathrm{m}^3$

Solution:
Given: Amplitude of electric field,
$$
E_0=4 \mathrm{v} / \mathrm{m}
$$
Absolute permitivity,
$$
\varepsilon_0=8.8 \times 10^{-12} \mathrm{c}^2 / \mathrm{N}-\mathrm{m}^2
$$
Average energy density $u_E=$ ?
Applying formula,
Average energy density $u_E=\frac{1}{4} \varepsilon_0 E^2$
$$
\begin{aligned}
&\Rightarrow \quad u_E=\frac{1}{4} \times 8.8 \times 10^{-12} \times(4)^2 \\
&=35.2 \times 10^{-12} \mathrm{~J} / \mathrm{m}^3
\end{aligned}
$$

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2014 (11 Apr Online)