datavorous/entrance-exam-dataset huggingface dataset:
The angle of incidence at which reflected light totally polarized for reflection from air to glass (refractive index $n$ ), is

Options:
A
$\sin ^{-1}(n)$

B
$\sin ^{-1}(1 / n)$

C
$\tan ^{-1}(1 / n)$

D
$\tan ^{-1}(n)$

The correct answer is:
$\tan ^{-1}(n)$

Solution:
Brewster's law: According to this law the ordinary light is completely polarised in the plane of incidence when it gets reflected from transparent medium at a particular angle known as the angle of polarisation.
$$
\mathrm{n}=\tan \mathrm{i}_{\mathrm{p}}
$$

Topic: Physics, Wave Optics, JEE Main, JEE Main 2004
The angle of intersection of the circles $x^{2}+y^{2}$ $-x+y-8=0$ and $x^{2}+y^{2}+2 x+2 y-11=0$ is

Options:
A $\tan ^{-1}\left(\frac{19}{9}\right)$
B $\tan ^{-1}(19)$
C $\tan ^{-1}\left(\frac{9}{19}\right)$
D $\tan ^{-1}(9)$

The correct answer is:
$\tan ^{-1}\left(\frac{9}{19}\right)$

Solution:
Angle of intersection between two circles is given by
$\cos \theta=\frac{r_{1}^{2}+r_{2}^{2}-d^{2}}{2 r_{1} r_{2}}=\frac{\frac{17}{2}+13-\frac{10}{4}}{2 \sqrt{\frac{17}{2}} \cdot \sqrt{13}}$
$\left\{\begin{aligned} \text { here, } r_{1} &=\sqrt{\left(\frac{1}{2}\right)^{2}+\left(-\frac{1}{2}\right)^{2}+8} \\ &=\sqrt{\frac{17}{2}} \\ r_{2} &=\sqrt{13} \end{aligned}\right.$
and $d=c_{1} c_{2}=\sqrt{\frac{10}{2}}$
$\Rightarrow \cos \theta=\left(\frac{19}{\sqrt{442}}\right)$
or $\tan \theta=\left(\frac{9}{19}\right)$
$\Rightarrow \theta=\tan ^{-1}\left(\frac{9}{19}\right)$

Topic: Mathematics, Circle, JEE Main
The angle of intersection of the curves $y=x^2$ and $x=y^2$ at $(1,1)$ is :

Options:
A $\tan ^{-1}\left(\frac{4}{3}\right)$
B $\tan ^{-1}(1)$
C $90^{\circ}$
D $\tan ^{-1}\left(\frac{3}{4}\right)$

The correct answer is:
$\tan ^{-1}\left(\frac{3}{4}\right)$

Solution:
$\begin{aligned} & y=x^2 \Rightarrow \frac{d y}{d x}=m_1=2 x \\ & \Rightarrow\left(\frac{d y}{d x}\right)_{(1,1)}=2=m_1 \quad \text { and } x=y^2 \Rightarrow 1=2 y \frac{d y}{d x} \\ & \Rightarrow \frac{d y}{d x}=m_2=\frac{1}{2 y} \Rightarrow\left(\frac{d y}{d x}\right)_{(1,1)}=\frac{1}{2}\end{aligned}$
$\therefore$ Angle of intersection, $\tan \theta=\frac{m_1-m_2}{1+m_1 m_2}=\frac{2-\frac{1}{2}}{1+2 \times \frac{1}{2}}=\frac{3}{4}$ $\Rightarrow \theta=\tan ^{-1}(3 / 4)$

Topic: Mathematics, Application of Derivatives, JEE Main
The angle of intersection of two given circles \( x^{2}+y^{2}-2 x+8 y+13=0 \) and \( x^{2}+y^{2}-4 x+6 y+11=0 \) is

Options:
A \( 60^{\circ} \)
B \( \tan ^{-1}\left(\frac{1}{2}\right) \)
C \( \tan ^{-1}\left(\frac{3}{5}\right) \)
D \( 45^{\circ} \)

The correct answer is:
\( 45^{\circ} \)

Solution:
Given circles are x2+y2-2x+8y+13=0 and x2+y2-4x+6y+11=0

Here,C1=1,-4, C2=2,-3

⇒r1=1+16-13=2

And r2=4+9-11=2

Now, d=C1C2=2-12+-3+42=2

∴ cosθ=d2-r12-r222r1r2=2-4-22×2×2=12

⇒ θ=45°

Topic: Mathematics, Circle, JEE Main
The angle of polarisation for any medium is $60^{\circ}$, what will be critical angle for this

Options:
A $\sin ^{-1} \sqrt{3}$
B $\tan ^{-1} \sqrt{3}$
C $\cos ^{-1} \sqrt{3}$
D $\sin ^{-1} \frac{1}{\sqrt{3}}$

The correct answer is:
$\sin ^{-1} \frac{1}{\sqrt{3}}$

Solution:
By using $\mu=\tan \theta_p \Rightarrow \mu=\tan 60=\sqrt{3}$
$\begin{aligned} & C=\sin ^{-1}\left(\frac{1}{\mu}\right) \Rightarrow C=\sin ^{-1}\left(\frac{1}{\sqrt{3}}\right)\end{aligned}$

Topic: Physics, Wave Optics, JEE Main
The angle of projection for a projectile to have same horizontal range and maximum height is :

Options:
A $\tan ^{-1}(4)$
B $\tan ^{-1}\left(\frac{1}{4}\right)$
C $\tan ^{-1}\left(\frac{1}{2}\right)$
D $\tan ^{-1}(2)$

The correct answer is:
$\tan ^{-1}(4)$

Solution:
$\begin{aligned} & \frac{\mathrm{u}^2 \sin 2 \theta}{\mathrm{g}}=\frac{\mathrm{u}^2 \sin ^2 \theta}{2 \mathrm{~g}} \\ & 4 \sin \theta \cos \theta=\sin ^2 \theta \\ & 4=\tan \theta\end{aligned}$

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2024 (08 Apr Shift 2)
The angle of the top of a vertical tower standing on a horizontal plane is observed to be 45° from a point A on the plane. Let B be the point 30 m vertically above the point A. If the angle of elevation of the top of the tower from B be 30° , then the distance (in m) of the foot of the tower from the point A is:

Options:
A 153+3
B 151+3
C 155-3
D 153-3

The correct answer is:
153+3

Solution:
tan45o=1=x+30y⇒x+30=y ......(i)
tan30o=13=xy⇒x=y3 ......(ii)
from (i) and (ii) y=153+3

Topic: Mathematics, Heights and Distances, JEE Main, JEE Main 2019 (12 Apr Shift 2)
The angles $A, B, C$ of a triangle $A B C$ are in AP. If $A B=6, B C=7$, then $A C=$

Options:
A $\sqrt{40}$
B $\sqrt{41}$
C $\sqrt{43}$
D 6

The correct answer is:
$\sqrt{43}$

Solution:
$\begin{aligned} & \text { (c) Given In } \triangle A B C \text {, } \\ & \angle A, \angle B, \angle C \text { are in AP } \\ & \Rightarrow \\ & \text { In } \triangle A B C,\end{aligned}$
$$
\begin{aligned}
\angle A+\angle B+\angle C & =180^{\circ} \\
2 \angle B+\angle B & =180^{\circ} \\
\Rightarrow \quad \angle B & =60^{\circ}
\end{aligned}
$$
From cosine rule,
$$
\begin{aligned}
\cos B & =\frac{A B^2+B C^2-A C^2}{2 A B \times B C} \\
\cos 60^{\circ} & =\frac{36+49-x^2}{2 \times 6 \times 7} \\
\frac{1}{2} & =\frac{85-x^2}{84} \\
41 & =85-x^2
\end{aligned}
$$
$$
\begin{aligned}
x^2 & =85-42 \\
x^2 & =43 \\
x & =\sqrt{43}
\end{aligned}
$$
Hence, option (3) is correct.

Topic: Mathematics, Sequences and Series, JEE Main
The angular depressions of the top and the foot of a tower, as seen from the top of a second tower which is 150 meters high and standing on the same level as the first, are tan-143 and tan-152 respectively. If the distance between their tops is d, then d20 is

Options:

The correct answer is:
5

Solution:
tanâ¡Î±=150x=52âx=60
tanâ¡Î²=150-hx=150-h60=43â150-h=80âh=70
â AE=80 units,CE=60 units
â AC=100 units (Pythagoras theorem)

Topic: Mathematics, Heights and Distances, JEE Main
The angular elevation of tower CD at a point A due south of it is 60 o and at a point B due west of A, the elevation is 3 0 o . If AB=5 km, then the height of the tower is (where, C,B and A are on the same ground level)

Options:
A 23 km
B 26 km
C 532km
D 5322km

The correct answer is:
5322km

Solution:
From ΔCDA, x=hcot60o=h3
From ΔCDB, y=hcot30o=3h
x2+52=y2 (in ∆ACB)
⇒h32+52=3h2⇒h=5322km.

Topic: Mathematics, Heights and Distances, JEE Main
The angular frequency of alternating current in a L-C-R circuit is 100 rad s-1. The components connected are shown in the figure. Find the value of inductance of the coil and capacity of condenser.

Options:
A 1.33 H and 150 μF
B 0.8 H and 150 μF
C 1.33 H and 250 μF
D 0.8 H and 250 μF

The correct answer is:
0.8 H and 250 μF

Solution:
Current through 60 Ω resistance =1560=14A

thus capacitor current =14 A

∵ VC=I XC

10=14×1ωC

∴ C=140ω=14000=250 μF

Now,

current through 40 Ω resistance =2040=12 A

thus current through inductor =12-14=14 A

VL=IXL=14×ωL

20=14×100×L

⇒L=0.8 H

Topic: Physics, Alternating Current, JEE Main, JEE Main 2021 (25 Feb Shift 1)
The angular frequency of the damped oscillator is given by, $\omega=\sqrt{\left(\frac{\mathrm{k}}{\mathrm{m}}-\frac{\mathrm{r}^2}{4 \mathrm{~m}^2}\right)}$ where $\mathrm{k}$ is the spring constant, $\mathrm{m}$ is the mass of the oscillator and $r$ is the damping constant. If the ratio $\frac{r^2}{\mathrm{mk}}$ is $8 \%$, the change in time period compared to the undamped oscillator is approximately as follows:

Options:
A
increases by $1 \%$

B
increases by $8 \%$

C
decreases by $1 \%$

D
decreases by $8 \%$

The correct answer is:
increases by $1 \%$

Solution:
The change in time period compared to the undamped oscillator increases by $1 \%$.

Topic: Physics, Oscillations, JEE Main, JEE Main 2014 (11 Apr Online)
The angular momentum for the electron in Bohr's orbit is L. If the electron is assumed to revolve in second orbit of hydrogen atom, then the change in angular momentum will be:

Options:
A $\frac{\mathrm{L}}{2}$
B
zero
C
L
D
2L

The correct answer is:
L

Solution:
From the Bohr's quantization rule
Angular momentum $\mathrm{L}=\frac{\mathrm{nh}}{2 \pi}, \mathrm{L}_1=\frac{1 . \mathrm{h}}{2 \pi}=\mathrm{L}$ ( $n=1$ for first orbit)
in ( $n=2$ for second orbit) $L_2=\frac{2 h}{2 \pi}=2 L$
Hence, change in angular momentum $\mathrm{L}_2-\mathrm{L}_1=2 \mathrm{~L}-$ $\mathrm{L}=\mathrm{L}$

Topic: Physics, Atomic Physics, JEE Main
The angular momentum for the electron in Bohr’s orbit is L. If the electron is assumed to revolve in second orbit of hydrogen atom, then the change in angular momentum will be

Options:
A Zero
B 2L
C L
D L2

The correct answer is:
L

Solution:
Bohr's formula for angular momentum of an electron in an orbit is given by

L=nh2π

For the first orbit, n=1 and the angular momentum is L1=h2π=L

For 2nd orbit

L2=2h2π=2L

Hence, the change in angular momentum is

ΔL=2L-L=L

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2023 (10 Apr Shift 1)
The angular momentum of a moving body remains constant if

Options:
A net external force is applied
B net external torque is applied
C net pressure is applied
D net external torque is not applied.

The correct answer is:
net external torque is not applied.

Solution:
When the net external torque acting on a system is zero, the total angular momentum of the system remains constant. Therefore the angular momentum of a moving body remains constant if net external torque is not applied.

Topic: Physics, Rotational Motion, AIIMS, AIIMS 2002
The angular momentum of a planet of mass M moving around the sun in an elliptical orbit is L→. The magnitude of the areal velocity of the planet is :

Options:
A 4LM
B LM
C 2LM
D L2M

The correct answer is:
L2M

Solution:
For small displacement ds of the planet its area can be written as

dA=12rdl

=12rdsinθ

A.vel =dAdt=12rsinθdsdt=Vrsinθ2

dAdt=12mVrsinθm=L2 m

Topic: Physics, Gravitation, JEE Main, JEE Main 2021 (18 Mar Shift 2)
The angular momentum of an electron in a hydrogen atom is proportional to :
(Where $r$ is the radius of orbit of electron)

Options:
A r
B $\sqrt{\mathrm{r}}$
C $\frac{1}{\sqrt{\mathrm{r}}}$
D $\frac{1}{\mathrm{r}}$

The correct answer is:
$\sqrt{\mathrm{r}}$

Solution:
$\begin{aligned} & \mathrm{F}_{\mathrm{C}}=\frac{\mathrm{mv}^2}{\mathrm{r}} \\ & \frac{\mathrm{Kq}_1 \mathrm{q}_2}{\mathrm{r}^2}=\frac{\mathrm{mv}^2}{\mathrm{r}} \\ & \mathrm{mv}^2 \mathrm{r}^2=K \mathrm{~K}_1 \mathrm{q}_2 \mathrm{r} \\ & \frac{\mathrm{L}^2}{\mathrm{~m}}=K \mathrm{Kq}_1 \mathrm{q}_2 \mathrm{r} \\ & \mathrm{L} \propto \sqrt{\mathrm{r}}\end{aligned}$

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2024 (05 Apr Shift 2)
The angular momentum of an electron in Bohr's hydrogen atom whose energy is - 3.4 eV, is

Options:
A 5h2π
B h2π
C hπ
D 2h3π

The correct answer is:
hπ

Solution:
The energy of an electron in nth orbit of a hydrogen atom is given by
En= -13.6n2 eV
⇒ 3.4=-13.6n2
⇒ n2=4
⇒ n=2
The angular momentum of electron is given by
L=nh2π=2h2π=hπ

Topic: Physics, Atomic Physics, JEE Main
The angular momentum of an electron in He+ moving in an orbit is h/π . The de-brogile wavelength associated with electron is: [ a0 is radius of first bohr's orbit of H -atom]

Options:
A 2πa0
B πa0
C 4πa0
D πa02

The correct answer is:
2πa0

Solution:
nh2π=hπ
n=2
r=a0×n2z=a0×222=2a0
nλ=2πr
λ=πr=π2a0

Topic: Chemistry, Structure of Atom, JEE Main
The angular momentum of an electron moving in an orbit of hydrogen atom is $1.5\left(\frac{\mathrm{h}}{\pi}\right)$. The energy in the same orbit is nearly

Options:
A $-1.3 \mathrm{eV}$
B $-1.4 \mathrm{eV}$
C $-1.5 \mathrm{eV}$
D $-1.6 \mathrm{eV}$

The correct answer is:
$-1.5 \mathrm{eV}$

Solution:
Angular momentum of electron moving in $n^{\text {th }}$ orbit of hydrogen atom
Hence, $n=3$
Now, energy of electron in $3^{\text {rd }}$ orbit
$\begin{aligned} E & =\frac{-13.6 \times z^2}{n^2} \\ & =\frac{-13.6 \times 1^2}{3^2} \\ & =-1.51 \mathrm{eV} \\ & \approx-1.5 \mathrm{eV}\end{aligned}$

Topic: Physics, Atomic Physics, JEE Main
The angular momentum of electron in Li2+ is found to be 14h11. Calculated the potential energy (in eV) of system.

Options:
A 13.6×382
B -13.6×382
C -2×13.6×382
D -2×13.6×832

The correct answer is:
-2×13.6×382

Solution:
The angular momentum of an electron in an nth orbit is given by
m e vr=n⋅ h 2π ;n=1.2.3.......
nh2Ï=14h11ân=8
P.E=-2Ã13.6Z2n2eV
Potential energy =-2Ã13.6ÃZn2
Potential energy =-2Ã13.6Ã382
âµfor Li2+,Z=3

Topic: Chemistry, Structure of Atom, JEE Main
The angular momentum of electron in \(\mathrm{n}^{\text {th }}\) orbit is given by

Options:
A nh
B \(\mathrm{n} \frac{\mathrm{h}}{2 \pi}\)
C \(\frac{\mathrm{h}}{2 \pi \mathrm{n}}\)
D \(\mathrm{n}^2 \frac{\mathrm{h}}{2 \pi}\)

The correct answer is:
\(\mathrm{n} \frac{\mathrm{h}}{2 \pi}\)

Solution:
According to Bohr's second postulate, angular momentum is integral multiple of \(\frac{\mathrm{h}}{2 \pi}\).
So, \(\mathrm{mvr}=\mathrm{n}\left(\frac{\mathrm{h}}{2 \pi}\right)\)

Topic: Physics, Atomic Physics, JEE Main
The angular resolution of a $10 \mathrm{~cm}$ diameter telescope at a wavelength of $5000 Å$ is of the order of:

Options:
A $10^6 \mathrm{rad}$
B $10^{-2} \mathrm{rad}$
C $10^{-4} \mathrm{rad}$
D $10^{-6} \mathrm{rad}$

The correct answer is:
$10^{-6} \mathrm{rad}$

Solution:
Here $R \cdot P .=\frac{1}{\Delta \theta}$
$$
\begin{aligned}
& \therefore \text { Angular resolution } \Delta \theta=\frac{1.22 \lambda}{\mathrm{D}} \\
&=\frac{1.22 \times 5000 \times 10^{-10}}{0.1} \\
&=6.10 \times 10^{-6} \\
& \cong 10^{-6}
\end{aligned}
$$

Topic: Physics, Ray Optics, JEE Main
The angular shape of molecule $\left(\mathrm{O}_3\right)$ consists of

Options:
A 2 singma and 1 pi bond
B 1 sigma and 2 pi bonds
C 2 sigma and 2 pi bonds
D 1 sigma and 1 pi bond

The correct answer is:
2 singma and 1 pi bond

Solution:
No Solution Available

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET, NEET 2008 (Mains)
The angular speed of a fly wheel making 120 revolutions/minute is

Options:
A $2 \pi \mathrm{rad} / \mathrm{s}$
B $4 \pi^2 \mathrm{rad} / \mathrm{s}$
C $\pi \mathrm{rad} / \mathrm{s}$
D $4 \pi \mathrm{rad} / \mathrm{s}$

The correct answer is:
$4 \pi \mathrm{rad} / \mathrm{s}$

Solution:
$120 \mathrm{rev} / \mathrm{min}=120 \times \frac{2 \pi}{60} \mathrm{rad} / \mathrm{sec}=4 \pi \mathrm{rad} / \mathrm{sec}$

Topic: Physics, Motion In Two Dimensions, JEE Main
The angular speed of a fly wheel moving with uniform angular acceleration changes from 1200 rpm to 3120 rpm in 16 seconds. The angular acceleration in rad s-2 is:

Options:
A 4π
B 12π
C 104π
D 2π

The correct answer is:
4π

Solution:
For uniform angular acceleration we can write,

ω=ω0+αt⇒α=ω-ω0t

Given, ω=3120 rpm=3120×2π60=104π rad s-1

and ω0=1200 rpm=1200×2π60=40π rad s-1

Therefore, α=104π rad s-1-40π rad s-116=4π rad s-2

Topic: Physics, Laws of Motion, JEE Main
The angular speed of truck wheel is increased from 900 rpm to 2460 rpm in 26 seconds. The number of revolutions by the truck engine during this time is ______.

(Assuming the acceleration to be uniform).

Options:

The correct answer is:
728

Solution:
We know, θ=ω1+ω22t

Let number of revolutions be N

∴2πN=2π900+246060×2×26

N=728

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2021 (17 Mar Shift 1)
The angular velocity of a particle is Ïâ=4i^+j^-2k^ about the origin. If the position vector of the particle is 2i^+3j^-3k^, then its linear velocity is

Options:
A 5i^+8j^-14k^
B 3i^+8j^+10k^
C 8i^+3j^-10k^
D -8i^+3j^-2k^

The correct answer is:
3i^+8j^+10k^

Solution:
v→=ω→×r→
=3i^+8j^+10k^

Topic: Physics, Laws of Motion, JEE Main
The angular width of a fringe in a double slit experiment is found to be 0.2° on a screen 1 m away. The wavelength of light used is 600 nm. The change in angular width of the fringe if the entire measurement system is immersed in water is [Use refractive index of water as 43]

Options:
A 0.05°
B 0.10°
C 0.15°
D 0.20°

The correct answer is:
0.05°

Solution:
Wavelength of light in a denser medium will decrease.

So,

λ2=λ1μ=λ143=34

Angular fringe width is given by,

βθ=λd⇒βθ∝λ

Therefore,

βθ,1βθ,2=λ1λ2⇒βθ,2=λ2λ1×βθ,1=34×0.2°=0.15°

Hence, change in angular frigne width

=0.2°-0.15°=0.05°

Topic: Physics, Wave Optics, JEE Main
The angular width of the central maximum in a single slit diffraction pattern is 60° . The width of the slit is 1 μm. The slit is illuminated by monochromatic plane waves. If another slit of the same width is made near it, Young's fringes can be observed on a screen placed at a distance 50 cm from the slits. If the observed fringe width is 1 cm, what is slit separation distance? (i.e., the distance between the centres of each slit.)

Options:
A 100 Î¼m
B 25 Î¼m
C 50 Î¼m
D 75 μm

The correct answer is:
25 Î¼m

Solution:
In diffraction, dsin30°=λ, λ=d2
Young’s fringe width [ d'- the separation between two slits]
β=λ×Dd'⇒10-2=d2×50×10-2d'
⇒10-2=10-6×50×10-22×d'⇒d'=25 μm.

Topic: Physics, Wave Optics, JEE Main, JEE Main 2018 (08 Apr)
The annular and spirally thickened conducting elements generally develop in the protoxylem when the root or stem is -

Options:
A Differentiating
B Maturing
C Elongating
D Widening

The correct answer is:
Differentiating

Solution:
No Solution Available

Topic: Biology, Anatomy of Flowering Plants, NEET, NEET 2009 (Mains)
The anode half-reaction occurring during the discharge of a lead storage battery is

Options:
A $\mathrm{Pb}(\mathrm{s})+\mathrm{SO}_2+\mathrm{O}_2 \rightarrow \mathrm{PbSO}_4(\mathrm{~s})$
B $\mathrm{Pb}(\mathrm{s})+\mathrm{SO}_4^{2-}(\mathrm{aq}) \rightleftharpoons \mathrm{PbSO}_4(\mathrm{~s})+2 \mathrm{e}^{-}$
C $\mathrm{PbO}_2(\mathrm{~s})+4 \mathrm{H}^{+}(\mathrm{aq})+2 \mathrm{e}^{-}+\mathrm{SO}_4^{2-}(\mathrm{aq}) \rightleftharpoons \mathrm{PbSO}_4(\mathrm{~s})+2 \mathrm{H}_2 \mathrm{O}$
D $\mathrm{Pb}^{2+}(\mathrm{aq})+\mathrm{SO}_4^{2-}(\mathrm{aq}) \rightarrow \mathrm{PbSO}_4(\mathrm{~s})$

The correct answer is:
$\mathrm{Pb}(\mathrm{s})+\mathrm{SO}_4^{2-}(\mathrm{aq}) \rightleftharpoons \mathrm{PbSO}_4(\mathrm{~s})+2 \mathrm{e}^{-}$

Solution:
$\mathrm{Pb}(\mathrm{s})+\mathrm{SO}_4^{2-}(\mathrm{aq}) \rightleftharpoons \mathrm{PbSO}_4(\mathrm{~s})+2 e^{-}$

Topic: Chemistry, Electrochemistry, JEE Main
The anode mud in the electrolytic refining of silver contains

Options:
A Zn, Cu, Ag, Au
B Zn, Ag, Au
C Cu, Ag, Au
D Au only

The correct answer is:
Au only

Solution:
The anode mud in the electrolytic refining of silver contains Au only.
Metals, which are less reactive than Ag, remain in anode mud, and metals which are more reactive than Ag metal will be in electrolyte solution.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, NEET
The anode voltage of a photocell is kept fixed. The wavelength $\lambda$ of the light falling on the cathode is gradually changed. The plate current $\mathrm{I}$ of the photocell varies as follows :

Options:
A

B

C

D

The correct answer is:

Solution:
No Solution Available

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2006
The anode voltage of a photocell is kept fixed. The wavelength λ of the light falling on the cathode is gradually changed. The plate current I of the photocell varies as follows :

Options:
A
B
C
D

The correct answer is:

Solution:
As λ is increased, there will be a value of λ above which photo-electron will be cease to come out, so photo-current will becomes zero.

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2013 (07 Apr)
The anode voltage of a photocell is kept fixed. The wavelength λ of the light falling on the cathode is gradually changed. The plate current I of the photocell varies as follows :

Options:
A
B
C
D

The correct answer is:

Solution:
As λ is increased, there will be a value of λ above which photo-electron will be cease to come out, so photo-current will becomes zero.

Topic: Physics, Dual Nature of Matter, JEE Main
The anodic half-cell of lead-acid battery is recharged using electricity of 0.05 Faraday. The amount of PbSO4 electrolyzed in g during the process is: (Molar mass of PbSO4=303g mol-1)

Options:
A 22.8
B 15.2
C 11.4
D 7.6

The correct answer is:
7.6

Solution:
Charge of one mole of e-=1 faraday
Pb(s)+SO4-2→PbSO4+2e-
If 2F charge is passed then amount of PbSO4gm deposited =303 g
So if 0.05 F faraday current passed then PbSO4 deposited =3032×.05
=7.575
≈7.6 g

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2019 (09 Jan Shift 1)
The antenna current of an AM broadcast transmitter which is modulated by 50% is 11 A. The carrier current is

Options:
A 9.25 A

B 22 A

C 10.3 A

D 5.5 A

The correct answer is:
10.3 A

Solution:
IT=IC1+m22

IC=111+0.522=10.3 A

Topic: Physics, Communication System, JEE Main
The antenna current of an AM transmitter is 8 A when only the carrier is sent, but it increases to 8.93 A when the carrier is modulated by a single sine wave. Find the percentage modulation.

Options:
A 60.1 %

B 70.1 %
C 80.1 %
D 50.1 %

The correct answer is:
70.1 %

Solution:
The power dissipated in any circuit is a function of the square of voltage across the circuit and the effective resistance of the circuit. Equation of AM wave reveals that it has three components of amplitude Clearly; power output must be distributed among these components.
Carrier power, Pc=Ec/22R =Ec22R
Total power of side bands,
PS= mEc/222R+mEc/222R= m2Ec24R
â´PSPC=12 m2
and PT=PC+PS=PC 1+m22
â´PTPC=1+m22âITIC2=1+m22
Given that, IT=8.93 A, IC=8 A, m=?
â´8.9382=1+m22
Or 1.246=1+m22 or m22=0.246
Or m=2Ã0.246=0.701=70.1%

Topic: Physics, Communication System, JEE Main
The antibiotic used for curing tuberculosis is:

Options:
A Penicillin
B Streptomycin
C Tetracycline
D Chloromycetin

The correct answer is:
Streptomycin

Solution:
Streptomycin is the most effective antibacterial agent known for Tuberculosis. In vitro it has a marked bacteriostatic action on the tubercle bacillus and in vivo it tends to exert a deterrent effect on the disease in both animals and man.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main
The antifertility drug "Novestrol" can react with :

Options:
A ZnCl2/HCl;FeCl3; Alcoholic HCN
B Br2/water,ZnCl2/HCl;FeCl3
C Alcoholic HCN;NaOCl;ZnCl2/HCl
D Br2/water,ZnCl2/HCl;NaOCl

The correct answer is:
Br2/water,ZnCl2/HCl;FeCl3

Solution:
Novestrol has phenolic functional group, alcoholic functional group and Terminal alkyne.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main, JEE Main 2020 (03 Sep Shift 1)
The antiseptic action of Dettol is due to

Options:
A Chlorobenzene
B Chloroxylenol
C Chloroquine
D Chloramphenicol

The correct answer is:
Chloroxylenol

Solution:
Chloroxylenol, also known as para-chloro-meta-xylenol (PCMX), is an antiseptic and disinfectant which is used for skin disinfection and cleaning surgical instruments. It is also used within a number of household disinfectants and wound cleaners. It is less effective than some other available agents. It is available as a liquid and sold with name Dettol.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main
The aperture of a telescope is made large, because

Options:
A To increase the intensity of image
B To decrease the intensity of image
C To have greater magnification
D To have lesser resolution

The correct answer is:
To increase the intensity of image

Solution:
For greater aperture of lens, light passing through lens is more and so intensity of image increases.

Topic: Physics, Ray Optics, JEE Main
The aperture of the objective is 24.4 cm. The resolving power of this telescope, if a light of wavelength 2440 A∘ is used to see the object will be

Options:
A 8.1×106
B 10.0×107
C 8.2×105
D 1.0×10-8

The correct answer is:
8.2×105

Solution:
Given,

The diameter of the objective lens, D=24.4 cm

The wavelength of the light, λ=2440 A∘

The resolving power of a telescope is given by,

R.P=D1.22λ ⇒R.P=24.4×10-21.22×2440×10-10=8.2×105

Hence, the resolving power of a telescope is 8.2×105.

Topic: Physics, Ray Optics, JEE Main, JEE Main 2022 (28 Jun Shift 1)