datavorous/entrance-exam-dataset huggingface dataset:
A hyperbola passing through a focus of the ellipse $\frac{x^2}{169}+\frac{y^2}{25}=1$. Its transverse and conjugate axes coincide respectively with the major and minor axes of the ellipse. The product of eccentricities is 1 . Then, the equation of the hyperbola is

Options:
A $\frac{x^2}{144}-\frac{y^2}{9}=1$
B $\frac{x^2}{169}-\frac{y^2}{25}=1$
C $\frac{x^2}{144}-\frac{y^2}{25}=1$
D $\frac{x^2}{25}-\frac{y^2}{9}=1$

The correct answer is:
$\frac{x^2}{144}-\frac{y^2}{25}=1$

Solution:
Let the equation of hyperbola be
$$
\frac{x^2}{a^2}-\frac{y^2}{b^2}=1
$$
Given equation of ellipse is
$$
\frac{x^2}{(13)^2}+\frac{y^2}{(5)^2}=1
$$
Here,
$$
\begin{aligned}
& \text { Here, } \quad a=13, b=5 \\
& \therefore \quad e=\sqrt{1-\frac{b^2}{a^2}} \\
& \therefore \quad=\sqrt{1-\frac{25}{169}}=\sqrt{\frac{144}{169}}=\frac{12}{13} \\
& \therefore \text { Focus }( \pm a e, 0)=\left( \pm 13 \times \frac{12}{13}, 0\right) \\
& \quad=( \pm 12,0)
\end{aligned}
$$
Since, Eq. (i) passes through $( \pm 12,0)$.
$$
\begin{array}{rrrl}
\therefore & \frac{144}{a^2}-\frac{0}{b^2}=1 \\
\Rightarrow & a^2=144 \\
\Rightarrow & a= \pm 12
\end{array}
$$
Now eccentricity of hyperbola
$$
\begin{aligned}
e^{\prime} & =\sqrt{1+\frac{b^2}{a^2}} \\
& =\sqrt{1+\frac{b^2}{144}}
\end{aligned}
$$
According to the equation,
$$
\begin{array}{rlrl}
& e^{\prime}=1 \\
\Rightarrow & & \frac{12}{13} \times \sqrt{1+\frac{b^2}{144}} & =1 \\
\Rightarrow & & \sqrt{1+\frac{b^2}{144}} & =\frac{13}{12} \\
\Rightarrow & & 1+\frac{b^2}{144} & =\frac{169}{144} \\
\Rightarrow & & \frac{b^2}{144} & =\frac{169}{144}-1 \\
\Rightarrow & & \frac{b^2}{144} & =\frac{25}{144} \\
b^2 & =25
\end{array}
$$
$\therefore$ Equation of hyperbola is
$$
\frac{x^2}{144}-\frac{y^2}{25}=1
$$

Topic: Mathematics, Hyperbola, JEE Main
A hyperbola whose transverse axis is along the major axis of the conic x23+y24=4 and has vertices at the foci of the conic. If the eccentricity of the hyperbola is 32, then which of the following points does not lie on the hyperbola ?

Options:
A 5,22
B ( 0,2 )
C 5,23
D 10,23

The correct answer is:
5,23

Solution:
Given conic is x212+y216=1
e= 1-1216=12

Foci 0,2 and 0,-2.

So, transverse axis of hyperbola=2b=4.

⇒b=2

Given e=32

And a2=b2e2-1

⇒a2=4 94-1

⇒a2=5

∴ The equation of hyperbola is x25-y24=-1.

Now, Check options.

Clearly, the point 5,23 does not satisfy above equation.

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2016 (10 Apr Online)
A hypothetical atom has only three energy levels. The ground level has energy, $E_1=-8 \mathrm{eV}$. The two excited states have energies, $E_2=-6 \mathrm{eV}$ and $E_3=-2 \mathrm{eV}$. Then which of the following wavelengths will not be present in the emission spectrum of this atom?

Options:
A
$207 \mathrm{~nm}$

B
$465 \mathrm{~nm}$

C
$310 \mathrm{~nm}$

D
$620 \mathrm{~nm}$

The correct answer is:
$465 \mathrm{~nm}$

Solution:
$E=\frac{h c}{\lambda}$

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2012 (12 May Online)
A hypothetical electromagnetic wave is show below.
The frequency of the wave is \(x \times 10^{19} \mathrm{~Hz}\).
\(x =\) ______ (nearest integer)

Options:

The correct answer is:
5

Solution:
$\begin{aligned} & \lambda=1.5 \times 4 \mathrm{pm} \\ &=6 \times 10^{-12} \text { meter } \\ & \lambda v=\mathrm{C} \\ & 6 \times 10^{-12} \times v=3 \times 10^8 \\ & v=5 \times 10^{19} \mathrm{~Hz}\end{aligned}$

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2024 (08 Apr Shift 1)
A hypothetical planet in the shape of a sphere is completely made of an incompressible fluid and has a mass M and radius R. If the pressure at the surface of the planet is zero, then the pressure at the centre of the planet is [G = universal constant of gravitation]

Options:
A P=3GM28πR4
B P=3GM24πR4
C P=3GM28π2R4
D P=3GM24π2R4

The correct answer is:
P=3GM28πR4

Solution:
The pressure gradient at a distance r from the centre of the spherical planet is

dPdr=-ρg, where g is the acceleration due to gravity at that point

dPdr=-3M4πR3GMR3r

dP=-3GM24πR6rdr

∫P0dP=-3GM24πR6∫0Rrdr

P=3GM28πR4

Topic: Physics, Gravitation, JEE Main
$a+i b \gt c+i d$ can be explained only when

Options:
A $b=0, c=0$
B $b=0, d=0$
C $a=0, c=0$
D $a=0, d=0$

The correct answer is:
$b=0, d=0$

Solution:
$a+i b \gt c+i d$, it is defined if and only if imaginary parts must be equal to zero.
Therefore $i b=i d=0 \Rightarrow b=d=0 \quad(i \neq 0)$

Topic: Mathematics, Complex Number, JEE Main
(a) In a metre bridge, the balance point is found to be at $39.5 \mathrm{~cm}$ from the end $A$. When the resistor $Y$ is of $12.5 \Omega$ determine the resistance of $X$. Why are the connections between resistors in a Wheatstone or meter bridge made of thick copper strips?
(b) Determine the balance point of the bridge above if $X$ and $Y$ are interchanged.
(c) What happens if the galvanometer and cell are interchanged at the balance point of the bridge? Would the galvanometer show any current?

Options:

None

Solution:
Fig. (a) $\quad$ Fig. (b)
Given that $\mathrm{Y}=12.5 \Omega, l_1=39.5 \mathrm{~cm}$, $\ell_2=100-39.5=60.5 \mathrm{~cm}, \mathrm{X}=$ ?
By formula, $\frac{\mathrm{X}}{\mathrm{Y}}=\frac{\ell_1}{\ell_2}$ and
$$
\mathrm{X}=\mathrm{Y} \frac{\ell_1}{\ell_2}=\frac{12.5 \times 39.5}{60.5}=8.1612 \Omega \text {. }
$$
Thick copper strips are used for connections to minimise resistance of connection which are not accounted for in bridge formula.
(b) When $\mathrm{X}$ and $\mathrm{Y}$ are interchanged, values of $\ell_1$ and $\ell_2$ also become interchanged. Then $\ell_1=60.5 \mathrm{~cm}$. Hence balance point will be at $60.5 \mathrm{~cm}$.
(c) When galvanometer and cell are interchanged, condition for balance of bridge remains satisfied. Hence galvanometer will not show any current.

Topic: Physics, Current Electricity
A is :

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
A is :

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main, JEE Main 2015 (10 Apr Online)
A is :

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
A is a square matrix and Î is an identity matrix of the same order. If A3=O, then inverse of matrix Î-A is

Options:
A Ι+A
B Ι-A+A2
C A+A2
D Ι+A+A2

The correct answer is:
Ι+A+A2

Solution:
We know,

I-AΙ+A+A2=Ι-A3=Ι
Hence, Ι-A-1=Ι+A+A2

Topic: Mathematics, Matrices, JEE Main
A is a square matrix of order 3×3. The matrices A, 4A-1, ATall have the same value of the determinant. If adjA=2AT, then 3A-1 is equal to

Options:
A 32
B 94
C 278
D 3

The correct answer is:
278

Solution:
A=4A-1=AT⇒A=64A=A
⇒A2=64⇒A=8 or -8
∵adjA=2AT⇒adjA=2AT⇒A2=8A⇒A=0,8
⇒A=8
So, 3A-1=27A=278

Topic: Mathematics, Determinants, JEE Main
A→ is a vector quantity such that A→= non-zero constant. Which of the following expression is true for A→?

Options:
A A→·A→=0
B A→×A→<0
C A→×A→=0
D A→×A→>0

The correct answer is:
A→×A→=0

Solution:
Since both the vectors are same, they will be parallel to each other. We know that vector product of two parallel vectors is zero.

Hence,

A→×A→=0 is the correct answer.

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2022 (25 Jun Shift 1)
$A$ is an aqueous acid; $B$ is an aqueous base. They are diluted separately, then

Options:
A $\mathrm{pH}$ of $A$ increases and $\mathrm{pH}$ of $B$ decreases
B $\mathrm{pH}$ of $A$ increases and $p H$ of $B$ decreases till $p H$ in each case is 7
C $p H$ of $A$ and $B$ increase
D $\mathrm{pH}$ of $B$ and $A$ decrease

The correct answer is:
$\mathrm{pH}$ of $A$ increases and $\mathrm{pH}$ of $B$ decreases

Solution:
On dilution the $p H$ of acid $A$ increases while $p H$ of base $B$ decreases.

Topic: Chemistry, Ionic Equilibrium, JEE Main
A is

Options:
A
B
C
D

The correct answer is:

Solution:
It is a case of nucleophilic substitution:

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
(a) It is known that density $\rho$ of air decreases with heighty as $\rho=\rho_0 \mathrm{e}^{-\mathrm{y} / \mathrm{y}_0}$
Where $\rho_0=1.25 \mathrm{~kg} \mathrm{~m}^{-3}$ is the density at sea level, and $y_0$ is a constant. This density variation is called the law of atmospheres. obtain this law assuming that the temperature of atmosphere remains a constant (isothermal conditions). Also assume that the value of $g$ remains constant.
(b) A large He balloon of volume $1425 \mathrm{~m}^3$ is used to lift a payload of $400 \mathrm{~kg}$. Assume that the balloon maintains constant radius as it rises. How high does it rise? [Take $y_0=8000 \mathrm{~m}$ and $\rho_{H e}=0.18 \mathrm{~kg} \mathrm{~m}^{-3}$ ]

Options:

None

Solution:
(a) We know that rate of decrease of density $\rho$ of air is directly proportional to the height $y$. It is given as $\frac{\mathrm{d} \rho}{\mathrm{dy}}=-\frac{\rho}{\mathrm{y}_0}$,
where $y$ is a constant of proportionality and $-v e$ sign signfies that density is decreasing with increase in height. on integration, we get $\int_{\rho_0}^\rho \frac{d \rho}{\rho}=-\int_0^y \frac{1}{y_0} d y$ $\Rightarrow[\log \rho]_{\rho_0}^\rho=-\left[\frac{\mathrm{y}}{\mathrm{y}_0}\right]_0^{\mathrm{y}}$, where, $\rho_0=$ density of air at sea level i.e., $y=0$
or $\log _e \frac{\rho}{\rho_0}=\frac{y}{y_0}$ or $\rho=\rho_0 e^{-\frac{y}{y_0}}$
Hence dimensions and units of constant $y_0$ are same as of $\mathrm{y}$.
(b) Here volume of $H_{e^{\prime}}$ ballon, $V=1425 \mathrm{~m}^3$, mass of payload, $m=400 \mathrm{~kg}$
$y_0=8000 \mathrm{~m}$, density of $\mathrm{He} \rho_{\mathrm{He}}=0.18 \mathrm{kgm}^{-3}$
Mean density of balloon,
$\rho=\frac{\text { Total mass of balloon }}{\text { Volume }}=\frac{\mathrm{m}+\mathrm{V} \cdot \rho_{\mu \mathrm{e}}}{\mathrm{V}}$
$=\frac{400+1425 \times 0.18}{1425}=0.4608=0.46 \mathrm{kgm}^{-3}$
As density of air at sea level $\rho_0=1.25 \mathrm{~kg} \mathrm{~m}^{-3}$. The balloon will rise up to a height $y$ where density of air $=$ density of balloon $\rho=0.46 \mathrm{kgm}^{-3}$
$$
\begin{aligned}
\text { As } & \rho=\rho_0 \mathrm{e}^{-\frac{\mathrm{y}}{y_0}} \text { or } \frac{\rho_0}{\rho}=\mathrm{e}^{\frac{y_0}{\mathrm{y}}} \\
\therefore \log _{\mathrm{e}}\left(\frac{\rho_0}{\rho}\right) &=\frac{\mathrm{y}_0}{\mathrm{y}} \text { or } \mathrm{y}=\frac{\mathrm{y}_0}{\log _{\mathrm{e}}\left(\frac{\rho_0}{\rho}\right)}=\frac{8000}{\log _{\mathrm{e}}\left(\frac{1.25}{0.46}\right)} \\
=& 8002 \mathrm{~m} \text { or } 8.0 \mathrm{~km} .
\end{aligned}
$$

Topic: Physics, Mechanical Properties of Fluids
A jar filled with two non mixing liquids 1 and 2 having densities $\rho_1$ and $\rho_2$ respectively. A solid ball, made of a material of density $\rho_3$, is dropped in the jar. It comes to equilibrium in the position shown in the figure. Which of the following is true for $\rho_1, \rho_2$ and $\rho_3$ ?

Options:
A
$\rho_3 < \rho_1 < \rho_2$
B
$\rho_1 < \rho_3 < \rho_2$
C
$\rho_1 < \rho_2 < \rho_3$
D
$\rho_1 < \rho_3 < \rho_2$

The correct answer is:
$\rho_1 < \rho_3 < \rho_2$

Solution:
As liquid 1 floats above liquid 2 ,
$$
\rho_1 < \rho_2
$$
The ball is unable to sink into liquid 2 ,
$\rho_3 < \rho_2$ The ball is unable to rise over liquid 1 ,
$\begin{array}{ll} & \rho_1 < \rho_3 \\ \text { Thus, } & \rho_1 < \rho_3 < \rho_2\end{array}$

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2008
A jet plane of wing span \( 20 \mathrm{~m} \) is travelling towards west at a speed of
\( 400 \mathrm{~ms}^{-1} \). If the earth's total magnetic field is \( 4 \times 10^{-4} \mathrm{~T} \) and the dip angle is \( 30^{\circ} \), at that
place, the voltage difference developed across the ends of the wing is

Options:
A \( 1.6 \mathrm{~V} \)
B \( 3.2 \mathrm{~V} \)
C \( 0.8 \mathrm{~V} \)
D \( 6.4 \mathrm{~V} \)

The correct answer is:
\( 1.6 \mathrm{~V} \)

Solution:
Given, wing span, $1=20 \mathrm{~m}$; speed of jet plane, $v=400 \mathrm{~ms}^{-1}$, Earth's magnetic field, $B=4 \times 10^{-4} \mathrm{~T}$; angle of dip, $\theta=$
$30^{\circ}$.
Now, emf induced
$e=B_{v} l v$
Here,
$B_{v}=B \sin \theta=4 \times 10^{-4} \times \sin 30^{\circ}=4 \times 10^{-4} \times \frac{1}{2}=2 \times 10^{-4} \mathrm{~T}$
Therefore,
$e=2 \times 10^{-4} \times 400 \times 20=16 \times 10^{-1} \mathrm{~V}=1.6 \mathrm{~V}$
Thus, voltage difference developed across the ends of the wing is $1.6 \mathrm{~V}$

Topic: Physics, Mathematics in Physics, JEE Main
A junction diode has a resistance of $25 \Omega$ when forward biased and $2500 \Omega$ when reverse biased. The current in the diode, for the arrangement shown will be

Options:
A $\frac{1}{15} \mathrm{~A}$
B $\frac{1}{7} \mathrm{~A}$
C $\frac{1}{25} \mathrm{~A}$
D $\frac{1}{180} \mathrm{~A}$

The correct answer is:
$\frac{1}{7} \mathrm{~A}$

Solution:
$\mathrm{R}_{\mathrm{eq}}=25+10=35 \Omega$
Because diode is forward biased. So $I=\frac{V}{R_{e q}}=\frac{5}{35}=\frac{1}{7} \mathrm{~A}$

Topic: Physics, Semiconductors, JEE Main
A KCl solution of conductivity 0.14 S m-1 shows a resistance of 4.19Ω in a conductivity cell. If the same cell is filled with an HCl solution, the resistance drops to 1.03Ω. The conductivity of the HCl solution is ____×10-2 S m-1. (Round off to the Nearest Integer).

Options:

The correct answer is:
57

Solution:
The thermal conductivity of a material is a measure of its ability to conduct heat. It is commonly denoted by κ, \lambda, or \kappa. Heat transfer occurs at a lower rate in materials of low thermal conductivity than in materials of high thermal conductivity

κ=1R·G*

For same conductivity cell, G* is constant and hence κ·R.= constant.

∴0.14×4.19=κ×1.03

or, =0.14×4.191.03

=0.5695Sm-1

=56.95×10-2Sm-1≈57×10-2Sm-1

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2021 (17 Mar Shift 2)
Al2(SO4)3 is used in the following but not

Options:
A as a coagulant in treating drinking water and sewage
B in plastic industry
C as a mordant in dyeing
D both C and D

The correct answer is:
in plastic industry

Solution:
Aluminium sulphate is mainly used a coagulating agent in purification of drinking water and waste water treatment, as mordant in dying and printing textile and also used in calico printing and sizing of paper industry.

Topic: Chemistry, p Block Elements (Group 13 & 14), NEET
AlCl3 achieves stability by forming a dimer. In trivalent state the compound is hydrolysed in water. AlCl3 in acidified aqueous solution forms

Options:
A AlOH3+HCl
B [AlH2O63++3Cl-
C AlCl3.2H2O
D Al2O3+HCl

The correct answer is:
[AlH2O63++3Cl-

Solution:
AlCl3 in acidified aqueous solution forms AlH2O63+ ion.
AlCl3+9H2O→AlH2O6OH3+3H++3Cl-

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
AlCl3 forms dimer in vapour phase but BCl3 does not because

Options:
A In Al there are vacant d orbitals in which it accommodates lone pair from chlorine atoms
B In BCl3 there is back bonding
C There is hydrogen bonding in between two AlCl3 molecules in vapour phase
D None of the above

The correct answer is:
In Al there are vacant d orbitals in which it accommodates lone pair from chlorine atoms

Solution:
The formation of stronger backbond ( 3p→2p overlap in BCl3 and weaker 3p→3p overlap in AlCl3 ) reduces the electron deficiency of the central atom and thus dimerisation tendency is reduced. So AlCl3 can form a dimer but BF3 and BCl3 cannot.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
A ladder 5 m long leans against a vertical wall. The bottom of the ladder is 3 m from the wall. If the bottom of the ladder is pulled 1 m farther from the wall, how much does the top of the ladder slide down the wall

Options:
A 1âm
B 4âm
C 2 m
D 3âm

The correct answer is:
1 m

Solution:
From first case
From second case,
â AB=4 m andBD=3 m
â´ AD=4-3=1 m

Topic: Mathematics, Heights and Distances, JEE Main
A ladder rests against a vertical wall at an angle Î± to the horizontal. If the foot is pulled away through a distance 2 m, then it slides a distance 5 m down the wall, finally making an angle Î² with the horizontal. The value of tanâ¡Î±+Î²2 is equal to

Options:
A 25
B 52
C 10
D None of these

The correct answer is:
25

Solution:
Let, the length of the ladder be l and AB is the wall.
∴AD=EC=l
Now, BE-BD=2
⇒lcos⁡β-lcos⁡α=2
⇒lcos⁡β-cos⁡α=2
⇒2lsin⁡α+β2sin⁡α-β2=2……i
Also, BA-BC=5
⇒lsin⁡α-lsin⁡β=lsin⁡α-sin⁡β=5
⇒2lcos⁡α+β2sin⁡α-β2=5……ii
Dividing equation i & ii, we get,
∴tan⁡α+β2=25

Topic: Mathematics, Heights and Distances, JEE Main
A lake near a village suffered heavy mortality of fishes within a few days. Consider the following reasons for this
(1) Lots of urea and phosphate fertilizer were used in the crops in the vicinity.
(2) The area was sprayed with DDT by an aircraft.
(3) The lake water turned green and stinky.
(4) Phytoplankton populations in the lake declined initially thereby greatly reducing photosynthesis.
Which two of above were the main causes of fish mortality in the lake?

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

The correct answer is:
1,3

Solution:
The heavy mortality of fishes in the lake within a span of few days is due to Eutrophication.
Eutrophication is the phenomenon of nutrient enrichment of nitrogen and phosphorus in a water body that supports a dense growth of plants and animal life. Due to nutrient enrichment, excess growth of algae occurs which is called algal bloom. When these algae die, their organic matter gets decomposed and this results in the depletion of oxygen leading to death of aquatic animals. Hence, this indicates that lot of urea and phosphate fertilizers are were used in nearby crops. The lake water could have turned green and stinky due to excess growth of Algae.

Topic: Biology, Ecosystem, NEET, NEET 2008 (Screening)
A lake near a village suffered heavy mortality of fishes within a few days. Consider the following reasons for this:

(1) Lots of urea and phosphate fertilizer were used in the crops in the vicinity.
(2) The area was sprayed with DDT by an aircraft.
(3) The lake water turned green and stinky.
(4) Phytoplankton populations in the lake declined initially thereby greatly reducing photosynthesis.

Which two of the above were the main causes of fish mortality in the lake?

Options:
A (2) and (3)
B (3) and (4)
C (1) and (3)
D (1) and (2)

The correct answer is:
(1) and (3)

Solution:
Lots of urea and phosphate fertilizer were used in the crops in the vicinity and the lake water turned green and stinky. Due to this, lake near a village suffered heavy mortality of fishes within a few days.

Topic: Biology, Environmental Issues, NEET
A lake near a village suffered heavy mortality of fishes within a few days. Consider the following reasons for this
(a) Lots of urea and phosphate fertilizer were used in the crops in the vicinity
(b) The area was sprayed with DDT by an aircraft
(c) The lake water turned green and stinky
(d) Phytoplankton populations in the lake declined initially thereby greatly reducing photosynthesis
Which two of the above were the main causes of fish mortality in the lake?

Options:
A a, b
B b, c
C $\mathrm{c}, \mathrm{d}$
D a, c

The correct answer is:
a, c

Solution:
Lots of urea and phosphate fertilizers used in crops in the vicinity of lake will lead to eutrophication that means the lake will turn green and stinky.

Topic: Biology, Environmental Issues, NEET, NEET 2008 (Mains)
A lamina is made by removing a small disc of diameter 2R from a bigger disc of uniform mass density and radius 2R, as shown in the figure. The moment of inertia of this lamina about axes perpendicular to the plane of the lamina and passing through the points O and P is IO and IPâ respectively. If the ratio IPIO=mn, where m and n are the smallest integers, then what is the value of m+n?

Options:

The correct answer is:
50

Solution:
I P = 4 m R 2 2 + m 4 R 2 - m 4 R 2 2 + m 4 5 R 2
I P = mR 2 2 + 4 - 1 8 + 5 4
I P = mR 2 6 - 1 1 8 = 3 7 8 mR 2 … 1
IO=4mR22-m4R22+m4R2
IO=mR22-18+14=mR22-38=mR2138…2
SoIPIO=37/813/8=3713

⇒m+n=37+13=50

Topic: Physics, Rotational Motion, JEE Main
A lamp emits monochromatic green light uniformly in all directions. The lamp is $3 \%$ efficient in converting electrical power to electromagnetic waves and consumes $100 \mathrm{~W}$ of power. The amplitude of the electric field associated with the electromagnetic radiation at a distance of $5 \mathrm{~m}$ from the lamp will be nearly:

Options:
A
$1.34 \mathrm{~V} / \mathrm{m}$

B
$2.68 \mathrm{~V} / \mathrm{m}$

C
$4.02 \mathrm{~V} / \mathrm{m}$

D
$5.36 \mathrm{~V} / \mathrm{m}$

The correct answer is:
$2.68 \mathrm{~V} / \mathrm{m}$

Solution:
Wavelength of monochromatic green light
$$
\begin{aligned}
&=5.5 \times 10^{-5} \mathrm{~cm} \\
&\text { Intensity I }=\frac{\text { Power }}{\text { Area }} \\
&=\frac{100 \times(3 / 100)}{4 \pi(5)^2} \\
&=\frac{3}{100 \pi} \mathrm{Wm}^{-2}
\end{aligned}
$$
Now, half of this intensity (I) belongs to electric field and half of that to magnetic field, therefore,
$$
\begin{aligned}
&\frac{\mathrm{I}}{2}=\frac{1}{4} \varepsilon_0 \mathrm{E}_0^2 \mathrm{C} \\
&\text { or } \mathrm{E}_0=\sqrt{\frac{2 \mathrm{I}}{\varepsilon_0 \mathrm{C}}}
\end{aligned}
$$
$$
\begin{aligned}
&=\sqrt{\frac{2 \times\left(\frac{3}{100} \pi\right)}{\left(\frac{1}{4 \pi \times 9 \times 10^9}\right) \times\left(3 \times 10^8\right.}} \\
&=\sqrt{\frac{6}{25} \times 30}=\sqrt{7.2} \\
&\therefore \mathrm{E}_0=2.68 \mathrm{~V} / \mathrm{m}
\end{aligned}
$$

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2014 (12 Apr Online)
A lamp is connected is series with a capacitor and an AC source, what happens if the capacity of the capacitor is reduced?

Options:
A The lamp shines more brightly
B The lamp shines less brightly
C There is no change in the brightness of the lamp
D Brightness may increase or decrease depending on the frequency of the AC

The correct answer is:
The lamp shines less brightly

Solution:
The lamp shines less brightly because, when capacity of the capaictor is reduced, the capacitive reactance $\left(X_{C}=\frac{1}{\omega C}\right)$ increases and hence, the current through the lamp decreases.

Topic: Physics, Alternating Current, JEE Main
A large block of wood of mass M=5.99 kg is hanging from two long massless cords. A bullet of mass m=10 g is fired into the block and gets embedded in it. The (block + bullet) then swing upwards, their center of mass rising a vertical distance h=9.8 cm before the (block + bullet) pendulum comes momentarily to rest at the end of its arc. The speed of the bullet just before the collision is: (Take g=9.8 m s-2)

Options:
A 841.4 m s-1
B 811.4 m s-1
C 831.4 m s-1
D 821.4 m s-1

The correct answer is:
831.4 m s-1

Solution:
From energy conservation,

[after bullet gets embedded till the system comes momentarily at rest]

M+mgh=12M+mv12
[v1 is velocity after collision]

∴ v1=2σh

Applying momentum conservation, (just before and just after collision)

mv=M+mv1

v=M+mmv1=610×10-3×2×9.8×9.8×10-2

≈831.55 m s-1

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2021 (16 Mar Shift 2)
A large cylindrical rod of length $L$ is made by joining two identical rods of copper and steel of length $\left(\frac{L}{2}\right)$ each. The rods are completely insulated from the surroundings. If the free end of copper rod is maintained at $100^{\circ} \mathrm{C}$ and that of steel at $0^{\circ} \mathrm{C}$ then the temperature of junction is (Thermal conductivity of copper is 9 times that of steel)

Options:
A
$90^{\circ} \mathrm{C}$

B
$50^{\circ} \mathrm{C}$

C
$10^{\circ} \mathrm{C}$

D
$67^{\circ} \mathrm{C}$

The correct answer is:
$90^{\circ} \mathrm{C}$

Solution:
Let conductivity of steel $K_{\text {steel }}=k$ then from question
Conductivity of copper $K_{\text {copper }}=9 k$
$$
\begin{aligned}
& \theta_{\text {copper }}=100^{\circ} \mathrm{C} \\
& \theta_{\text {steel }}=0^{\circ} \mathrm{C} \\
& l_{\text {steel }}=l_{\text {copper }}=\frac{L}{2}
\end{aligned}
$$
From formula temperature of junction;
$$
\begin{aligned}
\theta & =\frac{K_{\text {copper }} \theta_{\text {copper }} l_{\text {steel }}+K_{\text {steel }} \theta_{\text {steel }} l_{\text {copper }}}{K_{\text {copper }} l_{\text {steel }}+K_{\text {steel }} l_{\text {copper }}} \\
& =\frac{9 k \times 100 \times \frac{L}{2}+k \times 0 \times \frac{L}{2}}{9 k \times \frac{L}{2}+k \times \frac{L}{2}} \\
& =\frac{\frac{900}{2} k L}{\frac{10 k L}{2}}=90^{\circ} \mathrm{C}
\end{aligned}
$$

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2012 (19 May Online)