datavorous/entrance-exam-dataset huggingface dataset:
If a ray of light along the line $y=4$ gets reflected from a parabolic mirror whose equation is $(y-2)^2=4(x+1)$, then equation of reflected ray will be

Options:
A $y=0$
B $\mathrm{x}=0$
C $\mathrm{y}=2$
D $x=2$

The correct answer is:
$\mathrm{x}=0$

Solution:
The correct option is $\mathbf{B} \mathrm{x}=0$
As we know that any line which is moving parallel to axis of parabola after get reflected by the parabola passes through the focus of the parabola.
For the given parabola focus will be
$\mathrm{S}(\mathrm{o}, 2)$
Now let the ray of light meet the parabola at $P$ then
$(4-2)^2=4(x+1) \Rightarrow x=0$
Hence $P(0,4)$
So equation of the reflected ray will be line passing through $\mathrm{P}$ and $\mathrm{S}$ i.e., $\mathrm{x}=0$

Topic: Mathematics, Parabola, JEE Main
If a reaction follows the Arrhenius equation, the plot $\ln k$

vs $\frac{1}{(\mathrm{RT})}$ gives straight line with a gradient $(-\mathrm{y})$ unit.

The energy required to activate the reactant is:

Options:
A $y / R$ unit
B y unit
C yR unit
D $-y$ unit

The correct answer is:
y unit

Solution:
From Arrhenius equation,

$k=A e^{-E_{n} / R T}$

$\ln k=\ln A-\frac{E_{a}}{R T}$

slope $=-y$ (given) $-y=-E_{s}$

$\Rightarrow \mathrm{E}_{\mathrm{a}}=y$

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2019 (11 Jan Shift 1)
If a rectangle is inscribed in an equilateral triangle of side length 22 as shown in the figure, then the square of the largest area of such a rectangle is _____.

Options:

The correct answer is:
3

Solution:
In ΔDBF,

tan60°=2b22-ℓ

⇒b=322-ℓ2

A=Area of rectangle

=ℓ×b

⇒A=ℓ×3222-ℓ

On differentiating, we get

dAdℓ=3222-ℓ-ℓ·32=0

⇒ℓ=2

Now, A=ℓ×b

=2×322=3

⇒A2=3.

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2021 (25 Jul Shift 2)
If a refrigerator's door is left open, what happens to the temperature of the room?

Options:
A The room becomes cooler.
B The room remains the same temperature.
C The room becomes hotter.
D The room temperature fluctuates.

The correct answer is:
The room becomes hotter.

Solution:
Correct Option is : (C)
The room becomes hotter.

Topic: Chemistry, Thermodynamics, NEET
If $A$ represents Boltzmann constant, $B$ represents Planck's constant and $C$ represents speed of light in vacuum, then the quantity having the dimensions of $A^4 B^{-3} C^{-2}$ is

Options:
A universal gas constant
B specific heat capacity
C stefan’s constant
D heat energy

The correct answer is:
stefan’s constant

Solution:
Dimension of Boltzmann's constant
$A=\frac{\mathrm{J}}{\mathrm{K}}=\frac{\mathrm{ML}^2 \mathrm{~T}^{-2}}{\mathrm{~K}^1}=\left[\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~K}^{-1}\right]$
Dimension of Planck constant
$B=\mathrm{Js}=\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]$
Dimension of speed of light,
$C=\left[\mathrm{LT}^{-1}\right]$
Desired dimension $=A^4 B^{-3} C^{-2}$
$\Rightarrow\left[\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~K}^{-1}\right]^4\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]^{-3}\left[\mathrm{LT}^{-1}\right]^{-2}$
$\Rightarrow\left[\mathrm{M}^1 \mathrm{~L}^0 \mathrm{~T}^{-3} \mathrm{~K}^{-4}\right]$
Stefan constant $(\sigma)$ is given by
$E=\sigma T^4 \Rightarrow \sigma=\frac{E}{T^4}$
where, $E=$ energy per unit area per unit time and $\quad T=$ absolute temperature.
$\Rightarrow$ Dimension, $\sigma=\frac{\left[\mathrm{M} \mathrm{L}^2 \mathrm{~T}^{-2}\right]}{\left.\left[\mathrm{L}^2\right][\mathrm{T}] \mathrm{K}^4\right]}$
$\sigma=\left[\mathrm{ML}^0 \mathrm{~T}^{-3} \mathrm{~K}^{-4}\right]$
So, desired dimension is of Stefan constant.

Topic: Physics, Units and Dimensions, JEE Main
If a right circular cone, having maximum volume, is inscribed in a sphere of radius 3 cm, then the curved surface area (in cm2) of this cone is :

Options:
A 82π
B 62π
C 83π
D 63π

The correct answer is:
83π

Solution:
In △AMC, AM=3sin2θ & MC=3cos2θ

V=13πr2h where, r is radius and h is height of cone

⇒V=13π3sin⁡2θ23+3cos2θ

(since, radius of cone =AM and height of cone =MC)

⇒V=π36sin2⁡θcos2⁡θ2cos2⁡θ ∵sin⁡2θ=2sin⁡θcos⁡θ & cos⁡2θ=2cos2⁡θ-1

=72πsin2⁡θcos4⁡θ

Differentiating both sides with respect to θ, we get

dvdθ=72π2sin⁡θcos5⁡θ-4sin3⁡θcos3⁡θ

For maximum value, dVdθ=0

⇒72π2sin⁡θcos5⁡θ-4sin3⁡θcos3⁡θ=0⇒tan2⁡θ=12

Thus, volume is maximum when tan⁡θ=12

Hence, curved surface area S=πrl

=πr3+3cos⁡2θ2+(3sin⁡2θ)2 ∵l=r2+h2

=π3sin⁡2θ36cos2⁡θ=18π2sin⁡θcos2⁡θ

=36π13.23=24π3=83π

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2018 (15 Apr)
If a right circularcone having maximum volume, is inscribed in a sphere of radius $3 \mathrm{~cm}$, then the curved surface area (in $\mathrm{cm}^2$ ) of this cone is

Options:
A
$8 \sqrt{3} \pi$

B
$6 \sqrt{2} \pi$

C
$6 \sqrt{3} \pi$

D
$8 \sqrt{2} \pi$

The correct answer is:
$8 \sqrt{3} \pi$

Solution:
Sphere of radius $r=3 \mathrm{~cm}$
Let $b, h$ be base radius and height of cone respectively.
So, volume of cone $=\frac{1}{2} \pi b^2 h$
In right angled $\triangle A B C$ by Pythagoras theorem
$$
\begin{aligned}
&(h-r)^2+b^2=r^2 \\
\Rightarrow & b^2=r^2-(h-r)^2=r^2-\left(h^2-2 h r+\mathrm{r}^2\right)=2 h r-h^2 \\
\therefore & \text { Volume }(v) \frac{1}{3} \pi h\left[2 h r-h^2\right]=\frac{1}{3}\left[2 h r^2-h^3\right] \\
& \frac{d v}{d h}=\frac{1}{3}\left[4 h r-3 h^2\right]=0 \Rightarrow h(4 r-3 h)=0 \\
& \frac{d^2 v}{d h^2}=\frac{1}{3}[4 r-6 h]
\end{aligned}
$$
At $h=\frac{4 r}{3}, \frac{d^2 v}{d h^2}=\frac{1}{3}\left[4 r-\frac{4 r}{3} \times 6\right]=\frac{1}{3}[4 r-8 r < ] 0$
$\Rightarrow$ maximum volume ocurs at $h$
$$
=\frac{4 r}{3}=\frac{4}{3} \times 3=4 \mathrm{~cm}
$$
As from (1),
$$
(h-r)^2+b^2=r^2
$$
$$
\Rightarrow b^2=2 h r-h^2=2 \cdot \frac{4 r}{3} r-\frac{16 r^2}{9}
$$
$$
\begin{aligned}
&=\frac{8 r^2}{3}-\frac{16 r^2}{9} \\
&=\frac{(24-16) r^2}{9}=\frac{8 r^2}{9} \\
&\Rightarrow b=\frac{2 \sqrt{2}}{3} r=2 \sqrt{2} \mathrm{~cm}
\end{aligned}
$$
Therefore curved surface area
$$
\begin{aligned}
&=\pi b l=\pi b \sqrt{h^2+r^2} \\
&=\pi 2 \sqrt{2} \sqrt{4^2+8}=8 \sqrt{3} \pi \mathrm{cm}^2
\end{aligned}
$$

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
If a root of the equation $a x^2+b x+c=0$ be reciprocal of a root of the equation then $a^{\prime} x^2+b^{\prime} x+c^{\prime}=0$, then

Options:
A $\left(c c^{\prime}-a a^{\prime}\right)^2=\left(b a^{\prime}-c b^{\prime}\right)\left(a b^{\prime}-b c^{\prime}\right)$
B $\left(b b^{\prime}-a a^{\prime}\right)^2=\left(c a^{\prime}-b c^{\prime}\right)\left(a b^{\prime}-b c^{\prime}\right)$
C $\left(c c^{\prime}-a a^{\prime}\right)^2=\left(b a^{\prime}+c b^{\prime}\right)\left(a b^{\prime}+b c^{\prime}\right)$
D None of these

The correct answer is:
$\left(c c^{\prime}-a a^{\prime}\right)^2=\left(b a^{\prime}-c b^{\prime}\right)\left(a b^{\prime}-b c^{\prime}\right)$

Solution:
Let $\alpha$ be a root of first equation, then $\frac{1}{\alpha}$ be a root of second equation.
Therefore $a \alpha^2+b \alpha+c=0$ and $a^{\prime} \frac{1}{\alpha^2}+b^{\prime} \frac{1}{\alpha}+c^{\prime}=0$
or $c^{\prime} \alpha^2+b^{\prime} \alpha+a^{\prime}=0$
Hence $\frac{\alpha^2}{b a^{\prime}-b^{\prime} c}=\frac{\alpha}{c c^{\prime}-a a^{\prime}}=\frac{1}{a b^{\prime}-b c^{\prime}}$
$\left(c c^{\prime}-a a^{\prime}\right)^2=\left(b a^{\prime}-c b^{\prime}\right)\left(a b^{\prime}-b c^{\prime}\right)$

Topic: Mathematics, Quadratic Equation, JEE Main
If a rubber ball is taken at the depth of $200 \mathrm{~m}$ in a pool, its volume decreases by $0.1 \%$. If the density of the water is $1 \times 10^3 \mathrm{~kg} / \mathrm{m}^3$ and $g=10 \mathrm{~m} / \mathrm{s}^2$, then the volume elasticity in $N / \mathrm{m}^2$ will be

Options:
A $10^8$
B $2 \times 10^8$
C $10^9$
D $2 \times 10^9$

The correct answer is:
$2 \times 10^9$

Solution:
$K=\frac{\Delta P}{\Delta V / V}=\frac{h \rho g}{\Delta V / V}=\frac{200 \times 10^3 \times 10}{0.1 / 100}=2 \times 10^9$

Topic: Physics, Mechanical Properties of Solids, JEE Main
If a salt bridge is removed from the two half cells, the voltage

Options:
A drops to zero.
B does not change.
C increases gradually.
D increases rapidly.

The correct answer is:
drops to zero.

Solution:
If a salt bridge is removed from the two half cells, the voltage drops to zero. Salt bridge permits the flow of current by completing the circuit. No current will flow and the voltage will drop to zero if the salt bridge is removed, due to accumulation of charges.

Topic: Chemistry, Electrochemistry, JEE Main
If a satellite is revolving around a planet of mass M in an elliptical orbit of semi-major axis a, find the orbital speed of the satellite when it is at a distance r from the focus

Options:
A v2=GM2r-1a
B v2=GM2r2-1a
C v2=GM2r2-1a2
D v2=G2r-1a

The correct answer is:
v2=GM2r-1a

Solution:
As in case of elliptic orbit of a satellite mechanical energy
remains constant, at any position of satellite in the orbit,
KE + PE =-GMm2a ...(i)
Now, if at position r,v is the orbital speed of satellite
KE = 12 mv2 and PE = -GMmr ...(ii)
So, from Eqs. (i) and (ii), we have
12 mv2-GMmr=-GMm2a, i.e., v2=GM2r-1a

Topic: Physics, Gravitation, JEE Main
If a semiconductor photo diode can detect a photon with a maximum wavelength of 400 nm, then its band gap energy is: Planck's constant h=6.63×10-34 J.s
Speed of light c=3×108 m s-1

Options:
A 1.1 eV
B 2.0 eV
C 1.5 eV
D 3.1 eV

The correct answer is:
3.1 eV

Solution:
λ=400 nm

Band gap E0=hcλ=1237.5400=3.09 eV

Topic: Physics, Semiconductors, JEE Main, JEE Main 2020 (03 Sep Shift 2)
If a set $A$ has $m$ elements and set $B$ has $n$ elements and the number of injections from $A$ to $B$ is 2520 . Then, $m$ is equal to

Options:
A 2
B 7
C 6
D 5

The correct answer is:
5

Solution:
Given, $|A|=m$ and $|B|=n$
Total number of injective function from $A$ to $B$ is $\frac{n !}{(n-m) !}$
Given, $\frac{n !}{(n-m) !}=2520$
Total number of injections from $A$ to $B={ }^n P_m=2520$
$\begin{array}{ll}
\Rightarrow & { }^n P_m=7 \times 6 \times 5 \times 4 \times 3={ }^7 P_5 \\
\Rightarrow & n=7, m=5
\end{array}$

Topic: Mathematics, Sets and Relations, JEE Main
If a simple harmonic motion is represented by $\frac{\mathrm{d}^2 \mathrm{x}}{\mathrm{dt}^2}+\alpha \mathrm{x}=0$, its time period is

Options:
A
$\frac{2 \pi}{\alpha}$

B
$\frac{2 \pi}{\sqrt{\alpha}}$

C
$2 \pi \alpha$

D
$2 \pi \sqrt{\alpha}$

The correct answer is:
$\frac{2 \pi}{\sqrt{\alpha}}$

Solution:
$\omega^2=\alpha$
$\omega=\sqrt{\alpha}$
$T=\frac{2 \pi}{\sqrt{\alpha}}$

Topic: Physics, Oscillations, JEE Main, JEE Main 2005
If a simple pendulum has significant amplitude (up to a factor of $1 / e$ of original) only in the period between $t=0s$ to $t=\tau s$, then $\tau$ may be called the average life of the pendulum. When the spherical bob of the pendulum suffers a retardation (due to viscous drag) proportional to its velocity, with '$b$' as the constant of proportionality, the average life time of the pendulum is (assuming damping is small) in seconds:

Options:
A
$\frac{0.693}{b}$

B
$b$

C
$\frac{1}{b}$

D
$\frac{2}{b}$

The correct answer is:
$\frac{2}{b}$

Solution:
As retardation $=\mathrm{bv}$
$\therefore$ retarding force $=\mathrm{mbv}$
$\therefore$ net restoring torque when angular displacement is $\theta$ is given by $=-\operatorname{mg} \ell \sin \theta+\operatorname{mbv} \ell$
$\therefore \quad \mathrm{I} \alpha=-m g \ell \sin \theta+\operatorname{mbv} \ell$
where, $\mathrm{I}=\mathrm{m} \ell^2$
$\therefore \quad \frac{\mathrm{d}^2 \theta}{\mathrm{dt}^2}=\alpha=-\frac{\mathrm{g}}{\ell} \sin \theta+\frac{\mathrm{bv}}{\ell}$
for small damping, the solution of the above differential equation will be
$\therefore \quad \theta=\theta_0 \mathrm{e}^{-\frac{\mathrm{bt}}{2}} \sin (w t+\phi)$
$\therefore \quad$ angular amplitude will be $=\theta . e^{\frac{-b t}{2}}$
According to question, in $\tau$ time (average life-time),
angular amplitude drops to $\frac{1}{e}$ value of its original value
$(\theta)$
$\therefore \quad \frac{\theta_0}{\mathrm{e}}=\theta_0 \mathrm{e}^{-\frac{6 \tau}{2}}$
$\frac{6 \tau}{2}=1$
$\therefore \quad \tau=\frac{2}{b}$

Topic: Physics, Oscillations, JEE Main
If a simple pendulum has significant amplitude (up to a factor of $1 / e$ of original) only in the period between $t=0s$ to $t=\tau s$, then $\tau$ may be called the average life of the pendulum. When the spherical bob of the pendulum suffers a retardation (due to viscous drag) proportional to its velocity, with '$b$' as the constant of proportionality, the average life time of the pendulum is (assuming damping is small) in seconds:

Options:
A
$\frac{0.693}{b}$

B
$b$

C
$\frac{1}{b}$

D
$\frac{2}{b}$

The correct answer is:
$\frac{2}{b}$

Solution:
As retardation $=\mathrm{bv}$
$\therefore$ retarding force $=\mathrm{mbv}$
$\therefore$ net restoring torque when angular displacement is $\theta$ is given by $=-\operatorname{mg} \ell \sin \theta+\operatorname{mbv} \ell$
$\therefore \quad \mathrm{I} \alpha=-m g \ell \sin \theta+\operatorname{mbv} \ell$
where, $\mathrm{I}=\mathrm{m} \ell^2$
$\therefore \quad \frac{\mathrm{d}^2 \theta}{\mathrm{dt}^2}=\alpha=-\frac{\mathrm{g}}{\ell} \sin \theta+\frac{\mathrm{bv}}{\ell}$
for small damping, the solution of the above differential equation will be
$\therefore \quad \theta=\theta_0 \mathrm{e}^{-\frac{\mathrm{bt}}{2}} \sin (w t+\phi)$
$\therefore \quad$ angular amplitude will be $=\theta . e^{\frac{-b t}{2}}$
According to question, in $\tau$ time (average life-time),
angular amplitude drops to $\frac{1}{e}$ value of its original value
$(\theta)$
$\therefore \quad \frac{\theta_0}{\mathrm{e}}=\theta_0 \mathrm{e}^{-\frac{6 \tau}{2}}$
$\frac{6 \tau}{2}=1$
$\therefore \quad \tau=\frac{2}{b}$

Topic: Physics, Oscillations, JEE Main, JEE Main 2012 (Offline)
If a simple pendulum of length, L has maximum angular displacement α, then the maximum kinetic energy of bob of mass M is

Options:
A 12MLg
B Mg2L
C MgL(1-cos⁡α)
D MgL sin α2

The correct answer is:
MgL(1-cos⁡α)

Solution:
Given, mass of bob=M
Length of simple pendulum=L
Now, in âOAB, cosâ¡Î±=OAOB
or cosâ¡Î±=OAL
or AO=L cos Î±
and AP=OP-AO
AP=L-L cos Î±
or h=L(1-cos Î±)
The maximum kinetic energy of bob
= Maximum potential energy
KE=Mgh
KE=MgL(1-cos Î±)

Topic: Physics, Oscillations, JEE Main
If a=sin−1sin5 and b=cos−1cos5, then a2+b2 is equal to

Options:
A 4π2+25
B 8π2−40π+50
C 4π2−20π+50
D 25

The correct answer is:
8π2−40π+50

Solution:
Given: a=sin-1sin5 and b=cos-1cos5

⇒a=5-2π and b=2π-5

⇒a2+b2=5-2π2+2π-52

⇒a2+b2=25+4π2-20π+4π2+25-20π

⇒a2+b2=8π2-40π+50

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main, JEE Main 2024 (31 Jan Shift 2)
If asin2â¡x+bcos2â¡x=c, bsin2â¡y+acos2â¡y=d and atanâ¡x=btanâ¡y, then a2b2 is equal to (Here a, b, c and d are distinct)

Options:
A b-cd-ba-da-b
B a-dc-ab-cd-b
C d-ac-ab-cd-b
D b-cb-da-ca-d

The correct answer is:
a-dc-ab-cd-b

Solution:
Dividing the given equations by cos2â¡x & cos2â¡y respectively. We get,
tan2x=câbaâc and tan2y=dâabâd

âa2b2=tan2ytan2x=(aâd)â(câa)(bâc)â(dâb)

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
If $A=\sin ^2 x+\cos ^4 x$, then for all real $x$

Options:
A
$\frac{13}{16} \leq \mathrm{A} \leq 1$

B
$1 \leq \mathrm{A} \leq 2$

C
$\frac{3}{4} \leq \mathrm{A} \leq \frac{13}{16}$

D
$\frac{3}{4} \leq \mathrm{A} \leq 1$

The correct answer is:
$\frac{3}{4} \leq \mathrm{A} \leq 1$

Solution:
$A=\sin ^2 x+\cos ^4 x=\frac{7+\cos 4 x}{8} \Rightarrow \frac{3}{4} \leq A \leq 1$

Topic: Mathematics, Trigonometric Equations, JEE Main, JEE Main 2011
If a small sphere of mass m and charge q is hanged from a silk thread at an angle Î¸ with the surface of a vertical charged conducting plate, then for equilibrium of sphere the surface charge density of the plate -

Options:
A ε0mgtanθq
B 2ε0mgtanθq
C ε0 mg q tan θ
D ε0mgtanθ3q

The correct answer is:
2ε0mgtanθq

Solution:
T sin Î¸ = qEâ¦(1)
T cos Î¸ = mgâ¦(2)
tan Î¸= q Emg = q Ï2 Îµ0mg
Ï = 2 Îµ0mgtanÎ¸q

Topic: Physics, Electrostatics, JEE Main
If a soap bubble expands, the pressure inside the bubble

Options:
A increases
B remains the same
C is equal to the atmospheric pressure
D decreases

The correct answer is:
decreases

Solution:
A soap bubble has two liquid surfaces in contact with air, one inside the bubble and the other outside the bubble.

For pressure, we can write:

Pi=Po+4Tr, so as r increases, pressure inside the bubble decreases.

Topic: Physics, Mechanical Properties of Fluids, JEE Main
If a soap bubble expands, the pressure inside the bubble

Options:
A increases
B remains the same
C is equal to the atmospheric pressure
D decreases

The correct answer is:
decreases

Solution:
A soap bubble has two liquid surfaces in contact with air, one inside the bubble and the other outside the bubble.

For pressure, we can write:

Pi=Po+4Tr, so as r increases, pressure inside the bubble decreases.

Topic: Physics, Mechanical Properties of Fluids, NEET, NEET 2022 (Phase 1)
If a solid sphere of mass 5 kg and a disc of mass 4 kg have the same radius, then the ratio of moment of inertia of the disc about a tangent in its plane to the moment of inertia of the sphere about its tangent will be x7. The value of x is ______.

Options:

The correct answer is:
5

Solution:
Moment of inertia of the solid sphere about any diameter is given by, 2m1R25.

Using parallel axis theorem, we get

Isolid sphere =25m1R2+m1R2=75m1R2

=75×5R2

=7R2

Moment of inertia of the disc about the diameter is given by, m2R24.

Now using parallel axis theorem, we get

Idisc =m2R24+m2R2=54m2R2

=54×4R2

=5R2

Therefore,Idisc Isoild sphere =57

Thus, x=5.

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2023 (25 Jan Shift 2)
If a solute associates in a solvent, its experimentally calculated molar mass using boiling point elevation method will be

Options:
A half of the actual value
B will remain same as actual value
C one fourth of the actual value
D higher than the actual value

The correct answer is:
higher than the actual value

Solution:
If a solute associates in a solvent, its experimentally calculated molar mass using boiling point elevation method will be higher than the actual value. As colligative property are inversely proportional to molecular mass of solute.

Topic: Chemistry, Solutions, TS EAMCET, TS EAMCET 2019 (04 May Shift 2)
If a source of electromagnetic radiation having power 15 kW produces 1016 photons per second, the radiation belongs to a part of spectrum is :

(Take Planck constant h = 6×10-34 J s)

Options:
A Micro waves
B Ultraviolet rays
C Gamma rays
D Radio waves

The correct answer is:
Gamma rays

Solution:
As energy of one photon is given by E=PowerPhoton frequency

Or, E=hν=15×103 1016

So, frequency of ray is ν=15×10-136×10-34=2.5×1021 Hz.

Thus, the radiation belongs to gamma rays.

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2023 (31 Jan Shift 1)
If a source of power $4 \mathrm{~kW}$ produces $10^{20}$ photons/second, the radiation belong to a part of the spectrum called

Options:
A
X-rays

B
ultraviolet rays

C
microwaves

D
$\gamma$-rays

The correct answer is:
X-rays

Solution:
$$
\begin{aligned}
& 4 \times 10^3=10^{20} \mathrm{xhf} \\
& f=\frac{4 \times 10^3}{10^{20} \times 6.023 \times 10^{-34}} \\
& f=6.03 \times 10^{16} \mathrm{~Hz}
\end{aligned}
$$
The obtained frequency lies in the band of X-rays.

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2010
If a species has 16 protons, 18 electrons and 16 neutrons, find the species and its charge

Options:
A $\mathrm{S}^{1-}$
B $\mathrm{Si}^{2-}$
C $\mathrm{P}^{3-}$
D $\mathrm{S}^{2-}$

The correct answer is:
$\mathrm{S}^{2-}$

Solution:
Hints : $16 \mathrm{p}$ means $\mathrm{z}=16$
$18 \mathrm{e}^{-}$means, 2 unit negative charge is present.
Hence species is $\mathrm{S}^{-2}$

Topic: Chemistry, Structure of Atom, JEE Main
If a spring has time period T, and is cut into n equal parts, then the time period of each part will be

Options:
A
$\mathrm{T} \sqrt{\mathrm{n}}$

B
$\mathrm{T} / \sqrt{\mathrm{n}}$

C
$\mathrm{nT}$

D
$\mathrm{T}$

The correct answer is:
$\mathrm{T} / \sqrt{\mathrm{n}}$

Solution:
Spring constant becomes $n$ times for each piece. $T=2 \pi \sqrt{\mathrm{m} / \mathrm{k}}$
$$
\frac{T_1}{T_2}=\frac{\sqrt{n K}}{K} \text { or } T_2=T / \sqrt{n}
$$

Topic: Physics, Oscillations, JEE Main, JEE Main 2002
If 'a' stands for the edge length of the cubic systems : simple cubic, body centred cubic and face centred cubic, then the ratio of radii of the spheres in these systems will be respectively

Options:
A $1 a: \sqrt{3} a: \sqrt{2} a$
B $\frac{1}{2} a: \frac{\sqrt{3}}{4} a: \frac{1}{2 \sqrt{2}}^a$
C $\frac{1}{2} a: \sqrt{3} a: \frac{1}{\sqrt{2}} \mathrm{a}$
D $\frac{1}{2} \mathrm{a}: \frac{\sqrt{3}}{2} \mathrm{a}: \frac{\sqrt{2}}{2} \mathrm{a}$

The correct answer is:
$\frac{1}{2} a: \frac{\sqrt{3}}{4} a: \frac{1}{2 \sqrt{2}}^a$

Solution:
For simple cube, $a=2 r$
or $r=\frac{a}{2}$
For BCC, $4 r=\sqrt{3} a$
or $r=\frac{\sqrt{3}}{4} a$
For FCC, $4 r=\sqrt{2} a$
or $r=\frac{a}{2 \sqrt{2}}$
Thus, the ratio is $\frac{1}{2} a: \frac{\sqrt{3}}{4} a: \frac{1}{2 \sqrt{2}} a$

Topic: Chemistry, Solid State, NEET, NEET 2008 (Mains)
If ' $a$ ' stands for the edge length of the cubic systems : simple cubic, body centred cubic and face centred cubic, then the ratio of radii of the spheres in these systems will be respectively,

Options:
A $\frac{1}{2} a : \frac{\sqrt{3}}{4} a: \frac{1}{2 \sqrt{2}} a$
B $\frac{1}{2} a: \sqrt{3} a: \frac{1}{\sqrt{2}} a$
C $\frac{1}{2} a: \frac{\sqrt{3}}{2} a: \frac{\sqrt{2}}{2} a$
D $1 a: \sqrt{3} a: \sqrt{2} a$

The correct answer is:
$\frac{1}{2} a : \frac{\sqrt{3}}{4} a: \frac{1}{2 \sqrt{2}} a$

Solution:
For simple cubic,
$\begin{aligned}
\therefore & a & =2 r \\
\therefore & r & =\frac{a}{2}
\end{aligned}$
For body centred cubic,
$\begin{aligned}
& a=\frac{4 r}{\sqrt{3}} \\
& r=\frac{\sqrt{3} a}{4}
\end{aligned}$
For face centred cubic,
$\begin{aligned}
& a=2 \sqrt{2} r \\
& r=\frac{a}{2 \sqrt{2}}
\end{aligned}$
Hence, the ratio of radii in simple cubic, body centred cubic and face centred cubic is $\frac{a}{2}: \frac{\sqrt{3} a}{4}: \frac{a}{2 \sqrt{2}}$

Topic: Chemistry, Chemical Kinetics, JEE Main
If a stone is to hit at a point which is at a distance d away and at a height h above the point from where the stone starts, then what is the value of initial speed u, if the stone is launched at an angle Î¸?

Options:
A gcosθd2dtanθ-h
B dcosθg2dtanθ-h
C gd2hcos2θ
D gd2d-h

The correct answer is:
dcosθg2dtanθ-h

Solution:
Using equation of motion for projectile in x and y-direction.
h=usinθt-12gt2
d=ucosθt or t=ducosθ
⟹h=usinθ.ducosθ-12g.d2u2cos2θ
⟹u=dcosθg2dtanθ-h

Topic: Physics, Motion In Two Dimensions, JEE Main
If a straight line passing through the point P-3, 4 is such that its intercepted portion between the coordinate axes is bisected at P, then its equation is :

Options:
A 4x+3y=0
B 4x-3y+24=0
C 3x-4y+25=0
D x-y+7=0

The correct answer is:
4x-3y+24=0

Solution:
Let the line passing through -3, 4 intersect the coordinate axes at A & B.

By mid-point formula, co-ordinates of Pα+02, 0+β2.

Comparing with given co-ordinate values, we get

α+02=-3⇒α=-6

and β+ 02=4⇒β=8

Hence, equation of line is x-6+y8=1

⇒4x-3y+24=0

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2019 (12 Jan Shift 2)
If a straight line perpendicular to $2 x-3 y+7=0$ forms a triangle with the co-ordinate axes whose area is 3 sq. units, then the equation of the straight line is :

Options:
A $3 x+2 y= \pm 2$
B $3 x+2 y= \pm 6$
C $3 x+2 y= \pm 4$
D $3 x+2 y= \pm 8$

The correct answer is:
$3 x+2 y= \pm 6$

Solution:
The given line is $2 x-3 y+7=0$
A perpendicular line to given line is
$$
3 x+2 y+k=0
$$
Put $y=0 \Rightarrow 3 x+k=0$
$$
x=-\frac{k}{3}
$$
Put $x=0 \Rightarrow 0+2 y+k=0$
$$
y=-\frac{k}{2}
$$
Area of the triangle $O A B=\frac{1}{2} \cdot O A \cdot O B$
$$
\begin{aligned}
\Rightarrow & 3 & =\frac{1}{2} \cdot\left(-\frac{k}{3}\right)\left(-\frac{k}{2}\right) \\
\Rightarrow & k^2 & =36 \Rightarrow k= \pm 6
\end{aligned}
$$
Putting in (i)
$$
3 x+2 y= \pm 6
$$

Topic: Mathematics, Straight Lines, JEE Main