datavorous/entrance-exam-dataset huggingface dataset:
The differential equation of the family of circles passing through the origin and having centres on the $\mathrm{x}$ -axis is

Options:
A $\quad 2 x y \frac{d y}{d x}=x^{2}-y^{2}$
B $2 x y \frac{d y}{d x}=y^{2}-x^{2}$
C $\quad 2 x y \frac{d y}{d x}=x^{2}+y^{2}$
D $2 x y \frac{d y}{d x}+x^{2}+y^{2}=0$

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

Solution:
Eq. of family of circles passing through the origin \& having centres on the $\mathrm{x}$ -axis is:
$\mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{gx}=0$
$2 \mathrm{x}+2 \mathrm{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}+2 \mathrm{~g}=0 \quad$ [on differentiating]
$\Rightarrow g=-\left[x+y \frac{d y}{d x}\right]$
Putting the value of $(\mathrm{g})$ in eq. $(1)$ we get;
$2 x y \frac{d y}{d x}=y^{2}-x^{2}$

Topic: Mathematics, Differential Equations, NDA, NDA 2015 (Phase 2)
The differential equation of the family of curves $\mathrm{y}=\mathrm{p} \cos$ $(\mathrm{ax})+\mathrm{q} \sin (\mathrm{ax})$, where $\mathrm{p}, \mathrm{q}$ are arbitrary constants, is

Options:
A $\frac{d^{2} y}{d x^{2}}-a^{2} y=0$
B $\frac{d^{2} y}{d x^{2}}-a y=0$
C $\frac{d^{2} y}{d x^{2}}+a y=0$
D $\frac{d^{2} y}{d x^{2}}+a^{2} y=0$

The correct answer is:
$\frac{d^{2} y}{d x^{2}}+a^{2} y=0$

Solution:
$y=p \cos a x+q \sin a x$
$\Rightarrow \quad \frac{d y}{d x}=-p a \sin a x+q a \cos a x$
$\Rightarrow \quad \frac{d^{2} y}{d x^{2}}=-p a^{2} \cos a x-q a^{2} \sin a x=-a^{2} y$
$\Rightarrow \quad \frac{d^{2} y}{d x^{2}}+a^{2} y=0$

Topic: Mathematics, Differential Equations, NDA, NDA 2018 (Phase 2)
The differential equation of the system of circles touching the y-axis at the origin is

Options:
A $x^{2}+y^{2}-2 x y \frac{d y}{d x}=0$
B $x^{2}+y^{2}+2 x y \frac{d y}{d x}=0$
C $x^{2}-y^{2}+2 x y \frac{d y}{d x}=0$
D $x^{2}-y^{2}-2 x y \frac{d y}{d x}=0$

The correct answer is:
$x^{2}-y^{2}+2 x y \frac{d y}{d x}=0$

Solution:
The equation of circle touching $\mathrm{y}$ -axis at origin is $(\mathrm{x}-\alpha)^{2}+(\mathrm{y}-0)^{2}=\alpha^{2}$
$\Rightarrow x^{2}+\alpha^{2}-2 \alpha x+y^{2}=\alpha^{2}$
$\Rightarrow x^{2}+y^{2}-2 \alpha x=0$
$\Rightarrow x+\frac{y^{2}}{x}-2 \alpha=0$
Differentiating, $1+\frac{x \cdot 2 y \cdot \frac{d y}{d x}-y^{2}}{x^{2}}=0$
$\Rightarrow x^{2}+2 x y \cdot \frac{d y}{d x}-y^{2}=0$

Topic: Mathematics, Differential Equations, NDA, NDA 2019 (Phase 1)
The differential equation representing the family of curves $\mathrm{y}=\mathrm{a} \sin (\lambda \mathrm{x}+\alpha)$ is: $\quad$

Options:
A $\frac{d^{2} y}{d x^{2}}+\lambda^{2} y=0$
B $\frac{d^{2} y}{d x^{2}}-\lambda^{2} y=0$
C $\frac{d^{2} y}{d x^{2}}+\lambda y=0$
D None of the above

The correct answer is:
$\frac{d^{2} y}{d x^{2}}+\lambda^{2} y=0$

Solution:
Let $\mathrm{y}=\mathrm{a} \sin (\lambda \mathrm{x}+\alpha)$
$\Rightarrow \quad \frac{d y}{d x}=\lambda a \cos (\lambda x+\alpha)$
Again differentiating on both side we get
$$
\begin{aligned}
& \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-\lambda^{2} \text { a } \sin (\mathrm{x}+\alpha) \\
\Rightarrow & \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}+\lambda^{2} \mathrm{y}=0 \text { Required equation. }
\end{aligned}
$$

Topic: Mathematics, Differential Equations, NDA, NDA 2012 (Phase 2)
The differential equation $\mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{x}=\mathrm{a}$ where 'a' is any
constant represents : $\quad$

Options:
A A set of straight lines
B A set of ellipses
C A set of circles
D None of the above

The correct answer is:
A set of circles

Solution:
Given diff. equation is
$$
y \frac{d y}{d x}+x=a
$$
$\begin{aligned} \Rightarrow & y \frac{d y}{d x}=a-x \\ \Rightarrow & y d y=(a-x) d x \\ & \int y d y=\int(a-x) d x \\ \Rightarrow & \frac{y^{2}}{2}=a x-\frac{x^{2}}{2}+k \\ \Rightarrow & x^{2}+y^{2}-2 a x=2 k \end{aligned}$
Which represents a set of circles.

Topic: Mathematics, Differential Equations, NDA, NDA 2012 (Phase 2)
The diffraction pattern of a single slit is shown in the figure. The point at which the path difference of the extreme rays is two times the wavelength is

Options:
A point 1
B point 2
C point 4
D point 5

The correct answer is:
point 5

Solution:
In diffraction of light at a single slit, the pattern obtained consists of a central bright band having alternate dark and bright bands of decreasing intensity on both sides.
For nth minima (secondary),
sinâ¡Î¸n=nÎ»a
For nth maxima (secondary),
sinâ¡Î¸n=2n+1Î»2a
Thus, point 5 corresponds to the path difference of extreme rays equal to 2Î».

Topic: Physics, Wave Optics, NEET
The digital signals are used for transmission in communication system. They are

Options:
A only of discrete stepwise values.
B sound and picture signals in television.
C continuous variations of current or voltage.
D fundamental sine waves.

The correct answer is:
only of discrete stepwise values.

Solution:
Only for discrete stepwise values.

Topic: Physics, Communication System, MHT CET, MHT CET 2020 (13 Oct Shift 2)
The dircction ratios of the linc perpendicular to the lincs with direction ratios $ < 1,-2,-2>$ and $ < 0,2,1>$ are

Options:
A $ < 2,-1,2>$
B $ < -2,1,2>[2015-I I]$
C $ < 2,1,-2>$
D $ < -2,-1,-2\rangle$

The correct answer is:
$ < 2,-1,2>$

Solution:
Let the direction ratio be $ < a, b, c\rangle$ $\cos 90^{\circ}=\frac{(\mathrm{a})(1)+(\mathrm{b})(-2)+(\mathrm{c})(-2)}{\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}} \sqrt{(1)^{2}+(-2)^{2}+(-2)^{2}}}$
$a-2 b-2 c=0$
$\cos 90^{\circ}=\frac{(\mathrm{a})(0)+\mathrm{b}(2)+(\mathrm{c})(1)}{\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}} \sqrt{(0)^{2}+(-2)^{2}+(1)^{2}}}$
$2 b+c=0$
From eq. $(1) \&(2)$ $a=-2 b ; c=-2 b$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2015 (Phase 2)
The direction cosines of a line are proportional to $(2,1,2)$ and the line intersects a plane perpendicularly at the point $(1,-2,4)$. What is the distance of the plane from the point $(3,$, $2,3) ? \quad$

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

The correct answer is:
2

Solution:
Equation of plane passing through $(1,-2,4)$ and whose normal $(2,1,2)$ is $2(x-1)+1(y+2)+2(z-4)=0$
$\Rightarrow 2 x-2+y+2+2 z-8=0$
$\Rightarrow 2 x+y+2 z-8=0$
So, distance of the plane $2 x+y+2 z-8=0$ from the point $(3,2,3)$ is
$=\frac{2(3)+1(2)+(2)(3)-8}{\sqrt{4+1+4}}=\frac{6}{3}=2$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2009 (Phase 2)
The displacement of a particle executing SHM is given by
$y=5 \sin \left(4 t+\frac{\pi}{3}\right) .$
If $T$ is the time period and the mass of the particle is $2 \mathrm{~g}$, the kinetic energy of the particle when $t=\frac{T}{4}$ is given by

Options:
A 0.4 J
B 0.5 J
C 3 J
D 0.3 J

The correct answer is:
0.4 J

Solution:
The displacement of particle, executing SHM
Velocity of particle
$\begin{aligned}
\left(\frac{d y}{d t}\right) & =\frac{5 d}{d t} \sin \left(4 t+\frac{\pi}{3}\right) \\
& =5 \cos \left(4 t+\frac{\pi}{3}\right) \cdot 4 \\
& =20 \cos \left(4 t+\frac{\pi}{3}\right)
\end{aligned}$
Velocity at $t=\left(\frac{T}{4}\right)$
$\left(\frac{d y}{d t}\right)_{t=\frac{T}{4}}=20 \cos \left(4 \times \frac{T}{4}+\frac{\pi}{3}\right)$
Comparing the given equation with standard equation of SHM, given by
We get
$\begin{aligned}
& y=a \sin (\omega t+\phi) \\
& \omega=4
\end{aligned}$
As
$\omega=\frac{2 \pi}{T}$
$\begin{aligned} & \Rightarrow \quad T=\frac{2 \pi}{\omega} \\ & \text { or } & T=\frac{2 \pi}{4} \\ & \text { or } & T=\left(\frac{\pi}{2}\right) \\ & \end{aligned}$
Now, putting value of $T$ in Eq. (ii), we get
$\begin{aligned}
u & =20 \cos \left(\frac{\pi}{2}+\frac{\pi}{3}\right) \\
& =-20 \sin \frac{\pi}{3} \\
& =-20 \times \frac{\sqrt{3}}{2} \\
& =-10 \times \sqrt{3}
\end{aligned}$
The kinetic energy of particle,
$\begin{aligned}
& \mathrm{KE}=\frac{1}{2} m u^2 \\
& \because \quad m=2 \mathrm{~g}=2 \times 10^{-3} \mathrm{~kg} \\
& =\frac{1}{2} \times 2 \times 10^{-3} \times(-10 \sqrt{3})^2 \\
& =10^{-3} \times 100 \times 3 \\
& =3 \times 10^{-1} \\
& \mathrm{KE}=0.3 \mathrm{~J} \\
&
\end{aligned}$

Topic: Physics, Oscillations, TS EAMCET, TS EAMCET 2009
The displacement of a particle executing SHM is given by
$y=5 \sin \left(4 t+\frac{\pi}{3}\right) .$
If $T$ is the time period and the mass of the particle is $2 \mathrm{~g}$, the kinetic energy of the particle when $t=\frac{T}{4}$ is given by

Options:
A 0.4 J
B 0.5 J
C 3 J
D 0.3 J

The correct answer is:
0.4 J

Solution:
The displacement of particle, executing SHM
Velocity of particle
$\begin{aligned}
\left(\frac{d y}{d t}\right) & =\frac{5 d}{d t} \sin \left(4 t+\frac{\pi}{3}\right) \\
& =5 \cos \left(4 t+\frac{\pi}{3}\right) \cdot 4 \\
& =20 \cos \left(4 t+\frac{\pi}{3}\right)
\end{aligned}$
Velocity at $t=\left(\frac{T}{4}\right)$
$\left(\frac{d y}{d t}\right)_{t=\frac{T}{4}}=20 \cos \left(4 \times \frac{T}{4}+\frac{\pi}{3}\right)$
Comparing the given equation with standard equation of SHM, given by
We get
$\begin{aligned}
& y=a \sin (\omega t+\phi) \\
& \omega=4
\end{aligned}$
As
$\omega=\frac{2 \pi}{T}$
$\begin{aligned} & \Rightarrow \quad T=\frac{2 \pi}{\omega} \\ & \text { or } & T=\frac{2 \pi}{4} \\ & \text { or } & T=\left(\frac{\pi}{2}\right) \\ & \end{aligned}$
Now, putting value of $T$ in Eq. (ii), we get
$\begin{aligned}
u & =20 \cos \left(\frac{\pi}{2}+\frac{\pi}{3}\right) \\
& =-20 \sin \frac{\pi}{3} \\
& =-20 \times \frac{\sqrt{3}}{2} \\
& =-10 \times \sqrt{3}
\end{aligned}$
The kinetic energy of particle,
$\begin{aligned}
& \mathrm{KE}=\frac{1}{2} m u^2 \\
& \because \quad m=2 \mathrm{~g}=2 \times 10^{-3} \mathrm{~kg} \\
& =\frac{1}{2} \times 2 \times 10^{-3} \times(-10 \sqrt{3})^2 \\
& =10^{-3} \times 100 \times 3 \\
& =3 \times 10^{-1} \\
& \mathrm{KE}=0.3 \mathrm{~J} \\
&
\end{aligned}$

Topic: Physics, Oscillations, AP EAMCET, AP EAMCET 2009
The displacement of a particle executing SHM is given by
$y=5 \sin \left(4 t+\frac{\pi}{3}\right)$
If $T$ is the time period and the mass of the particle is $2 \mathrm{~g}$, the kinetic energy of the particle when $\mathrm{t}=\mathrm{T} / 4$ is given by

Options:
A $0.4 \mathrm{~J}$
B $0.5 \mathrm{~J}$
C $3 \mathrm{~J}$
D $0.3 \mathrm{~J}$

The correct answer is:
$0.3 \mathrm{~J}$

Solution:
Particle executing SHM.
Displacement $y=5 \sin \left(4 t+\frac{\pi}{3}\right)...(i)$
Velocity of particle
$$
\begin{array}{l}
\left(\frac{d y}{d t}\right)=\frac{5 d}{d t} \sin \left(4 t+\frac{\pi}{3}\right) \\
=5 \cos \left(4 t+\frac{\pi}{3}\right) \cdot 4=20 \cos \left(4 t+\frac{\pi}{3}\right)
\end{array}
$$
Velocity at $t=\left(\frac{T}{4}\right)$
$$
\left(\frac{d y}{d t}\right)_{t=\frac{T}{4}}=20 \cos \left(4 \times \frac{T}{4}+\frac{\pi}{3}\right)
$$
or $\quad u=20 \cos \left(T+\frac{\pi}{3}\right)...(ii)$
Comparing the given equation with standard equation of SHM.
$y=a \sin (\omega t+\phi)$
We get, $\omega=4$
As $\omega=\frac{2 \pi}{T} \Rightarrow T=\frac{2 \pi}{\omega}$
or $T=\frac{2 \pi}{4}=\left(\frac{\pi}{2}\right)$
Now, putting value of $T$ in Eq. (ii), we get
$$
\begin{aligned}
u &=20 \cos \left(\frac{\pi}{2}+\frac{\pi}{3}\right)=-20 \sin \frac{\pi}{3} \\
&=-10 \times \sqrt{3}
\end{aligned}
$$
The kinetic energy of particle,
$$
\begin{aligned}
\mathrm{KE} &=\frac{1}{2} m u^{2}=\frac{1}{2} \times 2 \times 10^{-3} \times(-10 \sqrt{3})^{2} \\
&=10^{-3} \times 100 \times 3=0.3 \mathrm{~J}
\end{aligned}
$$

Topic: Physics, Oscillations, VITEEE, VITEEE 2009
The displacement of a particle is represented by the equation $y=3 \cos \left(\frac{\pi}{4}-2 \omega t\right)$. The motion of the particle is

Options:
A
simple harmonic with period $2 \pi / \omega$

B
simple harmonic with period $\pi / \omega$

C
periodic but not simple harmonic

D
non-periodic

The correct answer is:
simple harmonic with period $\pi / \omega$

Solution:
As given that, $y=3 \cos \left(\frac{\pi}{4}-2 \omega t\right)$
Velocity of the particle
$$
\begin{aligned}
v &=\frac{d y}{d t}=\frac{d}{d t}\left[3 \cos \left(\frac{\pi}{4}-2 \omega t\right)\right] \\
&=3(-2 \omega)\left[-\sin \left(\frac{\pi}{4}-2 \omega t\right)\right] \\
&=6 \omega \sin \left(\frac{\pi}{4}-2 \omega t\right)
\end{aligned}
$$
So, acceleration,
$$
\begin{aligned}
a &=\frac{d v}{d t}=\frac{d}{d t}\left[6 \omega \sin \left(\frac{\pi}{4}-2 \omega t\right)\right] \\
&=(6 \omega) \times(-2 \omega) \cos \left(\frac{\pi}{4}-2 \omega t\right) \\
&=-4 \omega^2\left[3 \cos \left(\frac{\pi}{4}-2 \omega t\right)\right] \\
\Rightarrow a=-4 \omega^2 y
\end{aligned}
$$
In simple harmonic motion acceleration (or force) is directly proportional to the negative of displacement of particle
$\Rightarrow$ as acceleration, $a \propto-y$
Hence, due to negative sign motion is simple harmonic motion (SHM.)
A simple harmonic motion is always periodic. So motion is periodic simple harmonic.
From the given equation,
$$
y=3 \cos \left(\frac{\pi}{4}-2 \omega t\right)
$$
Compare it by standard equation
$$
y=a \cos (\omega t+\phi)
$$
So, $\omega^{\prime}=2 \omega$
$$
\frac{2 \pi}{T^{\prime}}=2 \omega \Rightarrow T^{\prime}=\frac{2 \pi}{2 \omega}=\frac{\pi}{\omega}
$$
Hence, the motion is SHM with period $\frac{\pi}{\omega}$.

Topic: Physics, Oscillations
The displacement of a particle performing S.H.M is given by $x=5 \sin (3 t+3)$, where $x$ is in $\mathrm{cm}$ and $t$ is in second. The maximum acceleration of the particle will be

Options:
A $15 \mathrm{~cm} \mathrm{~s}^{-2}$
B $30 \mathrm{~cm} \mathrm{~s}^{-2}$
C $45 \mathrm{~cm} \mathrm{~s}^{-2}$
D $90 \mathrm{~cm} \mathrm{~s}^{-2}$

The correct answer is:
$45 \mathrm{~cm} \mathrm{~s}^{-2}$

Solution:
$x=5 \sin (3 t+3)$
Standard equation of S.H.M. is $x=A \sin (\omega t+\alpha)$
$\therefore \mathrm{A}=5 \mathrm{~cm}, \omega=3 \mathrm{rad} / \mathrm{s}$
Maximum acceleration $\mathrm{a}_{\mathrm{m}}=\mathrm{A} \omega^2=5 \times(3)^2$
$=5 \times 9=45 \mathrm{~cm} \mathrm{~s}^{-2}$

Topic: Physics, Oscillations, MHT CET, MHT CET 2021 (24 Sep Shift 2)
The displacement of a string is given by $y(x, t)=0.06 \sin \left(\frac{2 \pi x}{3}\right) \cos (120 \pi t)$
where $x$ and $y$ are in metre and $t$ in second. The length of the string is $1.5 \mathrm{~m}$ and its mass is $3.0 \times 10^{-2} \mathrm{~kg}$.

Options:
A
It represents a progressive wave of frequency $60 \mathrm{~Hz}$.

B
It represents a stationary wave of frequency $60 \mathrm{~Hz}$

C
It is the result superposition of two waves of wavelength $3 \mathrm{~m}$, frequency $60 \mathrm{~Hz}$ each travelling with a speed of $180 \mathrm{~m} / \mathrm{s}$ in opposite direction

D
Amplitude of this wave is constant.

The correct answers are:
It represents a stationary wave of frequency $60 \mathrm{~Hz}$
,
It is the result superposition of two waves of wavelength $3 \mathrm{~m}$, frequency $60 \mathrm{~Hz}$ each travelling with a speed of $180 \mathrm{~m} / \mathrm{s}$ in opposite direction

Solution:
As we know that, standard equation of stationary wave
$$
y(x, t)=a \sin (k x) \cos (\omega t)
$$
As given that the equation is
$$
y(x, t)=0.06 \sin \left(\frac{2 \pi x}{3}\right) \cos (120 \pi t)
$$
Comparing with a standard equation of stationary wave with given equation. So, $a=0.06, k=\frac{2 \pi}{3}$, $\omega=120 \pi$
(a) Clearly, the given equation belongs to stationary wave. Hence, option (a) is not verifies.
(b) Given, $\omega=120 \pi$
$$
2 \pi v=120 \pi \Rightarrow v=60 \mathrm{~Hz}
$$
(c) $k=\frac{2 \pi}{3}=\frac{2 \pi}{\lambda} \quad\left(\because k=\frac{2 \pi}{\lambda}\right)$
So, $\lambda$ (wavelength) $=3 \mathrm{~m}$
Frequency $(v)=60 \mathrm{~Hz}$
Speed $(v)=v \lambda=(60 \mathrm{~Hz}) \times(3 \mathrm{~m})=180 \mathrm{~m} / \mathrm{s}$
(d) As waves are stationary so the amplitude of particles varies from 0 to $0.06 \mathrm{~m}$ from nodes to antinodes i.e. amplitude is not constant always.
Since in stationary wave, all particles of the medium execute SHM with varying amplitude nodes. So option
(d) is not verifies.

Topic: Physics, Waves and Sound
The displacement of the particle executing linear S.H.M. is $\mathrm{x}=0.25 \mathrm{sin}(11 \mathrm{t}+0.5) \mathrm{m}$. The period of S.H.M. is $\left(\pi=\frac{22}{7}\right)$

Options:
A $\frac{2}{7} \mathbf{S}$
B $\frac{4}{7} \mathrm{~S}$
C $\frac{3}{7} \mathrm{~S}$
D $\frac{1}{7}$ S

The correct answer is:
$\frac{4}{7} \mathrm{~S}$

Solution:
Comparing with standard equation of S.H.M.
$x=A \sin (\omega t+\phi)$
we get $\omega=11$
$\therefore T=\frac{2 \pi}{\omega}=\frac{2 \times 22}{11 \times 7}=\frac{4}{7} \kappa$

Topic: Physics, Oscillations, MHT CET, MHT CET 2020 (14 Oct Shift 1)
The displacement x (in meter) of a particle performing simple harmonic motion is related to time t (in second) as x=0.05cos4Ït+Ï4.The frequency of the motion will be

Options:
A 0.5 Hz
B 1.0 Hz
C 1.5 Hz
D 2.0 Hz

The correct answer is:
2.0 Hz

Solution:
Compare the given equation with the standard form
y=rcos⁡2πtT-2πxλ
Coefficient of t=2πT=2πn=4π,n=2 Hz

Topic: Physics, Oscillations, NEET
The displacement x(in metre) of a particle in simple harmonic motion is related to time t (in second) as

x=0.01cos⁡πt+π4

The frequency of the motion will be

Options:
A 0.5 Hz

B 1.0 Hz

C π2 Hz

D π Hz

The correct answer is:
0.5 Hz

Solution:
The standard equation in SHM is
x=acos⁡(ωt+ϕ) ...(i)
Where a is amplitude, ω the angular velocity and (ϕ) the phase difference.
Also, ω=2πT where T is periodic time.
So, Eq. (i) becomes
x=acos⁡2πtT+ϕ ...(ii)
Given, the equation is
x=0.01cos⁡2πt2+π4 ...(iii)
Comparing Eq. (ii) with Eq. (iii), we get
2πtT=2πt2
⇒ T=2 s
So, frequency n=1T=12=0.5 Hz

Topic: Physics, Oscillations, NEET
The distance between pair of parallel lines represented by
$$
x^2+2 x y+y^2-8 a x-8 a y-9 a^2=0
$$
is. units

Options:
A $5 \sqrt{2}$
B $5 \sqrt{2} a$
C $2 \sqrt{5} a$
D $a$

The correct answer is:
$5 \sqrt{2} a$

Solution:
Given, $\because x^2+2 x y+y^2-8 a x-8 a y-9 a^2=0$ represents a pair of parallel lines.
Compare with standard form
$$
\begin{aligned}
& a x^2+b y^2+2 h x y+2 g x+2 f y+c=0 \\
& a=1, b=1, h=1, g=-4 a, f=-4 a, c=-9 a^2
\end{aligned}
$$
$\because$ We know that distance between the parallel lines.
$$
\begin{aligned}
D & =2 \sqrt{\frac{g^2-a c}{a(a+b)}} \\
& =2 \sqrt{\frac{(-4 a)^2-(1)\left(-9 a^2\right)}{1(1+1)}} \\
& =2 \sqrt{\frac{16 a^2+9 a^2}{2}}=2 \sqrt{\frac{25 a^2}{2}}=2 \times \frac{5 a}{\sqrt{2}}=5 \sqrt{2} a
\end{aligned}
$$

Topic: Mathematics, Straight Lines, AP EAMCET, AP EAMCET 2021 (24 Aug Shift 1)
The distance between the parallel planes $4 x-2 y+4 z+9=0$ and $8 x-4 y+8 z+21=0$ is

Options:
A $\frac{1}{4}$
B $\frac{1}{2}$
C $\frac{3}{2}$
D $\frac{7}{4}$

The correct answer is:
$\frac{1}{4}$

Solution:
Given planes $: 4 x-2 y+4 z+9=0$ $\Rightarrow 8 x-4 y+8 z+18=0$
$8 x-4 y+8 z+21=0$
Distance $=\frac{\left|d_{1}-d_{2}\right|}{\sqrt{a^{2}+b^{2}+c^{2}}}=\frac{|18-21|}{\sqrt{64+16+64}}$
$=\frac{3}{\sqrt{144}}=\frac{3}{12}=\frac{1}{4}$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2019 (Phase 1)
The distance of the point $(1,3)$ from the line $2 \mathrm{x}+3 \mathrm{y}=6$, measured parallel to the line $4 \mathrm{x}+\mathrm{y}=4$, is

Options:
A $\frac{5}{\sqrt{13}}$ units
B $\frac{3}{\sqrt{17}}$ unit
C $\sqrt{17}$ units
D $\frac{\sqrt{17}}{2}$ units

The correct answer is:
$\frac{\sqrt{17}}{2}$ units

Solution:
The line $4 \mathrm{x}+\mathrm{y}=4$ can be written as $\mathrm{y}=-4 \mathrm{x}+4$. So, slope is $-4$. The line parallel to $4 \mathrm{x}+\mathrm{y}=4$ will have slope $-4$ only. Given point $=(1,3)$ Equation of line passing through $(1,3)$ and slope $-4$ is $\mathrm{y}-3=-4(\mathrm{x}-1)$
$\Rightarrow \mathrm{y}-3=-4 \mathrm{x}+4 \Rightarrow 4 \mathrm{x}+\mathrm{y}=7$
Solving the two equations, we get
$2 x+3 y=6 \Rightarrow 2 x+3(1)=6$
$\Rightarrow 2 \mathrm{x}=3 \Rightarrow \mathrm{x}=\frac{3}{2}$
Distance between the points $(1,3)$ and $\left(\frac{3}{2}, 1\right)$ is
$\sqrt{\left(\frac{3}{2}-1\right)^{2}+(1-3)^{2}}=\sqrt{\left(\frac{1}{2}\right)^{2}+(-2)^{2}}=\sqrt{\frac{1}{4}+4}=\sqrt{\frac{17}{4}}=\frac{\sqrt{17}}{2}$

Topic: Mathematics, Straight Lines, NDA, NDA 2017 (Phase 2)
The distribution of a random variable $X$ is given below
\begin{array}{lllll}
\hlineX=x & 1 & 2 & 3 & 4 \\
\hlineP(X=x) & 20 & 40 & 60 & 80 \\
\hline
\end{array}
Then, the standard deviation of $X$ is

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

The correct answer is:
1

Solution:
\begin{array}{ccccc}
\hlinex & 1 & 2 & 3 & 4 \\
\hlineP_x=x & 2 c & 4 c & 6 c & 8 c \\
\hline
\end{array}
So, standard deviation of $X$ is
Now, $2 c+4 c+6 c+8 c=1 \Rightarrow c=\frac{1}{20}$ for standard deviation
$$
\begin{aligned}
& \sigma=\sqrt{\Sigma x^2 \cdot P x-(\Sigma x \cdot P x)^2} \\
& \Rightarrow \quad \sigma=\sqrt{10-(3)^2} \Rightarrow 1 \\
&
\end{aligned}
$$

Topic: Mathematics, Statistics, TS EAMCET, TS EAMCET 2018 (04 May Shift 2)
The distributiong $X$ and $Y$ with total number of
combined. What is the mean of the resulting distribution $x+Y ?$

Options:
A $3.26$
B $3.32$
C 336
D $3.42$

The correct answer is:
336

Solution:
Required mean $=\frac{36 \times 4+64 \times 3}{36+64}=\frac{144+192}{100}=\frac{336}{100}$
$=3.36$

Topic: Mathematics, Statistics, NDA, NDA 2010 (Phase 2)
The $E^{\circ}$ values for Mn and Zn are more negative than expected because

Options:
A they have either half-filled and fully-filled configurations
B they can easily donate electrons
C it is Quite easy to remove electrons from their orbitals
D None of the above

The correct answer is:
they have either half-filled and fully-filled configurations

Solution:
The values of standard electrode potentials for Mn, and Zn are more negative than expected. The reason behind this is that they have exactly half-filled and fully-filled configuration ons.
$\begin{aligned} & { }_{25} \mathrm{Mn}: \left[\mathrm{Ar}\right] 3 d^5 4 s^2 \\ & { }_{30} \mathrm{Zn}:[\mathrm{Ar}] 3 d^{10} 4 s^2\end{aligned}$

Topic: Chemistry, d and f Block Elements, JIPMER, JIPMER 2016
The Earth Summit held in Rio de Janeiro in 1992 was called:

Options:
A To reduce CO2 emission and global warming
B For conservation of biodiversity and sustainable utilization of its benefits
C To assess threat posed to native species by invasive weed species
D For immediate steps to discontinue use of CFCs that were damaging the ozone layer

The correct answer is:
For conservation of biodiversity and sustainable utilization of its benefits

Solution:
In 1992, leaders from more than 100 countries participated in the first International Earth Summit at Rio de Janeiro in Brazil. The major objective was that every country should portray its own Agenda 21 for attaining sustainable development. The other objectives of the summit are (i) To achieve global sustainable development in the 21st century (ii) To conserve biological diversity (iii) To eliminate diseases with the help of shared welfare, requirements and obligations amongst various nations (iv) To prevent poverty and environmental damage.

Topic: Biology, Biodiversity and its Conservation, NEET, NEET 2019 (All India)
The ease of dehydrohalogenation of alkyl halide with alcoholic KOH is

Options:
A 3° < 2° < 1°
B 3Â° > 2Â° > 1Â°
C 3° < 2° > 1°
D 3° > 2° < 1°

The correct answer is:
3Â° > 2Â° > 1Â°

Solution:
Such dehydrohalogenation follow E2 mechanism. The driving force of such reaction is the stability of alkene produced. Since, tertiary alkyl halide can give more substituted alkene, it reacts fastest followed by secondary and primary i.e., 3Â°>2Â°>1Â°.

Topic: Chemistry, General Organic Chemistry, NEET
The ease of dehydrohalogenation of alkyl halide with alcoholic \(\mathrm{KOH}\) is

Options:
A \(3^{\circ} < 2^{\circ} < 1^{\circ}\)
B \(3^{\circ} > 2^{\circ} > 1^{\circ}\)
C \(3^{\circ} < 2^{\circ} > 1^{\circ}\)
D \(3^{\circ} > 2^{\circ} < 1^{\circ}\)

The correct answer is:
\(3^{\circ} > 2^{\circ} > 1^{\circ}\)

Solution:
Hints : Such dehydrohalogenation follows \(\mathrm{E}_2\) mechanism. The driving force of such reactions is the stability of alkene produced. Since tetriary alkyl halide can give more substituted alkene, it reacts fastest followed by secondary and primary i.e. \(3^{\circ} > 2^{\circ} > 1^{\circ}\).

Topic: Chemistry, Hydrocarbons, WBJEE, WBJEE 2011
The eccentricity e of an ellipse satisfies the condition

Options:
A e < 0
B $0 < \mathrm{e} < 1$
C e=1
D e>1

The correct answer is:
$0 < \mathrm{e} < 1$

Solution:
The eccentricity e of an ellipse satisties the condition $0 < \mathrm{e} < 1$

Topic: Mathematics, Ellipse, NDA, NDA 2012 (Phase 2)