datavorous/entrance-exam-dataset huggingface dataset:
The equation of a plane through the line of intersection of the planes $x+2 y=3, y-2 \mathrm{z}+1=0$, and perpendicular to the first plane is :

Options:
A
$2 x-y-10 z=9$

B
$2 x-y+7 z=11$

C
$2 x-y+10 z=11$

D
$2 x-y-9 z=10$

The correct answer is:
$2 x-y+10 z=11$

Solution:
Equation of a plane through the line of intersection of the planes $x+2 y=3, y-2 z+1=0$ is $(x+2 y-3)+\lambda(y-2 z+1)=0$
$$
\Rightarrow x+(2+\lambda) y-2 \lambda(z)-3+\lambda=0
$$
Now, plane (i) is $\perp$ to $x+2 y=3$
$\therefore \quad$ Their dot product is zero
i.e. $1+2(2+\lambda)=0 \Rightarrow \lambda=-\frac{5}{2}$
Thus, required plane is
$$
\begin{aligned}
& x+\left(2-\frac{5}{2}\right) y-2 \times \frac{-5}{2}(z)-3-\frac{5}{2}=0 \\
\Rightarrow & x-\frac{y}{2}+5 z-\frac{11}{2}=0 \\
\Rightarrow & 2 x-y+10 z-11=0
\end{aligned}
$$

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2013 (25 Apr Online)
The equation of a progressive wave can be given by $y=15 \sin (660 \pi t-0.02 \pi x) \mathrm{cm}$. The frequency of the wave is

Options:
A $330 \mathrm{~Hz}$
B $342 \mathrm{~Hz}$
C $365 \mathrm{~Hz}$
D $660 \mathrm{~Hz}$

The correct answer is:
$330 \mathrm{~Hz}$

Solution:
Given, equation of progressive wave,
$$
y=15 \sin (660 \pi t-0.02 \pi x)
$$
Comparing this equation with standard equation of progressive wave as $y=A \sin \left(\frac{2 \pi}{T}, t-\frac{2 \pi}{\lambda}, x\right)$
We get,
$\begin{aligned} \frac{2 \pi}{T} &=660 \pi \\ \Rightarrow \quad T &=\frac{2 \pi}{660 \pi} \\ \Rightarrow \quad T &=\frac{1}{330} \\ \therefore \text { Frequency of the wave, } \\ v &=\frac{1}{T} \\ &=\frac{1}{1 / 330}=330 \mathrm{~Hz} \end{aligned}$

Topic: Physics, Waves and Sound, JEE Main
The equation of a progressive wave is $y=4 \sin (4 \pi t-0.04 x+\pi / 3)$ where $x$ is in meter and $t$ is in second. The velocity of the wave is

Options:
A $100 \pi \mathrm{m} / \mathrm{s}$
B $50 \pi \mathrm{m} / \mathrm{s}$
C $25 \pi \mathrm{m} / \mathrm{s}$
D $\pi \mathrm{m} / \mathrm{s}$

The correct answer is:
$100 \pi \mathrm{m} / \mathrm{s}$

Solution:
Hints : Velocity of wave $=\frac{\omega}{\mathrm{K}}=\frac{4 \pi}{0.04}=100 \pi \mathrm{m} / \mathrm{sec}$

Topic: Physics, Waves and Sound, JEE Main
The equation of a simple harmonic progressive wave is given by y=Asinâ¡100Ït-3x. Find the distance between 2 particles having a phase difference of Ï3.

Options:
A Ï9 m
B Ï18Â m
C Ï6Â m
D Ï3Â m

The correct answer is:
Ï9 m

Solution:
Given, y=Asinâ¡100 Ït-3x
The general equation,
y=Asinâ¡Ït-kx
â´ k=3
Phase difference, Ï=Ï3
Kx =Ï3
x= π 3k
x= π 9
Distance, x=Ï9 m

Topic: Physics, Waves and Sound, JEE Main
The equation of a simple harmonic wave is given by
$$
y=3 \sin \frac{\pi}{2}(50 t-x)
$$
where $x$ and $y$ are in metres and $t$ is in seconds. The ratio of maximum particle velocity to the wave velocity is

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

The correct answer is:
$\frac{3}{2} \pi$

Solution:
We know that
$$
\begin{aligned}
& v_{\text {max }}=a \omega \\
& v=n \lambda \\
& \therefore \quad \frac{v_{\max }}{v}=\frac{a \omega}{n \lambda} \\
& =\frac{a(2 \pi n)}{n \pi}=\frac{2 \pi a}{\lambda} \\
& =\frac{2 \pi a}{2 \pi / k} \\
& =k a=\frac{\pi}{2} \times 3 \\
& =\frac{3 \pi}{2} \\
&
\end{aligned}
$$

Topic: Physics, Oscillations, NEET, NEET 2012 (Mains)
The equation of a simple harmonic wave is given by $y=6 \sin 2 \pi(2 t-0 . l x)$, where $x$ and $y$ are in $\mathrm{mm}$ and $t$ is in seconds. The phase difference between two particles $2 \mathrm{~mm}$ apart at any instant is

Options:
A $18^{\circ}$
B $36^{\circ}$
C $54^{\circ}$
D $72^{\circ}$

The correct answer is:
$72^{\circ}$

Solution:
Equalion of simple harmonic wave is given as
$$
y=6 \sin 2 \pi(2 t-0.1 x)
$$
Path difference, $\Delta x=2 \mathrm{~mm}$
Comparing with standard equation of wave,
$$
\begin{aligned}
y &=A \sin (\omega t-k x) \\
&=A \sin 2 \pi\left(v t-\frac{k}{2 \pi} \cdot x\right) \\
\frac{k}{2 \pi} &=0.1 \\
k &=0.2 \pi
\end{aligned}
$$
Wavelength, $\lambda=\frac{2 \pi}{k}=\frac{2 \pi}{0.2 \pi}=10 \mathrm{~mm}$
$\therefore$ We know that, phase difference,
$$
\begin{aligned}
\phi &=\frac{2 \pi}{\lambda} \times \text { path difference }(\Delta x) \\
&=\frac{2 \pi}{10} \times 2=\frac{2 \pi}{5}=\frac{2 \times 180}{5}=72^{\circ}
\end{aligned}
$$

Topic: Physics, Waves and Sound, JEE Main
The equation of a state of a gas is given by \( P(V-b)=n R T \). If \( 1 \) mole of a gas is isothermally expanded from volume \( V \) and \( 2 V \), the work done during the process is

Options:
A \( R T \ln \left|\frac{2 V-b}{V-b}\right| \)
B \( R T \ln \left|\frac{V-b}{V}\right| \)
C \( R T \ln \left|\frac{V-b}{2 V-b}\right| \)
D \( R T \ln \left|\frac{V}{V-b}\right| \)

The correct answer is:
\( R T \ln \left|\frac{2 V-b}{V-b}\right| \)

Solution:
Given:

The equation of a state of a gas,

P(V-b)=nRT

⇒P=nRT(V-b)

Number of moles, n =1 and change in volume, ∆V=2V-V=V

We know that,

Work done, W=∫P dV

From the given equation of state of a gas,

W=∫V2VnRT(V-b) dV

⇒W=nRT∫V2VdV(V-b)

⇒W=nRT ln2V-bV-b

Given that number of moles, n=1.

Therefore,

W=RT ln2V-bV-b

Topic: Physics, Thermodynamics, JEE Main
The equation of a stationary wave in a metal rod is given by y=0.002 sinÏx3 sin1000t, where x is in cm and t is in second. The maximum tensile stress at a point x=1 cm is nÏ3Ã108 dyne cm-2. What is the value of n? [Young's modulus of the material of rod is = 8Ã1011 dyne cm-2]

Options:

The correct answer is:
8

Solution:
y=0.002 sinπx3 sin1000t
dydx=0.002π3 cosπx3 sin1000t
∵ Tensile stress =Ydydx
The maximum value of the stress at x=1 cm
=Y×0.002π3 cosπ3
=8×1011×0.002×π3 cosπ3
=8π3×108 dyne cm-2

⇒n=8

Topic: Physics, Waves and Sound, JEE Main
The equation of a stationary wave is
$y=0.8 \cos \left(\frac{\pi x}{20}\right) \sin 200 \pi t$ , where $x$ is in $\mathrm{cm}$ and $t$ is in sec.
The separation between consecutive nodes will be

Options:
A 20 cm
B 10 cm
C 40 cm
D 30 cm

The correct answer is:
20 cm

Solution:
On comparing the given equation with standard equation $y=2 a \sin \frac{2 \pi x}{\lambda} \cos \frac{2 \pi v t}{\lambda}$
We get $\frac{2 \pi}{\lambda}=\frac{\pi}{20} \Rightarrow \lambda=40$
Separation between two consecutive nodes $=\frac{\lambda}{2}=\frac{40}{2}=20 \mathrm{~cm}$

Topic: Physics, Waves and Sound, JEE Main
The equation of a straight line passing through $(-3,2)$ and cutting an intercept equal in magnitude but opposite in sign from the axes is given by

Options:
A $x-y+5=0$
B $x+y-5=0$
C $x-y-5=0$
D $x+y+5=0$

The correct answer is:
$x-y+5=0$

Solution:
Let the equation be $\frac{x}{a}+\frac{y}{-a}=1 \Rightarrow x-y=a$
But it passes through $(-3,2)$, hence $a=-3-2=-5$. Hence the equation is $x-y+5=0$

Topic: Mathematics, Straight Lines, JEE Main
The equation of a straight line passing through the point 3,6 and normal to the curve y=x is

Options:
A 4x+y-18=0
B x+y-9=0
C 4x-y-6=0
D x-y+3=0

The correct answer is:
4x+y-18=0

Solution:
Normal to the parabola y2=x is
y=mx-m2-m34

Since, it passes through the point 3,6

âm3-10m+24=0; m=-4 is a root
Hence, the required equation is 4x+y-18=0

Topic: Mathematics, Parabola, JEE Main
The equation of a straight line passing through the points $(-5,-6)$ and $(3,10)$, is

Options:
A $x-2 y=4$
B $2 x-y+4=0$
C $2 x+y=4$
D None of these

The correct answer is:
$2 x-y+4=0$

Solution:
Slope $=\frac{10+6}{3+5}=2$. Hence the line is $2 x-y=-4$.

Topic: Mathematics, Straight Lines, JEE Main
The equation of a tangent to the hyperbola, 4x2-5y2=20, parallel to the line x-y=2, is

Options:
A x-y+7=0
B x-y-3=0
C x-y+1=0
D x-y+9=0

The correct answer is:
x-y+1=0

Solution:
The slope of a line ax+by+c=0 is m=-ab.

Thus, the slope of the line x-y=2 is m=-1-1=1 and we know that if two lines are parallel then their slopes are equal, hence the slope of the tangent is also m=1.

The given hyperbola is x25-y24=1

The equation of the tangent to the hyperbola x2a2-y2b2=1 having slope m is y=mx±a2m2-b2.

Hence, the equation of its tangent having slope 1 is y=x±5×12-4

⇒y=x±1

Thus, the tangents are x-y+1=0 & x-y-1=0.

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2019 (10 Jan Shift 1)
The equation of a tangent to the parabola, x2=8y, which makes an angle θ with the positive direction of x- axis, is

Options:
A y=x tanÎ¸+2 cotÎ¸
B y=x tanÎ¸-2 cotÎ¸
C x=y cotÎ¸+2 tanÎ¸
D x=y cotÎ¸-2 tanÎ¸

The correct answer is:
x=y cotÎ¸+2 tanÎ¸

Solution:
As we know, if the line y=mx+c is a tangent to x2=8y=4×2×y, then c=-2m2.

So, equation of tangent with slope m is

y=mx-2m2 ...i

Since, tangent is inclined at θ with the positive x- axis, i.e. m=tan⁡θ.

Equation of tangent is y=tanθ.x-2tan2⁡θ

⇒ycot⁡θ=x-2 tanθ

⇒x=y cotθ+2 tanθ

Topic: Mathematics, Parabola, JEE Main, JEE Main 2019 (12 Jan Shift 2)
The equation of a tangent to the parabola $y^2=4 a x$ making an angle $\theta$ with $x$-axis is

Options:
A $y=x \cot \theta+a \tan \theta$
B $x=y \tan \theta+a \cot \theta$
C $y=x \tan \theta+a \cot \theta$
D None of these

The correct answer is:
$y=x \tan \theta+a \cot \theta$

Solution:
$m=\tan \theta$. The tangent to $y^2=4 a x$ is $y=x \tan \theta+c$
Hence $\quad c=\frac{a}{\tan \theta}=a \cot \theta$
$\therefore$ The equation of tangent is $y=x \tan \theta+a \cot \theta$

Topic: Mathematics, Parabola, JEE Main
The equation of a tangent to the parabola $y^2=8 x$ is $y=x+2$. The point on this line from which the other tangent to the parabola is perpendicular to the given tangent is

Options:
A
$(-1,1)$

B
$(0,2)$

C
$(2,4)$

D
$(-2,0)$

The correct answer is:
$(-2,0)$

Solution:
Point must be on the directrix of the parabola.
Hence the point is $(-2,0)$.

Topic: Mathematics, Parabola, JEE Main, JEE Main 2007
The equation of a wave is given by $Y=A \sin \omega\left(\frac{x}{v}-k\right)$ where $\omega$ is the angular velocity and $v$ is the linear velocity. The dimension of $k$ is

Options:
A $LT$
B $T$
C $T^{-1}$
D $T^2$

The correct answer is:
$T$

Solution:
From the principle of homogenity $\left(\frac{x}{v}\right)$ has dimensions of T.

Topic: Physics, Units and Dimensions, JEE Main
The equation of a wave is $y=2 \sin \pi(0.5 x-200 t)$, where $x$ and $y$ are expressed in $\mathrm{cm}$ and $t$ in sec.
The wave velocity is

Options:
A $100 \mathrm{~cm} / \mathrm{sec}$
B $200 \mathrm{~cm} / \mathrm{sec}$
C $300 \mathrm{~cm} / \mathrm{sec}$
D $400 \mathrm{~cm} / \mathrm{sec}$

The correct answer is:
$400 \mathrm{~cm} / \mathrm{sec}$

Solution:
Cemparing given equation with standard equation of progressive wave. The velocity of wave
$v=\frac{\omega(\text { Co }- \text { efficientof } t)}{k(\text { Co }- \text { efficientof } x)}=\frac{200 \pi}{0.5 \pi}=400 \mathrm{~cm} / \mathrm{s}$

Topic: Physics, Waves and Sound, JEE Main
The equation of a wave on a string of linear mass density 0.04 kg m-1 is given by
y=0.02msinâ¡2Ït0.04(s)-x0.50(m)
The tension in the string (in N) is:

Options:

The correct answer is:
6.25

Solution:
From the given equation v=Î»T=0.500.04
Now v=TÎ¼ âT=v2Î¼
=12.52Ã0.04
=6.25 N

Topic: Physics, Waves and Sound, JEE Main
The equation of a wave on a string of linear mass density $0.04 \mathrm{~kg} \mathrm{~m}^{-1}$ is given by $y=0.02(\mathrm{~m}) \sin \left[2 \pi\left(\frac{\mathrm{t}}{0.04(\mathrm{~s})}-\frac{\mathrm{x}}{0.50(\mathrm{~m})}\right)\right]$. The tension in the string is

Options:
A
$4.0 \mathrm{~N}$

B
$12.5 \mathrm{~N}$

C
$0.5 \mathrm{~N}$

D
$6.25 \mathrm{~N}$

The correct answer is:
$6.25 \mathrm{~N}$

Solution:
$\mathrm{T}=\mu \mathrm{v}^2=\mu \frac{\omega^2}{\mathrm{k}^2}=0.04 \frac{(2 \pi / 0.004)^2}{(2 \pi / 0.50)^2}=6.25 \mathrm{~N}$

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2010
The equation of an ex-circle of a triangle formed by the common tangents to the circle x2+y2=4 and x2+y2-6x+8=0 is

Options:
A x2+y2=4
B x2+y2-6x+8=0
C x2+y2-6x+9=0
D x2+y2=1

The correct answer is:
x2+y2=4

Solution:
For x2+y2=22, the centre is 0,0 and radius is 2 units

and for x2+y2-6x+8=0, the centre is 3,0 and radius is 1 unit
Now, note that the circles touch each other externally, hence the larger circle is an excircle of the triangle formed by the common tangents.

Topic: Mathematics, Circle, JEE Main
The equation of auxiliary circle of the ellipse $16 x^{2}+25 y^{2}+32 x-100 y=284$ is

Options:
A $x^{2}+y^{2}+2 x-4 y-20=0$
B $x^{2}+y^{2}+2 x-4 y=0$
C $(x+1)^{2}+(y-2)^{2}=400$
D $(x+1)^{2}+(y-2)^{2}=225$

The correct answer is:
$x^{2}+y^{2}+2 x-4 y-20=0$

Solution:
Given equation of ellipse is
$$
16 x^{2}+25 y^{2}+32 x-100 y=284
$$
Simplifying the given equation, we have ellipse as
$$
\frac{(x+1)^{2}}{25}+\frac{(y-2)^{2}}{16}=1
$$
So, the auxiliary equation of circle is
$$
\begin{array}{r}
(x+1)^{2}+(y-2)^{2}=25 \\
\Rightarrow \quad x^{2}+2 x+1+y^{2}-4 y+4=25
\end{array}
$$
$\therefore$
$x^{2}+y^{2}+2 x-4 y-20=0$

Topic: Mathematics, Ellipse, JEE Main
The equation of current in a purely inductive circuit is 5sin49πt-30°. If the inductance is 30mH then the equation for the voltage across the inductor, will be

Options:
A 1.47sin49πt-30°
B 1.47sin49πt+60°
C 23.1sin49πt-30°
D 23.1sin49πt+60°

The correct answer is:
23.1sin49πt+60°

Solution:
In inductive circuit, the voltage leads the current by 90°. Therefore,

ϕ0V-ϕ0I=90°⇒ϕ0V=60°

From the given equation 5sin49πt-30°, angular frequency ω=49π rad s-1

The amplitude of the voltage across the inductor will be,

V0=i0XL

⇒V0=i0ωL

⇒V0=549π30×10-3=23.1

⇒V=23.1sin49πt+60°

Topic: Physics, Alternating Current, JEE Main
The equation of current in a purely inductive circuit is 5sin49πt-30°. If the inductance is 30mH then the equation for the voltage across the inductor, will be

Options:
A 1.47sin49πt-30°
B 1.47sin49πt+60°
C 23.1sin49πt-30°
D 23.1sin49πt+60°

The correct answer is:
23.1sin49πt+60°

Solution:
In inductive circuit, the voltage leads the current by 90°. Therefore,

ϕ0V-ϕ0I=90°⇒ϕ0V=60°

From the given equation 5sin49πt-30°, angular frequency ω=49π rad s-1

The amplitude of the voltage across the inductor will be,

V0=i0XL

⇒V0=i0ωL

⇒V0=549π30×10-3=23.1

⇒V=23.1sin49πt+60°

Topic: Physics, Alternating Current, JEE Main, JEE Main 2022 (28 Jul Shift 1)
The equation of incircle of the triangle formed by common tangents to the circles x2+y2=4 and x2+y2-6x+8=0 is

Options:
A x2+y2=1
B x2+y2-6x+8=0
C x2+y2-6x+9=0
D x2-y2=1

The correct answer is:
x2+y2-6x+8=0

Solution:
Note that circles touch each other externally, hence the smaller circle is incircle of the triangle formed by common tangents.

Topic: Mathematics, Circle, JEE Main
The equation of latus rectum of a parabola is $x+y=8$ and the equation of the tangent at the vertex is $x+y=12$, then length of the latus rectum is

Options:
A $4 \sqrt{2}$
B $2 \sqrt{2}$
C $8$
D $8 \sqrt{2}$

The correct answer is:
$8 \sqrt{2}$

Solution:
Clearly, $a=\left|\frac{-8}{\sqrt{1+1}}\right|-\left|\frac{-12}{\sqrt{1+1}}\right|=\frac{4}{\sqrt{2}}$
Length of latus rectum $=4 a \quad=4 \times \frac{4}{\sqrt{2}}=8 \sqrt{2}$

Topic: Mathematics, Parabola, JEE Main
The equation of line passing through ( 3,−1,2 ) and perpendicular to the lines r → =( i ^ + j ^ − k ^ )+λ( 2 i ^ −2 j ^ + k ^ ) and r → =( 2 i ^ + j ^ −3 k ^ )+μ( i ^ −2 j ^ +2 k ^ ) is

Options:
A x+32=y+13=z-22
B x-33=y+12=z-22
C x-32=y+13=z-22
D x-32=y+12=z-23

The correct answer is:
x-32=y+13=z-22

Solution:
A vector perpendicular to given lines is
â´| i ^ j ^ k ^ 2 â2 1 1 â2 2 |=â2 i ^ â3 j ^ â2 k ^
â´ drs of line
a=â2,ââb=â3,ââc=â2,
Or a=2,ââb=3,ââc=2,
Equation of line is x â 3 2 = y + 1 3 = z â 2 2

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The equation of line, which bisect the line joining two points $(2,-19)$ and $(6,1)$ and perpendicular to the line joining two points $(-1,3)$ and $(5,-1)$, is

Options:
A $3 x-2 y=30$
B $2 x-y-3=0$
C $2 x+3 y=20$
D None of these

The correct answer is:
$3 x-2 y=30$

Solution:
$\text { Mid point } \equiv(4,-9) \text {; Slope } \frac{\frac{-1}{3+1}}{\frac{-1-5}{-1-5}}=\frac{3}{2}$
Hence the required line is $3 x-2 y=30$

Topic: Mathematics, Straight Lines, JEE Main
The equation of line whose mid point is $\left(x_1, y_1\right)$ in between the axes, is

Options:
A $\frac{x}{x_1}+\frac{y}{y_1}=2$
B $\frac{x}{x_1}+\frac{y}{y_1}=\frac{1}{2}$
C $\frac{x}{x_1}+\frac{y}{y_1}=1$
D None of these

The correct answer is:
$\frac{x}{x_1}+\frac{y}{y_1}=2$

Solution:
Intersection point on $x$-axis is $\left(2 x_1, 0\right)$ and on $y$-axis is $\left(0,2 y_1\right)$. Thus equation of line passes through these points is
$\frac{x}{x_1}+\frac{y}{y_1}=2$

Topic: Mathematics, Straight Lines, JEE Main
The equation of motion of a body in S.H.M. is \( x=4 \sin \left(\pi t+\frac{\pi}{3}\right) . \) The acceleration, in \( \mathrm{cm} \mathrm{s}^{-2} \) at the end of \( 4 \mathrm{~s} \) will be

Options:
A \( \frac{3}{2} \pi^{2} \)
B \( \frac{2}{3} \pi^{2} \)
C \( -\frac{\sqrt{2}}{3} \pi^{2} \)
D \( -2 \sqrt{3} \pi^{2} \)

The correct answer is:
\( -2 \sqrt{3} \pi^{2} \)

Solution:
Given, x=4sinπ×t+π3, now compare the equation of S.H.M. of displacement, x=Asinωt+ϕ, it means A=4 cm and ω=π,

Now recall the formula of acceleration in terms of time,

a=-xω2=-4π2sinπ×4+π3=-4π2sinπ×4cosπ3+cos4πsinπ3=-4π20+32=-23π2 cm s-2

Topic: Physics, Oscillations, JEE Main
The equation of motion of a particle executing simple harmonic motion is given by $x=3 \sin \left(6 t+\frac{\pi}{6}\right)$, where $x$ is in metres and $t$ is in seconds. The ratio of the potential and kinetic energies of the particle at time $t=0$ is

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

The correct answer is:
$1: 3$

Solution:
Equation of motion is given as
$x=3 \sin \left(6 t+\frac{\pi}{6}\right), A=3$
At time $\mathrm{t}=0$
$x=3 \sin \left(\frac{\pi}{6}\right)=\frac{3}{2}=1.5$
Potential Energy of SHM is given as:
$\mathrm{V}=\frac{1}{2} \mathrm{kx}^2$ $\ldots$ (1)
Kinetic Energy of SHM is given as:
$K=\frac{1}{2} K\left(A^2-x^2\right)$ ...(2)
Dividing Equation (1) and (2)
$\begin{aligned} & \frac{V}{K}=\frac{1}{2} \frac{K x^2}{K\left(A^2-x^2\right)} \times 2 \\ & =\frac{x^2}{A^2-x^2}\end{aligned}$
Substitute, $x=1.5$ and $A=3$
$\begin{aligned} & \frac{\mathrm{V}}{\mathrm{K}}=\frac{(1.5)^2}{(3)^2-(1.5)^2}=\frac{2.25}{9-2.25} \\ & =\frac{225}{675}=\frac{45}{135}=\frac{9}{27}=\frac{1}{3} \\ & \frac{\mathrm{V}}{\mathrm{K}}=\frac{1}{3}\end{aligned}$

Topic: Physics, Oscillations, JEE Main
The equation of motion of a projectile is y=12x-34x2, where x and y are in metre. What is the range (in m) of the projectile?

Options:

The correct answer is:
16

Solution:
y = 12 x - 3 4 x 2

Range is the point where y = 0
thus 0 = 12 x - 3 4 x 2
â x = 0
or x = 16 m

Topic: Physics, Motion In Two Dimensions, JEE Main
The equation of normal at the point $(0,3)$ of the ellipse $9 x^2+5 y^2=45$ is

Options:
A $y-3=0$
B $y+3=0$
C $x$-axis
D $y$-axis

The correct answer is:
$y$-axis

Solution:
$9 x^2+5 y^2=45$
Diff. w.r.t ' $x$ ', we have
$\begin{aligned} & 18 x+10 y \frac{d y}{d y}=0 \\ & \Rightarrow \frac{d y}{d x}=-\frac{18 x}{10 y}=-\frac{9 x}{5 y} \\ & \left.\therefore \frac{d y}{d x}\right|_{(0,3)}=0\end{aligned}$
$\therefore$ Equation of normal at $(0,3)$ is
$\begin{aligned} & y-3=\frac{\frac{-1}{2}}{\left.x\right|_{(0,1)}} 0(x-0) \\ \Rightarrow & 0 \times(y-3)=-(x-0) \\ \Rightarrow & x=0 \\ \Rightarrow & y \text {-axis }\end{aligned}$

Topic: Mathematics, Ellipse, JEE Main
The equation of normal to the curve $y=(1+x)^{y}+\sin ^{-1}\left(\sin ^{2} x\right)$ at $x=0$ is

Options:
A $x+y=1$
B $x-y=1$
C $x+y=-1$
D $x-y=-1$

The correct answer is:
$x+y=1$

Solution:
Given curve is
$y=(1+x)^{y}+\sin ^{-1}\left(\sin ^{2} x\right)$
On differentiating w.r.t $x$, we get
$\begin{array}{l}
\frac{\mathrm{dy}}{\mathrm{dx}}=(1+\mathrm{x})^{\mathrm{y}}\left[\frac{\mathrm{y}}{1+\mathrm{x}}+\log (1+\mathrm{x}) \frac{\mathrm{dy}}{\mathrm{dx}}\right] \\
\Rightarrow\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\mathrm{at}(0,1)}=\frac{2 \sin \mathrm{x} \cos \mathrm{x}}{\sqrt{1-\sin ^{4} \mathrm{x}}}
\end{array}$
Slope of normal at $(x=0)=-1$
$\therefore \quad$ Equation of normal at $x=0$ and $y=1$ is $\begin{aligned} y-1 &=-1(x-0) \\ \Rightarrow y-1 &=-x \Rightarrow x+y=1 \end{aligned}$

Topic: Mathematics, Differentiation, JEE Main
The equation of normal to the curve $\mathrm{y}=\sin \left(\frac{\pi x}{4}\right)$ at the point $(2,5)$ is

Options:
A x+y=5
B y=5
C x=2
D x+y=2

The correct answer is:
x=2

Solution:
(D)
we have, $y=\sin \frac{\pi x}{4}$
$\begin{array}{l}
\frac{d y}{d x}=\frac{\pi}{4} \cdot \cos \frac{\pi x}{4} \\
\left(\frac{d y}{d x}\right)_{(2,5)}=\frac{\pi}{4} \cos \frac{2 \pi}{4}=0
\end{array}$
Since slope of tangent is zero, it is parallel to $\mathrm{X}$ axis. So normal is parallel to $\mathrm{Y}$ axis. Hence required equation of normal is $\mathrm{x}=2$

Topic: Mathematics, Application of Derivatives, JEE Main
The equation of normal to the parabola at the point $\left(\frac{a}{m^2}, \frac{2 a}{m}\right)$,is

Options:
A $y=m^2 x-2 m x-a m^3$
B $m^3 y=m^2 x-2 a m^2-a$
C $m^3 y=2 a m^2-m^2 x+a$
D None of these

The correct answer is:
$m^3 y=2 a m^2-m^2 x+a$

Solution:
$y-\frac{2 a}{m}=-\frac{2 a / m}{2 a}\left(x-\frac{a}{m^2}\right)$
$\Rightarrow y-\frac{2 a}{m}=\frac{-1}{m}\left(x-\frac{a}{m^2}\right)$
$\Rightarrow m^3 y+m^2 x-2 a m^2-a=0$

Topic: Mathematics, Parabola, JEE Main
The equation of one of the curves whose slope at any point is equal to $\mathrm{y}+2 \mathrm{x}$ is

Options:
A $y=2\left(e^x+x-1\right)$
B $y=2\left(e^x-x-1\right)$
C $y=2\left(e^x-x+1\right)$
D $y=2\left(e^x+x+1\right)$

The correct answer is:
$y=2\left(e^x-x-1\right)$

Solution:
$$
\begin{aligned}
& \text { Hints: } \frac{d y}{d x}=y+2 x \quad \text { Put } y+2 x=z \Rightarrow \frac{d y}{d x}+z=\frac{d z}{d x} \\
& \frac{d z}{d x}-2=z, \quad \frac{d z}{d x}=z+2 \Rightarrow \int \frac{d z}{z+2}=\int d x \\
& \log (z+2)=x+c, \log (y+2 x+2)=x+c \\
& y+2 x+2=x+c, y=2\left(e^x-x-1\right)
\end{aligned}
$$

Topic: Mathematics, Differential Equations, JEE Main
The equation of one of the straight lines which passes through the point (1,3) and makes an angles tan-1(2) with the straight line, y+1=32x is

Options:
A 42x+5y-(15+42)=0
B 52x+4y-(15+42)=0
C 42x+5y-42=0
D 42x-5y-(5+42)=0

The correct answer is:
42x+5y-(15+42)=0

Solution:
y=mx+c

3=m+c

2=m-321+32m

⇒6m+2=±m-32

from Positive sign

⇒5m=-42 ⇒m=-425

from negative sign

⇒6m+2=-m+32

⇒7 m=22⇒m=227

According to options take m=-425

So y=-42x5+3+425

42x+5y-(15+42)=0

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2021 (18 Mar Shift 1)
The equation of parabola which cuts the parabola y2=4bx orthogonally having the same axis and length of latus rectum as 8b, is

Options:
A y2=-8bx-8b
B y2=-8bx-4b
C y2=-8bx-32b
D y2=-8bx-3b

The correct answer is:
y2=-8bx-3b

Solution:
Focus of parabola y2=4bx is b,0
Focus of parabola in option A is 6b,0
Focus of parabola in option B is 2b,0
Focus of parabola in option C is 30b,0
Focus of parabola in option D is b,0
Hence, option D is correct

Topic: Mathematics, Parabola, JEE Main
The equation of plane containing line x-y=1,z=1 and parallel to x2-z3=1,y=3 is

Options:
A 3x+3y-2z=1
B 3x-3y-2z=1
C 3x+3y+2z=1
D 3x+3y+2z=-1

The correct answer is:
3x-3y-2z=1

Solution:
Equation of plane containing line x-y-1=0 and z-1=0 is (x-y-1)+λ(z-1)=0

⇒ x-y+λz-1-λ=0 ...(i)

Since, this plane is parallel to line x2-z3=1 and y=3

x-22=y-30=z3 ...(ii)

∴ 2(1)+(0)(-1)+3(λ)=0

⇒ λ=-2/3

Putting the value of λ in Eq. (i), we get

x-y-23z-1+23=0⇒3x-3y-2z=1

Topic: Mathematics, Three Dimensional Geometry, JEE Main