datavorous/entrance-exam-dataset huggingface dataset:
Equation of a progressive wave is given by
$y=4 \sin \left\{\pi\left(\frac{t}{5}-\frac{x}{9}\right)+\frac{\pi}{6}\right\}$
Then which of the following is correct

Options:
A $v=5 \mathrm{~m} / \mathrm{sec}$
B $\lambda=18 \mathrm{~m}$
C $a=0.04 \mathrm{~cm}$
D $n=50 \mathrm{~Hz}$

The correct answer is:
$\lambda=18 \mathrm{~m}$

Solution:
$\begin{aligned} & \text { From the given equation amplitude } a=0.04 \mathrm{~m} \\ & \text { Frequency } \\ & \qquad \frac{\text { Co-efficientof } t}{2 \pi}=\frac{\pi / 5}{2 \pi}=\frac{1}{10} \mathrm{~Hz} \\ & \text { Wave length } \quad \lambda=\frac{2 \pi}{\text { Co-efficientof } x} \quad=\frac{2 \pi}{\pi / 9}=18 \mathrm{~m} \text {. } \\ & \text { Wave speed } \quad v=\frac{\text { Co-efficientof } t}{\text { Co-efficientof } x} \quad=\frac{\pi / 5}{\pi / 9}=1.8 \mathrm{~m} / \mathrm{s} .\end{aligned}$

Topic: Physics, Waves and Sound, JEE Main
Equation of angle bisector between the lines \( 3 x+4 y-7=0 \) and \( 12 x+5 y+17=0 \) are

Options:
A \( \frac{3 x+4 y-7}{\sqrt{25}}=\pm \frac{12 x+5 y+17}{\sqrt{169}} \)
B \( \frac{3 x+4 y+7}{\sqrt{25}}=\frac{12 x+5 y+17}{\sqrt{169}} \)
C \( \frac{3 x+4 y+7}{\sqrt{25}}=\pm \frac{12 x+5 y+17}{\sqrt{169}} \)
D None of these

The correct answer is:
\( \frac{3 x+4 y-7}{\sqrt{25}}=\pm \frac{12 x+5 y+17}{\sqrt{169}} \)

Solution:
By direct formula
a1x+b1y+c1a12+b12= ±a2x+b2y+c2a22+b22
3x+4y-732+42= ±12x+5y+17122+52
3x+4y-75= ±12x+5y+1713 .

Topic: Mathematics, Straight Lines, JEE Main
Equation of any normal to the parabola $y^2=4 a(x-a)$ is

Options:
A $y=m x-2 a m-a m^3$
B $y=m(x+a)-2 a m-a m^3$
C $y=m(x-a)+\frac{a}{m}$
D $y=m(x-a)-2 a m-a m^3$

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

Solution:
Let normal at $(h, k)$ be $y=m x+c$ then, $k=m h+c$ also $k^2=4 a(h-a)$
slope of tangent at $(h, k)$ is $m_1$ then on differentiating equation of parabola.
$2 y m_1=4 a \Rightarrow m_1=\frac{2 a}{k}$ also $m m_1=-1$
$\Rightarrow \quad m=-\frac{k}{2 a}$, solving and replacing $(h, k)$ by $(x, y)$
$\Rightarrow y=m(x-a)-2 a m-a m^3$

Topic: Mathematics, Parabola, JEE Main
Equation of curve passing through $(3,9)$ which satisfies the differential equation $\frac{d y}{d x}=x+\frac{1}{x^2}$, is

Options:
A $6 x y=3 x^2-6 x+29$
B $6 x y=3 x^3-29 x+6$
C $6 x y=3 x^3+29 x-6$
D None of these

The correct answer is:
$6 x y=3 x^3+29 x-6$

Solution:
$\begin{aligned} & \frac{d y}{d x}=x+\frac{1}{x^2} \\ & \Rightarrow \int d y=\int\left(x+\frac{1}{x^2}\right) d x \Rightarrow y=\frac{x^2}{2}-\frac{1}{x}+c\end{aligned}$
Since it passes through $(3,9)$, therefore
$\begin{aligned} & 9=\frac{9}{2}-\frac{1}{3}+c \Rightarrow c=\frac{29}{6} \\ & y=\frac{x^2}{2}-\frac{1}{x}+\frac{29}{6} \Rightarrow 6 x y=3 x^3+29 x-6\end{aligned}$

Topic: Mathematics, Differential Equations, JEE Main
Equation of curve through point $(1,0)$ which satisfies the differential equation $\left(1+y^2\right) d x-x y d y=0$, is

Options:
A $x^2+y^2=1$
B $x^2-y^2=1$
C $2 x^2+y^2=2$
D None of these

The correct answer is:
$x^2+y^2=1$

Solution:
We have $\frac{d x}{x}=\frac{y d y}{1+y^2}$
Integrating, we get $\log |x|=\frac{1}{2} \log \left(1+y^2\right)+\log c$
or $|x|=c \sqrt{\left(1+y^2\right)}$
But it passes through $(1,0)$, so we get $c=1$
$\therefore$ Solution is $x^2=y^2+1$ or $x^2-y^2=1$

Topic: Mathematics, Differential Equations, JEE Main
Equation of diameter of parabola $y^2=x$ corresponding to the chord $x-y+1=0$ is

Options:
A $2 y=3$
B $2 y=1$
C $2 y=5$
D $y=1$

The correct answer is:
$2 y=1$

Solution:
Equation of diameter of parabola is $y=\frac{2 a}{m}$ Here $a=\frac{1}{4}, m=1 \Rightarrow y=\frac{2 \times 1 / 4}{1} \Rightarrow 2 y=1$

Topic: Mathematics, Parabola, JEE Main
Equation of one of the tangents passing through $(2,8)$ to the hyperbola $5 x^2-y^2=5$ is

Options:
A $3 x+y-14=0$
B $3 x-y+2=0$
C $x+y+3=0$
D $x-y+6=0$

The correct answer is:
$3 x-y+2=0$

Solution:
Given hyperbola is $5 x^2-y^2=5$ or It can be rewritten as
$\frac{x^2}{1}-\frac{y^2}{5}=1$
Here,
$\therefore$ Equation of tangent is
$\begin{aligned} & y=m x \pm \sqrt{a^2 m^2-b^2} \\ & y=m x \pm \sqrt{1 m^2-5}\end{aligned}$
But point $(2,8)$ lies on it.
$\begin{aligned} & \therefore \quad 8=2 m \pm \sqrt{m^2-5} \\ & \Rightarrow(8-2 m)= \pm \sqrt{m^2-5}\end{aligned}$
On squaring both sides, we get
$\begin{array}{rlrl} & 64+4 m^2-32 m & =m^2-5 \\ \Rightarrow \quad 3 m^2-32 m+69 & =0 \\ \Rightarrow \quad & (3 m-23)(m-3) & =0 \\ \Rightarrow & m & =3, \frac{23}{3}\end{array}$
On putting $m=3$ in Eq. (i), we get
$\begin{aligned} y & =3 x \pm \sqrt{3^2-5}=3 x \pm 2 \\ \Rightarrow & y=3 x+2 \text { and } y=3 x-2\end{aligned}$

Topic: Mathematics, Hyperbola, JEE Main
Equation of the chord of contact, drawn to the ellipse $4 x^2+9 y^2=36$ from the point $(m, n)$ where $m \cdot n=m+n$ and $m, n \in I^{+}$is

Options:
A $4 x+9 y=9$
B $2 x+2 y=1$
C $4 x+9 y=18$
D $4 x+9 y=36$

The correct answer is:
$4 x+9 y=18$

Solution:
The correct option is C $4 x+9 y=18$
Given that $m \cdot n=m+n$
$\begin{aligned}
& \Rightarrow m(n-1)=n \\
& \Rightarrow m=\frac{n}{n-1}
\end{aligned}$
As given that $n, m$ are integres. So, in above equation for being $m$ an integer $n$ must be divisible by $n-1$ that is possible only when $n=2$
$\therefore \mathrm{n}=2 \Rightarrow \mathrm{m}=2$
Hence, chord of contact drawn from $(2,2)$ to $\frac{x^2}{9}+\frac{y^2}{4}=1$ is,
$\begin{aligned}
& \Rightarrow \frac{2 x}{9}+\frac{2 y}{4}=1 \\
& \Rightarrow 4 x+9 y=18
\end{aligned}$

Topic: Mathematics, Ellipse, JEE Main
Equation of the ellipse whose axes are the axes of coordinates and which passes through the point $(-3,1)$ and has eccentricity $\sqrt{\frac{2}{5}}$ is

Options:
A
$5 x^2+3 y^2-48=0$

B
$3 x^2+5 y^2-15=0$

C
$5 x^2+3 y^2-32=0$

D
$3 x^2+5 y^2-32=0$

The correct answer is:
$3 x^2+5 y^2-32=0$

Solution:
$$
\begin{aligned}
& \mathrm{b}^2=\mathrm{a}^2\left(1-\mathrm{e}^2\right)=\mathrm{a}^2\left(1-\frac{2}{5}\right)=\mathrm{a}^2 \frac{3}{5}=\frac{3 \mathrm{a}^2}{5} \\
& \frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1 \Rightarrow \frac{9}{\mathrm{a}^2}+\frac{5}{3 \mathrm{a}^2}=1 \\
& \mathrm{a}^2=\frac{32}{3} \\
& \mathrm{~b}^2=\frac{32}{5}
\end{aligned}
$$
$\therefore$ Required equation of ellipse $3 x^2+5 y^2-32=0$

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2011
Equation of the ellipse with eccentricity ${ }^{\frac{1}{2}}$ and foci at $( \pm 1,0)$ is

Options:
A $\frac{x^2}{3}+\frac{y^2}{4}=1$
B $\frac{x^2}{4}+\frac{y^2}{3}=1$
C $\frac{x^2}{3}+\frac{y^2}{4}=\frac{4}{3}$
D None of these

The correct answer is:
$\frac{x^2}{4}+\frac{y^2}{3}=1$

Solution:
Given that, $e=\frac{1}{2}$ and $( \pm a e, 0)=( \pm 1,0)$
$\begin{aligned} & \Rightarrow a e=1 \Rightarrow a=2 . \text { Now } b^2=a^2\left(1-e^2\right) \\ & \Rightarrow b^2=4\left(1-\frac{1}{4}\right) \Rightarrow b^2=3\end{aligned}$
Hence, equation of ellipse is $\frac{x^2}{4}+\frac{y^2}{3}=1$.

Topic: Mathematics, Ellipse, JEE Main
Equation of the line of the shortest distance between the lines x⁡ 1 = y⁡ - 1 = z⁡ 1 and x⁡ - 1 0 = y⁡ + 1 - 2 = z⁡ 1 is

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

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

Solution:
The equation of a line of the shortest distance between the given lines will be along the perpendicular to both the lines.

A line perpendicular to L1: x1=y-1=z1 and L2: x-10=y+1-2=z1 is,

i^j^k^1-110-21=i^-1+2-j^1-0+k^-2+0

=i^−j^−2k^

Let, x1=y-1=z1=α

⇒x=α, y=-α, z=α

Thus, a point on the line L1 is P(α, −α, α)

Similarly, let x-10=y+1-2=z1=λ

⇒x=1, y=-2λ-1, z=λ

Thus, a point on the line L2 is Q(1, −1−2λ, λ)

Now, a vector joining the points P & Q is (α−1)i^+(2λ−α+1)j^+(α−λ)k^

Let, the vector joining the points P & Q is along perpendicular to the two lines.

Hence, (α−1)i^+(2λ−α+1)j^+(α−λ)k^ and i^−j^−2k^ are

proportional and hence on comparing, α-11=α-2λ-11=λ-α2

⇒α-1=α-2λ-1

⇒λ=0

Hence, point Q is 1, -1, 0

The equation of a line passing through a point x1, y1, z1 and parallel to a vector ai^+bj^+ck^ is x-x1a=y-y1b=z-z1c

So, the equation of required line is x−11=y+1−1=z−2.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2014 (19 Apr Online)
Equation of the line passing through $(-1,1)$ and perpendicular to the line $2 x+3 y+4=0$, is

Options:
A $2(y-1)=3(x+1)$
B $3(y-1)=-2(x+1)$
C $y-1=2(x+1)$
D $3(y-1)=x+1$

The correct answer is:
$2(y-1)=3(x+1)$

Solution:
The gradient of line $2 x+3 y+4=0$ is $\quad-\frac{2}{3}$. Now the equation of line passing through $(-1,1)$ is
$y-1=m(x+1) \text {, but } m=-\frac{1}{-2 / 3}=\frac{3}{2}$
Therefore, required equation is $2(y-1)=3(x+1)$

Topic: Mathematics, Straight Lines, JEE Main
Equation of the line passing through $(1,2)$ and parallel to the line $y=3 x-1$ is

Options:
A $y+2=x+1$
B $y+2=3(x+1)$
C $y-2=3(x-1)$
D $y-2=x-1$

The correct answer is:
$y-2=3(x-1)$

Solution:
The required equation which passes through $(1,2)$ and its gradient is $m=3$, is $(y-2)=3(x-1)$

Topic: Mathematics, Straight Lines, JEE Main
Equation of the line passing through the intersection of the plane \(x+2 y+3 z=4\) and the line \(\frac{x-1}{2}=\frac{y+1}{1}=\frac{z-1}{-1}\) and parallel to the vector \((2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}) \times(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}})\) is

Options:
A \(\frac{x-5}{3}=\frac{y-1}{2}=\frac{z+1}{-7}\)
B \(\frac{x-5}{-3}=\frac{y-1}{-2}=\frac{z-1}{7}\)
C \(\frac{x-5}{-3}=\frac{y-1}{-2}=\frac{z+1}{-7}\)
D \(\frac{x-5}{-3}=\frac{y-1}{2}=\frac{z+1}{7}\)

The correct answer is:
\(\frac{x-5}{-3}=\frac{y-1}{-2}=\frac{z+1}{-7}\)

Solution:
The general point \(p\) on the line
\(\frac{x-1}{2}=\frac{y+1}{1}=\frac{z-1}{-1}=r \quad \text{(Let)} \quad \ldots (i)\)
is \(P(2 r+1, r-1,1-r)\).
Let the point \(P\) is the intersection of line (i) and the plane \(x+2 y+3 z=4\), so
\(2 r+1+2 r-2+3-3 r=4 \Rightarrow r=2\)
So, point \(P(5, \cdot 1,-1)\)
\(\begin{aligned}
& \text{Now, } (2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}) \times(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}})=\left|\begin{array}{ccc}
\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\
2 & -3 & 0 \\
1 & 2 & -1
\end{array}\right| \\
& =\hat{\mathbf{i}}(3)-\hat{\mathbf{j}}(-2)+\hat{\mathbf{k}}(4+3)=3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+7 \hat{\mathbf{k}}
\end{aligned}\)
\(\therefore\) Equation of required line is
\(\frac{x-5}{3}=\frac{y-1}{2}=\frac{z+1}{7}\)
or \(\quad \frac{x-5}{-3}=\frac{y-1}{-2}=\frac{z+1}{-7}\)

Topic: Mathematics, Three Dimensional Geometry, JEE Main
Equation of the line passing through the points of intersection of the parabola $x^2=8 y$ and the ellipse $\frac{x^2}{3}+y^2=1$ is :

Options:
A
$y-3=0$

B
$y+3=0$

C
$3 y+1=0$

D
$3 y-1=0$

The correct answer is:
$3 y-1=0$

Solution:
$$
x^2=8 y
$$
$$
\frac{x^2}{3}+y^2=1
$$
From (i) and (ii),
$$
\frac{8 y}{3}+y^2=1 \Rightarrow y=-3, \frac{1}{3}
$$
When $y=-3$, then $x^2=-24$, which is not possible.
When $y=\frac{1}{3}$, then $x=\pm \frac{2 \sqrt{6}}{3}$
Point of intersection are
$$
\left(\frac{2 \sqrt{6}}{3}, \frac{1}{3}\right) \text { and }\left(-\frac{2 \sqrt{6}}{3}, \frac{1}{3}\right)
$$
Required equation of the line,
$$
y-\frac{1}{3}=0 \Rightarrow 3 y-1=0
$$

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2013 (09 Apr Online)
Equation of the plane passing through $(-2,2,2)$ and $(2,-2,-2)$ and perpendicular to the plane $9 x-13 y-3 z=0$ is

Options:
A $5 x+3 y+2 z=0$
B $5 x-3 y+2 z=0$
C $5 x-3 y-2 z=0$
D $5 x+3 y-2 z=0$

The correct answer is:
$5 x+3 y+2 z=0$

Solution:
Any plane passing through $(-2,2,2)$ is $A(x+2)+B(y-2)+C(z-2)=0$
$\because$ It passes through $(2,-2,-2)$ $\Rightarrow \quad 4 A-4 B-4 C=0$
It is parallel to $9 x-13 y-3 z=0$ $\begin{array}{lll}\therefore & 9 A-13 B-3 C=0 & \ldots \text { (ii) }\end{array}$
Solving Eqs. (i) and (ii),
$$
\begin{aligned}
& \frac{A}{12-52}=\frac{B}{-36+12}=\frac{C}{-52+36} \\
\Rightarrow \quad \frac{A}{-40}=\frac{B}{-24}=\frac{C}{-16}
\end{aligned}
$$
$\therefore$ Required equation of plane is $-40(x+2)-24(y-2)-16(z-2)=0$
$\begin{array}{lr}\Rightarrow & -40 x-80-24 y+48-16 z+32=0 \\ \Rightarrow & 40 x+24 y+16 z=0 \\ \Rightarrow & 5 x+3 y+2 z=0\end{array}$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
Equation of the plane passing through the point 1,-1,3, parallel to the vector i^+2j^+4k^ and perpendicular to the plane x-2y+z=6 is given by ax+by+cz+8=0, then the value of 2a-5b+7c is equal to

Options:
A 32
B 31
C -1845
D 725

The correct answer is:
-1845

Solution:
∵ required plane is passing through point 1,-1,3
and parallel to vector i^+2j^+4k^ and i^-2j^+k^
Therefore, the equation of plane is x-1y+1z-31241-21=0
⇒x-110-y+1-3+z-3-4=0
⇒10x+3y-4z+5=0
⇒16x+245y-325z+8=0
a=16,b=245,c=-325

Topic: Mathematics, Three Dimensional Geometry, JEE Main
Equation of the plane passing through the point of intersection of the line x-13=y-21=z-32 and x-31=y-12=z-23 and perpendicular to the line x+52=y-33=z+11 is

Options:
A 2x+3y+z+7=0
B 2x-3y-z+22=0
C 2x+3y+z-22=0
D 2x+3y+z+13=0

The correct answer is:
2x+3y+z-22=0

Solution:
Any general point on the line
x-13=y-21=z-32 is 3Î»+1,Î»+2,2Î»+3
For the point of intersection, this point must satisfy x-31=y-12=z-23
Hence, 3Î»-21=Î»+12=2Î»+13
6Î»-4=Î»+1âÎ»=1
So, the point is 4,3,5
Also, a vector parallel to x+52=y-33=z+11 is 2i^+3j^+k^
Hence, the equation of the required plane is
2x-4+3y-3+1z-5=0
2x+3y+z-22=0

Topic: Mathematics, Three Dimensional Geometry, JEE Main
Equation of the plane which passes through the point of intersection of lines x - 1 3 = y - 2 1 = z - 3 2 and x - 3 1 = y - 1 2 = z - 2 3 and has the largest distance from the origin is:

Options:
A 4x+3y+5z=50
B 3x+4y+5z=49
C 5x+4y+3z=57
D 7x+2y+4z=54

The correct answer is:
4x+3y+5z=50

Solution:
Line

L1:x-13=y-21=z-32=λ

L2:x-31=y-12=z-23=μ

⇒ x=3λ+1andx=μ+3 y=λ+2 y=2μ+1 z=2λ+3 z=3μ+2

⇒3λ+1=μ+3⇒3λ-μ=2

λ+2=2μ+1⇒λ-2μ=-1

2λ+3=3μ+2⇒2λ-3μ=-1

Solving μ=1, λ=1

∴ Point of intersection P4,3,5 and D.R. of normal is (largest distance is normal distance which is OP) 4,3,5.

Equation of plane 4x+3y+5z+d=0 this passes through the point 4,3,5

⇒ 4×4+3×3+5×5+d=0

⇒ d=-50

So, equation of plane is 4x+3y+5z−50=0

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2014 (09 Apr Online)
Equation of the straight line which meets the circle x2+y2=8 at two points where these points are at a distance of 2 units from the point A2,2 is

Options:
A x+y=2
B x+y=3
C x+y=1
D x+y=0

The correct answer is:
x+y=3

Solution:
Equation of circle assuming A2,2 as
centre and radius equal to 2 units is x-22+y-22=22 … 1
equation of given circle is x2+y2=8 … 2
equation 2 – equation 1 gives
4x+4y-8=4 ⇒x+y=3
which is the required equation

Topic: Mathematics, Circle, JEE Main
Equation of the tangent to the circle, at the point 1, -1, whose center, is the point of intersection of the straight lines x-y=1 and 2x+y=3 is:

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

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

Solution:
Point of intersection of lines

x-y=1 and 2x+y=3

O is 43,13

Slope of OP=13+143-1=4313=4

Slope of tangent =-14

⇒ Equation of tangent :y+1= -14x-1

⇒4y+4=-x+1

⇒x+4y+3=0

Topic: Mathematics, Circle, JEE Main, JEE Main 2016 (10 Apr Online)
Equation of $x$-axis is

Options:
A $\frac{x}{1}=\frac{y}{1}=\frac{z}{1}$
B $\frac{x}{0}=\frac{y}{1}=\frac{z}{1}$
C $\frac{x}{1}=\frac{y}{0}=\frac{z}{0}$
D $\frac{x}{0}=\frac{y}{0}=\frac{z}{1}$

The correct answer is:
$\frac{x}{1}=\frac{y}{0}=\frac{z}{0}$

Solution:
$\ \frac{\mathrm{X}}{1}=\frac{\mathrm{Y}}{0}=\frac{\mathrm{Z}}{0}$
The equation of a line with direction vector $\vec{d}=(1, m, n)$ that passes through the point $\left(\mathrm{x}_1, \mathrm{y}_1, \mathrm{z}_1\right)$ is given by
$\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}$
For the $\mathrm{x}$ axis, the line passes through the origin i.e. $\left(\mathrm{x}_1, \mathrm{y}_1, \mathrm{z}_1\right)$ corresponds to $(0,0,0)$
Also, it is parallel to the $\mathrm{x}$ axis and perpendicular to $\mathrm{y}$ and $\mathrm{z}$ axes and so the direction ratio becomes $(1,0,0)$
$\Rightarrow$ The line equation can be written as $\frac{X}{1}=\frac{Y}{0}=\frac{Z}{0}$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
Equation to the straight line cutting off an intercept \( 2 \) from the negative direction of the axis of \( y \) and inclined at \( 30^{\circ} \) to the positive direction of axis of \( x \), is

Options:
A \( y+x-\sqrt{3}=0 \)
B \( y-x+2=0 \)
C \( y-\sqrt{3} x-2=0 \)
D \( \sqrt{3} y-x+2 \sqrt{3}=0 \)

The correct answer is:
\( \sqrt{3} y-x+2 \sqrt{3}=0 \)

Solution:
Given that

Straight line having negative Y intercept of 2 that means straight line will passes through

x1, y1=0, –2

y+2=13x-0

because line is making an angle of 30° with X axis ( m=tan⁡30°=13 )

x-3y-2 3=0 is the required line equation.

Topic: Mathematics, Straight Lines, JEE Main
Equations of the latus rectum of the ellipse $9 x^2+4 y^2-18 x-8 y-23=0$ are :

Options:
A $y= \pm \sqrt{5}$
B $x= \pm \sqrt{5}$
C $y=1 \pm \sqrt{5}$
D $x=-1 \pm \sqrt{5}$

The correct answer is:
$y=1 \pm \sqrt{5}$

Solution:
Equation of ellipse is
$9 x^2+4 y^2-18 x-8 y-23=0$
$\Rightarrow \quad 9\left(x^2-2 x+1\right)-9+4\left(y^2-2 y+1\right)$ $-4-23=0$
$\Rightarrow \quad 9(x-1)^2+4(y-1)^2=36$
$\Rightarrow \quad \frac{(x-1)^2}{4}+\frac{(y-1)^2}{9}=1$
Here $\quad a < b$
Also$e=\sqrt{1-\frac{4}{9}}=\sqrt{\frac{5}{9}}=\frac{\sqrt{5}}{3}$
Equation of latus rectum are
$y-1= \pm 3 \cdot \frac{\sqrt{5}}{3}$
$y=1 \pm \sqrt{5}$

Topic: Mathematics, Ellipse, JEE Main
Equations of two diameters of a circle are 2x-3y=5 and 3x-4y=7. The line joining the points -227,-4 and -17,3 intersects the circle at only one point P(α,β). Then 17β-α is equal to

Options:

The correct answer is:
2

Solution:
Given diameters of circle are 2x-3y=5 and 3x-4y=7.

We know that, the intersection point of diameters is centre.

⇒x=5+3y2=7+4y3

⇒15+9y=14+8y

⇒y=-1

⇒x=5-32=1

So, centre of circle is 1,-1

Equation of AB is given by,

y+4=3+4-17+227x+227

⇒y+4=737x+227

⇒3y+12=7x+22

⇒7x–3y+10=0 ...i

Slope of AB, m1=73

So, slope of CP, m2=-1m1=-37

Equation of CP is given by,

y+1=-37x-1

⇒7y+7=-3x+3

⇒3x+7y+4=0 ...ii

Solving i and ii,

⇒x=3y-107=-7y-43

⇒9y-30=-49y-28

⇒58y=2

⇒y=129

⇒x=3129-107=3-2907×29

⇒x=-4129

So, α=-4129, β=129

⇒17β-α=1729+4129

⇒17β-α=2

Topic: Mathematics, Circle, JEE Main, JEE Main 2024 (29 Jan Shift 1)
Equilibrium constant for reaction NH4OHaq+dH⊕aq⇌NH4+aq+H2Ol is 1.8 ×109 . Hence equilibrium constant for ionization NH3+H2O⇌NH4+aq+OH-aqx×10-6 is.The value of ‘x’ is.

Options:

The correct answer is:
18

Solution:
NH4OHaq+H+aqâNH4+aq+H2Ol

K1=N+H4NH4OHH+=1.8Ã109

NH4OHaqâNH4+aq+OH-aq

K2=NH4+OH-NH4OH

Multiply this by H+ and divide also

K2=NH4+OH-H+NH4OH H+=K1ÃKw

=1.8Ã109Ã1.0Ã10-14=18Ã10-6

value of x = 18.

Topic: Chemistry, Chemical Equilibrium, JEE Main
Equilibrium constant for reaction NH4OHaq+H+aqâNH4+aq+H2Ol is 1.8Ã199 .
Hence equilibrium constant for ionization
NH3+ H2OâNH4+aq+OH-aq is xÃ10-6 . The value of ' x ' is.

Options:

The correct answer is:
1

Solution:
NH4OHaq + H+aqâNH4+aq+H2Ol
K1=NH4+NH4OHH+=1.8Ã109;
NH4OH aqâNH4+aq+OH-aq
K2=NH4+OH-NH4OH
Multiply this by H+ and divide also
K2=NH4+OH-H+NH4OHH+=K1ÃKw
= 1.8Ã109Ã1.0Ã10-14=18Ã10-6

Topic: Chemistry, Chemical Equilibrium, JEE Main
Equilibrium constant for the reaction 4A(g)⇌B(g)+2C(g) is –
Given:
2Ag⇌Bg+Yg;kc1=8
Cg⇌Ag+12Yg;kc2=14

Options:
A 32
B 16
C 2
D 128

The correct answer is:
128

Solution:
2Ag⇌Bg+Ygkc1=8 ……. (1)
2Ag+Yg⇌2Cgkc2=(4)2=16 ……. (2)
Add equation (1) and (2)
4Ag+Yg⇌Bg+2Cg
kC=kc1×kc2=8×16=128

Topic: Chemistry, Chemical Equilibrium, JEE Main
Equilibrium constant for the reaction $\mathrm{H}_2 \mathrm{O}(\mathrm{g})+\mathrm{CO}(g) \rightleftharpoons \mathrm{H}_2(g)+\mathrm{CO}_2(g)$ is 81 . If velocity constant of the forward reaction is $162 \mathrm{~L} \mathrm{~mol}^{-1} \mathrm{~s}^{-1}$. What is the velocity constant (in $\mathrm{L} \mathrm{mol}^{-1} \mathrm{~s}^{-1}$ ) for the backward reaction ?

Options:
A $13122$
B $2$
C $261$
D $243$

The correct answer is:
$2$

Solution:
$K_c=\frac{k_f}{k_b}=\frac{162}{81}=2$

Topic: Chemistry, Chemical Equilibrium, JEE Main
Equilibrium constant KC for the following reaction at 800 K is, 4
NH3(g)⇌12N2(g)+32H2(g)
The value of KP for the following reaction will be
N2(g)+3H2(g)⇌2NH3(g)

Options:
A 800R42-
B 16×(800R)2
C 14×800R2
D 800R1/24

The correct answer is:
14×800R2

Solution:
NH 3 (g)â 1 2 N 2 (g)â+â 3 2 H 2 (g);ââ K C â=â4
N2(g)+3H2(g)â2NH3(g) Îng=2-4=-2
So KC for the following reaction will be equal to 142
KP=KC(RT)Îng
=â 1 4 2 âÃâ (800âÃâR) â2
= 1 4×800 R 2

Topic: Chemistry, Chemical Equilibrium, JEE Main
Equilibrium constant K is related to Ecello and not Ecell because

Options:
A Ecello is easier to measure than Ecell
B Ecell becomes zero at equilibrium point but Ecello remains constant under all conditions
C at a given temperature, Ecell changes hence value of K can not be measured
D any of the terms Ecell or Ecello can be used

The correct answer is:
Ecell becomes zero at equilibrium point but Ecello remains constant under all conditions

Solution:
At equilibrium, Ecell becomes zero which means no current is generated by the cell at this point. Ecello remains constant hence equilibrium constant is related to Ecello .

Topic: Chemistry, Electrochemistry, NEET
Equilibrium constant Kp for the reaction CaCO 3 ⇌CaO+ CO 2 is 0.82 atm at 7 2 7 ∘ C.
If 1 mole of CaCO3 is placed in a closed container of 20 L and heated to this temperature, what amount of CaCO3 would dissociate at equilibrium?

Options:
A 0.2 g
B 80 g
C 20 g
D 50 g

The correct answer is:
20 g

Solution:
CaCO 3 âCaOâ+â CO 2
K p = p CO 2 =0.82 atm
0.82 atmÃ20âL= n CO 2 Ã0.082Ã1000
2 0 0 1 0 0 0 = 1 5 = n CO 2
No. of moles CO2 = no. of moles of CaCO3 decomposed = 1 5 mole
Amount of CaCO3 decomposed = 1 5 âÃâ100â=â20âg

Topic: Chemistry, Chemical Equilibrium, JEE Main
Equilibrium constants for the following reactions at $1200 \mathrm{K}$ are given
$2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g)$
$K_{1}=6.4 \times 10^{-8}$
$2 \mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g)$
$K_{2}=1.6 \times 10^{-6}$
The equilibrium constant for the reaction? $\mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g)$
at $1200 \mathrm{K}$ will be

Options:
A 0.05
B 20
C 0.2
D 5

The correct answer is:
5

Solution:
For the given reactions, $2 \mathrm{H}_{2} \mathrm{O} \rightleftharpoons 2 \mathrm{H}_{2}+\mathrm{O}_{2}, K_{1}=6.4 \times 10^{-8}$
or, $\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{2}+\frac{1}{2} \mathrm{O}_{2}, K_{1}^{\prime}=\sqrt{K_{1}}$
or $\mathrm{H}_{2}+\frac{1}{2} \mathrm{O}_{2}, \longrightarrow \mathrm{H}_{2} \mathrm{O}, K_{1}^{\prime \prime \prime}=1 / \sqrt{K_{1}} \ldots$
$2 \mathrm{CO}_{2} \rightleftharpoons 2 \mathrm{CO}+\mathrm{O}_{2}, K_{2}=1.6 \times 10^{-6}$
or $\mathrm{CO}_{2} \rightleftharpoons \mathrm{CO}+\frac{1}{2} \mathrm{O}_{2}, K_{2}^{\prime}=\sqrt{K_{2}}$
From (i) and (ii) [on adding] $K=\frac{\sqrt{K_{2}}}{\sqrt{K_{1}}}=\frac{\sqrt{1.6 \times 10^{-6}}}{\sqrt{6.4 \times 10^{-8}}}$
$K=\sqrt{\frac{10^{2}}{4}}=\sqrt{25}=5$

Topic: Chemistry, Chemical Equilibrium, JEE Main
Equimolar solutions in the same solvent have

Options:
A
Same boiling point but different freezing point

B
Same freezing point but different boiling point

C
Same boiling and same freezing points

D
Different boiling and different freezing points

The correct answer is:
Same boiling and same freezing points

Solution:
No Solution Available

Topic: Chemistry, Solutions, JEE Main, JEE Main 2005