datavorous/entrance-exam-dataset huggingface dataset:
If the line \( \mathrm{y}+\mathrm{x}=0 \) bisects two chords drawn from a point \( \left(\frac{1+a \sqrt{2}}{2}, \frac{1-a \sqrt{2}}{2}\right) \) to the circle \( 2 x^{2}+2 y^{2}-(1+a \sqrt{2}) x-(1-a \sqrt{2}) y=0 \), then \( a \) lies in the interval \( (-\infty,-\lambda) \cup(\lambda, \infty) \), the numerical quantity \( \lambda \) should be equal to

Options:

The correct answer is:
2

Solution:
Equation of chord AB whose mid point is h, -h is
T=S1
2 xh - 2 yh - 1 + 2 a x + h 2 - 1 - 2 a y - h 2
= 2 h 2 + 2 h 2 - 1 + 2 a h + 1 - 2 a h
⇒ 4 xh - 4 yh - 1 + 2 a x + h - 1 - 2 a y - h
= 8 h 2 - 2 1 + 2 a h + 2 1 - 2 a h

⇒ x 4 h - 1 + 2 a - y 4 h + 1 - 2 a - h 1 + 2 a + h 1 - 2 a = 8 h 2 - 2 1 + 2 a h + 2 1 - 2 a h
or 8 h 2 - 1 + 2 a h + 1 - 2 a h - x 4 h - 1 + 2 a + y 4 h + 1 - 2 a = 0
It passes through A 1 + 2 a 2 , 1 - 2 a 2 then
8 h 2 - 2 2 ah - 1 + 2 a 2 4 h - 1 + 2 a + 1 - 2 a 2 4 h + 1 - 2 a
or 8 h 2 - 2 2 ah - 2 h 1 + 2 a + 1 + 2 a 2 2 + 2 h 1 - 2 a + 1 - 2 a 2 2 = 0
or 8 h 2 - 6 2 ah + 1 + 2 a 2 = 0
Hence for two real and different values of h, we must have - 6 2 a 2 - 4 · 8 · 1 + 2 a 2 > 0

or a 2 - 4 > 0
a + 2 a - 2 > 0
∴ A ∈ - ∞ , - 2 ∪ 2 , ∞

Topic: Mathematics, Circle, JEE Main
If the line of intersection of the planes ax+by=3 and ax+by+cz=0,a>0 makes an angle 30° with the plane y-z+2=0, then the direction cosines of the line are

Options:
A 12,12,0
B 12,-12,0
C 15,-25,0
D 12,-32,0

The correct answer is:
12,-12,0

Solution:
The vector along the line of intersection will be n→=i^j^k^ab0abc

=bci^-acj^

i.e. direction ratios of line are b,-a,0

Direction ratio of normal to the plane y-z+2=0 are 0,1,-1

Since the line makes 30° with the plane, so it will make 60° with its normal.

i.e. cos60°=-a2b2+a2

⇒a2a2+b2=12

⇒b=±a

So, direction ratios of the line can be ±a,-a,0

∴ direction cosines can be ±±12,-12,0

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2022 (27 Jul Shift 2)
If the line segment joining P2,3 and Q5,7 subtends a right angle at Rx,y and the area of ΔPQR=2 sq. units, then the maximum number of such points R in xy-plane are

Options:

The correct answer is:
4

Solution:
PQ=32+42=5
Let, length of altitude through R is h

then 12×h×5=2
⇒h=45
Maximum possible length of altitude through R with right angle at R of ΔPQR (without the given area) is equal to 12PQ=52
Now, 45<52
So, 4 triangles are possible

Topic: Mathematics, Straight Lines, JEE Main
If the line segment joining the points $(5,2)$ and $(2, a)$ subtends an angle $\frac{\pi}{4}$ at the origin, then the absolute value of the product of all possible values of $a$ is :

Options:
A 6
B 8
C 2
D -4

The correct answer is:
-4

Solution:
$\begin{aligned} & \mathrm{m}_{\mathrm{OA}}=\frac{2}{5} \\ & \mathrm{~m}_{\mathrm{OB}}=\frac{\mathrm{a}}{2}\end{aligned}$
$\begin{aligned} & \tan \frac{\pi}{4}=\left|\frac{2}{5}-\frac{a}{2}\right| \\ & 1=\left|\frac{4-5 a}{10+2 a}\right|\end{aligned}$

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2024 (08 Apr Shift 2)
If the line x-1=0, is a directrix of the hyperbola kx2-y2=6, then the hyperbola passes through the point

Options:
A -25,6
B -5,3
C 5,-2
D 25,36

The correct answer is:
5,-2

Solution:
Given equation of hyperbola is x26k-y26=1

Equation of directrix will be x=±ae,

Where e=1+b2a2=1+66k=1+k

i.e. k≥-1

⇒x=±6k1+k is the equation of directrix

i.e. ±6k1+k=1

∴ k1+k2=6⇒k2+k-6=0

⇒ k=2 as k=-3 is rejected

So equation of hyperbola will be H≡x23-y26=1

∴ 5,-2 satisfy the given hyperbola

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2022 (26 Jul Shift 2)
If the line x-11=y-k-2=z-3λ lies in the plane 3x+4y-2z=6, then 5k+3λ is equal to

Options:
A 75
B 754
C 15
D 52

The correct answer is:
754

Solution:
If the line x-11=y-k-2=z-3Î» lies in the plane 3x+4y-2z=6
â1,k,3 satisfies the equation of the plane
â3+4k-6=6âk=94
And b â â n â =0 b â = 1,â2,Î» and n â = 3,4,â2
â3-8-2Î»=0
âÎ»=-52
5k+3Î»=454+152=754

Topic: Mathematics, Three Dimensional Geometry, JEE Main
If the line, x-12=y+13=z-24 meets the plane, x+2y+3z=15 at a point P, then the distance of P from the origin is,

Options:
A 25
B 92
C 52
D 72

The correct answer is:
92

Solution:
Let x-12=y+13=z-24=λ

Let, coordinate of any point on line is P=2λ+1, 3λ-1, 4λ+2

If P is intersection point of plane & line then P will lie on plane x+2y+3z=15

∴ 2λ+1+23λ-1+34λ+2=15

⇒λ=12
∴ P=2×12+1, 3×12-1, 4×12+2=2, 12, 4

The distance of a point x1, y1, z1 from origin O is x12+y12+z12

∴ OP=22+122+42=814=92 units.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2019 (09 Apr Shift 1)
If the line x-12=y-23=z-44 intersect the xy and yz plane at points A and B respectively. If the volume of the tetrahedron OABC is V cubic units (where, O is the origin) and point C is 1,0,4, then the value of 102V is equal to

Options:

The correct answer is:
68

Solution:
On the xy plane, z=0
⇒x-12=y-23=-1⇒x=-1,y=-1
⇒ Coordinates of point A are -1,-1,0
On the yz plane, x=0
⇒-12=y-23=z-44⇒y=12,z=2
⇒ Coordinates of point B are 0,12,2
OA→=-i^-j^
OB→=12j^+2k^
OC→=i^+4k^
The volume of the tetrahedron OABC is V=16OA→ OB→ OC→
=16-1-100122104
=16-12+1-2=46=23 cubic units

Hence, 102V=68

Topic: Mathematics, Vector Algebra, JEE Main
If the line x-15=y-32=z-32 intersects the curve x2-y2=k2, z=0, then the value of 2k can be equal to

Options:
A -13
B -132
C 132
D 134

The correct answer is:
-13

Solution:
Put z=0 in the given equation of line
x-15=y-32=-32
⇒x=-132,y=0
So, -1322-0=k2⇒k=±132

Topic: Mathematics, Three Dimensional Geometry, JEE Main
If the line x-23=y+12=z-1-1 intersects the plane 2x+3y-z+13=0 at a point P and the plane 3x+y+4z=16 at a point Q, then PQ is equal to

Options:
A 27
B 14
C 214
D 14

The correct answer is:
214

Solution:
x-23=y+12=z-1-1=λ
x=3λ+2, y=2λ-1, z=-λ+1
Intersection with plane 2x+3y-z+13=0
Hence, 23λ+2+32λ-1--λ+1+13=0
13λ+13=0
λ=-1
∴ P-1,-3,2
Intersection with plane 3x+y+4z=16
Hence, 33λ+2+2λ-1+4-λ+1=16
λ=1
Q5,1,0
PQ=62+42+22=56=214

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2019 (12 Apr Shift 1)
If the line x-2y=12 is a tangent to the ellipse x2a2+y2b2=1 at the point 3,-92, then the length of the latus rectum of the ellipse is

Options:
A 5 units
B 122 units
C 9 units
D 83 units

The correct answer is:
9 units

Solution:
∵ 3,-92 lies on x2a2+y2b2=1⇒9a2+814b2=1 ……1

Equation of the tangent at 3,-92 is 3xa2+-92yb2=1

& given equation of the tangent is:

x-2y=12 ⇒x12+ -y6=1

On comparing these equations:

a23=12⇒a2=36⇒a=6

2b29=6⇒b2=27⇒b=33

Therefore, the length of latus rectum

=2b2a=2×276=9

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2019 (10 Apr Shift 1)
If the line x-2y=12 is a tangent to the ellipse x2a2+y2b2=1 at the point 3,-92, then the length of the latus rectum of the ellipse is

Options:
A 5 units
B 122 units
C 9 units
D 83 units

The correct answer is:
9 units

Solution:
∵3,-92 lies on x2a2+y2b2=1⇒9a2+814b2=1 ……(1)
Equation of the tangent at 3,-92 is 3xa2+-92yb2=1
& given equation of the tangent is: x-2y=12 ⇒x12+ -y6=1
On comparing these equations: a23=12⇒a2=36⇒a=6
2b29=6⇒b2=27⇒b=33
∴Length of L.R.=2b2a=2×276=9

Topic: Mathematics, Ellipse, JEE Main
If the line, x-31=y+2-1=z+λ-2 lies in the plane, 2x-4y+3z=2 , then the shortest distance between this line and the line, x-112=y9=z4 is

Options:
A 1
B 2
C 3
D 0

The correct answer is:
0

Solution:
The points 3, -2, -λ lie on Plane 2x-4y+3z-2=0
⇒6+8-3λ-2=0
⇒3λ=12
⇒λ=4
Now,
x-31=y+2-1=z+4-2.......i
x-112=y9=z4........ii
Shortest distance formula
d=x1-x2y2-y1z2-z1a1a2a3b1b2b3i^j^k^a1a2a3b1b2b3
Calculating we get the value as 0
Hence lines are intersecting (As b→∦d→ )
∴ Shortest distance =0

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2017 (09 Apr Online)
If the line, x-32=y+2-1=z+43 lies in the plane lx+my-z=9, then l2+m2 is equal to

Options:
A 5
B 2
C 26
D 18

The correct answer is:
2

Solution:
x-32=y+2-1=z+43 lies in lx+my-z=9
The point ( 3,â2,â4 ) lies on the plane. So it should satisfy the equation of the plane.
3l-2m+4=9
3l-2m=5 .......(i)
â´ Line lies in plane
â´ line is perpendicular to the normal to the plane
2l-1.m-3=0
2l-m=3 .......(ii)
l=1 and m= -1
â´ l2+m2=1+1=2

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2016 (03 Apr)
If the line x-41=y-21=z-m2 lies in the plane 2x+ly+z=7, then the value of m+2l is equal to

Options:
A 1
B 2
C -1
D -2

The correct answer is:
-1

Solution:
A vector parallel to the line is i^+j^+2k^
A vector normal to the plane is 2i^+lj^+k^
∵ the line lies in the plane

⇒21+1l+21=0
⇒l=-4
∵ Point (4,2,m) also lies in the plane
⇒24+l2+m=7
⇒ m=7
Hence, m+2l=7-8=-1

Topic: Mathematics, Three Dimensional Geometry, JEE Main
If the line x-41=y-21=z-qp lies completely in the plane 2x-4y+z=7, then the ordered pair p,q is

Options:
A 2,7
B 7,2
C 2,4
D 1,1

The correct answer is:
2,7

Solution:
Since the line lie in the plane, so 4,2,q also lie in the plane and vector along the line is perpendicular to normal to the plane

Therefore, 2-4+p=0⇒p=2 and 8-8+q=7⇒q=7

Hence, the ordered pair is 2,7

Topic: Mathematics, Three Dimensional Geometry, JEE Main
If the line $x=k$ touches the circle $x^2+y^2=9$, then the value of $k$ is

Options:
A $2$ but not $-2$
B $-2$ but not $2$
C $3$
D None of these

The correct answer is:
$3$

Solution:
$k=3$, as perpendicular from centre on line $=$ radius.

Topic: Mathematics, Circle, JEE Main
If the line \( x-1=0 \) is the directrix of the parabola \( y^{2}-k x+8=0 \), then one of the values of \( k \) is

Options:
A \( \frac{1}{8} \)
B \( 8 \)
C \( \frac{1}{4} \)
D \( 4 \)

The correct answer is:
\( 4 \)

Solution:
Given, y2=kx-8

⇒y2=kx-8k

Shifting the origin

Y2=kX where Y=y, X=x-8k

Directrix of standard parabola is X=-k4

Directrix of original parabola is x=8k-k4

Now, x=1 also coincides with x=8k-k4

On solving, we get k=4,-8

Topic: Mathematics, Parabola, JEE Main
If the line y-2=0 is the directrix of the parabola x2-ky+32=0,kâ 0 and the parabola intersects the circle x2+y2=8 at two real distinct points, then the absolute value of k is

Options:

The correct answer is:
16

Solution:
x2-ky+32=0
âx2=ky-32k
Put, x=X,y-32k=Y
The equation of directrix is Y+k4=0
i.e. y-32k+k4=0
But, y-2=0 is the directrix.
â32k-k4=2
âk2+8k-128=0
âk=-16 or k=8
For k=8, the parabola is x2=8y-4 which does not intersect the circle.
For k=-16, the parabola is x2=-16y+2 which intersects the circle at two real distinct points.
â Absolute value of k=-16=16

Topic: Mathematics, Parabola, JEE Main
If the line $y=2 x+\lambda$ be a tangent to the hyperbola $36 x^2-25 y^2=3600$, then $\lambda=$

Options:
A $16$
B $-16$
C $\pm 16$
D None of these

The correct answer is:
$\pm 16$

Solution:
If $y=2 x+\lambda$ is tangent to given hyperabola, then
$\lambda= \pm \sqrt{a^2 m^2-b^2}= \pm \sqrt{(100)(4)-144}= \pm \sqrt{256}= \pm 16$

Topic: Mathematics, Hyperbola, JEE Main
If the line $y=2 x+c$ be a tangent to the ellipse $\frac{x^2}{8}+\frac{y^2}{4}=1$, then $c=$

Options:
A $\pm 4$
B $\pm 6$
C $\pm 1$
D $\pm 8$

The correct answer is:
$\pm 6$

Solution:
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \quad y=m \underline{x+c}$
Condition for tangency :- $c^2=a^2 m^2+b^2$
$\frac{x^2}{8}+\frac{y^2}{4}=1$
$\begin{aligned} & a^2=8 \\ & b^2=4\end{aligned}$
$y=2 x+c$
$m=-2$
$c^2=8(2)^2+4 \quad \Rightarrow \quad c^2=8 \times 4+4$
$\begin{aligned} & c^2=8(2)^2+4 \quad \Rightarrow c^2=8 \times 4+4 \\ & c^2=32+4 \\ & c^2=36 \\ & c= \pm \sqrt{36} \\ & c= \pm 6\end{aligned}$

Topic: Mathematics, Ellipse, JEE Main
If the line $y=\sqrt{3} x+k$ touches the circle $x^2+y^2=16$, then $k=$

Options:
A $0$
B $2$
C $4$
D $8$

The correct answer is:
$8$

Solution:
$k= \pm 4 \sqrt{1+3}= \pm 8$

Topic: Mathematics, Circle, JEE Main
If the line y=4+kx, k>0, is the tangent to the parabola y=x-x2 at the point P and V is the vertex of the parabola, then the slope of the line through P and V is

Options:
A 32
B 269
C 52
D 236

The correct answer is:
52

Solution:
Plotting the given data we have,

Now by diagram, Slope of tangent at P= Slope of line AP

dydxα,α-α2=1−2α=α−α2−4α

On solving we get α=-2⇒P-2,-6

So, slope of PV=52

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2022 (25 Jun Shift 2)
If the line y=-72 is the directrix of the parabola x2-ky+8=0, then the sum of all the possible values of k is equal to

Options:

The correct answer is:
14

Solution:
Given equation of the parabola is x2-ky+8=0⇒x2=ky-8k
⇒ Directrix is y=8k-k4
⇒y=-72 is same as y=8k-k4
⇒32-k24k=-72⇒-k2+14k+32=0
⇒k1+k2=14 (sum of possible values of k)

Topic: Mathematics, Parabola, JEE Main
If the line \( y \cos \alpha=x \sin \alpha+a \cos \alpha \) be a tangent to the circle \( x^{2}+y^{2}=a^{2} \), then

Options:
A \( \sin ^{2} \alpha=1 \)
B \( \cos ^{2} \alpha=1 \)
C \( \sin ^{2} \alpha=a^{2} \)
D \( \cos ^{2} \alpha=a^{2} \)

The correct answer is:
\( \cos ^{2} \alpha=1 \)

Solution:
Given, y=xtanâ¡Î±+a
Condition for tangency is a2=a2(1+tan2â¡Î±)
[âµ c2=a2(1+m2)]
â sec2â¡Î±=1
â cos2â¡Î±=1

Topic: Mathematics, Circle, JEE Main
If the line $y=m x+1$ meets the circle $x^2+y^2+3 x=0$ in two points equidistant from and on opposite sides of $x$-axis, then

Options:
A
$3 m+2=0$

B
$3 m-2=0$

C
$2 m+3=0$

D
$2 m-3=0$

The correct answer is:
$3 m-2=0$

Solution:
Circle: $x^2+y^2+3 x=0$
Centre, $B=\left(-\frac{3}{2}, 0\right)$
$y$-intercept of the line $=1$
$$
\therefore \quad \mathrm{A}=(0,1)
$$
Slope of line, $m=\tan \theta=\frac{O A}{O B}$
$$
\begin{aligned}
& \Rightarrow m=\frac{1}{\frac{3}{2}}=\frac{2}{3} \\
& \Rightarrow 3 m-2=0
\end{aligned}
$$

Topic: Mathematics, Circle, JEE Main, JEE Main 2012 (19 May Online)
If the line y=mx+73 is normal to the hyperbola x224-y218=1, then a value of m is:

Options:
A 52
B 35
C 152
D 25

The correct answer is:
25

Solution:
The given equation of the hyperbola is x224-y218=1, which is of the form x2a2-y2b2=1.

⇒a2=24; b2=18

We know that the equation of a normal of slope m to the hyperbola x2a2-y2b2=1 is y=mx±ma2+b2a2-b2m2

Thus, the equation of the normal to the given hyperbola is y=mx±m24+1824-18m2

y=mx±m4224-18m2

Given, the normal is y=mx+73

Hence, comparing the two equations, we get 42m24-18m2=73

⇒ 42m224-18m2=49×3

⇒ 42×42m2=49×324-18m2

⇒ 12m2=24-18m2

⇒ m2=2430=45

⇒ m=±45=±25.

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2019 (09 Apr Shift 1)
If the line y=mx bisects the area enclosed by the lines x=0, y=0, x=32 and the curve y=1+4x-x2, then 12m is equal to

.

Options:

The correct answer is:
26

Solution:
Given, the equation of line, y=mx

And the line x=0, y=0, x=32

The intersection point 32,0

Area of triangle is half the total area

12∫03/21+4x-x2dx=∫03/2mxdx

12x+2x2-x3303/2=m2x203/2

32+92-98=9m4

398=9m4

m=3918

Then, 12m

12 m=3918×12

12m=26

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2021 (31 Aug Shift 2)
If the line $y=m x+c$ be a tangent to the circle $x^2+y^2=a^2$, then the point of contact is

Options:
A $\left(\frac{-a^2}{c}, a^2\right)$
B $\left(\frac{a^2}{c}, \frac{-a^2 m}{c}\right)$
C $\left(\frac{-a^2 m}{c}, \frac{a^2}{c}\right)$
D $\left(\frac{-a^2 c}{m}, \frac{a^2}{m}\right)$

The correct answer is:
$\left(\frac{-a^2 m}{c}, \frac{a^2}{c}\right)$

Solution:
Find points of intersection by simultaneously solving for $x$ and $y$ from $y=m x+c$ and $x^2+y^2=a^2$ which comes out as $\left(-\frac{a^2 m}{c}, \frac{a^2}{c}\right)$.

Topic: Mathematics, Circle, JEE Main
If the line $y=m x+c$ is a tangent to the parabola $y^2=4 a(x+a)$ then $m a+\frac{a}{m}$ is equal to

Options:
A $c$
B $2 c$
C $-c$
D $3 \mathrm{c}$

The correct answer is:
$c$

Solution:
Tangent at $y^2=4 a x$ is $y=m x+\frac{a}{m}$
Therefore, tangent at $y^2=4 a(x+a)$ is,
$y=m(x+a)+\frac{a}{m}$
or $y=m x+m a+\frac{a}{m} \Rightarrow m a+\frac{a}{m}=c$.

Topic: Mathematics, Parabola, JEE Main
If the line $y=m x+c$ touches the ellipse $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$, then $c=$

Options:
A $\pm \sqrt{b^2 m^2+a^2}$
B $\pm \sqrt{a^2 m^2+b^2}$
C $\pm \sqrt{b^2 m^2-a^2}$
D $\pm \sqrt{a^2 m^2-b^2}$

The correct answer is:
$\pm \sqrt{b^2 m^2+a^2}$

Solution:
Since, here $a$ and $b$ are interchanged, So, \(c= \pm \sqrt{b^2 m^2+a^2}\)

Topic: Mathematics, Ellipse, JEE Main
If the line y=mx+c touches the parabola y2=12x+3 exactly for one value of m m>0, then the value of c+mc-m is equal to

Options:

The correct answer is:
1.4

Solution:
Equation of the tangent to y2=12x+3 with slope m is
y=mx+3+3m, which is same as y=mx+c
⇒c=3m+3m
⇒3m2-mc+3=0 which is quadratic in ‘m’

Since, it has both roots equal
⇒c2-36=0⇒c=±6
⇒c=6 m>0⇒c>0
⇒m=1 & c=6
⇒c+mc-m=75=1.4

Topic: Mathematics, Parabola, JEE Main
If the line y=x-1 bisects two chords of the parabola y2=4bx which are passing through the point b,-2b, then the length of the latus rectum can be equal to

Options:
A 3
B 5
C 6
D 8

The correct answer is:
3

Solution:
Any point on y=xâ1 can be taken as h,hâ1
Taking h,h-1 as the midpoint of chord, the equation of chord is yh-1-4bx+h2=h-12-4bh
Which passes through the point b,-2b
â -2bh-1-2bb+h=h2-2h+1-4bh
â h2-2h+2b2-2b+1=0
As, above equation has two real roots, hence,
-22-412b2-2b+1>0
â 2b2-2b<0âb<1â4b<4
â length of latus rectum is less than 4

Topic: Mathematics, Parabola, JEE Main
If the line y=x+c touches the hyperbola x29-y25=1 at the point Ph,k, then h,k can be equal to

Options:
A 92,-52
B 92,52
C -52,-92
D 52,-92

The correct answer is:
92,52

Solution:
Tangent at Ph,k is hx9-ky5=1 which is same as y=x+c (where c=±91-5=±2 )
⇒hx9-ky5-1=0 and x-y±2=0 are identical
⇒h91=-k5-1=-1±2
⇒h,k=-92,-52 or 92,52

Topic: Mathematics, Hyperbola, JEE Main
If the linear charge density of a cylinder is $4 \mu \mathrm{Cm}^{-1}$ then electric field intensity at point $3.6 \mathrm{~cm}$ from axis is

Options:
A $4 \times 10^{5} \mathrm{NC}^{-1}$
B $2 \times 10^{6} \mathrm{NC}^{-1}$
C $8 \times 10^{7} \mathrm{NC}^{-1}$
D $12 \times 10^{7} \mathrm{NC}^{-1}$

The correct answer is:
$2 \times 10^{6} \mathrm{NC}^{-1}$

Solution:
For a charged cylinder
$$
\begin{aligned}
E &=\frac{\lambda}{2 \pi \varepsilon_{0} r}=\frac{4 \times 10^{-6} \times 18 \times 10^{9}}{3.6 \times 10^{-2}} \\
&=2 \times 10^{6} \mathrm{NC}^{-1}
\end{aligned}
$$

Topic: Physics, Electrostatics, JEE Main
If the linear momentum is increased by $50 \%$, then kinetic energy will increase by

Options:
A $50 \%$
B $100 \%$
C $125 \%$
D $25 \%$

The correct answer is:
$125 \%$

Solution:
Kinetic energy of a body, $K=\frac{p^2}{2 m}$ where $p$ is the momentum of the body and $m$ is the mass of the body
$p^{\prime}=p+\frac{50}{100} p=\frac{3}{2} p$
Assuming $m$ remains constant
$\therefore \frac{K^{\prime}}{K}=\left(\frac{p^{\prime}}{p}\right)^2=\frac{9}{4}$
$\%$ increase in kinetic energy $=\frac{K^{\prime}-K}{K} \times 100 \%$
$=\left(\frac{K^{\prime}}{K}-1\right) \times 100 \%=\left(\frac{9}{4}-1\right) \times 100 \%=125 \%$

Topic: Physics, Center of Mass Momentum and Collision, AIIMS, AIIMS 2016
If the linear momentum of a body is increased by $50 \%$, then the kinetic energy of that body increases by

Options:
A $100 \%$
B $125 \%$
C $225 \%$
D $25 \%$

The correct answer is:
$125 \%$

Solution:
Kinetic energy of the body, $K=\frac{\mathrm{p}^{2}}{2 \mathrm{~m}}$ Since, the mass remains constant
$\therefore$
$$
\mathrm{K} \propto \mathrm{p}^{2}
$$
$$
\begin{gathered}
\frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}=\frac{\mathrm{p}_{2}^{2}}{\mathrm{p}_{1}^{2}}=\left[\frac{150}{100}\right]^{2}=\frac{9}{4} \\
100\left[\frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}-1\right]=\left(\frac{9}{4}-1\right) \times 100=125 \%
\end{gathered}
$$

Topic: Physics, Center of Mass Momentum and Collision, KCET, KCET 2010
If the lines $\frac{1-x}{2}=\frac{y-8}{\lambda}=\frac{z-5}{2}$ and $\frac{x-11}{5}=\frac{y-3}{3}=\frac{z-1}{1}$ are perpendicular, then $\lambda=$

Options:
A 4
B $-4$
C $\frac{8}{3}$
D $\frac{-8}{3}$

The correct answer is:
$\frac{8}{3}$

Solution:
We have $\frac{1-x}{2}=\frac{y-8}{\lambda}=\frac{z-5}{2} \Rightarrow \frac{x-1}{-2}=\frac{y-8}{\lambda}=\frac{z-5}{2}$
Direction ratio of given lines are $-2, \lambda, 2$ and $5,3,1$.
Given lines are $\perp$ er. $\therefore-2(5)+\lambda(3)+2(1)=0$
$-10+3 \lambda+2=0 \Rightarrow \lambda=\frac{8}{3}$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
If the lines $\frac{x+1}{2}=\frac{y-1}{1}=\frac{z+1}{3}$ and $\frac{x+2}{2}=\frac{y-k}{3}=\frac{z}{4}$ are coplanar, then the value of $k$ is :

Options:
A
$\frac{11}{2}$

B
$-\frac{11}{2}$

C
$\frac{9}{2}$

D
$-\frac{9}{2}$

The correct answer is:
$\frac{11}{2}$

Solution:
Two given planes are coplanar, if
$$
\begin{aligned}
& \left|\begin{array}{ccc}
-2-(-1) & k-1 & 0-(-1) \\
2 & 1 & 3 \\
2 & 3 & 4
\end{array}\right|=0 \\
\Rightarrow & \left|\begin{array}{ccc}
-1 & k-1 & 1 \\
2 & 1 & 3 \\
2 & 3 & 4
\end{array}\right|=0 \\
\Rightarrow & (-1)(4-9)-(k-1)(8-6)+6-2=0 \\
\Rightarrow & k=\frac{11}{2}
\end{aligned}
$$

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2013 (09 Apr Online)
If the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$ intersect, then $k$ is equal to

Options:
A
$-1$

B
$\frac{2}{9}$

C
$\frac{9}{2}$

D
$0$

The correct answer is:
$\frac{9}{2}$

Solution:
Any point on $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}=t$ is $(2 t+1,3 t-1,4 t+1)$
And any point on $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}=\mathrm{s}$ is $(s+3,2 s+k, s)$
Given lines are intersecting
$\Rightarrow \mathrm{t}=-\frac{3}{2} \text { and } \mathrm{s}=-5 \quad \therefore \mathrm{k}=\frac{9}{2}$

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2012 (Offline)
If the lines $2 x+3 y+1=0$ and $3 x-y-4=0$ lie along diameters of a circle of circumference $10 \pi$, then the equation of the circle is

Options:
A
$x^2+y^2-2 x+2 y-23=0$

B
$x^2+y^2-2 x-2 y-23=0$

C
$x^2+y^2+2 x+2 y-23=0$

D
$x^2+y^2+2 x-2 y-23=0$

The correct answer is:
$x^2+y^2-2 x+2 y-23=0$

Solution:
Intersection of given lines is the centre of the circle i.e. $(1,-1)$ Circumference $=10 \pi \Rightarrow$ radius $r=5$
$\Rightarrow$ equation of circle is $x^2+y^2-2 x+2 y-23=0$

Topic: Mathematics, Circle, JEE Main, JEE Main 2004
If the lines $\frac{x-1}{-3}=\frac{y-2}{2 k}=\frac{z-3}{2}, \frac{x-1}{3 k}=\frac{y-5}{1}=\frac{z-6}{-5}$ are at right angles, then $k=$

Options:
A -10
B $\frac{10}{7}$
C $\frac{-10}{7}$
D $\frac{-7}{10}$

The correct answer is:
$\frac{-10}{7}$

Solution:
Lines are perpendicular if $a_1 a_2+b_1 b_2+c_1 c_2=0$
Hence,
$-3(3 k)+2 k(1)+2(-5)=0 \Rightarrow k=-\frac{10}{7} \text {. }$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
If the lines $3 x-4 y-7=0$ and $2 x-3 y-5=0$ are two diameters of a circle of area $49 \pi$ square units, the equation of the circle is

Options:
A
$x^2+y^2+2 x-2 y-47=0$

B
$x^2+y^2+2 x-2 y-62=0$

C
$x^2+y^2-2 x+2 y-62=0$

D
$x^2+y^2-2 x+2 y-47=0$

The correct answer is:
$x^2+y^2-2 x+2 y-47=0$

Solution:
Point of intersection of $3 x-4 y-7=0$ and $2 x-3 y-5=0$ is $(1,-1)$, which is the centre of the circle and radius $=7$.
$\therefore$ Equation is $(x-1)^2+(y+1)^2=49 \Rightarrow x^2+y^2-2 x+2 y-47=0$

Topic: Mathematics, Circle, JEE Main, JEE Main 2006
If the lines $4 x+3 y-1=0, x-y+5=0$ and $k x+5 y-3=0$ are concurrent, then $k$ is equal to :

Options:
A 4
B 5
C 6
D 7

The correct answer is:
6

Solution:
Since, the given three lines are concurrent, then
$\left|\begin{array}{rrr}4 & 3 & -1 \\ 1 & -1 & 5 \\ k & 5 & -3\end{array}\right|=0$
$\Rightarrow 4(3-25)-3(-3-5 k)-1(5+k)=0$
$\Rightarrow-88+9+15 k-5-k=0$
$\Rightarrow \quad 14 k=84$
$\therefore \quad k=6$

Topic: Mathematics, Straight Lines, JEE Main
If the lines $a x+b y+c=0, b x+c y+a=0$ and $c x+a y+b=0$ be concurrent, then

Options:
A $a^3+b^3+c^3+3 a b c=0$
B $a^3+b^3+c^3-a b c=0$
C $a^3+b^3+c^3-3 a b c=0$
D None of these

The correct answer is:
$a^3+b^3+c^3-3 a b c=0$

Solution:
Here the given lines are
$\begin{aligned}
& a x+b y+c=0 \\
& b x+c y+a=0 \\
& c x+a y+b=0
\end{aligned}$
The lines will be concurrent, if
$\begin{aligned}
& \qquad\left|\begin{array}{lll}
a & b & c \\
b & c & a \\
c & a & b
\end{array}\right|=0 \\
& \Rightarrow a^3+b^3+c^3-3 a b c=0 .
\end{aligned}$

Topic: Mathematics, Straight Lines, JEE Main