datavorous/entrance-exam-dataset huggingface dataset:
A tangent and a normal are drawn at the point P(2,-4) on the parabola y2=8x, which meet the directrix of the parabola at the points A and B respectively. If Q(a, b) is a point such that AQBP is a square, then 2a+b is equal to

Options:
A -12
B -20
C -16
D -18

The correct answer is:
-16

Solution:
Equation of tangent at 2, -4 T=0

-4y=4x+2

x+y+2=0 ...(1)

Equation of normal

x-y+λ=0

2, -4

λ=-6

Thus,

x-y=6 ...(2)

Equation of normal

POI of (1) & x=-2 is A-2, 0

POI of 2 & x=-2 is B-2, -8

Given

AQBP is a square

mAQ·mAP=-1

ba+2×4-4=-1

a+2=b

Also PQ must be parallel to x-axis

thus,

b=-4

a=-6

Thus, 2a+b=-16

Topic: Mathematics, Parabola, JEE Main, JEE Main 2021 (27 Aug Shift 1)
A tangent and a normal are drawn at the point P8,8 on the parabola y2=8x which cuts the axis of the parabola at the points A and B respectively. If the centre of the circle through P, A and B is C, then the sum of sinâ PCB and cotâ PCB is equal to

Options:

The correct answer is:
1.55

Solution:
Let, 2t2,4t=8,8=P
ât=2
Equation of the tangent at 8,8 is 2y=x+8
Equation of the normal at 8,8 is y=-2x+24
{Tangent & normal at at2,2at are ty=x+at2 & y=-tx+2at+at3 }
âA=-8,0 & B=12,0
Because ÎPAB is a right-angled triangle
Hence, the centre of the circumcircle is the mid-point of the hypotenuse AB is C=2,0
The slope of line PC is 8-08-2=43
âtanâ PCB=43, sinâ PCB=45
âcotâ PCB+sinâ PCB=0.75+0.8=1.55

Topic: Mathematics, Parabola, JEE Main
A tangent drawn to hyperbola \( \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 \) at \( P\left(\frac{\pi}{6}\right) \) forms a triangle of area \( 3 a^{2} \) square units, with coordinate axes, then the square of its eccentricity.

Options:

The correct answer is:
17.00

Solution:
Pasecπ6, b tanπ6≡P2a3,b3

Therefore, equation of tangent at P is X3a2-

y3b=1

∴ Area of the triangle =12×3a2×3 b=3a2

⇒ba=4 ∴e2=1+b2a2=17

Topic: Mathematics, Hyperbola, JEE Main
A tangent drawn to the hyperbola x2a2-y2b2=1 at Pasec⁡π6,btan⁡π6 form a triangle of area 3a2 sq. units with the coordinate axes. The eccentricity of the conjugate hyperbola of x2a2-y2b2=1 is

Options:
A 17
B 174
C 172
D 817

The correct answer is:
174

Solution:
P=2a3,b3
Equation of the tangent at P is 2x3a-y3b=1
x intercept & y intercept of the tangent are 3a2 & -3b respectively
Area of the triangle formed by the tangent with the coordinate axes is
12×3a2×3b=3a2
⇒ba=4⇒ab=14
The eccentricity of the conjugate hyperbola equals to
1+116=174

Topic: Mathematics, Hyperbola, JEE Main
A tangent having slope -43 to the ellipse x218 + y232=1 intersects the major and minor axes at A and B. If O is the origin, then the area of ÎOAB is

Options:
A 48Â sq.Â units
B 9Â sq.Â units
C 24 sq. units
D 16Â sq.Â units

The correct answer is:
24 sq. units

Solution:
Any point on the ellipse x2322 + y2422=1 can be taken as 32cosâ¡Î¸, 42sinâ¡Î¸ and the slope of the tangent as - b2xa2y
Hence, slope=- 3232cosâ¡Î¸18 42sinâ¡Î¸=- 43cotâ¡Î¸â¦. 1
Given, the slope of the tangent =-43â¦. 2
From equations (1) and (2), we get,
cotâ¡Î¸=1âÎ¸=Ï4
Hence, the equation of the tangent is x.1232 + y. 1242=1
(i.e.) x6 + y8=1
Hence, A=6, 0, B=0, 8
Area of ÎOAB=12Ã 6 Ã 8=24 sq. units

Topic: Mathematics, Ellipse, JEE Main
A tangent is drawn to the parabola y2=6x which is perpendicular to the line 2x+y=1. Which of the following points does NOT lie on it?

Options:
A 0,3
B 4,5
C 5,4
D -6,0

The correct answer is:
5,4

Solution:
Slope of tangent =mT=m

So, m-2=-1⇒m=12

Equation: y=mx+am

⇒ y=12x+32×12 ∵a=64=32

⇒ y=x2+3

⇒ 2y=x+6

Point 5, 4 will not lie on it

Topic: Mathematics, Parabola, JEE Main, JEE Main 2021 (25 Feb Shift 1)
A tangent is drawn to the parabola y2=8x at P2,4 to intersect the x-axis at Q, from which another tangent is drawn to the parabola to touch it at R. If the normal at R intersects the parabola again at S, then the coordinates of S are

Options:
A 6,4
B 18,12
C 8,8
D 8,6

The correct answer is:
18,12

Solution:
P is one of the end points of the latus rectum.

So, Q lies on the directrix.
Hence, R=2,-4.
Equation of the normal at R is y=x-6
âS=18,12

Topic: Mathematics, Parabola, JEE Main
A tangent line L is drawn at the point 2, -4 on the parabola y2=8x. If the line L is also tangent to the circle x2+y2=a, then a is equal to

.

Options:

The correct answer is:
2

Solution:
Given, Parabola y2=8x and circle x2+y2=a2

Parabola y2=8x at point 2,-4

Differentiating w .r. t. x''

2ydydx=8

dydx=4y

dydx2,-4=-1

Equation of tangent

y-y1=dydxx-x1

y+4=-1x-2

x+y+2=0

Line L is also tangent to the circle x2+y2=a2

So, OA=a at 0,0

1×0+1×0+22=a

2=a

⇒a=2

Topic: Mathematics, Parabola, JEE Main, JEE Main 2021 (31 Aug Shift 2)
A tangent of slope 2 of the ellipse x2a2+y21=1 passes through -2,0. Then, three times the square of the eccentricity of the ellipse is equal to

Options:

The correct answer is:
2.2

Solution:
A tangent of slope 2 to the ellipse x2a2+y21=1 is
y=2xÂ±4a2+1 which passes through -2,0
â0=2-2Â±4a2+1
â4a2+1=16âa2=154
âe2=1-415=1115
â3e2=2.2

Topic: Mathematics, Ellipse, JEE Main
A tangent to the circle $x^2+y^2=5$ at the point $(1,-2) \ldots .$. the circle $x^2+y^2-8 x+6 y+20=0$

Options:
A Touches
B Cuts at real points
C Cuts at imaginary point
D None of these

The correct answer is:
Cuts at real points

Solution:
equation of tangent to the circle $x^2+y^2=5$ at $\left(x_1, y_1\right)$will be $\mathrm{xx}_1+\mathrm{yy}_1=5$
$\begin{aligned}& x \times 1+y(-2)=5 \\& x-2 y=5 \text { or } x=(2 y+5)\end{aligned}$
now this is tangent at
circle $x^2+y^2-8 x+6 y+20=0$
or $(2 y+5)^2+y^2-8(2 y+5)+6 y+20=0$
or $4 y^2+20 y+25+y^2-16 y-40+6 y+20=0$
or $5 y^2+10 y+5=0$
or $\mathrm{y}^2+2 \mathrm{y}+1=0$
$(y+1)^2=0$
$\mathrm{y}=-1$
$\therefore \mathrm{x}=2(-1)+5=3$
$\therefore$ point at which tangent is $(3,-1)$

Topic: Mathematics, Circle, JEE Main
A tangent to the curve $9 b^2 x^2-4 a^2 y^2=36 a^2 b^2$ makes intercepts of unit length on each of the coordinate axes, then the point $(a, b)$ lies on

Options:
A $x^2-y^2=1$
B $x^2+y^2=1$
C $4 x^2-9 y^2=1$
D $4 x^2+9 y^2=1$

The correct answer is:
$4 x^2-9 y^2=1$

Solution:
Equation of given curve is,
$$
\frac{x^2}{4 a^2}-\frac{y^2}{9 b^2}=1
$$
Let at point $(2 a \sec \theta, 3 b \tan \theta)$ on the curve (i).
So, equation of tangent at point is
$$
\frac{x}{\frac{2 a}{\sec \theta}}+\frac{y}{-\frac{3 b}{\tan \theta}}=1
$$
According to the question,
$$
2 a=\sec \theta \text { and } 3 b=-\tan \theta
$$
So,
$$
4 a^2-9 b^2=1
$$
On taking locus of point $(a, b)$, we are getting
$$
4 x^2-9 y^2=1 \text {. }
$$

Topic: Mathematics, Hyperbola, JEE Main
A tangent to the curve, y=fx at Px, y meets x-axis at A and y-axis at B. If AP:BP=1:3 and f1=1, then the curve also passes through the point

Options:
A 13, 24
B 12, 4
C 2,18
D 3,128

The correct answer is:
2,18

Solution:
Equation of tangent,
Y−y= dy dx (X−x) A( x−y dx dy ,0 )  ​:  B( 0,y−x dy dx )
Given, AP:BP=1:3,

Using section formula, we get

y=y−xdydx4

⇒y−xdydx=4y

⇒dyy+3dxx=0
By integrating w.r.t x, we get
ln|y|+3ln|x|=lnc

|yx3|=c

∵f(1)=1

∴c=1

⇒|yx3|=1

Hence, the curve passes throught the point 2,18

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2017 (09 Apr Online)
A tangent to the hyperbola $\frac{x^2}{4}-\frac{y^2}{2}=1$ meets $x$-axis at $\mathrm{P}$ and $y$-axis at $\mathrm{Q}$. Lines $\mathrm{PR}$ and $\mathrm{QR}$ are drawn such that OPRQ is a rectangle (where $\mathrm{O}$ is the origin). Then $\mathrm{R}$ lies on :

Options:
A
$\frac{4}{x^2}+\frac{2}{y^2}=1$

B
$\frac{2}{x^2}-\frac{4}{y^2}=1$

C
$\frac{2}{x^2}+\frac{4}{y^2}=1$

D
$\frac{4}{x^2}-\frac{2}{y^2}=1$

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

Solution:
Equation of the tangent at the point ' $\theta$ ' is
$$
\begin{aligned}
& \frac{x \sec \theta}{a}-\frac{y \tan \theta}{b}=1 \\
& \Rightarrow P=(a \cos \theta, 0) \text { and } Q=(0,-b \cot \theta) \\
& \text { Let } \mathrm{R} \text { be }(\mathrm{h}, \mathrm{k}) \Rightarrow \mathrm{h}=a \cos \theta, \mathrm{k}=-b \cot \theta \\
& \Rightarrow \quad \frac{k}{h}=\frac{-b}{a \sin \theta} \Rightarrow \sin \theta=\frac{-b h}{a k} \quad \text { and } \\
& \cos \theta=\frac{h}{a}
\end{aligned}
$$
By squaring and adding,
$$
\frac{b^2 h^2}{a^2 k^2}+\frac{h^2}{a^2}=1
$$
$$
\begin{aligned}
& \Rightarrow \frac{b^2}{k^2}+1=\frac{a^2}{h^2} \\
& \Rightarrow \frac{a^2}{h^2}-\frac{b^2}{k^2}=1
\end{aligned}
$$
Now, given $\mathrm{q}^{\mathrm{n}}$ of hyperbola is $\frac{x^2}{4}-\frac{y^2}{2}=1$
$$
\Rightarrow a^2=4, b^2=2
$$
$\therefore$ R lies on $\frac{a^2}{x^2}-\frac{b^2}{y^2}=1$ i.e., $\frac{4}{x^2}-\frac{2}{y^2}=1$

Topic: Mathematics, Hyperbola, JEE Main
A tangent to the hyperbola $\frac{x^2}{4}-\frac{y^2}{2}=1$ meets $x$-axis at $\mathrm{P}$ and $y$-axis at $\mathrm{Q}$. Lines $\mathrm{PR}$ and $\mathrm{QR}$ are drawn such that OPRQ is a rectangle (where $\mathrm{O}$ is the origin). Then $\mathrm{R}$ lies on :

Options:
A
$\frac{4}{x^2}+\frac{2}{y^2}=1$

B
$\frac{2}{x^2}-\frac{4}{y^2}=1$

C
$\frac{2}{x^2}+\frac{4}{y^2}=1$

D
$\frac{4}{x^2}-\frac{2}{y^2}=1$

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

Solution:
Equation of the tangent at the point ' $\theta$ ' is
$$
\begin{aligned}
& \frac{x \sec \theta}{a}-\frac{y \tan \theta}{b}=1 \\
& \Rightarrow P=(a \cos \theta, 0) \text { and } Q=(0,-b \cot \theta) \\
& \text { Let } \mathrm{R} \text { be }(\mathrm{h}, \mathrm{k}) \Rightarrow \mathrm{h}=a \cos \theta, \mathrm{k}=-b \cot \theta \\
& \Rightarrow \quad \frac{k}{h}=\frac{-b}{a \sin \theta} \Rightarrow \sin \theta=\frac{-b h}{a k} \quad \text { and } \\
& \cos \theta=\frac{h}{a}
\end{aligned}
$$
By squaring and adding,
$$
\frac{b^2 h^2}{a^2 k^2}+\frac{h^2}{a^2}=1
$$
$$
\begin{aligned}
& \Rightarrow \frac{b^2}{k^2}+1=\frac{a^2}{h^2} \\
& \Rightarrow \frac{a^2}{h^2}-\frac{b^2}{k^2}=1
\end{aligned}
$$
Now, given $\mathrm{q}^{\mathrm{n}}$ of hyperbola is $\frac{x^2}{4}-\frac{y^2}{2}=1$
$$
\Rightarrow a^2=4, b^2=2
$$
$\therefore$ R lies on $\frac{a^2}{x^2}-\frac{b^2}{y^2}=1$ i.e., $\frac{4}{x^2}-\frac{2}{y^2}=1$

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2013 (23 Apr Online)
A tangent to the parabola $y^2=8 x$ makes an angle of $45^{\circ}$ with the straight line $y=3 x+5$, then the equation of tangent is

Options:
A $2 x+y-1=0$
B $x+2 y-1=0$
C $2 x+y+1=0$
D None of these

The correct answer is:
$2 x+y+1=0$

Solution:
The equation of any tangent to the parabola
\(y^2=8 x \text { is } y=m x+\frac{2}{m}\)
Clearly,
\(\begin{aligned}
& \tan \left(45^{\circ}\right)=\left|\frac{\mathrm{m}-3}{1+3 \mathrm{~m}}\right| \\
& \Rightarrow \frac{\mathrm{m}-3}{1+3 \mathrm{~m}}= \pm 1 \\
& \Rightarrow \frac{\mathrm{m}-3}{1+3 \mathrm{~m}}=1 \text { and } \\
& \frac{\mathrm{m}-3}{1+3 \mathrm{~m}}=-1 \\
& \Rightarrow \mathrm{m}=-2, \frac{1}{2}
\end{aligned}\)
Hence, the equations of tangents are
\(\begin{aligned}
& y=-2 x-1, y=\frac{x}{2}+4 \\
& \Rightarrow y=-2 x-1, x-2 y+8=0
\end{aligned}\)

Topic: Mathematics, Parabola, JEE Main
A tangential force of $20 \mathrm{~N}$ is applied on a cylinder of mass $4 \mathrm{~kg}$ and moment of inertia $\mid 0.02 \mathrm{~kg} \mathrm{~m}^2 \mid$ about its own axis. If the cylinder rolls without slipping, then linear acceleration of its centre of mass will be

Options:
A $6.7 \mathrm{~m} / \mathrm{s}^2$
B $10 \mathrm{~m} / \mathrm{s}^2$
C $3.3 \mathrm{~m} / \mathrm{s}^2$
D None of these

The correct answer is:
$6.7 \mathrm{~m} / \mathrm{s}^2$

Solution:
Let friction force $=\mathrm{f} F+f=m a \mid$
(i) $(F-f) R=I \alpha \mid$
From $e q^n s .(i)$ and (ii),$
$\left|2 F=m a+\frac{I \alpha}{R}\right|$
Use $\left.\alpha=\frac{a}{R} \right\rvert\,$ (for pure rolling)
$2 F=m a+\frac{I a}{R^2}\left|40=4 a+\frac{0.02 a}{(0.1)^2} ; a=\frac{40}{6}=6.7 \mathrm{~m} / \mathrm{s}^2\right|$

Topic: Physics, Rotational Motion, JEE Main
A tank is filled with water up to a height $H$. Water is allowed to come out of a hole $P$ in one of the walls at a depth $D$ below the surface of water. Express the horizontal distance $x$ in terms of

Options:
A $x=\sqrt{D(H-D)}$
B $x=\sqrt{\frac{D(H-D)}{2}}$
C $x=2 \sqrt{D(H-D)}$
D $x=4 \sqrt{D(H-D)}$

The correct answer is:
$x=2 \sqrt{D(H-D)}$

Solution:
Time taken by water to reach the bottom
$=t=\sqrt{\frac{2(H-D)}{g}}$
and velocity of water coming out of hole, $v=\sqrt{2 g D}$
$\therefore$ Horizontal distance covered $x=v \times t$
$=\sqrt{2 g D} \times \sqrt{\frac{2(H-D)}{g}}=2 \sqrt{D(H-D)}$

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A tank with a rectangular base and rectangular sides, open at the top is to be constructed so that its depth is 4 meter and volume is 36 cubic meters. If building of the tank costs $₹ 100$ per square meter for the base and $₹ 50$ per square meter for the sides, then the cost of least expensive tank is

Options:
A $₹ 3000$
B $₹ 3300$
C $₹ 2400$
D $₹ 3500$

The correct answer is:
$₹ 3300$

Solution:
Let length and breadth of the tank be ' $x$ ' $\mathrm{m}$ and ' $y$ ' $m$ respectively.
Height of the tank is $4 \mathrm{~m}$.
Volume $=36 \mathrm{~m}^3$
$$
\begin{array}{ll}
\therefore & 4 x y=36 \\
\therefore & x y=9 \\
\therefore & y=\frac{9}{x}
\end{array}
$$
$\therefore \quad$ Total area of the tank including sides and base $=x y+2(4 x)+2(4 y)$
....[From (i) and (ii)]
$\therefore \quad \mathrm{f}(x)=9+8 x+8\left(\frac{9}{x}\right)$
$$
=9+8 x+\frac{72}{x}
$$
$$
\therefore \quad \mathrm{f}^{\prime}(x)=8-\frac{72}{x^2}
$$
$\therefore \quad \mathrm{f}^{\prime}(x)=0 \Rightarrow x=3$
$$
\Rightarrow y=3
$$
$\therefore \quad$ Required cost $=100 \times(3 \times 3)+50$
$$
\begin{aligned}
& \times(2 \times 4 \times 3+2 \times 4 \times 3) \\
= & 900+2400 \\
= & ₹ 3300
\end{aligned}
$$

Topic: Mathematics, Application of Derivatives, JEE Main
A tank with a small hole at the bottom has been filled with water and kerosene (specific gravity $0.8$ ). The height of water is $3 \mathrm{~m}$ and that of kerosene $2 \mathrm{~m}$. When the hole is opened the velocity of fluid coming out from it is nearly: (take $\mathrm{g}=10 \mathrm{~ms}^{-2}$ and density of water $\left.=10^3 \mathrm{~kg} \mathrm{~m}^{-3}\right)$

Options:
A
$10.7 \mathrm{~ms}^{-1}$

B
$9.6 \mathrm{~ms}^{-1}$

C
$8.5 \mathrm{~ms}^{-1}$

D
$7.6 \mathrm{~ms}^{-1}$

The correct answer is:
$9.6 \mathrm{~ms}^{-1}$

Solution:
$$
\begin{aligned}
&h_1 \rho_1 g+h_2 \rho_2 g=\frac{1}{2} \rho_1 \mathrm{v}^2 \\
&\Rightarrow 1000 \times 10 \times 3+800 \times 10 \times 2 \\
&=\frac{1}{2} \times 1000 \times \mathrm{v}^2 \\
&\Rightarrow \mathrm{v}^2=2 \times \frac{46000}{1000} \\
&\Rightarrow \mathrm{v}=\sqrt{92}=9.6 \mathrm{~m} / \mathrm{s}
\end{aligned}
$$

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2014 (11 Apr Online)
A target is made of two plates, one of wood and the other of iron. The thickness of the wooden plate is 4 cm and that of the iron plate is 2 cm. A bullet fired goes through the wood first and then penetrates 1 cm into iron. A similar bullet fired with the same velocity from the opposite direction goes through iron first and then penetrates 2 cm into the wood. If a1 and a2 be the retardation offered to the bullet by wood and iron plates respectively, then

Options:
A a1=2a2
B a2=2a1
C a1=a2
D Data insufficient

The correct answer is:
a2=2a1

Solution:
Let a1and a2be the retardations offered to be bullet by wood and iron respectively.
For AâBâC,
v12-u2=2a14, and 02-v12=2a2(1)
Adding, we get
-u2=2(4a1+a2) ...(i)
For A′âB′âC′,
v22-u2=2a22
and 02-v22=2a12
Adding, we get
-u2=2(2a1+2a2) ....(ii)
Equating Eqs. (i) and (ii) and solving, we get
4a1+a2=2a1+2a2
âÂ a2=2a1

Topic: Physics, Motion In One Dimension, JEE Main
A teacher in his physics laboratory allotted an experiment to determine the resistance G a galvanometer. Students took the observations for 13 deflection in the galvanometer. Which of the below is true for measuring value of G?

Options:
A 13 deflection method cannot be used for determining the resistance of the galvanometer.
B 13 deflection method can be used and in this case the G equals to twice the value of shunt resistance(s).
C 13 deflection method can be used and in this case, the G equals to three times the value of shunt resistance(s).
D 13 deflection method can be used and in this case the G value equals to the shunt resistance(s)

The correct answer is:
13 deflection method can be used and in this case the G equals to twice the value of shunt resistance(s).

Solution:
If a large resistance R is connected with a battery, then current will be approximately εR, and t will remain almost constant.

In 1st case, only G is connected ⇒ig=i

In 2nd case,

igR0=I-igS⇒ig2=SRg+Si

If ig2=13ig,⇒SRg+Si=13i

Rg=2S

Topic: Physics, Current Electricity, JEE Main, JEE Main 2022 (25 Jun Shift 1)
A teacher was trying to show the different stages of a type of cell division to the students. At a particular stage, it was observed that the chromosomes were separating but there was no splitting of centromeres. This stage would be

Options:
A metaphase I
B metaphase II
C anaphase I
D anaphase II

The correct answer is:
anaphase I

Solution:
Meiosis is a reductional division process where the cell undergoes two successive division processes to reduce chromosome number to half. It is divided into the meiotic stages I and II. Anaphase I is part of the meiotic stage I, where the sister chromatids separate from its homologous counterpart. The chromatids centromere remains intact in this stage and do not separate. Their separation occurs in anaphase II in meiotic stage II.

Topic: Biology, Cell Cycle and Cell Division, NEET
A telephone wire of length 200 km has a capacitance of 0.014 Î¼F km-1. If it carries an AC frequency of 5âkHz, what should be the value of an inductor required to be connected in series so that the impedance of the circuit is minimum?

Options:
A 0.35âmH
B 3.5 mH
C 2.5 mH
D zero

The correct answer is:
0.35 mH

Solution:
Capacitance of wire
C=0.014×10-6×200
=2.8×10-6 F=2.8 μF
For impedance of the circuit to be minimum
XL=XC
⇒2πfL=12πfC
L=14π2f2C
=143.142×5×1032×2.8×10-6
=0.35×10-3 H=0.35 mH

Topic: Physics, Alternating Current, JEE Main
A telephonic communication service is working at a carrier frequency of 10 GHz. Only 10% of it is utilized for transmission. How many telephonic channels can be transmitted simultaneously if each channel requires a bandwidth of 5 kHz?

Options:
A 2×106
B 2×103
C 2×104
D 2×105

The correct answer is:
2×105

Solution:
Number of telephonic channels that can be transmitted simultaneously,
=0.1×10×1095×103=2×105

Topic: Physics, Communication System, JEE Main, JEE Main 2018 (08 Apr)
A telescope has an objective lens of 10 $\mathrm{cm}$ diameter and is situated at a distance of one kilometers from two objects. The minimum distance between these two objects, which can be resolved by the telescope, when the mean wavelength of light is $5000 Å$, is of the order of:

Options:
A $0.5 \mathrm{~m}$
B $5 \mathrm{~m}$
C $5 \mathrm{~mm}$
D $5 \mathrm{~cm}$

The correct answer is:
$5 \mathrm{~mm}$

Solution:
According to the question:

$\begin{array}{ll}
\theta =\frac{1.22}{a} \lambda \\
\therefore \frac{d}{D} =\frac{\lambda}{\mathrm{a}} \\
\Rightarrow a^{\prime}=\frac{\lambda D}{a} \\
a =\frac{5000 \times 10^{-10} \times 10^3}{10 \times 10^{-2}}=5 \mathrm{~mm}
\end{array}$

Topic: Physics, Ray Optics, JEE Main
A telescope has an objective lens of focal length 150 cm and an eyepiece of focal length 5 cm. If a 50 m tall tower at a distance of 1 km is observed through this telescope in a normal setting, the angle formed by the image of the tower is θ, then θ is close to

Options:
A 30°
B 15°
C 1°
D 86°

The correct answer is:
86°

Solution:
Angular magnification of a telescope in normal adjustment is

m=βα=f0fe⇒β=f0feα
According to the question

α=yx=501000=120

⇒α=120

⇒β=1505×120=32 rad

=32×180°3.14=86°

Topic: Physics, Ray Optics, JEE Main, JEE Main 2015 (10 Apr Online)
A telescope of aperture $3 \times 10^{-2} \mathrm{~m}$ diameter is focused on a window at $80 \mathrm{~m}$ distance fitted with a wire mesh of spacing $2 \times 10^{-3} \mathrm{~m}$. Given: $\lambda=5.5 \times 10^{-7} \mathrm{~m}$, which of the following is true for observing the mesh through the telescope?

Options:
A
Yes, it is possible with the same aperture size.

B
Possible also with an aperture half the present diameter.

C
No, it is not possible.

D
Given data is not sufficient.

The correct answer is:
Yes, it is possible with the same aperture size.

Solution:
Given : $d=3 \times 10^{-2} \mathrm{~m}$
$$
\lambda=5.5 \times 10^{-7} \mathrm{~m}
$$
Limit of resolution, $\Delta \theta=\frac{1.22 \lambda}{d}$
$$
=\frac{1.22 \times 5.5 \times 10^{-7}}{3 \times 10^{-2}}=2.23 \times 10^{-5} \mathrm{rad}
$$
At a distance of $80 \mathrm{~m}$, the telescope is able to resolve between two points which are separated by $2.23 \times 10^{-5} \times 80 \mathrm{~m}$ $=1.78 \times 10^{-3} \mathrm{~m}$

Topic: Physics, Ray Optics, JEE Main, JEE Main 2012 (26 May Online)
A temporary family planning method based on the natural effect of breastfeeding on fertility is due to the high level of:

Options:
A Progesterone
B Prolactin
C Estrogen
D FSH

The correct answer is:
Prolactin

Solution:
Progesterone is a female sex hormone. It's produced mainly in the ovaries and it helps in maintenance of pregnancy.
Prolactin causes the breasts to grow and make milk during pregnancy and after birth. Prolactin levels are normally high for pregnant women and new mothers.
Estrogen also helps regulate the menstrual cycle, controlling the growth of the uterine lining during the first part of the cycle and also control secondary sexual characters in female.
FSH helps control the menstrual cycle and stimulates the growth of eggs in the ovaries

Topic: Biology, Reproductive Health, NEET
A tennis ball hits the floor with a speed $v$ at an angle $\theta$ with the normal to the floor. If the collision is inelastic and the co-efficient of restitution is $\varepsilon$, what will be the angle of reflection?

Options:
A $\tan ^{-1}\left(\frac{\tan \theta}{\varepsilon}\right)$
B $\sin ^{-1}\left(\frac{\sin \theta}{\varepsilon}\right)$
C $\theta \varepsilon$
D $\theta \frac{2 \varepsilon}{\varepsilon+1}$

The correct answer is:
$\tan ^{-1}\left(\frac{\tan \theta}{\varepsilon}\right)$

Solution:
Hint:

$\tan \theta^{\prime}=\frac{\operatorname{usin} \theta}{\operatorname{eucos} \theta} \quad \tan \theta^{\prime}=\frac{1}{\mathrm{e}} \tan \theta$
Since the floor is smooth. Hence tangential component of velocity remains unchanged.

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A tennis ball hits the floor with a speed $v$ at an angle $\theta$ with the normal to the floor. If the collision is inelastic and the co-efficient of restitution is $\varepsilon$, what will be the angle of reflection?

Options:
A $\tan ^{-1}\left(\frac{\tan \theta}{\varepsilon}\right)$
B $\sin ^{-1}\left(\frac{\sin \theta}{\varepsilon}\right)$
C $\theta \varepsilon$
D $\theta \frac{2 \varepsilon}{\varepsilon+1}$

The correct answer is:
$\tan ^{-1}\left(\frac{\tan \theta}{\varepsilon}\right)$

Solution:
Hint:

$\tan \theta^{\prime}=\frac{\operatorname{usin} \theta}{\operatorname{eucos} \theta} \quad \tan \theta^{\prime}=\frac{1}{\mathrm{e}} \tan \theta$
Since the floor is smooth. Hence tangential component of velocity remains unchanged.

Topic: Physics, Center of Mass Momentum and Collision, WBJEE, WBJEE 2020
A tennis ball is dropped on to the floor from a height of 9.8 m. It rebounds to a height 5.0 m. Ball comes in contact with the floor for 0.2 s. The average acceleration during contact is ______ m s-2. [Given g=10 m s-2]

Options:

The correct answer is:
120

Solution:
The speed of ball just before collision with ground is vi2=0+2ghi

⇒vi=2ghi

=2×10×9.8

=14 m s-1 Downward

The speed of ball just after collision is 0=vf2-2ghf

⇒vf=2ghf

=2×10×5

=10 m s-1 Upward

Average acceleration of ball is a→avg=Δv→Δt=vf--vit=10+140.2=240.2=120 m s-2.

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2023 (29 Jan Shift 1)
A tennis ball is released from height $h$ above ground level. If the ball makes inelastic collision with the ground, to what height will it rise after third collision

Options:
A $h e^6$
B $e^2 h$
C $e^3 h$
D None of these

The correct answer is:
$h e^6$

Solution:
$h_n$ he $2 n$
after third
collision $h_3$ he 6 [as $n=3$ ]

Topic: Physics, Center of Mass Momentum and Collision, JEE Main