datavorous/entrance-exam-dataset huggingface dataset:
The locus of the point of intersection of any two perpendicular tangents to the hyperbola is a circle which is called the director circle of the hyperbola, then the $\mathrm{eq}^{\mathrm{n}}$ of this circle is

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

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

Solution:
Equation of hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
Any tangent to hyperbola are $y=m x \pm \sqrt{a^2 m^2-b^2}$
Also tangent perpendicular to this is $y=\frac{-1}{m} \times \pm \sqrt{\frac{a^2}{m^2}-b^2}$
Eliminating $m$, we get $x^2+y^2=a^2-b^2$.

Topic: Mathematics, Hyperbola, JEE Main
The locus of the point of intersection of any two perpendicular tangents to the hyperbola is a circle which is called the director circle of the hyperbola, then the egn of this circle is

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

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

Solution:
Equation of hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
Any tangent to hyperbola are $y=m x \pm \sqrt{a^2 m^2-b^2}$
Also tangent perpendicular to this is $y=\frac{-1}{m} x \pm \sqrt{\frac{a^2}{m^2}-b^2}$ Eliminating $m$, we get $x^2+y^2=a^2-b^2$.

Topic: Mathematics, Hyperbola, JEE Main
The locus of the point of intersection of any two perpendicular tangents to the hyperbola is a circle which is called the director circle of the hyperbola, then the equn of this circle is

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

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

Solution:
Equation of hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
Any tangent to hyperbola are $y=m x \pm \sqrt{a^2 m^2-b^2}$
Also tangent perpendicular to this is $y=\frac{-1}{m} \times \pm \sqrt{\frac{a^2}{m^2}-b^2}$ Eliminating $m$, we get $x^2+y^2=a^2-b^2$.

Topic: Mathematics, Hyperbola, JEE Main
The locus of the point of intersection of lines $(x+y) t=a$ and $x-y=a t$, where $t$ is the parameter, is

Options:
A A circle
B An ellipse
C A rectangular hyperbola
D None of these

The correct answer is:
A rectangular hyperbola

Solution:
Multiplying both, we get $x^2-y^2=a^2$. This is equation of rectangular hyperbola as $a=b$

Topic: Mathematics, Hyperbola, JEE Main
The locus of the point of intersection of lines $(x+y) t=a$ and $x-y=a t$, where $t$ is the parameter, is

Options:
A A circle
B An ellipse
C A rectangular hyperbola
D None of these

The correct answer is:
A rectangular hyperbola

Solution:
Multiplying both, we get $x^2-y^2=a^2$. This is equation of rectangular hyperbola as $a=b$.

Topic: Mathematics, Hyperbola, JEE Main
The locus of the point of intersection of mutually perpendicular tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, is

Options:
A A straight line
B A parabola
C A circle
D None of these

The correct answer is:
A circle

Solution:
A circle is the locus of the point of intersection of mutually perpendicular tangent to the ellipse will form director circle.

Topic: Mathematics, Ellipse, JEE Main
The locus of the point of intersection of tangents to the circle \( x=a \cos \theta, y=a \sin \theta \) at a point whose parametric angles differ by \( \pi / 3 \) is \( x^{2}+y^{2}=k a^{2} \). Find the value of \( 3 k \).

Options:

The correct answer is:
4

Solution:
x2 + y2 = a2
A(a cosθ, a sinθ)
B acos⁡θ+π3,asin⁡θ+π3
Equation of tangent at A
x cosθ + y sinθ = a ……… (i)
Equation of tangent at B
x cos θ+π3 + y sin θ+π3 = a
xcos⁡θ2-3sin⁡θ2+ysin⁡θ2+3cos⁡θ2=a
12[xcos⁡θ+ysin⁡θ)-32(xsin⁡θ-ycos⁡θ)=a
a2-32(xsin⁡θ-ycos⁡θ)=a
3[xsin⁡θ-ycos⁡θ)=-a
x sinθ – y cosθ = – a/ 3 … (ii)
on squaring and adding Equation (i) and (ii)
x2 + y2 = 43 a2 ⇒ k = 43 = 3k = 4

Topic: Mathematics, Circle, JEE Main
The locus of the point of intersection of the lines 2x-y+42k=0 and 2kx+ky-42=0 (k is any non-zero real parameter) is

Options:
A an ellipse whose eccentricity is 13
B a hyperbola whose eccentricity is 3
C a hyperbola with length of its transverse axis 82
D an ellipse with length of its major axis 82

The correct answer is:
a hyperbola with length of its transverse axis 82

Solution:
Given equation of lines are 2x-y+42k=0 ...1

and 2kx+ky-42=0 ...2

Eliminating k from 1 and 2,

⇒2x+y2x-y-42=42

⇒2x2-y2=-32

⇒y232-x216=1, which is a hyperbola.

⇒e=1+1632=32 and length of transverse axis

=82.

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2018 (16 Apr Online)
The locus of the point of intersection of the lines 3kx+ky-43=0 and 3x-y-43k=0 is a conic, whose eccentricity is

Options:

The correct answer is:
2

Solution:
The point of intersection of the lines 3kx+ky-43=0 and 3x-y-43k=0.

k=433x+y=3x-y43

⇒ 3x2-y2=48

⇒ x216-y248=1

Which is a hyperbola.

b2=a2e2-1

Now, 48=16e2-1

⇒ e=4=2

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2021 (25 Feb Shift 1)
The locus of the point of intersection of the lines $\sqrt{3} x-y-4 \sqrt{3} k=0$ and $\sqrt{3} k x+k y-4 \sqrt{3}=0$ for different value of $k$ is

Options:
A Circle
B Parabola
C Hyperbola
D Ellipse

The correct answer is:
Hyperbola

Solution:
Multiplying both, we get $3 x^2-y^2=48$ or $\frac{x^2}{(48 / 3)}-\frac{y^2}{48}=1$, which is a hyperbola.

Topic: Mathematics, Hyperbola, JEE Main
The locus of the point of intersection of the lines $a x \sec \theta+b y \tan \theta=a$ and $a x \tan \theta+b y \sec \theta=b$ where $\theta$ is the parameter, is

Options:
A A straight line
B A circle
C An ellipse
D A hyperbola

The correct answer is:
A hyperbola

Solution:
Squaring and subtracting, we get $a^2 x^2-b^2 y^2=a^2-b^2$, which is the equation of hyperbola.

Topic: Mathematics, Hyperbola, JEE Main
The locus of the point of intersection of the lines $b x t-a y t=a b$ and $b x+a y=a b t$ is

Options:
A A parabola
B An ellipse
C A hyperbola
D None of these

The correct answer is:
A hyperbola

Solution:
Multiplying both, we get
$(b x)^2-(a y)^2=(a b)^2 \Rightarrow \frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
which is the standard equation of hyperbola

Topic: Mathematics, Hyperbola, JEE Main
The locus of the point of intersection of the perpendicular tangents to the parabola $x^2=4 a y$ is

Options:
A Axis of the parabola
B Directrix of the parabola
C Focal chord of the parabola
D Tangent at vertex to the parabola

The correct answer is:
Directrix of the parabola

Solution:
It is a fundamental property.

Topic: Mathematics, Parabola, JEE Main
The locus of the point of intersection of the straight lines, tx-2y-3t=0 and x-2ty+3=0 t∈R, is:

Options:
A A hyperbola with the length of conjugate axis 3
B A hyperbola with eccentricity 5
C An ellipse with the length of major axis 6
D An ellipse with eccentricity 25

The correct answer is:
A hyperbola with the length of conjugate axis 3

Solution:
tx-2y-3t=0.......1
x-2ty+3=0......2 ,

Multiply by t and subtract from above equation, we get

y2t2-2=6t

Now multiply by t in first equation and subtract second, we get

t2-1x=3t2-1

⇒x=3(t2+1)t2-1
⇒x29=t2+1t2-12 & 2y3=2tt2-1
⇒4y29=4t2t2-12
⇒ x29-4y29=1
⇒x29-y294=1
a2=9;⇒a=3

b2=94⇒b=32
∴ Length of conjugate axis =2b=2×32=3
Length of transverse axis=2a= 6

e2=1+949=1+14; e=52

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2017 (08 Apr Online)
The locus of the point of intersection of the tangents at the extremities of a chord of the circle x2+y2=r2 which touches the circle x2+y2+2rx=0 is

Options:
A y2=2rx-r2
B y2=-2rx+r2
C y2=2rx+r2
D y2=-2rx-r2

The correct answer is:
y2=2rx+r2

Solution:
Let, point of intersection of the tangents is h,k.
Chord of contact through h,k for the circle x2+y2=r2 is
hx+ky-r2=0 which touches the circle x2+y2+2rx=0
⇒ The distance of hx+ky-r2=0 from the centre -r,0 of circle x2+y2+2rx=0
is equal to the radius r of the circle x2+y2+2rx=0
⇒h-r+k0-r2h2+k2=r
⇒h+r=h2+k2
⇒x2+2rx+r2=x2+y2⇒y2=2rx+r2 is the required locus

Topic: Mathematics, Circle, JEE Main
The locus of the point of intersection of two tangents of the hyperbola x22-y24=1, if the product of their slopes is 1, is

Options:
A y2-4=x2+2
B y2+4=x2-2
C y2+2=x2-4
D y2-2=x2+4

The correct answer is:
y2+4=x2-2

Solution:
Equation of any tangent of the hyperbola with slope m is y=mx±2m2-4
If it passes through x1,y1 then
y1-mx12=2m2-4⇒x12-2m2-2x1y1m+y12+4=0
If n=m1,m2 then as given m1m2=1

⇒y12+4x12-2=1
Hence, the required locus is y2+4=x2-2

Topic: Mathematics, Hyperbola, JEE Main
The locus of the point $\mathrm{P}=(\mathrm{a}, \mathrm{b})$ where $\mathrm{a}, \mathrm{b}$ are real numbers such that the roots of $\mathrm{x}^{3}+\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{a}=0$ are in arithmetic progression is-

Options:
A an ellipse
B a circle
C a parabola whose vertex in on the $y$-axis
D a parabola whose vertex is on the $\mathrm{x}$-axis

The correct answer is:
a parabola whose vertex in on the $y$-axis

Solution:
Let roots $\alpha-\mathrm{d}, \alpha, \alpha+\mathrm{d}$
$\operatorname{Sum} 3 \alpha=-\mathrm{a} \Rightarrow \alpha=-\frac{a}{3}$
product
$\alpha\left(\alpha^{2}-\mathrm{d}^{2}\right)=-\mathrm{a}$
$\begin{aligned} \text { pair product } b &=\alpha^{2}-\alpha d+\alpha^{2}+\alpha d+\alpha^{2}-d^{2} \quad \alpha^{2}-d^{2}=3 \\ b &=2 \alpha^{2}+3 \\ & b-3=\frac{2}{9} a^{2} \Rightarrow \text { locus } x^{2}=\frac{9}{2}(y-3) \text { parabola } \end{aligned}$

Topic: Mathematics, Parabola, JEE Main
The locus of the point which moves such that the ratio of its distances from two fixed points in the plane is always a constant $k( < 1)$ is

Options:
A circle
B straight line
C ellipse
D hyperbola

The correct answer is:
ellipse

Solution:
The locus of the point which such that the ratio of its distance from two fixed point in the plane is always a constant $k(k < 1)$ is an ellipse.

Topic: Mathematics, Ellipse, JEE Main
The locus of the point x,y whose distance from the line y=2x+2 is equal to the distance from 2,0, is a parabola with the length of latus rectum same as that of the parabola y=Kx2, then the value of K is equal to

Options:
A 512
B 54
C 45
D 125

The correct answer is:
512

Solution:
Length of LR=2SM
=2 22+2-022+1=125
⇒1K=125⇒K=512

Topic: Mathematics, Parabola, JEE Main
The locus of the trisection point of any arbitrary double ordinate of the parabola x2=4y, is

Options:
A 9x2=y
B 3x2=2y
C 9x2=4y
D 9x2=2y

The correct answer is:
9x2=4y

Solution:
Let A≡2t,t2,B≡-2t,t2 be the extremities on the double ordinate AB.
Let Ch,k be it’s trisection point, then
6h=4t and k=t2
⇒t=3h2,t2=k⇒k=9h24
Thus, locus of C is 9x2=4y

Topic: Mathematics, Parabola, JEE Main
The locus of the vertices of the family of parabolas $y=\frac{a^3 x^2}{3}+\frac{a^2 x}{2}-2 a$ is

Options:
A
$x y=\frac{105}{64}$

B
$x y=\frac{3}{4}$

C
$x y=\frac{35}{16}$

D
$x y=\frac{64}{105}$

The correct answer is:
$x y=\frac{105}{64}$

Solution:
Parabola: $y=\frac{a^3 x^2}{3}+\frac{a^2 x}{2}-2 a$
Vertex: $(\alpha, \beta)$
$$
\begin{aligned}
& \alpha=\frac{-a^2 / 2}{2 a^3 / 3}=-\frac{3}{4 a}, \beta=\frac{-\left(\frac{a^4}{4}+4 \cdot \frac{a^3}{3} \cdot 2 a\right)}{4 \frac{a^3}{3}}=-\frac{-\left(\frac{1}{4}+\frac{8}{3}\right) a^4}{\frac{4}{3} a^3} \\
& =-\frac{35}{12} \frac{a}{4} \times 3=-\frac{35}{16} a \\
& \alpha \beta=-\frac{3}{4 a}\left(-\frac{35}{16}\right) a=\frac{105}{64} .
\end{aligned}
$$

Topic: Mathematics, Parabola, JEE Main
The locus of the vertices of the family of parabolas $y=\frac{a^3 x^2}{3}+\frac{a^2 x}{2}-2 a$ is

Options:
A
$x y=\frac{105}{64}$

B
$x y=\frac{3}{4}$

C
$x y=\frac{35}{16}$

D
$x y=\frac{64}{105}$

The correct answer is:
$x y=\frac{105}{64}$

Solution:
Parabola: $y=\frac{a^3 x^2}{3}+\frac{a^2 x}{2}-2 a$
Vertex: $(\alpha, \beta)$
$$
\begin{aligned}
& \alpha=\frac{-a^2 / 2}{2 a^3 / 3}=-\frac{3}{4 a}, \beta=\frac{-\left(\frac{a^4}{4}+4 \cdot \frac{a^3}{3} \cdot 2 a\right)}{4 \frac{a^3}{3}}=-\frac{-\left(\frac{1}{4}+\frac{8}{3}\right) a^4}{\frac{4}{3} a^3} \\
& =-\frac{35}{12} \frac{a}{4} \times 3=-\frac{35}{16} a \\
& \alpha \beta=-\frac{3}{4 a}\left(-\frac{35}{16}\right) a=\frac{105}{64} .
\end{aligned}
$$

Topic: Mathematics, Parabola, JEE Main, JEE Main 2006
The logarithm of equilibrium constant for the reaction Pd2++4Cl-⇌PdCl42- is
(Nearest integer)

Given: 2.303RTF=0.06 V

Pd(aq)2++2e-⇌Pd(s) Eo=0.83 V

PdCl42-(aq)+2e-⇌Pd(s)+4Cl-(aq)

Eo=0.65 V

Options:

The correct answer is:
6

Solution:
Given, 2.303RTF=0.06 V

Using gibbs free energy equation,

ΔG°=-RT lnK

-nFEcell o=-RT×2.303log10 K

ECell00.06×n=log K---------(i)

Pd2+ (aq) + 2e-⇌ Pd(s), Ecathode, reduction 0 = 0.83 V

Pd (s)+4Cl- (aq.) ⇌ PdCl42- (aq) + 2e-, Eanode, oxidationo = 0.65V

Net Reaction →Pd2+ (aq.) + 4Cl- (aq.) ⇌ PdCl42- (aq.)

Ecell o = Ecathode,reductiono - Eanode, oxidation o

⇒ Ecello = 0.83 - 0.65⇒Ecello = 0.18--------(ii)

Also n=2--------(iii)

Using equation (i), (ii) & (iii), we get

ECell00.06×n=log K

⇒ 0.180.06× 2 = log K

⇒ logK = 6

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2023 (31 Jan Shift 1)
The logic circuit shown above is equivalent to :

Options:
A
B
C
D

The correct answer is:

Solution:
Truth table of the given gate:

A B C 000011100110

Truth table of option 1

A B C 001011100111

Truth table of option 2

A B C 001010101111

Truth table of option 3

A B C001010100110

Truth table of option 4

A B C000011100110

Since option 1 has same truth table, hence answer is option 4 is correct.

Given Boolean expression can be written as

A+B¯¯=C

∴ C=A¯·B¯=A¯·B

Topic: Physics, Semiconductors, JEE Main, JEE Main 2021 (24 Feb Shift 2)
The logic circuit shown below has the input wave forms ' $A$ ' and ' $B$ 'as shown. Pick out the correct output waveform.
Output is -

Options:
A

B

C

D

The correct answer is:

Solution:
\begin{tabular}{|c|c|c|c|c|}
\hline $\mathrm{A}$ & $\mathrm{B}$ & $\overline{\mathrm{A}}$ & $\overline{\mathrm{B}}$ & $(\mathrm{Y}=\overline{\overline{\mathrm{A}}+\overline{\mathrm{B}}})$ \\
\hline 1 & 1 & 0 & 0 & 1 \\
\hline 0 & 0 & 1 & 1 & 0 \\
\hline 0 & 1 & 1 & 0 & 0 \\
\hline 1 & 0 & 0 & 1 & 0 \\
\hline 1 & 1 & 0 & 0 & 1 \\
\hline 0 & 0 & 1 & 1 & 0 \\
\hline 0 & 1 & 1 & 0 & 0 \\
\hline
\end{tabular}

Topic: Physics, Semiconductors, JEE Main
The logic circuit shown below has the input waveforms 'A' and 'B' as shown. Pick out the correct output waveform.

Options:
A

B

C

D

The correct answer is:

Solution:
Truth Table
\begin{array}{|l|l|l|}
\hlineA & B & Y \\
\hline 1 & 1 & 1 \\
\hline 1 & 0 & 0 \\
\hline 0 & 1 & 0 \\
\hline 0 & 0 & 0 \\
\hline
\end{array}

Topic: Physics, Semiconductors, JEE Main, JEE Main 2009
The logic gate equivalent to the given circuit diagram is :

Options:
A OR
B NAND
C NOR
D AND

The correct answer is:
NAND

Solution:
When both switches are open, the current will flow through the bulb and it will glow.

When switch A is closed and switch B is open, the current will flow through the bulb and it will glow.

When switch A is open and switch B is closed, the current will flow through bulb and it will glow.

When both the switches are closed, since the resistance of left arm of circuit is negligible, the current will flow through the left arm and hence the bulb will not glow.

The truth table is given as:

A	B	Bulb

0	0	1

1	0	1

0	1	1

1	1	0

Which shows that the given circuit is a NAND gate.

Topic: Physics, Semiconductors, JEE Main, JEE Main 2023 (24 Jan Shift 2)
The logic gate equivalent to the given logic circuit is:

Options:
A NAND
B AND
C NOR
D OR

The correct answer is:
OR

Solution:
From the diagram,
y=A-.B-____
From De Morgan’s law,
y=A--+B--
=A+ B
∴ Given circuit represents ‘OR’ gate

Topic: Physics, Semiconductors, JEE Main, JEE Main 2019 (09 Apr Shift 2)
The logic operations performed by the given digital circuit is equivalent to:

Options:
A NOR
B AND
C OR
D NAND

The correct answer is:
AND

Solution:
As shown in the figure the various logic gates have been mentioned.

Let the output before the NOT gate be Z.

Using the expression for the logic gates, it can be written

Z=A+B·(A·B)

Hence, the output Y is

Y=Z=A+B·(A·B)⇒Y=A·B

Hence the digital circuit is equivalent to an AND gate.

Topic: Physics, Semiconductors, JEE Main, JEE Main 2023 (11 Apr Shift 2)
The logic performed by the circuit shown in figure is equivalent to

Options:
A AND
B NOR
C OR
D NAND

The correct answer is:
AND

Solution:
All the gates used in the given configuration are NOR gates. Thus, the output can be found as follows-

Y=a+b=ab=ab

The output of the given configuration can also be verified by the following truth table:

a	b	Output from first gate	Output from second gate	Y

0	0	1	1	0

0	1	1	0	0

1	0	0	1	0

1	1	0	0	1

The calculation and the truth table clearly shows that the configuration represents AND gate.

Topic: Physics, Semiconductors, JEE Main, JEE Main 2023 (11 Apr Shift 1)
The logical statement ~~p∨q∨p∧r∧~q∧r is equivalent to

Options:
A ~p∧~q∧r
B p∧r∧~q
C p∧~q∨r
D ~pâ¨r

The correct answer is:
p∧r∧~q

Solution:
~~p∨q∨p∧r∧~q∧r

=p∧~q∨p∧r∧~q∧r

By distributive law

=p∧~q∨r∧~q∧r

p∧~q∧r≡p∧r∧~q

=p∧∼q∧r

By Associative law

=p∧r∧~q

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2019 (09 Jan Shift 2)
The logical statement â¼â¼pâ¨â¼qâ¨pâ§râ§qâ§r is equivalent to

Options:
A p∧q∨r
B ∼p∨r
C ∼p∧q∧r
D p∧r∧q

The correct answer is:
p∧r∧q

Solution:
\((\sim(\sim p \vee q) \vee(p \wedge r)) \wedge(\sim q \wedge r)\)
\(=((p \wedge \sim q) \vee(p \wedge r)) \vee(\sim q \wedge r)\)
\(=(p \wedge(\sim q \vee r)) \wedge(\sim q \wedge r)\)
\(=p \wedge(\sim q \wedge r)=p \wedge(\sim q) \wedge r\)

Topic: Mathematics, Mathematical Reasoning, JEE Main
The logically equivalent preposition of $p \Leftrightarrow q$ is

Options:
A
$(p \Rightarrow q \wedge) q \Rightarrow p \quad)$

B
$p \wedge q$

C
$(p \wedge q \vee) q \neq p \quad)$

D
$(p \wedge q \Rightarrow q \vee(p \quad)$

The correct answer is:
$(p \Rightarrow q \wedge) q \Rightarrow p \quad)$

Solution:
$(p \Rightarrow q) \wedge(q \Rightarrow p)$ means $p \Leftrightarrow q$

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2012 (12 May Online)
The logically equivalent proposition of p⇔q is

Options:
A p∧q∨p∧q
B p⇒q∧q⇒p
C p∧q∨q⇒p
D p∧q⇒q∨p

The correct answer is:
p⇒q∧q⇒p

Solution:
p⇒q∧q⇒p means p⇔q.
Hence the logically equivalent proposition of p⇔q is p⇒q∧q⇒p

Topic: Mathematics, Mathematical Reasoning, JEE Main
The logistic population growth is expressed by the equation

Options:
A $d t / d N=N r\left(\frac{K-N}{K}\right)$
B $d N / d t=r N\left(\frac{K-N}{K}\right)$
C $d N / d t=r N$
D $d N / d t=r N\left(\frac{N-K}{N}\right)$

The correct answer is:
$d N / d t=r N\left(\frac{K-N}{K}\right)$

Solution:
A population growing in a habitat with limited resources show initially a lag phase, followed by phases of acceleration and deceleration and finally an asymptote, when the population density reaches the carrying capacity. A plot of $\mathrm{N}$ (population density at time $t$ ) in relation to time (t) results in sigmoid curve. This type of population grown is called Verhulst-Pearl Logistic Growth and is describes by the following equation : $d N / d t=r N\left(\frac{K-N}{K}\right)$ where, $N=$ Population density at time $t$, $r=$ Intrinsic rate of natural increase $K$ = Carrying capacity
Since resources for growth for most animal populations are finite and become limiting soner or later, the logistic growth model is considered a more realistic one.

Topic: Biology, Organisms and Populations, NEET, NEET 2011 (Mains)
The long form of periodic table has

Options:
A Eight horizontal rows and seven vertical columns
B Seven horizontal rows and eighteen vertical columns
C Seven horizontal rows and seven vertical columns
D Eight horizontal rows and eight vertical columns

The correct answer is:
Seven horizontal rows and eighteen vertical columns

Solution:
The long form of periodic table (modern periodic table) has 7 horizontal rows called periods and 18 vertical columns called groups.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
The longest pollen grain is found in the species of angiosperm which produces

Options:
A Entomophilous flowers
B Anemophilous flowers
C Hydrophilous flowers
D Ornithophilous flowers

The correct answer is:
Hydrophilous flowers

Solution:
The longest pollen grain is found in a type of sea grass called Zostera. In this plants the flowers are pollinated by water (hypohydrophily). Their pollen grains have starch granules which maintains their specific gravity and prevents them from sinking in water.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
The longest wavelength associated with Paschen series is : (Given $\mathrm{R}_{\mathrm{H}}=1.097 \times 10^7 \mathrm{SI}$ unit)

Options:
A $3.646 \times 10^{-6} \mathrm{~m}$
B $1.876 \times 10^{-6} \mathrm{~m}$
C $2.973 \times 10^{-6} \mathrm{~m}$
D $1.094 \times 10^{-6} \mathrm{~m}$

The correct answer is:
$1.876 \times 10^{-6} \mathrm{~m}$

Solution:
For longest wavelength in Paschen's series:
$\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right]$
For longest $\mathrm{n}_1=3$
$\mathrm{n}_2=4$
$\begin{aligned} & \frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{(3)^2}-\frac{1}{(4)^2}\right] \\ & \frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{9}-\frac{1}{16}\right] \\ & \frac{1}{\lambda}=\mathrm{R}\left[\frac{16-9}{16 \times 9}\right] \\ & \Rightarrow \lambda=\frac{16 \times 9}{7 \mathrm{R}}=\frac{16 \times 9}{7 \times 1.097 \times 10^7} \\ & \lambda=1.876 \times 10^{-6} \mathrm{~m}\end{aligned}$

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2024 (06 Apr Shift 2)
The longest wavelength of light that can be used for the ionisation of lithium ion $\mathrm{(Li^{2+})}$ is x×10-8 m. The value of x is (Nearest Integer)

(Given : Energy of the electron in the first shell of the hydrogen atom is -2.2×10-18 J; h=6.63×10-34Js and c=3×108 ms-1)

Options:

The correct answer is:
4

Solution:
Electronic configuration of Li=1s22s1

ELin=2=EHZ2n2=-2.2×10-18×94

6.63×10-34×3×108λ=2.2×10-18×94

λ=4×10-8 m

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2022 (25 Jun Shift 1)
The longitudinal stress in a rod (with rectangular cross-section) is 1% of the Young's modulus of its material. What is the approximate percentage of change of its volume? (Poisson's ratio of the material of the rod is 0.3)

Options:
A 3%
B 1%
C 0.7%
D 0.4%

The correct answer is:
0.4%

Solution:
∵ Stress, F∆A=1% of Y=Y100 But Young′s modulus, Y=stressstrain=F∆A∆ll
∴ Y=Y100∆ll puttingF∆A=Y100
∴ ∆ll=1100
Poisson′s ratio, σ=-∆rr∆ll
∴ ∆rr=-σ⋅∆ll=-0.3100
∆rr=-0.3100
∴ Change in volume, ∆VV=2∆rr+∆ll
=2×-0.3100+1100
=1-0.6100=0.4100

So the percentage change in volume = 0.4%

Topic: Physics, Mechanical Properties of Solids, JEE Main
The loss of biodiversity in a region may lead to

Options:
A Decline in plant production
B Lowered resistance to environmental perturbations such as drought
C Increased variability in certain ecosystem processes such as plant productivity, water use and pest and disease cycles
D All of the above

The correct answer is:
All of the above

Solution:
Biodiversity includes all the flora and fauna on Earth. All the varieties of organisms come under biodiversity. Ecologist say that without any doubt there is decline in the biological wealth which is more rapid due to the activities of us humans. Basically, it is a way to measure the differences and variations among the organisms at genetic, species and ecological level. Biodiversity plays a very important role in maintaining species richness and stability of the ecosystem. Loss in the biodiversity will eventually result in decreased plant production and reduce the resistance of the flora-fauna to environmental change. Biodiversity loss will also bring about high rate of changes in ecosystem processes like plant productivity, pest and diseases cycles.

Topic: Biology, Biodiversity and its Conservation, NEET