datavorous/entrance-exam-dataset huggingface dataset:
List the hormones secreted by the followings.
(a) Hypothalamus
(b) Pituitary
(c) Thyroid
(d) Parathyroid
(e) Adrenal
(f) Pancreas
(g) Testis
(h) Ovary
(i) Thymus
(j) Atrium
(k) Kidney
(I) G-ITract

Options:

None

Solution:
(a) Hypothalamus: Gonadotrophic releasing hormone (GnRH), somatostatin.
(b) Pituitary : Growth hormone, prolactin, thyroid stimulating hormone, luteinizing hormone and follicle Stimulating hormone, oxytocin, vasopressin.
(c) Thyroid : triiodothyronine ( $\left.\mathrm{T}_3\right)$ and tetraiodo thyronine $\left(\mathrm{T}_4\right)$.
(d) Parathyroid: Parathyroid hormone.
(e) Adrenal : Adrenaline or epinephrine and noradrenaline or norepinephrine.
(f) Pancreas: Insulin and glucagon.
(g) Testis : A group of androgens mainly testesterone.
(h) Ovary : Estrogens and progesterone.
(i) Thymus : Thymosins.
(j) Atrium : Atrial Natriuretic factor (ANF)
(k) Kidney : Erythropoietin.
(1) G-I tract : Gastrin, secretin, cholecystokinin (CCK).

Topic: Biology, Chemical Coordination and Integration
List the important sources of sulphur.

Options:

None

Solution:
Sulphur mainly occurs in the combined states in earth's crust in the form of sulphates and sulphides.
Sulphates : gypsum \(\left(\mathrm{CaSO}_4 \cdot 2 \mathrm{H}_2 \mathrm{O}\right)\); epsom \(\left(\mathrm{MgSO}_4 \cdot 7 \mathrm{H}_2 \mathrm{O}\right)\); baryte \(\left(\mathrm{BaSO}_4\right)\), etc.
Sulphides: Galena (PbS); zinc blende (ZnS); copper pyrites \(\left(\mathrm{CuFeS}_2\right)\); iron pyrites \(\left(\mathrm{FeS}_2\right)\), etc.
Traces of sulphur occur as \(\mathrm{H}_2 \mathrm{~S}\) and in organic materials such as eggs, proteins, garlic, onion, mustard, hair and wool.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18)
List the main differences between mitosis and meiosis.

Options:

None

Solution:
The main differences between mitosis and meiosis are as follows:

Topic: Biology, Cell Cycle and Cell Division
List the organs of human alimentary canal and name the major digestive glands with their location.

Options:

None

Solution:
Human digestive system consists of two main parts: alimentary canal and digestive glands. The organ of human alimentary canal are mouth, pharynx oesophagus, stomach, small intestine, large intestine, rectum and anus. Major digestive glands with their locations are as follows :

Topic: Biology, Digestion and Absorption
List the uses of neon and argon gases.

Options:

None

Solution:
Uses of Neon
Neon is used in discharge tubes and fluorescent bulbs for advertisement display purposes.
Glow of different colours 'neon signs' can be produced by mixing neon with other gases.
Neon bulbs and used in botanical gardens and in green houses.
Uses of Argon
Argon is used mainly to provide an inert atmosphere in high temperature metallurgical processes such as arc welding of metals and alloys. In the laboratory, it is used for handling substance which are air sensitive.
It is used in filling incandescent and fluorescent lamps where its presence retards the sublimation of the filament and thus increases the life of the lamp.
It is also used in "neon signs" for obtaining lights of different colours.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18)
Listed below are four respiratory capacities (1 - 4) and four jumbled respiratory volumes of a normal human adult
(1) Residual volume $2500 \mathrm{~mL}$
(2) Vital capacity $3500 \mathrm{~mL}$
(3) Inspiratory reserve $1200 \mathrm{~mL}$ volume
(4) Inspiratory capacity $4500 \mathrm{~mL}$
Which one of the following is the correct matching of two capacities and volumes?

Options:
A (2) $2500 \mathrm{~mL}$,
(3) $4500 \mathrm{~mL}$
B (3) $1200 \mathrm{~mL}$
(4) $2500 \mathrm{~mL}$
C (4) $3500 \mathrm{~mL}$
(1) $1200 \mathrm{~mL}$
D (1) $4500 \mathrm{~mL}$,
(2) $3500 \mathrm{~mL}$

The correct answer is:
(4) $3500 \mathrm{~mL}$
(1) $1200 \mathrm{~mL}$

Solution:
Inspiratory Capacity (IC) is the maximum amount of air that can be inspired after a normal expiration, IC = TV + IRV . It is $3600 \mathrm{~mL}$ in adult male and $2400 \mathrm{~mL}$ in adult female. Residual Volume (RV) is the amount of air remaining in the lungs after a forced exhalation. Its average value is $1200 \mathrm{~mL}$ and $1100 \mathrm{~mL}$ in adult male and female respectively.

Topic: Biology, Breathing and Exchange of Gases, NEET, NEET 2010 (Screening)
Lithium aluminium hydride reacts with silicon tetrachloride to form:

Options:
A
$\mathrm{LiCl}, \mathrm{AlH}_3$ and $\mathrm{SiH}_4$

B
$\mathrm{LiCl}, \mathrm{AlCl}_3$ and $\mathrm{SiH}_4$

C
$\mathrm{LiH}, \mathrm{AlCl}_3$ and $\mathrm{SiCl}_2$

D
$\mathrm{LiH}, \mathrm{AlH}_3$ and $\mathrm{SiH}_4$

The correct answer is:
$\mathrm{LiCl}, \mathrm{AlCl}_3$ and $\mathrm{SiH}_4$

Solution:
$\mathrm{SiCl}_4+\mathrm{LiAlH}_4 \rightarrow \mathrm{LiCl}+\mathrm{AlCl}_3+\mathrm{SiH}_4$

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
Lithium forms body centred cubic structure. The length of the side of its unit cell is $351 \mathrm{pm}$. Atomic radius of the lithium will be:

Options:
A
$75 \mathrm{~pm}$

B
$300 \mathrm{~pm}$

C
$240 \mathrm{~pm}$

D
$152 \mathrm{~pm}$

The correct answer is:
$152 \mathrm{~pm}$

Solution:
For BCC, $\sqrt{3} a=4 r$
$r=\frac{\sqrt{3} \times 351}{4}=152 \mathrm{~pm}$

Topic: Chemistry, Solid State, JEE Main, JEE Main 2012 (Offline)
Lithium hydride reacts with aluminium chloride to form a complex. The geometry of the complex and the ligand present in the complex is

Options:
A Octahedral, chloride
B Tetrahedral, hydride
C Octahedral, bridging chloride
D Tetrahedral, Chloride and hydride

The correct answer is:
Tetrahedral, hydride

Solution:
4 LiH +  AlCl 3 →Li AlH 4  + 3 LiCl
Tetrahedral and Hydride ligand

Topic: Chemistry, s Block Elements, JEE Main
Lithium is the strongest reducing agent though it has highest ionistion energy in its group. Which of the following factors is responsible for making Li the strongest reducing agent?

Options:
A Large heat of atomisation
B Smaller size
C Large sublimation energy
D Large amount of hydration enthalpy

The correct answer is:
Large amount of hydration enthalpy

Solution:
Large amount of hydration energy makes it strongest reducing agent inspite of its highest ionisation enthalpy.

Topic: Chemistry, s Block Elements, JEE Main
Lithium metal crystallizes in a body centred cubic crystal. If the length of the side of the unit cell of lithium is $351 \mathrm{pm}$, the atomic radius of lithium will be :

Options:
A $300.5 \mathrm{pm}$
B $240.8 \mathrm{pm}$
C $151.8 \mathrm{pm}$
D $75.5 \mathrm{pm}$

The correct answer is:
$151.8 \mathrm{pm}$

Solution:
For bcc type of unit cell
$$
\begin{aligned}
\sqrt{3} \mathrm{a} & =4 \mathrm{r} \\
\mathrm{r} & =\frac{\sqrt{3}}{4} \mathrm{a} \\
& =\frac{1.732 \times 351}{4} \\
& =151.98
\end{aligned}
$$

Topic: Chemistry, Solid State, NEET, NEET 2009 (Mains)
Lithium shows similarities to magnesium in its chemical behaviour because

Options:
A Similar size, greater electronegativity and similar polarizing power.
B Similar size same electronegativity and lower polarizing power
C Similar size, same electronegativity and similar high polarizing power
D None of these

The correct answer is:
Similar size, same electronegativity and similar high polarizing power

Solution:
Lithium shows digonal relationships with $M g$

Topic: Chemistry, s Block Elements, JEE Main
Lithium shows similarities to magnesium in its chemical behaviour because

Options:
A Similar size, same electronegativity and lower polarizing power
B Similar size, greater electronegativity and similar polarizing power
C Similar size, same electronegativity and similar high polarizing power
D None of these

The correct answer is:
Similar size, same electronegativity and similar high polarizing power

Solution:
Lithium shows similarities with magnesium in its chemical behaviour because they have similar size and charge/size ratio. They have similar electronegativity.
This type of diagonal similarity is commonly referred to as diagonal relationship in the periodic table.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
Liver is the largest gland and is associated with various functions, choose one which is not correct.

Options:
A
Metabolism of carbohydrate

B
Digestion of fat

C
Formation of bile

D
Secretion of hormone called gastrin

The correct answer is:
Secretion of hormone called gastrin

Solution:
Liver is involved in the production of bile, which helps in the digestion of fats in the small intestine by the emulsification process (conversion of large fat droplets into small ones).
Liver also plays a critical role in controlling rate metabolism by maintaining the glucose concentration in the normal range. Gastrin is secreted by G-cells in the pyrolic region of stomach. It stimulates gastric glands to secrete and release gastric juices.

Topic: Biology, Digestion and Absorption
Living state is a _____________, __________ state able to perform work.

Options:
A equilibrium, steady
B non-equilibrium, unsteady
C equilibrium, unsteady
D non-equilibrium, steady

The correct answer is:
non-equilibrium, steady

Solution:
The living systems are in a metabolic flux and thus they maintain the concentration of biomolecules. Thus they are in non-equilibrium state. This is important because if they reach the equilibrium state, no work will be performed. Hence the living organisms receive an input of energy to prevent reaching an equilibrium state. This energy is received through metabolism (anabolism and catabolism).

Topic: Biology, The Living World, NEET
â«lnâ¡x+1-lnâ¡xxx+1dx is equal to (where C is an arbitarary constant)

Options:
A -12lnâ¡x+1x2+C
B C - lnâ¡x+12- lnâ¡x2
C -lnâ¡lnâ¡x+1x+C
D -lnâ¡x+1x+C

The correct answer is:
-12ln⁡x+1x2+C

Solution:
Put lnâ¡x+1-lnâ¡x=t
â 1x+1-1x=dtdxâ x-x+1xx+1=dtdx
â -dxxx+1=dtâ dxxx+1=- dt
so question becomes
- â«ât dt=-t22+C

Topic: Mathematics, Indefinite Integration, JEE Main
∫ ln x - 1 x + 1 x 2 - 1 dx is equal to

Options:
A 1 2 ln x - 1 x + 1 2 + C
B 1 2 ln x + 1 x - 1 2 + C
C 1 4 ln x - 1 x + 1 2 + C
D 14 ln x+1x-1+ C

The correct answer is:
1 4 ln x - 1 x + 1 2 + C

Solution:
I = â« ln x - 1 x + 1 x 2 - 1 dx ,
Let t = ln x - 1 x + 1
â dt dx = x + 1 x - 1 x + 1 - x - 1 x + 1 2 = 2 x 2 - 1
â dx x 2 - 1 = dt 2
â I = 1 2 â« tdt = 1 4 t 2 + C = 1 4 ln x - 1 x + 1 2 + C

Topic: Mathematics, Indefinite Integration, JEE Main
$\left|\begin{array}{cc}\log _2 512 & \log _4 3 \\ \log _3 8 & \log _4 9\end{array}\right| \times\left|\begin{array}{ll}\log _2 3 & \log _8 3 \\ \log _3 4 & \log _3 4\end{array}\right|=$

Options:
A 7
B 10
C 13
D 17

The correct answer is:
10

Solution:
$\left|\begin{array}{cc}\log _3 512 & \log _4 3 \\ \log _3 8 & \log _4 9\end{array}\right| \times\left|\begin{array}{ll}\log _2 3 & \log _8 3 \\ \log _3 4 & \log _3 4\end{array}\right|$
$=\left(\frac{\log 512}{\log 3} \times \frac{\log 9}{\log 4}-\frac{\log 3}{\log 4} \times \frac{\log 8}{\log 3}\right)$ $\times\left(\frac{\log 3}{\log 2} \times \frac{\log 4}{\log 3}-\frac{\log 3}{\log 8} \times \frac{\log 4}{\log 3}\right)$
$\begin{array}{r}=\left(\frac{\log 2^9}{\log 3} \times \frac{\log 3^2}{\log 2^2}-\frac{\log 2^3}{\log 2^2}\right) \times\left(\frac{\log 2^2}{\log 2}-\frac{\log 2^2}{\log 2^3}\right) \\ =\left(\frac{9 \times 2}{2}=\frac{3}{2}\right)\left(2-\frac{2}{3}\right)=\frac{15}{2} \times \frac{4}{3}-10\end{array}$

Topic: Mathematics, Determinants, JEE Main
$\int \log x d x=$

Options:
A $x+x \log x+c$
B $x \log x-x+c$
C $x^2 \log x+c$
D $\frac{1}{x} \log x+x+c$

The correct answer is:
$x \log x-x+c$

Solution:
$\int \log x d x=x \log x-\int x \cdot \frac{1}{x} d x+c=x \log x-x+c$
Note : Students should remember this question as a formula.

Topic: Mathematics, Indefinite Integration, JEE Main
Local maximum and local minimum values of the function $(x-1)(x+2)^2$ are

Options:
A $-4,0$
B $0,-4$
C 4,0
D None of these

The correct answer is:
$0,-4$

Solution:
$\begin{aligned}& f(x)=2(x-1)(x+2)+(x+2)^2=3 x^2+6 x \\
& f^{\prime}(x)=0 \Rightarrow x=0,-2 \\
& f(-2)=(-2-1)(-2+2)^2=0\end{aligned}$
(Maximum value)
and
$f(0)=(0-1)(0+2)^2=-4$ (Minimum value).

Topic: Mathematics, Application of Derivatives, JEE Main
Local maximum and local minimum values respectively of the function $f(x)=(x-1)(x+2)^2$ are

Options:
A $-4,0$
B $0,-4$
C $-4,4$
D $4,-4$

The correct answer is:
$0,-4$

Solution:
$\begin{aligned} & f(x)=(x-1)(x+2)^2 \\ & f^{\prime}(x)=(x+2)^2+(x-1) 2(x+2)=(x+2)(x+2+2 x-2)=3 x(x+2) \\ & \stackrel{+}{+}-0_{-2}^{+} \\ & f_{\max }=f(-2)=0 \\ & f_{\min }=f(0)=-4\end{aligned}$

Topic: Mathematics, Application of Derivatives, JEE Main
Locations of cerebral aqueduct and corpora quadrigemina in the human brain are respectively

Options:
A Midbrain, midbrain
B Forebrain, midbrain
C Hindbrain, hindbrain
D Hindbrain, midbrain

The correct answer is:
Midbrain, midbrain

Solution:
Cerebral aqueduct or Iter connects the lateral ventricles with 4th ventricle. It passes through the midbrain. The dorsal portion of the midbrain mainly consists of 4 round swellings (lobes). These are termed as corpora quadrigemina. Midbrain receives tactile and visual responses.

Topic: Biology, Neural Control and Coordination, NEET
Locus of a centriod of the triangle whose vertices are $(a \cos t, a \sin t),(b \sin t,-b \cos t)$ and $(1,0)$, where $t$ is a parameter, is

Options:
A
$(3 x+1)^2+(3 y)^2=a^2-b^2$

B
$(3 x-1)^2+(3 y)^2=a^2-b^2$

C
$(3 x-1)^2+(3 y)^2=a^2+b^2$

D
$(3 x+1)^2+(3 y)^2=a^2+b^2$

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

Solution:
$x=\frac{\cos t+b \sin t+1}{3} \Rightarrow a \cos t+b \sin t=3 x-1$
$y=\frac{a \sin t-b \cos t}{3} \Rightarrow a \sin t-b \sin t=3 y$
Squaring \& adding, $(3 \mathrm{x}-1)^2+(3 \mathrm{y})^2=\mathrm{a}^2+\mathrm{b}^2$

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2003
Locus of mid point of the portion between the axes of $x \cos \alpha+y \sin \alpha=p$ where $p$ is constant is

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

B
$x^2+y^2=4 p^2$

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

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

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

Solution:
Equation of $A B$ is $x \cos \alpha+y \sin \alpha=p \Rightarrow \frac{x \cos \alpha}{p}+\frac{y \sin \alpha}{p}=1 \Rightarrow \frac{x}{p / \cos \alpha}+\frac{y}{p / \sin \alpha}=1$
So co-ordinates of $A$ and $B$ are $\left(\frac{p}{\cos \alpha}, 0\right)$ and $\left(0, \frac{p}{\sin \alpha}\right)$; So coordinates of mid point of $A B$ are
$\left(\frac{p}{2 \cos \alpha}, \frac{p}{2 \sin \alpha}\right)=\left(x_1, y_1\right)($ let $) ; x_1=\frac{p}{2 \cos \alpha} \& y_1=\frac{p}{2 \sin \alpha}$
$\Rightarrow \cos \alpha=p / 2 x_1$ and $\sin \alpha=p / 2 y_1 ; \cos ^2 \alpha+\sin ^2 \alpha=1 \Rightarrow \frac{p^2}{4}\left(\frac{1}{x_1^2}+\frac{1}{y_1^2}\right)=1$
Locus of $\left(x_1, y_1\right)$ is $\frac{1}{x^2}+\frac{1}{y^2}=\frac{4}{p^2}$.

Topic: Mathematics, Circle, JEE Main
Locus of mid point of the portion between the axes of $x \cos \alpha+y \sin \alpha=p$ where $p$ is constant is

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

B
$x^2+y^2=4 p^2$

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

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

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

Solution:
Equation of $A B$ is $x \cos \alpha+y \sin \alpha=p \Rightarrow \frac{x \cos \alpha}{p}+\frac{y \sin \alpha}{p}=1 \Rightarrow \frac{x}{p / \cos \alpha}+\frac{y}{p / \sin \alpha}=1$
So co-ordinates of $A$ and $B$ are $\left(\frac{p}{\cos \alpha}, 0\right)$ and $\left(0, \frac{p}{\sin \alpha}\right)$; So coordinates of mid point of $A B$ are
$\left(\frac{p}{2 \cos \alpha}, \frac{p}{2 \sin \alpha}\right)=\left(x_1, y_1\right)($ let $) ; x_1=\frac{p}{2 \cos \alpha} \& y_1=\frac{p}{2 \sin \alpha}$
$\Rightarrow \cos \alpha=p / 2 x_1$ and $\sin \alpha=p / 2 y_1 ; \cos ^2 \alpha+\sin ^2 \alpha=1 \Rightarrow \frac{p^2}{4}\left(\frac{1}{x_1^2}+\frac{1}{y_1^2}\right)=1$
Locus of $\left(x_1, y_1\right)$ is $\frac{1}{x^2}+\frac{1}{y^2}=\frac{4}{p^2}$.

Topic: Mathematics, Circle, JEE Main, JEE Main 2002
Locus of the centre of a family of circles cutting the family of circles \( x^{2}+y^{2}+4 x\left(\lambda-\frac{3}{2}\right)+3 y\left(\lambda-\frac{4}{3}\right)-6(\lambda+2)=0 \) is

Options:
A \( x-y-1=0 \)
B \( 4 x+3 y-6=0 \)
C \( 4 x+3 y+7=0 \)
D \( 3 x-4 y-1=0 \)

The correct answer is:
\( 4 x+3 y-6=0 \)

Solution:
x2+y2+4xλ-32+3yλ-43-6λ+2=0

i.e. x2+y2-6x-4y-12+λ4x+3y-6=0

So, the family of circles is passing through the points of intersection of circle x2+y2-6x-4y-12=0 and the line 4x+3y-6=0.

The centre of the family of circles cutting this family of circles at A and B lies on the line AB.

Hence, the required locus of the centre of the family of circles is the common chord 4x+3y-6=0.

Topic: Mathematics, Circle, JEE Main
Locus of the image of the point ( 2,3 ) in the line 2 x - 3 y + 4 + k x - 2 y + 3 = 0 , k ∈ R , is a

Options:
A Circle of radius 3
B Straight line parallel to x-axis.
C Straight line parallel to y-axis.
D Circle of radius 2

The correct answer is:
Circle of radius 2

Solution:
Let image is ( α,β )
∴ point ( 2+α 2 , 3+β 2 ) lies on the given variable line.
∴  2( 2+α 2 )−3( 3+β 2 )+4+k( 2+α 2 −2× 3+β 2 +3 )=0
⇒2α−3β+3+k(α−2β+2)=0 .......1
again, product of two slopes =−1
⇒ 3−β 2−α × 2+k 3+2k =−1
⇒6−2β+k( 3−β )=−( 6−3α )−k( 4−2α )
⇒k(7−2α−β)=3α+2β−12
⇒k= 3α+2β−12 7−2α−β
Substituting value of k in ,1 we get
2α−3β+3+ 3α+2β+−12 7−2α−β ( α−2β+2 )=0
⇒14α−4 α 2 −2αβ−21β+6αβ+3 β 2 +21−6α−3β+3 α 2 −6αβ+6α+2αβ−4 β 2 +4β−12α+24β−24=0
⇒ α 2 + β 2 −2a−4β+3=0
∴ Locus is x 2 + y 2 −2x−4y+3=0
Which is a circle with radius = 1+4−3 = 2

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2015 (04 Apr)
Locus of the mid-points of the chords of contact of x2+y2=2 from the points on the line 3x+4y=10 is a circle with centre C. If O be the origin, then 2OC is equal to

Options:
A 2
B 3
C 1
D 13

The correct answer is:
1

Solution:
Equation of the chord with mid point h,k is hx+ky=h2+k2….i
Equation of the chord with mid point (x1,y1) is xx1+yy1=2……ii
On comparing (i) & (ii), we get
x1=2hh2+k2 and y1=2kh2+k2
x1,y1 lies on 3x+4y=10⇒6h+8k=10h2+k2
∴ locus of h,k is x2+y2-35x-45y=0
Which is a circle with centre C=310,410
∴OC=12

Topic: Mathematics, Circle, JEE Main
Locus of the point of intersection of straight lines $\frac{x}{a}-\frac{y}{b}=m$ and $\frac{x}{a}+\frac{y}{b}=\frac{1}{m}$ is

Options:
A An ellipse
B A circle
C A hyperbola
D A parabola

The correct answer is:
A hyperbola

Solution:
$\begin{aligned}
& \frac{x}{a}-\frac{y}{b}=m ....(i) \\
& \frac{x}{a}+\frac{y}{b}=\frac{1}{m} ....(ii)
\end{aligned}$
Multiplying equation (i) and (ii),
$\begin{aligned}
& \left(\frac{x}{a}-\frac{y}{b}\right)\left(\frac{x}{a}+\frac{y}{b}\right)=m \cdot \frac{1}{m} \\
& \frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \text { which is, which is the equation of hyperbola. }
\end{aligned}$

Topic: Mathematics, Hyperbola, JEE Main
Locus of the points which are at equal distance from $3 x+4 y-11=0$ and $12 x+5 y+2=0$ and which is near the origin is

Options:
A $21 x-77 y+153=0$
B $99 x+77 y-133=0$
C $7 x-11 y=19$
D None of these

The correct answer is:
$99 x+77 y-133=0$

Solution:
Let point be $\left(x_1, y_1\right)$, then according to the condition $\frac{3 x_1+4 y_1-11}{5}=-\left(\frac{12 x_1+5 y_1+2}{13}\right)$ Since the given lines are on opposite sides with respect to origin, hence the required locus is $99 x+77 y-133=0$

Topic: Mathematics, Straight Lines, JEE Main
Locus of the poles of focal chord of a parabola is

Options:
A the axis
B a focal chord
C the directrix
D the tangent at the vertex

The correct answer is:
the directrix

Solution:
Equation of $P Q$ is
$$
\begin{aligned}
y-2 a t_1 & =\frac{2 a t_2-2 a t_1}{a t_2^2-a t_1^2}\left(x-a t_1^2\right) \\
\Rightarrow \quad\left(t_2+t_1\right)\left(y-2 a t_1\right) & =2\left(x-a t_1^2\right) \\
\Rightarrow \quad\left(t_1+t_2\right) y-2 x & =2 a t_1 t_2
\end{aligned}
$$
This line is passing through $(a, 0)$
$$
\begin{aligned}
\Rightarrow & & \left(t_1+t_2\right)\left(-2 a t_1\right)=\frac{2}{t_1+t_2}\left(a-a t_1^2\right) \\
\Rightarrow & & t_1 t_2=-1
\end{aligned}
$$
Let $P\left(x_1, y_1\right)$ be the pole of (i) w.r.t. $y^2=4 a x$
Its polar is $y y_1=2 a\left(x+x_1\right)$
From equaiton (i) and (iii), we get
$$
\frac{t_1+t_2}{y_1}=\frac{1}{a}=\frac{2 a t_1 t_2}{2 a x_1}
$$
From last two relations, we get
$$
\begin{array}{ll}
& x_1=a t_1 t_2 \\
\Rightarrow & x_1=-a \\
\therefore \text { locus is } x=-a &
\end{array}
$$

Topic: Mathematics, Parabola, JEE Main
Locus of the poles of focal chords of a parabola is of parabola

Options:
A The tangent at the vertex
B The axis
C A focal chord
D The directrix

The correct answer is:
The directrix

Solution:
Let equation of the parabola is $\mathrm{y}^2=4 \mathrm{ax}$ Polar of the pole $\mathrm{P}\left(\mathrm{x}_1, \mathrm{y}_1\right)$ w.r.t the given parabola is $\mathrm{yy}_1-2 \mathrm{ax}-2 \mathrm{ax}_1=0$, which is identical to the focal chord and passes through focus $(a, 0)$ of the parabola, then
$0 \times y_1-2 a \times a-2 a x_1=0$
$-2 a^2-2 a x_1=0$
$\mathrm{x}_1=-\mathrm{a}$
$x=-a$, which is equation of the directrix. Therefore, the desired locus is directrix of the parabola.

Topic: Mathematics, Parabola, JEE Main
lodoform can be prepared from all except :

Options:
A
Ethyl methyl ketone

B
Isopropyl alcohol

C
$3-$Methyl$-2-$ butanone

D
Isobutyl alcohol

The correct answer is:
Isobutyl alcohol

Solution:
Iodoform is given by
$1)$ methyl ketones $\mathrm{R-CO-CH_3}$
$2)$ alcohols of the type $\mathrm{R-CH(OH) CH}_3$ where $\mathrm{R}$ can be hydrogen also

Topic: Chemistry, Aldehydes and Ketones, JEE Main
lodoform can be prepared from all except :

Options:
A
Ethyl methyl ketone

B
Isopropyl alcohol

C
$3-$Methyl$-2-$ butanone

D
Isobutyl alcohol

The correct answer is:
Isobutyl alcohol

Solution:
Iodoform is given by
$1)$ methyl ketones $\mathrm{R-CO-CH_3}$
$2)$ alcohols of the type $\mathrm{R-CH(OH) CH}_3$ where $\mathrm{R}$ can be hydrogen also

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2012 (Offline)
Long filamentous threads protruding at the end of a young cob of maize are:

Options:
A hairs
B anthers
C styles
D ovaries

The correct answer is:
styles

Solution:
In a cob of maize, each ovary has a long silky (hair) style, called corn silk. Collectively, these styles protrude at the end of a young cob.
The grains are formed on the cob which remains covered by leafy bracts.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET, NEET 2006
Long-range radio transmission is possible when the radio waves are reflected from the ionosphere. For this to happen the frequency of the radio waves must be in the range:

Options:
A 150-500 kHz
B 80-150 MHz
C 1-3 MHz
D 8-25 MHz

The correct answer is:
8-25 MHz

Solution:
Skywaves get reflected from the ionosphere having frequency 3 MHz to 40 MHz.

Topic: Physics, Communication System, JEE Main, JEE Main 2014 (19 Apr Online)
Look at the drawing given in the figure which has been drawn with ink of uniform line-thickness. The mass of ink used to draw each of the two inner circles, and each of the two line segments is m. The mass of the ink used to draw the outer circle is 6m. The coordinates of the centres of the different parts are: outer circle (0, 0) left inner circle (âa, a), right inner circle (a, a), vertical line (0, 0) and horizontal line (0, âa). The y-coordinate of the centre of mass of the ink in this drawing is
â

Options:
A a 1 0
B a 8
C a 1 2
D a 3

The correct answer is:
a 1 0

Solution:
y CM = m 1 y 1 + m 2 y 2 + m 3 y 3 + m 4 y 4 + m 5 y 5 m 1 + m 2 m 3 + m 4 + m 5
= 6 m 0 + m a + m a + m 0 + m - a 6 m + m + m + m + m
= a 1 0

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Lorentz force can be calculated by using the formula

Options:
A $\vec{F}=q(\vec{E}+\vec{v} \times \vec{B})$
B $\vec{F}=q(\vec{E}-\vec{v} \times \vec{B})$
C $\vec{F}=q(\vec{E}+\vec{v} \cdot \vec{B})$
D $\vec{F}=q(\vec{E} \times \vec{B}+\vec{v})$

The correct answer is:
$\vec{F}=q(\vec{E}+\vec{v} \times \vec{B})$

Solution:
Lorentz force is given by
\(\begin{aligned}
& \vec{F}=\vec{F}_e+\vec{F}_m \\
& =q \vec{E}+q(\vec{v} \times \vec{B}) \\
& =q[\vec{E}+(\vec{v} \times \vec{B})]
\end{aligned}\)

Topic: Physics, Magnetic Effects of Current, JEE Main