datavorous/entrance-exam-dataset huggingface dataset:
Seven unbiased coins are tossed 128 times. In how many

Options:
A 99
B 102
C 103
D 104

The correct answer is:
99

Solution:
Let $\mathrm{X}$ denote the no. of coins showing 3 or more heads in a set of 7 coins. $\mathrm{X}$ follows binomial distribution with $\mathrm{n}=7$ $\mathrm{p}=$ probability of getting a head in a single toss of a coin
$\Rightarrow \mathrm{p}=\frac{1}{2} ;$ thus $\mathrm{q}=1-\mathrm{p}=\frac{1}{2}$
$\therefore$ Probability of getting at least 3 heads $=\mathrm{P}(\mathrm{x} \geq 3)$
$=1-\mathrm{P}(\mathrm{x} < 3)$
$=1-[\mathrm{P}(\mathrm{x}=0)+\mathrm{P}(\mathrm{x}=1)+\mathrm{P}(\mathrm{x}=2)]$
$=1-\left[{ }^{7} \mathrm{C}_{0}+{ }^{7} \mathrm{C}_{1}+{ }^{7} \mathrm{C}_{2}\right] \frac{1}{2^{7}}$
$=\frac{128}{128}-\frac{29}{128}=\frac{99}{128}$
$\therefore$ No. of throws $=\frac{99}{128} \times 128=99$

Topic: Mathematics, Probability, JEE Main
Several blocks of magnesium are fixed to the bottom of a ship to

Options:
A
make the ship lighter

B
prevent action of water and salt

C
prevent puncturing by under-sea rocks

D
keep away the sharks

The correct answer is:
prevent action of water and salt

Solution:
Magnesium provides cathodic protection and prevent rusting or corrosion.

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2003
Several hormones like hCG, hPL, estrogen, progesterone are produced by

Options:
A Ovary
B Placenta
C Fallopian tube
D Pituitary

The correct answer is:
Placenta

Solution:
Hormones secreted by the placenta are hCG, hPL, oestrogen, and progesterone. Human chorionic gonadotropin hormone (hCG). This hormone is only produced during pregnancy almost exclusively in the placenta. HCG hormone levels found in maternal blood and urine increase dramatically during the first trimester and may contribute to causing nausea and vomiting often associated with pregnancy.
Human placental lactogen (hPL). This hormone, produced by the placenta, helps in the process of providing nutrition for the foetus and plays a role in stimulating milk glands in the breasts in anticipation of breastfeeding.
Oestrogen. This group of hormones is responsible for the development of female sexual characteristics. Normally formed in the ovaries, oestrogen is also produced by the placenta during pregnancy to help maintain a healthy pregnancy.
Progesterone. This hormone is produced by the ovaries and by the placenta during pregnancy. Progesterone stimulates the thickening of the uterine lining in anticipation of the implantation of a fertilised egg.

Topic: Biology, Human Reproduction, NEET, NEET 2016 (Phase 2)
Several hormones like hCG, hPL, estrogen, progesterone are produced by

Options:
A Ovary
B Placenta
C Fallopian tube
D Pituitary

The correct answer is:
Placenta

Solution:
Hormones secreted by the placenta are hCG, hPL, estrogen, and progesterone. Human chorionic gonadotropin hormone (hCG). This hormone is only produced during pregnancy almost exclusively in the placenta. HCG hormone levels found in maternal blood and urine increase dramatically during the first trimester and may contribute to causing nausea and vomiting often associated with pregnancy.
Human placental lactogen (hPL). This hormone, produced by the placenta, helps in the process of providing nutrition for the fetus and plays a role in stimulating milk glands in the breasts in anticipation of breastfeeding.
Estrogen. This group of hormones is responsible for the development of female sexual characteristics. Normally formed in the ovaries, estrogen is also produced by the placenta during pregnancy to help maintain a healthy pregnancy.
Progesterone. This hormone is produced by the ovaries and by the placenta during pregnancy. Progesterone stimulates the thickening of the uterine lining in anticipation of implantation of a fertilized egg.

Topic: Biology, Human Reproduction, NEET
Sexual stage (gametocytes) of Plasmodium occurs in

Options:
A Salivary glands of mosquito
B Human RBC
C Intestine of mosquito
D Human liver

The correct answer is:
Human RBC

Solution:
Inside the red blood cells, oval-shaped merozoites stop proceeding with erythrocytic cycle to increase in size and become rounded gametocytes. Male gametocytes or microgametocytes are smaller and contain a large diffused nucleus. Female gametocytes (or mega gametocytes) are larger with a small compact peripheral nucleus. These do not divide but remain as intracellular parasites within their host's blood corpuscles, until they either die or are ingested by the vectors. They give rise to gametes in insects.

Topic: Biology, Human Health and Diseases, AIIMS, AIIMS 2013
Shapes of certain interhalogen compounds are stated below. Which one of them is not correctly stated?

Options:
A
$\mathrm{IF}_7$ : pentagonal bipyramid

B
$\mathrm{BrF}_5$ : trigonal bipyramid

C
$\mathrm{BrF}_3$ : planar T-shaped

D
$\mathrm{ICI}_3$ : planar dimeric

The correct answer is:
$\mathrm{BrF}_5$ : trigonal bipyramid

Solution:
The molecular geometry of $\mathrm{BrF}_5$ is square pyramidal with asymmetric charge distribution on the central atom.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Shapes of certain interhalogen compounds are stated below. Which one of them is not correctly stated?

Options:
A
$\mathrm{IF}_7$ : pentagonal bipyramid

B
$\mathrm{BrF}_5$ : trigonal bipyramid

C
$\mathrm{BrF}_3$ : planar T-shaped

D
$\mathrm{ICI}_3$ : planar dimeric

The correct answer is:
$\mathrm{BrF}_5$ : trigonal bipyramid

Solution:
The molecular geometry of $\mathrm{BrF}_5$ is square pyramidal with asymmetric charge distribution on the central atom.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2014 (11 Apr Online)
Shark is

Options:
A Oviparous
B Viviparous
C Ovoviviparous
D Both (A) and (C)

The correct answer is:
Ovoviviparous

Solution:
Ovoviviparity, ovoviviparous, ovivipary, or aplacental viviparity is an outmoded term used as a 'bridging' form of reproduction between egg-laying oviparous and live-bearing viviparous reproduction. Ovoviviparous animals have the embryos develop inside eggs that remain in the mother's body until they are ready to hatch, e.g., sharks.

Topic: Biology, Reproduction in Organisms, NEET
Ship A is sailing towards north-east with velocity) v→=30i^+50j^ km h-1 where i^ points east and j^, north. The ship B is at a distance of 80 km east and 150 km north of Ship A and is sailing towards the west at 10 km h-1. A will be at the minimum distance from B in:

Options:
A 4.2 h
B 3.2 h
C 2.6 h
D 2.2 h

The correct answer is:
2.6 h

Solution:
v→rel=v→1-v2=40i+50j
r→rel=r→1-r→2=-80i-150j
tmin=v→rel· r→relv→rel2=-3200-7500402+5022
tmin=107004100=10741=2.6 h

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2019 (08 Apr Shift 1)
Shoot bud formation in tissue culture is promoted by

Options:
A high auxin - low sucrose
B high GA3 - high cytokinin
C low auxin - high cytokinin
D high auxin - low cytokinin

The correct answer is:
low auxin - high cytokinin

Solution:
Shoot or stem development is mainly induced by high cytokinin.

Topic: Biology, Plant - Growth and Development, NEET
Shortest distance between the lines
x-12=y+8-7=z-45 and x-12=y-21=z-6-3 is

Options:
A 23
B 43
C 33
D 53

The correct answer is:
43

Solution:
Given lines are

x-12=y+8-7=z-45 and

x-12=y-21=z-6-3

Corresponding vectors are:

a→=i^-8j^+4k^, p→=2i^-7j^+5k^

b→=i^+2j^+6k^, q→=2i^+j^-3k^

S.D.=|a→-b→ p→ q→ ||p→×q→|

a→-b→=-10j^-2k^

[a→-b→ p→ q→]=0-10-22-7521-3

=0-(-10)(-6-10)+(-2)(2+14)

=0-160-32=-192

|[a→-b→ p→ q→]|=192

p→×q→=i^j^k^2-7521-3

=i^16-j^-16+k^16

=16i^+j^+k^

|p→×q→|=163

S.D.=|a→-b→ p→ q→ ||p→×q→|=192163=123=43

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2023 (29 Jan Shift 2)
Show that the function $f: R \rightarrow R$ defined by $f(x)=\frac{1}{x}$ is one-one onto, where $R$ is the set of all non-zero real numbers. Is the result true, if the domain $R$ is replaced by $\mathrm{N}$ with co-domain being same as $\mathrm{R}$ ?

Options:

None

Solution:
(a) We observe the following properties of $f$ :
(i) $f(x)=\frac{1}{x}$, if $f\left(x_1\right)=f\left(x_2\right)$ $\Rightarrow \frac{1}{x_1}=\frac{1}{x_2} \Rightarrow x_1=x_2$
Each $x \in R$ has a unique image in codomain $\Rightarrow \mathrm{f}$ is one-one.
(ii) For each $y$ belonging codomain then $y=\frac{1}{x}$ or $x=\frac{1}{y}$ there is a unique pre-image of $y$. $\Rightarrow \mathrm{f}$ is onto.
(b) When domain $\mathrm{R}$ is replaced by $\mathrm{N}$. codomain remaining the same, then $\mathrm{f}: \mathrm{N} \rightarrow \mathrm{R}$ If $\mathrm{f}\left(\mathrm{x}_1\right)=\mathrm{f}\left(\mathrm{x}_2\right)$ $\Rightarrow \frac{1}{\mathrm{n}_1}=\frac{1}{\mathrm{n}_2} \Rightarrow \mathrm{n}_1=\mathrm{n}_2$ where $\mathrm{n}_1, \mathrm{n}_2 \in \mathrm{N}$ $\Rightarrow \mathrm{f}$ is one-one.
But for every real number belonging to codomain may not have a pre-image in $\mathrm{N}$.
e.g. $\frac{1}{2}, \frac{3}{2}, \mathrm{~N} \quad \therefore \mathrm{f}$ is not onto.

Topic: Mathematics, Relations and Functions (Class 12)
Show that the function $f: R \rightarrow\{x \in R:-1 < x < 1\}$ defined by $f(x)=\frac{x}{1+|x|}, x \in R$ is one one and onto function.

Options:

None

Solution:
(i) Let $x \geq 0, f(x)=\frac{x}{1+x}, f\left(x_1\right)=f\left(x_2\right)$
$$
\Rightarrow \frac{\mathrm{x}_1}{1+\mathrm{x}_1}=\frac{\mathrm{x}_2}{1+\mathrm{x}_2}
$$
or $\mathrm{x}_1\left(1+\mathrm{x}_2\right)=\mathrm{x}_2\left(1+\mathrm{x}_1\right)$
or $\mathrm{x}_1+\mathrm{x}_1 \mathrm{x}_2=\mathrm{x}_2+\mathrm{x}_2 \mathrm{x}_1 \Rightarrow \mathrm{x}_1=\mathrm{x}_2$,
when $x < 0, f(x)=\frac{x}{1-x}$
$$
\mathrm{f}\left(\mathrm{x}_1\right)=\mathrm{f}\left(\mathrm{x}_2\right) \Rightarrow \frac{\mathrm{x}_1}{1-\mathrm{x}_1}=\frac{\mathrm{x}_2}{1-\mathrm{x}_2}
$$
or $x_1\left(1-x_2\right)=x_2\left(1-x_1\right)$
or $x_1-x_1 x_2=x_2-x_2 x_1$
$\Rightarrow \mathrm{x}_1=\mathrm{x}_2 \Rightarrow \mathrm{f}$ is one one
(ii) As $-1 < \mathrm{x} < 1, \mathrm{f}(\mathrm{x})=\frac{\mathrm{x}}{1+|\mathrm{x}|}$ lies betwen $-\frac{1}{2}$ and $\frac{1}{2}$ when $x \geq 0$,
Let $y=\frac{x}{1+x} \therefore(1+x) y=x$ or $x y-x=y$
$$
\therefore \mathrm{x}(\mathrm{y}-1)=\mathrm{y} \Rightarrow \mathrm{x}=\frac{\mathrm{y}}{\mathrm{y}-1}
$$
when $x < 0, f(x)=\frac{x}{1-x}=y$
$$
\therefore x=y-x y \therefore x=\frac{y}{1+y}
$$
In both cases such value of $y$ in codomain of has a unique value in its domain. Hence $f$ is onto.
Thus, $\mathrm{f}$ is one one and onto.

Topic: Mathematics, Relations and Functions (Class 12)
Show that the Modulus Function $f: R \rightarrow R$ given by $f(x)$ $=|\mathbf{x}|$, is neither one-one nor onto, where $|\mathbf{x}|$ is $\mathbf{x}$, if $\mathbf{x}$ is positive or 0 and $|x|$ is $-x$, if $x$ is negative.

Options:

None

Solution:
$f: R \rightarrow R$ given by $f(x)=|x|$
(a) $\mathrm{f}(-1)=|-1|=1, \mathrm{f}(1)=|1|=1$
$\Rightarrow-1$ and 1 have the same image
$\therefore \mathrm{f}$ is not one-one
(b) No negative value belonging to codomain of f has any pre-image in its domain
$\therefore \mathrm{f}$ is not onto. Hence, $\mathrm{f}$ is neither one-one nor onto.

Topic: Mathematics, Relations and Functions (Class 12)
Show that the relation $R$ in the set $A$ of all the books in a library of a college, given by $R=\{(x, y): x$ and $y$ have same number of pages $\}$ is an equivalence relation.

Options:

None

Solution:
(i) The number of pages in a book remain the same $\Rightarrow$ Relation $\mathrm{R}$ is reflexive.
(ii) The book $x$ has the same number of pages as the book $y$.
$\Rightarrow$ Book y has the same number of pages as the book $x$.
$\Rightarrow$ The relation $R$ is symmetric.
(iii) Book $x$ and $y$ have the same number of pages. Also book $y$ and $z$ have the same number of pages.
$\Rightarrow$ Books $x$ and $z$ also have the same number of pages.
$\Rightarrow R$ is transitive also
Thus, $\mathrm{R}$ is an equivalence relation.

Topic: Mathematics, Relations and Functions (Class 12)
Show that the Signum Function $f: R \rightarrow R$ given by $f(x)$ $=\left\{\begin{array}{l}1, \text { if } x>0 \\ 0, \text { if } x=0 \\ -1, \text { if } x < 0\end{array}\right.$ is neither one-one nor onto.

Options:

None

Solution:
$f: R \rightarrow R$ given by $f(x)=\left\{\begin{array}{ccc}1, & \text { if } & x>0 \\ 0, & \text { if } & x=0 \\ -1, & \text { if } & x < 0\end{array}\right.$
(a) $f(1)=f(2)=1$
$\therefore 1$ and 2 have the same image i.e.
$\mathrm{f}\left(\mathrm{x}_1\right)=\mathrm{f}\left(\mathrm{x}_2\right)=1$ for $\mathrm{x}>0 \Rightarrow \mathrm{x}_1 \neq \mathrm{x}_2$
Similarly $f\left(x_1\right)=f\left(x_2\right)=-1$, for $x < 0$ where $x_1 \neq x_2$ $\Rightarrow \mathrm{f}$ is not one-one.
(b) Except $-1,0,1$ no other member of codomain of $\mathrm{f}$ has any pre-image in its domain.
$\therefore \mathrm{f}$ is not onto. $\Rightarrow \mathrm{f}$ is neither into nor onto.

Topic: Mathematics, Relations and Functions (Class 12)
Show that $y=\log (1+x)-\frac{2 x}{2+x} \quad x>-1$, is an increasing function of $x$ throughout its domain.

Options:

None

Solution:
Let $\mathrm{f}(\mathrm{x})=\log (1+\mathrm{x})-\frac{2 \mathrm{x}}{2+\mathrm{x}}$
$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{x}^2}{(\mathrm{x}+1)(\mathrm{x}+2)^2}$
For $f(x)$ to be increasing $f^{\prime}(x)>0$
$\Rightarrow \frac{1}{x+1}>0 \Rightarrow x>-1$
Hence, $y=\log x-\frac{2 x}{x+2}$ is an increasing function of $x$ for all values of $x>-1$.

Topic: Mathematics, Application of Derivatives
Shown in the figure are two point charges +Q and -Q inside the cavity of a spherical shell. The charges are kept near the surface of the cavity on opposite sides of the centre of the shell. If σ1 is the surface charge on the inner surface and Q1 net charge on it and σ2 the surface charge on the outer surface and Q2 net charge on it then:

Options:
A Ï1=0, Q1=0 , Ï2=0, Q2=0
B Ï1â 0,Â Q1=0 , Ï2â 0,Â Q2=0
C Ï1â 0,Q1â 0 , Ï2â 0,Â Q2â 0
D Ï1â 0, Q1=0 , Ï2=0, Q2=0

The correct answer is:
Ï1â 0, Q1=0 , Ï2=0, Q2=0

Solution:
By the property of electrostatic shielding in the conductors E=0 in the conductor. So, electric flux,=0 through a dotted Gaussian surface as shown
The net enclosed charge through the Gaussian surface=0,⇒ The net charge Q1 on the inner surface=0, but the equal and opposite induced charge on the surface will be distributed non uniformly on the inner surface
So, σ1=0
∵Q1=0 on the inner surface
So, net charge Q2=0 on the outer surface as conductor is neutral but ∵ outer surface is free from any electric field so no charge density exists on the outer surface. So, σ2=0 .

Topic: Physics, Electrostatics, JEE Main, JEE Main 2015 (10 Apr Online)
Shown in the figure are two point charges +Q and -Q inside the cavity of a spherical shell. The charges are kept near the surface of the cavity on opposite sides of the centre of the shell. If σ1 is the surface charge on the inner surface and Q1 net charge on it and σ2 the surface charge on the outer surface and Q2 net charge on it then:

Options:
A Ï1=0, Q1=0 , Ï2=0, Q2=0
B Ï1â 0,Â Q1=0 , Ï2â 0,Â Q2=0
C Ï1â 0,Q1â 0 , Ï2â 0,Â Q2â 0
D Ï1â 0, Q1=0 , Ï2=0, Q2=0

The correct answer is:
Ï1â 0, Q1=0 , Ï2=0, Q2=0

Solution:
By the property of electrostatic shielding in the conductors E=0 in the conductor. So, electric flux,=0 through a dotted Gaussian surface as shown
The net enclosed charge through the Gaussian surface=0,⇒ The net charge Q1 on the inner surface=0, but the equal and opposite induced charge on the surface will be distributed non uniformly on the inner surface
So, σ1=0
∵Q1=0 on the inner surface
So, net charge Q2=0 on the outer surface as conductor is neutral but ∵ outer surface is free from any electric field so no charge density exists on the outer surface. So, σ2=0 .

Topic: Physics, Electrostatics, JEE Main
Shown in the figure below is a meter-bridge set up with null deflection in the galvanometer.
The value of the unknown resistor $R$ is

Options:
A
$13.75 \Omega$

B
$220 \Omega$

C
$110 \Omega$

D
$55 \Omega$

The correct answer is:
$220 \Omega$

Solution:
$$
\frac{55}{20}=\frac{R}{80} \Rightarrow R=\frac{55 \times 8}{2}=220 \Omega
$$

Topic: Physics, Experimental Physics, JEE Main, JEE Main 2008
Shown in the figure is a block at rest having an inclined smooth groove in a vertical plane. The horizontal surface is smooth. A ball of the same mass as that of the block at rest is released from the top end. The time when the ball will leave groove is nL23g. Find n.

Options:

The correct answer is:
7

Solution:
âmv cos 60o + 2mu=0 â v=4u
12m [v2 +u2 +2uv cos 120Â°]+12mu2 =mgx sin60Â°
âv2=873 gx â ar = 437g
â´ t=2L Ã 743 g

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Shown in the figure is rigid and uniform one meter long rod AB held in horizontal position by two strings tied to its ends and attached to the ceiling. The rod is off mass 'm' and has another weight of mass 2 m hung at a distance of 75 cm from A. The tension in the string at A is:

Options:
A 0.5 mg
B 2 mg
C 0.75 mg
D 1 mg

The correct answer is:
1 mg

Solution:
Net torqueτnet aboutB is zero at equilibrium

TA×100-mg×50-2mg×25=0

TA×100=100×mg

∴TA=1 mg.

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2020 (02 Sep Shift 1)
Shubham has 75% chance of attending the annual meet. Shikha has a 90% chance if Shubham also attends otherwise she has a 40% chance of attending the meet. If I go to the annual meet and see Shikha there, then the probability that Shubham is also there, is

Options:
A 2731
B 1930
C 15
D 910

The correct answer is:
2731

Solution:
A= Shubham attends annual meet, PA=34
B= Shikha attends annual meet
PBA=910,PBA-=410=25
PAB=PA⋅PBAPA⋅PBA+PA¯PBA-=34×91034×910+14×410=2731

Topic: Mathematics, Probability, JEE Main
SI unit of permittivity is

Options:
A $C^2 m^2 N^{-1}$
B $C^{-1} m^2 N^{-2}$
C $C^2 m^2 N^2$
D $C^2 m^{-2} N^{-1}$

The correct answer is:
$C^2 m^{-2} N^{-1}$

Solution:
$F=\frac{1}{4 \pi \in} \frac{q_1 q_2}{r^2}$
$\Rightarrow \epsilon=\frac{1}{4 \pi} \frac{q_1 q_2}{F r^2}=C^2 m^{-2} N^{-1}$

Topic: Physics, Units and Dimensions, JEE Main
SI unit of universal gas constant is

Options:
A \( \mathrm{Cal}{ }^{\circ} \mathrm{C}^{-1} \)
B \( \mathrm{J} \mathrm{mol}^{-1} \)
C \( \mathrm{J} \mathrm{mol}^{-1} \mathrm{~K}^{-1} \)
D \( \mathrm{J} \mathrm{kg}^{-1} \)

The correct answer is:
\( \mathrm{J} \mathrm{mol}^{-1} \mathrm{~K}^{-1} \)

Solution:
The ideal gas equation is given by, PV=nRT, where P is pressure, V is volume, n is number of moles, R is gas constant and T is temperature. Gas constant is given by, R=PVnT. After putting the units of all physical quantities. We get, ⇒R=N m-2×m3mol×K

⇒R=N mmol×K N m=J, so the unit of gas constant,

R=J mol-1 K-1.

Topic: Physics, Units and Dimensions, JEE Main
Sickel cell anemia is :

Options:
A Characterized by elongated sickle like RBCs with a nucleus
B An autosomal linked dominant trait
C Caused by substitute of valine by glutamic acid in the beta globin chain of haemoglobin
D Caused by a change in a single base pair of DNA

The correct answer is:
Caused by a change in a single base pair of DNA

Solution:
No Solution Available

Topic: Biology, Principles of Inheritance and Variation, NEET, NEET 2009 (Mains)
Sickle cell anaemia has not been eliminated from the African population because:

Options:
A it is controlled by dominant genes
B it is controlled by recessive genes
C it is not a fatal disease
D it provides immunity against malaria

The correct answer is:
it provides immunity against malaria

Solution:
Sickle cell anaemia in which RBCs become sickle-shaped and stiff is a genetic disorder which is autosomal and linked to a recessive allele. It has not been eliminated from the African population because it provides immunity against malaria. People who are heterozygous for the sickle cell allele are much less susceptible to falciparum malariaone of the main causes of illness and death. Since sickle-shaped RBC effectively kills the malaria parasite, the individuals heterozygous for this variant are able to cope with malaria infections much better than normal persons. The natural selection thus maintains this variant form of haemoglobin along with the normal haemoglobin in malaria prone areas.

Topic: Biology, Principles of Inheritance and Variation, NEET, NEET 2006
Sickle cell anaemia is favoured by nature in a malaria-prone area. Which of the following category will be favoured and what type of selection is it?

Options:
A Normal individual, balancing selection
B Homozygous individual sickle cell, disruptiveselection
C Heterozygous female, balancing selection
D Homozygous male, normalising selection

The correct answer is:
Heterozygous female, balancing selection

Solution:
Balancing selection refers to a number of selective processes by which multiple alleles (different versions of a gene) are actively maintained in the gene pool of a population at frequencies above that of gene mutation.
Sickle cell anaemia is an abnormality of haemoglobin first reported by an American physician, James B. Herrick (1949) in Africa. In sufferers, RBCs become crescentic under oxygen deficiency conditions and clog blood capillaries causing death. The incidence of sickle cell anaemia is more widespread in regions malaria caused by Plasmodium falciparum and earlier it was believed that it was the only well-established case of balanced polymorphism. When forests were cleared to provide land for agriculture in Africa, it produced more breeding areas for mosquito, Anopheles gambiae, and caused spread of malaria. In this changed situation heterozygotes having both normal and sickle genes got selective advantage as their sickled RBCs could not support schizogony of malarial parasite but still carried enough oxygen under hypoxia conditions through normal RBCs. Dominant homozygotes suffered from malaria and died while recessive homozygotes carrying sickle gene died before reaching sexual maturity because of the inability of blood to carry enough oxygen and clogging of blood capillaries by sickled cells. Therefore, about 25% superiority of heterozygotes prevented elimination of the sickle gene and its frequency increased in malaria-prone areas.

Topic: Biology, Evolution, NEET
$\mathrm{SiCl}_4$ on hydrolysis forms ' $X$ ' and $\mathrm{HCl}$. Compound ' $X$ ' loses water at $1000^{\circ} \mathrm{C}$ and gives ' $Y$ '. Compounds ' $X$ ' and ' $Y$ ' respectively are

Options:
A $\mathrm{H}_2 \mathrm{SiCl}_6, \mathrm{SiO}_2$
B $\mathrm{H}_4 \mathrm{SiO}_4, \mathrm{Si}$
C $\mathrm{SiO}_2, \mathrm{Si}$
D $\mathrm{H}_4 \mathrm{SiO}_4, \mathrm{SiO}_2$

The correct answer is:
$\mathrm{H}_4 \mathrm{SiO}_4, \mathrm{SiO}_2$

Solution:
$\begin{array}{r}\mathrm{SiCl}_4+4 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{H}_4 \mathrm{SiO}_4+4 \mathrm{HCl} \\ \mathrm{H}_4 \mathrm{SiO}_4 \frac{\Delta}{1000^{\circ} \mathrm{C}} \mathrm{SiO}_2+2 \mathrm{H}_2 \mathrm{O}\end{array}$

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
Signals from fully developed foetus and placenta ultimately lead to parturition which requires the release of

Options:
A oestrogen from placenta
B oxytocin from maternal pituitary
C oxytocin from foetal pituitary
D relaxin from placenta

The correct answer is:
oxytocin from maternal pituitary

Solution:
The process of giving birth to a baby is called parturition. Parturition is induced by a complex neuroendocrine mechanism. The signals for parturition originate from the fully-developed foetus and the placenta which require the release of oxytocin from posterior lobe of maternal pituitary.

Topic: Biology, Human Reproduction, NEET, NEET 2010 (Mains)
Silanes are silicon hydrides of general formula SinH2n+2 and have several applications. From thedata given below, the bond dissociation enthalpy of Si-Si bond (in kJmol-1 ) is

Given:

âH of the reaction 2Sis+3H2gâ Si2H6g is 80.3 kJmol-1

Bond dissociation enthalpy for H - H = 436 kJmol-1

Bond dissociation enthalpy for Si - H = 304 kJmol-1

ÎfH Sig=450 kJmol-1

Options:

The correct answer is:
304

Solution:
8.03=2Ã405+3Ã436-Si-Si+6304
(Si - Si) Bond energy =304 kJmol-1

Topic: Chemistry, Thermodynamics (C), JEE Main