datavorous/entrance-exam-dataset huggingface dataset:
The equation of trajectory of a projectile is $y=10 x-\left(\frac{5}{9}\right) x^2$.
If we assume $g=10 \mathrm{~ms}^{-2}$, the range of projectile (in metre) is

Options:
A 36
B 24
C 18
D 9

The correct answer is:
18

Solution:
Equation of projectile is
$y=10 x-\left(\frac{5}{9}\right) x^2$
Standard equation is
$y=x \tan \theta-\frac{g}{2 u^2 \cos ^2 \theta} \cdot x^2$
On comparing,
$\tan \theta=10$
and
$\begin{array}{rlrl}
& \frac{g}{2 u^2 \cos ^2 \theta} & =\frac{9}{9} \\
\Rightarrow & 10 u^2 \cos ^2 \theta & =9 \mathrm{~g} \\
& \ddots & g & =10 \mathrm{~m} / \mathrm{s}^2 \\
& \therefore & u^2 \cos ^2 \theta & =9
\end{array}$
(given)
$\therefore$ Range of projectile $R=\frac{2 u^2 \sin \theta \cos \theta}{g}$
$\begin{aligned}
& =\frac{2 u^2 \tan \theta \cdot \cos ^2 \theta}{g} \\
& =\frac{2\left(u^2 \cos ^2 \theta\right) \cdot \tan \theta}{g} \\
& =\frac{2 \times 9 \times 10}{10}=18 \mathrm{~m}
\end{aligned}$

Topic: Physics, Motion In Two Dimensions, TS EAMCET, TS EAMCET 2005
The equation $p x^{2}+q x+r=0$ (where $p, q, r$, all are positive) has distinct real roots a and $\mathrm{b}$. Which one of the following is correct?

Options:
A $\mathrm{a}>0, \mathrm{~b}>0$
B $\mathrm{a} < 0, \mathrm{~b} < 0$
C $\mathrm{a}>0, \mathrm{~b} < 0$
D $\mathrm{a} < 0, \mathrm{~b}>0$

The correct answer is:
$\mathrm{a} < 0, \mathrm{~b} < 0$

Solution:
$\mathrm{px}^{2}+\mathrm{qx}+\mathrm{r}=0,(\mathrm{p}, \mathrm{q}, \mathrm{r}$ arepositive)
Whenever the coefficients and constant are positive in quadratic equation, its roots are always negative $\therefore \mathrm{a} < 0, \mathrm{~b} < 0$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2019 (Phase 1)
The equation $\sin ^{-1}\left(3 x-4 x^{3}\right)=3 \sin ^{-1}(x)$ is true for all
values of $x$ lying in which one of the following intervals?

Options:
A $\left[-\frac{1}{2}, \frac{1}{2}\right]$
B $\left[\frac{1}{2}, 1\right]$
C $\left[-1,-\frac{1}{2}\right]$
D $[-1,1]$

The correct answer is:
$[-1,1]$

Solution:
Let $\sin ^{-1} x=\theta \Rightarrow x=\sin \theta$
$\sin ^{-1}\left(3 \sin \theta-4 \sin ^{3} \theta\right)=\sin ^{-1} \sin 3 \theta=3 \theta=3 \sin ^{-1} x$
Equaion $\sin ^{-1}\left(3 x-4 x^{3}\right)=3 \sin ^{-1} x$ is true for all values
of $\mathrm{x}$ lying in the interval $[-1,1]$.

Topic: Mathematics, Inverse Trigonometric Functions, NDA, NDA 2008 (Phase 1)
The equation $\tan ^{-1}(1+x)+\tan ^{-1}(1-x)=\frac{\pi}{2}$ is satisfied by

Options:
A $x=1$
B $x=-1$
C $x=0$
D $x=\frac{1}{2}$

The correct answer is:
$x=0$

Solution:
$\tan ^{-1}(1+\mathrm{x})+\tan ^{-1}(1-\mathrm{x})=\frac{\pi}{2}$
$\tan ^{-1}\left[\frac{(1+x)+(1-x)}{1-(1+x)(1-x)}\right]=\frac{\pi}{2}$
$\Rightarrow \quad \frac{1+x+1-x}{1-(1+x)(1-x)}=\tan \frac{\pi}{2}$
$\Rightarrow \quad \frac{2}{1-(1+\mathrm{x})(1-\mathrm{x})}=\frac{1}{0}$
$\Rightarrow \quad 1-(1+\mathrm{x})(1-\mathrm{x})=0$
$\Rightarrow(1+\mathrm{x})(1-\mathrm{x})=1$
$1-\mathrm{x}^{2}=1$
$\mathrm{x}^{2}=0$
$\mathrm{x}=0$
$\therefore$ Option (c) is correct.

Topic: Mathematics, Inverse Trigonometric Functions, NDA, NDA 2015 (Phase 1)
The equation $\tan ^{2} \phi+\tan ^{6} \phi=\tan ^{3} \phi \cdot \sec ^{2} \phi$ is

Options:
A identity for only one value of $\phi$
B not an identity
C identity for all values of $\phi$
D None of the above

The correct answer is:
not an identity

Solution:
Given equation is $\tan ^{2} \phi+\tan ^{6} \phi=\tan ^{3} \phi \cdot \sec ^{2} \phi$
$\Rightarrow \tan ^{2} \phi\left(1+\tan ^{4} \phi\right)=\tan ^{3} \phi \cdot \sec ^{2} \phi$
$\Rightarrow\left(1+\tan ^{4} \phi\right)=\tan \phi \cdot \sec ^{2} \phi$
Now, $\sec ^{2} \phi=1+\tan ^{2} \phi$
$\therefore\left(1+\tan ^{4} \phi\right)=\tan \phi\left(1+\tan ^{2} \phi\right)$
$\Rightarrow 1+\tan ^{4} \phi=\tan \phi+\tan ^{3} \phi$
It is not an identity

Topic: Mathematics, Trigonometric Equations, NDA, NDA 2008 (Phase 2)
The equation $\tan ^{4} x-2 \sec ^{2} x+a^{2}=0$ will have at least one
real solution if

Options:
A $|a| \leq 4$
B $|a| \leq 2$
C $|a| \leq \sqrt{3}$
D None of the above

The correct answer is:
$|a| \leq \sqrt{3}$

Solution:
$\Rightarrow\left(\tan ^{2} x\right)^{2}-2 \tan ^{2} x+a^{2}-2=0$
This is the quadratic equation in $\tan x$. So, roots of this will be real when discriminant $\geq 0$ a. $b^{2}-4 a c \geq 0$
$\Rightarrow 12-4 a^{2} \geq 0 \Rightarrow 4\left(3-a^{2}\right) \geq 0$
$\Rightarrow 3-a^{2} \geq 0 \quad(\because 4 \neq 0)$
$\Rightarrow a^{2} \leq 3$
$\Rightarrow a \leq \pm \sqrt{3} \Rightarrow|a| \leq \sqrt{3}$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2011 (Phase 2)
The equation to sphere passing through origin and the points $(-1,0,0),(0,-2,0)$ and $(0,0,-3)$ is $\mathrm{x}^{2}+y^{2}+z^{2}+f(x,$,
$y, z)=0$. What is $f(x, y, z)$ equal to? $\quad$

Options:
A $-x-2 y-3 z$
B $x+2 y+3 z$
C $x+2 y+3 z-1$
D $x+2 y+3 z+1$

The correct answer is:
$x+2 y+3 z$

Solution:
As we know, general equation of sphere is given as
$x^{2}+y^{2}+z^{2}+2 u x+2 v y+2 w z+d=0$
Given equation of sphere is $x^{2}+y^{2}+z^{2}+f(x, y, z)=0$
On comparing both the equations $(1)$ and $(2)$, we get $f(x, y, z)=2 t x+2 v y+2 w z+d$
Since, sphere passing through $(0,0,0),(-1,0,0)$, $(0,-2,0),(0,0,-3)$
we have from (1),
$d=0,1-2 u=0 \Rightarrow u=\frac{1}{2}$,
$4-4 v=0 \Rightarrow v=1,9-6 w=0 \Rightarrow w=\frac{3}{2}$
Hence,
$f(x, y, z)=2 x\left(\frac{1}{2}\right)+2 y(1)+2 z\left(\frac{3}{2}\right)+0$
$=x+2 y+3 z$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2011 (Phase 2)
The equation to the locus of a point which is always equidistant from the points $(1,0)$ and $(0,-2)$ is :

Options:
A $\quad 2 x+4 y+3=0$
B $\quad 4 x+2 y+3=0$
C $\quad 2 x+4 y-3=0$
D $4 x+2 y-3=0$

The correct answer is:
$\quad 2 x+4 y+3=0$

Solution:
Let $\mathrm{P}(\mathrm{x}, \mathrm{y})$ be the point. Let $\mathrm{A}=(1,0)$ and $\mathrm{B}=(0,-2)$ then $\mathrm{PA}=\mathrm{PB}$
$\Rightarrow(\mathrm{PA})^{2}=(\mathrm{PB})^{2}$
$\Rightarrow(\mathrm{x}-1)^{2}+\mathrm{y}^{2}=\mathrm{x}^{2}+(\mathrm{y}+2)^{2}$
$\Rightarrow \quad 1-2 x=4 y+4$
$\Rightarrow \quad 2 \mathrm{x}+4 \mathrm{y}+3=0$

Topic: Mathematics, Straight Lines, NDA, NDA 2012 (Phase 2)
The equation $x^{2}-4 x+29=0$ hasone root $2+5 i$. What is the other root?
$(i=\sqrt{-1})$

Options:
A 2
B 5
C $2+5$
D $2-5$

The correct answer is:
$2-5$

Solution:
Since, complex roots occur in pairs therefore other root is $2-5 i$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2011 (Phase 2)
The equation $x-2(x-1)^{-1}=1-2(x-1)^{-1}$ has

Options:
A no roots
B one root
C two equal roots
D infinite roots

The correct answer is:
no roots

Solution:
Given, $x-2(x-1)^{-1}=1-2(x-1)^{-1}$
$x-\frac{2}{x-1}=1-\frac{2}{x-1}$
$\Rightarrow \frac{x(x-1)-2}{x-1}=\frac{x-1-2}{x-1}$
$\Rightarrow x^{2}-x-2=x-1-2$
$\Rightarrow x^{2}-x-2-x+3=0$
$\Rightarrow x^{2}-2 x+1=0$
$\Rightarrow(x-1)^{2}=0$
$\Rightarrow x=1$
But $x=1$ is not sat isfied in the given equation.
Hence, no roots exist.

Topic: Mathematics, Quadratic Equation, NDA, NDA 2009 (Phase 2)
The equations of motion of a projectile are given by $x=36 t$ metre and $2 y=96 t-9.8 t^2$ metre. The angle of projection is :

Options:
A $\sin ^{-1}\left(\frac{4}{5}\right)$
B $\sin ^{-1}\left(\frac{3}{5}\right)$
C $\sin ^{-1}\left(\frac{4}{3}\right)$
D $\sin ^{-1}\left(\frac{3}{4}\right)$

The correct answer is:
$\sin ^{-1}\left(\frac{4}{5}\right)$

Solution:
$x=36 t$ metre
$2 y=\left(96 t-9.8 t^2\right)$ metre
Standard equations are,
$x=(u \cos \theta) t$
$y=u \sin \theta t-\frac{1}{2} g t^2$
$u \cos =36$
$u \sin \theta=\frac{96}{2}=48$
$\therefore \quad \tan \theta=\frac{u \sin \theta}{u \cos \theta}=\frac{48}{36}=\frac{4}{3}$
$\theta=\tan ^{-1} \frac{4}{3}$
or $\theta=\sin ^{-1} \frac{4}{5}$

Topic: Physics, Motion In Two Dimensions, AP EAMCET, AP EAMCET 2003
The equations of motion of a projectile are given by $x=36 t$ metre and $2 y=96 t-9.8 t^2$ metre. The angle of projection is :

Options:
A $\sin ^{-1}\left(\frac{4}{5}\right)$
B $\sin ^{-1}\left(\frac{3}{5}\right)$
C $\sin ^{-1}\left(\frac{4}{3}\right)$
D $\sin ^{-1}\left(\frac{3}{4}\right)$

The correct answer is:
$\sin ^{-1}\left(\frac{4}{5}\right)$

Solution:
$x=36 t$ metre
$2 y=\left(96 t-9.8 t^2\right)$ metre
Standard equations are,
$x=(u \cos \theta) t$
$y=u \sin \theta t-\frac{1}{2} g t^2$
$u \cos =36$
$u \sin \theta=\frac{96}{2}=48$
$\therefore \quad \tan \theta=\frac{u \sin \theta}{u \cos \theta}=\frac{48}{36}=\frac{4}{3}$
$\theta=\tan ^{-1} \frac{4}{3}$
or $\theta=\sin ^{-1} \frac{4}{5}$

Topic: Physics, Motion In Two Dimensions, TS EAMCET, TS EAMCET 2003
The equations
$x+2 y+3 z=1$
$2 x+y+3 z=2$
$5 x+5 y+9 z=4$

Options:
A have the unique solution
B have infinitely many solutions
C are inconsistent
D None of the above

The correct answer is:
have the unique solution

Solution:
$\begin{aligned} & x+2 y+3 z=1 \\ & 2 x+y+3 z=2 \\ & 5 x+5 y+9 z=4 \end{aligned}$
Writing in matrix form, $\mathrm{Ax}=\mathrm{B}$
$A=\left[\begin{array}{lll}1 & 2 & 3 \\ 2 & 1 & 3 \\ 5 & 5 & 9\end{array}\right]$
$|\mathrm{A}|=1(9-15)-2(18-15)+3(10-5)$
$=-6-2(3)+3(5)=-12+15=3 \neq 0$
So, these equations have unique solution

Topic: Mathematics, Determinants, NDA, NDA 2017 (Phase 1)
The equilibrium constant Kp for a homogenous gaseous reaction is 10-3 . The standard Gibbs free energy change ΔG° for the reaction at 27°C (using R = 2 cal K-1mol-1 ) is

Options:
A 0
B -1.8 kcal
C -4.1454 kcal
D +4.1454 kcal

The correct answer is:
+4.1454 kcal

Solution:
ÎGâ=ââ2.303âRTâlogâ K p
=â2.303Ã2Ã300âlogâ 10 â3
=â2.303âÃâ2âÃâ300âÃâ(â3)
= 1445.4 cal = 4.1454 kcal

Topic: Chemistry, Chemical Equilibrium, NEET
The equilibrium constant of a reaction is $0.008$ at $298 \mathrm{~K}$. The standard free energy change of the reaction at the same temperature is

Options:
A $-11.96 \mathrm{~kJ}$
B $-5.43 \mathrm{~kJ}$
C $-8.46 \mathrm{~kJ}$
D + $11.96 \mathrm{~kJ}$

The correct answer is:
+ $11.96 \mathrm{~kJ}$

Solution:
Standard free energy change,
$\begin{aligned} \Delta \mathrm{G}^{\circ} &=-2.303 \mathrm{RT} \log \mathrm{K} \\ &=-2.303 \times 8.314 \times 298 \times \log 0.008 \\ &=-2.303 \times 8.314 \times 298 \times(3 \log 2-3 \log 10) \\ &=-2.303 \times 8.314 \times 298 \times(0.903-3.0) \\ &=-2.303 \times 8.314 \times 298 \times(-2.097) \\ &=+11965.16 \mathrm{~J}=+11.96 \mathrm{~kJ} \end{aligned}$

Topic: Chemistry, Electrochemistry, KCET, KCET 2012
The evolutionary story of moths in England during industrialisation reveals, that 'evolution is apparently reversible'. Clarify this statement.

Options:

None

Solution:
The peppered moths were initially white coloured, then black coloured due to industrialisation. In recent years, the light coloured moths are increasing in population.
During the 20th century a light coloured peppered moth Biston betularia was found on the bark of trees in the industrial regions of England. The tree bark was covered by whitish lichens, so the light coloured moths escaped unnoticed from predatory birds.
The barks after industrialisation got covered by smoke, so the white moths were selectively picked up by birds. The black coloured moths however, escaped unnoticed against a dark background and became abundant gradually.
However, in recent years, reduction in industrial pollution has led to the growth of lichens again and thus, the population of light coloured moths is increasing again. The evolutionary story of moths in England, thus reveals, that 'evolution is apparently reversible'.

Topic: Biology, Evolution
The expansion of $(\mathrm{x}-\mathrm{y})^{\mathrm{n}}, \mathrm{n} \geq 5$ is done in the descending powers of $x$. If the sum of the fifth and sixth terms is zero,
then $\frac{\mathrm{x}}{\mathrm{y}}$ is equal to

Options:
A $\frac{\mathrm{n}-5}{6}$
B $\frac{\mathrm{n}-4}{5}$
C $\frac{5}{n-4}$
D $\frac{6}{n-5}$

The correct answer is:
$\frac{\mathrm{n}-4}{5}$

Solution:
$(x-y)^{n}, n \geq 5$
General term, $T_{r+1}=n_{c_{r}} x^{n-r}(-y)^{r}$
$T_{5}+T_{6}=0$
$\Rightarrow\left[n_{c_{4}} x^{n-4}(-y)^{4}\right]+\left[n_{c_{5}} x^{n-5}(-y)^{5}\right]=0$
$\Rightarrow n_{c_{4}} x^{n-4} y^{4}-n_{c_{5}} x^{n-5} y^{5}=0$
$\Rightarrow n_{c_{4}} x^{n-4} y^{4}=n_{c_{5}} x^{n-5} y^{5}$
$\Rightarrow \quad \frac{x^{n-4-n+5}}{y}=\frac{n_{c_{5}}}{n_{c_{4}}} \Rightarrow \frac{x}{y}=\frac{x !}{5 !(n-5) !} \times \frac{4 !(n-4) !}{\not x !}$
$=\frac{\not !(n-4)(n-5) !}{5 \times \not 1 !(n-5) !}=\frac{n-4}{5}$

Topic: Mathematics, Binomial Theorem, NDA, NDA 2017 (Phase 1)
The experimental data for decomposition of N2O5: [ 2N2O5→ 4NO2+ O2] in gas phase at 318K are given below:
Which of the following options is closest to the half-life period for the reaction?

Options:
A 864 seconds
B 1450 seconds
C 2863 seconds
D 3320 seconds

The correct answer is:
1450 seconds

Solution:
Correct Option is : (B)
1450 seconds

Topic: Chemistry, Chemical Kinetics, NEET
The expression $\frac{\sin \alpha+\sin \beta}{\cos \alpha+\cos \beta}$ is equal to

Options:
A $\tan \left(\frac{\alpha+\beta}{2}\right)$
B $\cot \left(\frac{\alpha+\beta}{2}\right)$
C $\sin \left(\frac{\alpha+\beta}{2}\right)$
D $\cos \left(\frac{\alpha+\beta}{2}\right)$

The correct answer is:
$\tan \left(\frac{\alpha+\beta}{2}\right)$

Solution:
$\begin{aligned} & \frac{\sin \alpha+\sin \beta}{\cos \alpha+\cos \beta}=\frac{\not 2 \sin \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\alpha-\beta}{2}\right)}{\not 2 \cos \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\alpha-\beta}{2}\right)} \\ &=\tan \left(\frac{\alpha+\beta}{2}\right) \end{aligned}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2017 (Phase 1)
The expression $\frac{\cot x+\operatorname{cosecx}-1}{\cot x-\operatorname{cosecx}+1}$ is equal to:

Options:
A $\frac{\sin x}{1-\cos x}$
B $\frac{1-\cos x}{\sin x}$
C $\frac{1+\cos x}{\sin x}$
D $\frac{\sin x}{1+\cos x}$
$\frac{1-\tan ^{2} \frac{x}{2}}{2}$

The correct answer is:
$\frac{1+\cos x}{\sin x}$

Solution:
$\frac{\cot x+\operatorname{cosecx}-\left(\operatorname{cosec}^{2} x-\cot ^{2} x\right)}{\cot x-\operatorname{cosec} x+1}$
$=\frac{\cot x+\operatorname{cosecx}-[(\operatorname{cosec} x-\cot x)(\operatorname{cosec} x+\cot x)]}{\cot x-\operatorname{cosecx}+1}$
$=\frac{(\cot x+\operatorname{cosecx})(1+\cot x-\operatorname{cosec} x)}{(\cot x-\operatorname{cosec} x+1)}$
$=\cot x+\operatorname{cosec} x$
$=\frac{\cos x}{\sin x}+\frac{1}{\sin x}=\frac{1+\cos x}{\sin x}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2013 (Phase 1)
The extinct subspecies of zebra, Quagga, was originally found in

Options:
A Russia
B South Africa
C Mauritius
D India

The correct answer is:
South Africa

Solution:
The quagga was a subspecies of the plain zebras which was found abundantly in the areas of Cape Province and South Africa.

Topic: Biology, Biodiversity and its Conservation, NEET
The ΔfH° for CO2(g),CO(g) and H2O(g) are -393.5, -110.5 and -241.8 kJmol-1 respectively. The standard enthalpy change (in kJ) for the reaction CO2(g)+H2(g)→CO(g)+H2O(g) is

Options:
A 524.1
B 41.2
C -262.5
D -41.2

The correct answer is:
41.2

Solution:
ÎrH=Î£ÎfHoProducts-Î£ÎfHoReactants
=ÎfHCOgo+ÎfHH2Ogo-ÎfHCO2go
=-110.5+-241.5--393.5
=+41.2 kJ

Topic: Chemistry, Thermodynamics (C), NEET
The false statement about the Terror of Bengal is:

Options:
A It has been a native of India since hundreds of years.
B It is commonly called water hyacinth and have mauve flowers.
C Its scientific name is Eichhornia crassipes.
D It causes disturbance in ecosystem dynamics of water body in a short span of time.

The correct answer is:
It has been a native of India since hundreds of years.

Solution:
Plant of water hyacinth Eichhornia crassipes is known by 'Terror of Bengal'. It grows abundantly in eutrophic water bodies and can cause blocks in waterways. They give distinct colour to water, deteriorate the quality of water and are extremely harmful to fishes. They grow rapidly and are responsible for imbalance in water body ecosystem. They are native to Brazil and were introduced in India by the British

Topic: Biology, Environmental Issues, NEET
The false statements among the following are
I. A primary carbocation is less stable than a tertiary carbocation.
II. A secondary propyl carbocation is less stable than allyl carbocation.
III. A tertiary free radical is more stable than a primary free radical.
IV. Isopropyl carbanion is more stable than ethyl carbanion.

Options:
A $\overline{\mathrm{I}}$ and $\overline{\mathrm{II}}$
B $\bar{\Pi}$ and III
C I and IV
D II and IV

The correct answer is:
II and IV

Solution:
Since $2^{\circ}$ propyl carbocation is little more stable than allyl carbocation and ethyl carbanion is more stable than isopropyl carbanion.

Topic: Chemistry, General Organic Chemistry, VITEEE, VITEEE 2015
The ferns, conifers, gnetales and Ginkgo all are direct descendants of

Options:
A Bryophytes
B Psilophytons
C Tracheophytes
D Chlorophytes

The correct answer is:
Psilophytons

Solution:
Psilophyton is a genus of extinct vascular plants. Described in 1859, it was one of the first fossil plants to be found which was of Devonian age. Plants lacked leaves or true roots; spore-forming organs or sporangia were born on the ends of branched clusters.

Topic: Biology, Evolution, NEET
The fifth term of an APof n terms, whose sum is $\mathrm{n}^{2}-2 \mathrm{n}$, is

Options:
A 5
B 7
C 8
D 15

The correct answer is:
7

Solution:
Given, $\mathrm{S}_{\mathrm{n}}=\mathrm{n}^{2}-2 \mathrm{n}$.
We know, $T_{n}=S_{n}-S_{n-1}$
So, $T_{5}=S_{5}-S_{4}$
$S_{5}=5^{2}-2(5)=25-10=15$
$S_{4}=4^{2}-2(4)=16-8=8$
So, $T_{5}=15-8=7$

Topic: Mathematics, Sequences and Series, NDA, NDA 2017 (Phase 1)
The figure shows a human blood cell. Identify it and give its characteristics.

Options:
A $\begin{array}{|l|l|} \hline \text{ Basophil } & \text{ Secretes } \\ \text{ inflammatory } & \text{ serotonin }, \end{array}$
B $\begin{array}{|c|c|} \hline \begin{array}{l}\text { B-lymphocyte } \\ \text { cells }\end{array} & \begin{array}{l}\text { Forms about } 20 \% \\ \text { of blood involved } \\ \text { immune }\end{array} \\ \hline \end{array}$
C $\begin{array}{|l|l|} \hline \text{ Neutrophil } & \begin{array}{l}\text { Most abundant } \\ \text { blood cells }, \\ \text { phagocytic }\end{array} \\ \hline \end{array}$
D $\begin{array}{|l|l|} \hline \text{ Monocyte } & \begin{array}{l}\text { Life span of } 3 \\ \text { days, produces } \\ \text { antibodies }\end{array} \\ \hline \end{array}$

The correct answer is:
$\begin{array}{|l|l|} \hline \text{ Basophil } & \text{ Secretes } \\ \text{ inflammatory } & \text{ serotonin }, \end{array}$

Solution:
Basophils have three lobed nucleus and a smaller number of coarse granules. They secrete histamine, serotonin, heparin, etc., and are involved in inflammatory reactions. Lymphocytes are smaller in size having round nucleus. They are of two major types - ' $B$ ' and ' $T$ ' lymphocytes. Both $\mathrm{B}$ and $T$ lymphocytes are responsible for immune responses of the body. Neutrophils are the most abundant cells and are phagocytic in nature which destroy foreign organisms entering the body. They are polymorphic and have multi-lobed nucleus. Monocytes are the largest of all WBCs. They are motile, phagocytic in nature. Their life span is about $10-20 \mathrm{hrs}$.
Related Theory
There are two types of $W B C$ - granulocytes and agranulocytes. Neutrophils, eosinophils and basophils are different types of granulocytes, while lymphocytes and monocytes are the Agranulocytes.

Topic: Biology, Body Fluids and Circulation, NEET, NEET 2013 (Karnataka)
The figure shows a section of human ovary. Select the option which gives the correct identification of cither A or B with function/characteristic.

Options:
A b- Corpus luteum - Secretes progesterone
B a-Tertiary follicle - Forms Graafian follicle
C b-Corpus luteum-Secretes estrogen
D a- Primary oocyte - It is in the prophase-I of the meiotic division

The correct answer is:
b- Corpus luteum - Secretes progesterone

Solution:
The zona pellucida and thecal cells of Graafian follicle in the ovary after ovulation forms corpus luteum, which produce progesterone. The tertiary follicle changes into the mature follicle or Graafian follicle and forms a new membrane called zona pellucida surrounding it. The Graafian follicle now ruptures to release the secondary oocyte (ovum) from the ovary. During oogenesis in embryonic development, oogonia start division and enter into prophase-I of the meiotic division and get temporarily arrested at that stage, called primary oocytes. Each primary oocyte then gets surrounded by a layer of granulosa cells and is called the primary follicle
Related Theory
The secondary follicle transforms into a tertiary follicle characterized by a fluid filled cavity called antrum. The theca layer organize into an inner theca interna and an outer theca externa. At this stage (puberty) the primary oocyte within the tertiary follicle grows in size and completes its first meiotic division. It is an unequal division resulting in the formation of a large haploid secondary oocyte and a tiny first polar body.

Topic: Biology, Human Reproduction, NEET, NEET 2013 (Karnataka)
The figure shows four animals A, B, C, and D. Select the incorrect answer with respect to a common characteristic of these animals.

A B
C D

Options:
A Tapeworm- False segmentation, triploblastic
B Aurelia-Polyp form, stinging capsules are present on body.
C Octopus- Body is unsegmented
D Prawn-They have jointed appendages

The correct answer is:
Aurelia-Polyp form, stinging capsules are present on body.

Solution:
Tapeworms belong to Platyhelminthes, which are triploblastic and show false segmentation. Octopus belongs to Molluscs which have an unsegmented body. Prawns belong to Arthropods and thus have jointed appendages. Aurelia have umbrella shaped body and belong to phylum Colenterates. Umbrella shaped body is the medusa form. Being a Coelenterate, it shows the presence of cnidocytes, which are stinging cells present on the body and are used for anchorage, defense, capturing the prey.

Topic: Biology, Animal Kingdom, NEET
The first biosphere reserve established in India is

Options:
A Pachmarhi Biosphere Reserve
B Nokrek Biosphere Reserve
C Sundarbans Biosphere Reserve
D Nilgiri Biosphere Reserve

The correct answer is:
Nilgiri Biosphere Reserve

Solution:
In 1986, India established its first biosphere reserve called the Nilgiri Biosphere Reserve which covers Tamilnadu, Kerala and Karnataka. Gulf of Mannar biosphere reserve in Tamilnadu is India's largest biosphere reserve spanning an area of 10,500 sq. km.

Topic: Biology, Biodiversity and its Conservation, NEET
The first meiotic division during spermatogenesis takes place when:

Options:
A Primary spermatocytes divide to form secondary spermatocytes
B Secondary spermatocytes divide to form spermatids
C Spermatogonia divide to form primary spermatocytes
D Spermatids to spermatozoa

The correct answer is:
Primary spermatocytes divide to form secondary spermatocytes

Solution:
Spermatogenesis is a highly specialized process of cellular differentiation resulting in the formation of functional spermatozoa. Spermatogenesis begins with a diploid spermatogonium in the seminiferous tubules, which divides mitotically to produce two diploid primary spermatocytes.
The primary spermatocyte then undergoes meiosis I to produce two haploid secondary spermatocytes.
The haploid secondary spermatocytes undergo meiosis II to produce four haploid spermatids.

Topic: Biology, Human Reproduction, NEET
The focal distance of a point on the parabola $y^{2}=12 x$ is 4 . What is the abscissa of the point?

Options:
A 1
B $-1$
C 2 $\sqrt{3}$
D -2

The correct answer is:
1

Solution:
Focal distance of a point $\left(x_{1}, y_{1}\right)$ on the parabola is $y^{2}=4 a x$ is equal to its distance from directrix $x+a=0$ is
$\mathrm{x}_{1}+\mathrm{a}$
For $y^{2}=12 x$, comparing with $y^{2}=4 a x$.
$4 a=12 \Rightarrow a=3$
so, $x_{1}+3=4$
$\Rightarrow \quad x_{1}=1$

Topic: Mathematics, Parabola, NDA, NDA 2006 (Phase 1)