datavorous/entrance-exam-dataset huggingface dataset:
The set of hormones which lead to the formation of secondary messengers like cyclic AMP during their mode of action is:

Options:
A LH, FSH, ACTH
B Estrogen, oxytocin, FSH
C Cortisol, estrogen, progesterone
D Thyroxine, estradiol, LH

The correct answer is:
LH, FSH, ACTH

Solution:
All water-soluble hormones must bind to their specific membrane-bound or extracellular receptors on target cells. This results in the formation of the hormone-receptor complex. Hormones which interact with membrane-bound receptors normally do not enter the target cell, but generate second messengers (e.g., cyclic AMP, IP3, Ca++ etc) which in turn regulate cellular metabolism. E.g. peptide, polypeptide, protein hormones (e.g., insulin, glucagon, pituitary hormones, hypothalamic hormones, etc.)

Topic: Biology, Chemical Coordination and Integration, NEET
The shadow of a tower standing on a level plane is found to be $50 \mathrm{~m}$ longer when the Sun's elevation is $30^{\circ}$ than when it is $60^{\circ}$. The height of the tower is:

Options:
A $25 \mathrm{~m}$
B $25 \sqrt{3} \mathrm{~m}$
C $50 \mathrm{~m}$
D None of these

The correct answer is:
$25 \sqrt{3} \mathrm{~m}$

Solution:
Let $\mathrm{AB}$ be the tower and $\mathrm{AC} \& \mathrm{AD}$ be the shadows of the tower. Let $\mathrm{h}$ be the height of the tower. $\operatorname{In} \Delta \mathrm{ABC}, \tan 60^{\circ}=\frac{\mathrm{h}}{\mathrm{x}}$
$\operatorname{In} \Delta \mathrm{ABD}, \tan 30^{\circ}=\frac{\mathrm{h}}{\mathrm{x}+50}$
(i) $\div$ (ii) $\Rightarrow \frac{\sqrt{3}}{\frac{1}{\sqrt{3}}}=\frac{\mathrm{x}+50}{\mathrm{x}}$
$\Rightarrow 3 \mathrm{x}=\mathrm{x}+50$
$\Rightarrow \quad \mathrm{x}=25^{\circ}$
$\Rightarrow \quad \mathrm{h}=\mathrm{x} \sqrt{3}=25 \sqrt{3}$

Topic: Mathematics, Heights and Distances, NDA, NDA 2013 (Phase 1)
The shells of crabs found around a particular coastal location in Japan seem mostly to resemble the legendary face of a Samurai. Given below are two explanations of this observed fact. Which of these strikes you as a scientific explanation?
(a) A tragic sea accident several centuries ago drawned a young Samurai. As a tribute to his bravery, nature through its inscrutable ways immortalised his face by imprinting it on the crab shells in that area.
(b) After the sea tragedy, fishermen in that area, in a gesture to honour to their dead hero, let free any crab shell caught by them which accidentally had a shape resembling the face of a samurai. Consequently the particular shape of the crab shell survived longer and therefore in course of time the shape was genetically propagated. This is an example of evalution by artificial selection.
[Note : This interesting illustration taken from Carl Sagan's 'The Cosmos' highlights the fact that often strange and inexplicable facts which on the first sight appear 'supernatural' actually turn out to have simple scientific explanations. Try to think out other examples of this kind].

Options:

None

Solution:
(b)

Topic: Physics, Mathematics in Physics
The sides $a, b, c$ of a triangle $\mathrm{ABC}$ are in arithmetic progression and ' $a$ ' is the smallest side. What is $\cos A$ equal to?

Options:
A $\frac{3 c-4 b}{2 c}$
B $\frac{3 c-4 b}{2 b}$
C $\frac{4 c-3 b}{2 c}$
D $\frac{3 b-4 c}{2 c}$

The correct answer is:
$\frac{4 c-3 b}{2 c}$

Solution:
Given a, b, c are in arithmetic progression.
$\therefore 2 b=a+c$ ...(i)
Now, we know
$\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}$
$=\frac{b^{2}+c^{2}-(2 b-c)^{2}}{2 b c}($ from 1$)$
$=\frac{b^{2}+c^{2}-4 b^{2}-c^{2}+4 b c}{2 b c}$
$=\frac{-3 b^{2}+4 b c}{2 b c}=\frac{4 c-3 b}{2 c}$

Topic: Mathematics, Properties of Triangles, NDA, NDA 2011 (Phase 2)
The similarity of bone structure in the forelimbs of many vertebrates is an example of:

Options:
A Convergent evolution
B Analogy
C Homology
D Adaptive radiation

The correct answer is:
Homology

Solution:
The similarity of bone structure in the forelimbs of many vertebrates is an example of homology, The homologous organs have the same fundamental structure but are adapted to perform different functions, eg. forelimbs of man, cheetah, whale, and bat.
Analogous organs show convergent evolution. These organs have similar functions but are different in their structural details and origin. Development of different functional structures from a common ancestral form is called adaptive radiation.

Topic: Biology, Evolution, NEET, NEET 2018
The simultaneous equations $3 x+5 y=7$ and $6 x+10 y=18$ have

Options:
A no solution
B infinitely many solutions
C unique solution
D any finite number of solutions

The correct answer is:
no solution

Solution:
Given system of equations $3 x+5 y=7$ and $6 x+10 y=18$
This system can be written as $A X=B$ where
$A=\left(\begin{array}{cc}3 & 5 \\ 6 & 10\end{array}\right), X=\left(\begin{array}{l}x \\ y\end{array}\right), B=\left(\begin{array}{c}7 \\ 18\end{array}\right)$
Now, $|A|=30-30=0$
and $(\operatorname{adj} A) B=\left(\begin{array}{cc}10 & -5 \\ -6 & 3\end{array}\right)\left(\begin{array}{c}7 \\ 18\end{array}\right)=\left(\begin{array}{c}-20 \\ -96\end{array}\right) \neq 0$
$\therefore$ system of equations is inconsistent. $\Rightarrow$ system of equations have no solution.

Topic: Mathematics, Determinants, NDA, NDA 2011 (Phase 2)
The sine of the angle between vectors $\vec{a}=2 \hat{i}-6 \hat{j}-3 \hat{k}$ and $\vec{b}=4 \hat{i}+3 \hat{j}-\hat{k}$

Options:
A $\frac{1}{\sqrt{26}}$
B $\frac{5}{\sqrt{26}}$
C $\frac{5}{26}$
D $\frac{1}{26}$

The correct answer is:
$\frac{5}{\sqrt{26}}$

Solution:
Given, $\vec{a}=2 \hat{i}-6 \hat{j}-3 \hat{k}$
$\vec{b}=4 \hat{i}+3 \hat{j}-\hat{k}$
$\vec{a} \cdot \vec{b}=(2 \hat{i}-6 \hat{j}-3 \hat{k}) \cdot(4 \hat{i}+3 \hat{j}-\hat{k})$
$=(2)(4)+(-6)(3)+(-3)(-1)$
$=8-18+3=-7$
$\Rightarrow|\vec{a}||\vec{b}| \cos \theta=-7 \Rightarrow \cos \theta=\frac{-7}{|\vec{a}||\vec{b}|}$
$=\frac{-7}{\sqrt{49} \sqrt{26}}=\frac{-1}{\sqrt{26}}$
$\sin \theta=\sqrt{1-\cos ^{2} \theta}=\sqrt{1-\left(\frac{-1}{\sqrt{26}}\right)^{2}}=\sqrt{1-\frac{1}{26}}$
$=\sqrt{\frac{25}{26}}=\frac{5}{\sqrt{26}}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2019 (Phase 1)
The sines of two angles of a triangle are equal to $5 / 13$ and $99 / 101 .$ What is the cosine of the third angle?

Options:
A $255 / 1313$
B $265 / 1313$
C $275 / 1313$
D 7701313

The correct answer is:
$255 / 1313$

Solution:
Let $\sin \theta=\frac{5}{13}$ and $\sin \phi=\frac{99}{101}$
$\therefore \cos \{\pi-(\theta+\phi)\}$
$=-\cos (\theta+\phi)$
$=-\{\cos \theta \cos \phi-\sin \theta \sin \phi\}$
$=-\left\{\sqrt{1-\frac{25}{169}} \sqrt{1-\left(\frac{99}{101}\right)^{2}}-\frac{5}{13} \times \frac{99}{101}\right\}$
$=-\left\{\frac{12}{13} \times \frac{20}{101}-\frac{5}{13} \times \frac{99}{101}\right\}$
$=-\left\{\frac{240}{1313}-\frac{495}{1313}\right\}=\frac{255}{1313}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2009 (Phase 1)
The sixth term of an AP is 2 and its common difference is greater than 1
What is the common difference of the AP so that the product of the first, fourth and fifth terms is greatest? $\quad[2016-I I]$

Options:
A $8 / 5$
B $9 / 5$
C 2
D $11 / 5$

The correct answer is:
$8 / 5$

Solution:
Let first term $=\mathrm{a} \&$ common difference $=\mathrm{x}$
$\therefore \quad a+5 x=2 \Rightarrow a=2-5 x$
$\begin{aligned} & \text { Let } \mathrm{P}=\mathrm{T}_{1} \times \mathrm{T}_{4} \times \mathrm{T}_{5} \\ \Rightarrow & \mathrm{P}=\mathrm{a}(\mathrm{a}+3 \mathrm{x})(\mathrm{a}+4 \mathrm{x}) \\ \Rightarrow & \mathrm{P}=(2-5 \mathrm{x})(2-5 \mathrm{x}+3 \mathrm{x})(2-5 \mathrm{x}+4 \mathrm{x}) \\ \Rightarrow & \mathrm{P}=-10 \mathrm{x}^{3}+34 \mathrm{x}^{2}-32 \mathrm{x}+8 \\ & \frac{d p}{d x}=0 \Rightarrow 15 \mathrm{x}^{2}-34 \mathrm{x}+16=0 \\ & \Rightarrow(5 \mathrm{x}-8)(3 \mathrm{x}-2)=0 \\ \Rightarrow & \mathrm{x}=\frac{8}{5},\left[\because x=\frac{2}{3} < 1\right] \end{aligned}$

Topic: Mathematics, Sequences and Series, NDA, NDA 2016 (Phase 2)
The sixth term of an AP is 2 and its common difference is greater than 1
What is the first term of the AP so that the product of the first, fourth and fifth terms is greatest? $\quad[2016-I I]$

Options:
A $-4$
B $-6$
C $-8$
D $-10$

The correct answer is:
$-6$

Solution:
Since, $\mathrm{a}=2-5 \mathrm{x}$
$\Rightarrow \quad a=2-5\left(\frac{8}{5}\right)$
$\Rightarrow \quad a=-6$

Topic: Mathematics, Sequences and Series, NDA, NDA 2016 (Phase 2)
The smallest positive divisor greater than 1 of a composite number $a$ is

Options:
A $ < \sqrt{a}$
B $=\sqrt{a}$
C $>\sqrt{a}$
D $\leq \sqrt{a}$

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

Solution:
By property, the smallest positive divisor greater than 1 of a composite number $a$ is less than or equal to $\sqrt{a}$, which is standard result.

Topic: Mathematics, Basic of Mathematics, KCET, KCET 2013
The smallest positive integer $n$ for which $\left(\frac{1+i}{1-i}\right)^{n}=1$, is

Options:
A 1
B 4
C 8
D 16

The correct answer is:
4

Solution:
$\left(\frac{1+\mathrm{i}}{1-\mathrm{i}}\right)^{\mathrm{n}}=1$
We know, $\mathrm{i}^{2}=-1$ and $\mathrm{i}^{4}=1$. Now, rationalise the denominator.
$\left(\frac{(1+i)(1+i)}{(1-i)(1+i)}\right)^{n}=\left(\frac{1+2 i+i^{2}}{1-i^{2}}\right)^{n}=\left(\frac{1+2 i-i}{1-(-1)}\right)^{n}=\left(\frac{2 i}{2}\right)^{n}=-i^{n}$
The smallest positive integer for which $\mathrm{i}^{\mathrm{n}}=1$ is 4

Topic: Mathematics, Complex Number, NDA, NDA 2017 (Phase 2)
The smallest positive integral value of $n$ for which $\left(\frac{1-i}{1+i}\right)^{n}$ is purely imaginary with positive imaginary part is

Options:
A 1
B 3
C 4
D 5

The correct answer is:
3

Solution:
$\begin{aligned} &\left(\frac{1-i}{1+i}\right)^{n}=\left[\frac{(1-i)^{2}}{(1+i)(1-i)}\right]^{n} \\ &=\left[\frac{1+i^{2}-2 i}{1-i^{2}}\right]^{n}=\left(\frac{-2 i}{2}\right)^{n}=(-i)^{n} \\ & \text { For } n=3 \end{aligned}$
$(-i)^{n}=(-i)^{3}=-i \times-i \times-i=-i^{3}=-(-i)=i$
Which is purely imaginary with positive imaginarypart.
Hence, $n=3$.

Topic: Mathematics, Complex Number, NDA, NDA 2011 (Phase 2)
The solubility product of a sparingly soluble $A B_2$ salt is $2.56 \times 10^{-4} \mathrm{M}^3$ at $25^{\circ} \mathrm{C}$. The $K_f$ of water is $1.8 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$. The depression in freezing point of a standard solution of $A B_2$ is

Options:
A 0.432 K
B 0.216 K
C 0.108 K
D 13.824 K

The correct answer is:
0.216 K

Solution:
Let, solubility of $A B_2$ (1:2 type electrolyte) is pure water $=5 \mathrm{~mol} \mathrm{~L}^{-1}=5 \mathrm{M}$
$\Rightarrow K_{\mathrm{sp}}=4 S^3=2.56 \times 10^{-4} \mathrm{M}^3$ (given)
$\therefore S=0.04 \mathrm{M}=0.04 \mathrm{~m}$ (molal) under standard condition of the solution.
Depression of freezing point,
$\Delta T_f=K_f \times m \times i=1.8 \times 0.04 \times 3=0.216 \mathrm{~K}$
$\left[\because\right.$ Assuming complete dissociation of $A B_2$,
van't Hoff factor, $i=3]$

Topic: Chemistry, Solutions, TS EAMCET, TS EAMCET 2020 (14 Sep Shift 2)
The solution of $\frac{d y}{d x}=|x|$ is: $\quad$

Options:
A $y=\frac{x|x|}{2}+c$
B $y=\frac{|x|}{2}+c$
C $y=\frac{x^{2}}{2}+c$
D $y=\frac{x^{3}}{2}+c$
Where $c$ is an arbitary constant

The correct answer is:
$y=\frac{x|x|}{2}+c$

Solution:
$\frac{d y}{d x}=|x|$
$\frac{d y}{d x}=x$ for $x \geq 0 ; \frac{d y}{d x}=-x$ for $x < 0$
$\int \mathrm{dy}=\int \mathrm{x} \mathrm{dx}$
$\mathrm{y}=\frac{\mathrm{x}^{2}}{2}+\mathrm{C}_{1} \quad \ldots(\mathrm{i}) ; \int \mathrm{dy}=-1 \mathrm{x} \mathrm{d} \mathrm{x}$
$\mathrm{y}=-\frac{\mathrm{x}^{2}}{2}+\mathrm{C}_{1}$
From (i) and (ii)
$\mathrm{y}=\frac{\mathrm{x}|\mathrm{x}|}{2}+\mathrm{C}$

Topic: Mathematics, Differential Equations, NDA, NDA 2014 (Phase 1)
The solution of $\frac{\mathrm{dy}}{\mathrm{dx}}=\sqrt{1-\mathrm{x}^{2}-\mathrm{y}^{2}+\mathrm{x}^{2} \mathrm{y}^{2}}$ is

Options:
A $\sin ^{-1} \mathrm{y}=\sin ^{-1} \mathrm{x}+\mathrm{c}$
B $2 \sin ^{-1} \mathrm{y}=\sqrt{1-\mathrm{x}^{2}}+\sin ^{-1} \mathrm{x}+\mathrm{c}$
C $2 \sin ^{-1} \mathrm{y}=\mathrm{x} \sqrt{1-\mathrm{x}^{2}}+\sin ^{-1} \mathrm{x}+\mathrm{c}$
D $2 \sin ^{-1} \mathrm{y}=\mathrm{x} \sqrt{1-\mathrm{x}^{2}}+\cos ^{-1} \mathrm{x}+\mathrm{c}$
where $\mathrm{c}$ is an arbitrary constant.

The correct answer is:
$2 \sin ^{-1} \mathrm{y}=\mathrm{x} \sqrt{1-\mathrm{x}^{2}}+\sin ^{-1} \mathrm{x}+\mathrm{c}$

Solution:
$\quad \because \frac{d y}{d x}=\sqrt{1-x^{2}-y^{2}+x^{2} y^{2}}$
$\frac{d y}{d x}=\sqrt{\left(1-x^{2}\right)\left(1-y^{2}\right)}$
$\Rightarrow \frac{d y}{\sqrt{1-y^{2}}}=\sqrt{1-x^{2}} \cdot d x$
$=\int \frac{\mathrm{dy}}{\sqrt{1-\mathrm{y}^{2}}}=\int \sqrt{1-\mathrm{x}^{2}} \cdot \mathrm{dx}$
[integrating $\mathrm{b} / \mathrm{s}$ ]
$=\sin ^{-1}\left(\frac{\mathrm{y}}{1}\right)=\frac{\mathrm{x}}{2} \sqrt{1-\mathrm{x}^{2}}+\frac{1}{2} \sin ^{-1}\left(\frac{\mathrm{x}}{1}\right)+\mathrm{c}$
$=2 \sin ^{-1} \mathrm{y}=\mathrm{x} \sqrt{1-\mathrm{x}^{2}}+\sin ^{-1} \mathrm{x}+\mathrm{c}$

Topic: Mathematics, Differential Equations, NDA, NDA 2015 (Phase 2)
The solution of the differential equation
$\frac{d y}{d x}=\cos (\mathrm{y}-\mathrm{x})+1$ is

Options:
A $\mathrm{e}^{\mathrm{x}}[\sec (\mathrm{y}-\mathrm{x})-\tan (\mathrm{y}-\mathrm{x})]=\mathrm{c}$
B $\mathrm{e}^{\mathrm{x}}[\sec (\mathrm{y}-\mathrm{x})+\tan (\mathrm{y}-\mathrm{x})]=\mathrm{c}$
C $\mathrm{e}^{\mathrm{x}} \sec (\mathrm{y}-\mathrm{x}) \tan (\mathrm{y}-\mathrm{x})=\mathrm{c}$
D $\mathrm{e}^{\mathrm{x}}=\mathrm{c} \sec (\mathrm{y}-\mathrm{x}) \tan (\mathrm{y}-\mathrm{x})$

The correct answer is:
$\mathrm{e}^{\mathrm{x}}[\sec (\mathrm{y}-\mathrm{x})-\tan (\mathrm{y}-\mathrm{x})]=\mathrm{c}$

Solution:
$\frac{d y}{d x}=\cos (y-x)+1$
Let $y-x=t \Rightarrow \frac{d y}{d x}-1=\frac{d t}{d x}$
$\therefore(1) \Rightarrow \frac{d t}{d x}+1=\cos t+1$
$\Rightarrow \frac{d t}{d x}=\cos t$
$\Rightarrow \sec t . d t=1 . d x$
$\Rightarrow \int \sec t \cdot d t=\int 1 \cdot d x$
$\Rightarrow \log |\operatorname{sect}+\tan t|=x+c$
$\Rightarrow-\log |\sec t-\tan t|=x+c$
$\Rightarrow \log |\sec t-\tan t|=-x-c$
$\Rightarrow \log (\sec (y-x)-\tan (y-x)=-x-c$
$\Rightarrow \sec (y-x)-\tan (y-x)=e^{-x} \cdot e^{-c}$
$\Rightarrow e^{x}(\sec (y-x)-\tan (y-x))=c$

Topic: Mathematics, Differential Equations, NDA, NDA 2019 (Phase 1)
The solution of the differential equation $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{y} \phi^{\prime}(\mathrm{x})-\mathrm{y}^{2}}{\phi(\mathrm{x})}$
is

Options:
A $\mathrm{y}=\frac{\mathrm{x}}{\phi(\mathrm{x})+\mathrm{c}}$
B $\mathrm{y}=\frac{\phi(\mathrm{x})}{\mathrm{x}}+\mathrm{c}$
C $\mathrm{y}=\frac{\phi(\mathrm{x})+\mathrm{c}}{\mathrm{x}}$
D $\mathrm{y}=\frac{\phi(\mathrm{x})}{\mathrm{x}+\mathrm{c}}$

The correct answer is:
$\mathrm{y}=\frac{\phi(\mathrm{x})}{\mathrm{x}+\mathrm{c}}$

Solution:
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{y} \phi^{\prime}(\mathrm{x})-\mathrm{y}^{2}}{\phi(\mathrm{x})}=\frac{\mathrm{y} \phi^{\prime}(\mathrm{x})}{\phi(\mathrm{x})}-\frac{\mathrm{y}^{2}}{\phi(\mathrm{x})}$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{y}^{2} \cdot \phi^{\prime}(\mathrm{x})}{\mathrm{y} \cdot \phi(\mathrm{x})}-\frac{\mathrm{y}^{2}}{\phi(\mathrm{x})}$
$\Rightarrow \frac{1}{\mathrm{y}^{2}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)=\frac{\phi^{\prime}(\mathrm{x})}{\mathrm{y} \cdot \phi(\mathrm{x})}-\frac{1}{\phi(\mathrm{x})}$
$\Rightarrow \frac{1}{\mathrm{y}^{2}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)-\frac{1}{\mathrm{y}} \cdot \frac{\phi^{\prime}(\mathrm{x})}{\phi(\mathrm{x})}=\frac{-1}{\phi(\mathrm{x})}$
Let $\mathrm{t}=\frac{-1}{\mathrm{y}} \Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=\frac{1}{\mathrm{y}^{2}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)$
$\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}+\mathrm{t} \cdot \frac{\phi^{\prime}(\mathrm{x})}{\phi(\mathrm{x})}=\frac{-1}{\phi(\mathrm{x})}$
$\mathrm{IF}=\mathrm{e}^{\int \frac{\phi^{\prime}(\mathrm{x})}{\phi(\mathrm{x})} \cdot \mathrm{dx}}=\mathrm{e}^{\log \phi(\mathrm{x})}=\phi(\mathrm{x})$
Solution of differential equation is
$\mathrm{t} \cdot \phi(\mathrm{x})=\int \frac{-1}{\phi(\mathrm{x})} \times \phi(\mathrm{x}) \cdot \mathrm{dx}$
$\Rightarrow \frac{-1}{y} \phi(x)=-x-c \Rightarrow \frac{\phi(x)}{y}=x+c \Rightarrow y=\frac{\phi(x)}{x}+c$

Topic: Mathematics, Differential Equations, NDA, NDA 2017 (Phase 2)
The solution of the simultaneous linear equations $2 \mathrm{x}+\mathrm{y}=6$ and $3 \mathrm{y}=8+4 \mathrm{x}$ will also be satisfied by which one of the following linear equations?

Options:
A $\quad x+y=5$
B $2 x+y=5$
C $\quad 2 x-3 y=10$
D $2 x+3 y=6$

The correct answer is:
$\quad x+y=5$

Solution:
Given equations $2 \mathrm{x}+\mathrm{y}=6 \Rightarrow 2 \mathrm{x}=6-\mathrm{y}$...(1)
and $3 y=8+2(2 x)$
$\Rightarrow 3 y=8+2(6-y)$ (from(1))
$\Rightarrow 3 y=8+12-2 y$
$\Rightarrow \mathrm{y}=4$ and $\mathrm{x}=1$
Now, equation given in option (a) is satisfied by $\mathrm{x}=1$ and $\mathrm{y}=4$

Topic: Mathematics, Basic of Mathematics, NDA, NDA 2012 (Phase 1)
The solution of which of the following will be non-conducting?

Options:
A
B
C
D

The correct answer is:

Solution:
Using Werner's theory in this complex all the chlorine atoms are showing secondary valency so they cannot be precipitated.

Topic: Chemistry, Coordination Compounds, NEET
The species of organisms which plays a vital role in controlling the relative abundance of other species in a community are called

Options:
A Edge species
B Keystone species
C Pioneer species
D Seral species

The correct answer is:
Keystone species

Solution:
The species that plays an important role in the functioning and stability of the ecosystem by controlling the relative abundance of other species in a community is called keystone species.

Topic: Biology, Organisms and Populations, NEET
The species of plants that plays a vital role in controlling the relative abundance of other species in a community is called

Options:
A Exotic species
B Keystone species
C Alien species
D Endemic species

The correct answer is:
Keystone species

Solution:
A species of plant that play vital role in controlling relative abundance of other species in a community is called keystone species. They are important for overall structure and function of an ecosystem. Exotic species or Alien species are organisms that have been introduced into an area outside their normal distribution. Endemic species exist only in one particular geographical region or area.

Topic: Biology, Biodiversity and its Conservation, NEET, NEET 2023 (Manipur)
The species, which is going to become extinct due to lack of proper care would be called

Options:
A rare
B endangered
C vulnerable
D extinct

The correct answer is:
endangered

Solution:
Taxa whose numbers have been reduced to a critical level or whose habitats have been so drastically reduced that they are deemed to be in immediate danger of extinction are called endangered animals. e.g. Lion tailed macaque, Crocodile, Musk deer, Rhino, etc.

Topic: Biology, Biodiversity and its Conservation, JIPMER, JIPMER 2005
The speed of light c, gravitational constant G and Planck’s constant h are taken as fundamental units in a system. The dimensions of time in this new system should be:

Options:
A

G 1 2 â h 1 2 â c â5 2

B

G −1 2   h 1 2   c 1 2

C

G 1 2   h 1 2   c −3 2

D

G 1 2   h 1 2   c 1 2

The correct answer is:

G 1 2 â h 1 2 â c â5 2

Solution:
Let time, Tâ c X G Y h Z

âT=k c X G Y h Z

Taking dimensions on both sides

[ M 0 L 0 T 1 ]= [L T â1 ] X [ M â1 L 3 T â2 ] Y [M L 2 T â1 ] Z

i.e.,

[ M 0 L 0 T 1 ]=[ M âY+Z L X+3Y+2Z T âXâ2YâZ ]

Equating powers of M, L, and T, on both sides, we get â

âY+Z=0âââââ...(1)

X+3Y+2Z=0ââââââ...(2)

âXâ2YâZ=1âââââââ...(3)

From equation (i)

Adding (2) and (3)

Y+Z=1

or 2Y=1ââââ( âµââZ=Y )

â´Z=Y= 1 2 âââââ

Putting these values in (2), we get â

X+ 3 2 +1=0ââorââX= â5 2

Hence, [T]=[ G 1 2 â h 1 2 â c â5 2 ]

Topic: Physics, Units and Dimensions, NEET
The sporophylls forms a distinct compact structures called ________ .

Options:
A calyptra
B indusium
C stobili cones
D sporangium

The correct answer is:
stobili cones

Solution:
The sporophytes bear sporangia that are subtended by leaf-like appendages called sporophylls. In some cases, sporophylls may form distinct compact structures called strobili or cones, especially in Selaginella, Equisetum. The sporangia produce spores by meiosis in spore mother cells. The spores germinate to give rise to the inconspicuous, small but multicellular prothallus.

Topic: Biology, Plant Kingdom, NEET
The stability of a lyophobic colloid is due to

Options:
A adsorption of covalent molecules on the colloid
B the size of the particles
C the charge on the particles
D Tyndall effect

The correct answer is:
the charge on the particles

Solution:
The stability of a lyophobic colloidal solution is due to the fact that the colloidal particles in the sol are electrically charged. The particles, therefore repel one another and do not coalesce (come close together) to form large non-colloidal particles Colloidal particles carry either positive or negative charge. All the dispersed particles in a colloidal solution carry the same charge while the dispersion medium has an equal and opposite charge.

Topic: Chemistry, Surface Chemistry, KCET, KCET 2013
The standard cell potential for $Z n\left|Z n^{2+} \| C u^{2+}\right| C u$ is $1.10 \mathrm{~V}$. When the cell is completely discharged, $\log \left[\mathrm{Zn}^{2+}\right] /\left[\mathrm{Cu}^{2+}\right]$ is closest to

Options:
A $37.3$
B $0.026$
C $18.7$
D $0.052$

The correct answer is:
$37.3$

Solution:

Topic: Chemistry, Electrochemistry, KVPY, KVPY 2015 (SB/SX)
The standard cell potential of the following cell $\mathrm{Zn}\left|\mathrm{Zn}^{2+}(\mathrm{aq})\right| \mathrm{Fe}^{2+}(\mathrm{aq}) \mid \mathrm{Fe}$ is 0.32 V . Calculate the standard Gibbs energy change for the reaction :
$\mathrm{Zn}(\mathrm{s})+\mathrm{Fe}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{Fe}(\mathrm{s})$
(Given : $1 \mathrm{~F}=96487 \mathrm{C}$ )

Options:
A $-61.75 \mathrm{~kJ} \mathrm{~mol}^{-1}$
B $+5.006 \mathrm{~kJ} \mathrm{~mol}^{-1}$
C $-5.006 \mathrm{~kJ} \mathrm{~mol}^{-1}$
D $+61.75 \mathrm{~kJ} \mathrm{~mol}^{-1}$

The correct answer is:
$-61.75 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Solution:
$\Delta_r G^{\ominus}=-n F E_{\text {cell }}^{\ominus}$
For the given reaction, $n=2$
$\therefore \quad \Delta_{\mathrm{r}} \mathrm{G}^{\ominus}=-2 \times 96487 \times 0.32$
$=-61751.68 \mathrm{~J} \mathrm{~mol}^{-1}$
$=-61.751 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Topic: Chemistry, Electrochemistry, NEET, NEET 2024 (Re-NEET)
The standard deviation and mean of five observations are 0 and 9 respectively. If one of the observations is changed such that the mean of the new set of five observations becomes 10 , then their standard deviation is

Options:
A 1
B 2
C 3
D 0

The correct answer is:
2

Solution:
Let the five observations are $x_1, x_2, x_3, x_4, x_5$.
So, mean $=\frac{x_1+x_2+x_3+x_4+x_5}{5}=9$
(given)
$\Rightarrow \quad x_1+x_2+x_3+x_4+x_5=45$
and standard deviation $=\sum_{i=1}^5 \sqrt{\frac{\left(9-x_i\right)^2}{5}}=0$ (given)
$\Rightarrow \quad \sum_{i=1}^5\left(9-x_i\right)^2=0 \Rightarrow x_i=9 \forall i$
Now, let observation $x_5$ is changed with $y$, so
$$
\begin{array}{lrr}
\text { mean }=\frac{x_1+x_2+x_3+x_4+y}{5}=10 & \text { (given) } \\
\Rightarrow & x_1+x_2+x_3+x_4+y=50 & \\
\Rightarrow & 45-x_5+y=50 & \text { (from Eq. (i)) } \\
\Rightarrow & x_5+5=y \Rightarrow y=14 & \left\{\because x_5=9\right\}
\end{array}
$$
(given)
Now, the changed standard deviation is
$$
\begin{aligned}
& \sqrt{\frac{\left(10-x_1\right)^2+\left(10-x_2\right)^2+\left(10-x_3\right)^2}{\frac{+\left(10-x_4\right)^2+(10-y)^2}{5}}} \\
= & \sqrt{\frac{1+1+1+1+16}{5}}=\sqrt{\frac{20}{5}}=\sqrt{4}=2
\end{aligned}
$$

Topic: Mathematics, Statistics, AP EAMCET, AP EAMCET 2020 (22 Sep Shift 1)
The standard deviation is to the mean deviation taken from the arithmetic mean.

Options:

None

Solution:
greater than or equal

Topic: Mathematics, Statistics