datavorous/entrance-exam-dataset huggingface dataset:
Let $X$ denote the sum of the numbers obtained when two fair dice are rolled. The variance and standard deviation of $X$ are

Options:
A $\frac{31}{6}$ and $\sqrt{\frac{31}{6}}$
B $\frac{35}{6}$ and $\sqrt{\frac{35}{6}}$
C $\frac{17}{6}$ and $\sqrt{\frac{17}{6}}$
D $\frac{31}{6}$ and $\sqrt{\frac{35}{6}}$

The correct answer is:
$\frac{35}{6}$ and $\sqrt{\frac{35}{6}}$

Solution:
Let $\mathbf{x}$ denote the sum of the numbers obtained when two fair dice are rolled. So, X may have values $2,3,4,5,6,7,8,9$, 10,11 and 12 .
$\begin{array}{l}
\mathrm{P}(\mathrm{X}=2)=\mathrm{P}(1,1)=\frac{1}{36} \\
\mathrm{P}(\mathrm{X}=3)=\mathrm{P}\{(1,2),(2,1)\}=\frac{2}{36} \\
\mathrm{P}(\mathrm{X}=4)=\frac{3}{36}, \mathrm{P}(\mathrm{X}=5)=\frac{4}{36}: \\
\mathrm{P}(\mathrm{X}=6)=\frac{5}{36} ; \mathrm{P}(\mathrm{X}=7)=\frac{6}{36} ; \\
\mathrm{P}(\mathrm{X}=9)=\frac{4}{36} ; \mathrm{P}(\mathrm{X}=10)=\frac{3}{36} ; \mathrm{P}(\mathrm{X}=11) \\
=\frac{2}{36} ; \\
\mathrm{P}(\mathrm{X}=12)=\frac{1}{36}
\end{array}$
$\therefore \quad$ Probability distribution table is given below
\begin{array}{c|c|c|c|c|c|c|c|c|c|c|c}
\hline \mathbf{X} & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\
\hline \mathbf{P}(\mathbf{X}) & \frac{1}{36} & \frac{2}{36} & \frac{3}{36} & \frac{4}{36} & \frac{5}{36} & \frac{6}{36} & \frac{5}{36} & \frac{4}{36} & \frac{3}{36} & \frac{2}{36} & \frac{1}{36} \\
\hline
\end{array}
Mean $\bar{X}=\sum X P(X)$
$=\frac{\left[\begin{array}{l}2 \times 1+3 \times 2+4 \times 3+5 \times 4+6 \times 5+7 \times 6+7 \\ 8 \times 5+9 \times 4+10 \times 3+11 \times 2+12 \times 1\end{array}\right]}{36}$
$=\frac{252}{36}=7$
Variance $=\sum X^{2} P(X)-\bar{X}^{2}$
$\begin{aligned}
&\left[\begin{array}{l}
2^{2} \times 1+3^{2} \times 2+4^{2} \times 3+5^{2} \times 4 \\
+6^{2} \times 5+7^{2} \times 6+8^{2} \times 5+9^{2} \times \\
4+10^{2} \times 3+11^{2} \times 2+12^{2} \times 1
\end{array}\right] \\
=& \frac{36}{1974}-7^{2} \\
=& \frac{36}{1974-1764} \\
=& \frac{36}{36}=\frac{35}{6} \\
\therefore & \text { Variance }=\frac{35}{6}
\end{aligned}$
And, $S D=\sqrt{\frac{35}{6}}$

Topic: Mathematics, Probability, VITEEE, VITEEE 2014
Let $x>y$ be two real numbers and $z \in R, z \neq 0$. Consider the following:
1.$\quad x+z>y+z$ and $x z>y z$
$2 \quad x+z>y-z$ and $x-z>y-z$
3.$\mathrm{xz}>$ yzand $\frac{\mathrm{x}}{\mathrm{z}}>\frac{\mathrm{y}}{\mathrm{z}}$
4.$x-z>y-z$ and $\frac{x}{z}>\frac{y}{z}$
Which of the above is/are correct?

Options:
A 1 only
B 2 only
C 1 and 2 only
D $1,2,3$ and 4

The correct answer is:
$1,2,3$ and 4

Solution:
All statements are correct.

Topic: Mathematics, Sets and Relations, NDA, NDA 2006 (Phase 1)
Let $x, y, z$ be positive real numbers such that $x, y, z$ are in GP and $\tan ^{-1} x, \tan ^{-1} y$ and $\tan ^{-1} z$ are in AP. Then which one of the following is correct?

Options:
A $x=y=z$
B $\mathrm{xz}=1$
C $x \neq y$ and $y=z$
D $x=y$ and $y \neq z$

The correct answer is:
$x=y=z$

Solution:
$2 \tan ^{-1} \mathrm{y}=\tan ^{-1} \mathrm{x}+\tan ^{-1} \mathrm{z}$
$\left(\therefore \tan ^{-1} \mathrm{x}, \tan ^{-1} \mathrm{y}, \tan ^{-1} \mathrm{z}\right.$ are in A.P)
and $\mathrm{x}, \mathrm{y}, \mathrm{z}$ are in $\mathrm{G} \mathrm{P} \quad \therefore \quad \mathrm{y}^{2}=\mathrm{xz}$ ...(i)
$\Rightarrow \frac{2 y}{1-y^{2}}=\frac{x+z}{1-x z}$
$\Rightarrow 2 \mathrm{y}=\mathrm{x}+\mathrm{z}$
$\mathrm{x}, \mathrm{y}, \mathrm{z}$ are in A.P.
Given $\mathrm{x}, \mathrm{y}, \mathrm{z}$ are also in G.P
So, $x=y=z$

Topic: Mathematics, Sequences and Series, NDA, NDA 2017 (Phase 1)
Let $y(x)=a x^{n}$ and $\delta y$ denote small change in $y$. What is limit
of $\frac{\delta y}{\delta x}$ as $\delta x \rightarrow 0 ?$

Options:
A 0
B 1
C $\operatorname{anx}^{n-1}$
D $a x^{n} \log (a x)$

The correct answer is:
0

Solution:
We know $\lim _{\delta x \rightarrow 0} \frac{\delta y}{\delta x}=\left(\frac{d y}{d x}\right)_{\text {at } x=0}$
$\quad=\frac{d}{d x}\left(a x^{n}\right)_{\text {at } x=0}=\left(\operatorname{an} x^{n-1}\right)_{\mathrm{at} x=0}=0$

Topic: Mathematics, Limits, NDA, NDA 2009 (Phase 1)
Let $\mathrm{z}_{1}$ and $\mathrm{z}_{2}$ be two non-zero complex numbers such that
$\left|z_{1}\right|=\left|z_{2}\right|=\left|\frac{1}{z_{1}}+\frac{1}{z_{2}}\right|=2$
What is the value of $\left|z_{1}+z_{2}\right|$ ?

Options:
A 8
B 4
C 2
D 1

The correct answer is:
8

Solution:
As given $\left|z_{1}\right|=\left|z_{2}\right|=\left|\frac{1}{z_{1}}+\frac{1}{z_{2}}\right|=2$
$\left|\frac{1}{z_{1}}+\frac{1}{z_{2}}\right|=2 \Rightarrow \frac{z_{2}+z_{1}}{z_{1} z_{2}}=2$
$\Rightarrow \frac{\left|z_{1}+z_{2}\right|}{\left|z_{1}\right|\left|z_{2}\right|}=2$
$\Rightarrow\left|z_{1}+z_{2}\right|=2\left|z_{1} \| z_{2}\right|$
$\Rightarrow\left|z_{1}+z_{2}\right|=2.2 .2$
$\Rightarrow\left|z_{1}+z_{2}\right|=8$

Topic: Mathematics, Complex Number, NDA, NDA 2006 (Phase 2)
Let $\mathrm{z}_{1}, \mathrm{z}_{2}$ and $\mathrm{z}_{3}$ be non-zero complex numbers satisfying $\mathrm{z}^{2}=\mathrm{i} \overline{\mathrm{z}}$, where $\mathrm{i}=\sqrt{-1}$.
Consider the following statements:
1- $\quad \mathrm{z}_{1} \mathrm{z}_{2} \mathrm{z}_{3}$ is purely imaginary.
2- $\quad z_{1} z_{2}+z_{2} z_{3}+z_{3} z_{1}$ is purelyreal. Which of the above statements is/are correct?

Options:
A 1 only
B 2 only
C Both 1 and 2
D Neither 1 nor 2

The correct answer is:
Both 1 and 2

Solution:
$z_{1} z_{2} z_{3}=(-i)\left(\frac{\sqrt{3}+i}{2}\right)\left(\frac{-\sqrt{3}+i}{2}\right)$
$=\frac{-i}{4}\left(i^{2}-(\sqrt{3})^{2}\right)$
$=\frac{-i}{4}(-3-1)=i$
Hence $z_{1} z_{2} z_{3}$ is purely imaginary. $z_{1} z_{2}=-i\left(\frac{\sqrt{3}+i}{2}\right)=\frac{-\sqrt{3} i+1}{2}$
$\begin{aligned} z_{2} z_{3} &=\frac{(\sqrt{3}+i)(-\sqrt{3}+i)}{4} \\ &=\left(\frac{-3-\sqrt{3} i+\sqrt{3} i+i^{2}}{4}\right)=-1 \end{aligned}$
$z_{3} z_{1}=\frac{(-\sqrt{3}+i)(-i)}{2}=\frac{+\sqrt{3} i+1}{2}$
$z_{1} z_{2}+z_{2} z_{3}+z_{3} z_{1}=\left(\frac{-\sqrt{3} i+1}{2}\right)+(-1)+\left(\frac{\sqrt{3} i+1}{2}\right)$
$=\left(\frac{-\sqrt{3} i+1+\sqrt{3} i+1}{2}\right)-1$
$=0 \in R$
Hence $z_{1} z_{2}+z_{2} z_{3}+z_{3} z_{1}=0$ is purelyreal. Hence both statements are correct.

Topic: Mathematics, Complex Number, NDA, NDA 2016 (Phase 1)
Let $\mathrm{z}_{1}, \mathrm{z}_{2}$ and $\mathrm{z}_{3}$ be non-zero complex numbers satisfying $\mathrm{z}^{2}=\mathrm{i} \overline{\mathrm{z}}$, where $\mathrm{i}=\sqrt{-1}$.
What is $\mathrm{z}_{1}+\mathrm{z}_{2}+\mathrm{z}_{3}$ equal to?

Options:
A 1
B $-\mathrm{i}$
C 0
D 1

The correct answer is:
0

Solution:
Given $z^{2}=i \bar{z}$
Let us suppose that $z=x+i y$
$\Rightarrow(x+i y)^{2}=i(x-i y)$
$\Rightarrow x^{2}-y^{2}+2 x y i=i x+y$
Comparing real and imaginary part of both sides
$x^{2}-y^{2}=y$ and $2 x y=x$.
Taking $2 x y=x$ $(2 y-1) x=0$
$\Rightarrow x=0$ and $y=\frac{1}{2}$
if $x=0$ $\therefore y+y^{2}=0 \quad\left(\because x^{2}-y^{2}=y\right)$
$y(y+1)=0$
$y=0$ and $-1$ $x=0 \quad y=0$ and $-1$
If $y=\frac{1}{2}$
$x^{2}-\frac{1}{4}=\frac{1}{2}$
$\Rightarrow x^{2}=\frac{3}{4}$
$\Rightarrow x=\pm \frac{\sqrt{3}}{2}$
Since given numbers are non zero complex numbers.
So, $z_{1}=0+(-1) i=-i$
$z_{2}=\frac{\sqrt{3}}{2}+\frac{1}{2} i \quad z_{3}=\frac{-\sqrt{3}}{2}+\frac{1}{2} i$
$z_{1}+z_{2}+z_{3}=(-i)+\left(\frac{\sqrt{3}}{2}+\frac{1}{2} i\right)+\left(-\frac{\sqrt{3}}{2}+\frac{1}{2} i\right)=0$
Hence $z_{1}+z_{2}+z_{3}=0$

Topic: Mathematics, Complex Number, NDA, NDA 2016 (Phase 1)
Let $z$ be a complex number satisfying $\left|\frac{z-4}{z-8}\right|=1$ and $\left|\frac{z}{z-2}\right|=\frac{3}{2}$
What is $\left|\frac{z-6}{z+6}\right|$ equal to?

Options:
A 3
B 2
C 1
D 0

The correct answer is:
0

Solution:
$\left|\frac{z-6}{z+6}\right|=\left|\frac{6-6}{6+6}\right|=0$

Topic: Mathematics, Complex Number, NDA, NDA 2016 (Phase 1)
Let $z$ be a complex number satisfying$\left|\frac{z-4}{z-8}\right|=1$ and $\left|\frac{z}{z-2}\right|=\frac{3}{2}$
What is $|z|$ equal to?

Options:
A 6
B 12
C 18
D 36

The correct answer is:
6

Solution:
$\left|\frac{z-4}{z-8}\right|=1$ and $\left|\frac{z}{z-2}\right|=\frac{3}{2}$
$\Rightarrow|z-4|=|z-8|$
Let $z=x+i y$ $|x+i y-4|=|x+i y-8|$
Squaring both sides, we get
$\left[(x-4)^{2}+y^{2}\right]=\left[(x-8)^{2}+y^{2}\right]$
$(x-4)^{2}=(x-8)^{2}$
$\Rightarrow x^{2}+16-8 x=x^{2}+64-16 x$
$\Rightarrow 8 x=48 \Rightarrow x=6$
when $\left|\frac{z}{z-2}\right|=\frac{3}{2}$
$\Rightarrow 2|z|=3|z-2|$
Squaring both sides, we get $4\left(x^{2}+y^{2}\right)=9\left[(x-2)^{2}+y^{2}\right]$
$\Rightarrow 4 x^{2}+4 y^{2}=9 x^{2}+36-36 x+9 y^{2}$
$\Rightarrow 5 x^{2}+5 y^{2}-36 x+36=0$
as we know $x=6$
$5(6)^{2}+5 y^{2}-36 \times 6+36=0$
$\Rightarrow 5 y^{2}=0 \Rightarrow y=0$
Hence $x=6$ and $y=0$. $\Rightarrow z=6$
$|z|=6$

Topic: Mathematics, Complex Number, NDA, NDA 2016 (Phase 1)
Let $z$ be a complex number such that $|z|=4$ and arg $z=\frac{5 \pi}{6}$.
Where $i=\sqrt{-1}$. What is $z$ equal to ?

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

The correct answer is:
$-2 \sqrt{3}+2 i$

Solution:
$\begin{aligned} \text { Let } z &=\mathrm{K}(\cos \theta+i \sin \theta) \\ &|\mathrm{z}|=\sqrt{\mathrm{K}^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)}=4 \\ & \therefore \mathrm{K}=4 \\ & \text { Again } \operatorname{Arg}(\mathrm{z})=\frac{5 \pi}{6} \\ \text { So, } \theta=\frac{5 \pi}{6} \\ & \text { Now, } \mathrm{Z}=4\left(\cos \left(\frac{5 \pi}{6}\right)+\mathrm{i} \sin \left(\frac{5 \pi}{6}\right)\right) \\ &=4\left(\frac{-\sqrt{3}}{2}+\mathrm{i} \frac{(1)}{2}\right)=-2 \sqrt{3}+2 \mathrm{i} \end{aligned}$

Topic: Mathematics, Complex Number, NDA, NDA 2014 (Phase 2)
Let $z$ be a non-zero complex number. Then, what is $z^{-1}$ (multiplicative inverse of z) equal to?

Options:
A $\frac{\overline{\mathrm{z}}}{|\mathrm{z}|^{2}}$
B $\frac{z}{|z|^{2}}$
C $\frac{\overline{\mathrm{z}}}{|\mathrm{z}|}$
D $\frac{|z|}{z}$

The correct answer is:
$\frac{\overline{\mathrm{z}}}{|\mathrm{z}|^{2}}$

Solution:
Let $z$ be a non-zero complex number, such that $\mathrm{z}=\mathrm{x}+$ iy where $\mathrm{x}, \mathrm{y} \in \mathrm{R}$
Then $\mathrm{z}^{-1}=\frac{1}{\mathrm{x}+\mathrm{iy}}$
So, $z^{-1}=\frac{x-i y}{(x+i y)(x-i y)}=\frac{x-i y}{x^{2}+y^{2}}=\frac{\bar{z}}{|z|^{2}}$

Topic: Mathematics, Complex Number, NDA, NDA 2006 (Phase 2)
Let $Z$ be the set of integers and $a R b$, where $a, b \in Z$ if anc only if $(a-b)$ is divisible by $5 .$ Consider the following statements:
1.The relation $R$ partitions $Z$ into five equivalentclasses
2.Any twoequivalent classes are either equal or disjoint Which of the above statements is/are correct?

Options:
A 1 only
B 2 only
C Both 1 and 2
D Neither 1 nor 2

The correct answer is:
Both 1 and 2

Solution:
No Solution

Topic: Mathematics, Sets and Relations, NDA, NDA 2015 (Phase 1)
Let $z=i^{3}(1+i)$ be a complex number. What is its argument?

Options:
A $\pi$
B $\frac{\pi}{4}$
C $-\frac{\pi}{4}$
D $\frac{5 \pi}{4}$

The correct answer is:
$-\frac{\pi}{4}$

Solution:
The complex number is $z=i^{3}(1+i)=-i(1+i)$ $\begin{aligned} &=-i-\mathrm{i}^{2} \\ \Rightarrow \mathrm{z} &=1-\mathrm{i} \end{aligned}$
This gives, $\arg (z)=\tan ^{-1}\left(\frac{-1}{1}\right)=\tan ^{-1}(-1)=-\frac{\pi}{4}$

Topic: Mathematics, Complex Number, NDA, NDA 2006 (Phase 2)
Let $z=x+i y$ Where $x, y$ are real variables $i=\sqrt{-1}$. If
$|2 z-1|=|z-2|$, then the point $z$ describes:

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

The correct answer is:
A circle

Solution:
$|2 z-1|=|z-2|$
$|2(\mathrm{x}+\mathrm{iy})-1|=|\mathrm{x}+\mathrm{iy}-2|$
$|(2 x-1)+2 y i|=|(x-2)+i y|$
$\sqrt{(2 \mathrm{x}-1)^{2}+\mathrm{y}^{2}}=\sqrt{(\mathrm{x}-2)^{2}+\mathrm{y}^{2}}$
Squaring both sides $4 x^{2}+1-4 x+4 y^{2}=x^{2}+4-4 x+y^{2}$
$\Rightarrow 3 x^{2}+3 y^{2}=3$
$\Rightarrow x^{2}+y^{2}=1$
It is the equation of a circle. $\therefore$ The point $z$ describes a circle.

Topic: Mathematics, Complex Number, NDA, NDA 2014 (Phase 1)
Li occupies higher position in the electrochemical series of metals as compared to Cu since

Options:
A the standard reduction potential of $\mathrm{Li}^{+} / \mathrm{Li}$ is lower than that of $\mathrm{Cu}^{2+} / \mathrm{Cu}$
B the standard reduction potential of $\mathrm{Cu}^{2+} / \mathrm{Cu}$ is lower than that of $\mathrm{Li}^{+} / \mathrm{Li}$
C the standard oxidation potential of $\operatorname{Li} / \mathrm{Li}^{+}$ is lower than that of $\mathrm{Cu} / \mathrm{Cu}^{2+}$
D $\mathrm{L} \mathrm{i}$ is smaller in size as compared to $\mathrm{Cu}$

The correct answer is:
the standard reduction potential of $\mathrm{Li}^{+} / \mathrm{Li}$ is lower than that of $\mathrm{Cu}^{2+} / \mathrm{Cu}$

Solution:
In the electrochemical series, metals are arranged in increasing order of their standard reduction potential. The standard reduction potentials of $L^{+} / L$ and $C u^{2+} / C u$ are -3.05 V and $+.034 \mathrm{V}$ respectively. Therefore, Li occupies higher position in series as compared to Cu.

Topic: Chemistry, Electrochemistry, WBJEE, WBJEE 2012
$\lim _{x \rightarrow 0} e^{-1 / x}$ is equal to

Options:
A 0
B $\infty$
C e
D does not exist

The correct answer is:
does not exist

Solution:
$\begin{aligned} \mathrm{LHL} &=\lim _{h \rightarrow 0^{-}} e^{\frac{-1}{(0-h)}}=\lim _{h \rightarrow 0} \mathrm{e}^{\frac{1}{h}} \\=\mathrm{e}^{\infty}=\infty \end{aligned}$

Topic: Mathematics, Limits, NDA, NDA 2008 (Phase 2)
$\lim _{x \rightarrow 0} \frac{1-\cos ^{3} 4 x}{x^{2}}$ is equal to

Options:
A 0
B 12
C 24
D 36

The correct answer is:
24

Solution:
$\lim _{x \rightarrow 0} \frac{1-\cos ^{3} 4 x}{x^{2}}$
The given limit is in $\frac{0}{0}$ form. So, apply L'hospital rule.
$\lim _{x \rightarrow 0} \frac{-3 \cos ^{2} 4 x(-\sin 4 x) 4}{2 x}$
$=\lim _{x \rightarrow 0} \frac{12 \cos ^{2} 4 x \sin 4 x}{(2 x) 2} \times 2$
$=24 \cdot \lim _{x \rightarrow 0} \cos ^{2} 4 x \cdot \frac{\sin 4 x}{4 x}$
$=24(1)=24$

Topic: Mathematics, Limits, NDA, NDA 2019 (Phase 1)
$\lim _{x \rightarrow 0} \frac{\sin x-\tan x}{x}$ equal to?

Options:
A 0
B 1
C -1
D 1/2

The correct answer is:
0

Solution:
Consider $\lim _{x \rightarrow 0} \frac{\sin x-\tan x}{x}$
$=\lim _{x \rightarrow 0} \frac{\cos x-\sec ^{2} x}{1}\left(\right.$ By $L^{\prime}$ Hospital Rule $)$
$=\frac{\cos 0-\sec ^{2} 0}{1}=\frac{1-1}{1}=0$

Topic: Mathematics, Limits, NDA, NDA 2013 (Phase 1)
Lipids are insoluble in water because lipids molecules are:

Options:
A hydrophilic
B hydrophobic
C neutral
D zwitter ions

The correct answer is:
hydrophobic

Solution:
Lipid have long nonpolar chains of carbon atoms which adversely affects its solubility in water. Thus, they are insoluble in water because they are hydrophobic.
Related Theory
Allofthelipidmoleculesincellmembranesare amphipathic that is, they have a hydrophilic or polar end and a hydrophobic or nonpolar end. The most abundant membrane lipids are the phospholipids. These have a polar head group and two hydrophobic hydrocarbon tails.

Topic: Biology, Biomolecules (B), NEET, NEET 2002
List of endangered species was released by

Options:
A WWF
B FOAM
C IUCN
D GEAC

The correct answer is:
IUCN

Solution:
List of endangered species was released by - IUCN.

Topic: Biology, Biodiversity and its Conservation, NEET, NEET 2024
List the harmful effects caused by alcohol/drug abuse.

Options:

None

Solution:
Harmful effects caused by alcohol abuse are-
(i) Alcohol generates more energy mostly in the form of heat, but at the same time, it dilates the blood vessels. Consequently the heat generated is rapidly lost. Due to constant dilation, the arterial walls soon become brittle $\&$ rigid. Such a change in the property of blood vessels $\&$ deposition of alcoholic fat affect the working of heart.
(ii) Alcoholism leads to gastric ulcers \& gastritis.
(iii) In chronic alcoholism, the axon of the nerve inflame thus causing neuritis.
(iv) Permanent damage to liver cells occur due to deposition of fats. The liver dries up \& harden (cirrhosis).
Harmful effects caused by drug abuse are-
(i) Excessive doses of drugs may lead to coma \& death due to respiratory failure, heart failure \& cerebral haemorrhage.
(ii) Lack of interest in personal hygiene, withdrawal, isolation, depression, fatigue aggressive \& rebellious behaviour etc.
(iii) Acquire serious infections like AIDS \& hepatitis B, who take drugs intravenously.
(iv) The adverse effects of drugs are manifested in the form of reckless behaviour, vandalism \& violence.

Topic: Biology, Human Health and Diseases
lodoform reaction is answered by all, except

Options:
A \( \mathrm{CH}_{3} \mathrm{CHO} \)
B (1) \( \mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2} \mathrm{OH} \)
C
D \( \mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{OH} \)

The correct answer is:
(1) \( \mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2} \mathrm{OH} \)

Solution:
For alcohols secondary and tertiary alcohols gives iodoform reaction and ethanol is the only primary alcohol which gives
iodoform reaction, therefore, propanol will not give iodoform reaction.
(1) \( \underset{\text { Acetaldehyde }}{\mathrm{C}_{3} \mathrm{CH} \mathrm{O}}+3 \mathrm{NaOI} \rightarrow \underset{\text { Yellow ppt }}{\mathrm{CH}}+ \) HCOON \( a+2 \mathrm{NaOH} \)
(2) \( \begin{array}{c}\text { Acetaldehyde } \\ \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\end{array}+\mathrm{NaOI}_{2}-\mathrm{NaOH}_{\mathrm{S}_{2}}> \) do not give iodoform reaction.
(3) \( \mathrm{CH}_{3}-\stackrel{\mathrm{OH}}{\mathrm{CH}}-\mathrm{CH}_{2} \mathrm{COOH} \) NaOI \( \underset{\text { Yellowppt. }}{\mathrm{CHI}_{3}}+\mathrm{NaOCCH}_{2} \mathrm{COOH} \)
\( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}-\mathrm{NaOH}_{\mathrm{I}_{2}} \rightarrow \mathrm{CH}_{3} \mathrm{CHO}+\mathrm{NaI}+\mathrm{H}_{2} \mathrm{O} \)
\( \mathrm{CH}_{3} \mathrm{CHO}+3 \mathrm{NaOI} \rightarrow \mathrm{CHI}_{3}+\mathrm{HCOON} a+2 \mathrm{NaOH} \)

Topic: Chemistry, Chemical Bonding and Molecular Structure, KCET, KCET 2014
Looking from the top of a $20 \mathrm{~m}$ high building, the angle of elevation of the top of a tower is $60^{\circ}$ and the angle of depression of its bottom is $30^{\circ}$. What is the height of the tower? $\quad$

Options:
A $50 \mathrm{~m}$
B $60 \mathrm{~m}$
C $70 \mathrm{~m}$
D $80 \mathrm{~m}$

The correct answer is:
$80 \mathrm{~m}$

Solution:
$\operatorname{In} \Delta A B D, \tan 30^{\circ}=\frac{A D}{A B}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{A D}{A B}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{20}{A B}$
$\Rightarrow A B=20 \sqrt{3} \mathrm{~m} .$
$\operatorname{In} \Delta D C F, \tan 60^{\circ}=\frac{h}{D C} \Rightarrow \sqrt{3}=\frac{b}{D C}$
$\Rightarrow h=(\sqrt{3})(20 \sqrt{3})$
$(\because A B=D C=20 \sqrt{3})$
$\Rightarrow h=60 \mathrm{~m}$
$\therefore$ Height of tower, $B F=60+20=80 \mathrm{~m}$

Topic: Mathematics, Heights and Distances, NDA, NDA 2009 (Phase 2)
Lucas reagent is

Options:
A anhy. \(\mathrm{ZnCl}_2\) and \(\mathrm{NH}_3\)
B anhy. \(\mathrm{ZnCl}_2\) and \(\mathrm{CaCl}_2\)
C anhy. \(\mathrm{ZnCl}_2\) and conc. HCl
D anhy. \(\mathrm{ZnCl}_2\) and HCl gas

The correct answer is:
anhy. \(\mathrm{ZnCl}_2\) and conc. HCl

Solution:
Anhy. \(\mathrm{ZnCl}_2\) and conc. HCl is called Lucas reagent. It is used to distinguish the primary, secondary and tertiary alcohol.

Topic: Chemistry, Alcohols Phenols and Ethers, JIPMER, JIPMER 2008
Lucas test is associated with

Options:
A aldehydes
B phenols
C carboxylic acids
D alcohols

The correct answer is:
alcohols

Solution:
Lucas test is used to distinguish primary, secondary and tertiary alcohols.

Topic: Chemistry, Alcohols Phenols and Ethers, KCET, KCET 2009
Lucas test is used for the determination of

Options:
A alcohols
B alkyl halides
C phenols
D aldehydes.

The correct answer is:
alcohols

Solution:
Lucas test-Alcohol reacts with concentrated hydrochloric acid in presence of anhydrous $\mathrm{ZnCl}_2$ to form alkyl halides. The three type of alcohols undergo this reaction at different rates. Order of rate of reaction is : tertiary $>$ secondary $>$ primary.
$$
R-\mathrm{OH}+\mathrm{HCl} \underset{\mathrm{ZnCl}_2}{\longrightarrow} R-\mathrm{Cl}+\mathrm{H}_2 \mathrm{O}
$$
Alkyl chloride

Topic: Chemistry, Alcohols Phenols and Ethers, AIIMS, AIIMS 2002
Lucas test is used for the determination of

Options:
A aldehydes
B phenols
C carboxylic acid
D alcohols

The correct answer is:
alcohols

Solution:
Lucas test is used to distinguish alcohols.
Lucas reagent is a solution of anhydrous zinc chloride in concentrated $\mathrm{HCl}$.
Lucas reagent converts alcohols to alkyl chlorides. Tertiary alcohols give immediate reaction with Lucas reagent, secondary alcohols give reaction within five minutes, while primary alcohol do not give reaction with it.

Topic: Chemistry, Alcohols Phenols and Ethers, AP EAMCET, AP EAMCET 2021 (23 Aug Shift 1)
$\mathrm{M}$ is a matrix with real entries given by
$\mathrm{M}=\left[\begin{array}{lll}
4 & \mathrm{k} & 0 \\
6 & 3 & 0 \\
2 & \mathrm{t} & \mathrm{k}
\end{array}\right]$
Which of the following conditions guarantee the invertibility of $\mathrm{M}$ ?
$1. k \neq 2$
$2. \mathrm{k} \neq 0$
$3. \mathrm{t} \neq 0$
$4. \mathrm{t} \neq 1$
Select the correct answer using the code given below:

Options:
A 1 and 2
B 2 and 3
C 1 and 4
D 3 and 4

The correct answer is:
1 and 2

Solution:
As given $\mathrm{M}=\left[\begin{array}{lll}4 & \mathrm{k} & 0 \\ 6 & 3 & 0 \\ 2 & \mathrm{t} & \mathrm{k}\end{array}\right]$
$\mathrm{M}$ will be invertible, if
$\left[\begin{array}{lll}4 & \mathrm{k} & 0 \\ 6 & 3 & 0 \\ 2 & \mathrm{t} & \mathrm{k}\end{array}\right] \neq 0$
$\Rightarrow \mathrm{k} \neq 0$ or $\mathrm{k}(12-6 \mathrm{k}) \neq 0$
$\Rightarrow \mathrm{k} \neq 0, \mathrm{k} \neq 2$
Thus, statement (1) and (2) are correct.

Topic: Mathematics, Matrices, NDA, NDA 2006 (Phase 1)
Magnesium reacts with an element X to form an ionic compound. If the ground state electronic configuration of X is 1s2 , 2s2, 2p3, the simplest formula for this compound is

Options:
A Mg2X
B MgX2
C Mg2X3
D Mg3X2

The correct answer is:
Mg3X2

Solution:
Magnesium ion =Mg+2

The electron configuration of an element describes how electrons are distributed in its atomic orbitals. Electron configurations of atoms follow a standard notation in which all electron-containing atomic subshells (with the number of electrons they hold written in superscript) are placed in a sequence.

Topic: Chemistry, s Block Elements, NEET, NEET 2018
Major product of the given reaction is

Options:
A

B

C

D

The correct answer is:

Solution:
In Williamson's synthesis, due to steric hindrance secondary halides prefer to undergo elimination reaction and, as per Saytzeff's rule a more stable alkene is formed.

Topic: Chemistry, Alcohols Phenols and Ethers, AIIMS, AIIMS 2018 (27 May)
Malarial parasite is

Options:
A Polygenetic
B Digenetic
C Monogenetic
D Monomorphic

The correct answer is:
Digenetic

Solution:
Malarial parasite is digenetic in nature because its life cycle gets completed in two different hosts. Primary or principal host is man and secondary or intermediate or vector host is female Anopheles mosquito.

Topic: Biology, Human Health and Diseases, NEET
Male gametophyte in angiosperms produces :

Options:
A Single sperm and a vegetative cell
B Single sperm and two vegetative cells
C Three sperms
D Two sperms and a vegetative cell

The correct answer is:
Two sperms and a vegetative cell

Solution:
In Angiosperms, pollen grain is the first male gametophyte.

Microspore divides mitotically forming two cells.
1. Vegetative cell: Bigger in size having abundant food reserve and large irregularly shaped nucleus.
2. Generative cell: Small and floats in the cytoplasm of the vegetative cell. It is spindle-shaped with dense cytoplasm and nucleus.
Shedding of pollen grains takes place at this two-celled stage in 60 % of angiosperms. In rest, the generative cell divides to form the male gametes (sperms) and pollen is shed at the three-celled stage.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET, NEET 2015 (Phase 2)