datavorous/entrance-exam-dataset huggingface dataset:
Along the X-axis, three charges $\frac{q}{2},-q$ and $\frac{q}{2}$ are placed at $x=0, x=a$ and $x=2 a$ respectively. The resultant electric potential at $x=a+r$ (if $a< < r)$ is : ( $\varepsilon_0$ is the permittivity of free space)

Options:
A $\frac{q a}{4 \pi \varepsilon_0 r^2}$
B $\frac{q a^2}{4 \pi \varepsilon_0 r^3}$
C $\frac{q\left(\frac{a^2}{4}\right)}{4 \pi \varepsilon_0 r^3}$
D $\frac{q}{4 \pi \varepsilon_0 r}$

The correct answer is:
$\frac{q a^2}{4 \pi \varepsilon_0 r^3}$

Solution:
We have to find the electric potential at point $P$.
$V_P=\left[\frac{1}{4 \pi \varepsilon_0} \frac{q / 2}{(r+a)}-\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r}+\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q / 2}{(r-a)}\right]$
$=\frac{q}{4 \pi \varepsilon_0}\left[\frac{1}{2(r+a)}-\frac{1}{r}+\frac{1}{2(r-a)}\right]$
$=\frac{q}{4 \pi \varepsilon_0}\left[\frac{r(r-a)-2\left(r^2-a^2\right)+(r+a) r}{2 r\left(r^2-a^2\right)}\right]$
$=\frac{q}{4 \pi \varepsilon_0}\left[\frac{r^2-a r-2 r^2+2 a^2+r^2+a r}{2 r\left(r^2-a^2\right)}\right]$
$=\frac{q a^2}{4 \pi \varepsilon_0 r\left(r^2-a^2\right)}$
Since, $r>>a$, so we have $r^2-a^2 \approx r^2$
$\therefore \quad V_P=\frac{q a^2}{4 \pi \varepsilon_0 r^3}$

Topic: Physics, Electrostatics, AP EAMCET, AP EAMCET 2006
Along the X-axis, three charges $\frac{q}{2},-q$ and $\frac{q}{2}$ are placed at $x=0, x=a$ and $x=2 a$ respectively. The resultant electric potential at $x=a+r$ (if $a< < r)$ is : ( $\varepsilon_0$ is the permittivity of free space)

Options:
A $\frac{q a}{4 \pi \varepsilon_0 r^2}$
B $\frac{q a^2}{4 \pi \varepsilon_0 r^3}$
C $\frac{q\left(\frac{a^2}{4}\right)}{4 \pi \varepsilon_0 r^3}$
D $\frac{q}{4 \pi \varepsilon_0 r}$

The correct answer is:
$\frac{q a^2}{4 \pi \varepsilon_0 r^3}$

Solution:
We have to find the electric potential at point $P$.
$V_P=\left[\frac{1}{4 \pi \varepsilon_0} \frac{q / 2}{(r+a)}-\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r}+\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q / 2}{(r-a)}\right]$
$=\frac{q}{4 \pi \varepsilon_0}\left[\frac{1}{2(r+a)}-\frac{1}{r}+\frac{1}{2(r-a)}\right]$
$=\frac{q}{4 \pi \varepsilon_0}\left[\frac{r(r-a)-2\left(r^2-a^2\right)+(r+a) r}{2 r\left(r^2-a^2\right)}\right]$
$=\frac{q}{4 \pi \varepsilon_0}\left[\frac{r^2-a r-2 r^2+2 a^2+r^2+a r}{2 r\left(r^2-a^2\right)}\right]$
$=\frac{q a^2}{4 \pi \varepsilon_0 r\left(r^2-a^2\right)}$
Since, $r>>a$, so we have $r^2-a^2 \approx r^2$
$\therefore \quad V_P=\frac{q a^2}{4 \pi \varepsilon_0 r^3}$

Topic: Physics, Electrostatics, TS EAMCET, TS EAMCET 2006
Alternating current of peak value $\left(\frac{2}{\pi}\right)$ A flows through the primary coil of transformer. The coefficient of mutual inductance between primary and secondary coil is $1 \mathrm{H}$. The peak e.m.f. induced in secondary coil is (Frequency of a.c. $=50 \mathrm{~Hz}$ )

Options:
A $400 \mathrm{~V}$
B $200 \mathrm{~V}$
C $300 \mathrm{~V}$
D $100 \mathrm{~V}$

The correct answer is:
$200 \mathrm{~V}$

Solution:
Given \(: \mathrm{I}_{0}=\frac{2}{\pi}\) ampere \(v=50 \mathrm{~Hz} \mathrm{~L}=1 \mathrm{H}\)
Thus \(w=2 \pi v=2 \pi(50)=100 \pi\)
Alternating current flowing through the coil is given by \(I=I_{0} \sin w t\)
Differentiating it wr.t. time we get \(\frac{\mathrm{dI}}{\mathrm{dt}}=\mathrm{I}_{0} \mathrm{w} \cos w \mathrm{t}\) \(\left.\therefore \frac{\mathrm{dI}}{\mathrm{dt}}\right|_{\max }=\mathrm{I}_{0} \mathrm{~W}=\frac{2}{\pi} \times 100 \pi=200\) ampere per second
Peak e.m.f induced \(\mathcal{E}=\mathrm{L} \frac{\mathrm{dI}}{\mathrm{dt}}\)
\(\Longrightarrow \mathcal{E}=1 \times 200=200 \mathrm{~V}\

Topic: Physics, Alternating Current, MHT CET, MHT CET 2020 (15 Oct Shift 2)
Alternating current of peak value 2π ampere flows through the primary coil of the transformer. The coefficient of mutual inductance between primary and secondary coil is 1 henry. The peak emf induced in secondary coil is (Frequency of AC = 50 Hz)

Options:
A 100 V
B 200 V
C 300 V
D 400 V

The correct answer is:
200 V

Solution:
Peak value of current I0=IrmsÃ2=2Ï Amp
Co-efficient of mutual inductance is M = 1 henry
Induced emf in secondary is given by
e2=Mâdidt where i=i0 .sinâ¡Ït+Ï
Here Ï=2Ïn=100 Ï.
â´ e2=1Ãddt(i0 .sinâ¡ Ït)
i0 .Ï .cosâ¡ Ït=2ÏÃ2ÏÃ50 .cosâ¡ (100 Ït)
For, t=0 , we have e2=4Ã50=200V .

Topic: Physics, Alternating Current, MHT CET, MHT CET 2016
Amino terminal and carboxyl terminal are observed in

Options:
A primary structure of the protein
B secondary structure of DNA
C sugar molecule
D primary structure of RNA

The correct answer is:
primary structure of the protein

Solution:
The sequence of amino acids, i.e., the positional information in a protein – which is the first amino acid, which is second, and so on – is called the primary structure of a protein. A protein is imagined as a line, the left end represented by the first amino acid, and the right end represented by the last amino acid. The first amino acid is also called as N-terminal amino acid. The last amino acid is called the C-terminal amino acid. A protein thread does not exist throughout as an extended rigid rod. The thread is folded in the form of a helix (similar to a revolving staircase).

Topic: Chemistry, Biomolecules, NEET
Ammonolysis of R-X give

Options:
A Only p-amine
B Only sec amine
C Only quaternary ammonium halide
D Mixture of Primary, Secondary, Tertiary and quaternary ammonium halide

The correct answer is:
Mixture of Primary, Secondary, Tertiary and quaternary ammonium halide

Solution:
Ammonolysis of R-X give mixture of Primary, Secondary, Tertiary and quaternary ammonium halide

Topic: Chemistry, Amines, NEET
Among all cyclic quadrilaterals inscribed in a circle of radius $\mathrm{R}$ with one of its angles equal to $120^{\circ}$. Consider the one with maximum possible area. Its area is-

Options:
A $\sqrt{2} \mathbb{R}^{2}$
B $\sqrt{3} \mathrm{R}^{2}$
C $2 \mathbb{R}^{2}$
D $2 \sqrt{3} \mathrm{R}^{2}$

The correct answer is:
$\sqrt{3} \mathrm{R}^{2}$

Solution:
$\mathrm{A}=2 \times \frac{1}{2} \times \sqrt{3} \mathrm{R} \times \mathrm{R}=\sqrt{3} \mathrm{R}^{2}$

Topic: Mathematics, Circle, KVPY, KVPY 2010 (SB/SX)
Among all the isomeric alcohols present of molecular formula C5H12O,what are the number of primary, secondary and tertiary alcohols?

Options:
A 5 Primary,3 Secondary and 0 Tertiary Alcohol
B 4 Primary,2 Secondary and 2 Tertiary Alcohol
C 4 Primary,3 Secondary and 1 Tertiary Alcohol
D 6 Primary,1 Secondary and 1 Tertiary Alcohol

The correct answer is:
4 Primary,3 Secondary and 1 Tertiary Alcohol

Solution:
Correct Option is : (C)
4 Primary,3 Secondary and 1 Tertiary Alcohol

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
Among $\mathrm{Mg}, \mathrm{Cu}, \mathrm{Fe}, \mathrm{Zn}$, the metal that does not produce hydrogen gas in reaction with hydrochloric acid is.

Options:
A $\mathrm{Cu}$
B $\mathrm{Zn}$
C $\mathrm{Mg}$
D $\mathrm{Fe}$

The correct answer is:
$\mathrm{Cu}$

Solution:
Metals having more standard reduction potential than $\mathrm{H}^{+} / \mathrm{H}_{2}(\mathrm{~g})$ can't produce $\mathrm{H}_{2}(\mathrm{~g})$ in acidic medium.

Topic: Chemistry, Electrochemistry, KVPY, KVPY 2015 (SA)
Among primary, secondary, and tertiary halides, which is the least reactive in SN2 reactions?

Options:
A Primary halide
B Secondary halide
C Tertiary halide
D They all have the same reactivity.

The correct answer is:
Tertiary halide

Solution:
Tertiary halides are the least reactive in SN2 reactions because they have bulky groups that hinder the approaching nucleophiles.

Topic: Chemistry, Haloalkanes and Haloarenes, NEET
Among the following elements, $\mathrm{X}$ exhibits maximum catenation and $\mathrm{Y}$ is the least abundant on earth. $\mathrm{X}$ and $\mathrm{Y}$ elements are

Options:
A C, Ge
B Si, Ge
C C, Pb
D Ge, C

The correct answer is:
C, Ge

Solution:
Due to smaller size of carbon it shows maximum catenation and form strong $\mathrm{C}-\mathrm{C}$ bond. $\mathrm{Ge}$ is the least abundant on earth.

Topic: Chemistry, p Block Elements (Group 13 & 14), TS EAMCET, TS EAMCET 2022 (19 Jul Shift 2)
Among the following equations, which are linear?
1.$\quad 2 x+y-z=5$
2.$\pi x+y-e z=\log 3$
3.$\quad 3^{x}+2 y=7$
4.$\quad \sin x-y-5 z=4$
Select the correct answer using the code given below

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

The correct answer is:
1 and 2 only

Solution:
An equation of the form $a x+b y+c z=d$, where $a, b, c$, d are real number, not all zero, is linear.
$\Rightarrow 2 x+y-z=5$
and $\pi x+y-e z=\log 3$ are linear.

Topic: Mathematics, Sets and Relations, NDA, NDA 2007 (Phase 2)
Among the following given substances, the one with zero ΔfH° is

Options:
A Diamond
B Graphite
C Fullerene
D Bituminous coal

The correct answer is:
Graphite

Solution:
The standard enthalpy of formation or standard heat of formation of a compound is the change of enthalpy during the formation of 1 mole of the substance from its constituent elements, with all substances in their standard states.

The reference state of an element is its most stable state of aggregation at 25°C and 1 bar pressure. For example, the reference state of dihydrogen is H2 gas and those of dioxygen, carbon and sulphur are O2 gas, Cgraphite and Srhombic respectively.

graphite is thermodynamically most stable allotrope of carbon and, therefore, the enthalpy of formation of graphite is taken as zero.Enthalpy of formation of diamond and fullerene, C60 are 1.90 and 38.1 kJ mol–1, respectively.

Topic: Chemistry, Thermodynamics (C), TS EAMCET, TS EAMCET 2022 (18 Jul Shift 2)
Among the following set of reactions, the most suitable
method for preparing secondary amine is

Options:
A A
B B
C C
D D

The correct answer is:
C

Solution:

Topic: Chemistry, Amines, TS EAMCET, TS EAMCET 2022 (19 Jul Shift 1)
Among the following, the equation representing a progressive wave is
(A) $y=2 \cos 3 x \sin 10 t$
(B) $\mathrm{y}=2 \sqrt{\mathrm{x}-\mathrm{vt}}$
(C) $y=3 \sin (5 x-0.5 t)+4 \cos (5 x-0.5 t)$
(D) $y=\cos x \sin t+\cos 2 x \cdot \sin 2 t$

Options:
A $A$ and $D$
B $\mathrm{C}$
C $\mathrm{A}, \mathrm{C}, \mathrm{D}$
D B

The correct answer is:
$\mathrm{C}$

Solution:
Only equation C follows standard wave condition

Topic: Physics, Waves and Sound, AP EAMCET, AP EAMCET 2022 (05 Jul Shift 2)
Among the following, the incorrect statement about the formation of female gamete in humans is:

Options:
A A layer of granulosa cells surrounds primary oocytes forming primary follicles.
B Ovum is released in the secondary oocyte stage from the secondary follicle.
C The ploidy level of oogonia and primary oocyte is the same.
D A fluid filled cavity called antrum develops which leads to the formation of tertiary follicle.

The correct answer is:
Ovum is released in the secondary oocyte stage from the secondary follicle.

Solution:
The sequence of oogenesis (formation of ovum) is as follows:
Millions of oogonia are added in a female fetal ovary after birth. These oogonia (2n) undergo division and lead to the formation of primary oocytes (2n). A layer of granulosa cells surrounds these oocytes, which leads to the formation of primary follicles. More and more layer of granulosa cells and a new theca surround these primary follicles which leads to the formation of secondary follicles A fluid filled cavity called antrum develops which leads to the formation of tertiary follicle. An unequal division of the primary oocyte in the tertiary follicle leads to the formation of a haploid secondary oocyte i.e. ovum and a tiny first polar body. The largets follicle formed is Graafian follicle. The ovum is released from the Graafian follicle in the secondary oocyte stage

Topic: Biology, Human Reproduction, NEET
Among the fundamental forces, which one of the following is the strongest force

Options:
A Electromagnetic force
B Strong Nuclear force
C Gravitational force
D Weak nuclear force

The correct answer is:
Strong Nuclear force

Solution:
Among all fundamental forces, "strong nuclear force" is strongest.

Topic: Physics, Laws of Motion, TS EAMCET, TS EAMCET 2022 (20 Jul Shift 2)
Among the heavier members of group 14, which element shows the highest tendency to exhibit a +2 oxidation state?

Options:
A Germanium
B Silicon
C Tin
D Carbon

The correct answer is:
Tin

Solution:
In heavier members, the tendency to show +2 oxidation state increases in the sequence Ge

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Topic: Chemistry, p Block Elements (Group 13 & 14), NEET
Amongst the animal groups given below, which one has the highest percentage of endangered species?

Options:
A
Insects

B
Mammals

C
Amphibians

D
Reptiles

The correct answer is:
Amphibians

Solution:
At present amphibians have the highest percentage, $32 \%$ of all amphibian species in the world face the threat of extinction. Other than these, $23 \%$ of all mammal species and $12 \%$ of all bird species also facing the risk.

Topic: Biology, Biodiversity and its Conservation
Amongst the animal groups given below, which one has the highest percentage of endangered species?

Options:
A Insects
B Mammals
C Amphibians
D Reptiles

The correct answer is:
Amphibians

Solution:
Presently, 12% all the birds species, 23% all mammals species, 31% all gymnosperms species and 32% all amphibian species in world face the threat of extinctions.

Topic: Biology, Biodiversity and its Conservation, NEET
Amongst the animal groups given below, which one has the highest percentage of endangered species?

Options:
A Insects
B Mammals
C Amphibians
D Reptiles

The correct answer is:
Amphibians

Solution:
As per the IUCN RED list -
32% of amphibian species, 12% birds species, 23% mammal species, 31% of gymnosperm species are considered to be facing threat of extinctions and have highest endangered species percentage. The threats including habitat loss, fragmentation and degradation, pollution, climate change, and alien species invasion.

Topic: Biology, Biodiversity and its Conservation, NEET
An AC current is given by the expression, It=50sin200πt in amperes. The frequency and RMS value of the current respectively are

Options:
A 100 Hz,502 A
B 100 Hz,252 A
C 200 Hz,502 A
D 200 Hz,252 A

The correct answer is:
100 Hz,252 A

Solution:
Standard equation of the current variation in an AC supply can be written as,

I=I0sinωt, where I0=peak current and ω=angular frequency.

Comparing it with the given expression of current, we get

I0=50 A.

Therefore, RMS current

IRMS=I02=252 A

Similarly, ω=200π

⇒2πf=200π⇒f=100 Hz

Topic: Physics, Alternating Current, TS EAMCET, TS EAMCET 2022 (18 Jul Shift 1)
An aeroplane flying at a height of $300 \mathrm{~m}$ above the ground passes vertically above another plane at an instant when the angles of elevation of two planes from the same point on the ground are $60^{\circ}$ and $45^{\circ}$ respectively. What is the height of the lower plane from the ground? $\quad$

Options:
A $50 \mathrm{~m}$
B $\frac{100}{\sqrt{3}} \mathrm{~m}$
C $100 \sqrt{3} \mathrm{~m}$
D $150(\sqrt{3}+1) \mathrm{m}$

The correct answer is:
$100 \sqrt{3} \mathrm{~m}$

Solution:
Let $\mathrm{P}$ and $\mathrm{Q}$ be the positions of two aeroplanes when $\mathrm{Q}$ is vertically below $\mathrm{P}$ and $\mathrm{OP}=300 \mathrm{~m}$ Let the angles of elevation of $\mathrm{P}$ and $\mathrm{Q}$ at a point $\mathrm{A}$ on the ground be $60^{\circ}$ and $45^{\circ}$ respectively. - In $\Delta \mathrm{AOQ}$
$\tan 45^{\circ}=\frac{\mathrm{OQ}}{\mathrm{OA}} \Rightarrow \mathrm{OA}=\mathrm{OQ}$
In $\Delta \mathrm{AOP}$
$\tan 60^{\circ}=\frac{\mathrm{OP}}{\mathrm{OA}}=\frac{300}{\mathrm{OA}}=\sqrt{3}$
$\Rightarrow \mathrm{OA}=\frac{300}{\sqrt{3}}=100 \sqrt{3}$
Hence, $\mathrm{OQ}=100 \sqrt{3} \mathrm{~m}$

Topic: Mathematics, Heights and Distances, NDA, NDA 2011 (Phase 2)
An air column in a pipe, which is closed at one end will be in resonance with a vibrating tuning fork of frequency $264 \mathrm{~Hz}$ for various lengths. Which one of the following lengths is not possible? $(\mathrm{V}=330 \mathrm{~m} / \mathrm{s})$

Options:
A $62.50 \mathrm{~cm}$
B $93.75 \mathrm{~cm}$
C $156.25 \mathrm{~cm}$
D $31.25 \mathrm{~cm}$

The correct answer is:
$62.50 \mathrm{~cm}$

Solution:
$$
\mathrm{n}=264 \mathrm{~Hz}, \mathrm{~V}=300 \mathrm{~m} / \mathrm{s}
$$
For fundamental frequency, $\mathrm{n}=\frac{\mathrm{V}}{4 \ell}$
$$
\ell=\frac{\mathrm{V}}{4 \mathrm{n}}=\frac{330}{4 \times 264}=0.3125 \mathrm{n}=31.25 \mathrm{~cm}
$$
For fundamental mode, $\ell=\frac{\lambda}{4}$ other possible lengths are $\frac{3 \lambda}{4}, \frac{5 \lambda}{4}, \ldots \ldots$ Hence, $62.50 \mathrm{~m}$ is not possible.

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2021 (20 Sep Shift 1)
An air column in a tube $32 \mathrm{~cm}$ long, closed at one end, is in resonance with a tuning fork. The air column in another tube, open at both ends, of length $66 \mathrm{~cm}$ is in resonance with
another tuning fork. When these two tuning forks are sounded together, they produce 8 beats per second. Then the frequencies of the two tuning forks are, (Consider fundamental frequencies only)

Options:
A $250 \mathrm{~Hz}, 258 \mathrm{~Hz}$
B $240 \mathrm{~Hz}, 248 \mathrm{~Hz}$
C $264 \mathrm{~Hz}, 256 \mathrm{~Hz}$
D $280 \mathrm{~Hz}, 272 \mathrm{~Hz}$

The correct answer is:
$264 \mathrm{~Hz}, 256 \mathrm{~Hz}$

Solution:
We knows frequency of a closed end an column
$$
n_1=\frac{v}{4 / 1}
$$
We knows frequency of a open end an column
$$
n_2=\frac{v}{2 l_2}
$$
Given, $l_1=32 \mathrm{~cm}, I_2=66 \mathrm{~cm}$
and $\quad n_1-n_2=8$ heat $/ \mathrm{s}$
So, $\quad n_1=\frac{v}{4 \times 32}=\frac{v}{128}$
and $\quad n_2=\frac{v}{2 \times 66}=\frac{v}{132}$
In given condition,
$$
\begin{aligned}
& \frac{v}{128}-\frac{v}{132}=8 \\
& v=8448 \times 4 \\
& v=33792 \\
& \text { Hence, } \quad n_1=\frac{33792}{128} \\
& n_1=264 \mathrm{~Hz} \\
& \text { and } \\
& n_2=\frac{33792}{132} \\
& n_2=256 \mathrm{~Hz} \\
&
\end{aligned}
$$

Topic: Physics, Waves and Sound, AP EAMCET, AP EAMCET 2013
An air column in a tube $32 \mathrm{~cm}$ long, closed at one end, is in resonance with a tuning fork. The air column in another tube, open at both ends, of length $66 \mathrm{~cm}$ is in resonance with
another tuning fork. When these two tuning forks are sounded together, they produce 8 beats per second. Then the frequencies of the two tuning forks are, (Consider fundamental frequencies only)

Options:
A $250 \mathrm{~Hz}, 258 \mathrm{~Hz}$
B $240 \mathrm{~Hz}, 248 \mathrm{~Hz}$
C $264 \mathrm{~Hz}, 256 \mathrm{~Hz}$
D $280 \mathrm{~Hz}, 272 \mathrm{~Hz}$

The correct answer is:
$264 \mathrm{~Hz}, 256 \mathrm{~Hz}$

Solution:
We knows frequency of a closed end an column
$$
n_1=\frac{v}{4 / 1}
$$
We knows frequency of a open end an column
$$
n_2=\frac{v}{2 l_2}
$$
Given, $l_1=32 \mathrm{~cm}, I_2=66 \mathrm{~cm}$
and $\quad n_1-n_2=8$ heat $/ \mathrm{s}$
So, $\quad n_1=\frac{v}{4 \times 32}=\frac{v}{128}$
and $\quad n_2=\frac{v}{2 \times 66}=\frac{v}{132}$
In given condition,
$$
\begin{aligned}
& \frac{v}{128}-\frac{v}{132}=8 \\
& v=8448 \times 4 \\
& v=33792 \\
& \text { Hence, } \quad n_1=\frac{33792}{128} \\
& n_1=264 \mathrm{~Hz} \\
& \text { and } \\
& n_2=\frac{33792}{132} \\
& n_2=256 \mathrm{~Hz} \\
&
\end{aligned}
$$

Topic: Physics, Waves and Sound, TS EAMCET, TS EAMCET 2013
An aircraft has three engines $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$. The aircraft crashes if all the three engines fail. The probabilities of failure are $0.03,0.02$ and $0.05$ for engines $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ respectively. What is the probability that the aircraft will not crash?

Options:
A $0.00003$
B $0.90$
C $0.99997$
D $0.90307$

The correct answer is:
$0.99997$

Solution:
Since, probabilities of failure for engines A, B and C $\mathrm{P}(\mathrm{A}), \mathrm{P}(\mathrm{B})$ and $\mathrm{P}(\mathrm{C})$ are $0.03,0.02$ and $0.05$ respectively.
The aircraft will crash only when all the three engine fail. So, probability that it crashes $=\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B}) \cdot \mathrm{P}(\mathrm{C})$
$=0.03 \times 0.02 \times 0.05$
$=0.00003$
Hence, the probability that the aircraft will not crash, $=1-0.00003$
$=0.99997$

Topic: Mathematics, Probability, NDA, NDA 2006 (Phase 2)
An alkene $A\left(\mathrm{C}_4 \mathrm{H}_8\right)$ exhibits cis/trans isomerism. $A$ on ozonolysis gives $B$, which when reacted with $\mathrm{NaOH}$ followed by hydroxylamine gave $C$. What are $B$ and $C$ ?

Options:
A $\begin{array}{cc}\mathbf{B} & \mathbf{C} \\ \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CHO} & \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}==\mathrm{NOH}\end{array}$
B $\begin{array}{cc}\mathbf{B} & \mathbf{C} \\ \mathrm{CH}_3 \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CHO} & \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}=\mathrm{CHCH} \mathrm{CH}_2=\mathrm{NOH}\end{array}$
C $\begin{array}{cc}\mathbf{B} & \mathbf{C} \\ \mathrm{CH}_3 \mathrm{CHO} & \mathrm{CH}_3 \mathrm{CH}=\mathrm{CHCH}=\mathrm{NNH}_2\end{array}$
D $\begin{array}{cc}\mathbf{B} & \mathbf{C} \\ \mathrm{CH}_3 \mathrm{CHO} & \mathrm{CH}_3 \mathrm{CH}=\mathrm{CHCH}=\mathrm{NOH}\end{array}$

The correct answer is:
$\begin{array}{cc}\mathbf{B} & \mathbf{C} \\ \mathrm{CH}_3 \mathrm{CHO} & \mathrm{CH}_3 \mathrm{CH}=\mathrm{CHCH}=\mathrm{NOH}\end{array}$

Solution:
Here, $(B)$ is $\mathrm{CH}_3 \mathrm{CHO}$
and $(C)$ is $\mathrm{CH}_3 \mathrm{CH}==\mathrm{CHCH}=\mathrm{NOH}$.
(i) On ozonolysis double bond get broken to give $\mathrm{CH}_3 \mathrm{CHO}$.
(ii) On reaction with $\mathrm{NaOH}$ followed by hydroxyl amine, we get
$\mathrm{CH}_3-\mathrm{CH}=\mathrm{CHCH}=\mathrm{NOH}$

Topic: Chemistry, Hydrocarbons, TS EAMCET, TS EAMCET 2020 (11 Sep Shift 2)
An alkene (A) on ozonolysis gives a mixture of two carbonyl compounds. Mixture on Clemmensen reduction gives just one alkane (B) . (B) is the lowest alkane which in pure form can not be prepared by standard Wurtz method. (A) is

Options:
A MeCH=CHMe
B MeCH2CH=CMe2
C MeCH2CH2CH2CH=CEt2
D MeCH2CH2CH=C(Me)Et

The correct answer is:
MeCH2CH=CMe2

Solution:
Simplest alkane which cannot be prepared by wurtz reaction is propane. So the alkene must be unsymmetrical having six carbon, which can give two different carbonyl each contain three carbons.

Topic: Chemistry, Hydrocarbons, NEET
An alkyl halide reacts with alcoholic ammonia in a sealed tube, the product formed will be

Options:
A a primary amine
B a secondary amine
C a tertiary amine
D a mixture of all the three

The correct answer is:
a mixture of all the three

Solution:
When an alkyl halide reacts with alcoholic ammonia in a sealed tube then a mixture of primary, secondary and tertiary amine is formed
$R \mathrm{X}+\mathrm{NH}_{3} \longrightarrow R \mathrm{NH}_{2}+\mathrm{H} X$
pri-amine
$\mathrm{RNH}_{2}+X R \longrightarrow R_{2} \mathrm{NH}+\mathrm{H} X$
$R_{2} \mathrm{NH}+X R \longrightarrow R_{3} \mathrm{~N}+\mathrm{H} X$

Topic: Chemistry, Amines, KCET, KCET 2007
An alkyl halide reacts with alcoholic ammonia in a sealed tube, the product formed will be

Options:
A primary amine
B a secondary amine
C a tertiary amine
D a mixture of all the three

The correct answer is:
a mixture of all the three

Solution:
When an alkyl halide reacts with alcoholic ammonia in at sealed tube, the product formed will be mixture of primaty, secondary and tertiary amine.

Topic: Chemistry, Amines, COMEDK, COMEDK 2020
An amine $\mathrm{C}_{3} \mathrm{H}_{9} \mathrm{N}$ reacts with benzene sulphonyl chloride to form a white precipitate which is insoluble in $\mathrm{aq} . \mathrm{NaOH}$. The amine is

Options:
A
B
C
D

The correct answer is:

Solution:
Since, the amine gives white precipitate with benzene sulphonyl chloride which is insoluble in aq. $\mathrm{Na} \mathrm{OH},$ so Ti must be a secondary amine lie. amine having RNHR group).
Among the given compound, only
is a secondary amine, so it represents the structure of given amine.

Topic: Chemistry, Amines, WBJEE, WBJEE 2014