datavorous/entrance-exam-dataset huggingface dataset:
The tissue whose cells acts like a diffusion boundary is:

Options:
A Simple Cuboidal epithelium
B Simple Columnar epithelium
C Simple Squamous epithelium
D All of the above

The correct answer is:
Simple Squamous epithelium

Solution:
The squamous epithelium is made up of a single thin layer of flattened cells with irregular boundaries. They are found in the walls of blood vessels and air sacs of lungs and are involved in functions like forming diffusion boundary for efficient diffusion process.

Topic: Biology, Structural Organisation in Animals, NEET
The top of a hill observed from the top and bottom of a
building of height $\mathrm{h}$ is at angles of elevation $\frac{\pi}{6}$ and $\frac{\pi}{3}$ respectively. What is the height of the hill?

Options:
A $2 \mathrm{~h}$
B $\frac{3 \mathrm{~h}}{2}$
C $\mathrm{h}$
D $\frac{\mathrm{h}}{2}$

The correct answer is:
$\frac{3 \mathrm{~h}}{2}$

Solution:
From question, In $\Delta$ PQRand $\Delta$ PST.
$$
\begin{aligned}
& \mathrm{B}=\frac{\mathrm{H}}{\tan \frac{\pi}{3}}=\frac{\mathrm{H}-h}{\tan \frac{\pi}{6}} \\
\Rightarrow & \frac{\mathrm{H}}{3}=\mathrm{H}-h \\
\Rightarrow & \frac{2}{3} \mathrm{H}=h \\
\Rightarrow & \mathrm{H}=\frac{3}{2} h
\end{aligned}
$$

Topic: Mathematics, Heights and Distances, NDA, NDA 2018 (Phase 2)
The top of a hill observed from the top and bottom of a building of height $\mathrm{h}$ is at angles of elevation $\alpha$ and $\beta$ respectively. The height of the hill is: $\quad$

Options:
A $\frac{h \cot \beta}{\cot \beta-\cot \alpha}$
B $\frac{\mathrm{h} \cot \alpha}{\cot \alpha-\cot \beta}$
C $\frac{\mathrm{h} \tan \alpha}{\tan \alpha-\tan \beta}$
D None of the above

The correct answer is:
$\frac{\mathrm{h} \cot \alpha}{\cot \alpha-\cot \beta}$

Solution:
$\operatorname{In} \Delta \mathrm{BDC}$,
$\tan \beta=\frac{\mathrm{CD}}{\mathrm{BD}}=\frac{\mathrm{H}}{\mathrm{BD}}$
$\mathrm{BD}=\frac{\mathrm{H}}{\tan \beta}=\mathrm{H} \cot \beta$
In $\triangle$ ALC, $\tan \alpha=\frac{\mathrm{CL}}{\mathrm{AL}}=\frac{\mathrm{H}-\mathrm{h}}{\mathrm{BD}}$
$\mathrm{BD}=(\mathrm{H}-\mathrm{h}) \cot \alpha$
from (i) \& (ii), $(\mathrm{H}-\mathrm{h}) \cot \alpha=\mathrm{H} \cot \beta$
$\mathrm{H} \cot \alpha-\mathrm{h} \cot \alpha=\mathrm{H} \cot \beta$
$\mathrm{H}[\cot \alpha-\cot \beta]=\mathrm{h} \cot \alpha$
$\mathrm{H}=\frac{\mathrm{h} \cot \alpha}{\cot \alpha-\cot \beta}$

Topic: Mathematics, Heights and Distances, NDA, NDA 2012 (Phase 2)
The top of a hill when observed from the top and bottom of a building of height $\mathrm{h}$ is at angles of elevation $\mathrm{p}$ and $\mathrm{q}$ respectively. What is the height of the hill?

Options:
A $\frac{\text { hcot } q}{\cot q-\cot p}$
B $\frac{\text { hcot } p}{\cot p-\cot q}$
C $\frac{2 \text { htanp }}{\tan p-\tan q}$
D $\frac{2 \text { htan } q}{\tan q-\tan p}$

The correct answer is:
$\frac{\text { hcot } p}{\cot p-\cot q}$

Solution:
Let height of hill $=\mathrm{H}$ \& horizontal distance between building \& hill = d
$\tan \mathrm{q}=\frac{\mathrm{H}}{\mathrm{d}} \Rightarrow \mathrm{d}=\frac{\mathrm{H}}{\tan \mathrm{q}}=\mathrm{H} \cot \mathrm{q}$
$\tan \mathrm{p}=\frac{(\mathrm{H}-\mathrm{h})}{\mathrm{d}} \Rightarrow \mathrm{d}=(\mathrm{H}-\mathrm{h}) \cot \mathrm{p}$
$\Rightarrow \mathrm{H} \cot \mathrm{q}=(\mathrm{H}-\mathrm{h}) \cot \mathrm{p}$
$H=\frac{h \cot p}{\cot p-\cot q}$

Topic: Mathematics, Heights and Distances, NDA, NDA 2016 (Phase 2)
The total dry weight accessible to heterotrophs in an ecosystem will form the:

Options:
A Gross primary productivity
B Net primary productivity
C Secondary productivity
D Tertiary productivity

The correct answer is:
Net primary productivity

Solution:
Gross primary productivity of an ecosystem is the rate of production of organic matter during photosynthesis. A considerable amount of GPP is utilised by plants in respiration. Gross primary productivity minus respiration losses (R), is the net primary productivity (NPP). It is the available biomass for the consumption to heterotrophs (herbivores and decomposers)

Topic: Biology, Ecosystem, NEET
The total number of subsets of a finite set $A$ has 56 more elements than the total number of subsets of another finite set $B$. What is the number of elements in the set A ?

Options:
A 5
B 6
C 7
D 8

The correct answer is:
6

Solution:
Let sets $A$ and $B$ have $m$ and $n$ elements respectively. The set made by subsets of a finite sets $A$ and $B_{\underline{1}}$ s known as power set i.e. $\mathrm{P}(\mathrm{A})$ and $\mathrm{P}(\mathrm{B})$ We know, if set $\mathrm{A}$ has $n$ elements then $\mathrm{P}(\mathrm{A})$ has $2^{n}$ elements. Thus, The total no. of subsets of a finite set $\mathrm{A}=2^{m}$
and set $B=2^{n}$
So, According to the question $2^{m}-2^{n}=56$
$\Rightarrow 2^{n}\left(2^{m-n}-1\right)=8 \times 7=2^{3} \times\left(2^{3}-1\right)$
On comparing the powers, both side $n=3$ and $m-n=3$
$\Rightarrow m=6$ and $n=3$

Topic: Mathematics, Sets and Relations, NDA, NDA 2009 (Phase 1)
The transparent media A and B are separated by a plane boundary. The speed of light in the medium A is 2Ã108 m s-1 and in the medium B is 2.5Ã108 m s-1. The critical angle for which a ray of light going from A to B is totally internally reflected is

Options:
A sin-1⁡12
B sin-1⁡12
C sin-1⁡45
D 90

The correct answer is:
sin-1⁡45

Solution:
Refractive index of the medium A relative B is given by
BÎ¼A=Î¼AÎ¼B=VBVA=2.52=54
Since, critical angle for the light ray going from medium A to B is given by
C=sin-1â¡1 BÎ¼A=sin-1â¡45

Topic: Physics, Ray Optics, NEET
The transverse displacement of a string of a linear density $0.01 \mathrm{~kg} \mathrm{~m}^{-1}$, clamped at its ends is given by $Y_{(x, t)}=0.03 \sin \left(\frac{2 \pi x}{3}\right) \cos (60 \pi t)$, where $x$ and $y$ are in metres and time $t$ is in seconds. Tension in the string is

Options:
A 9 N
B 36 N
C 162 N
D 81 N

The correct answer is:
81 N

Solution:
Transverse displacement
$$
y_{(x, t)}=0.03 \sin \left(\frac{2 \pi x}{3}\right) \cos 60 \pi t \text {. }
$$
This is equation of standing wave.
The standard equation of standing wave is $y=a \sin k x \cos \omega t$
Here, $\omega=60 \pi, k=2 \pi / 3$
So, $V=\frac{60 \pi}{2 \pi} \times 3=90$
So, $\sqrt{\frac{T}{\mu}}=90$
$$
\Rightarrow \quad T=90^2 \mu \Rightarrow T=90^2 \times 10=81 \mathrm{~N}
$$

Topic: Physics, Waves and Sound, AP EAMCET, AP EAMCET 2017 (26 Apr Shift 1)
The transverse section of a plant shows following anatomical features:

(a) Large number of scattered vascular bundles surrounded by bundle sheath
(b) Large conspicuous parenchymatous ground tissue
(c) Vascular bundles conjoint and closed
(d) Phloem parenchyma absent

Identify the category of plant and its part:

Options:
A Monocotyledonous root
B Dicotyledonous stem
C Dicotyledonous root
D Monocotyledonous stem

The correct answer is:
Monocotyledonous stem

Solution:
MONOCOT STEM:

a. Epidermis is uniseriate with compactly arranged barrel shaped cells covered with thick cuticle trichomes are absent.
b. Hypodermis: It is made up of a few layers of compactly arranged sclerenchyma cells. It provides mechanical support.
c. Ground tissue is the major component and is formed by an undifferentiated mass of loosely arranged parenchyma cells enclosing intercellular spaces. It is not differentiated into cortex or pith.
d. Many vascular bundles are found scattered in the ground tissue and this condition is called attactostele. From periphery to center there is a gradual decrease in the number of vascular bundles and increase in size. The vascular bundles are conjoint, collateral and closed with endarch xylem. Water filled lysigenous cavity present in association with the protoxylem. It is formed by the breakdown of inner protoxylem during earlier stages of growth. Phloem parenchyma is absent. The protophloem forms a thin layer above the functional metaphloem present towards the metaxylem. Each vascular bundle has a prominent bundle sheath made up of sclerenchyma fibres.

Topic: Biology, Anatomy of Flowering Plants, NEET, NEET 2020 (Phase 1)
The trophic level of Cuscuta, Orobanche, and Albugo is

Options:
A Decomposer
B Primary producer
C Primary consumer
D Secondary consumer

The correct answer is:
Primary consumer

Solution:
A parasitic plant is a plant that derives some or all of its nutritional requirement from another living plant. They make up about 1% of angiosperms and are in almost every biome in the world. All parasitic plants have modified roots, named haustoria (singular: haustorium), which penetrates the host plants, connecting them to the conductive system – either the xylem, the phloem, or both. This provides them with the ability to extract water and nutrients from the host. All these plants are parasitic.

Topic: Biology, Ecosystem, NEET
The T-wave of a standard ECG represents:

Options:
A Repolarisation of both ventricles
B Returning of ventricles from excited to normal state
C Ends of T-wave marks the end of systole
D All of the above

The correct answer is:
All of the above

Solution:
In an standard ECG, the T-wave represents the repolarisation of both ventricles means repolarisation of the ventricles or returning of ventricles from excited to normal state. The end of the T-wave marks the end of systole.

Topic: Biology, Body Fluids and Circulation, NEET
The two circles $x^{2}+y^{2}=r^{2}$ and $x^{2}+y^{2}-10 x+16=0$ intersect at two distinct points. Then which one of the following is correct?

Options:
A $\quad 2 < \mathrm{r} < 8$
B $\mathrm{r}=2$ or $\mathrm{r}=8$
C $\quad \mathrm{r} < 2$
D $\mathrm{r}>2$

The correct answer is:
$\quad 2 < \mathrm{r} < 8$

Solution:
For the circle, $\mathrm{x}^{2}+\mathrm{y}^{2}=\mathrm{r}^{2}$, centre $(0,0)$ radius $=\mathrm{r}$ If the circle $x^{2}+y^{2}-10 x+16=0$ is compared with general form $x^{2}+y^{2}+2 g+2 f y+c=0$, we get
$2 \mathrm{~g}=-10,2 \mathrm{f}=0, \mathrm{c}=16$
$\Rightarrow \mathrm{g}=-5, \mathrm{f}=0$
$\therefore$ centre $=(+5,0)$
radius $=\sqrt{g^{2}+f^{2}-c}=\sqrt{25+0-16}=\sqrt{9}=3$
Given, two circles intersect at two distinct points,
To interset at two points, $\mathrm{r}$ should be greater than 2 So, $2 < \mathrm{r} < 8$

Topic: Mathematics, Circle, NDA, NDA 2017 (Phase 1)
The two electrons have the following set of quantum numbers :
\[
\begin{array}{l}
P=3,2,-2,+\frac{1}{2} \\
Q=3,0,0,+\frac{1}{2}
\end{array}
\]
Which of the following statement is true?

Options:
A \( P \) and \( Q \) have same energy
B \( P \) has greater energy than \( Q \)
C \( P \) has lesser energy than \( Q \)
D \( \mathrm{P} \) and \( \mathrm{Q} \) represent same electron

The correct answer is:
\( P \) has greater energy than \( Q \)

Solution:
For electron $P$, the values of quantum numbers are $n=3, I=2, m l=-2$ and $s=+1 / 2$. For electron $Q$ quantum numbers are $n=3, I=0, m l=0$ and $s=+1 / 2$. Therefore, from the quantum number values, electron $P$ belongs to $3 d-$ orbital and electron $Q$ to $3 s$ - orbital. Energy of the orbital is $s < p < d < f$.
Hence, energy of electron $P$ is higher than electron $Q$.

Topic: Chemistry, Structure of Atom, KCET, KCET 2015
The two enantiomers of secondary butyl chloride differ from each other in which one of the following properties?

Options:
A Boiling point
B Specific rotation
C Density
D $\mathrm{Cl}-\mathrm{Cl}$ bond length

The correct answer is:
Specific rotation

Solution:
Enantiomers have same physical properties but differ in specific rotation. Thus, the two enantiomers of secondary butyl chloride differ in specific rotation.

Topic: Chemistry, General Organic Chemistry, AP EAMCET, AP EAMCET 2010
The two enantiomers of secondary butyl chloride differ from each other in which one of the following properties?

Options:
A Boiling point
B Specific rotation
C Density
D $\mathrm{Cl}-\mathrm{Cl}$ bond length

The correct answer is:
Specific rotation

Solution:
Enantiomers have same physical properties but differ in specific rotation. Thus, the two enantiomers of secondary butyl chloride differ in specific rotation.

Topic: Chemistry, General Organic Chemistry, TS EAMCET, TS EAMCET 2010
$\begin{array}{lllll}\text {The two } & \text { half-cell } & \text { reactions } & \text { of } & \text { an }\end{array}$ electrochemical cell is given as $\mathrm{Ag}^{+}+e^{-} \longrightarrow \mathrm{Ag} ; \quad E_{\mathrm{Ax}^{+} / \mathrm{Ag}}^{\circ}=-0.3995 \mathrm{V}$ $\mathrm{Fe}^{3+} \longrightarrow \mathrm{Fe}^{3 *}+e^{-} ; E_{\mathrm{Fe}^{3+} \mathrm{Fv}^{2+}}=-0.7120 \mathrm{V}$. The value of cell EMF will be

Options:
A $-0.3125 \mathrm{V}$
B $0.3125 \mathrm{V}$
C $1.114 \mathrm{V}$
D $-1.114 \mathrm{V}$

The correct answer is:
$0.3125 \mathrm{V}$

Solution:
Species with more negative $E^{\circ}$ (standard reduction potential) generally acts as reducing agent while with less negative $E^{\circ}$ acts as oxidising agent. Thus, the overall reaction is
$\begin{aligned} \mathrm{Ag}^{+}+\mathrm{Fe}^{2+} & \longrightarrow \mathrm{Fe}^{3+}+\mathrm{Ag} \\ \Delta E^{\circ} &=E_{\mathrm{OA}}^{\circ}-E_{\mathrm{RA}}^{\circ} \\ &=-0.3995-(-0.7120) \mathrm{V} \\ &=-0.3995+0.7120 \mathrm{V} \\ &=+0.3125 \mathrm{V} \end{aligned}$

Topic: Chemistry, Electrochemistry, WBJEE, WBJEE 2014
The two lines of regression are $8 x-10 y=66$ and $40 x-18 y=214$ and variance of $x$ series is 9 . What is the standard deviation of y series? $\quad$

Options:
A 3
B 4
C 6
D 8

The correct answer is:
4

Solution:
Let us consider lines $\begin{array}{lll} & 8 x-10 y=66 & \text { and } & 40 x-18 y=214 \\ 10 y=8 x-66 & \Rightarrow & 40 x=18 y+214\end{array}$
$\Rightarrow \quad b_{y x}=\frac{8}{10}=\frac{4}{5} \quad \Rightarrow \quad b_{x y}=\frac{18}{40}=\frac{9}{20}$
Thus, $r=\pm \sqrt{\frac{4}{5} \times \frac{9}{20}}=\frac{3}{5}$
Also, $\sigma_{x}=\sqrt{9}=3$
$\therefore \sigma_{y}=\frac{b y x \times \sigma_{x}}{r}=\frac{\frac{4}{5} \times 3}{r}=\frac{12}{5 r}=\frac{12}{5} \times \frac{5}{3}=4$

Topic: Mathematics, Statistics, NDA, NDA 2010 (Phase 2)
The two nearest harmonics of a tube closed at one end and open at other end are 220 Hz and 260 Hz. What is the fundamental frequency of the system?

Options:
A 10 Hz
B 20 Hz
C 30 Hz
D 40 Hz

The correct answer is:
20 Hz

Solution:
Two successive frequencies of closed pipe
nv4l=220 ....(i)
n+2v4l=260 ......(ii)
Dividing (ii) by (i), we get
n+2n=260220=1311
11n+22=13n
n=11
So, 11v4l=220
v4l=20
So fundamental frequency is 20 Hz.

Topic: Physics, Waves and Sound, NEET, NEET 2017
The two parabolas $y^{2}=4 a x$ and $x^{2}=4$ ay intersect

Options:
A at two points on the line $y=x$
B only at the origin
C at three points one of which lies on $y+x=0$
D only at $(4 a, 4 a)$

The correct answer is:
at two points on the line $y=x$

Solution:
The parabolas $\mathrm{y}^{2}=4 \mathrm{ax}$ and $\mathrm{x}^{2}=4$ ay They intersect at $(0,0)$ and $(4 a, 4 a)$ These points lie on $\mathrm{y}=\mathrm{x}$

Topic: Mathematics, Parabola, NDA, NDA 2019 (Phase 1)
The two planes $a x+b y+c z+d=0$ and $a x+b y+c z+d_{1}$ $=0$, where $d \neq d_{1}$, have

Options:
A one point only in common
B three points in common
C infinite points in common
D no points in common

The correct answer is:
no points in common

Solution:
Two planes $a x+b y+c z+d=0$ and $a x+b y+c z+d_{1}$ $=0$ are parallel to each other.
They have no common point.

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2010 (Phase 2)
The type of epithelial cells which line the inner surface of Fallopian tubes, bronchioles and small bronchi are known as

Options:
A squamous epithelium
B columnar epithelium
C ciliated epithelium
D cuboidal epithelium.

The correct answer is:
ciliated epithelium

Solution:
Ciliated epithelium is a region of epithelium consisting of columnar or cuboidal cells bearing hairlike appendages that are capable of beating rapidly. Ciliated epithelium performs the functions of moving particles of fluids over the epithelial surface. Cuboidal ciliated epithelium occurs in certain parts of uriniferous tubules whereas columnar ciliated epithelium occurs in fallopian tubules, nasal passages, bronchioles, small bronchi and buccopharyngeal cavity of frog. It often occurs in the vicinity of mucus secreting goblet cells.

Topic: Biology, Structural Organisation in Animals, AIIMS, AIIMS 2009
The type of epithelial cells which line the inner surface of Fallopian tubes, bronchioles and small bronchi are known as

Options:
A squamous epithelium
B columnar epithelium
C ciliated epithelium
D cubical epithelium.

The correct answer is:
ciliated epithelium

Solution:
Ciliated epithelium is a region of epithelium consisting of columnar or cuboidal cells bearing hairlike appendages called cilia, that are capable of beating rapidly. Ciliated epithelium performs the functions of moving particles of fluids over the epithelial surface. Cubical ciliated epithelium occurs in certain parts of uriniferous tubules whereas columnar ciliated epithelium occurs in fallopian tubules, nasal passages, bronchioles, small bronchi and buccopharyngeal cavity of frog. It often occurs in the vicinity of mucus secreting goblet cells.
Squamous epithelium are found in linning of alveoli in the lungs and Bowman's capsule in the kidney. Columnar epithelium are found in the lining layer of gastrointestinal tract, gall bladder, uterus etc. and cubical or cuboidal epithelium lines the salivary ducts, pancreated ducts, gastric glands, thyroid follicles, germinal epithelium of ovary, proximal part of uriniferous tubules etc.

Topic: Biology, Structural Organisation in Animals, AIIMS, AIIMS 2006
The type of placentation in which ovary is syncarpous, unilocular and ovules on sutures is called

Options:
A Apical placentation
B Parietal placentation
C Marginal placentation
D Superficial placentation

The correct answer is:
Parietal placentation

Solution:
Screw pine (Pandanus) bear stilt roots. Trapa bears assimilatory roots. Orchid bear clinging roots.

Topic: Biology, Morphology of Flowering Plants, NEET
The typical body plan of a chordate is given in the following diagram. Given labellings A and B, respectively, are:

Options:
A Nerve cord and notochord
B Nerve cord and myotomes
C Notochord and vertebral column
D Notochord and nerve cord

The correct answer is:
Nerve cord and notochord

Solution:
Members of Chordata possess 3 fundamental characters at any stage of life:
1. Notochord - Solid, mesodermal and later on develop into vertebral column.
2. Dorsal hollow nerve cord (or dorsal tubular nerve cord or neural tube)
3. Pharyngeal gill slits.
Post anal tail may or may not be present.

Topic: Biology, Animal Kingdom, NEET
The unit of energy if fundamental units are taken as force F, acceleration A and time T is

Options:
A FAT2
B FA2T
C FA2T2
D [FAT]

The correct answer is:
FAT2

Solution:
Energy = Work = Force × Displacement× Time2Time2
=F1A1T2

Topic: Physics, Units and Dimensions, NEET
"The use of bio-resources by multinational companies and other organisation, without proper authorisation from the countries and people concerned, without compensatory payment" will be termed as

Options:
A biopatent.
B bioethics.
C bioweapon.
D biopiracy.

The correct answer is:
biopiracy.

Solution:
Bioethics may be viewed as a set of standards that may be used to regulate our activities in relation to the biological world. Patents which are being granted for biological entities and for products derived from them are known as biopatents.

Topic: Biology, Biotechnology and Its Applications, NEET
The use of bio-resources by multinational companies and other organisations without proper authorisation from the countries and people concerned, is known as:

Options:
A Biopatent
B Biopiracy
C Biowar
D Biodiversity

The correct answer is:
Biopiracy

Solution:
Biopiracy is exploitation of bioresources of any company. This is done by other companies for commercial use with or without patent rights. Bioresouces mean organisms that can provide benefits commercially. These are abundant in developing countries but less in developed countries. But the developed countries have an advantage as they have good technologies. Developing countries have resources and also traditional methods and these are used by developed countries and with the help of their rich technology these countries get patent rights easily without proper laws.

Topic: Biology, Biotechnology and Its Applications, NEET
The value of $\int_{0}^{\frac{\pi}{4}} \sqrt{\tan x} d x+\int_{0}^{\frac{\pi}{4}} \sqrt{\cot x} d x$ is equal to

Options:
A $\frac{\pi}{4}$
B $\frac{\pi}{2}$
C $\frac{\pi}{2 \sqrt{2}}$
D $\frac{\pi}{\sqrt{2}}$

The correct answer is:
$\frac{\pi}{\sqrt{2}}$

Solution:
$\int_{0}^{\frac{\pi}{4}} \sqrt{\tan x} d x+\int_{0}^{\frac{\pi}{4}} \sqrt{\cot x} \cdot \mathrm{dx}$
$=\int_{0}^{\frac{\pi}{4}}(\sqrt{\tan x}+\sqrt{\cot x}) \mathrm{dx}$
$=\int_{0}^{\frac{\pi}{4}}\left(\frac{\sqrt{\sin x}}{\sqrt{\cos x}}+\frac{\sqrt{\cos x}}{\sqrt{\sin x}}\right) \mathrm{dx}$
$=\sqrt{2} \int_{0}^{\frac{\pi}{4}} \frac{(\sin x+\cos x)}{\left(\sqrt{1-\left(\sin ^{2} x+\cos ^{2} x-2 \sin x \cos x\right)}\right)} \cdot d x$
$\quad=\int_{0}^{4} \frac{\sin x+\cos x}{\sqrt{\sin x} \cdot \sqrt{\cos x}} \cdot \mathrm{dx}$
$\quad$ Put $\sin x-\cos x=t \Rightarrow(\cos x+\sin x) d x=d t$
When $x=0, \mathrm{t}=-1$ and $\mathrm{x}=\frac{\pi}{4}, \mathrm{t}=0 .$
$=\sqrt{2} \int_{-1}^{0} \frac{1}{\sqrt{1-\mathrm{t}^{2}}} \cdot \mathrm{dt}=\sqrt{2}\left(\sin ^{-1} \mathrm{t}\right)_{-1}^{0}$
$=\sqrt{2}\left[\sin ^{-1}(0)-\sin ^{-1}(-1)\right]$
$=\sqrt{2}\left[0-\left(\frac{-\pi}{2}\right)\right]$
$=\frac{\pi}{\sqrt{2}} .$

Topic: Mathematics, Definite Integration, NDA, NDA 2017 (Phase 2)
The value of
$\left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & 1+x & 1 \\ 1 & 1 & 1+y\end{array}\right|$ is

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

The correct answer is:
$x y$

Solution:
$\left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & 1+x & 1 \\ 1 & 1 & 1+y\end{array}\right|$
$=[(1+x)(1+y)-1]-1(1+y-1)+1(1-1-x)$
$=1+x+y+x y-1-y-x$
$=x y$
$\therefore$ Option (c) is correct.

Topic: Mathematics, Determinants, NDA, NDA 2015 (Phase 1)
The value of 1 mole of any pure gas at standard temperature and pressure is always equal to

Options:
A 0.022414m3
B 22.414m3
C 2.2414m3
D 0.22414m3

The correct answer is:
0.022414m3

Solution:
The value of 1 mole of any pure gas at standard and pressure (STP) is equal to 22.414 L 1L=10-3m3
∴      22.4L=0.022414m3
So, from Eq. (i), we get
p∝1R or  p∝R-1

Topic: Chemistry, Some Basic Concepts of Chemistry, MHT CET, MHT CET 2019 (Shift 2)
The value of
$\left(\frac{-1+\mathrm{i} \sqrt{3}}{2}\right)^{\mathrm{n}}+\left(\frac{-1-\mathrm{i} \sqrt{3}}{2}\right)^{\mathrm{n}}$
where $\mathrm{n}$ is not a multiple of 3 and $i=\sqrt{-1}$, is

Options:
A 1
B -1
C i
D -i

The correct answer is:
-1

Solution:
$\left(\frac{-1+\mathrm{i} \sqrt{3}}{2}\right)^{\mathrm{n}}+\left(\frac{-1-\mathrm{i} \sqrt{3}}{2}\right)^{\mathrm{n}}$
we know that cube roots of a unit use $1, \omega, \omega^{2}$ $\omega=\frac{-1+\mathrm{i} \sqrt{3}}{2}, \omega^{2}=\frac{-1-\mathrm{i} \sqrt{3}}{2}$
So, $\left(\frac{-1+i \sqrt{3}}{2}\right)^{n}+\left(\frac{-1-i \sqrt{3}}{2}\right)^{n}=\omega^{n}+\omega^{2 n}$
Given, $\mathrm{n}$ is not multiple of $3 .$ So, $\mathrm{n}=1,2,4,5 \ldots$ In any case, $\omega^{\mathrm{n}}+\omega^{2 \mathrm{n}}=\omega+\omega^{2}=-1$

Topic: Mathematics, Complex Number, NDA, NDA 2017 (Phase 1)
The value of $\int_{-2}^{2}\left(a x^{3}+b x+c\right) d x$ depends on which of the following? $\quad\mathrm{}$

Options:
A Values of $x$ only
B Values of each of $a, b$ and $c$
C Value of $c$ only
D Value of $b$ only

The correct answer is:
Value of $c$ only

Solution:
$\begin{aligned} & \int_{-2}^{2}\left(a x^{3}+b x+c\right) d x \\ &=\left[\frac{a x^{4}}{4}+\frac{b x^{2}}{2}+\frac{c x}{1}\right]_{-2}^{2} \\ &=\left[\frac{a(16)}{4}+\frac{b(4)}{2}+2 c\right]-\left[\frac{a(16)}{4}+\frac{b(4)}{2}-2 c\right]=4 c \end{aligned}$
So, the value of given integral depends on the value of $c$ only

Topic: Mathematics, Definite Integration, NDA, NDA 2008 (Phase 2)