datavorous/entrance-exam-dataset huggingface dataset:
A variable plane passes through a fixed point $(a, b, c)$ and cuts the axes in $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ respectively. The locus of the centre of the sphere $\mathrm{OABC}$, $\mathrm{O}$ being the origin $\mathrm{k}$

Options:
A $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$
B $\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=$.
C $\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=2$
D $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=2$

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

Solution:
Equation of plane passing through points
$(\mathrm{p}, 0,0),(0, \mathrm{q}, 0)$ and $(0,0, \mathrm{r})$ is $\frac{\mathrm{x}}{\mathrm{p}}+\frac{\mathrm{y}}{\mathrm{q}}+\frac{\mathrm{z}}{\mathrm{r}}=1$
Given that this plane passes through $(\mathrm{a}, \mathrm{b}, \mathrm{c})$.
$\therefore \frac{a}{p}+\frac{b}{q}+\frac{c}{r}=1$
Equation of sphere is $\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2}-\mathrm{px}-\mathrm{q} y-\mathrm{rz}=0$.
Centre of the sphere $=(1, \mathrm{~m}, \mathrm{n})=\left(\frac{\mathrm{p}}{2}, \frac{\mathrm{q}}{2}, \frac{\mathrm{r}}{2}\right)$
$\Rightarrow \mathrm{p}=2 \ell, \mathrm{q}=2 \mathrm{~m}, \mathrm{r}=2 \mathrm{n}$
$\therefore$ locus of the centre $\Rightarrow \frac{\mathrm{a}}{\mathrm{x}}+\frac{\mathrm{b}}{\mathrm{y}}+\frac{\mathrm{c}}{\mathrm{z}}=2$.

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2017 (Phase 2)
A variate $X$ takes values $2,3,4,2,5,4,3,2,1$. What is the mode? $\quad$

Options:
A 2
B 3
C 4
D 5

The correct answer is:
2

Solution:
Given data is $2,3,4,2,5,4,3,2,1$
Mode $=2$
$[\because 2$ occurs maximum number of times].

Topic: Mathematics, Statistics, NDA, NDA 2011 (Phase 2)
A variate $\mathrm{X}$ takes values $3,7,5,4,3,2,10$. What is the median?

Options:
A $-2$
B 4
C 7
D 9

The correct answer is:
4

Solution:
First we arrange the data in ascending order $2,2,3,3,4,5,7,9,10$
Since number of observation is odd
$:$ median $=\left(\frac{9+1}{2}\right)^{\mathrm{th}}$ observation $=5$ th observation

Topic: Mathematics, Statistics, NDA, NDA 2012 (Phase 1)
A vascular bundle which lacks cambium is called

Options:
A Conjoint, and collateral
B Conjoint, but not collateral
C Closed, and collateral
D Bicollateral

The correct answer is:
Conjoint, but not collateral

Solution:
Vascular bundles having cambium in between xylem and phloem are called open bundles, due to cambium they carries secondary growth. Vascular bundles which lacks cambium are called closed bundles. Open bundles occurs in dicot and closed bundles occurs in monocots.

Topic: Biology, Anatomy of Flowering Plants, NEET
A vector $\vec{b}$ is collinear with the vector $\vec{a}=(2,1,-1)$ and
satisfies the condition $\vec{a} \cdot \vec{b}=3 .$ What is $\vec{b}$ equal to?

Options:
A $(1,1 / 2,-1 / 2)$
B $\quad(2 / 3,1 / 3,-1 / 3)$
C $(1 / 2,1 / 4,-1 / 4)$
D $(1,1,0)$

The correct answer is:
$(1,1 / 2,-1 / 2)$

Solution:
Let $\vec{b}=x \vec{i}+y \vec{j}+z \vec{k}$
Since, $\vec{b}$ is collinear with vetor $\vec{a}$
therefore $\vec{a}=k \vec{b}$ where $k$ is a scalar.
Given $\vec{a}=(2,1,-1)$
$\therefore \quad(2,1,-1)=k(x, y, z)$
$\Rightarrow x=\frac{2}{k}, y=\frac{1}{k}, z=\frac{-1}{k}$
Also, $\vec{a} \cdot \vec{b}=3$
$\Rightarrow 2 x+y-z=3$
$\Rightarrow \quad 2\left(\frac{2}{k}\right)+\frac{1}{k}+\frac{1}{k}=3$
$\Rightarrow \frac{4}{k}+\frac{1}{k}+\frac{1}{k}=3$
$\Rightarrow \frac{6}{k}=3 \Rightarrow k=2$
$\therefore \quad x=1, y=\frac{1}{2}$ and $z=\frac{-1}{2}$
Hence $\vec{b}=\left(1, \frac{1}{2}, \frac{-1}{2}\right)$

Topic: Mathematics, Vector Algebra, NDA, NDA 2010 (Phase 1)
A vertical pole with height more than $100 \mathrm{~m}$ consists of two parts, the lower being one-third of the whole. At a point on a horizontal plane through the foot and $40 \mathrm{~m}$ fromit, the upper
part subtends an angle uhose tangent is $\frac{1}{2}$. What is the height of the pole ?

Options:
A $110 \mathrm{~m}$
B $200 \mathrm{~m}$
C 120 m
D 150 m

The correct answer is:
120 m

Solution:
Let h be the height of pole, upper portion CD subtend angle $\theta$ at $A$.
Then, $\tan \theta=\frac{1}{2}$
Let lower part $\mathrm{BC}$ subtend angle $\phi$ at A then $\operatorname{In} \Delta \mathrm{ABC}$
$\tan \phi=\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{\mathrm{h} / 3}{40}=\frac{\mathrm{h}}{120}$
$\operatorname{In} \Delta \mathrm{ABD}$,
$\tan (\theta+\phi)=\frac{\mathrm{BD}}{\mathrm{AB}}$
$\Rightarrow \frac{\tan \theta+\tan \phi}{1-\tan \theta \tan \phi}=\frac{\mathrm{h}}{40}$
$\Rightarrow \frac{\frac{1}{2}+\frac{\mathrm{h}}{120}}{1-\frac{\mathrm{h}}{240}}=\frac{\mathrm{h}}{40}$
$\Rightarrow \frac{2(60+\mathrm{h})}{(240-\mathrm{h})}=\frac{\mathrm{h}}{40}$
$\Rightarrow 80(60+\mathrm{h})=240 \mathrm{~h}-\mathrm{h}^{2} \Rightarrow 4800+80 \mathrm{~h}=240 \mathrm{~h}-\mathrm{h}^{2}$
$\Rightarrow \mathrm{h}^{2}-160 \mathrm{~h}+4800=0 \quad \Rightarrow(\mathrm{h}-120)(\mathrm{h}-40)=0$
$\Rightarrow \mathrm{h}=120$
$[\mathrm{h}=40$ is discarded, since $\mathrm{h}>100$ is given]

Topic: Mathematics, Heights and Distances, NDA, NDA 2006 (Phase 1)
A vertical tower standing on a levelled field is mounted with a vertical flag staff of length $3 \mathrm{~m}$. From a point on the field. the angles of elevation of the bottom and tip of the flag staff are $30^{\circ}$ and $45^{\circ}$ respectively. Which one of the following gives the best approximation to the height of the tower?

Options:
A $3.90 \mathrm{~m}$
B $4.00 \mathrm{~m}$
C $4.10 \mathrm{~m}$
D $4.25 \mathrm{~m}$

The correct answer is:
$4.10 \mathrm{~m}$

Solution:
as $\angle \mathrm{CPA}=45^{\circ}$
so $\mathrm{AC}=\mathrm{AP}=\mathrm{x}+3$
$\tan 30^{\circ}=\frac{\mathrm{AB}}{\mathrm{AP}}=\frac{\mathrm{x}}{\mathrm{x}+3}$
$\frac{1}{\sqrt{3}}=\frac{\mathrm{x}}{\mathrm{x}+3}$
$x+3=\sqrt{3} x$
$\mathrm{x}=\frac{3}{\sqrt{3}-1} \times \frac{(\sqrt{3}+1)}{(\sqrt{3}+1)}$
$\mathrm{x}=\frac{3 \times 2.73}{2}=\frac{8.19}{2}=4.095 \mathrm{~m} \approx 4 . \mathrm{lm}$

Topic: Mathematics, Heights and Distances, NDA, NDA 2015 (Phase 2)
A vertical tower stands on a horizontal plane and is surmounted by a vertical flag staff of height $h$. At a point $P$ on the plane, the angle of elevation of the bottom of the flag staff is $\beta$ and that of the top is $\alpha$. What is the height of the tower?

Options:
A $\frac{h \tan \beta}{\tan \alpha-\tan \beta}$
B $\frac{h \tan \beta}{\tan \alpha+\tan \beta}$
C $\frac{h \cos \beta}{\cos \alpha-\cos \beta}$
D $\frac{h}{\cos (\alpha-\beta)}$

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

Solution:
Let $\mathrm{BC}$ be the vertical tower and $\mathrm{CD}$ be the flagstaff so that $\mathrm{CD}=h$
Let $\mathrm{P}$ be the point of observation on the plane. Then, $\angle \mathrm{BPC}=\beta$ and $\angle \mathrm{BPD}=\alpha$
Let $\mathrm{BC}=x$
Now, $\frac{\mathrm{PB}}{x}=\cot \beta \Rightarrow \mathrm{PB}=x \cot \beta$
and $\frac{\mathrm{PB}}{x+h}=\cot \alpha \Rightarrow \mathrm{PB}=(x+h) \cot \alpha$
From (1) and (2), we get $x \cot \beta=(\mathrm{x}+\mathrm{h}) \cot \alpha \Rightarrow x(\cot \beta-\cot \alpha)=h \cot \alpha$
$\therefore$ Height of the tower,
$\mathrm{x}=\frac{h \cot \alpha}{\cot \beta-\cot \alpha}=\frac{h \tan \beta}{\tan \alpha-\tan \beta}$

Topic: Mathematics, Heights and Distances, NDA, NDA 2011 (Phase 2)
A wave is represented by the equation $y=0.5 \sin (10 t-x)$ metre
It is a travelling wave propagating along $+x$ direction with velocity

Options:
A $10 \mathrm{~m} \mathrm{~s}^{-1}$
B $20 \mathrm{~m} \mathrm{~s}^{-1}$
C $5 \mathrm{~m} \mathrm{~s}^{-1}$
D None of these

The correct answer is:
$10 \mathrm{~m} \mathrm{~s}^{-1}$

Solution:
Compare the given eqn $y=0.5 \sin (10 t-x)$ with the standard form
$\begin{aligned} & y=r \sin \left[\frac{2 \pi t}{T}+\frac{2 \pi x}{\lambda}\right] \\ & \frac{2 \pi}{T}=10, \frac{2 \pi}{\lambda}=1 \\ & v=\frac{\lambda}{T}=\frac{10}{1}=10 \mathrm{~m} / \mathrm{s}\end{aligned}$

Topic: Physics, Waves and Sound, AIIMS, AIIMS 2016
$A$ whistle whose air column is open at both ends has a fundamental frequency of $5100 \mathrm{Hz}$. If the speed of sound in air is $340 \mathrm{ms}^{-1}$, the length of the whistle, in $\mathrm{cm}$, is

Options:
A $5 / 3$
B $10 / 3$
C 5
D $20 / 3$

The correct answer is:
$10 / 3$

Solution:
For an open pipe, the frequency
$$
\begin{array}{l}
f=\frac{v}{2 l} \Rightarrow 5100=\frac{340}{2 \times 1} \\
l=\frac{340}{5100 \times 2} \Rightarrow l=\frac{2}{30 \times 2}=\frac{1}{30} \mathrm{m} \\
l=\frac{100}{30}=\frac{10}{3} \mathrm{cm}
\end{array}
$$

Topic: Physics, Waves and Sound, WBJEE, WBJEE 2014
A wire $34 \mathrm{~cm}$ long is to be bent in the form of a quadrilateral of which each angle is $90^{\circ}$. What is the maximum area which can be enclosed inside the quadrilateral?

Options:
A $68 \mathrm{~cm}^{2}$
B $70 \mathrm{~cm}^{2}$
C $71.25 \mathrm{~cm}^{2}$
D $\begin{array}{lll}\text { } & 72.25 \mathrm{~cm}^{2} & {}\end{array}$

The correct answer is:
$\begin{array}{lll}\text { } & 72.25 \mathrm{~cm}^{2} & {}\end{array}$

Solution:
Let one side of quadrilateral be $x$ and another side be $y$ so, $2(x+y)=34$
or, $(\mathrm{x}+\mathrm{y})=17$
We know from the basic principle that for a given perimeter square has the maximum area, so, $x=y$ and putting this value in equation (i)
$\mathrm{x}=\mathrm{y}=\frac{17}{2}$
Area $=\mathrm{x} \cdot \mathrm{y}=\frac{17}{2} \times \frac{17}{2}=\frac{289}{4}=72.25$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2007 (Phase 1)
A wire loop enclosing a semi-circle of radius R is located on the boundary of a uniform magnetic field of induction B→. At time t=0, the loop is set into rotation with velocity ω? about its axis 0, coinciding with a line of vector B→ on the boundary as shown in the figure. The emf induced in the loop is

Options:
A BR22ω
B BRω
C BR2ω
D BR22ω

The correct answer is:
BR22ω

Solution:
dϕ=B→·ds→=Bds

=B12ϕR2 (ϕ=Sector angle)

Magnitude of induced emf =-dϕdt

=B12R2dϕdt

=BR2ω2

Topic: Physics, Electromagnetic Induction, TS EAMCET, TS EAMCET 2018 (04 May Shift 1)
A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of $45 \mathrm{~Hz}$. The mass of the wire is $3.5 \times 10^{-2} \mathrm{~kg}$ and its linear mass density is $4 \times 10^{-2} \mathrm{~kg} / \mathrm{m}$. What is (a) the speed of the transverse wave on the string and (b) the tension in the string?

Options:

None

Solution:
(a) Given, $v=45 \mathrm{~Hz}, M=3.5 \times 10^{-2} \mathrm{~kg}$,
$$
\begin{aligned}
&m=4 \times 10^{-2} \mathrm{~kg} / \mathrm{m} \\
&\therefore l=\frac{M}{m}=\frac{3.5 \times 10^{-2}}{4 \times 10^{-2}}=\frac{7}{8} \mathrm{~m}
\end{aligned}
$$
As it is vibrating in fundamental mode
$$
\frac{\lambda}{2}=l=\frac{7}{8} \quad \therefore \quad \lambda=\frac{7}{4}=1.75 \mathrm{~m} .
$$
Speed $=v=v \lambda=45 \times 1.75=78.75 \mathrm{~m} / \mathrm{s}$
(b)
$$
\begin{aligned}
v &=\sqrt{\frac{T}{m}} \Rightarrow T=v^2 \times m \\
&=(78.5)^2 \times 4 \times 10^{-2}=248.06 \mathrm{~N}
\end{aligned}
$$

Topic: Physics, Waves and Sound
A wire under tension vibrates with a fundamental frequency of $600 \mathrm{~Hz}$. If the length of the wire is doubled, the radius is halved and the wire is made to vibrate under one-ninth the tension. Then the fundamental frequency will become

Options:
A $200 \mathrm{~Hz}$
B $300 \mathrm{~Hz}$
C $600 \mathrm{~Hz}$
D $400 \mathrm{~Hz}$

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

Solution:
Frequency of wire, $f=\frac{1}{2 l r} \sqrt{\frac{T}{\pi \rho}}$
So here,
$$
\begin{aligned}
\frac{f_{1}}{f_{2}} &=\frac{l_{2}}{l_{1}} \times \frac{r_{2}}{r_{1}} \times \sqrt{\frac{T_{1}}{T_{2}}} \\
\frac{600}{f_{2}} &=\frac{2}{1} \times \frac{1}{2} \times \sqrt{\frac{T}{T / 9}} \\
\frac{600}{f_{2}} &=3 \\
f_{2} &=\frac{600}{3} \\
f_{2} &=200 \mathrm{~Hz}
\end{aligned}
$$

Topic: Physics, Waves and Sound, KCET, KCET 2011
A wire vibrates at a fundamental frequency of $500 \mathrm{~Hz}$. A second identical wire produces 5 beats per second with it when the tension in the first wire is slightly decreased. The ratio of the tension in the second wire to the tension in the first wire is approximately equal to

Options:
A 1.04
B 1.01
C 1.05
D 1.02

The correct answer is:
1.02

Solution:
(d) Given that, fundamental frequency of first wire, $f_1=500 \mathrm{~Hz}$
Let frequency of 2 nd wire be $f_2$.
Beat frequency, $f_b=5$
$$
\begin{aligned}
& f_b=f_2-f_1 \\
& f_2=f_b+f_1=500+5=505 \mathrm{~Hz}
\end{aligned}
$$
Using, expression for fundamental frequency of wire,
$$
\begin{aligned}
f & =\frac{1}{2 L} \sqrt{\frac{T}{\mu}} \Rightarrow f \propto \sqrt{T} \\
\Rightarrow \quad \frac{f_2}{f_1} & =\sqrt{\frac{T_2}{T_1}} \Rightarrow \frac{505}{500}=\sqrt{\frac{T_2}{T_1}} \\
\Rightarrow \quad \frac{T_2}{T_1} & =\left(\frac{505}{500}\right)^2=1.02
\end{aligned}
$$

Topic: Physics, Waves and Sound, TS EAMCET, TS EAMCET 2021 (06 Aug Shift 1)
$\mathrm{A}=\left[\begin{array}{cc}\mathrm{x}+\mathrm{y} & \mathrm{y} \\ \mathrm{x} & \mathrm{x}-\mathrm{y}\end{array}\right], \mathrm{B}=\left[\begin{array}{l}3 \\ -2\end{array}\right]$ and $\mathrm{C}=\left[\begin{array}{l}4 \\ -2\end{array}\right]$.
If $\mathrm{AB}=\mathrm{C}$, then what is $\mathrm{A}^{2}$ equal to?

Options:
A $\left[\begin{array}{cc}4 & 8 \\ -4 & -16\end{array}\right]$
B $\left[\begin{array}{cc}4 & -4 \\ 8 & -16\end{array}\right]$
C $\left[\begin{array}{cc}-4 & -8 \\ 4 & 12\end{array}\right]$
D $\left[\begin{array}{cc}-4 & -8 \\ 8 & 12\end{array}\right]$

The correct answer is:
$\left[\begin{array}{cc}-4 & -8 \\ 8 & 12\end{array}\right]$

Solution:
$A=\left[\begin{array}{cc}x+y & y \\ x & x-y\end{array}\right], B=\left[\begin{array}{c}3 \\ -2\end{array}\right], c=\left[\begin{array}{c}4 \\ -2\end{array}\right]$
$A B=C$
$\Rightarrow\left[\begin{array}{l}3 x+3 y-2 y \\ 3 x-2 x+2 y\end{array}\right]=\left[\begin{array}{c}4 \\ -2\end{array}\right]$
$\Rightarrow 3 x+y=4 ; x+2 y=-2$
Solving these equations,
\begin{array}{l}
3 x+y=4 \\
3 x+6 y=-6 \\
(-)(-) \quad(+) \\
\hline-5 y=10
\end{array}
$\Rightarrow \mathrm{y}=-2$
So, $3 \mathrm{x}-2=4 \Rightarrow 3 \mathrm{x}=6 \Rightarrow \mathrm{x}=2$
$\therefore \mathrm{A}=\left[\begin{array}{cc}0 & -2 \\ 2 & 4\end{array}\right], \quad \mathrm{A}^{2}=\left[\begin{array}{cc}0 & -2 \\ 2 & 4\end{array}\right]\left[\begin{array}{cc}0 & -2 \\ 2 & 4\end{array}\right]$
$=\left[\begin{array}{cc}-4 & -8 \\ 8 & 12\end{array}\right]$

Topic: Mathematics, Matrices, NDA, NDA 2017 (Phase 1)
Aand $\mathrm{B}$ are two matrices such that $\mathrm{AB}=\overline{\mathrm{A}}$ and $\mathrm{BA}=\mathrm{B}$ then what is $\mathrm{B}^{2}$ equal to?

Options:
A $\mathrm{B}$
B $\mathrm{A}$
C I
D $-I$
where I is the identity matrix

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

Solution:
Given $\mathrm{AB}=\mathrm{A}$ and $\mathrm{BA}=\mathrm{B}$
Consider $\mathrm{B}=\mathrm{BA}=\mathrm{B}(\mathrm{AB})(\because \mathrm{AB}=\mathrm{A})$
$=(\mathrm{BA}) \cdot \mathrm{B}=\mathrm{B} \cdot \mathrm{B}=\mathrm{B}^{2}$
Hence, $\mathrm{B}^{2}=\mathrm{B}$

Topic: Mathematics, Matrices, NDA, NDA 2012 (Phase 1)
$\mathrm{ABC}$ is a triangle inscribed in a circle with centre $\mathrm{O}$. Let $\alpha=$ $\angle \mathrm{BAC}$, where $45^{\circ} < \alpha < 90^{\circ} .$ Let $\beta=\angle \mathrm{BOC}$. Which one of
the following is correct?

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

The correct answer is:
$\cos \beta=\frac{1-\tan ^{2} \alpha}{1+\tan ^{2} \alpha}$

Solution:
$\angle \mathrm{BAC}=\alpha, \angle \mathrm{BOC}=\beta$
We know, from figure, $\beta=2 \alpha$
$\therefore \cos \beta=\cos 2 \alpha$
$=\frac{1-\tan ^{2} \alpha}{1+\tan ^{2} \alpha}$

Topic: Mathematics, Properties of Triangles, NDA, NDA 2018 (Phase 1)
ABCD is a parallelogram and P is the point of intersection of the diagonals. If $\mathrm{O}$ is the origin, then $\overrightarrow{\mathrm{OA}}+\overrightarrow{\mathrm{OB}}+\overrightarrow{\mathrm{OC}}+\overrightarrow{\mathrm{OD}}$ is equal to

Options:
A $\quad 4 \overrightarrow{\mathrm{OP}}$
B $2 \overrightarrow{\mathrm{OP}}$
C $\overrightarrow{\mathrm{OP}}$
D Null vector

The correct answer is:
$\quad 4 \overrightarrow{\mathrm{OP}}$

Solution:
$\because \overrightarrow{\mathrm{OA}}=\overrightarrow{\mathrm{OP}}-\overrightarrow{\mathrm{AP}} ; \overrightarrow{\mathrm{OB}}=\overrightarrow{\mathrm{OP}}+\overrightarrow{\mathrm{PB}}$
$\overrightarrow{\mathrm{OC}}=\overrightarrow{\mathrm{OP}}+\overrightarrow{\mathrm{PC}} \& \overrightarrow{\mathrm{OD}}=\overrightarrow{\mathrm{OP}}-\overrightarrow{\mathrm{DP}}$
Also, $\overrightarrow{\mathrm{AP}}=\overrightarrow{\mathrm{PC}} \& \overrightarrow{\mathrm{DP}}=\overrightarrow{\mathrm{PB}}$
$\therefore \overrightarrow{\mathrm{OA}}+\overrightarrow{\mathrm{OB}}+\overrightarrow{\mathrm{OC}}+\overrightarrow{\mathrm{OD}}=4 \overrightarrow{\mathrm{OP}}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2015 (Phase 2)
$\mathrm{ABCD}$ is a parallelogram. If $\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{b}}$, then what is
$\overrightarrow{\mathrm{BD}}$ equal to ?

Options:
A $\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}$
B $\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}$
C $\quad-\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}$
D $\quad-\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}$

The correct answer is:
$\quad-\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}$

Solution:
$\overrightarrow{\mathrm{BD}}=-\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2012 (Phase 2)
$\mathrm{ABCD}$ is a quadrilateral. Forces $\overrightarrow{\mathbf{A B}}, \overrightarrow{\mathbf{C B}}, \overrightarrow{\mathbf{C D}}$ and $\overrightarrow{\mathbf{D A}}$ act along its sides. What is their resultant?

Options:
A $2 \overrightarrow{\mathrm{CD}}$
B $2 \overline{\mathbf{D A}}$
C $2 \overrightarrow{\mathrm{BC}}$
D $2 \overrightarrow{\mathbf{C B}}$

The correct answer is:
$2 \overrightarrow{\mathbf{C B}}$

Solution:
Let $\mathrm{ABCD}$ be a quadrilateral
$\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{CB}}+\overrightarrow{\mathrm{CD}}+\overrightarrow{\mathrm{DA}}$
$=\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{CB}}+\overrightarrow{\mathrm{CA}}(\because \overrightarrow{\mathrm{CD}}+\overrightarrow{\mathrm{DA}}=\overrightarrow{\mathrm{CA}})$
$=\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{CA}}+\overrightarrow{\mathrm{CB}}$
$=\overrightarrow{\mathrm{CB}}+\overrightarrow{\mathrm{CB}} \quad(\because \overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{CA}}=\overrightarrow{\mathrm{CB}})$
$=2 \overline{\mathrm{CB}}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2010 (Phase 2)
$\mathrm{ABCD}$ is a quadrilateral whose diagonals are $\mathrm{AC}$ and $\mathrm{BD}$. Which one of the following is correct?

Options:
A $\overrightarrow{\mathrm{BA}}+\overrightarrow{\mathrm{CD}}=\overrightarrow{\mathrm{AC}}+\overrightarrow{\mathrm{DB}}$
B $\overrightarrow{\mathrm{BA}}+\overrightarrow{\mathrm{CD}}=\overrightarrow{\mathrm{BD}}+\overline{\mathrm{CA}}$
C $\overrightarrow{\mathrm{BA}}+\overrightarrow{\mathrm{CD}}=\overrightarrow{\mathrm{AC}}+\overrightarrow{\mathrm{BD}}$
D $\overrightarrow{\mathrm{BA}}+\overrightarrow{\mathrm{CD}}=\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{AD}}$

The correct answer is:
$\overrightarrow{\mathrm{BA}}+\overrightarrow{\mathrm{CD}}=\overrightarrow{\mathrm{BD}}+\overline{\mathrm{CA}}$

Solution:
$\overrightarrow{\mathrm{BA}}+\overrightarrow{\mathrm{AD}}=\overrightarrow{\mathrm{BD}}$...(i)
$\overrightarrow{\mathrm{CD}}+\overrightarrow{\mathrm{DA}}=\overrightarrow{\mathrm{CA}}$...(ii)
$(\mathrm{i})+(\mathrm{ii}) \Rightarrow \overrightarrow{\mathrm{BA}}+\overrightarrow{\mathrm{CD}}=\overrightarrow{\mathrm{BD}}+\overrightarrow{\mathrm{CA}}(\because \overrightarrow{\mathrm{AD}}, \overrightarrow{\mathrm{DA}}$ cancel $)$

Topic: Mathematics, Vector Algebra, NDA, NDA 2017 (Phase 1)
According to Hess's Law, what is the relationship between the standard reaction enthalpy of a reaction that occurs in several steps and the standard enthalpies of the intermediate reactions?

Options:
A The standard reaction enthalpy is always greater than the sum of the standard enthalpies of the intermediate reactions.
B The standard reaction enthalpy is always equal to the sum of the standard enthalpies of the intermediate reactions.
C The standard reaction enthalpy is always less than the sum of the standard enthalpies of the intermediate reactions.
D The standard reaction enthalpy has no relationship with the standard enthalpies of the intermediate reactions.

The correct answer is:
The standard reaction enthalpy is always equal to the sum of the standard enthalpies of the intermediate reactions.

Solution:
According to Hess's Law, the standard reaction enthalpy of a reaction that occurs in several steps is equal to the sum of the standard enthalpies of the intermediate reactions into which the overall reaction may be divided at the same temperature.

Topic: Chemistry, Thermodynamics (C), NEET
According to IUCN Red List, what is the status of Red Panda (Ailurus fulgens)?

Options:
A critically endangered species
B vulnerable species
C extinct species
D endangered species

The correct answer is:
endangered species

Solution:
Red panda is listed as Endangered on the IUCN Red List.

Topic: Biology, Biodiversity and its Conservation, NEET, NEET 2005
According to IUCN red list what is the status of red Panda (Athurus fulgens)?

Options:
A Vulnerable species
B Critically endangered species
C Extinct species
D Endangered species

The correct answer is:
Endangered species

Solution:
Approximately 300 species and sub species of mammals are considered as endangered by the International Union for the Conservation of Nature and Natural Resources (IUCN). Red Panda is an endangered species because it is facing a very high risk of extinction in near future.

Topic: Biology, Biodiversity and its Conservation, JIPMER, JIPMER 2007
According to Robert Constanza, the ________ accounts for about 50% in ecosystem services.

Options:
A soil formation
B nutrient cycling
C recreation
D climate regulation

The correct answer is:
soil formation

Solution:
The product of ecosystem processes are called as ecosystem services. Robert Constanza and his colleagues have put a price tag of US $ 33 trillion a year on these fundamental ecosystem services. Out of the total cost of various ecosystem services, the soil formation accounts for about 50%.

Topic: Biology, Ecosystem, NEET
According to Robert Constanza, what percentage of the total cost of various ecosystem services is used for nutrient recycling and recreation activity?

Options:
A

Nutrient recycling	Recreation

Less than 10%	More than 10%

B

Nutrient recycling	Recreation

Less than 10%	Less than 10%

C

Nutrient recycling	Recreation

More than 10%	More than 10%

D

Nutrient recycling	Recreation

6%	6%

The correct answer is:

Nutrient recycling	Recreation

Less than 10%	Less than 10%

Solution:
Robert Constanza and his colleagues have very recently tried to put price tags on nature’s life-support services. Researchers have put an average price tag of USD 33 trillion a year on these fundamental ecosystem services, which are largely taken for granted because they are free. This is near twice the value of the global gross national product GNP which is USD 18 trillion.
Out of the total cost of various ecosystem services, the soil formation accounts for about 50%, and contributions of other services like recreation and nutrient cycling, are less than 10% each. The cost of climate regulation and habitat for wildlife is about 6% each.

Topic: Biology, Ecosystem, NEET
According to the recommendation of IUPAC, how are the groups in the Periodic Table numbered?

Options:
A Roman numerals
B From 1 to 18
C From IA to VIIIA
D From IA to VIIIB

The correct answer is:
From 1 to 18

Solution:
According to the recommendation of International Union of Pure and Applied Chemistry (IUPA

Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET