datavorous/entrance-exam-dataset huggingface dataset:
The variance of 20 observations is $5 .$ If each observatiot multiplied by 2 , then what is the new variance of the resultin:
observations

Options:
A 5
B 10
C 40
D 20

The correct answer is:
40

Solution:
Detailed Method:
Variance $=\frac{\sum \mathrm{d}^{2}}{\mathrm{n}}$
$5=\frac{\sum \mathrm{d}^{2}}{20} \Rightarrow \sum \mathrm{d}^{2}=100$
According to question $\sum\left(\mathrm{d}_{1}\right)^{2}=\sum(2 \mathrm{~d})^{2}=4 \sum \mathrm{d}^{2}=4 \times 100=400$
New variance $=\frac{400}{20}=20$
Shortcut Method:
If each observation is multiplied by 2 New variance $=2^{2} \times 5=20$

Topic: Mathematics, Statistics, NDA, NDA 2013 (Phase 2)
The variance of 25 observations is 4 . If 2 is added to each observation, then the new variance of the resulting observations is $\quad$

Options:
A 2
B 4
C 6
D $\underline{8}$

The correct answer is:
4

Solution:
Variance will not change by adding or subtracting a fixed value to all the elements

Topic: Mathematics, Statistics, NDA, NDA 2018 (Phase 2)
The variance of numbers $x_{1}, x_{2} x_{3, \cdots \ldots} x_{n,}$ is $V$. Consider the following statements: $\quad$
1 . If every $x_{1}$ is increased by 2, the variance of the new set of the new set of numbers is $V$
2 . If the numbers $x_{i}$ is squared, the variance of the new set is $V^{2}$. Which of the following statements is/are correct?

Options:
A 1 only
B 2 only
C Both 1 and 2
D Neither $\operatorname{lnor} 2$

The correct answer is:
1 only

Solution:
I : Variance is not dependent on change of origin. Therefore, if every $x_{i}$ is increased by 2, the variance of the new set of numbers is not changed.

Topic: Mathematics, Statistics, NDA, NDA 2014 (Phase 1)
The vascular bundles in a dicot stem are

Options:
A Closed, collateral, exarch
B Open, collateral, exarch
C Closed, collateral, endarch
D Open, collateral, endarch

The correct answer is:
Open, collateral, endarch

Solution:
In dicot stem, vascular bundles are conjoint, collateral in which xylem is present towards inner side and phloem towards center. Cambium is present in between xylem and phloem thus vascular bundles are open. Xylem is endarch as protoxylem is present towards center and metaxylem towards periphery.

Topic: Biology, Anatomy of Flowering Plants, NEET
The vascular cambium is generally more active on the .....A.... and produce more ....B.......
What is A and B, respectively?

Options:
A A- Outer side
B- secondary xylem
B A- Inner side
B- secondary xylem
C A- Inner side
B- secondary phloem
D A- Outer side
B- Secondary phloem

The correct answer is:
A- Inner side
B- secondary xylem

Solution:
Vascular cambium when active divide and cut off cells towards outside, giving secondary phloem and towards inside giving secondary xylem, but since inner side is more active than outer side, secondary xylem is more produced.

Topic: Biology, Anatomy of Flowering Plants, NEET
The vascular cambium normally gives rise to

Options:
A Phelloderm
B Primary phloem
C Secondary xylem
D Periderm

The correct answer is:
Secondary xylem

Solution:
During secondary growth in Dicotyledonous plants, vascular cambium and extrasteler cambium contribute together. While the vascular cambium gives rise to secondary xylem and secondary phloem. Phelloderm is formed by cork cambium which is extrasteler cambium.

Topic: Biology, Anatomy of Flowering Plants, NEET, NEET 2017
The vascular cambium of dicot root originates from the

Options:
A tissue located just below the phloem bundles & a portion of pericycle tissue.
B tissue located just below the xylem bundles & a portion of pericycle tissue.
C tissue located just below the xylem bundles & a portion of endodermis tissue.
D tissue located just below the phloem bundles & a portion of endodermis tissue.

The correct answer is:
tissue located just below the phloem bundles & a portion of pericycle tissue.

Solution:
The vascular cambium is a lateral meristem in the vascular tissue of plants. It is a cylinder of unspecialized meristem cells that divide to give rise to cells that further divide, differentiate and specialize to form the secondary vascular tissues. Vascular cambium of dicot root originates from the tissue located just below the phloem bundles & a portion of pericycle tissue.

Topic: Biology, Anatomy of Flowering Plants, NEET
The vector $2 \hat{\mathrm{j}}-\hat{\mathrm{k}}$ lies:

Options:
A in the plane of $\mathrm{XY}$
B in the plane of $\mathrm{YZ}$
C in the plane of $\mathrm{XZ}$
D along the X-axis

The correct answer is:
in the plane of $\mathrm{YZ}$

Solution:
The vector $2 \hat{\mathrm{j}}-\hat{\mathrm{k}}$ lies in the plane of $\mathrm{YZ}$.

Topic: Mathematics, Vector Algebra, NDA, NDA 2012 (Phase 2)
The vector $\overrightarrow{\mathrm{a}} \times(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{a}})$ is coplanar with :

Options:
A a only
B $\overrightarrow{\mathrm{b}}$ only
C Both $\vec{a}$ and $\vec{b}$
D Neither a nor $\overrightarrow{\mathrm{b}}$

The correct answer is:
Neither a nor $\overrightarrow{\mathrm{b}}$

Solution:
$\overrightarrow{\mathrm{a}} \times(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{a}})$ is coplanar with neither a nor $\overrightarrow{\mathrm{b}}$.

Topic: Mathematics, Vector Algebra, NDA, NDA 2012 (Phase 2)
The vector a lies in the plane of vectors $\overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{c}}$. Which one of the following is correct?

Options:
A $\overrightarrow{\mathbf{a}} \cdot(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}})=0$
B $\vec{a} \cdot(\vec{b} \times \vec{c})=1$
C $\overrightarrow{\mathbf{a}} \cdot(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}})=-1$
D $\vec{a} \cdot(\vec{b} \times \vec{c})=3$

The correct answer is:
$\overrightarrow{\mathbf{a}} \cdot(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}})=0$

Solution:
Let the vector $\vec{a}$ lies in the plane of vectors $\vec{b}$ and $\vec{c}$.
$\Rightarrow \vec{a}, \vec{b}$ and $\vec{c}$ are coplanar.
$\therefore \quad \vec{a} \cdot(\vec{b} \times \vec{c})=0$

Topic: Mathematics, Vector Algebra, NDA, NDA 2011 (Phase 1)
The vectors $\vec{a}, \vec{b}, \vec{c}$ and $\vec{d}$ are such that $\vec{a} \times \vec{b}=\vec{c} \times d$ and
$\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{d}}$. Which of the following is/ are correct?
$1.$ $\quad(\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{d}}) \times(\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{c}})=\overrightarrow{0}$
$2.$ $\quad(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \times(\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{d}})=\overrightarrow{0}$
Select the correct answer using the code given below:

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:
$\begin{aligned} &(\vec{a}-\vec{d}) \times(\vec{b}-\vec{c}) \\ &=\vec{a} \times \vec{b}-\vec{d} \times \vec{b}-\vec{a} \times \vec{c}+\vec{d} \times \vec{c} \\ &=\vec{c} \times \vec{d}-\vec{d} \times \vec{b}-\vec{b} \times \vec{d}-\vec{c} \times \vec{d} \\ &=-\vec{d} \times \vec{b}+\vec{d} \times \vec{b} \\ &=0 \\ & \text { again }(\vec{a} \times \vec{b})=(\vec{c} \times \vec{d}) \quad \text { given } \end{aligned}$
$\Rightarrow(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \times(\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{d}})=(\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{d}}) \times(\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{d}})=0 \quad($ as $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{a}}=0)$
So both $(1)$ and $(2)$ are correct.

Topic: Mathematics, Vector Algebra, NDA, NDA 2015 (Phase 2)
The vectors $\vec{a}=x \vec{i}+y \vec{j}+z \vec{k}, \vec{b}=\hat{k}, \vec{c}$ are such that they
form a right handed system. What is $\vec{c}$ equal to?

Options:
A $\hat{j}$
B $y \hat{j}-x \hat{k}$
C $y \hat{\boldsymbol{i}}-x \hat{\boldsymbol{j}}$
D $x \hat{\boldsymbol{i}}-y \hat{\boldsymbol{j}}$

The correct answer is:
$y \hat{\boldsymbol{i}}-x \hat{\boldsymbol{j}}$

Solution:
We know, scalar triple product $(\vec{a} \times \vec{b}) \cdot \vec{c}$ is positive or
negative according as $\vec{a}, \vec{b}, \vec{c}$ form a right handed or
left handed system respectively. consider option (a)
Let $\vec{c}=\vec{j}$
$\therefore[\vec{a} \vec{b} \vec{c}]=\left|\begin{array}{lll}x & y & z \\ 0 & 0 & 1 \\ 0 & 1 & 0\end{array}\right|=x(-1)-y(0)+z(0)$
$=-x$
option (b)
Let $\vec{c}=y \hat{j}-x \hat{k}$
$\begin{aligned} \therefore \quad[\vec{a} \vec{b} \vec{c}]=\left|\begin{array}{ccc}x & y & z \\ 0 & 0 & 1 \\ 0 & y-x\end{array}\right| &=x(-y)-y(0)+z(0) \\ &=-x y \end{aligned}$
option (c)
Let $\vec{c}=y \hat{i}-x \hat{j}$
$\begin{aligned} \therefore \quad[\vec{a} \vec{b} \vec{c}] &=\left|\begin{array}{rrr}x & y & z \\ 0 & 0 & 1 \\ y & -x & 0\end{array}\right| \\ &=x(x)-y(-y)+z(0)=x^{2}+y^{2} \end{aligned}$
Since, scalar triple product is positive when $\vec{c}=y \hat{i}-x \hat{j}$
$\therefore$ option (c) is correct.

Topic: Mathematics, Vector Algebra, NDA, NDA 2010 (Phase 1)
The vectors $\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{c}}, \overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{CA}}=\overrightarrow{\mathrm{b}}$, are the sides of a
triangle $\mathrm{ABC}$. Which of the following vectors represent (s)
the median $\overline{\mathrm{AD}}$ ?
$1.$ $\frac{1}{2} \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{c}}$
$2.$ $\quad-\frac{1}{2} \overrightarrow{\mathrm{b}}+\frac{1}{2} \overrightarrow{\mathrm{c}}$
$3.$ $\frac{1}{2} \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}$
Select the correct answer using the code given below

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

The correct answer is:
1 only

Solution:
Refer to the figure which is self explanatory In $\Delta \mathrm{ABD}$
$\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{AD}}+\overrightarrow{\mathrm{DB}}$
$\overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{AD}}-\frac{1}{2} \overrightarrow{\mathrm{a}}$
$\left[\because \quad \overrightarrow{\mathrm{DB}}=\frac{1}{2} \overrightarrow{\mathrm{CB}}\right]$
$\overrightarrow{\mathrm{AD}}=\frac{1}{2} \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{c}}$
Also in $\triangle \mathrm{ACD}$.
$\overrightarrow{\mathrm{AD}}+\overrightarrow{\mathrm{DC}}=\overrightarrow{\mathrm{CA}}$
$\Rightarrow \quad \overrightarrow{\mathrm{AD}}+\frac{1}{2} \overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{b}}$
$\overrightarrow{\mathrm{AD}}=\overrightarrow{\mathrm{b}}-\frac{1}{2} \overrightarrow{\mathrm{a}}$
But this is not in the statement.
Hence, only (1) statement represent median $\overrightarrow{\mathrm{AD}}$.

Topic: Mathematics, Vector Algebra, NDA, NDA 2007 (Phase 2)
The vectors $\hat{i}-2 x \hat{j}-3 y \hat{k}$ and $\hat{i}+3 \mathrm{x} \hat{j}+2 y \hat{k}$ are orthogonal to each other. Then the locus of the point $(x, y)$ is

Options:
A hyperbola
B ellipse
C parabola
D circle

The correct answer is:
circle

Solution:
$\begin{array}{ll}\text {} & (\hat{\mathrm{i}}-2 \mathrm{x} \hat{\mathrm{j}}-3 \mathrm{y} \hat{\mathrm{k}}) \cdot(\hat{\mathrm{i}}+3 \hat{\mathrm{j}}+2 \mathrm{yk})=0 \\ & 1-6 \mathrm{x}^{2}-6 \mathrm{y}^{2}=0 \\ & -6 \mathrm{x}^{2}-6 \mathrm{y}^{2}=-1 \\ & \mathrm{x}^{2}+\mathrm{y}^{2}=\frac{1}{6} \\ & \mathrm{x}^{2}+\mathrm{y}^{2}=\left(\sqrt{\frac{1}{6}}\right)^{2}\end{array}$
Hence, locus of the point i.e. a circle.

Topic: Mathematics, Vector Algebra, NDA, NDA 2013 (Phase 2)
The velocity of a particle varies with its displacement as v=9-x2 m s-1. Find the magnitude of the maximum acceleration of the particle.

Options:
A 3 m s-2
B 4 m s-2
C 3.5 m s-2
D 5 m s-2

The correct answer is:
3 m s-2

Solution:
Comparing the given equation with standard velocity-displacement equation of simple harmonic motion, i.e., v = ω A 2 - x 2 , we get
ω=1 rad/s and A=3 m
The magnitude of maximum acceleration of the particle in SHM is=ω2A
=123 m/s2=3 m/s2

Topic: Physics, Oscillations, NEET
The velocity of telegraphic communication is given by $v=x^{2} \log (1 / x)$, where $x$ is the displacement. For maximum velocity, $x$ equals to?

Options:
A $e^{1 / 2}$
B $e^{-1 / 2}$
C $(2 e)^{-1}$
D $2 e^{-1 / 2}$

The correct answer is:
$e^{-1 / 2}$

Solution:
Given, velocity is $v=x^{2} \log \frac{1}{x}=-x^{2} \log x$ where $x$ is displacement.
For maximum velocity, $\quad \frac{d v}{d x}=0$
Now, $\frac{d v}{d x}=-x^{2} \frac{1}{x}+\log x(-2 x)$
$=-x-2 x \log x$
$\frac{d v}{d x}=0 \Rightarrow-x-2 x \log x=0 \Rightarrow x=-2 x \log x$
$\Rightarrow \frac{-1}{2}=\log x$
$\Rightarrow x=e^{-\frac{1}{2}}$
Hence, for maximum velocity $x=e^{-1 / 2}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2009 (Phase 2)
The velocity $v$ of a particle at any instant $t$ moving in a straight line is given by $v=s+1$ where $s$ metre is the distance travelled in $t$ second. What is the time taken by the particle to cover a distance of $9 \mathrm{~m} ? \quad$

Options:
A $1 s$
B $(\log 10) s$
C $2(\log 10) s$
D $10 \mathrm{~s}$

The correct answer is:
$(\log 10) s$

Solution:
Given velocity is $v=s+1$
Since, velocity $=\frac{d s}{d t}$
$\therefore \frac{d s}{d t}=s+1 \Rightarrow \frac{d s}{s+1}=d t$
Integrate both side we get $\log (s+1)=t$
At $s=9 \mathrm{~m}$
$t=\log (10)$ second

Topic: Mathematics, Application of Derivatives, NDA, NDA 2009 (Phase 2)
The venturi-meter works on :

Options:
A The principle of perpendicular axes
B Huygen's principle
C Bernoulli's principle
D The principle of parallel axes

The correct answer is:
Bernoulli's principle

Solution:
Venturi Meter works on the principle of Bernoulli's standard which expresses that the distinction in pressure made over the hindrance gives the proportion of the liquid stream rate. The stream rate is relative to the square base of the differential pressure.

Topic: Physics, Mechanical Properties of Fluids, NEET, NEET 2023 (All India)
The versatile anthers are helpful for

Options:
A Anemophily
B Entomophily
C Epihydrophily
D Hypohydrophily

The correct answer is:
Anemophily

Solution:
Some features of wind-pollinated flowers (anemophilous) are:
(i) Flowers are devoid of scent and nectar.
(ii) Inflorescence generally pendant for swinging purpose.
(iii) Anthers generally versatile.
(iv) Plants produce very large quantity of dusty pollen grains.
(v) Stigma may be large and feathery of brushy.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
The vertex of the parabola \(y^2+6 x-2 y+13=0\) is
\(\begin{aligned}
& (y-1)^2=-6 x-12 \\
& (y-1)^2=-6(x+2)=4\left(\frac{-6}{4}\right)(x+2) \\
& \text { Vertex } \rightarrow(-2,1)
\end{aligned}\)

Options:
A \((1,-1)\)
B \((-2,1)\)
C \(\left(\frac{3}{2}, 1\right)\)
D \(\left(-\frac{7}{2}, 1\right)\)

The correct answer is:
\((-2,1)\)

Solution:
Hints:

Topic: Mathematics, Parabola, WBJEE, WBJEE 2011
The vertices of a triangle $\mathrm{ABC}$ are $\mathrm{A}(2,3,1), \mathrm{B}(-2,2,0)$, and $\mathrm{C}(0,1,-1).$
What is the area of the triangle?

Options:
A $6 \sqrt{2}$ square unit
B $3 \sqrt{2}$ square unit
C $10 \sqrt{3}$ square unit
D None of these

The correct answer is:
$3 \sqrt{2}$ square unit

Solution:
Area of triangle $\mathrm{ABC}=\frac{1}{2}|\overrightarrow{\mathrm{BA}} \times \overrightarrow{\mathrm{BC}}|$
$\overrightarrow{\mathrm{BA}} \times \overrightarrow{\mathrm{BC}}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 4 & 1 & 1 \\ 2 & -1 & -1\end{array}\right|$
$=\hat{\mathrm{i}}(0)-\hat{\mathrm{j}}(-6)+\hat{\mathrm{k}}(-6)$
$|\overrightarrow{\mathrm{BA}} \times \overrightarrow{\mathrm{BC}}|=|6 \hat{\mathrm{j}}-6 \hat{\mathrm{k}}|=\sqrt{6^{2}+(-6)^{2}}$
$=6 \sqrt{2}$
Area of triangle $=\frac{1}{2} \times 6 \sqrt{2}=3 \sqrt{2}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2014 (Phase 1)
The vertices of a triangle $\mathrm{ABC}$ are $\mathrm{A}(2,3,1), \mathrm{B}(-2,2,0)$, and $\mathrm{C}(0,1,-1).$
What is the cosine of angle $A B C$ ?

Options:
A $\frac{1}{\sqrt{3}}$
B $\frac{1}{\sqrt{2}}$
C $\frac{2}{\sqrt{6}}$
D None of these

The correct answer is:
$\frac{1}{\sqrt{3}}$

Solution:
$\overrightarrow{\mathrm{BA}}=4 \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$
$\overrightarrow{\mathrm{BC}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}}$
$\cos B=\frac{\overline{\mathrm{BA}} \cdot \overrightarrow{\mathrm{BC}}}{|\overline{\mathrm{BA}}||\overrightarrow{\mathrm{BC}}|}$
$=\frac{(4 \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}) \cdot(2 \hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}})}{\left|\sqrt{4^{2}+1^{2}+1^{2}}\right|\left|\sqrt{2^{2}+(-1)^{2}+(-1)^{2}}\right|}$
$=\frac{6}{\sqrt{18} \sqrt{6}}=\frac{1}{\sqrt{3}}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2014 (Phase 1)
The vertices of a triangle $\mathrm{ABC}$ are $\mathrm{A}(2,3,1), \mathrm{B}(-2,2,0)$, and $\mathrm{C}(0,1,-1).$
What is the magnitude of the line joining mid points of the sides $A C$ and $B C$ ?

Options:
A $\frac{1}{\sqrt{2}}$ unit
B 1 unit
C $\frac{3}{\sqrt{2}}$ unit
D 2 unit

The correct answer is:
$\frac{3}{\sqrt{2}}$ unit

Solution:
Mid-point of $\mathrm{A}$ and $\mathrm{C},\left(\frac{2+0}{2}, \frac{3+1}{2}, \frac{1-1}{2}\right)=(1,2,0)$
Mid-point of $\mathrm{B}$ and $\mathrm{C},\left(\frac{-2+0}{2}, \frac{2+1}{2}, \frac{0-1}{2}\right)$
$=\left(-1, \frac{3}{2}, \frac{-1}{2}\right)$
Magnitude $=\sqrt{(1+1)^{2}+\left(2-\frac{3}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}}$
$=\sqrt{4+\frac{1}{4}+\frac{1}{4}}=\frac{3}{\sqrt{2}}$ units

Topic: Mathematics, Vector Algebra, NDA, NDA 2014 (Phase 1)
The wave described by $y=0.25 \sin (10 \pi x-2 \pi t)$, where $x$ and $y$ are in metre and $t$ in second, is a wave travelling along the

Options:
A - ve $x$ direction with frequency $1 \mathrm{~Hz}$
B +ve $x$ direction with frequency $\pi \mathrm{Hz}$ and wavelength $\lambda=0.2 \mathrm{~m}$
C + ve $x$ direction with frequency $1 \mathrm{~Hz}$ and wavelength $\lambda=0.2 \mathrm{~m}$
D -ve $x$ direction with amplitude $0.25 \mathrm{~m}$ and wavelength $\lambda=0.2 \mathrm{~m}$

The correct answer is:
+ ve $x$ direction with frequency $1 \mathrm{~Hz}$ and wavelength $\lambda=0.2 \mathrm{~m}$

Solution:
Key Idea : The sign between two terms in argument of sine will define its direction.
Writing the given wave equation
$y=0.25 \sin (10 \pi x-2 \pi t)$
The minus $(-)$ between $(10 \pi x)$ and $(2 \pi t)$ implies that the wave is travelling along positive $x$ direction.
Now comparing Eq. (i) with standard wave equation
$y=a \sin (k x-\omega t)$
We have
$\begin{array}{l}
a=0.25 \mathrm{~m}, \omega=2 \pi, k=10 \pi \mathrm{m} \\
\therefore \frac{2 \pi}{T}=2 \pi \\
\Rightarrow f=1 \mathrm{~Hz}
\end{array}$
Also, $\lambda=\frac{2 \pi}{k}=\frac{2 \pi}{10 \pi}=0.2 \mathrm{~m}$
Therefore, the wave is travelling along +ve $x$ direction with frequency $1 \mathrm{~Hz}$ and wavelength $0.2 \mathrm{~m}$

Topic: Physics, Waves and Sound, NEET, NEET 2008 (Screening)
The weights of coffee in 70 jars is shown in the following table:
$$
\begin{array}{|c|c|}
\hline \text { Weight (in grams) } & \text { Frequency } \\
\hline 200-201 & 13 \\
201-202 & 27 \\
202-203 & 18 \\
203-204 & 10 \\
204-205 & 1 \\
205-206 & 1 \\
\hline
\end{array}
$$
Determine variance and standard deviation of the above distribution.

Options:

None

Solution:
$$
\begin{array}{|c|c|l|c|l|l|}
\hline \text { Class } & \boldsymbol{x} & \boldsymbol{f} & \boldsymbol{d}=\boldsymbol{x}-\mathbf{2 0 2 . 5} & \boldsymbol{f d} & \boldsymbol{f d}^2 \\
\hline 200-201 & 200.5 & 13 & -2 & -26 & 52 \\
201-202 & 201.5 & 27 & -1 & -27 & 27 \\
202-203 & 202.5 & 18 & 0 & 0 & 0 \\
203-204 & 203.5 & 10 & 1 & 10 & 10 \\
204-205 & 204.5 & 1 & 2 & 2 & 4 \\
205-206 & 205.5 & 1 & 3 & 3 & 9 \\
\hline & & 70 & & -38 & 102 \\
\hline
\end{array}
$$
Let $A=$ Assumed Mean $=202.5$
$$
\Rightarrow \text { S.D. }=\sqrt{\frac{102}{70}-\left(\frac{-38}{70}\right)^2}
$$
$$
=\sqrt{1.162}=1.08 \mathrm{gm}
$$
$\&$ variance $=(\mathrm{SD})^2=1.162 \mathrm{gm}$

Topic: Mathematics, Statistics
The wood is actually a ________

Options:
A Primary xylem
B Primary phloem
C Secondary xylem
D Secondary Phloem

The correct answer is:
Secondary xylem

Solution:
Wood is secondary xylem developed from the lateral meristems or lateral cambium. The older xylem is filled with tyloses, resins, tannins and other excretory plant products that make it hard, thus, forming the wood.

Topic: Biology, Anatomy of Flowering Plants, NEET
The work done in blowing a soap bubble of radius $R$ is $W_1$ at room temperature. Now the soap solution is heated. From the heated solution another soap bubble of radius $2 R$ is blown and the work done is $W_2$. Then

Options:
A $W_2=0$
B $W_2=4 W_1$
C $W_2<4 W_1$
D $W_2=W_1$

The correct answer is:
$W_2<4 W_1$

Solution:
The work done in forming a soap in forming a soap bubble of radius $r$ is $W_1=8 \pi r^2 T_1$, where $T_1$ is the surface tension at room temperature and the work done to form a soap bubble of radius $2 r$ is
$\therefore \frac{W_2}{W_1}=\frac{328 \pi r^2 T_2}{8 \pi r^2 T_1}=\frac{4 T_2}{T_1}$
If $T_1=T_2$, then $W_2=4 W_1$
But it is given that the solution is heated. As the temperature is increased, the surface tension of the soap solution is decreases
$\therefore T_2

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Topic: Physics, Mechanical Properties of Fluids, MHT CET, MHT CET 2022 (10 Aug Shift 2)
The yellowish fluid "colostrum" secreted by mammary glands of mother during the initial days of lactation has abundant antibodies (IgA) to protect the infant. This type of immunity is called as:

Options:
A Passive immunity
B Active immunity
C Acquired immunity
D Autoimmunity

The correct answer is:
Passive immunity

Solution:
Immunity is said to be passive when antibodies produced in other organisms are injected into a person to counteract antigens such as snake venom.

The yellowish fluid ‘colostrum’ secreted by the mother during the initial days of lactation is a form of natural passive immunity. It has plenty of antibodies (IgA) to protect the infant during the initial months of life. Maternal antibodies (IgG) are also transmitted through the placenta. Such an immunity is called natural passive immunity.

Topic: Biology, Human Health and Diseases, NEET, NEET 2020 (Phase 2)
There are 4 candidates for the post of a lecturer in Mathematics and one is to be selected by votes of 5 men. What is the number of ways in which the votes can be given?

Options:
A 1048
B 1072
C 1024
D 625

The correct answer is:
625

Solution:
There are 4 candidates.
This means there are 4 blank spaces for 1 post. Now, that 1 post is to be selected by votes of 5 men.
So, All 4 places can be fill by each man's votes. $\therefore$ There is 5 ways for 1 place.. Hence, Required ways $=5 \times 5 \times 5 \times 5=625 \quad(\because$ we have 4 places)

Topic: Mathematics, Permutation Combination, NDA, NDA 2011 (Phase 2)
There are 4 letters and 4 directed envelopes. These 4 letters are randomly inserted into the 4 envelopes. What is the probability that the letters are inserted into the corresponding envelopes?

Options:
A $\frac{11}{12}$
B $\frac{23}{24}$
C $\frac{1}{24}$
D None of these

The correct answer is:
$\frac{1}{24}$

Solution:
Since, 4 letters are randomly inserted into the 4 envelopes therefore
Required probability $=\frac{1}{4 !}=\frac{1}{4 \times 3 \times 2}=\frac{1}{24}$

Topic: Mathematics, Probability, NDA, NDA 2011 (Phase 1)