datavorous/entrance-exam-dataset huggingface dataset:
Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.
$4 x^2+9 y^2=36$

Options:

None

Solution:
The given equation of the ellipse can be written in standard form as
$\frac{4 x^2}{36}+\frac{9 y^2}{36}=\frac{36}{36} \Rightarrow \frac{x^2}{9}+\frac{y^2}{4}=1$
Since the denominator of $x^2>$ denominator of $y^2$, the major axis is along the $x$-axis. Comparing the given equation with the standard equation
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
We have $a=3$ and $b=2$.
Also, $c=\sqrt{a^2-b^2}=\sqrt{9-4}=\sqrt{5}$
and $e=\frac{c}{a}=\frac{\sqrt{5}}{3}$
Hence, the foci, are $(\sqrt{5}, 0)$ and $(-\sqrt{5}, 0)$. vertices, are $(3,0)$ and $(-3,0)$; the length of the major axis, $2 a$ is 6 units, the length of the minor axis, $2 b$ is 4 units and the eccentricity of the ellipse, $e$ is $\frac{\sqrt{5}}{3}$.
The length of latus rectum
$=\frac{2 b^2}{a}=\frac{2\left(2^2\right)}{3}=\frac{8}{3} \text { units. }$

Topic: Mathematics, Conic Sections
Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.
$\frac{x^2}{4}+\frac{y^2}{25}=1$

Options:

None

Solution:
Since the denominator of $y^2$ is larger than the denominator of $x^2$, the major axis is along the $y$-axis. Comparing the given equation with the standard equation
$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$
We have $b=2$ and $a=5$
Also $c=\sqrt{a^2-b^2}=\sqrt{25-4}=\sqrt{21}$ and $e=\frac{c}{a}=\frac{\sqrt{21}}{5}$
Hence, the foci, $(0, c)$ and $(0,-c)$ are $(0, \sqrt{21})$ and $(0,-\sqrt{21})$; the vertices are $(0,5)$ and $(0,-5)$; length of the major axis, $2 a$ is 10 units, the length of the minor axis, $2 b$ is 4 units and the eccentricity of the ellipse, $e$ is $\frac{\sqrt{21}}{5}$. The length of latus rectum $=\frac{2 b^2}{a}=\frac{8}{5}$ units.

Topic: Mathematics, Conic Sections
Find the current through the primary coil \((P)\) of the transformer shown below.

Options:
A \(0.08 \mathrm{~A}\)
B \(0.04 \mathrm{~A}\)
C \(0.02 \mathrm{~A}\)
D \(0.01 \mathrm{~A}\)

The correct answer is:
\(0.02 \mathrm{~A}\)

Solution:
The given situation is shown in the figure
For the given transformer,
\(\begin{aligned}
& V_p=230 \mathrm{~V} \\
& V_S=23 \mathrm{~V} \\
& R_S=115 \Omega
\end{aligned}\)
Currrent in secondary coil,
\(I_S=\frac{V_S}{R_S}=\frac{23}{115}=0.2 \mathrm{~A}\)
We know that, in a transformer
\(\begin{aligned}
\frac{V_S}{V_P} & =\frac{I_P}{I_S} \\
\Rightarrow \quad & I_P=\frac{V_S I_S}{V_P}=\frac{23 \times 0.2}{230}=0.02 \mathrm{~A}
\end{aligned}\)

Topic: Physics, Alternating Current, AP EAMCET, AP EAMCET 2020 (21 Sep Shift 1)
Find the equation of the parabola which passes through (6,-2), has its vertex at the origin and its axis along the y-axis.

Options:
A y2=18x
B x2=18y
C y2=−18x
D x2=−18y

The correct answer is:
x2=−18y

Solution:
As given, the vertex of the parabola is origin and it's axis is along the y-axis,

So, the standard equation of the parabola with given conditions is, x2=-4ay ...1, as it passes through 6, -2

Hence, x2=-4ay ⇒ 62=-4×a×-2 ⇒ a=368=92

So, substituting the value of a in equation 1

The equation is x2=-18y

Topic: Mathematics, Parabola, AP EAMCET, AP EAMCET 2021 (19 Aug Shift 2)
Find the INCORRECT statement about human reproduction.

Options:
A Hormonal contraceptives are effective with lesser side effects and well accepted by females
B Primary oocyte completes meiosis-I when sperm enters the cytoplasm of the primary oocyte
C Spermatogenesis and oogenesis require different temperature
D Spermatogenesis takes place in seminiferous tubules of testes

The correct answer is:
Primary oocyte completes meiosis-I when sperm enters the cytoplasm of the primary oocyte

Solution:
The primary oocyte completes the first reduction division in the tertiary follicle and results in the formation of the secondary oocyte. The completion of the meiotic division in the secondary oocyte takes place when the sperm enters the cytoplasm of the ovum. The second meiotic division which is unequal results in the formation of a second polar body and an ootid(haploid ovum).

Topic: Biology, Human Reproduction, NEET
Find the mean and standard deviation for the following data:

Options:

None

Solution:
Calculation for Standard Deviation
Mean $=A+\frac{\Sigma f_i d_i}{N}=98+\frac{44}{22}=100$
Standard Deviation $=\sqrt{\frac{\Sigma f_i d_i^2}{N}-\left(\frac{\Sigma f_i d_i}{N}\right)^2}$
$\begin{aligned}
&=\sqrt{\frac{728}{22}-\left(\frac{44}{22}\right)^2}=\sqrt{33.0909-4} \\
&=\sqrt{29.0909}=5.09
\end{aligned}$
Hence $\bar{x}=100$, S.D. $=5.09$.

Topic: Mathematics, Statistics
Find the mean and standard deviation for the following data :

Options:

None

Solution:
Calculation for Mean and Standard Deviation
$\text { Mean }=\frac{\Sigma f_i x_i}{\Sigma f_i}=\frac{760}{40}=19$
$\begin{aligned}
\text { S.D. } &=\sqrt{\frac{\Sigma f_i x_i^2}{\Sigma f_i}-\left(\frac{\Sigma f_i x_i}{\Sigma f_i}\right)^2}=\sqrt{\frac{16176}{40}-\left(\frac{760}{40}\right)^2} \\
&=\sqrt{404.4-361}=\sqrt{43.4}=6.59 .
\end{aligned}$
Hence, Mean $=19$, S.D. $=6.59$.

Topic: Mathematics, Statistics
Find the mean and standard deviation using shortcut method

Options:

None

Solution:
Let $y_i=\frac{x_i-\mathrm{A}}{i}=\frac{x_i-64}{1}$
$[\because i=1$ and $\mathrm{A}$ is assumed to be 64$]$
$\operatorname{Mean} \bar{x}=\mathrm{A}+i \bar{y}=64+\frac{\sum f_i y_i}{\sum f_i}=64$
Variance, $\sigma^2=\frac{1}{\sum f_i}\left[\sum f_i y_i^2-n \bar{y}^2\right]$
$=\frac{1}{100}[286]=2.86$
$\therefore \quad \sigma=\sqrt{2.86}=1.69$

Topic: Mathematics, Statistics
Find the mean, variance and standard deviation using short cut method.

Options:

None

Solution:
$\begin{aligned} y_i &=\frac{x_i-92.5}{5} \quad \because h=5, A=92.5 \\ \therefore \bar{x} &=A+\frac{\sum f_i y_i}{\sum f_i} \times h \\ &=92.5+\frac{6}{60} \times 5=93 \end{aligned}$
$\begin{aligned}
&\text { Variance }=\sigma^2=\frac{h^2}{N^2}\left[N \sum f_i y_i^2-\left(\sum f_i y_i\right)^2\right] \\
&=\frac{25}{3600}\left[60 \times 254-(6)^2\right]=105.58
\end{aligned}$
Standard deviation $\sigma=\sqrt{105.58}=10.27$

Topic: Mathematics, Statistics
Find the standard deviation of the first $n$ natural numbers.

Options:

None

Solution:
We have $x=\mathrm{i}$, where $\mathrm{i}=1,2, \ldots ., \mathrm{n}$. Let $x$ be the mean and $\sigma$ be the S.D. Then,
$\overline{\mathrm{x}}=\frac{1}{\mathrm{n}} \sum_{\mathrm{i}=1}^{\mathrm{n}} \mathrm{x}_{\mathrm{i}}=\frac{1}{\mathrm{n}} \sum_{\mathrm{i}-1}^{\mathrm{n}}$ (i)
$=\frac{1}{n}(1+2+3+\ldots .+n)=\frac{n(n+1)}{2 n}=\frac{n+1}{2}$
$=$ Again, variance $\sigma^2=\frac{1}{n}\left(\sum_{\mathrm{i}=1}^{\mathrm{n}} \mathrm{x}_{\mathrm{i}}^2\right)-\left(\frac{1}{\mathrm{n}} \sum_{\mathrm{i}=1}^{\mathrm{n}} \mathrm{x}_{\mathrm{i}}\right)^2$
$=\frac{1}{n}\left(1^2+2^2+\ldots n^2\right)-\left(\frac{n+1}{2}\right)^2$
$=\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{(\mathrm{n})(6)}-\frac{(\mathrm{n}+1)^2}{4}$
$=\frac{(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}-\frac{(\mathrm{n}+1)^2}{4}$
$=\frac{2\left(2 n^2+3 n+1\right)-3\left(n^2+2 n+1\right)}{12}=\frac{n^2-1}{12}$
$$
\begin{aligned}
&\Rightarrow \text { Variance }=\frac{\mathrm{n}^2-1}{12} \\
&\Rightarrow \text { S.D. }=\sqrt{\frac{\mathrm{n}^2-1}{12}}
\end{aligned}
$$

Topic: Mathematics, Statistics
Find the time period of mass $M$ when displaced from its equilibrium position and then released for the system shown in figure.

Options:

None

Solution:
When mass $(M)$ is pulled and released then the mass $(M)$ oscillates up-down along with pulley.
In this situation we will neglect gravity because it is constant throughout and will not effect the net restoring force.
Let in the equilibrium position, the spring has extended by an amount $x_0$.
Now, if the mass is given a further displacement downwards by an amount $x$. The string and spring both should increase in length by $x$.
But, string is inextensible, hence the spring alone will contribute the total extension $x+x=2 x$, to lower the mass down by $x$ from initial equilibrium mean position $x_0$. So, total extension in spring $h=\left(2 x+x_0\right)$
When pulled downward by $x, T=k\left(x_0+2 x\right)$
The extension and compression of string from initial postion is larger and smaller respectively.
Now force on the mass before bulling (in the $x_0$ extension case)
$$
F=2 T
$$
When no pulling, $T=k x_0$
(where $k$ is spring constant)
$F=2 T \Rightarrow F=2 k x_0$
When the mass is lowered down further by $x$,
$$
F^{\prime}=2 T^{\prime} \quad \because T^{\prime}=k\left(x_0+2 x\right)
$$
So, $F^{\prime}=2 k\left(2 x+x_0\right)$
Restoring force on the system.
$$
F_{\text {restoring }}=-\left[F^{\prime}-F\right]
$$
From Eqs. (i) and (ii), then,
$$
\begin{aligned}
&\begin{aligned}
F_{\text {restoring }} &=-\left[2 k\left(2 x+x_0\right)-2 k x_0\right] \\
&=-\left[2 \times 2 k x+2 k x_0-2 k x_0\right]
\end{aligned} \\
&\begin{aligned}
F_{\text {restoring }} &=-4 k x \\
\text { or } M a &=-4 k x \quad \quad\left(\text { As, } F_{\text {res. }}=m a\right)
\end{aligned}
\end{aligned}
$$
The Acceleration is, $a=-\left(\frac{4 k}{M}\right) x$
$k, M$ being constant.
$$
\therefore a \propto-x
$$
Hence, motion is SHM.
Comparing the above acceleration expression with standard SHM equation $a=-\omega^2 x$, we get
$$
\begin{aligned}
&\omega^2=\frac{+4 k}{M} \Rightarrow \omega=\sqrt{\frac{4 k}{M}} \\
&\therefore \text { Time period } T=\frac{2 \pi}{\omega}=\frac{2 \pi}{\sqrt{\frac{4 k}{M}}} \\
&=2 \pi \sqrt{\frac{M}{4 k}} \\
&T=\pi \sqrt{\frac{M}{k}}
\end{aligned}
$$

Topic: Physics, Oscillations
Find the value of ' \(k\) ', if
\(\left|\begin{array}{ccc}
k-2 & 2 k-3 & 3 k-4 \\
k-4 & 2 k-9 & 3 k-16 \\
k-8 & 2 k-27 & 3 k-64
\end{array}\right|=0\)

Options:
A 1
B 2
C 3
D 4

The correct answer is:
4

Solution:
It is given that,
\(\left|\begin{array}{ccc}
k-2 & 2 k-3 & 3 k-4 \\
k-4 & 2 k-9 & 3 k-16 \\
k-8 & 2 k-27 & 3 k-64
\end{array}\right|=0\)
On applying, \(R_2 \longrightarrow R_2-R_1\) and \(R_3 \longrightarrow R_3-R_1\), we get
\(\begin{aligned}
& \Rightarrow \quad\left|\begin{array}{ccc}
k-2 & 2 k-3 & 3 k-4 \\
-2 & -6 & -12 \\
-6 & -24 & -60
\end{array}\right|=0 \\
& \Rightarrow \quad\left|\begin{array}{ccc}
k-2 & 2 k-3 & 3 k-4 \\
1 & 3 & 6 \\
1 & 4 & 10
\end{array}\right|=0 \\
& \Rightarrow \quad(k-2)[30-24]-(2 k-3)[10-6]+(3 k-4)[4-3]=0 \\
& \Rightarrow \quad 6 k-12-8 k+12+3 k-4=0 \Rightarrow k=4
\end{aligned}\)

Topic: Mathematics, Determinants, AP EAMCET, AP EAMCET 2020 (18 Sep Shift 1)
Find the value of $k$ in which the system of equations $k x+2 y=5$ and $3 x+y=1$ has no solution?

Options:
A 0
B 3
C 6
D 15

The correct answer is:
6

Solution:
We know,
System of a pair of linear equations in two variables are given as
$a_{1} x+b_{1} y+c_{1}=0$ ...(i)
and $a_{2} x+b_{2} y+c_{2}=0$ ...(ii)
This system has no solution if
$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$
Now, on comparing given equations with (i) and (ii), we get
$a_{1}=k, a_{2}=3, b_{1}=2, b_{2}=1, c_{1}=-5, c_{2}=-1$
For no solution,
$\frac{k}{3}=\frac{2}{1} \Rightarrow k=6$

Topic: Mathematics, Determinants, NDA, NDA 2011 (Phase 1)
Find the value of the equilibrium constant $(K)$ of a reaction at $300 \mathrm{~K}$, when standard Gibbs free energy change is $-25 \mathrm{~kJ} \mathrm{~mol}^{-1}$ ? (Consider $R=8.33 \mathrm{Jmol}^{-1} \mathrm{~K}^{-1}$ )

Options:
A $e^8$
B $e^9$
C $e^{10}$
D $e^{11}$

The correct answer is:
$e^{10}$

Solution:
Given,
Gibbs free energy $=-25 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Temperature $=300 \mathrm{~K}$
Equilibrium constant $(K)=$ ?
$\Delta G^{\circ}=-R T \ln K$
$-25 \mathrm{~kJ} \mathrm{~mol}^{-1}=\left(-8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \times 300 \mathrm{~K}\right) \ln k$
$\ln K=\left(-\frac{25 \mathrm{~kJ} \mathrm{~mol}^{-1}}{8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \times 300 \mathrm{~K}}\right)$
$=\frac{25 \times 10^3 \mathrm{~J} \mathrm{~mol}^{-1}}{8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \times 300 \mathrm{~K}}$
$\ln K=-10.02, K=e^{10.02}$

Topic: Chemistry, Thermodynamics (C), TS EAMCET, TS EAMCET 2020 (11 Sep Shift 1)
Find the values of $p$ so that the lines $\frac{1-x}{3}=\frac{7 y-14}{2 p}=\frac{z-3}{2}$ and $\frac{7-7 x}{3 p}=\frac{y-5}{1}=\frac{6-z}{5}$ are at right angles.

Options:

None

Solution:
The given equation are not in the standard form. The equation of the given lines
$$
\frac{x-1}{-3}=\frac{y-2}{2 p / 7}=\frac{z-3}{2}
$$
and $\frac{x-1}{3 p /-7}=\frac{y-5}{1}=\frac{z-6}{-5}$
The direction ratios of the given lines are $-3, \frac{2 \mathrm{P}}{7}, 2$ and $\frac{-3 \mathrm{P}}{7}, 1,-5$.
The lines are perpendicualr to each other
$$
\therefore(-3)\left(\frac{-3 \mathrm{P}}{7}\right)+\left(\frac{2 \mathrm{P}}{7}\right)(1)+2(-5)=0 \Rightarrow \mathrm{P}=\frac{70}{11}
$$

Topic: Mathematics, Three Dimensional Geometry
Fishes in eutrophic lake die due to

Options:
A lack of oxygen
B nutrient enrichment
C $\mathrm{CO}_2$
D none of these.

The correct answer is:
lack of oxygen

Solution:
Eutrophic lakes have high levels of biological productivity due to the rich nutrient constituents, especially nitrogen and phosphorus. An abundance of plants are supported in this lake. Initially, eutrophic lakes accelerate multiplication and growth of lake fauna. Plants and/ or algal blooms overcrowd the lake, they later die and decomposition of dead organic matter starts which leads to decrease in dissolved oxygen content. Decrease in dissolved oxygen suffocats lake fauna, e.g., fishes to death.

Topic: Biology, Strategies for Enhancement in Food Production, AIIMS, AIIMS 2018 (27 May)
Five coins whose faces are marked 2,3 are thrown. What is the probability of obtainining a total of $12 ?$

Options:
A $\frac{1}{16}$
B $\frac{3}{16}$
C $\frac{5}{16} \quad$
D $\frac{7}{16}$

The correct answer is:
$\frac{5}{16} \quad$

Solution:
Let $\mathrm{E}$ be the event of total of 12 . $\mathrm{E}=(2,2,2,3,3),(2,2,3,3,2),(2,3,3,2,2)$,
$(3,3,2,2,2),(3,2,3,2,2),(3,2,2,3,2)$,
$(3,2,2,2,3),(2,3,2,3,2),(2,3,2,2,3)$,
$(2,2,3,2,3)$
$\mathrm{n}(\mathrm{E})=10$
Sample sapce contain total possibility $=2^{5}=32$ Hence, $n(s)=32$
So, $\mathrm{P}(\mathrm{E})=\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{S})}=\frac{10}{32}=\frac{5}{16}$

Topic: Mathematics, Probability, NDA, NDA 2006 (Phase 1)
Five sticks of length $1,3,5,7$ and 9 feet are given. Three of these sticks are selected at random. What is the probability that the selected sticks can from a triangle?

Options:
A $0.5$
B $0.4$
C $0.3$
D 0

The correct answer is:
$0.3$

Solution:
$3$ sticks can be selected from 5 sticks in ${ }^{5} \mathrm{c}_{3}$ ways
${ }^{5} \mathrm{c}_{3}=10$.
Probability that selected sticks from a triangle is
$\mathrm{n}(\mathrm{E})={ }^{4} \mathrm{c}_{3}-1=\frac{4 \times 3 \times 2 \times 1}{3 \times 2 \times 1}-1=4-1=3$
$\therefore \mathrm{p}(\mathrm{E})=\frac{3}{10}=0.3$

Topic: Mathematics, Probability, NDA, NDA 2017 (Phase 2)
Flocs produced in the secondary treatment plant of the sewage comprises of,

Options:
A Algae and Fungi
B Only Algae
C Viruses
D Bacteria and fungi

The correct answer is:
Bacteria and fungi

Solution:
The primary effluent was passed into large aeration tanks with constant agitation and the air was pumped into it. This allowed the vigorous growth of useful aerobic microbes like bacteria and fungi to form mesh-like flocs. The microbes consumes a significant part of the organic matter in the effluent and reduced its BOD.

Topic: Biology, Microbes in Human Welfare, NEET
Flocs, which are used in the secondary treatment of sewage, are:

Options:
A masses of bacteria in symbiosis with algae
B foliose lichens
C masses of fungi associated with algae
D masses of bacteria associated with fungal filaments forming mesh like structures

The correct answer is:
masses of bacteria associated with fungal filaments forming mesh like structures

Solution:
Flocs are aerobic masses of bacteria associated with fungal filaments forming mesh like structures. These microbes consume the major part of the organic matter in the effluent reducing its biological oxygen demand.

Topic: Biology, Microbes in Human Welfare, NEET
Flower in angiosperms:

Options:
A Is a modified reproductive shoot
B Possess different floral appendages at successive nodes
C Have floral appendages which are modified leaves
D More than one option is correct

The correct answer is:
More than one option is correct

Solution:
Flower is defined as modified dwarf shoot which is meant for sexual reproduction. It is characteristic feature of angiosperm.
The flowers are termed pedicellate if they possess stalks and sessile if they lack them. The flower may be described as complete if it bears all the floral parts and incomplete, when one or more floral parts are absent. Flowers are called bisexual if they bear both androecium and gynoecium. The unisexual flowers have either androecium or gynoecium. The unisexual flowers may be male flowers or female flowers. The male flower are also called staminate flowers as they have stamens only.
The female flowers have only the carpels and hence called pistillate flowers. Flowers with sterile sex organs are described as neutral flowers. According to the distribution of male, female and bisexual flowers, various patterns are recognized.

Topic: Biology, Morphology of Flowering Plants, NEET
Folloming table gives the mean and variance of monthly demand for four products $A, B, C$ and $D$ in a supermarket \begin{array}{|l|c|c|c|c|}
\hline \multicolumn{1}{|c|} { Product } & A & B & C & D \\
\hline Mcan demand & 60 & 90 & 80 & 120 \\
\hline Variance & 12 & 25 & 36 & 16 \\
\hline For which nmduct the demand is cnncictent? & 77000-II
\end{array}

Options:
A Product $A$
B Product $B$
C Product $f$
D Product $D$

The correct answer is:
Product $D$

Solution:
To find the consistent demand we will calculate coeff of variance.
We know coefficient of variance $=\frac{\sqrt{\mathrm{SD}}}{\text { mean }}$
Also, we know, $\mathrm{S.D}=\sqrt{\text { variance }}$
Coefficient of variance of $A=\frac{\sqrt{12}}{60}=\frac{3.46}{60}=0.057$
Coefficient of variance of $B=\frac{\sqrt{25}}{90}=\frac{5}{90}=0.055$
Coefficient of variance of $C=\frac{\sqrt{36}}{80}=\frac{6}{80}=0.075$
Coefficient of variance of $D=\frac{\sqrt{16}}{120}=\frac{4}{120}=0.033$ We see that minimum coefficient of variance is of $D$. hence product $D$ is consistent.

Topic: Mathematics, Statistics, NDA, NDA 2009 (Phase 2)
Following are the differences between monocot and dicot stem. Which ones are correct?

Property	Monocot stem	Dicot stem

(i)	Hypodermis	Sclerenchymatous	Collenchymatous

(ii)	Xylem	Exarch	Endarch

(iii)	Pith	Absent	Present

(iv)	Bundle Sheath	Sclerenchymatous	Parenchymatous

Options:
A i and ii
B iii and iv
C i and iii
D ii and iv

The correct answer is:
i and iii

Solution:
In dicot stem, hypodermis consists of a few layers of collenchymatous cells just below the epidermis, which provide mechanical strength to the young stem. The monocot stem has a sclerenchymatous hypodermis.
In stems (monocots and dicots), the protoxylem lies towards the centre (pith) and the metaxylem lies towards the periphery of the organ. This type of primary xylem is called endarch. In roots (monocots and dicots), the protoxylem lies towards periphery and metaxylem lies towards the centre. Such arrangement of primary xylem is called exarch. In dicot stem, a large number of vascular bundles are arranged in a ring. It also has a large number of rounded, parenchymatous cells with large intercellular spaces which occupy the central portion of the stem which constitute the pith. Moncot stems have a large number of scattered vascular bundles, hence pith is absent. In monocot stem each vascular bundle is surrounded by a sclerenchymatous bundle sheath. Bundle sheaths are absent in dicot stem

Topic: Biology, Anatomy of Flowering Plants, NEET
Following data shows the information about marks obtained in Physics, Chemistry, Mathematics and Biology by 100 students in a class. Then
subject shows the highest variability in marks

Options:
A Mathematics
B Chemistry
C Biology
D Physics

The correct answer is:
Biology

Solution:
$\begin{aligned} & \text { We know that Coefficient of Variation }=\frac{\text { Standard Deviation }}{\text { mean }} \\ & \therefore \quad(\text { C.V. })_{\text {Physics }}=\frac{3}{20}=0.15 \text { and }(\text { C.V. })_{\text {Chemistry }}=\frac{2}{25}=0.08 \\ & \left.\text { (C.V. })_{\text {Mathematics }}=\frac{4}{23}=0.174 \text { and (C.V. }\right)_{\text {Biology }}=\frac{5}{27}=0.185\end{aligned}$

Topic: Mathematics, Statistics, MHT CET, MHT CET 2021 (22 Sep Shift 2)
Following is the frequency distribution oflife length in hours of 100 electric bulbs:
\begin{array}{|c|c|c|c|c|c|c|}
\hline Life length of bulbs (in hrs) & 8.5-13.5 & 13.5-18.5 & 18.5-23.5 & 23.5-28.5 & 28.5-33.5 & 33.5-38.5 \\
\hline No. of bulbs & 7 & \mathrm{x} & 40 & \mathrm{y} & 10 & 2 \\
\hline
\end{array} If the median of life length is 20 hours, then what are the missing frequencies $(\mathrm{x}, \mathrm{y})$ ?

Options:
A $(28,13)$
B $(23,18)$
C $(31,10)$
D $(25,16)$

The correct answer is:
$(31,10)$

Solution:
Let $x=31$ and $y=10$ C.I. $\quad \mathrm{x} \quad \mathrm{f} \quad \mathrm{cf}$
$8.5-13.5$
$\begin{array}{lll}\mathrm{x} & \mathrm{f} \\ 11 & 7\end{array}$
$\begin{array}{rrrr}13.5-18.5 & 16 & 31 & 38 \\ 18.5-23.5 & 21 & 40 & 78\end{array}$
$\begin{array}{llll}23.5-28.5 & 26 & 10 & 88\end{array}$
$\begin{array}{llll}28.5-33.5 & 31 & 10 & 98\end{array}$
$\begin{array}{llll}33.5-38.5 & 36 & 2 & 100\end{array}$
$\mathrm{N}=100, \therefore \frac{\mathrm{N}}{2}=50$
Median group is $18.5-23.5$
$\mathrm{~L}_{1}=18.5, \mathrm{~L}_{2}=23.5, \mathrm{C}=38, \mathrm{~h}=5, \mathrm{f}=40$
Median $=\mathrm{L}_{1}+\frac{\left(\frac{\mathrm{N}}{2}-\mathrm{C}\right)}{\mathrm{f}} \times \mathrm{h}$
$=18.5+\frac{50-38}{40} \times 5=18.5+\frac{12 \times 5}{40}=18.5+1.5=20$
Thus, our assumption is correct. Therefore missing numbers are 31 and 10 respectively.

Topic: Mathematics, Statistics, NDA, NDA 2006 (Phase 1)
Following lists contain reactions and their corresponding equilibrium constants at different temperatures:

List - I (Reaction)	List - II Kp

2SO2( g)+O2( g)⇌2SO3( g) at 298 K	4.0×1024

2SO2( g)+O2( g)⇌2SO3( g) at 700 K	3.0×104

N2O4( g)⇌2NO2( g) at 298 K	0.98

N2O4( g)⇌2NO2( g) at 500 K	1700

If ∆H10 and ∆H20 are the standard enthalpies for the reactions 2SO2 + O2 ⇌ 2SO3 and N2O4 ⇌ 2NO2 respectively, then:

Options:
A Both ∆H10 and ∆H20 are negative
B Both ∆H10 and ∆H20 are positive
C ∆H10 is negative and ∆H20 is positive
D ∆H10 is positive and ∆H20 is negative

The correct answer is:
∆H10 is negative and ∆H20 is positive

Solution:
From thermodynamics of equilibria, we have:

ln Kep = ln A - ∆H0RT

For endothermic reaction, the equilibrium constant increases with temperature while for the exothermic reaction, the equilibrium constant decreases when temperature is raised.

Topic: Chemistry, Thermodynamics (C), TS EAMCET, TS EAMCET 2019 (03 May Shift 2)
Following process is known as

hv↔e++e-

Options:
A Pair production
B Photoelectric effect
C Compton effect
D Zeeman effect

The correct answer is:
Pair production

Solution:
Pair production refers to the creation of an elementary particle and its anti-particle usually from a photon (or another neutral boson). This is allowed, provided these is enough energy available to created the pair.

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE &
NEET with chapter-wise PYQs, revision notes, formula sheets,
custom tests & much more.

Use on iOS or Desktop Download from Google Play

Topic: Physics, Nuclear Physics, NEET