datavorous/entrance-exam-dataset huggingface dataset:
The points $(1,3,4),(-1,6,10),(-7,4,7)$ and $(-5,1,1)$ are the vertices of a

Options:
A rhombus
B rectangle
C parallelogram
D square

The correct answer is:
rhombus

Solution:
Let $(1,3,4),(-1,6,10),(-7,4,7)$ and $(-5,1,1)$ be the
coordinates of points $A, B, C$ and $D$ respectively.
$\begin{aligned} A B &=\sqrt{(-1-1)^{2}+(6-3)^{2}+(10-4)^{2}} \\ &=\sqrt{4+9+36}=7 \\ B C &=\sqrt{(-7+1)^{2}+(4-6)^{2}+(7-10)^{2}} \\ &=\sqrt{36+4+9}=7 \\ C D &=\sqrt{(-5+7)^{2}+(1-4)^{2}+(1-7)^{2}} \\ &=\sqrt{4+9+36}=7 \\ D A &=\sqrt{(1+5)^{2}+(3-1)^{2}+(4-1)^{2}} \\ &=\sqrt{36+4+9}=7 \\ A C &=\sqrt{(-7-1)^{2}+(4-3)^{2}+(7-4)^{2}} \\ &=\sqrt{64+1+9}=\sqrt{74} \end{aligned}$
and $B D=\sqrt{(-5+1)^{2}+(1-6)^{2}+(1-10)^{2}}$
$=\sqrt{16+25+81}=\sqrt{122}$
$\therefore A B=B C=C D=D A$
But $B D \neq A C$
\therefore Points $A$, $B, C$ and $D$ are the vertices of a rhombus

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2009 (Phase 1)
The points $(1,3)$ and $(5,1)$ are two opposite vertices of a rectangle. The other two vertices lie on the line $\mathrm{y}=2 \mathrm{x}+\mathrm{c}$. What is the value of c?

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

The correct answer is:
$-4$

Solution:
Given, opposite vertices of rectangle are $\mathrm{A}(1,3)$ and $\mathrm{C}$ $(5,1)$
We know, diagonals of rectangle bisect each other. So, midpoint of AC lies on line $\mathrm{y}=2 \mathrm{x}+\mathrm{c}$.
Mid point of $\mathrm{AC}=\left(\frac{1+5}{2}, \frac{3+1}{2}\right)=\left(\frac{6}{2}, \frac{4}{2}\right)=(3,2)$
$\mathrm{y}=2 \mathrm{x}+\mathrm{c} \Rightarrow 2=2(3)+\mathrm{c}$
$\Rightarrow c=2-6=-4$

Topic: Mathematics, Straight Lines, NDA, NDA 2019 (Phase 1)
The points $(2,-2),(8,4),(4,6)$ and $(-1,1)$ in order are the vertices of which one of the following quadrilaterals?

Options:
A Square
B Rhombus
C Rectangle(but not square)
D Trapezium

The correct answer is:
Trapezium

Solution:
Let points be $\mathrm{A}(2,-2), \mathrm{B}(8,4), \mathrm{C}(4,6)$ and $\mathrm{D}(-1,1)$ in
order and are vertices of a quadrilateral. $\mathrm{AB}^{2}=(8-2)^{2}+(4+2)^{2}=36+36=72$
$\mathrm{BC}^{2}=(4-8)^{2}+(6-4)^{2}=16+4=20$
$\mathrm{CD}^{2}=(4+1)^{2}+(6-1)^{2}=25+25=50$
$\mathrm{AD}^{2}=(2+1)^{2}+(-2-1)^{2}=9+9=18$
Thus $\mathrm{AB} \neq \mathrm{BC} \neq \mathrm{CD} \neq \mathrm{AD}$
Hence, quadrilateral is a trapezium.

Topic: Mathematics, Straight Lines, NDA, NDA 2007 (Phase 1)
The points $(5,1),(1,-1)$ and $(11,4)$ are:

Options:
A collinear
B vertices of right angled triangle
C vertices of equilateral triangle
D vertices of an isosceles triangle

The correct answer is:
collinear

Solution:
Let $\mathrm{A}(5,1) ; \mathrm{B}(1,-1)$ and $\mathrm{C}(11,4)$ are the given points.
Slope of $A B=\frac{-1-1}{-4}=\frac{-2}{-4}=\frac{1}{2}$
Slope of $\mathrm{BC}=\frac{4-(-1)}{11-1}=\frac{5}{10}=\frac{1}{2}$
Slope of $\mathrm{AB}=$ Slope of $\mathrm{BC}$ $\Rightarrow \quad \mathrm{AB} \| \mathrm{BC}$ and $\mathrm{B}$ is a common point
$\Rightarrow$ Points, $A, B, C$ lie on a same straight line. $\Rightarrow \mathrm{A}(5,1) ; \mathrm{B}(1,-1)$ and $\mathrm{C}(11,4)$ are collinear.

Topic: Mathematics, Straight Lines, NDA, NDA 2012 (Phase 2)
The points $(a, b),(0,0),(-a,-b)$ and $\left(a b, b^{2}\right)$ are

Options:
A the vertices of a parallelogram
B the vertices of a rectangle
C the vertices of a square
D collinear

The correct answer is:
the vertices of a rectangle

Solution:
Given points, $\mathrm{A}(\mathrm{a}, \mathrm{b}), \mathrm{B}(0,0), \mathrm{C}(-\mathrm{a},-\mathrm{b}), \mathrm{D}\left(\mathrm{ab}, \mathrm{b}^{2}\right)$.
Slope of $A B=\frac{b-0}{a-0}=\frac{b}{a}$
Slope of $\mathrm{BC}=\frac{\mathrm{b}}{\mathrm{a}}$
Slope of $A C=\frac{b}{a}$
Slope of $\mathrm{BD}=\frac{\mathrm{b}}{\mathrm{a}}$.
So. the points are collinear.

Topic: Mathematics, Straight Lines, NDA, NDA 2017 (Phase 2)
The points of intersection of the perpendicular tangents drawn to the ellipse 4x2+9y2=36 lie on the curve

Options:
A x2+y2=13
B x2-y2=5
C x+y=5
D x29+y24=1

The correct answer is:
x2+y2=13

Solution:
Given:

4x2+9y2=36

⇒x29+y24=1

Comparing with the standard form x2a2+y2b2=1, we get

a2=9 & b2=4

Now, we know that the locus of two perpendicular tangents drawn to the ellipse is the director circle of the ellipse.

Now, equation of the director circle to the ellipse x2a2+y2b2=1 is given by x2+y2=a2+b2.

Therefore, required curve is

x2+y2=9+4

orx2+y2=13.

Topic: Mathematics, Ellipse, AP EAMCET, AP EAMCET 2018 (25 Apr Shift 1)
The points $\mathrm{P}(3,2,4), \mathrm{Q}(4,5,2), \mathrm{R}(5,8,0)$ and $\mathrm{S}(2,-1,6)$ are

Options:
A vertices of a rhombus which is not a square/2017-I]
B non-coplanar
C collinear
D coplanar but not collinear

The correct answer is:
collinear

Solution:
$\mathrm{P}(3,2,4), \mathrm{Q}(4,5,2), \mathrm{R}(5,8,0), \mathrm{S}(2,-1,6)$
$\mathrm{PQ}=\sqrt{(4-3)^{2}+(5-2)^{2}+(2-4)^{2}}$
$=\sqrt{1+9+4}=\sqrt{14}$
$\mathrm{QR}=\sqrt{(5-4)^{2}+(8-5)^{2}+(0-2)^{2}}$
$=\sqrt{1+9+4}=\sqrt{14}$
$\mathrm{RS}=\sqrt{(2-5)^{2}+(-1-8)^{2}+(6-0)^{2}}$
$=\sqrt{9+81+36}=\sqrt{126}=3 \sqrt{14}$
$\mathrm{PS}=\sqrt{(2-3)^{2}+(-1-2)^{2}+(6-4)^{2}}$
$=\sqrt{1+9+4}=\sqrt{14}$
\begin{array}{llll}
\multicolumn{2}{c} {\sqrt{14}} & & \sqrt{14} & \multicolumn{2}{c} {\sqrt{14}} \\
\hline & Q & R & S
\end{array}
Since, $\mathrm{PQ}+\mathrm{QR}+\mathrm{PS}=\mathrm{RS}$, points are collinear.

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2017 (Phase 1)
The points with position vectors
$10 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}, 12 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}, \hat{\mathbf{i}}+11 \hat{\mathbf{j}}$ are collinear, if the value of $a$ is

Options:
A $-8$
B 4
C 8
D 12

The correct answer is:
8

Solution:
Since, the points with position vectors $10 \hat{i}+3 \hat{j}$,
$12 \hat{i}-5 \hat{j}, a \hat{i}+11 \hat{j}$ are collinear.
$\therefore \quad\left|\begin{array}{ccc}10 & 3 & 1 \\ 12 & -5 & 1 \\ a & 11 & 1\end{array}\right|=0$
$\Rightarrow 10(-5-11)-3(12-a)+1(132+5 a)=0$
$\Rightarrow-160-36+3 a+132+5 a=0$
$\Rightarrow-196+132+8 a=0$
$\Rightarrow 8 a=64$
$\Rightarrow a=8$

Topic: Mathematics, Vector Algebra, NDA, NDA 2011 (Phase 1)
The pollutants which come directly in the air from sources are called primary pollutants. Primary pollutants are sometimes converted into secondary pollutants. Which of the following belongs to secondary air pollutants?

Options:
A
$\mathrm{CO}$

B
Hydrocarbon

C
Peroxyacetyl nitrate

D
NO

The correct answer is:
Peroxyacetyl nitrate

Solution:
Hydrocarbons present in atmosphere combine with oxygen atom produced by the photolysis of $\mathrm{NO}_2$ to form highly reactive intermediate called free radical. Free radical initiates a series of reaction.
Peroxyacetyl nitrates are formed, which can be said as secondary pollutants.

Topic: Chemistry, Environmental Chemistry
The poly dispersity index (PDI) of a polymer is ( $\bar{M}_W=$ weight average molecular mass and $\bar{M}_n=$ number average molecular mass)

Options:
A the product of $\bar{M}_n$ and $\bar{M}_w$
B the sum of $\bar{M}_n$ and $\bar{M}_w$
C the difference between $\bar{M}_w$ and $\bar{M}_n$
D the ratio between $\bar{M}_w$ and $\bar{M}_n$

The correct answer is:
the ratio between $\bar{M}_w$ and $\bar{M}_n$

Solution:
The polydispersity index (PDl) of a polymer is the ratio of $\bar{M}_w$ and $\bar{M}_n$.
$$
\mathrm{PDI}=\frac{\bar{M}_w}{\bar{M}_n}
$$
It is directly related to the standard deviation of polymer molecular weight distribution. Polydispersity index can be used as a measure of the width of the molecular mass distribution.

Topic: Chemistry, Polymers, TS EAMCET, TS EAMCET 2016
The populations of four towns $\mathrm{A}, \mathrm{B}, \mathrm{C}$ and $\mathrm{D}$ as on $2001 \mathrm{are}$ as follows
\begin{array}{ll}
\hline Town & Population \\
\hline A & 6863 \\
B & 519 \\
C & 12185 \\
D & 1755 \\
\hline
\end{array} What is the most appropriate diagram to present the above data?

Options:
A Pie diagram
B Bar chart
C Cubic chart
D Histogram

The correct answer is:
Bar chart

Solution:
Bar chart is most appropriate

Topic: Mathematics, Statistics, NDA, NDA 2008 (Phase 1)
The position of $\mathrm{Br}$ in the compound $\mathrm{CH}_3 \mathrm{CH}=\mathrm{CHC}(\mathrm{Br})$ $\left(\mathrm{CH}_3\right)_2$ can be classified as

Options:
A
allyl

B
aryl

C
vinyl

D
secondary

The correct answer is:
allyl

Solution:
Compounds in which the halogen atom is bonded to $s p^3$ hybridised carbon atom next to carbon carbon-double bond are known as allyl halides.
$$
\mathrm{CH}_3 \mathrm{CH}=\mathrm{CHC}(\mathrm{Br})\left(\mathrm{CH}_3\right)_2
$$

Topic: Chemistry, Haloalkanes and Haloarenes
The position of the point $(1,2)$ relative to the ellipse $2 x^{2}+7 y^{2}=20$ is

Options:
A outside the ellipse
B inside the ellipse but not at the focus
C on the ellipse
D at the focus

The correct answer is:
outside the ellipse

Solution:
Given ellipse, $2 \mathrm{x}^{2}+7 \mathrm{y}^{2}=20$. Given point, $(1,2)$ $2(1)^{2}+7(2)^{2}-20=2+28-20=38-20=18>0 .$
$\therefore$ Point is outside the ellipse.

Topic: Mathematics, Ellipse, NDA, NDA 2017 (Phase 2)
The position vectors of the points $A$ and $B$ are respectively $3 \hat{i}-5 \hat{j}+2 \hat{k}$ and $\hat{i}+\hat{j}-\hat{k} .$ What is the length of $A B$ ?

Options:
A 11
B 9
C 7
D 6

The correct answer is:
7

Solution:
$\quad \overrightarrow{\mathrm{AB}}=-2 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}$
$$
|\overrightarrow{\mathrm{AB}}|=\sqrt{(-2)^{2}+6^{2}+(-3)^{2}}=\sqrt{49}=7
$$

Topic: Mathematics, Vector Algebra, NDA, NDA 2013 (Phase 2)
The possible value of x if SD of the numbers 2, 3, 2x and 11 is 3.5 is equal to

Options:
A $4, \frac{5}{3}$
B $5, \frac{7}{3}$
C $3, \frac{7}{3}$
D $2, \frac{5}{3}$

The correct answer is:
$3, \frac{7}{3}$

Solution:
Given Standard deviation of 2, 3, 2x and 11 is 3.5.
$\begin{aligned} & \because \text { Mean }(\bar{x})=\frac{2+3+2 x+11}{4}=\frac{8+x}{2} \\ & \text { Now, } \sum_{i=1}^4\left(x_i-\bar{x}\right)^2\left\{2-\left(\frac{8+x}{2}\right)\right\}^2+\left\{3-\left(\frac{8+x}{2}\right)\right\}^2 \\ &+\left\{2 x-\left(\frac{8+x}{2}\right)\right\}^2+\left\{11-\left(\frac{8+x}{2}\right)\right\}^2 \\ &=\left(\frac{4+x}{2}\right)^2+\left(\frac{2+x}{2}\right)^2+\left(\frac{3 x-8}{2}\right)^2+\left(\frac{14-x}{2}\right)^2\end{aligned}$
$\begin{aligned} & \because \mathrm{SD}=\sqrt{\left[\begin{array}{l}{\left[\frac{4+x}{4}\right)^2+\left(\frac{2+x}{2}\right)^2+\left(\frac{3 x-8}{2}\right)^2} \\ +\left(\frac{14-x}{2}\right)^2\end{array}\right]} \\ & (3 \cdot 5)^2 \times 4=\frac{1}{4}\end{aligned}$
$\begin{aligned} & {\left[16+x^2+8 x+4+x^2+4 x+9 x^2+64-48 x\right.} \\ & \left.\quad+196+x^2-28 x\right] \\ & \\ & \Rightarrow 12 x^2-64 x+84=0 \\ & \Rightarrow \quad x=3,7 / 3\end{aligned}$

Topic: Mathematics, Statistics, AP EAMCET, AP EAMCET 2021 (25 Aug Shift 2)
The potential of hydrogen electrode of $\mathrm{pH}=10$ with respect to standard hydrogen electrode is

Options:
A $-0.0591 \mathrm{~V}$
B $-0.591 \mathrm{~V}$
C $0.2 \mathrm{~V}$
D 0

The correct answer is:
$-0.591 \mathrm{~V}$

Solution:
For, hydrogen electrode
$$
\mathrm{pH}=10 \quad\left[\mathrm{H}^{+}\right]=10^{-10} \mathrm{M} \quad \mathrm{H}^{+}+e^{-}=\frac{1}{2} \mathrm{H}_2
$$
$$
\begin{aligned}
& E_{\text {cell }}=E^{\circ}-\frac{0.0591}{1} \log \frac{\left[\mathrm{H}_2\right]}{\left[\mathrm{H}^{+}\right]} \\
& E^{\circ}=0 ; \log \mathrm{H}_2=0 \text { because } \\
& E_{\text {cell }}=-\frac{0.0591}{1} \log \left(\frac{1}{10^{-10}}\right)-0.0591 \times 10 \\
& E_{\text {cell }}=-0.591 \mathrm{~V}
\end{aligned}
$$

Topic: Chemistry, Electrochemistry, TS EAMCET, TS EAMCET 2018 (05 May Shift 2)
The powerhouse of cell is first discovered by

Options:
A C Benda in 1897
B Kolliker in 1850
C Claude in 1880
D Kingsburg in 1882

The correct answer is:
Kolliker in 1850

Solution:
Mitochondria which is also known as the power house of the cell was first discovered by Kolliker in 1850. It is the site of cellular respiration.

Topic: Biology, Cell - The Unit of Life, JIPMER, JIPMER 2016
The primary and secondary coils of a transformer have 50 and 1500 turns respectively. If the magnetic flux $\phi$ linked with the primary coil is given by $\phi=\phi_0+4 t$, where $\phi$ is in weber, $t$ is time in second and $\phi_0$ is a constant, the output voltage across the secondary coil is

Options:
A 90 V
B 120 V
C 220 V
D 30 V

The correct answer is:
120 V

Solution:
The magnetic flux linked with the primary $c$ is given by
$\phi=\phi_0+4 t$
So, voltage across primary
$\begin{aligned}
V_p & =\frac{d \phi}{d t}=\frac{d}{d t}\left(\phi_0+4 t\right) \\
& =4 \mathrm{volt} \quad\left(\text { as } \phi_0=\right.\text { consta }
\end{aligned}$
Also, we have
$N_p=50 \text { and } N_s=1500$
From relation,
or
$\begin{aligned}
\frac{V_s}{V_p} & =\frac{N_s}{N_p} \\
V_s & =V_p \frac{N_s}{N_p} \\
& =4\left(\frac{1500}{50}\right) \\
& =120 \mathrm{~V}
\end{aligned}$

Topic: Physics, Alternating Current, JIPMER, JIPMER 2009
The primary and secondary coils of a transformer have 50 and 1500 turns respectively. If the magnetic flux $\phi$ linked with the primary coil is given by $\phi=\phi_v+4 t$, where $\phi$ is in webers, $t$ is time in seconds and $\phi_v$ is a constant, the output voltage across the output voltage across the secondary coil is:

Options:
A 120 volts
B 220 volts
C 30 volts
D 90 volts.

The correct answer is:
120 volts

Solution:
The magnetic flux linked with the primary coil is given by
\(\phi=\phi_0+4 t\)
So, voltage across primary
\(\begin{aligned}
& V_p=\frac{d \phi}{d t}=\frac{d}{d t}\left(\phi_0+4 t\right) \\
& \left.=4 \text { volt (as } \phi_0=\text { constant }\right)
\end{aligned}\)
Also, we have
\(N_p=50 \text { and } N_s=1500\)
From relation,
\(\begin{aligned}
& \frac{V_s}{V_p}=\frac{N_s}{N_p} \\
& \text {or } V_s=V_p \frac{N_s}{N_p} \\
& =4\left(\frac{1500}{50}\right) \\
& =120 \mathrm{~V}
\end{aligned}\)

Topic: Physics, Alternating Current, NEET, NEET 2007
The primary and secondary voltage of an ideal step-down transformer is $200 \mathrm{~V}$ and $25 \mathrm{~V}$ respectively. The secondary is connected to a device, which draws a current of $2 \mathrm{~A}$. The current in the primary is

Options:
A $25 \mathrm{~mA}$
B $42 \mathrm{~mA}$
C $160 \mathrm{~mA}$
D $250 \mathrm{~mA}$

The correct answer is:
$250 \mathrm{~mA}$

Solution:
For a lossless transformer, with primary coil $\left(I_P, V_P\right)$ and secondary coil $\left(I_S, V_S\right)$, there has to be conservation of energy per unit time:
$\begin{aligned} & I_P V_P=I_S V_S \\ & \Rightarrow \frac{I_P}{I_S}=\frac{V_S}{V_P} \\ & \therefore \frac{I_P}{2}=\frac{25}{500} \\ & \therefore I_P=\frac{1}{4} \mathrm{~A}=250 \mathrm{~mA}\end{aligned}$

Topic: Physics, Alternating Current, MHT CET, MHT CET 2022 (10 Aug Shift 1)
The primary and the secondary coils of a transformer contain 10 and 100 turns, respectively. The primary coil is connected to a battery that supplies a constant voltage of $1.5$ volts. the voltage across the secondary coil is

Options:
A $1.5 \mathrm{~V}$
B $0.15 \mathrm{~V}$
C $0.0 \mathrm{~V}$
D $15 \mathrm{~V}$

The correct answer is:
$0.0 \mathrm{~V}$

Solution:
Since the voltage production is based upon A.C. supply and this voltage is D.C which is constant. Therefore, no flux will change in secondary and no voltage will be induced.
Answer is (C) $0 \mathrm{~V}$.

Topic: Physics, Alternating Current, KVPY, KVPY 2016 (SA)
The primary of a transformer has 100 turns and operates at \(100 \mathrm{~V}-200 \mathrm{~W}\). The number of turns in the secondary, if the output voltage is \(2000 \mathrm{~V}\), then

Options:
A 2000
B 200
C 100
D 500

The correct answer is:
2000

Solution:
For a transformer, it is given that, \(N_P=100\),
\(V_P=100 \mathrm{~V}, V_S=2000 \mathrm{~V}\)
We know that,
\(\begin{aligned}
& \frac{N_S}{N_P}=\frac{V_S}{V_P} \\
\Rightarrow \quad & N_S=\frac{V_S N_P}{V_P}=\frac{2000 \times 100}{100}=2000
\end{aligned}\)

Topic: Physics, Alternating Current, AP EAMCET, AP EAMCET 2020 (21 Sep Shift 2)
The primary of a transformer when connected to a DC battery of $10 \mathrm{~V}$ draws a current of $1 \mathrm{~mA}$. The number of turns of the primary and secondary windings are 50 and 100 , respectively. The voltage in the secondary and the current drawn by the circuit in the secondary are respectively

Options:
A $20 \mathrm{~V}$ and $2.0 \mathrm{~mA}$
B $10 \mathrm{~V}$ and $0.5 \mathrm{~mA}$
C $0$ and $0$
D $20 \mathrm{~V}$ and $0.5 \mathrm{~mA}$

The correct answer is:
$0$ and $0$

Solution:
A transformer does not work on DC supply because mutual induction does not take place due to constant value of current. So, $V_{\mathrm{S}}=0$ and $I_{\mathrm{S}}=0$.

Topic: Physics, Alternating Current, COMEDK, COMEDK 2013
The primary treatment basically involves the physical removal of particles from the sewage through

Options:
A filtration.
B sedimentation.
C microbial treatment.
D Both A and B

The correct answer is:
Both A and B

Solution:
Primary water treatment removes large pieces of floating debris, oily substances from the sewage. Secondary treatment is called bio-chemical or microbial treatment where many useful aerobic bacteria and fungus grow together to form flocs. These microbes consume major part of the organic matter in the effluent as they grow and reduces the BOD of water. BOD of pure water is 1-2 ppm and BOD of polluted water is 15-18 ppm.

Topic: Biology, Microbes in Human Welfare, NEET
The primary treatment of waste water involves the removal of

Options:
A
dissolved impurities

B
stable particles

C
toxic substances

D
harmful bacteria

The correct answer is:
harmful bacteria

Solution:
The primary treatment of waste water involves physical removal of both larger and small particles (stable particles) through filtration and sedimentation from the sewage. These are removed in stages; initially, floating debris is removed by sequential filtration.
Then by sedimentation the grit (Soil and small pebbles) are removed. All solids that settle from the secondary treatment.
The primary treatment does not remove the dissolved impurities (a), toxic substances (c) and harmful bacteria (d).

Topic: Biology, Microbes in Human Welfare
The primary winding of a transformer has 100 turns and its secondary winding has 200 turns. The primary winding is connected to an AC supply of 120 V and the current flowing in it is 10 A. The voltage and the current in the secondary winding is

Options:
A 240 V, 5 A
B 240Â V,Â 10Â A
C 60Â V,Â 20Â A
D 120Â V,Â 20Â A

The correct answer is:
240 V, 5 A

Solution:
NsNp=VsVpâ200100=Vs120âVs=240 V
also VsVp=ipisâ240120=10isâis=5 A

Topic: Physics, Alternating Current, NEET
The primary winding of transformer when connected to a $d c$ battery of 10 Volt draws a current of $1 \mathrm{~mA}$. The number of turns of the primary and secondary windings are 50 and 100 respectively. The voltage in the secondary and the current drawn by the circuit in the secondary are respectively:

Options:
A $20 \mathrm{~V}$ and $2.0 \mathrm{~mA}$
B $10 \mathrm{~V}$ and $0.5 \mathrm{~mA}$
C Zero volt and therefore no current
D $20 \mathrm{~V}$ and $0.5 \mathrm{~mA}$

The correct answer is:
Zero volt and therefore no current

Solution:
Transformer can not work on dc.

Topic: Physics, Alternating Current, NEET, NEET 2013 (Karnataka)
The principal value of $\sin ^{-1}$ x lies in the interval

Options:
A $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$
B $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
C $\left[0, \frac{\pi}{2}\right]$
D $[0, \pi]$

The correct answer is:
$\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$

Solution:
The principal value of $\sin ^{-1} x$ lies in it's range. The range of $\sin ^{-1} x$ is $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$.

Topic: Mathematics, Inverse Trigonometric Functions, NDA, NDA 2017 (Phase 2)
The probabilities of two events $A$ and $B$ are given as $P(A)=0.8$ and $P(B)=0.7$. What is the minimum value of $P(A \cap B) ?$

Options:
A 0
B 0.1
C 0.5
D 1

The correct answer is:
0.5

Solution:
As we know $P(A \cup B) \leq 1$ $\therefore P(A)+P(B)-P(A \cap B) \leq 1$
$\Rightarrow 0.8+0.7-P(A \cap B) \leq 1$
$\Rightarrow P(A \cap B) \geq 1.5-1$
$\Rightarrow P(A \cap B) \geq 0.5$
Hence, the minimum value of $P(A \cap B)$ is $0.5$.

Topic: Mathematics, Probability, NDA, NDA 2009 (Phase 2)
The probabilities that a student will solve Question $\mathrm{A}$ and Question $\mathrm{B}$ are $0.4$ and $0.5$ respectively. What is the probability that he solves at least one of the two questions?

Options:
A $0.6$
B $0.7$
C $0.8$
D $0.9$

The correct answer is:
$0.7$

Solution:
Probability of solving Question A, $\mathrm{P}(\mathrm{A})=0.4$ Probability of solving Question B, $\mathrm{P}(\mathrm{B})=0.5$ $\therefore \mathrm{P}\left(\mathrm{A}^{\prime}\right)=1-\mathrm{P}(\mathrm{A})=1-0.4=0.6$
$\mathrm{P}\left(\mathrm{B}^{\prime}\right)=1-\mathrm{P}(\mathrm{B})=1-0.5=0.5$
Probability to solve atleast one question $=\mathrm{P}(\mathrm{A} \cup \mathrm{B})$ $\mathrm{P}(\mathrm{A} \cup \mathrm{B})=1-\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\right)$
$=1-(0.6 \times 0.5)$
$=1-0.3$
$=0.7$

Topic: Mathematics, Probability, NDA, NDA 2018 (Phase 1)
The probability distribution of a discrete r. v. $X$ is \begin{array}{|c|c|c|c|c|c|}
\hline \mathrm{X}=x & 0 & 1 & 2 & 3 & 4 \\
\hline \mathrm{P}(\mathrm{X}=x) & k & 2 k & 4 \mathrm{k} & 2 k & k \\
\hline
\end{array}
then value of $\mathrm{P}(\mathrm{X} \leq 2)$ is

Options:
A $\frac{1}{10}$
B $\frac{7}{10}$
C $\frac{3}{10}$
D $\frac{9}{10}$

The correct answer is:
$\frac{7}{10}$

Solution:
Here $\mathrm{k}+2 \mathrm{k}+4 \mathrm{k}+2 \mathrm{k}+\mathrm{k}=10 \mathrm{k}=1 \Rightarrow \mathrm{k}=\frac{1}{10}$
Now $\mathrm{P}(\mathrm{X} \leq 2)=\mathrm{P}(0)+\mathrm{P}(1)+\mathrm{P}(2)=\frac{1}{10}+\frac{2}{10}+\frac{4}{10}=\frac{7}{10}$

Topic: Mathematics, Probability, MHT CET, MHT CET 2020 (14 Oct Shift 2)
The probability distribution of a random variable $X$ is given by
\begin{array}{l|r|r|r|r|c}
\hlineX=X & 0 & 1 & 2 & 3 & 4 \\
\hlineP(X=x) & 0.4 & 0.3 & 0.1 & 0.1 & 0.1 \\
\hline
\end{array}
The variance of $X$ is

Options:
A $1.76$
B $2.45$
C $3.2$
D $4.8$

The correct answer is:
$1.76$

Solution:
Given,
$\begin{aligned} \text { Mean } \bar{x} & =\sum_{i=0}^4 p_i x_i=1.2 \\ \text { Variance } & =\sum_{i=0}^4 p_i x_i^2-\bar{x}^2 \\ & =3.20-1.44=1.76\end{aligned}$

Topic: Mathematics, Probability, AP EAMCET, AP EAMCET 2007
The probability distribution of a random variable $X$ is given by
\begin{array}{l|r|r|r|r|c}
\hlineX=X & 0 & 1 & 2 & 3 & 4 \\
\hlineP(X=x) & 0.4 & 0.3 & 0.1 & 0.1 & 0.1 \\
\hline
\end{array}
The variance of $X$ is

Options:
A $1.76$
B $2.45$
C $3.2$
D $4.8$

The correct answer is:
$1.76$

Solution:
Given,
$\begin{aligned} \text { Mean } \bar{x} & =\sum_{i=0}^4 p_i x_i=1.2 \\ \text { Variance } & =\sum_{i=0}^4 p_i x_i^2-\bar{x}^2 \\ & =3.20-1.44=1.76\end{aligned}$

Topic: Mathematics, Probability, TS EAMCET, TS EAMCET 2007
The probability distribution of a random variable $\mathrm{X}$ is
Then the standard deviation of $\mathrm{X}$ is

Options:
A 1
B 2
C 4
D 7

The correct answer is:
2

Solution:
No solution. Refer to answer key.

Topic: Mathematics, Probability, AP EAMCET, AP EAMCET 2017 (26 Apr Shift 2)
The probability distribution of random variable $X$ with two missing probabilities $p_{1}$ and $p_{2}$ is given below
$\begin{array}{cc}\boldsymbol{X} & \boldsymbol{P}(\boldsymbol{X}) \\ 1 & k \\ 2 & p_{1} \\ 3 & 4 k \\ 4 & p_{2} \\ 5 & 2 k\end{array}$
It is further given that $P(X \leq 2)=0.25$ and $P(X \geq 4)=0.35$. Consider the following statements
$1.$ $p_{1}=p_{2}$
$2.$ $p_{1}+p_{2}=P(X=3)$
Which of the statements given above is/are correct?

Options:
A 1 only
B 2 only
C Both 1 and 2
D Neither 1 nor 2

The correct answer is:
Neither 1 nor 2

Solution:
Let $P(X \leq 2)=0.25$
$\Rightarrow \quad P(X=1)+P(X=2)=0.25$
$\Rightarrow k+p_{1}=0.25($ from the table $)$
$\Rightarrow p_{1}=0.25-k$
and $P(X \geq 4)=0.35$
$\Rightarrow \quad P(X=4)+P(X=5)=0.35$
$\Rightarrow p_{2}+2 k=0.35$ (from the table)
$\Rightarrow p_{2}=0.35-2 k$
From (1) and (2)
$p_{1} \neq p_{2}$
and $p_{1}+p_{2}=0.25-k+0.35-2 k=0.6-3 k$
$\neq P(X=3)$
Hence, neither 1 nor 2 is correct

Topic: Mathematics, Probability, NDA, NDA 2010 (Phase 1)