datavorous/entrance-exam-dataset huggingface dataset:
The numbers of turns in the primary and the secondary of a transformer are 1000 and 3000 respectively. If $80 \mathrm{~V}$ a.c. is applied to the primary coil of the transformer, then the potential difference per turn of the secondary coil would be

Options:
A $240 \mathrm{~V}$
B $2400 \mathrm{~V}$
C $0.24 \mathrm{~V}$
D $0.08 \mathrm{~V}$

The correct answer is:
$0.08 \mathrm{~V}$

Solution:
$\frac{E_p}{E_s}=\frac{N_p}{N_s}$
$\frac{80}{E_{\mathrm{s}}}=\frac{1000}{3000}$
$E_s=240 \mathrm{~V}$
$\therefore \quad E_{\mathrm{s}}$ per turn $=\frac{240}{3000}=0.08 \mathrm{~V}$

Topic: Physics, Alternating Current, MHT CET, MHT CET 2023 (13 May Shift 2)
The numerical value of the perimeter of a square exceeds that of its area by 4 . What is the side of the square?

Options:
A 1 unit
B 2 unit
C 3 unit
D 4 unit

The correct answer is:
2 unit

Solution:
Let the side of the square $=x$ units Area of square $=x^{2}$ unit and perimeter of square $=4 \mathrm{x}$ unit According to question, $x^{2}+4=4 x$
$\Rightarrow \quad x^{2}-4 x+4=0$
$\Rightarrow \quad(\mathrm{x}-2)^{2}=0$
$\Rightarrow \quad \mathrm{x}=2$
$\therefore \quad$ Side of square $=2$ unit

Topic: Mathematics, Straight Lines, NDA, NDA 2008 (Phase 1)
The oogonial cells start dividing and enter into __A__ of meiotic division and get temporarily arrested at that stage, called __B__. During oogenesis at puberty, this oocyte gets surrounded by granulosa cells and then called __C__.

Options:
A A–Metaphase II; B–Primary oocyte; C–Primary follicle
B A–Prophase I; B–Primary oocyte; C–Primary follicle
C A–Prophase I; B–Secondary oocyte; C–Secondary follicle
D A–Prophase II; B–Secondary oocyte; C–Secondary follicle

The correct answer is:
A–Prophase I; B–Primary oocyte; C–Primary follicle

Solution:
During embryonic development in females, the oogonia enter meiosis I and get halted in the pachytene stage of the prophase I (the most prolonged phase of the cell cycle) and are called a primary oocyte. During puberty, the halted meiosis I continue and forms the secondary oocyte and starts meiosis II that stops at metaphase II. In the course of development, the primary oocyte is surrounded by granulosa cells, which release estrogen and is called a primary follicle. During puberty, primary follicle matures and forms the Graafian follicle from which ovulation occurs.

Topic: Biology, Human Reproduction, NEET
The order and degree of the differential equation
$\left[1+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}\right]^{3}=\rho^{2}\left[\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\right]^{2} \quad$ are respectively

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

The correct answer is:
2 and 2

Solution:
Degree of a differential equation is power of the highest order derivative. In the given differential equation,
$\quad\left[1+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}\right]^{3}=\rho^{2}\left[\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\right]^{2}$

Topic: Mathematics, Differential Equations, NDA, NDA 2017 (Phase 2)
The order and degree of the differential equation of parabolas having vertex at the origin and focus at $(a, 0)$ where a $>0$, are respectively

Options:
A 1,1
B 2,1
C 1,2
D 2,2

The correct answer is:
1,1

Solution:
The eq. of parabolas having vertex at $(0,0) \&$ focus at $(a, 0)$, where $(a>0)$ is:
$$
\begin{array}{l}
y^{2}=4 a x \\
2 y \cdot \frac{d y}{d x}=4 a
\end{array}
$$
[on differentiating]
On putting the value of $(4 a)$ in eq. (1) we get,
$$
2 \mathrm{x} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{y}=0
$$
in order $=1 \&$ degree $=1$.

Topic: Mathematics, Differential Equations, NDA, NDA 2015 (Phase 2)
The order and degree of the differential equation $\mathrm{y}^{2}=4 \mathrm{a}(\mathrm{x}-\mathrm{a})$, where $\mathrm{a}$ ' is an arbitrary constant, are respectively

Options:
A 1,2
B 2,1
C 2,2
D 1,1

The correct answer is:
1,2

Solution:
No Solution

Topic: Mathematics, Differential Equations, NDA, NDA 2018 (Phase 1)
The order of a set $A$ is 3 and that of a set $B$ is $2 .$ What is the
number of relations from $A$ to B ?

Options:
A 4
B 6
C 32
D 64

The correct answer is:
6

Solution:
Since, order of a set $\mathrm{A}$ is 3 and order of set $\mathrm{B}$ is 2 therefore
$n(A)=3$ and $n(B)=2$
Number of relations from $A$ to $B$
$=n(A) \times n(B)=3 \times 2=6$

Topic: Mathematics, Sets and Relations, NDA, NDA 2010 (Phase 1)
The order of differential equation of all circles of given radius ' \( a \) ' is

Options:
A \( 04 \)
B \( 12 \)
C \( 11 \)
D \( 13 \)

The correct answer is:
\( 12 \)

Solution:
The standard equation of a circle is
\[
(x-b)^{2}+(y-k)=a^{2}
\]
There are two arbitrary constants.
So, the order of the differential equation is equal to number of arbitrary constants.
Therefore, order \( =2 \)

Topic: Mathematics, Binomial Theorem, KCET, KCET 2015
The order of organism in an aquatic food chain is

Options:
A Bacteria, Seal, Diatom, Fish, Crustacea
B Crustacea, Seal, Fish, Polar Bear, Diatom
C Polar Bear, Diatom, Seal, Crustacea
D Diatom, Crustacea, Fish, Seal, Bacteria

The correct answer is:
Diatom, Crustacea, Fish, Seal, Bacteria

Solution:
Diatoms are producer, crustacea is primary consumer, fish is secondary consumer, seal is top consumer and bacteria are decomposers.

Topic: Biology, Ecosystem, NEET
The order of reactivity of the following compounds towards dilute aqueous KOH in SN1 reaction is

CH3CH2CH2CH2BrI H3CCH2CHBrCH3II CH32CHCH2BrIII CH33CBrIV

Options:
A I<IV<III<II
B IV<II<III<I
C III<II<I<IV
D I<III<II<IV

The correct answer is:
I<III<II<IV

Solution:
The SN1 reaction rate depends on the stability of cabocation formed in the rate determining step.

CH3CH2CH2CH2+I H3CCH2CH+CH3II CH32CHCH2+III CH33C+IV

Tertiary carbocations are more stable than secondary and primary. Among the carbocations I and III, III is readily formed than I. Hence, the correct reactivity order is I<III<II<IV.

Topic: Chemistry, Haloalkanes and Haloarenes, TS EAMCET, TS EAMCET 2022 (18 Jul Shift 2)
The order of reactivity of the following compounds towards the esterification with acetic acid is

Options:
A I>II>III>IV
B IV>III>II>I
C I>IV>II>III
D I>IV>III>II

The correct answer is:
I>IV>II>III

Solution:
The chemical reaction in which alcohols and acids react with each other to form products as esters is called esterification.

Here, the organic acid group reacts with alcohol to form stable ester and water.

The mechanism of this reaction is nucleophilic acyl substitution.During this reaction the cleavage O-H bond of alcohols takes place.The reactivity for this cleavage goes down from primary to tertiary alcohol as the bulkiness of the substitutes on alcohol increases.
The alcohol acts as nucleophile here and also the steric effects support the reactivity order. Thus, tertiary alcohol is the least reactive than primary and secondary.

So, with acetic acid ethanol reacts the fastest followed by propanol and then 2-methyl propan-ol while the least reactive would be 2-3-dimethyl butan-1-ol.

Hence, the correct order of reactivity would be

Topic: Chemistry, Alcohols Phenols and Ethers, TS EAMCET, TS EAMCET 2022 (18 Jul Shift 1)
The organs involved in the reproductive system can be grouped as primary and secondary. Which of the following is not a secondary sex organ?

1 Prostate gland
2 Mammary gland
3 Seminal vesicle
4 Ovary

Options:
A Only 2
B Both 2 & 4
C Only 4
D 1, 2 & 3

The correct answer is:
Only 4

Solution:
Ovary is part of the reproductive system and is the primary organ as it is the one that produces the female gametic cell participating in the fertilization process. Other organs, prostate gland, mammary gland, seminal vesicle, come under the secondary group of reproductive organs.

Topic: Biology, Human Reproduction, NEET
The outcomes of 5 tosses of a coin are recorded in a single sequence as $\mathrm{H}$ (head) and T (tail) for each toss. What is the number of elementary events in the sample space?

Options:
A 5
B 10
C 25
D 32

The correct answer is:
10

Solution:
A coin has two faces and is tossed 5 times. So, number of elements in the sample space $=10$.

Topic: Mathematics, Probability, NDA, NDA 2008 (Phase 1)
The outermost limiting layer of mycoplasma is made up of

Options:
A cell wall
B cell membrane
C mucilaginous sheath
D slime layer.

The correct answer is:
cell membrane

Solution:
Mycoplasma are the smallest and simplest free living Gram negative prokaryotes or monerans. A cell wall is absent. Plasma membrane forms the outer boundary of the cell. A substantial amount of polysaccharides having acetyl glucosamine are associated with cell membrane.

Topic: Biology, Biological Classification, AIIMS, AIIMS 2009
The output of a step down transformer is measured to be \( 48 \mathrm{~V} \) when connected to a \( 12 \mathrm{~W} \)
bulb. The value of peak current is

Options:
A ( \( \frac{1}{\sqrt{2}} \mathrm{~A} \)
B \( \sqrt{2} A \)
C \( \frac{1}{2 \sqrt{2}} A \)
D \( \frac{1}{4} A \)

The correct answer is:
\( \frac{1}{2 \sqrt{2}} A \)

Solution:
Given, output of step down transformer is $48 \mathrm{~V}$
$\Rightarrow$ peak secondary voltage $V_{s}=48 \mathrm{v}$
Also, power at secondary $P_{s}=12 W$
Now, current at secondary, $I_{s}=\frac{P_{s}}{V_{s}}=\frac{12}{48}=0.25 \mathrm{~A}$
Thus, value of peak current is $I_{S} \sqrt{2}=0.25 \times \sqrt{2}=\frac{1}{2 \sqrt{2}} A$

Topic: Physics, Magnetic Effects of Current, KCET, KCET 2017
The output of a step-down transformer is measured to be $24 \mathrm{~V}$ when connected to a $12 \mathrm{~W}$ light bulb. The value of the peak current is

Options:
A
$1 / \sqrt{2} \mathrm{~A}$

B
$\sqrt{2} \mathrm{~A}$

C
$2 \mathrm{~A}$

D
$2 \sqrt{2} \mathrm{~A}$

The correct answer is:
$1 / \sqrt{2} \mathrm{~A}$

Solution:
As given that, Secondary voltage $\left(\mathrm{V}_{\mathrm{S}}\right)$ is :
$$
\mathrm{V}_{\mathrm{S}}=24 \text { Volt }
$$
Power associated with secondary is :
$$
\mathrm{P}_{\mathrm{S}}=12 \mathrm{Watt}
$$
As we know that $P_S=V_S I_S$
$$
\mathrm{I}_{\mathrm{S}}=\frac{\mathrm{P}_{\mathrm{S}}}{\mathrm{V}_{\mathrm{S}}}=\frac{12}{24}=\frac{1}{2} \mathrm{~A}=0.5 \mathrm{Amp}
$$
Peak value of the current in the secondary
$$
\begin{aligned}
I_0 &=I_S \sqrt{2}=0.5 \sqrt{2} \\
&=\frac{5}{10} \sqrt{2}\left[I_0=\frac{1}{\sqrt{2}} \text { Amp }\right]
\end{aligned}
$$

Topic: Physics, Alternating Current
The ovule of an angiosperm is technically equivalent to

Options:
A Megasporangium
B Megasporophyll
C Megaspore mother cell
D Megaspore

The correct answer is:
Megasporangium

Solution:
The angiosperms are heterosporous. The stamens of the flower are the male sexual parts and they are highly modified leaves called microsporophylls. The microsporophylls produce anthers or microsporangia. Within the anther, there are special diploid cells called pollen mother cell or microspore mother cells. The microspore mother cells undergo meiosis to produce haploid microspores or pollen grains. The microspore produces a few celled male gametophytes, each male gametophyte has two cells tube cell and a generative cell. Gynoecium or pistil is the female sexual parts. The gynoecium is made up of highly modified leaves called megasporophylls or carpels. The megasporophylls fused to form ovaries. The function of carpels is to produce ovules or megasporangia. Within the ovules, there are specialized cells called the megaspore mother cells (2n). The megaspores mother cells are diploid (2n) which undergo meiosis or reduction division to form haploid megaspores (n). The megaspores produce a multicellular female gametophyte or embryo sac. The mature embryo sac has a group of three antipodal cells, two synergid cells, one female gamete or egg cell. There are two haploid polar nuclei which fuse together to produce a diploid secondary nucleus.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET, NEET 2016 (Phase 2)
The part of the embryosac which helps in guiding the pollen tube towards it is __(P)___ and it is through ___(Q)____.

Options:
A P is antipodal cells and Q is thigmotaxis
B P is egg and Q is chemotaxis
C P is a secondary nucleus and Q is thigmotaxis
D P is synergids and Q is chemotaxis

The correct answer is:
P is synergids and Q is chemotaxis

Solution:
Synergids secrete a chemical to attract pollen tube towards the embryo sac. Thus, the movement of the pollen tube is under the influence of a chemical and is called chemotaxis. The pollen tube enters into the embryo sac with the help of the filiform apparatus of synergids.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
The percentage of abundance of three elements $X, Y$ and $Z$ in the earth's crust are about $27.7,8.3$ and 46.0 by mass respectively. $X, Y$ and $\dot{Z}$ are

Options:
A O, Si, Al
B Al, Si; O
C $\mathrm{Si}, \mathrm{Al}, \mathrm{O}$
D $\mathrm{Al}, \mathrm{O}, \mathrm{Si}$

The correct answer is:
$\mathrm{Si}, \mathrm{Al}, \mathrm{O}$

Solution:
Percentage abundance of elements in earth's crust are

Topic: Chemistry, p Block Elements (Group 13 & 14), AP EAMCET, AP EAMCET 2022 (07 Jul Shift 1)
The periderm is produced by:

Options:
A vascular cambium
B fascicular cambium
C phellogen
D intrafascicular cambium

The correct answer is:
phellogen

Solution:
A periderm is a group of cylindrical tissues that forms a protective layer and cover the stems and roots of many plants during the secondary growth. These tissues are usually found in perennial plants and consist of three different layers, the phelloderm (inner surface of the cork cambium), the phellogen (meristematic tissue called the cork cambium) which gives rise to the periderm and the phellem which is the suberized cells.

Topic: Biology, Anatomy of Flowering Plants, NEET
The perisperm is the

Options:
A degenerate part of synergids
B peripheral part of endosperm
C degenerate part of secondary nucleus
D remnant of nucellus

The correct answer is:
remnant of nucellus

Solution:
In some seeds such as black pepper and beet, remnants of nucellus are also persistent. This residual, persistent nucellus is the perisperm.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
The Perpendicular distance between the straight lines $6 x+8 y+15=0$ and $3 x+4 y+9=0$ is

Options:
A $\frac{3}{2}$ units
B $\frac{3}{10}$ unit
C $\frac{3}{4}$ unit
D $\frac{2}{7}$ unit

The correct answer is:
$\frac{3}{10}$ unit

Solution:
$6 \mathrm{x}+8 \mathrm{y}+15=0$...(i)
and $3 x+4 y+9=0$ ...(ii)
Multiply equation (ii) by 2 , we get $6 x+8 y+18=0$
Distance between the straight lines
$\frac{\left|c_{2}-c_{1}\right|}{\sqrt{a^{2}+b^{2}}}=\frac{18-15}{\sqrt{(6)^{2}+(8)^{2}}}=\frac{3}{10}$ unit
Option (b) is correct.

Topic: Mathematics, Straight Lines, NDA, NDA 2015 (Phase 1)
The perpendiculars that fall from any point of the straight line $2 \mathrm{x}+11 \mathrm{y}=5$ upon the two straight lines $24 \mathrm{x}+7 \mathrm{y}=20$ and $4 x-3 y=2$ are

Options:
A 12 and 4 respectively
B 11 and 5 respectively
C Equal to each other
D Not equal to each other

The correct answer is:
Equal to each other

Solution:
We know, the perpendicular distance (d) from point
$\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)$ to line $\mathrm{ax}+\mathrm{by}+\mathrm{c}=0$ is $\frac{\mid \mathrm{ax}_{1}+\mathrm{by}_{1}+\mathrm{c}}{\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}}}$
Let us find a point on line $2 x+11 y-5=0$. for $\mathrm{x}=0,2(0)+11 \mathrm{y}-5=0$
$\Rightarrow 11 \mathrm{y}-5=0$
$\Rightarrow \mathrm{y}=\frac{5}{11}$
So, $(\mathrm{x}, \mathrm{y})=\left(0, \frac{5}{11}\right)$
Let us find perpendicular distances of this point to the given lines.
$\therefore \quad d_{1}=d$

Topic: Mathematics, Straight Lines, NDA, NDA 2018 (Phase 1)
The pH at the equivalence point of a titration may differ from 7.0 because of

Options:
A the initial concentration of the standard solution
B the indicator used
C the self-ionization of H2O
D hydrolysis of the salt formed

The correct answer is:
hydrolysis of the salt formed

Solution:
The pH at the equivalence point of a titration may differ from 7.0 because of hydrolysis of the salt formed as due to salt-hydrolysis, the solution may become either acidic or alkaline at equivalence point.

Topic: Chemistry, Ionic Equilibrium, NEET
The phenomenon by which the undividing parenchyma cells start to divide mitotically during plant tissue culture is called as :

Options:
A Differentiation
B Dedifferentiation
C Redifferentiation
D Secondary growth

The correct answer is:
Dedifferentiation

Solution:
The process of conversion of a Non-dividing (permanent cell) to a dividing cell is called de-differentiation. This is an important process for secondary growth which taken place in plants.

Topic: Biology, Anatomy of Flowering Plants, NEET, NEET 2022 (Phase 2)
The phenomenon of dedifferentiation is

Options:
A regaining the capacity to divide mitotically by differentiated cells
B the event of losing the ability to divide by dedifferentiated cells
C regaining the capacity to divide meiotically by differentiated cells
D regaining the capacity to divide meiotically by redifferentiated cells

The correct answer is:
regaining the capacity to divide mitotically by differentiated cells

Solution:
The living differentiated cells that have lost the capability to divide can regain the capacity of the division under certain conditions; this phenomenon is called dedifferentiation. It occurs during secondary growth when permanent cell-like parenchyma regains the power of division to form vascular cambium and cork cambium. The redifferentiation is the event of losing the ability to divide by dedifferentiated cells.

Topic: Biology, Plant - Growth and Development, NEET
The planes $p x+2 y+2 z-3=0$ and $2 x-y+z+2=0$ intersect
at an angle $\frac{\pi}{4}$. What is the value of $p^{2}$ ?

Options:
A 24
B 12
C 6
D 3

The correct answer is:
24

Solution:
We know that the angle between the planes $a_{1} x+b_{1} y$ $+c_{1} z+d_{1}=0$ and $a_{2} x+b_{2} y+c_{2} z+d_{2}=0$ is given by
$\cos \theta=\left|\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}}\right|$
Given equation of planes are $p x+2 y+2 z-3=0$ and $2 x-y+z+2=0$
On comparing with standard equations, we get
$a_{1}=p, a_{2}=2, b_{1}=2, b_{2}=-1, \quad c_{1}=2, c_{2}=1$
Also, $\theta=\frac{\pi}{4} \quad$ (given)
$\cos \frac{\pi}{4}=\left|\frac{p \times 2+2 \times(-1)+2 \times 1}{\sqrt{p^{2}+4+4} \sqrt{4+1+1}}\right|$
$\Rightarrow \frac{1}{\sqrt{2}}=\frac{2 p}{\sqrt{p^{2}+8} \sqrt{6}} \Rightarrow \frac{1}{2}=\frac{4 p^{2}}{\left(p^{2}+8\right) 6}$
$\Rightarrow \frac{3}{4}=\frac{p^{2}}{p^{2}+8}$
$\Rightarrow 3 p^{2}+24=4 p^{2} \Rightarrow p^{2}=24$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2010 (Phase 1)
The plant part that possesses polyarch condition of vascular bundles with a well-developed pith is

Options:
A dicot root.
B monocot root.
C dicot stem.
D monocot stem.

The correct answer is:
dicot root.

Solution:
Dicot root possesses diarch to hexarch (2-6) vascular bundles. Vascular bundle refers to the xylem and phloem tissue which help in water and organic matter transport in different parts of plant. Dicot root has smaller pith and well-developed cambium tissue for secondary growth.

Topic: Biology, Anatomy of Flowering Plants, NEET
The plasma membrane is the boundary between a cell's internal and external environments. Both environments are aqueous.

To allow the separation of these aqueous environments, the regions of the plasma membrane indicated in the diagram have which of the following properties?

Options:
A (X) hydrophilic layers on the outside and (Y) hydrophobic layers on the inside of the membrane
B X and Y are both hydrophobic layers
C (X) hydrophobic layers on the outside and (Y) hydrophilic layers on the inside of the membrane
D X and Y are both hydrophilic layers.

The correct answer is:
(X) hydrophilic layers on the outside and (Y) hydrophobic layers on the inside of the membrane

Solution:
Fatty acid heads are hydrophilic in nature, and their hydrocarbon tails are hydrophobic in nature. Lipids found in plasma membrane form bilayer in such a way that their hydrophilic head faces outside and hydrophobic hydrocarbon tail is sandwich between hydrophilic heads.

Topic: Biology, Cell - The Unit of Life, NEET
The ploidy level is not the same in

Options:
A Integument and nucellus
B Root tip and shoot tip
C Secondary nucleus and endosperm
D Antipodals and synergids

The correct answer is:
Secondary nucleus and endosperm

Solution:
Double fertilization is a unique characteristic observed in the Angiosperm group of plants. In this, two fertilizations are observed - 1st between the female and male gametic cells forming zygote and 2nd is between the second male gamete and polar nuclei forming the primary endosperm nucleus. The ploidy of PEN is 3n (male gamete[1n] + polar nuclei[2n]). The polar nuclei are also called as secondary nucleus which is 2n. Endosperm (3n) and secondary nucleus do not have the same ploidy level.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
The point in the interval $(0,2 \pi)$ where $\mathrm{f}(\mathrm{x})=\mathrm{e}^{\mathrm{x}} \sin \mathrm{x}$ has
maximumslope is

Options:
A $\frac{\pi}{4}$
B $\frac{\pi}{2}$
C $\pi$
D $\frac{3 \pi}{2}$

The correct answer is:
$\frac{\pi}{4}$

Solution:
Given $f(x)=e^{x} \sin x$
$\Rightarrow f^{\prime}(x)=e^{x} \cos x+e^{x} \sin x$
$\Rightarrow$ slope $=e^{x}(\cos x+\sin x)$
Now, $\frac{d}{d x}(\cos x+\sin x)=0$
$\Rightarrow-\sin x+\cos x=0$
$\Rightarrow \sin x=\cos x \Rightarrow \tan x=1 \Rightarrow x=\frac{\pi}{4}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2011 (Phase 2)
The point of intersection of the line joining the points $(-3,4,-8)$ and $(5,-6,4)$ with the XY-plane is

Options:
A $\left(\frac{7}{3},-\frac{8}{3}, 0\right)$
B $\left(-\frac{7}{3},-\frac{8}{3}, 0\right)$
C $\left(-\frac{7}{3}, \frac{8}{3}, 0\right)$
D $\left(\frac{7}{3}, \frac{8}{3}, 0\right)$

The correct answer is:
$\left(\frac{7}{3},-\frac{8}{3}, 0\right)$

Solution:
The equation of the line joining the points $(-3,4,-8)$ and $(5,-6,4)$ is
$\frac{x+3}{8}=\frac{y-4}{-10}=\frac{z+8}{12}=k($ say $)$
$8 \quad-10 \quad 12$
$\Rightarrow x+3=8 k ; y-4=-10 k ; z+8=12 k$
$\Rightarrow x=8 k-3 ; y=-10 k+4 ; z=12 k-8$
Given that this line intersects with $x y$ plane. $S o, z=0$
$12 \mathrm{k}-8=0 \Rightarrow 12 \mathrm{k}=8 \Rightarrow \mathrm{k}=\frac{8}{12}=\frac{2}{3}$
$\mathrm{x}=8\left(\frac{2}{3}\right)-3 ; \mathrm{y}=-10\left(\frac{2}{3}\right)+4 ; \mathrm{z}=12\left(\frac{2}{3}\right)-8$
$\Rightarrow \mathrm{x}=\frac{16}{3}-3 ; \mathrm{y}=\frac{-20}{3}+4 ; \mathrm{z}=\frac{24}{3}-8$
$\Rightarrow x=\frac{7}{3} ; y=\frac{-8}{3} ; z=0$
$(x, y, z)=\left(\frac{7}{3}, \frac{-8}{3}, 0\right)$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2017 (Phase 2)
The point of intersection of the two lines $2 x+3 y+4=0$ and $4 x+3 y+2=0$ is at a distance $d$ from origin. What is the value of $d ?$

Options:
A $\sqrt{2}$
B $\sqrt{3}$
C $\sqrt{5}$
D $\sqrt{7}$

The correct answer is:
$\sqrt{5}$

Solution:
Given equations are
$$
2 x+3 y+4=0 ...(i)
$$
and $4 x+3 y+2=0$...(ii)
On solving (i) and (ii), the coordinates of the intersecting point are $(1,-2)$
Now, $\sqrt{(0-1)^{2}+\{0-(-2)\}^{2}}=d$
$\Rightarrow d=\sqrt{1+4}=\sqrt{5}$

Topic: Mathematics, Straight Lines, NDA, NDA 2008 (Phase 2)
The point on the parabola $y^{2}=4 a x$ nearest to the focus has its abscissa

Options:
A $x=0$
B $x=a$
C $x=\frac{a}{2}$
D $x=2 a$

The correct answer is:
$x=0$

Solution:
Here, 'S' represents focus $\mathrm{O}(0,0)$ is apoint which is on parabola $y^{2}=4 a x$ and nearest to focus $(a, 0)$
$\therefore$ abscissa of $\mathrm{O}(0,0)$ is $\mathrm{x}=0$
4. Option (a) is correct.

Topic: Mathematics, Parabola, NDA, NDA 2015 (Phase 1)
The point whose abscissa is equal to its ordinate and which is equidistant from $\mathrm{A}(-1,0)$ and $\mathrm{B}(0,5)$ is $\quad$

Options:
A $(1,1)$
B $(2,2)$
C $(-2,-2)$
D $(3,3)$

The correct answer is:
$(2,2)$

Solution:
Let the required point be $\mathrm{P}(\mathrm{x}, \mathrm{x})$ Since $\mathrm{PA}=\mathrm{PB}$
$\Rightarrow \quad(\mathrm{PA})^{2}=(\mathrm{PB})^{2}$
$\Rightarrow \quad(\mathrm{x}+1)^{2}+\mathrm{x}^{2}=\mathrm{x}^{2}+(\mathrm{x}-5)^{2}$
$\Rightarrow \quad x^{2}+1+2 x=x^{2}+25-10 x$
$\Rightarrow \quad 12 \mathrm{x}=24 \Rightarrow \mathrm{x}=2$
Hence, Required point is $(2,2)$.

Topic: Mathematics, Straight Lines, NDA, NDA 2013 (Phase 1)