datavorous/entrance-exam-dataset huggingface dataset:
The IUPAC name of is

Options:
A Ethyl acetylate
B Ethyl methyl butenoate
C Ethyl acetoethanoate
D Ethyl (3 methyl) but -3- enoate

The correct answer is:
Ethyl (3 methyl) but -3- enoate

Solution:
Ethyl (3 methyl) but -3-enoate

Topic: Chemistry, General Organic Chemistry, NEET
The kind of epithelium which forms the inner walls of blood vessels is

Options:
A Cuboidal epithelium
B Columnar epithelium
C Ciliated columnar epithelium
D Squamous epithelium

The correct answer is:
Squamous epithelium

Solution:
Squamous epithelium is made of a single thin layer of flattened cells with irregular boundaries. It is also called pavement epithelium. It is found in the walls of blood vessels and air sacs of lungs and involved in functions like forming a diffusion boundary. It is also found in descending limbs of loops of Henle, membranous labyrinth, coelomic cavities, and rete testis.

Topic: Biology, Structural Organisation in Animals, NEET
The kingdom Protista forms a link with kingdom _____.

Options:
A Plantae
B Fungi
C Animalia
D all of these

The correct answer is:
all of these

Solution:
Protista is a kingdom in which single-celled eukaryotic organisms are present. Protists don't have well-defined cell boundaries due to the presence of the nucleus. They form a link in different categories. The cell body of Protista contains a well-defined nucleus and membrane-bound organelles. Along with that some have cilia and some have flagella are also present that help in movement. Protist reproduces asexually as well as sexually depending on the external environment. So they show similarity with Plantae, fungi, and Animalia because they all are also eukaryotes. So the correct answer is D.

Topic: Biology, Biological Classification, NEET
The labelings 'a' and 'b' for the Mosses are

Options:
A a- Thallus and b-Roots
B a- Gemma cups and b-Rhizoids
C a- Thallus and b-Rhizoids
D a-Gemma cups and b-Thallus

The correct answer is:
a- Thallus and b-Rhizoids

Solution:
Mosses are the bryophytes with gametophytic plant body, e.g., Funaria. On the lower portion of leafy gametophore (thallus) of moss, numerous branched multicellular rhizoids with oblique septa are present. These rhizoids are meant for the purpose of attachment or anchorage to the substratum.

Topic: Biology, Plant Kingdom, NEET
The largest value of $2 x^{3}-3 x^{2}-12 x+5$ for $-2 \leq x \leq 2$ occurs when $\quad$

Options:
A $x=-2$
B $x=-1$
C $x=2$
D $x=0$

The correct answer is:
$x=-1$

Solution:
$\operatorname{Let} f(x)=2 x^{3}-3 x^{2}-12 x+5$
We find, $f(-2), f(-1), f(0), f(1), f(2)$.
Now, $f(-2)=-16-12+24+5=1$
$f(-1)=-2-3+12+5=12$
$f(0)=5$
$f(1)=2-3-12+5=-8$
and $f(2)=2(2)^{3}-3(2)^{2}-12(2)+5$
$=16-12-24+5=21-36=-15$
$\therefore \quad$ Largest value of $2 x^{3}-3 x^{2}-12 x+5$ is at $x=-1$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2011 (Phase 1)
The lateral roots develop from primordia originated by the division of

Options:
A Pericycle cells lying opposite to protoxylem points
B Pericycle cells lying between two protoxylem points
C Endodermal cells lying between two protoxylem points
D Endodermal cells lying opposite to protoxylem points

The correct answer is:
Pericycle cells lying between two protoxylem points

Solution:
Root is that part of plant body which grow and down into earth. The primary roots develop from radicle and gives secondary and tertiary roots. Lateral roots develop endogeneously i.e. from pericycle.

Topic: Biology, Anatomy of Flowering Plants, NEET
The left-hand derivative of $f(x)=[x] \sin (\pi x)$ at $x=k$ where k is an integer and $[\mathrm{x}]$ is the greatest integer function, is

Options:
A $(-1)^{\mathrm{k}}(\mathrm{k}-1) \pi$
B $(-1)^{\mathrm{k}-1}(\mathrm{k}-1)^{2} \pi$
C $\quad(-1)^{\mathrm{k}} \mathrm{k} \pi$
D $(-1)^{\mathrm{k}-1} \mathrm{k} \pi$

The correct answer is:
$(-1)^{\mathrm{k}}(\mathrm{k}-1) \pi$

Solution:
$\mathrm{f}(\mathrm{x})=[\mathrm{x}] \sin (\pi \mathrm{x})$ at $\mathrm{x}=\mathrm{k}$.
Left hand derivative, $\lim _{h \rightarrow 0} \frac{\mathrm{f}(\mathrm{k})-\mathrm{f}(\mathrm{k}-\mathrm{h})}{\mathrm{h}}$ (k-integer)
$=\lim _{h \rightarrow 0} \frac{[k] \sin k \pi-[k-h] \sin (k-h) \pi}{h}$
$=\lim _{h \rightarrow 0} \frac{-(k-1) \sin (k-h) \pi}{h}$
$\sin k \pi=0$ and $\sin (k \pi-\theta)=(-1)^{k-1} \sin \theta .$
$=\lim _{h \rightarrow 0} \frac{-(k-1)-(1)^{k-1} \sin \pi}{h \pi} \times \pi$
$=\lim _{h \rightarrow 0} \frac{-(k-1)(-1)^{k-1} \sin \pi}{h \pi} \times \pi$
$=\pi(k-1)(-1)^{k}$

Topic: Mathematics, Limits, NDA, NDA 2017 (Phase 2)
The length and diameter of a metal wire used in sonometer is doubled. The fundamental frequency will change from ' $n$ ' to

Options:
A $\frac{\mathrm{n}}{4}$
B n
C 2n
D $\frac{\mathrm{n}}{2}$

The correct answer is:
$\frac{\mathrm{n}}{4}$

Solution:
The fundamental frequency is given by
$\mathrm{n}=\frac{1}{2 \mathrm{Cr}} \sqrt{\frac{\mathrm{T}}{\pi \rho}}$
$\begin{aligned} & \therefore \mathrm{n} \propto \frac{1}{\ell_{\mathrm{r}}} \\ & \therefore \frac{\mathrm{n}_2}{\mathrm{n}_1}=\frac{\ell_1 \mathrm{r}_1}{\ell_2 \mathrm{r}_2}=\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}\end{aligned}$
$\therefore \mathrm{n}_2=\frac{\mathrm{n}_1}{4}=\frac{\mathrm{n}}{4}$

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2021 (24 Sep Shift 2)
The length of a wire of a potentiometer is 100 cm and the emf of its standard cell is E volt. It is employed to measured emf of a battery whose internal resistance is 0.5 Î©. If the balance point is obtained at l = 30 cm from the positive end, the emf of the battery is -

Options:
A 0.2Â E
B 0.3 E
C 0.4Â E
D 0.5Â E

The correct answer is:
0.3 E

Solution:
Îµ=xlÎµ =Îµ0lLâ â´âÎµ =E100Ã30
â´ Îµ = 0.3 El = balance lengthL = length of potentialmeter wire

Topic: Physics, Experimental Physics, NEET
The length of an open organ pipe is twice the length of another closed organ pipe. The fundamental frequency of the open pipe $100 \mathrm{Hz}$. The frequency of the third harmon of the closed pipe is

Options:
A $100 \mathrm{Hz}$
B $200 \mathrm{Hz}$
C $300 \mathrm{Hz}$
D $150 \mathrm{Hz}$

The correct answer is:
$300 \mathrm{Hz}$

Solution:
Let the length of closed organ pipe 1 be. Then,
$$
v_{n}=(2 n-1) \frac{v}{4 l}
$$
Its third harmonics (put $n=2$ ) in above question.
$$
v_{3}=\frac{3}{4} \frac{v}{1}
$$
Now, the length of open organ pipe is 21 , then
$$
v_{n}=n \frac{v}{2(2 l)}=\frac{n v}{4 l}
$$
Its fundamental frequency,
$$
v_{0}=\frac{v}{2 l}=100
$$
$$
v_{3}=\frac{3 v}{4 l}=3 \times 100=300 \mathrm{Hz}
$$

Topic: Physics, Waves and Sound, WBJEE, WBJEE 2015
The length of latus rectum of the ellipse $4 x^{2}+9 y^{2}=36$ is

Options:
A $\frac{4}{3}$
B $\frac{8}{3}$
C 6
D 12

The correct answer is:
$\frac{8}{3}$

Solution:
$4 \mathrm{x}^{2}+9 \mathrm{y}^{2}=36$
$\frac{x^{2}}{3^{2}}+\frac{y^{2}}{2^{2}}=1$
Length of latus rectum $=2 \times \frac{2^{2}}{3}=\frac{8}{3}$

Topic: Mathematics, Straight Lines, NDA, NDA 2013 (Phase 2)
The length of latus rectum of the parabola \( 4 y^{2}+3 x+3 y+1=0 \) is

Options:
A \( \frac{4}{3} \)
B \( 07 \)
C \( 12 \)
D \( \frac{3}{4} \)

The correct answer is:
\( \frac{3}{4} \)

Solution:
Given parabola
\[
\begin{array}{l}
4 y^{2}+3 x+3 y+1=0 \\
\Rightarrow y^{2}+\frac{3}{4} y+\frac{1}{4}=\frac{-3}{4} x \\
\Rightarrow\left(y+\frac{3}{8}\right)^{2}-\left(\frac{3}{8}\right)^{2}+\frac{1}{4}=\frac{-3}{4} x \\
\Rightarrow\left(y+\frac{3}{8}\right)^{2}+\frac{7}{64}=\frac{-3}{4} x \\
\Rightarrow\left(y+\frac{3}{8}\right)^{2}=-\frac{3}{4} x-\frac{7}{64} \\
\Rightarrow\left(y+\frac{3}{8}\right)^{2}=\frac{-3}{4}\left(x+\frac{7}{48}\right) \\
\Rightarrow\left(y+\frac{3}{8}\right)^{2}=\frac{-3}{4}\left(x+\frac{7}{48}\right) \rightarrow(1)
\end{array}
\]
Comparing Eq. (1) with standard equation of parabola,
\( y^{2}=4 a X \)
Therefore, length of latus rectum is \( \frac{3}{4} \)

Topic: Mathematics, Parabola, KCET, KCET 2016
The length of perpendicular from the origin to a line is 5 units and the line makes an angle $120^{\circ}$ with the positive direction of $x$ -axis. The equation of the line is

Options:
A $x+\sqrt{3} y=5$
B $\sqrt{3} x+y=10$
C $\quad \sqrt{3} x-y=10$
D None of these

The correct answer is:
$\sqrt{3} x+y=10$

Solution:
$\sin 60^{\circ}=\frac{5}{\mathrm{OB}} \Rightarrow \mathrm{OB}=\frac{5}{\sin 60^{\circ}}$
$\mathrm{OB}=\frac{5 \times 2}{\sqrt{3}}=\frac{10}{\sqrt{3}}$
In $\triangle \mathrm{ACO}$
$\angle \mathrm{OAC}=30^{\circ}$
$\sin 30^{\circ}=\frac{5}{\mathrm{AO}} \Rightarrow \mathrm{AO}=\frac{5 \times 2}{1}=10$
Normal form of line AB
$\frac{\mathrm{X}}{\mathrm{OB}}+\frac{\mathrm{Y}}{\mathrm{OA}}=1$
$\frac{\sqrt{3} \mathrm{X}}{10}+\frac{\mathrm{Y}}{10}=1$
$\Rightarrow \quad \sqrt{3} \mathrm{X}+\mathrm{Y}=10$

Topic: Mathematics, Straight Lines, NDA, NDA 2015 (Phase 1)
The length of the normal from origin to the plane $\mathrm{x}+2 \mathrm{y}-2 \mathrm{z}=9$ is equal to

Options:
A 2 units
B 3 units
C 4 units
D 5 units

The correct answer is:
3 units

Solution:
Given plane, $x+2 y-2 z=9$. $\frac{d}{\sqrt{a^{2}+b^{2}+c^{2}}}$
length of the normal $=\frac{9}{\sqrt{(1)^{2}+(2)^{2}+(-2)^{2}}}$
$=\frac{9}{\sqrt{9}}=\frac{9}{3}=3$ units

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2017 (Phase 2)
The length of the parabola $y^{2}=12 x$ cut off by the latus-rectum is

Options:
A $6(\sqrt{2}+\log (1+\sqrt{2}))$
B $3(\sqrt{2}+\log (1+\sqrt{2}))$
C $6(\sqrt{2}-\log (1+\sqrt{2}))$
D $3(\sqrt{2}-\log (1+\sqrt{2}))$

The correct answer is:
$6(\sqrt{2}+\log (1+\sqrt{2}))$

Solution:
On comparing the equation of the parabola
$y^{2}=12 x$ with the standard equation,
$y^{2}=4$ ax, we get $4 a=12$ or $a=3$
Hence, the focus, point $\mathrm{C}$ will be at $(3,0)$ and the extremities of the latus-rectum $\mathrm{AB}$ will be at $(a, 2 a)$ and $(a,-2 a)$. So the coordinates of A and $B$ are $(3,6)$ and $(3,-6)$ respectively. Now we need to find the length ( curve $\mathrm{AOB}$ ) of the parabola. As it is not a straight line so we cannot directly find the length of this curve as we cannot directly apply Pythagorous theorem. Let us consider a small length $\mathrm{ds}$ on the parabola. Using pythagorous theorem for this length,
$\mathrm{ds}=\sqrt{(\mathrm{dx})^{2}+(\mathrm{dy})^{2}}=\sqrt{\left(\frac{\mathrm{dx}}{\mathrm{dy}}\right)^{2}+1 \mathrm{dy}}$
$\Rightarrow \mathrm{s}=\int_{-6}^{6}\left[\sqrt{\left(\frac{\mathrm{dx}}{\mathrm{dy}}\right)^{2}+1}\right] \cdot \mathrm{dy}$ ...(1)
From $y^{2}=12 x \Rightarrow x=\frac{y^{2}}{12}$
$\frac{d x}{d y}=\frac{2 y}{12}=\frac{y}{6} \quad$ Putting in (1),
$s=\int_{-6}^{6}\left[\sqrt{\left(\frac{y}{6}\right)^{2}+1}\right] d y=2 \int_{0}^{6} \sqrt{\frac{y^{2}+36}{36}} d y$
$=\frac{2}{6} \int_{0}^{6} \sqrt{y^{2}+6^{2}} d y$
Using $\int \sqrt{x^{2}+a^{2}}=\frac{x}{2} \sqrt{a^{2}+x^{2}}$
$+\frac{a^{2}}{2} \log \left(x+\sqrt{a^{2}+x^{2}}\right)+C$
We get $s=\frac{1}{3}\left[\frac{y}{2} \sqrt{6^{2}+y^{2}}+\frac{6^{2}}{2}\right.$
$\left.\log \left(y+\sqrt{6^{2}+y^{2}}\right)+C\right]_{0}^{6}$
$=\frac{1}{3}\left[\frac{6}{2} \sqrt{6^{2}+6^{2}}+18 \log \left(6+\sqrt{6^{2}+6^{2}}\right)\right.$
$\left.\quad+C-0-18 \log \left(0+\sqrt{6^{2}+0}\right)-C\right]$
$=\frac{1}{3}[3.6 \sqrt{2}+18 \log (6+6 \sqrt{2})-18 \log 6]$
$=6 \sqrt{2}+6 \log \frac{6(1+\sqrt{2})}{6}$
$=6[(\sqrt{2}+\log (1+\sqrt{2})]$

Topic: Mathematics, Vector Algebra, VITEEE, VITEEE 2008
The length of the string of a musical instrument is 90 cm and has a fundamental frequency of 120 Hz. Where should it be pressed to produce fundamental frequency of 180 Hz?

Options:
A 75 cm
B 60 cm
C 45 cm
D 80 cm

The correct answer is:
60 cm

Solution:
The fundamental frequency in terms of speed of wave and length between two boundaries, ν=v2l, it means.

ν1ν2=l2l1

⇒l2=l1ν1ν2=90×120180=60 cm

Topic: Physics, Waves and Sound, NEET, NEET 2020 (Phase 2)
The length of the wire between two ends of sonometer is $100 \mathrm{~cm}$. What should be the positions of two bridges below the wire so that the three segments of the wire have their fundamental frequencies in the ratio $1: 3: 5$.

Options:
A $\frac{1500}{23} \mathrm{~cm}, \frac{500}{23} \mathrm{~cm}$
B $\frac{1500}{23} \mathrm{~cm}, \frac{300}{23} \mathrm{~cm}$
C $\frac{300}{23} \mathrm{~cm}, \frac{1500}{23} \mathrm{~cm}$
D $\frac{1500}{23} \mathrm{~cm}, \frac{2000}{23} \mathrm{~cm}$

The correct answer is:
$\frac{1500}{23} \mathrm{~cm}, \frac{2000}{23} \mathrm{~cm}$

Solution:
According to question
Length of wire $=L=100 \mathrm{~cm}$
And ratio of fundamental frequencies
$$
=1: 3: 5
$$
$$
\begin{aligned}
L_1: L_2: L_3 & =\frac{1}{1}: \frac{1}{3}: \frac{1}{5} \\
& =15: 5: 3
\end{aligned}
$$
Let $x=$ comman factor then
$$
\begin{aligned}
& 15 x+5 x+3 x=100 \\
& \Rightarrow \quad 2 x=100 \\
& \Rightarrow \quad x=\frac{100}{23} \\
& \therefore \quad L_1=15 \times \frac{100}{23} \\
& =\frac{1500}{23} \mathrm{~cm} \\
& \therefore \quad L_2=5 \times \frac{100}{23} \\
& =\frac{1500}{23} \mathrm{~cm} \\
&
\end{aligned}
$$
$$
\begin{aligned}
\therefore \quad L_3 & =3 \times \frac{100}{23} \\
& =\frac{300}{23} \mathrm{~cm}
\end{aligned}
$$
This shows that bridge should be placed from $A$ at $\frac{1500}{23} \mathrm{~cm}$ and $\frac{2000}{23} \mathrm{~cm}$ respectively

Topic: Physics, Waves and Sound, NEET, NEET 2013 (Karnataka)
The lengths of the intercepts on the co-ordinate axes made by the plane $5 x+2 y+z-13=0$ are

Options:
A $5,2,1$ unit
B $\frac{13}{5}, \frac{13}{2}, 13$ unit
C $\frac{5}{13}, \frac{2}{13}, \frac{1}{13}$ unit
D $1,2,5$ unit

The correct answer is:
$\frac{13}{5}, \frac{13}{2}, 13$ unit

Solution:
$5 x+2 y+z-13=0$
Putting $y=0 \& z=0$
$\mathrm{x}=\frac{13}{5}$
Putting $z=0 \& x=0$
$y=\frac{13}{2}$
Putting $x=0 \& y=0$
$z=13$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2015 (Phase 2)
The 'less than' ogive curve and the 'more than' ogive curve intersect at

Options:
A median
B mode
C arithmetic mean
D None of these

The correct answer is:
median

Solution:
The 'less than' ogive curve and the 'more than' ogive curve intersect at median. Option (a) is correct.

Topic: Mathematics, Statistics, NDA, NDA 2015 (Phase 1)
The lichen commonly known as the ‘reindeer moss’ is

Options:
A Marselia
B Cladonia
C Georgia
D Usnea

The correct answer is:
Cladonia

Solution:
Cladonia rangiferina is the lichen that is commonly known as the âreindeer mossâ and this is found growing luxuriantly in tundra region where the reindeer are also abundantly found. These are terricolour lichen found growing on soil.

Topic: Biology, Biological Classification, NEET
The line $\frac{x-1}{2}-\frac{y-2}{3}=\frac{z-3}{3}$ is given by

Options:
A $x+y+z=6, x+2 y-3 z=-4$
B $x+2 y-2 z=-1,4 x+4 y-5 z-3=0$
C $3 x+2 y-3 z=0,3 x-6 y+3 z=-2$
D $3 x+2 y-3 z=-2,3 x-6 y+3 z=0$

The correct answer is:
$3 x+2 y-3 z=-2,3 x-6 y+3 z=0$

Solution:
drs of line is $2,3,4$
going through option, $2(1)+3(2)+4(-2)=0$
$2(4)+3(4)-4(5)=0$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2018 (Phase 2)
The line $2 y=3 x+12$ cuts the parabola $4 y=3 x^{2}.$
What is the area enclosed by the parabola and the line?

Options:
A 27 square unit
B 36 square unit
C 48 square unit
D 54 square unit

The correct answer is:
27 square unit

Solution:
Equation of line $2 y=3 x+12, y=\frac{3 x+12}{2}$
Equation of parabola $4 y=3 x^{2}, y=\frac{3 x^{2}}{4}$
$=\int_{-2}^{4}\left[\frac{3 x+12}{2}-\frac{3 x^{2}}{4}\right] d x$
$=\frac{1}{2}\left[\frac{3 \mathrm{x}^{2}}{2}+12 \mathrm{x}\right]_{-2}^{4}-\frac{3}{4}\left[\frac{\mathrm{x}^{3}}{3}\right]_{-2}^{4}$
$=\frac{1}{2}\left[\left\{\frac{3(4)^{2}}{2}+12(4)\right\}-\left\{\frac{3(-2)^{2}}{2}+12(-2)\right\}\right]$
$-\frac{3}{4}\left[\frac{4^{3}}{3}-\frac{(-2)^{3}}{3}\right]$
$=\frac{1}{2}[24+48-6+24]-\frac{3}{4}\left[\frac{64+8}{3}\right]$
$=\frac{1}{2} \times 90-18=27$ sq. units.

Topic: Mathematics, Parabola, NDA, NDA 2014 (Phase 2)
The line $2 y=3 x+12$ cuts the parabola $4 y=3 x^{2}.$
What is the area enclosed by the parabola, the line and the $Y$ axis in the first quadrant

Options:
A 7 square unit
B 14 square unit
C 20 square unit
D 21 square unit

The correct answer is:
20 square unit

Solution:
Equation of line $2 \mathrm{y}=3 \mathrm{x}+12$ and equations of parabola $4 y=3 x^{2}$
$=\int_{0}^{4}\left(\frac{3 x+12}{2}-\frac{3 x^{2}}{4}\right) d x=\left(\frac{3}{4} x^{2}+6 x-\frac{x^{3}}{4}\right)_{0}^{4}$
$=3 \times 4+24-16=36-16=20$ sq. units.
in Area enclosed by the parabola, the line and the y axis in first quadrant $=20$ sq. units

Topic: Mathematics, Parabola, NDA, NDA 2014 (Phase 2)
The line $2 y=3 x+12$ cuts the parabola $4 y=3 x^{2}.$
Where does the line cut the parabola?

Options:
A At $(-2,3)$ only
B At $(4,12)$ only
C At both $(-2,3)$ and $(4,12)$
D Neither at $(-2,3)$ nor $(4,12)$

The correct answer is:
At both $(-2,3)$ and $(4,12)$

Solution:
Equation of line $2 y=3 x+12$
Equation of parabola $4 y=3 x^{2}$
From eqs. (i) and (ii), we get $2(3 x+12)=3 x^{2}$
$3 x^{2}-6 x-24=0$
$x^{2}-2 x-8=0$
$(x-4)(x+2)=0$
$\therefore \mathrm{x}=4$
and $x=-2$ Now putting the value of $\mathrm{x}$ in eqn (ii) We get $y=12$ and $y=3$ Thus, the points $(-2,3)$ and $(4,12)$

Topic: Mathematics, Parabola, NDA, NDA 2014 (Phase 2)
The line $3 \mathrm{x}+4 \mathrm{y}-24=0$ intersects the $\mathrm{x}$ -axis at $\mathrm{A}$ and $\mathrm{y}$ -axis at $\mathrm{B}$. Then the circumcentre of the triangle $\mathrm{OAB}$ where $\mathrm{O}$ is the origin is

Options:
A $(2,3)$
B $(3,3)$
C $(4,3)$
D None of the above

The correct answer is:
$(4,3)$

Solution:
Since circumcentre of right angled triangle lies on the midpoint of hypotenuse.
So mid point of $\mathrm{AB}$ is $\left(\frac{0+8}{2}, \frac{6+0}{2}\right)$ or $(4,3)$

Topic: Mathematics, Straight Lines, NDA, NDA 2015 (Phase 2)
The line joining the points $(2,1,3)$ and $(4,-2,5)$ cuts the plane $2 x+y-z=3 .$ What is the ratio in which the plane divideds the line?

Options:
A 1:1
B 2:3
C 3:4
D None of the above

The correct answer is:
None of the above

Solution:
Let the ratio is $\mathrm{k}: 1$
Then, $0=\frac{4 k+2}{k+1}$
$\Rightarrow 4 k+2=0 \Rightarrow k=-\frac{1}{2}$
and $4=\frac{-2 \mathrm{k}+1}{\mathrm{k}+1} \Rightarrow 4 \mathrm{k}+4=-2 \mathrm{k}+1 \Rightarrow \mathrm{k}=-\frac{1}{2}$
Ilcnce, plane divides the line in ratio 1 : 2 cxternally.

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2014 (Phase 2)
The line joining the points $(2,1,3)$ and $(4,-2,5)$ cuts the plane $2 x+y-z=3 .$ Where does the line cut the plane?

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

The correct answer is:
$(0,4,1)$

Solution:
Equation of line passing through the points $(2,1,3)$ and $(4,-2,5)$ is
$\frac{x-2}{4-2}=\frac{y-1}{-2-1}=\frac{z-3}{5-3}=\lambda$
$\Rightarrow \frac{\mathrm{x}-2}{2}=\frac{\mathrm{y}-1}{-3}=\frac{\mathrm{z}-3}{2}=\lambda$
$\Rightarrow \mathrm{x}=2 \lambda+2, \mathrm{y}=-3 \lambda+1$ and $\mathrm{z}=2 \lambda+3$
Since, this line cuts the plane $2 x+y-z=3$ So, $(2 \lambda+2,-3 \lambda+1,2 \lambda+3)$ satisfies the equation of
plane $\therefore 2 \lambda+2-3 \lambda+1-2 \lambda-3=3$
$\Rightarrow-3 \lambda=3$
$\Rightarrow \lambda=-1$
Hence, points are $[2(-1)+2,-3(-1)+1,2(-1)+3]$ i.e.,
$(0,4,1)$.

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2014 (Phase 2)
The line $m x+n y=1$ passes through the points $(1,2)$ and $(2,1) .$ What is the value of $m ?

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

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

Solution:
Sinceline $m x+n y=1$ passes through $(1,2)$ and $(2,1)$ therefore they satisfied the equation.
$\Rightarrow \quad m+2 n=1$ ...(i)
and $2 m+n=1$ ... (ii)
From eqs.
(i) and (ii),
we get $m=n=\frac{1}{3}$

Topic: Mathematics, Straight Lines, NDA, NDA 2011 (Phase 1)
The line making an angle $\left(-120^{\circ}\right)$ with $\mathrm{x}$ -axis is situated in the:

Options:
A first quadrant
B second quadrant
C third quadrant
D fourth quadrant

The correct answer is:
third quadrant

Solution:
The line making an angle $-120^{\circ}$ with $\mathrm{x}$ -axis is situated in the third quadrant.
$\therefore-120^{\circ}$ means $120^{\circ}$ clockwise which goes in 3rd quadrant.

Topic: Mathematics, Straight Lines, NDA, NDA 2012 (Phase 2)
The line passing through $(1,2,3)$ and having direction ratios given by $ < 1,2,3>$ cuts the $x$ -axis at a distance $k$ from origin. What is the value of $k ?$

Options:
A 0
B 1
C 2
D 3

The correct answer is:
0

Solution:
Let the equation of line which is passing through $(1,2,3)$ and having direction ratios $(1,2,3)$ is
$\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}=a$
$x-1=a$
$\begin{aligned} y &-2=2 a \text { and } z-3=3 a \\ \Rightarrow x &=a+1, y=2 a+2 \text { and } z=3 a+3 \end{aligned}$
At $x$ -axis, $y=0$ and $z=0$ $\Rightarrow 2 a+2=0$ and $3 a+3=0$
$\Rightarrow a=-1$ and $a=-1$
$x=(-1)+1=0$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2008 (Phase 2)
The line passing through the points $(1,2,-1)$ and $(3,-1,2)$ meets the yz-plane at which one of the following points?

Options:
A $\left(0,-\frac{7}{2}, \frac{5}{2}\right)$
B $\left(0, \frac{7}{2}, \frac{1}{2}\right)$
C $\left(0,-\frac{7}{2},-\frac{5}{2}\right)$
D $\left(0, \frac{7}{2},-\frac{5}{2}\right)$

The correct answer is:
$\left(0, \frac{7}{2},-\frac{5}{2}\right)$

Solution:
Eqn. of line $\Rightarrow \frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}}$
i.e, $\frac{x-1}{2}=\frac{y-2}{-3}=\frac{z+1}{3}=K($ say $)$
$\Rightarrow \mathrm{x}-1=2 \mathrm{~K} ; \mathrm{y}-2=-3 \mathrm{~K} ; \mathrm{z}+1=3 \mathrm{~K}$
$\Rightarrow \mathrm{x}=2 \mathrm{~K}+1 ; \mathrm{y}=-3 \mathrm{~K}+2 ; \mathrm{z}=3 \mathrm{~K}-1$
Since the line meets yz plane, $\mathrm{x}=0$
$2 \mathrm{~K}+1=0 \Rightarrow \mathrm{K}=\frac{-1}{2}$
$y=-3\left(\frac{-1}{2}\right)+2=\frac{3}{2}+2=\frac{7}{2}$
$z=3\left(\frac{-1}{2}\right)-1=\frac{-3}{2}-1=\frac{-5}{2}$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2017 (Phase 1)
The line through the points $(4,3)$ and $(2,5)$ cuts offintercepts of lengths $\lambda$ and $\mu$ on the axes. Which one of the following is correct?

Options:
A $\lambda>\mu$
B $\lambda < \mu$
C $\lambda>-\mu$
D $\quad \lambda=\mu$

The correct answer is:
$\quad \lambda=\mu$

Solution:
Let the equation of line be
$\frac{x}{\lambda}+\frac{y}{\mu}=1$
The line passes through $(4,3)$ and $(2,5)$.
$\therefore \frac{4}{\lambda}+\frac{3}{\mu}=1$, is possible when $\lambda=\mu=7$
and $\frac{2}{\lambda}+\frac{3}{\mu}=1$, is possible when $\lambda=\mu=5$
So, $\mu=\lambda$

Topic: Mathematics, Straight Lines, NDA, NDA 2008 (Phase 2)
The line $\mathrm{y}=0$ divides the line joining the points $(3,-5)$ and $(-4,7)$ in the ratio:

Options:
A $3: 4$
B $4: 5$
C $5: 7$
D $7: 9$

The correct answer is:
$5: 7$

Solution:
Let $\mathrm{P}(\mathrm{x}, \mathrm{y})$ be the point of division that divides the line joining $(3,-5)$ and $(-4,7)$ in the ratio of $\mathrm{k}: 1$
Now, $\mathrm{y}=\frac{7 \mathrm{k}-5}{\mathrm{k}+1}$ ...(i)
Since, P lies on $\mathrm{y}=0$ or $\mathrm{x}$ -axis then, from eq. (i) $0=\frac{7 \mathrm{k}-5}{\mathrm{k}+1} \Rightarrow 7 \mathrm{k}=5 \Rightarrow \mathrm{k}=\frac{5}{7}$

Topic: Mathematics, Straight Lines, NDA, NDA 2012 (Phase 2)
The lines $2 \mathrm{x}=3 \mathrm{y}=-\mathrm{z}$ and $6 \mathrm{x}=-\mathrm{y}=-4 \mathrm{z}$

Options:
A are perpendicular
B are parallel
C intersect at an angle $45^{\circ}$
D intersect at an angle $60^{\circ}$

The correct answer is:
are perpendicular

Solution:
$2 \mathrm{x}=3 \mathrm{y}=-\mathrm{z}$
or $\frac{\mathrm{x}}{3}=\frac{\mathrm{y}}{2}=\frac{\mathrm{z}}{-6}$
$6 \mathrm{x}=-\mathrm{y}=-4 \mathrm{z}$
or $\frac{\mathrm{x}}{2}=\frac{\mathrm{y}}{-12}=\frac{\mathrm{z}}{-3}$
$\cos \theta=\frac{\mathrm{x}_{1} \mathrm{x}_{2}+\mathrm{y}_{1} \mathrm{y}_{2}+\mathrm{z}_{1} \mathrm{z}_{2}}{\sqrt{\mathrm{x}_{1}^{2}+\mathrm{y}_{1}^{2}+\mathrm{z}_{1}^{2}} \cdot \sqrt{\mathrm{x}_{2}^{2}+\mathrm{y}_{2}^{2}+\mathrm{z}_{3}^{2}}}$
$=\frac{(6-24+18)}{\sqrt{3^{2}+2^{2}+(-6)^{2}} \cdot \sqrt{2^{2}+(-12)^{2}+(-3)^{2}}}$
$\cos \theta=0$
$\theta=90^{\circ}$
So lines are perpendicular

Topic: Mathematics, Straight Lines, NDA, NDA 2015 (Phase 2)