datavorous/entrance-exam-dataset huggingface dataset:
What is tan $(\mathrm{A}+2 \mathrm{~B}) \cdot \tan (2 \mathrm{~A}+\mathrm{B})$ equal to?

Options:
A $-1$
B $\underline{0}$
C 1
D 2

The correct answer is:
1

Solution:
$\tan (\mathrm{A}+2 \mathrm{~B}) \cdot \tan (2 \mathrm{~A}+\mathrm{B})$
Put $\mathrm{A}=60$ and $\mathrm{B}=30$ in above expression We get $\tan \left(120^{\circ}\right) \cdot \tan \left(150^{\circ}\right)$
$=\tan \left(90^{\circ}+30^{\circ}\right) \tan \left(90^{\circ}+60^{\circ}\right)$
$=\cot 30^{\circ} \cdot \cot 60^{\circ}=\sqrt{3} \cdot \frac{1}{\sqrt{3}}=1$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2012 (Phase 1)
What is tan $\left(\cos ^{-1} x\right)$ equal to?

Options:
A $\frac{\sqrt{1-x^{2}}}{x}$
B $\frac{\mathrm{x}}{1+\mathrm{x}^{2}}$
C $\frac{\sqrt{1+x^{2}}}{x}$
D $\sqrt{1-\mathrm{x}^{2}}$

The correct answer is:
$\frac{\sqrt{1-x^{2}}}{x}$

Solution:
Let $\cos ^{-1} \mathrm{x}=\theta$
$\Rightarrow \cos \theta=\mathrm{x} \Rightarrow \sin \theta=\sqrt{1-\mathrm{x}^{2}}$
$\Rightarrow \tan \theta=\frac{\sqrt{1-x^{2}}}{x}$ and $\theta=\cos ^{-1} x$
This can be represented by a triangle with hypotenuous $=1$ and sides $\mathrm{x}$ and $\sqrt{1-\mathrm{x}^{2}}$.
$\Rightarrow \tan \left(\cos ^{-1} x\right)=\frac{\sqrt{1-x^{2}}}{x}$

Topic: Mathematics, Inverse Trigonometric Functions, NDA, NDA 2008 (Phase 1)
What is the $15^{\text {th }}$ term of the series $3,7,13,21,31,43, \ldots . ?$

Options:
A 205
B 225
C 238
D 241

The correct answer is:
241

Solution:
Let
$\Rightarrow \mathrm{a}_{\mathrm{n}}=3+[4+6+8+10+12+\ldots(\mathrm{n}-1)]$
$=3+\frac{(n-1)}{2}[8+\{(n-1)-1\} \times 2]$
$=3+\frac{(n-1)}{2}[8+2 n-4]$
$=3+\frac{(n-1)}{2}(2 n+4)$
$=3+(\mathrm{n}-1)(\mathrm{n}+2)$
$\therefore 15$ th term $=\mathrm{a}_{15}=3+(15-1)(15+2)$
$=3+14 \times 17=241$

Topic: Mathematics, Sequences and Series, NDA, NDA 2008 (Phase 2)
What is the 10 th common term between the series $2+6+10+\ldots .$ and $1+6+11+\ldots . ?$

Options:
A 180
B 186
C 196
D 206

The correct answer is:
186

Solution:
Given two series are $2+6+10+14+18+22+26+30+\ldots$
and $1+6+11+16+21+26+31+\ldots$
So, series which are made to be common terms of both series is $6,26,-\mathrm{u}$
This is an A. P where $a=6, d=20$ $\begin{aligned} \therefore \mathrm{a}_{10} &=\mathrm{a}+(\mathrm{n}-1) \mathrm{d} \\ &=6+(10-1) 20=6+9 \times 20 \\ &=180+6=186 \end{aligned}$

Topic: Mathematics, Sequences and Series, NDA, NDA 2011 (Phase 2)
What is the abundance of boron in the Earth's crust by mass?

Options:
A 0.001%
B 0.01%
C 0.1%
D 0.0001%

The correct answer is:
0.0001%

Solution:
The abundance of boron in the Earth's crust is less than 0.0001% by mass.

Topic: Chemistry, p Block Elements (Group 13 & 14), NEET
What is the acid soluble pool representative of?

Options:
A The entire chemical composition of living tissues or organisms
B The chemical composition of living tissue from abundance point of view
C The cytoplasmic composition
D The macromolecules from cytoplasm and organelles

The correct answer is:
The cytoplasmic composition

Solution:
The acid soluble pool represents roughly the cytoplasmic composition.

Topic: Biology, Biomolecules (B), NEET
What is the acute angle between the lines $\mathrm{Ax}+\mathrm{By}=\mathrm{A}+\mathrm{B}$ and $A(x-y)+B(x+y)=2 B ?$

Options:
A $45^{\circ}$
B $\tan ^{-1}\left(\frac{A}{\sqrt{A^{2}+B^{2}}}\right)$
C $30^{\circ}$
D $60^{\circ}$

The correct answer is:
$45^{\circ}$

Solution:
Lines are $L_{1} \equiv A x+B y=A+B$ and $L_{2} \equiv A(x-y)+B(x+y)=2 B$
Slope of $L_{1}$ is $-\frac{A}{B}$
$\mathrm{m}_{1}=-\frac{\mathrm{A}}{\mathrm{B}} \quad\left[\mathrm{m}_{1}\right.$ is the side of line $\left.\mathrm{L}_{1}\right]$
For line $\mathrm{L}_{2}$ :
$\mathrm{Ax}-\mathrm{Ay}+\mathrm{Bx}+\mathrm{By}=2 \mathrm{~B}$
$(\mathrm{A}+\mathrm{B}) \mathrm{x}-(\mathrm{A}-\mathrm{B}) \mathrm{y}=2 \mathrm{~B}$
Slope of line $L_{2}$ in $\frac{(A+B)}{A-B}$
$\mathrm{m}_{2}=\frac{(\mathrm{A}+\mathrm{B})}{(\mathrm{A}-\mathrm{B})} \quad\left[\mathrm{m}_{2}\right.$ is the slope of line $\left.\mathrm{L}_{2}\right]$
If angle between line $\mathrm{L}_{1}$ and $\mathrm{L}_{2}$ is $\theta$ then
$\tan \theta=\frac{\mathrm{m}_{1}-\mathrm{m}_{2}}{1+\mathrm{m}_{1} \mathrm{~m}_{2}}$
$=\frac{-\frac{A}{B}-\frac{A+B}{A-B}}{1+\left(-\frac{A}{B}\right) \times\left(\frac{A+B}{A-B}\right)}=\frac{\frac{-A(A-B)-B(A+B)}{B(A-B)}}{\frac{B(A-B)-A(A+B)}{B(A-B)}}$
$=\frac{-A^{2}+A B-A B-B^{2}}{A B-B^{2}-A^{2}-A B}=\frac{-B^{2}-A^{2}}{-B^{2}-A^{2}}=1$
so, $\theta=\frac{\pi}{4}$

Topic: Mathematics, Straight Lines, NDA, NDA 2007 (Phase 1)
What is the acute angle between the lines represented by the equations $y-\sqrt{3} x-5=0$ and $\sqrt{3} y-x+6=0$ ?

Options:
A $30^{\circ}$
B $45^{\circ}$
C $60^{\circ}$
D $75^{\circ}$

The correct answer is:
$30^{\circ}$

Solution:
$y-\sqrt{3} x-5=0$
(i)
$\sqrt{3} y-x+6=0$
$y=m x+c$
From (i) and from (ii)
$y=\sqrt{3} x+5 \quad y=\frac{x}{\sqrt{3}}-\frac{6}{\sqrt{3}}$
$m_{1}=\sqrt{3}$
$m_{2}=\frac{1}{\sqrt{3}}$
Angle between two lines, $\tan \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right|=\left|\frac{\sqrt{3}-\frac{1}{\sqrt{3}}}{1+\sqrt{3} \frac{1}{\sqrt{3}}}\right|=\frac{1}{\sqrt{3}}$
$=\tan 30^{\circ}$
$\theta=30^{\circ}$

Topic: Mathematics, Pair of Lines, NDA, NDA 2016 (Phase 1)
What is the acute angle between the pair of straight lines
$\sqrt{2} \mathrm{x}+\sqrt{3} \mathrm{y}=1$ and $\sqrt{3} \mathrm{x}+\sqrt{2} \mathrm{y}=2$ ?

Options:
A $\tan ^{-1}\left(\frac{1}{2 \sqrt{6}}\right)$
B $\tan ^{-1}\left(\frac{1}{\sqrt{2}}\right)$
C $\tan ^{-1}(3)$
D $\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)$

The correct answer is:
$\tan ^{-1}\left(\frac{1}{2 \sqrt{6}}\right)$

Solution:
Given lines, $\sqrt{2} \mathrm{x}+\sqrt{3} \mathrm{y}=1$ and $\sqrt{3} \mathrm{x}+\sqrt{2} \mathrm{y}=2$
We know, Slope $(\mathrm{m})=\frac{-\mathrm{a}}{\mathrm{b}}$
$\therefore$ Slope of $\sqrt{2} \mathrm{x}+\sqrt{3} \mathrm{y}=1$ is $\mathrm{m}_{1}=\frac{-\sqrt{2}}{\sqrt{3}}$
Slope of $\sqrt{3} x+\sqrt{2} y=2$ is $m_{2}=\frac{-\sqrt{3}}{\sqrt{2}}$
$\tan \theta=\left|\frac{\mathrm{m}_{1}-\mathrm{m}_{1}}{1+\mathrm{m}_{1} \mathrm{~m}_{2}}\right|=\left|\frac{\frac{-\sqrt{3}}{\sqrt{2}}+\frac{\sqrt{2}}{\sqrt{3}}}{1+\left(\frac{-\sqrt{2}}{\sqrt{3}}\right)\left(\frac{-\sqrt{3}}{\sqrt{2}}\right)}\right|$
$=\left|\frac{\frac{-3+2}{\sqrt{6}}}{2}\right|=\left|\frac{-1}{2 \sqrt{6}}\right|=\frac{1}{2 \sqrt{6}}$
$\therefore \theta=\tan ^{-1}\left(\frac{1}{2 \sqrt{6}}\right)$

Topic: Mathematics, Pair of Lines, NDA, NDA 2017 (Phase 1)
What is the acute angle between the planes $\mathrm{x}+\mathrm{y}+2 \mathrm{z}=3$ and $-2 x+y-z=11 ?$

Options:
A $\pi / 5$
B $\pi / 4$
C $\pi / 6$
D $\pi / 3$

The correct answer is:
$\pi / 3$

Solution:
The given equation of the planes are $x+y+2 z=3$ and $-2 x+y-z=11$
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
$\cos \theta=\left|\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}\right|$
Here, $a_{1}=1, b_{1}=1, c_{1}=2, a_{2}=-2, b_{2}=1, c_{2}=-1$
$\therefore \quad \cos \theta=\left|\frac{1 \times(-2)+1 \times 1+2 \times(-1)}{\sqrt{1+1+4} \sqrt{4+1+1}}\right|$
$=\left|\frac{-2+1-2}{\sqrt{6} \sqrt{6}}\right|=\left|\frac{3}{6}\right|=\frac{1}{2}=\cos \frac{\pi}{3} \Rightarrow \theta=\frac{\pi}{3}$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2011 (Phase 1)
What is the adjoint of the matrix
$\left(\begin{array}{l}\cos (-\theta)-\sin (-\theta) \\ -\sin (-\theta) \cos (-\theta)\end{array}\right) ?$

Options:
A $\left(\begin{array}{l}\cos \theta-\sin \theta \\ -\sin \theta \cos \theta\end{array}\right)$
B $\left(\begin{array}{l}\cos \theta \sin \theta \\ \sin \theta \cos \theta\end{array}\right)$
C $\left(\begin{array}{ll}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right)$
D $\left(\begin{array}{l}\cos \theta-\sin \theta \\ \sin \theta \cos \theta\end{array}\right)$

The correct answer is:
$\left(\begin{array}{l}\cos \theta-\sin \theta \\ -\sin \theta \cos \theta\end{array}\right)$

Solution:
$\mathrm{A}=\left[\begin{array}{ll}\cos \theta & \sin \theta \\ \sin \theta & \cos \theta\end{array}\right]$
$(\because \cos (-\theta)=\cos \theta$ and $\sin (-\theta)=-\sin \theta)$
$\operatorname{adj} \mathrm{A}=\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]$

Topic: Mathematics, Matrices, NDA, NDA 2018 (Phase 2)
What is the amount of heat needed to raise the temperature of the gas in a 44.8-liter cylinder containing helium gas at standard temperature and pressure by 15.0 °C, assuming the gas constant (R) is 8.31 J mol⁻¹ K⁻¹?

Options:
A 449.76 J
B 66.45 J
C 374 J
D 977 J

The correct answer is:
374 J

Solution:
Correct Option is : (C)
374 J

Topic: Chemistry, Kinetic Theory of Gases, NEET
What is the amount of heat required to vaporize one mole of a liquid at constant temperature and under standard pressure (1 bar) called?

Options:
A Standard enthalpy of sublimation
B Standard enthalpy of fusion
C Standard enthalpy of vaporization or molar enthalpy of vaporization
D Standard enthalpy of condensation

The correct answer is:
Standard enthalpy of vaporization or molar enthalpy of vaporization

Solution:
The amount of heat required to vaporize one mole of a liquid at constant temperature and under standard pressure (1 bar) is called its standard enthalpy of vaporization or molar enthalpy of vaporization, ∆vapH.

Topic: Chemistry, Thermodynamics (C), NEET
What is the angle between the lines $\frac{x-2}{1}=\frac{y+1}{-2}=\frac{z+2}{1}$ and $\frac{x-1}{1}=\frac{2 y+3}{3}=\frac{z+5}{2} ?$

Options:
A $\frac{\pi}{2}$
B $\frac{\pi}{3}$
C $\frac{\pi}{6}$
D None of the above

The correct answer is:
$\frac{\pi}{2}$

Solution:
The given lines are:-
$\frac{\mathrm{x}-2}{1}=\frac{\mathrm{y}-(-1)}{-2}=\frac{\mathrm{z}-(-2)}{1}$ and
$\frac{x-1}{1}=\frac{y-\left(-\frac{3}{2}\right)}{\frac{3}{2}}=\frac{z-(-5)}{2}$
dr's of Ist line are:-
$a_{1}=1, b_{1}=-2, c_{1}=1$
dr's of IInd line are:$\mathrm{a}_{2}=2, \mathrm{~b}_{2}=3, \mathrm{c}_{2}=4$
Let $^{\prime} \theta^{\prime}$ be the angle $\mathrm{b} / \mathrm{w}$ two lines, then,
$\cos \theta=\frac{\left|\mathrm{a}_{1} \mathrm{a}_{2}+\mathrm{b}_{1} \mathrm{~b}_{2}+\mathrm{c}_{1} \mathrm{c}_{2}\right|}{\sqrt{\mathrm{a}_{1}^{2}+\mathrm{b}_{1}^{2}+\mathrm{c}_{1}^{2}} \cdot \sqrt{\mathrm{a}_{2}^{2}+\mathrm{b}_{2}^{2}+\mathrm{c}_{2}^{2}}}$
$\cos \theta=0$
$\Rightarrow \quad \theta=\frac{\pi}{2}$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2012 (Phase 2)
What is the angle between the lines whose direction cosines are proportional to $(2,3,4)$ and $(1,-2,1)$ respectively?

Options:
A $90^{\circ}$
B $60^{\circ}$
C $45^{\circ}$
D $30^{\circ}$

The correct answer is:
$90^{\circ}$

Solution:
Since, direction cosines are proportional to $(2,3,4)$ and $(1,-2,1)$ respectively $2 \times 1+3 \times(-2)+4 \times 1=0$
Angle between the lines is $90^{\circ}$.

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2011 (Phase 1)
What is the angle between the lines $x+y=1$ and $x-y=1$ ?

Options:
A $\frac{\pi}{6}$
B $\frac{\pi}{4}$
C $\frac{\pi}{3}$
D $\frac{\pi}{2}$

The correct answer is:
$\frac{\pi}{2}$

Solution:
Slope of $\mathrm{x}+\mathrm{y}=1$ is $-1$ Slope of $\mathrm{x}-\mathrm{y}=1$ is 1 Let $\tan A=-1, \tan B=1$
$\mathrm{A}=\frac{3 \pi}{4}, \mathrm{~B}=\frac{\pi}{4}$
$\mathrm{~A}-\mathrm{B}=\frac{\pi}{2}$

Topic: Mathematics, Straight Lines, NDA, NDA 2013 (Phase 2)
What is the angle between the lines $x+z=0, y=0$ and $20 x$ $=15 v=12 z ? \quad$

Options:
A $\cos ^{-1}(1 / 5)$
B $\cos ^{-1}(1 / 7)$
C $\cos ^{-1} \frac{45}{7 \sqrt{61}}$
D $\sin ^{-1}(1 / 7)$

The correct answer is:
$\cos ^{-1}(1 / 5)$

Solution:
Given, $x+z=0, y=0$ and $20 x=15 y=12 z$
$\Rightarrow \frac{x}{1}=\frac{y}{0}=\frac{z}{-1}$ and $\frac{x}{3}=\frac{y}{4}=\frac{z}{5}$
Let $\theta$ be angle between two lines. $\cos \theta=\left|\frac{(1)(3)+(0)(4)+(-1)(5)}{\sqrt{1+0+1} \sqrt{9+16+25}}\right|=\left|\frac{3+0-5}{\sqrt{2} \sqrt{50}}\right|$
$=\frac{2}{\sqrt{2} \cdot 5 \sqrt{2}}=\frac{1}{5}$
$\theta=\cos ^{-1}\left(\frac{1}{5}\right)$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2009 (Phase 1)
What is the angle between the planes $2 \mathrm{x}-\mathrm{y}-2 \mathrm{z}+1=0$ and $3 x-4 y+5 z-3=0 ? \quad$

Options:
A $\frac{\pi}{6}$
B $\frac{\pi}{4}$
C $\frac{\pi}{3}$
D $\frac{\pi}{2}$

The correct answer is:
$\frac{\pi}{2}$

Solution:
Given equation of plane are $2 x-y-2 z+1=0$
$\Rightarrow \mathrm{a}_{1}=2, \mathrm{~b}_{1}=-1, \mathrm{c}_{1}=-2, \mathrm{~d}_{1}=1$
and $3 x-4 y+5 z-3=0$
$\Rightarrow a_{2}=3, b_{2}=-4, c_{2}=5, d_{2}=-3$
$\therefore$ Required angle is
$\cos \theta=\pm \frac{\mathrm{a}_{1} \mathrm{a}_{2}+\mathrm{b}_{1} \mathrm{~b}_{2}+\mathrm{c}_{1} \mathrm{c}_{2}}{\sqrt{\mathrm{a}_{1}^{2}+\mathrm{b}_{1}^{2}+\mathrm{c}_{1}^{2}} \sqrt{\mathrm{a}_{2}^{2}+\mathrm{b}_{2}^{2}+\mathrm{c}_{2}^{2}}}$
$\Rightarrow \cos \theta=\frac{2(3)+(-1)(-4)+5(-2)}{\sqrt{2^{2}+1^{2}+2^{2}} \sqrt{3^{2}+4^{2}+5^{2}}}=0$
$\Rightarrow \theta=\frac{\pi}{2}$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2013 (Phase 1)
What is the angle between the planes $2 x-y+z=6$ and $x+y+2 z=3 ? \quad$

Options:
A $\pi / 2$
B $\pi / 3$
C $\pi / 4$
D $\pi / 6$

The correct answer is:
$\pi / 3$

Solution:
We know, if
$$
\begin{array}{l}
a_{1} x+b_{1} y+c_{1} z=d_{1} \text { and } \\
a_{2} x+b_{2} y+c_{2} z=d_{2}
\end{array}$$ are two planes then angle between them is $$
\cos \theta=\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}} $$
Let $\theta$ be the angle between given planes Here, $a_{1}=2, b_{1}=-1, c_{1}=1$ $$ \begin{aligned}
& a_{2}=1, b_{2}=1, c_{2}=2 \\
\therefore \quad & \cos \theta=\frac{2 \times 1+1 \times(-1)+1 \times 2}{\sqrt{4+1+1} \sqrt{1+1+4}} \\
=& \frac{3}{6}=\frac{1}{2}=\cos \frac{\pi}{3} \\
\Rightarrow \quad \theta=\frac{\pi}{3}
\end{aligned} $$ $\Rightarrow \theta=\frac{\pi}{3}$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2009 (Phase 1)
What is the angle between the straight lines $\left(\mathrm{m}^{2}-\mathrm{mn}\right)$ $y=\left(m n+n^{2}\right) x+n^{3}$ and $\left(m n+m^{2}\right) y=\left(m n-n^{2}\right) x+m^{3}$
where $\mathrm{m}>\mathrm{n} ?$

Options:
A $\tan ^{-1}\left(\frac{2 \mathrm{mn}}{\mathrm{m}^{2}+\mathrm{n}^{2}}\right)$
B $\tan ^{-1}\left(\frac{4 \mathrm{~m}^{2} \mathrm{n}^{2}}{\mathrm{~m}^{4}-\mathrm{n}^{4}}\right)$
C $\tan ^{-1}\left(\frac{4 \mathrm{~m}^{2} \mathrm{n}^{2}}{\mathrm{~m}^{4}+\mathrm{n}^{4}}\right)$
D $45^{\circ}$

The correct answer is:
$\tan ^{-1}\left(\frac{4 \mathrm{~m}^{2} \mathrm{n}^{2}}{\mathrm{~m}^{4}-\mathrm{n}^{4}}\right)$

Solution:
Given straight lines are
$\left(m^{2}-m n\right) y=\left(m n+n^{2}\right) x+n^{3}$ ...(1)
$\left(m n+m^{2}\right) y=\left(m n-n^{2}\right) x+m^{3}$ ...(2)
(1) $\Rightarrow \mathrm{y}=\frac{\left(\mathrm{mn}+\mathrm{n}^{2}\right)}{\mathrm{m}^{2}-\mathrm{mn}} \mathrm{x}+\frac{\mathrm{n}^{3}}{\mathrm{~m}^{2}-\mathrm{mn}}$
So, slope of $(1), \mathrm{m}_{1}=\frac{\mathrm{mn}+\mathrm{n}^{2}}{\mathrm{~m}^{2}-\mathrm{mn}}$
(2) $\Rightarrow \mathrm{y}=\frac{\mathrm{mn}-\mathrm{n}^{2}}{\mathrm{mn}+\mathrm{m}^{2}} \mathrm{x}+\frac{\mathrm{m}^{3}}{\mathrm{mn}+\mathrm{m}^{2}}$
So, slope of $(2), \mathrm{m}_{2}=\frac{\mathrm{mn}-\mathrm{n}^{2}}{\mathrm{mn}+\mathrm{m}^{2}}$
If $\alpha$ is the angle between lines $(1),(2)$, then
$\tan \alpha=\frac{\mathrm{m}_{1}-\mathrm{m}_{2}}{1+\mathrm{m}_{1} \mathrm{~m}_{2}}$
$=\frac{\frac{m n+n^{2}}{m^{2}-m n}-\frac{m n-n^{2}}{m n+m^{2}}}{1+\left(\frac{m n+n^{2}}{m^{2}-m n}\right)\left(\frac{m n-n^{2}}{m n+m^{2}}\right)}$
$=\frac{\left(m n+n^{2}\right)\left(m n+m^{2}\right)-\left(m n-n^{2}\right)\left(m^{2}-m n\right)}{\left(m^{2}-m n\right)\left(m n+m^{2}\right)+\left(m n+n^{2}\right)\left(m n-n^{2}\right)}$
$=\frac{\mathrm{m}^{2} \mathrm{n}^{2}+\mathrm{m}^{3} \mathrm{n}+\mathrm{mn}^{3}+\mathrm{m}^{2} \mathrm{n}^{2}-\mathrm{m}^{3} \mathrm{n}+\mathrm{m}^{2} \mathrm{n}^{2}+\mathrm{m}^{2} \mathrm{n}^{2}-\mathrm{mn}^{3}}{\mathrm{~m}^{3} \mathrm{n}+\mathrm{m}^{4}-\mathrm{m}^{2} \mathrm{n}^{2}-\mathrm{m}^{3} \mathrm{n}+\mathrm{m}^{2} \mathrm{n}^{2}-\mathrm{mn}^{3}+\mathrm{mn}^{3}-\mathrm{n}^{4}}$
$\tan \alpha=\frac{4 \mathrm{~m}^{2} \mathrm{n}^{2}}{\mathrm{~m}^{4}-\mathrm{n}^{4}} \Rightarrow \alpha=\tan ^{-1}\left(\frac{4 \mathrm{~m}^{2} \mathrm{n}^{2}}{\mathrm{~m}^{4}-\mathrm{n}^{4}}\right)$

Topic: Mathematics, Pair of Lines, NDA, NDA 2018 (Phase 1)
What is the angle between the two lines whose direction numbers are $(\sqrt{3}-1,-\sqrt{3}-1,4)$ and $(-\sqrt{3}-1, \sqrt{3}-1,4) ?$

Options:
A $\frac{\pi}{6}$
B $\frac{\pi}{4}$
C $\frac{\pi}{3}$
D $\frac{\pi}{2}$

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

Solution:
Let position vectors of two points be
$\overrightarrow{\mathrm{OA}}=(\sqrt{3}-1) \hat{\mathrm{i}}-(\sqrt{3}+1) \hat{\mathrm{j}}+4 \hat{\mathrm{k}}$ and
$\overrightarrow{\mathrm{OB}}=-(\sqrt{3}+1) \hat{\mathrm{i}}+(\sqrt{3}-1) \hat{\mathrm{j}}+4 \hat{\mathrm{k}}$
$|\overrightarrow{\mathrm{OA}}|=\sqrt{(\sqrt{3}-1)^{2}+(\sqrt{3}+1)^{2}+4^{2}}$
$=\sqrt{3+1-2 \sqrt{3}+3+1+2 \sqrt{3}+16}=\sqrt{24}$
Also, $|\overrightarrow{\mathrm{OB}}|=\sqrt{(\sqrt{3}+1)^{2}+(\sqrt{3}-1)^{2}+4^{2}}=\sqrt{24}$
$\overrightarrow{\mathrm{OA}} \cdot \overrightarrow{\mathrm{OB}}=|\overrightarrow{\mathrm{OA}}| \cdot|\overrightarrow{\mathrm{OB}}| \cdot \cos \theta$, where $\theta$ is the angle
between $\overrightarrow{\mathrm{OA}} \& \overrightarrow{\mathrm{OB}}$
$\overrightarrow{\mathrm{OA}} \cdot \overrightarrow{\mathrm{OB}}=-(\sqrt{3}-1)(\sqrt{3}+1)-(\sqrt{3}+1)(\sqrt{3}-1)+16$
$=-3+1-3+1+16=12$
So, $\cos \theta=\frac{\overrightarrow{\mathrm{OA}} \cdot \overrightarrow{\mathrm{OB}}}{|\overrightarrow{\mathrm{OA}}| \overrightarrow{\mid \mathrm{OB}} \mid}=\frac{12}{24}=\frac{1}{2}$
$\Rightarrow \theta=\frac{\pi}{3}$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2006 (Phase 1)
What is the angle between the two straight lines
$\mathrm{y}=(2-\sqrt{3}) \mathrm{x}+5$ and $\mathrm{y}=(2+\sqrt{3}) \mathrm{x}-7 ?$

Options:
A $60^{\circ}$
B $45^{\circ}$
C $30^{\circ}$
D $15^{\circ}$

The correct answer is:
$60^{\circ}$

Solution:
The given lines are $y=(2-\sqrt{3}) x+5$ and $y=(2+\sqrt{3}) x-7$
Therefore, slope of first line $\mathrm{m}_{1}=2-\sqrt{3}$ and slope of second line $\mathrm{m}_{2}=2+\sqrt{3}$ $\tan \theta=\left|\frac{\mathrm{m}_{2}-\mathrm{m}_{1}}{1+\mathrm{m}_{1} \mathrm{~m}_{2}}\right|=\left|\frac{2+\sqrt{3}-2+\sqrt{3}}{1+(4-3)}\right|$
$=\left|\frac{2 \sqrt{3}}{2}\right|=\sqrt{3}=\tan \frac{\pi}{3} \Rightarrow \theta=\frac{\pi}{3}=60^{\circ}$

Topic: Mathematics, Straight Lines, NDA, NDA 2006 (Phase 1)
What is the angle between two planes $2 x-y+z=4$ and $x+y+2 z=6 ? \quad$

Options:
A $\frac{\pi}{2}$
B $\frac{\pi}{3}$
C $\frac{\pi}{4}$
D $\frac{\pi}{6}$

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

Solution:
Given equations of two planes are $2 x-y+z=4$ and $x+y+2 z=6$
So, angle between them is
$\cos \theta=\frac{2(1)+(-1)(1)+(1)(2)}{\sqrt{4+1+1} \sqrt{4+1+1}}$
$=\frac{2-1+2}{\sqrt{6} \sqrt{6}}=\frac{3}{6}=\frac{1}{2}$
$\Rightarrow \theta=\frac{\pi}{3}$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2011 (Phase 2)
What is the angle (in circular measure) between the hour hand and the minute hand of a clock when thetime is half past $4 \mathrm{~m}$

Options:
A $\frac{\pi}{3}$
B $\frac{\pi}{4}$
C $\frac{\pi}{6}$
D None of these

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

Solution:
Angle traced by the hour hand in 12 hours $=360^{\circ}$ Angle traced by it in $4 \mathrm{hr} 30 \mathrm{~min}\left(4 \mathrm{~h}+\frac{30}{60} \mathrm{hr}\right)=\frac{9}{2} \mathrm{hr}$
$=\frac{9}{2} \times \frac{360}{12}=13$
Angle traced by minute hand is $60^{\circ} \mathrm{min}=360^{\circ}$
Angle traced by it in $30 \mathrm{~min}=\frac{30}{60} \times 360=180^{\circ}$
RequiredAngle $=180^{\circ}-135^{\circ}=45^{\circ} \Rightarrow 45 \times \frac{\pi}{100}=\frac{\pi}{4}$ radian

Topic: Mathematics, Sets and Relations, NDA, NDA 2013 (Phase 2)
What is the angle subtended by $1 \mathrm{~m}$ pole at a distance $1 \mathrm{~km}$ on the ground in sexagesimal measure?

Options:
A $\frac{9}{50 \pi}$ degrec
B $\frac{9}{5 \pi}$ degrec
C $3.4$ minute
D $3.5$ minute

The correct answer is:
$\frac{9}{50 \pi}$ degrec

Solution:
Let $\mathrm{AB}$ be the pole of $1 \mathrm{~m}$. $B C=1 \mathrm{~km}=1000 \mathrm{~m}$
Let ' $\theta$ ' be the required angle.
Now, $\tan \theta=\left(\frac{1}{1000}\right)^{\circ}$
Since ${ }^{\circ} \theta$ ' is very small
$\therefore \tan \theta=\theta=\left(\frac{1}{1000}\right)^{\circ}$
Now, consider option (a)
$\left(\frac{9}{50 \pi}\right)^{\circ}=\left(\frac{9}{50 \times 180}\right)^{\circ}=\left(\frac{9}{9000}\right)^{\circ}=\left(\frac{1}{1000}\right)^{\circ}$
Hence, required angle $=\left(\frac{9}{50 \pi}\right)^{\circ}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2012 (Phase 1)
What is the approximate standard free energy change per mole of $\mathrm{Zn}$ (in $\mathrm{kJ} \mathrm{mol}^{-1}$ ) for a Daniel cell at $298 \mathrm{~K}$ ?

Options:
A -212.3
B 230.0
C 0.0
D -1.10

The correct answer is:
-212.3

Solution:
For Daniell cell we have
$$
\mathrm{Zn}(s)+\mathrm{Cu}^{2+}(a q) \longrightarrow \mathrm{Zn}^{2+}(a q)+\mathrm{Cu}(s)
$$
Gibbs free energy $=-n F E$
$n=$ number of electrons $=2$
$F=$ Faraday's constant $=96500 \mathrm{C}$
$E=\operatorname{emf}$ of cell $=1.1 \mathrm{~V}$
$\therefore$ Gibbs free energy $=-2 \times 96500 \times 1.1 \mathrm{~J}=-212.3 \mathrm{~kJ}$

Topic: Chemistry, Electrochemistry, TS EAMCET, TS EAMCET 2018 (07 May Shift 1)
What is the approximate value of $(1.02)^{8}$ ?

Options:
A 1.171
B 1.175
C 1.177
D 1.179

The correct answer is:
1.171

Solution:
$(1.02)^{8}=(1+0.02)^{8}$
$(1+x)^{\mathrm{n}}=1+\mathrm{nx}+\frac{\mathrm{n}(\mathrm{n}-1)}{2 !} \mathrm{x}^{2}+\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)}{3 !} \mathrm{x}^{3}+\ldots .$
$\mathrm{n}=8, \mathrm{x}=0.02$
$(1+0.02)^{8}$
$=1+8 \times 0.02+\frac{8 \times 7}{2 !} \cdot(0.02)^{2}+\frac{8.7 .6}{3 !}(.02)^{3}$
Neglecting higher terms $=1+0.16+28 \times 0.0004+56 \times 0.000008$
$\simeq 1+0.16+0.0112=1.171$

Topic: Mathematics, Binomial Theorem, NDA, NDA 2008 (Phase 1)
What is the area between the curve
$y=\cos 3 x, 0 \leq x \leq \frac{\pi}{6}$ and the co-ordinate axes?

Options:
A 1 squale unit
B $\frac{1}{2}$ squaue unit
C $\frac{1}{3}$ square unit
D $\frac{1}{4}$ square unit

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

Solution:
Required Area $=\int_{0}^{\pi / 6} \cos 3 x d x$
$=\left.\frac{\sin 3 x}{3}\right|_{0} ^{\pi / 6}=\frac{\sin 3(\pi / 6)}{3}-\sin 0$
$=\frac{1}{3} \sin \pi / 2-0=\frac{1}{3}(1)=\frac{1}{3}$ sq. unit.

Topic: Mathematics, Area Under Curves, NDA, NDA 2011 (Phase 2)
What is the area bounded by the curve $y=4 x-x^{2}-3$ and the $x$ -axis? $\quad$

Options:
A $2 / 3$ sq unit
B $4 / 3$ sq unit
C $5 / 3 \mathrm{sq}$ unit
D $4 / 5 \mathrm{sq}$ unit

The correct answer is:
$4 / 3$ sq unit

Solution:
Given curve is $y=4 x-x^{2}-3$ Since, area bounded by $x$ -axis $\therefore y=0$
$\Rightarrow 4 x-x^{2}-3=0 \Rightarrow x^{2}-4 x+3=0$
$\Rightarrow x^{2}-3 x-x+3=0 \Rightarrow(x-3)(x-1)=0$
$\Rightarrow x=1,3$
$\therefore$ Required area $=\int_{1}^{3}\left(4 x-x^{2}-3\right) d x$
$=\frac{4 x^{2}}{2}-\frac{x^{3}}{3}-\left.3 x\right|_{1} ^{3}=\left(\frac{36}{2}-\frac{27}{3}-9\right)-\left(\frac{4}{2}-\frac{1}{3}-3\right)$
$=(18-9-9)-\left(2-\frac{10}{3}\right)=0-\left(\frac{-4}{3}\right)=\frac{4}{3}$ sq. unit.

Topic: Mathematics, Area Under Curves, NDA, NDA 2009 (Phase 2)
What is the area bounded by the curve $y=x^{2}$ and the line $y=16 ?$

Options:
A $32 /$
B $64 / 3$
C $256 / 3$
D $128 /$

The correct answer is:
$256 / 3$

Solution:
Given, equations of curves are $y=x^{2} \quad \ldots$ (i)
and $y=16$ $\ldots$ (ii) On solving Eqs. (i) and (ii), we get $x^{2}=16 \Rightarrow x=4,-4$
$\therefore \quad$ Points of intersection are $(4,16)$, and $(-4,16)$.
Required area $=\int_{-4}^{4}\left(16-x^{2}\right) d x=2 \int_{0}^{4}\left(16-x^{2}\right) d x$
$=2\left[16 x-\frac{x^{3}}{3}\right]_{0}^{4}=2\left[64-\frac{64}{3}\right]=2 \times 64 \times \frac{2}{3}$
$=\frac{256}{3}$ sq unit

Topic: Mathematics, Area Under Curves, NDA, NDA 2010 (Phase 2)