datavorous/entrance-exam-dataset huggingface dataset:
Suppose $f(x)$ is such a quadrant expression that it is positive for all real $\mathrm{x}$. If $g(x)=f(x)+f^{\prime}(x)+f^{\prime \prime}(x)$, then for any real $x$. Then for any real $\mathrm{x} .$

Options:
A $g(x) < 0$
B $g(x)>0$
C $g(x)=0$
D $\mathrm{g}(\mathrm{x}) \geq 0$

The correct answer is:
$g(x)>0$

Solution:
Let $f(x)=a x^{2}+b x+c, a>0, b^{2} < 4 a c \quad(\because f(x)>0)$
Now, $g(x)=a x^{2}+b x+c+2 a x+b+2 a$
$=a x^{2}+(b+2 a) x+2 a+b+c$
Now, $(b+2 a)^{2}-4 a(2 a+b+c)$
$=b^{2}+4 a b+4 a^{2}-8 a^{2}-4 a b-4 a c$
$=b^{2}-4 a c-8 a^{2} < 0$
$\Rightarrow g(x)>0$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2018 (Phase 2)
Suppose $\omega$ is a cube root of unity with $\omega \neq 1$. Suppose $\mathrm{P}$ and $Q$ are the points on the complex plane defined by $\omega$ and $\omega^{2}$. If $\mathrm{O}$ is the origin, then what is the angle between $\mathrm{OP}$ and $\mathrm{OQ} ?$

Options:
A $60^{\circ}$
B $90^{\circ}$
C $120^{\circ}$
D $150^{\circ}$

The correct answer is:
$120^{\circ}$

Solution:
$\because \omega=\frac{-1}{2}+i \frac{\sqrt{3}}{2}$
$\& \omega^{2}=\frac{-1}{2}-i \frac{\sqrt{3}}{2}$
$-\frac{1}{2}+\frac{\sqrt{3}}{2} i$ $(\because \omega \neq 1)$
$P$ and $Q$ are points on complex plane. Angle between $O P$ and $O Q$ is
$\theta=\tan ^{-1}\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right|$
$m_{1}$ for line $O P, \quad m_{2}$ for line $O Q$
$m_{1}=\frac{\frac{\sqrt{3}}{2}-0}{\frac{-1}{2}-0} \quad m_{2}=\frac{\frac{-\sqrt{3}}{2}-0}{\frac{-1}{2}-0}$
$\Rightarrow m_{1}=-\sqrt{3} \quad \Rightarrow m_{2}=\sqrt{3}$
$\theta=\tan ^{-1}\left[\frac{-\sqrt{3}-\sqrt{3}}{1+(-\sqrt{3})(\sqrt{3})}\right]$
$=\tan ^{-1}\left[\frac{-2 \sqrt{3}}{-2}\right]=\pi-\tan ^{-1} \sqrt{3}=\pi-\tan ^{-1} \tan \frac{\pi}{3}$
$\theta=\pi-\frac{\pi}{3}$
$\theta=120^{\circ}$

Topic: Mathematics, Complex Number, NDA, NDA 2016 (Phase 1)
Suppose that A denotes the collection of all complexnumbers whose square is a negative real number. Which one of the following statements is correct?

Options:
A $\mathrm{A} \subseteq \mathrm{R}$
B $\mathrm{A} \supseteq \mathrm{R}$
C $\mathrm{A}=\left\{\mathrm{x}+\mathrm{iyp}^{2} \in \mathrm{R} \psi, \mathrm{y} \in \mathrm{R}\right\}$
D $A=\{$ iy $y \in R$ \}

The correct answer is:
$A=\{$ iy $y \in R$ \}

Solution:
If A denotes the collection of all complex number whose square is a negative real number, then Square of a complex number is a negative real number only if it has no real partand has only imaginary part. Hence, $\mathrm{A}=\{\mathrm{iy}: \mathrm{y} \in \mathrm{R}\}$

Topic: Mathematics, Sets and Relations, NDA, NDA 2006 (Phase 1)
Suppose that the electric field part of an electromagnetic wave in vacuum is
$E=\left\{(3.1 \mathrm{~N} / \mathrm{C}) \cos \left[(1.8 \mathrm{rad} / \mathrm{m}) \mathrm{y}+\left(5.4 \times 10^6 \mathrm{rad} / \mathrm{s}\right) t\right]\right\} \hat{\mathrm{i}}$.
(a) What is the direction of propagation?
(b) What is the wavelength $\lambda$ ?
(c) What is the frequency $v$ ?
(d) What is the amplitude of the magnetic field part of the wave?
(e) Write an expression for the magnetic field part of the wave.

Options:

None

Solution:
$\mathrm{E}=\{(3.1 \mathrm{~N} / \mathrm{C}) \cos [1.8 \mathrm{rad} / \mathrm{m}) \mathrm{y}+(1.8 \mathrm{rad} / \mathrm{s}) \mathrm{t}]\} \mathrm{i}$
(a) The waves is propagating in the -ve y-direction.
(b) Comparing eqs.(1) with the standard equation for the e.m. wave. $E=E_0 \sin \left[2 \pi\left(\frac{y}{\lambda}-v t\right)\right]$ we have $\frac{2 \pi}{\lambda}=1.8 \Rightarrow \lambda=\frac{2 \pi}{1.8}=\frac{2 \times 22}{7 \times 1.8}=3.488 \Rightarrow \lambda=3.5 \mathrm{~m}$
(c) Frequency $v=\frac{\mathrm{c}}{\lambda}=\frac{3 \times 10^8}{3.5}=0.85 \times 10^8 \mathrm{~Hz}$ $=8.6 \times 10^7 \mathrm{~Hz}=86 \mathrm{MHz}$
(d) Amplitude of magnetic field
$$
=\mathrm{B}_0=\frac{\mathrm{E}_0}{\mathrm{c}}=\frac{3.1}{3 \times 10^8}=1.03 \times 10^{-8}
$$
$$
\approx 10.3 \times 10^{-9}=10 \mathrm{nT}
$$
(e) Expression for magnetic field
$$
\mathrm{B}=\left[(10 \mathrm{nT}) \cos \left\{(1.8 \mathrm{rad} / \mathrm{m}) \mathrm{y}+\left(5.4 \times 10^6 \mathrm{rad} / \mathrm{s}\right) \mathrm{t}\right\}\right] \hat{\mathrm{k}}
$$

Topic: Physics, Electromagnetic Waves
Suppose the function $f(\mathrm{x})=\mathrm{x}^{\mathrm{n}}, \mathrm{n} \neq 0$ is differentiable for all $x$. Then $n$ can be any element of the interval

Options:
A $(1, \infty)$
B $(0, \infty)$
C $\left(\frac{1}{2}, \infty\right)$
D None of the above

The correct answer is:
$(1, \infty)$

Solution:
$\mathrm{f}(\mathrm{x})=\mathrm{x}^{\mathrm{n}}, \mathrm{n} \neq 0$ This function is differentiable for all values of $\mathrm{n}$, except 0 . Hence $\mathrm{n} \in(1, \infty)$

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2017 (Phase 1)
Suppose the system of equations
$\begin{array}{l}
a_{1} x+b_{1} y+c_{1} z=d_{1} \\
a_{2} x+b_{2} y+c_{2} z=d_{2} \\
a_{3} x+b_{3} y+c_{3} z=d_{3}
\end{array}$
has a unique solution $\left(x_{0}, y_{0}, z_{0}\right)$. If $x_{0}=0$, then which one of the following is correct?

Options:
A $\left|\begin{array}{lll}\mathrm{a}_{1} & \mathrm{~b}_{1} & \mathrm{c}_{1} \\ \mathrm{a}_{2} & \mathrm{~b}_{2} & \mathrm{c}_{2} \\ \mathrm{a}_{3} & \mathrm{~b}_{3} & \mathrm{c}_{3}\end{array}\right|=0$
B $\left|\begin{array}{lll}\mathrm{d}_{1} & \mathrm{~b}_{1} & \mathrm{c}_{1} \\ \mathrm{~d}_{2} & \mathrm{~b}_{2} & \mathrm{c}_{2} \\ \mathrm{~d}_{3} & \mathrm{~b}_{3} & \mathrm{c}_{3}\end{array}\right|=0$
C $\left|\begin{array}{lll}\mathrm{d}_{1} & \mathrm{a}_{1} & \mathrm{c}_{1} \\ \mathrm{~d}_{2} & \mathrm{a}_{2} & \mathrm{c}_{2} \\ \mathrm{~d}_{3} & \mathrm{a}_{3} & \mathrm{c}_{3}\end{array}\right|=0$
D $\left|\begin{array}{lll}\mathrm{d}_{1} & \mathrm{a}_{1} & \mathrm{~b}_{1} \\ \mathrm{~d}_{2} & \mathrm{a}_{2} & \mathrm{~b}_{2} \\ \mathrm{~d}_{3} & \mathrm{a}_{3} & \mathrm{~b}_{3}\end{array}\right|=0$

The correct answer is:
$\left|\begin{array}{lll}\mathrm{d}_{1} & \mathrm{~b}_{1} & \mathrm{c}_{1} \\ \mathrm{~d}_{2} & \mathrm{~b}_{2} & \mathrm{c}_{2} \\ \mathrm{~d}_{3} & \mathrm{~b}_{3} & \mathrm{c}_{3}\end{array}\right|=0$

Solution:
The given system of equations is
$\mathrm{a}_{1} \mathrm{x}+\mathrm{b}_{1} \mathrm{y}+\mathrm{c}_{1} \mathrm{z}=\mathrm{d}_{1}$
$a_{2} x+b_{2} y+c_{2} z=d_{2}$
and $\mathrm{a}_{3} \mathrm{x}+\mathrm{b}_{3} \mathrm{y}+\mathrm{c}_{3} \mathrm{z}=\mathrm{d}_{3}$
Let $\Delta=\left|\begin{array}{lll}\mathrm{a}_{1} & \mathrm{~b}_{1} & \mathrm{c}_{1} \\ \mathrm{a}_{2} & \mathrm{~b}_{2} & \mathrm{c}_{2} \\ \mathrm{a}_{3} & \mathrm{~b}_{3} & \mathrm{c}_{3}\end{array}\right|$
This system has a unique solution $x_{0}, y_{0}, z_{0}$ if $\Delta \neq 0$
and $x_{0}=\frac{\Delta x}{\Delta}=0 \Rightarrow \Delta x=0$
$\Rightarrow\left|\begin{array}{lll}\mathrm{d}_{1} & \mathrm{~b}_{1} & \mathrm{c}_{1} \\ \mathrm{~d}_{2} & \mathrm{~b}_{2} & \mathrm{c}_{2} \\ \mathrm{~d}_{3} & \mathrm{~b}_{3} & \mathrm{c}_{3}\end{array}\right|=0$

Topic: Mathematics, Determinants, NDA, NDA 2008 (Phase 1)
Suppose there is a relation $*$ between the positive numbers
$x$ and $y$ given by $x * y$ if and only if $x \leq y^{2}$. Then which one of the following is correct?

Options:
A $*$ is reflexive but not transitive and symmetric
B $*$ is transitive but not reflexive and symmetric
C *is symmetric and reflexive but not tran sitive
D $*$ is symmetric and but not reflexive and transitive

The correct answer is:
$*$ is reflexive but not transitive and symmetric

Solution:
$x$ and $y$ are positive numbers.
$x \leq v^{2}$
Reflexive-
$x \leq x^{2} \forall$ positive numbers. Hence relation is reflexive. Transitive $-$ $x \leq y^{2} \quad y \leq z^{2}$
Let $x=5, y=3, z=2$
$5 \leq(3)^{2} \quad(3) \leq(2)^{2}$
but $5 \pm(2)^{2}$
Hence, $x \leq y^{2} \quad y \leq z^{2}$
but $x \leq z^{2}$
Thusrelation is not transitive. Symmetric
$1 \leq(2)^{2}$ while $2 \pm(1)^{2}$
Hence relation is not symmetric.
Thus $x \leq y^{2} \forall$ positivenumbersis reflexive, but not transitive and symmetric.

Topic: Mathematics, Sets and Relations, NDA, NDA 2016 (Phase 1)
Suppose $\mathrm{X}=\{1,2,3,4\}$ and $\mathrm{R}$ is a relation on $\mathrm{X}$. If $\mathrm{R}=\{(1,1),(2,2),(3,3),(1,2),(2,1),(2,3) \cdot(3,2)\}$, then which
one of the following is correct?

Options:
A $\mathrm{R}$ is reflexive and symmetric, but not transitive
B $\mathrm{R}$ is symmetric and transitive, but not reflexive
C $\mathrm{R}$ is reflexiveand transitive, but not symmetric
D $\mathrm{R}$ is neither reflexive nor transitive, but symmetric

The correct answer is:
$\mathrm{R}$ is neither reflexive nor transitive, but symmetric

Solution:
Given,
$x=\{1,2,3,4\}$ $\mathrm{R}=\{(1,1),(2,2),(3,3),(1,2),(2,1),(2,3),(3,2)\}$
$\mathrm{R}$ is reflexive, if aRa for all $\mathrm{a} \in \mathrm{x}$ $(4,4) \notin \mathrm{R}$
$\therefore \mathrm{R}$ is not reflexive ...(1)
$\mathrm{R}$ is transitive, if $\mathrm{R} \mathrm{R} \mathrm{b}, \mathrm{b} \mathrm{R} \mathrm{c} \Rightarrow \mathrm{aR} \mathrm{C}$ for all $\mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathrm{x}(1,2)$
$(2,3), \in \mathrm{R}$, but $(1,3) \notin \mathrm{R}$
$\therefore \mathrm{R}$ is not transitive...(2)
$\mathrm{R}$ is symmetric, if $\mathrm{a} \mathrm{R} \mathrm{b} \Rightarrow \mathrm{b} \mathrm{R} \mathrm{a}$, for all $\mathrm{a}, \mathrm{b} \in \mathrm{x}$
$\therefore \mathrm{R}$ is symmetric...(3)
(3) From $(1),(2),(3)$ we can say, $\mathrm{R}$, is neither reflexivenor transitive, but symmetric.

Topic: Mathematics, Sets and Relations, NDA, NDA 2019 (Phase 1)
Take one flower each of the families Fabaceae and Solanaceae and write its semi-technical description. Also draw their floral diagram after studying them.

Options:

None

Solution:
Fabaceae: This family was earlier called papilionoidae, a subfamily of family Leguminosae. It is distributed all over the world.
Example: Pisum sativum
Semi technical descritpion of $P$ isum sativum are as follows:
Vegetative characters :
Habit : An annual herb.
Root: Nodulated tap root.
Stem : Climber, leaflet tendrils
Leaves: Alternate, pinnately compound or simple; leaf base, pulvinate, stipulate, venation reticulate.
Floral characters :
Inflorescence : Racemose
Flower: Bisexual, zygomorphic, complete, irregular, hypogynous Calyx : Sepal 5, gamosepalous, valvate aestivation.
Corolla : Petals 5, polypetalous, papilionaceous consisting of a posterior standard, two lateral wings, two anterior ones forming a keel. Thus, flower becomes zygomorphic, with descending imbricate aestivation i.e. vexillary aestivation. Androecium : Stamen 10, diadelphous $[1+(9)]$, anther dithecous, introrse.
Gynoecium: Ovary superior, monocarpellary, unilocular with many ovules, marginal placentation.
Fruit : Legume
Seed : One to many, non-endospermic.
Economic importance
Many plants belonging to the family are sources of pulses (gram, arhar, sem, moong, soyabean, edible oil (soyabean, groundnut); fibres (sunhemp); fodder (Sesbania, Trifolium), ornamentals (lupin, sweet pea); medicine (muliathi).
Solanaceae :
It is large family, commonly called as the potato family. It is widely distributed in tropics, subtropics and temperate zones.
Example: Datura
Semi technical description of Solanum nigrum.
Vegetative characters :
Habit : Annual herb
Stem : Erect, cylindrical, hairy, slightly fistular.
Leaves : Alternate, simple, petiolate, ovate with acute apex, venation reticulate.
Floral characters :
Inflorescence : Solitary, axillary
Flower : Ebracteate, actinomorphic, hypogynous
Calyx : Sepals 5, gamosepalous, persistent, valvate astivation.
Corolla : Petals 5 , gamopetalous, valvate aestivation
Androecium : Stamen 5, epipetalous
Gynoecium : Bicarpellary, syncarpous, ovary superior, bilocular but four celled by formation of false septum, placenta swollen with many ovules.
Fruit: Spinous capsule with septifragal dehiscence.

Topic: Biology, Morphology of Flowering Plants
$\tan 54^{\circ}$ can be expressed as

Options:
A $\frac{\sin 9^{\circ}+\cos 9^{\circ}}{\sin 9^{\circ}-\cos 9^{\circ}}$
B $\frac{\sin 9^{\circ}-\cos 9^{\circ}}{\sin 9^{\circ}+\cos 9^{\circ}}$
C $\frac{\cos 9^{\circ}+\sin 9^{\circ}}{\cos 9^{\circ}-\sin 9^{\circ}}$
D $\frac{\sin 36^{\circ}}{\cos 36^{\circ}}$

The correct answer is:
$\frac{\cos 9^{\circ}+\sin 9^{\circ}}{\cos 9^{\circ}-\sin 9^{\circ}}$

Solution:
\tan 54^{\circ}=\tan \left(45^{\circ}+9^{\circ}\right) \\=& \frac{\tan 45^{\circ}+\tan 9^{\circ}}{1-\tan 45^{\circ} \cdot \tan 9^{\circ}}=\frac{1+\tan 9^{\circ}}{1-\tan 9^{\circ}} \\

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2019 (Phase 1)
\text { 1 L closed flask contains a mixture of } 4 \mathrm{~g} \text { of methane and } \\ 4.4 \mathrm{~g} \text { of carbon dioxide. The pressure inside the flask at }\end{array}$
$27^{\circ} \mathrm{C}$ is [Assume ideal behaviour of gases]

Options:
A $8.6 \mathrm{~atm}$
B $2.2 \mathrm{~atm}$
C $4.2 \mathrm{~atm}$
D $6.1 \mathrm{~atm}$

The correct answer is:
$8.6 \mathrm{~atm}$

Solution:
no. of moles of $\mathrm{CH}_4\left(\mathrm{n}_1\right)=4 / 16=0.25 \mathrm{~mol}$
no. of moles of $\mathrm{CO}_2\left(\mathrm{n}_2\right)=4.4 / 44=0.1 \mathrm{~mol}$
Total no. of moles $\left(\mathrm{n}_{\mathrm{T}}\right)=\mathrm{n}_1+\mathrm{n}_2=0.35 \mathrm{~mol}$
$\therefore \quad \mathrm{P}=\frac{0.35 \times 0.082 \times 300}{1}=8.6 \mathrm{~atm}$

Topic: Chemistry, States of Matter, TS EAMCET, TS EAMCET 2022 (20 Jul Shift 2)
\[
\text { If } f(x)=\left\{\begin{array}{cc}
\frac{3 \sin \Pi x}{5 x} ; & x \neq 0 \\
2 K ; & x=0
\end{array}\right.
\]
is continuous at \( x=0 \), then the value of \( K \) is

Options:
A \( \frac{3 \pi}{10} \)
B \( \frac{3 \pi}{5} \)
C \( \frac{\Pi}{10} \)
D \( \frac{3 \pi}{2} \)

The correct answer is:
\( \frac{3 \pi}{10} \)

Solution:
Given that \( f(x)=\left\{\begin{array}{cl}\frac{3 \sin \pi x}{5 x} & x \neq 0 \\ 2 K & x=0\end{array}\right. \)
Now,
\[
\begin{array}{l}
\lim _{x \rightarrow 0}\left(\frac{3 \sin \pi x}{5 x}\right)=2 K \\
\Rightarrow \frac{3 \pi}{5} \lim _{x \rightarrow 0} \frac{\sin \pi x}{x I I}=2 K \\
\Rightarrow \frac{3 \Pi}{5} \times 1=2 K \Rightarrow K=\frac{3 \Pi}{10}
\end{array}
\]

Topic: Mathematics, Trigonometric Ratios & Identities, KCET, KCET 2014
\text { What is the value of } \tan 75^{\circ}+\cot 75^{\circ} ?

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

The correct answer is:
4

Solution:
$\tan 75^{\circ}+\cot 75^{\circ}=2+\sqrt{3}+2-\sqrt{3}=4$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2019 (Phase 1)
The 59 th term of an $\mathrm{AP}$ is 449 and the 449 th term is 59 . Which term is equal to 0 (zero)?

Options:
A $501^{\text {st }}$ term
B $502^{\text {nd }}$ term
C $508^{\text {th }}$ term
D $509^{\text {th }}$ term

The correct answer is:
$508^{\text {th }}$ term

Solution:
Let a and $\mathrm{d}$ be the first term and common difference of an AP respectively
$\therefore \quad a+58 d=449$ and $a+448 d=59$
On solving Eqs. (i) and (ii), we get
$a=507$ and $d=-1$
$\mathrm{n}^{\text {th }}$ term of $\mathrm{AP}$ is $\mathrm{a}+(\mathrm{n}-1) \mathrm{d}$
Let us assume that $\mathrm{n}^{\text {th }}$ term will be zero.
$\therefore \quad a+(n-1) d=0$
$\Rightarrow 0=507+(n-1)(-1)$
$\Rightarrow 507=n-1 \Rightarrow n=508$
Hence, $508^{\text {th }}$ term will be zero.

Topic: Mathematics, Sequences and Series, NDA, NDA 2010 (Phase 1)
The abundantly distributed enzyme in germinating seeds and adipocytes is

Options:
A lipases
B proteases
C nucleases
D cellulases

The correct answer is:
lipases

Solution:
The germinating seeds and adipocytes contain fat as reserved food materials and while metabolic reactions in order to release the free fatty acids and glycerol the lipases acts on the stored lipids.

Topic: Chemistry, Biomolecules, NEET
The acidic oxide from the following is

Options:
A $\mathrm{SnO}_2$
B $\mathrm{SiO}_2$
C $\mathrm{PbO}_2$
D $\mathrm{SnO}$

The correct answer is:
$\mathrm{SiO}_2$

Solution:
On moving down the group, the tendancy to lose
electron increases due to which basic character of oxides
increases and acidic character decreases. Hence, $\mathrm{SiO}_2$
form acidic oxide.

Topic: Chemistry, p Block Elements (Group 13 & 14), TS EAMCET, TS EAMCET 2022 (19 Jul Shift 1)
The acromion process is a part of the

Options:
A humerus
B sternum
C clavicle
D scapula

The correct answer is:
scapula

Solution:
Human skeleton system is divided into Axial and Appendicular. Axial skeleton consists of 80 bones and Appendicular skeleton consists of 126 bones. It includes the bones of girdles and bones of appendages. Pectoral girdle is the upper girdle. It consists of two scapulae and two clavicles. Scapula is also known as shoulder blade. It articulates with the clavicle on one side and with the humerus on the other side. It shows the presence of two processes: Acromion process and Coracoid process. Acromion process articulates with the clavicle and coracoid process with the ligament.

Topic: Biology, Locomotion and Movement, NEET
The activation energy of a chemical reaction can be determined by,

Options:
A evaluating rate constants at two different temperatures.
B O changing the concentration of reactants.
C evaluating the concentration of reactants at two different temperatures.
D evaluating rate constant at standard temperature,

The correct answer is:
evaluating rate constants at two different temperatures.

Solution:
Activation energy of a chemical reaction can be determined by measuring the values of rate constant at two
differenttemperatures
Let \( k_{1} \) and \( k_{2} \) are the rate constant for the reaction at two different temperature, that is, \( T_{1} \) and \( T_{2} \) respectively.
\( \log k_{1}=\log A-\frac{E_{a}}{2.303 R T_{1}} \rightarrow(1) \) \( \log k_{2}=\log A-\frac{E_{a}}{2.303 R T_{2}} \rightarrow(2) \)
Subtract Eq. (1) from Eq. (2), we get
\[
\begin{array}{l}
\log k_{2}-\log k_{1}=\frac{E_{a}}{2.303 R}\left[\frac{1}{T_{1}}-\frac{1}{T_{2}}\right] \\
\log \frac{k_{2}}{k_{1}}=\frac{E_{a}}{2.303 R}\left[\frac{1}{T_{1}}-\frac{1}{T_{2}}\right] \\
\log \frac{k_{2}}{k_{1}}=\frac{E_{a}}{2.303 R}\left[\frac{T_{2}-T_{1}}{T_{1} T_{2}}\right] \\
E_{a}=2.303 R \log \frac{k_{2}}{k_{1}} \times\left[\frac{T_{1} T_{2}}{T_{2}-T_{1}}\right]
\end{array}
\]

Topic: Chemistry, Chemical Kinetics, KCET, KCET 2016
The actual 3D structure of tRNA molecule appears as

Options:
A L-shaped
B E-shaped
C Y-shaped
D S-shaped

The correct answer is:
L-shaped

Solution:
In actual or primary structure, the tRNA is a compact molecule which looks like inverted L-shaped structure. Kim et al., (1973) suggested the 3D, L-shaped model of t-RNA by X-Ray diffraction analysis. The secondary structure of tRNA is the clover-leaf structure.

Topic: Biology, Molecular Basis of Inheritance, NEET
The acute angle which the perpendicular from origin on the line $7 x-3 y=4$ makes with the $x$ -axis is

Options:
A zero
B positive but not $\pi / 4$
C negative
D $\pi / 4$

The correct answer is:
negative

Solution:
Given line, $7 x-3 y=4$ $\Rightarrow \quad 3 y=7 x-4$
$\Rightarrow \quad y=\left(\frac{7}{3}\right) x-\left(\frac{4}{3}\right)$
$\therefore \quad$ Slope $=\frac{7}{3}$
The slope of the line perpendicular to $7 \mathrm{x}-3 \mathrm{y}=4$ is $\left(\frac{-3}{7}\right)$
If $^{\prime} \theta$ ' is the angle between the perpendicular line with
slope $\frac{-3}{7}$ and $\mathrm{x}$ -axis, then $\frac{-3}{7}=\tan \theta \Rightarrow \theta$ is negative

Topic: Mathematics, Straight Lines, NDA, NDA 2012 (Phase 1)
The adjacent sides $\mathrm{AB}$ and $\mathrm{AC}$ of a triangle $\mathrm{ABC}$ are represented by the vectors $-2 \hat{i}+3 \hat{j}+2 \hat{k}$ and $-4 \hat{i}+5 \hat{j}+2 \hat{k}$ respectively. The area of the triangle $\mathrm{ABC}$ is

Options:
A 6 square units
B 5 square units
C 4 square units
D 3 square units

The correct answer is:
3 square units

Solution:
Area of $\Delta \mathrm{ABC}=\frac{1}{2}(\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}})$
$=\frac{1}{2}\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ -2 & 3 & 2 \\ -4 & 5 & 2\end{array}\right|$
$=\frac{1}{2}[\hat{\mathrm{i}}(6-10)-\hat{\mathrm{j}}(-4+8)+\hat{\mathrm{k}}(-10+12)]$
$=\frac{1}{2}[-4 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}]$
$=\frac{1}{2} \sqrt{16+16+4}=\frac{1}{2} \sqrt{36}=\frac{1}{2} \times 6$
$=3$ square units
$\therefore \quad$ Option (d) is correct.

Topic: Mathematics, Vector Algebra, NDA, NDA 2015 (Phase 1)
The adjoint of the matrix $\mathrm{A}=\left[\begin{array}{lll}1 & 0 & 2 \\ 2 & 1 & 0 \\ 0 & 3 & 1\end{array}\right]$ is

Options:
A $\left[\begin{array}{ccc}-1 & 6 & 2 \\ -2 & 1 & -4 \\ 6 & 3 & 1\end{array}\right]$
B $\left[\begin{array}{ccc}1 & 6 & -2 \\ -2 & 1 & 4 \\ 6 & -3 & 1\end{array}\right]$
C $\left[\begin{array}{rrr}6 & 1 & 2 \\ 4 & -1 & 2 \\ 6 & 3 & -1\end{array}\right]$
D $\left[\begin{array}{ccc}-6 & 2 & 1 \\ 4 & -2 & 1 \\ 3 & 1 & -6\end{array}\right]$

The correct answer is:
$\left[\begin{array}{ccc}1 & 6 & -2 \\ -2 & 1 & 4 \\ 6 & -3 & 1\end{array}\right]$

Solution:
$\mathrm{A}=\left[\begin{array}{lll}1 & 0 & 2 \\ 2 & 1 & 0 \\ 0 & 3 & 1\end{array}\right]$
$A_{11}=1-0=1 ; A_{12}=-(2-0)=-2 ; A_{13}=6-0=6$
$A_{21}=-(0-6)=6 ; A_{22}=1-0=1 ; A_{23}=-3-0=-3$
$A_{31}=0-2=-2 ; A_{32}=-(0-4)=4 ; A_{33}=1-0=1$
$\therefore \operatorname{Adj} A=\left[\begin{array}{ccc}1 & 6 & -2 \\ -2 & 1 & 4 \\ 6 & -3 & 1\end{array}\right]$

Topic: Mathematics, Matrices, NDA, NDA 2017 (Phase 2)
The air column in an organ pipe closed at one end is made to vibrate so that there are
2 nodes and antinodes each. The mode of vibration is called

Options:
A 3rd overtone
B 1st overtone
C fundamental
D 2nd overtone

The correct answer is:
1st overtone

Solution:
Theory question

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2020 (14 Oct Shift 2)
The alcohol having molecular formula $\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{OH}$, when shaken with a mixture of anhydrous $\mathrm{ZnCl}_{2}$ and conc. HCl gives an oily layer product after five minutes. The alcohol is

Options:
A $\mathrm{H}_{3} \mathrm{C}-\left(\mathrm{CH}_{2}\right)_{3}-\mathrm{OH}$
B $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH}-\mathrm{CH}_{2} \mathrm{OH}_{1}$
C $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{OH}$
D $\mathrm{H}_{3} \mathrm{C}-\mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{2}-\mathrm{CH}_{3}$

The correct answer is:
$\mathrm{H}_{3} \mathrm{C}-\mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{2}-\mathrm{CH}_{3}$

Solution:
Secondary alcohol, when shaken with a mixture of anthydrous $\mathrm{ZnCl}_{2}$ and conc. $\mathrm{HCl}$ (Lucas regent) gives an oily layer product after five minutes.
oily layer product after $5 \mathrm{~min}$.
The amines in which amino group is directly attached to benzene ring undergo diazot isat ion reaction.

Topic: Chemistry, Alcohols Phenols and Ethers, VITEEE, VITEEE 2011
The alcohol that produces turbidity immediately with $\mathrm{ZnCl}_{2} /$ conc. HCl at room temperature

Options:
A 1-hydroxy butane
B 2-hydroxy butane
C 2-hydroxy-2-methyl propane
D 1-hydroxy-2-methyl propane

The correct answer is:
2-hydroxy-2-methyl propane

Solution:
Mixture of anhydrous $\mathrm{ZnCl}_{2}$ and conc. HCl is known as Lucas reagent. Lucas test is used for the distinction between primary, secondary and tertiary alcohols. The tertiary alcohol reacts immediately with Lucas reagent producing turbidity. The secondary alcohol gives turbidity within 5-10 min and primary alcohol do not give turbidity at all at room temperature. In the given alternates, 2 -hydroxy-2-methyl propane is $3^{\circ}$ alcohol, so it is more reactive.

Topic: Chemistry, Alcohols Phenols and Ethers, MHT CET, MHT CET 2008
The algebraic sum of the deviations of 20 observations measured from 30 is 2 . What would be the mean of the $\begin{array}{ll}\text { observations? } & {}\end{array}$

Options:
A $\quad 30$
B 32
C $30.2$
D 30,1

The correct answer is:
30,1

Solution:
Given $\sum_{i=1}^{20}\left(x_{i}-30\right)=2$
$\Rightarrow \sum_{i=1}^{20} x_{i}-\sum_{i=1}^{20} 30=2$
$\Rightarrow \sum_{i=1}^{20} x_{1}=2+\sum_{i=1}^{20} 30=2+30 \times 20=602$
Now, mean $=\frac{\sum_{i=1}^{20} x_{i}}{20}$
$\therefore \quad$ Mean $=\frac{602}{20}=30.1$

Topic: Mathematics, Statistics, NDA, NDA 2012 (Phase 2)