datavorous/entrance-exam-dataset huggingface dataset:
The ratio of frequencies of fundamental harmonic produced by an open pipe to that of closed pipe having the same length is:

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

The correct answer is:
2:1

Solution:
Fundamental frequency of open pipe is given by, fopen=v2l.

Fundamental frequency of closed pipe is given by,

fclosed=v4l.

Therefore,

fopenfclosed=v2l×4lv⇒fopen:fclosed=2:1

Topic: Physics, Waves and Sound, NEET, NEET 2023 (All India)
The ratio of the strengths of the four fundamental forces in nature, $\mathrm{F}_{\mathrm{G}}: \mathrm{F}_{\mathrm{N}}: \mathrm{F}_{\mathrm{E}}: \mathrm{F}_{\mathrm{W}}=$

Options:
A $1: 10^{26}: 10^{36}: 10^{38}$
B $1: 10^{36}: 10^{26}: 10^{38}$
C $1: 10^{38}: 10^{36}: 10^{26}$
D $10^{38}: 1: 10^{26}: 10^{36}$

The correct answer is:
$1: 10^{38}: 10^{36}: 10^{26}$

Solution:
No solution. Refer to answer key.

Topic: Physics, Laws of Motion, AP EAMCET, AP EAMCET 2020 (07 Oct Special)
The ration of roots of the equations $a x^{2}+b x+c=0$ and $\mathrm{px}^{2}+\mathrm{qx}+\mathrm{r}=0$ areequal. IfD $_{1}$ and $\mathrm{D}_{2}$ are respective discriminates, then what is $\frac{\mathrm{D}_{1}}{\mathrm{D}_{2}}$ equal to?

Options:
A $\frac{\mathrm{a}^{2}}{\mathrm{p}^{2}}$
B $\frac{\mathrm{b}^{2}}{\mathrm{q}^{2}}$
C $\frac{\mathrm{c}^{2}}{\mathrm{r}^{2}}$
D None of these

The correct answer is:
$\frac{\mathrm{b}^{2}}{\mathrm{q}^{2}}$

Solution:
$\frac{\mathrm{D}_{1}}{\mathrm{D}_{2}}=(\text { ratio of coefficient of } x)^{2}=\frac{b^{2}}{q^{2}}$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2018 (Phase 2)
The region of the argand plane defined by $|z-i|+|z+i| \leq 4$ is

Options:
A interior of an ellipse
B exterior of a circle
C interior and boundary of an ellipse
D interior of a parabola

The correct answer is:
interior and boundary of an ellipse

Solution:
Given, $|z-i|+|z+i| \leq 4$
$\Rightarrow|z-(0+i)|+|z-(0-i)| \leq 0$
This equation represent the interior and boundary of ellipse with foci at $(0,1) \&$ $(0,-1)$, whose major axis is along the $y$-axis.

Topic: Mathematics, Circle, VITEEE, VITEEE 2006
The regression coefficients of a bivariate distribution are $=0.64$ and $-0.36$. Then the correlation coefficient of the distribution is $\quad$

Options:
A $0.48$
B $-0.48$
C $0.50$
D $-0.50$

The correct answer is:
$-0.48$

Solution:
Let $\mathrm{b}_{1}$ $b_{2}=-0.36$
$r=\pm \sqrt{-0.64 \times-0.36}=\pm 0.48$
$\mathrm{b}_{1} < 0 \& \mathrm{~b}_{2}$
$ < 0$
ㄹ $\mathrm{r}=-0.48$

Topic: Mathematics, Statistics, NDA, NDA 2015 (Phase 2)
The regression lines will be perpendicular to each other jf the coefficient of correlation $r$ is equal to

Options:
A 1 only
B 1 or $-1$
C $-1$ only
D 0

The correct answer is:
0

Solution:
When regression lines perpendicular to each other then angle will be
$\tan \theta=\left\{\left(\frac{1-r^{2}}{r}\right)\left(\frac{\sigma_{x} \cdot \sigma_{y}}{\sigma_{x}^{2}+\sigma_{y}^{2}}\right)\right\}$
$\Rightarrow \tan \frac{\pi}{2}=\left(\frac{1-r^{2}}{r}\right)\left(\frac{\sigma_{x} \cdot \sigma_{y}}{\sigma_{x}^{2}+\sigma_{y}^{2}}\right)$
$\Rightarrow \mathrm{r} .\left(\sigma_{\mathrm{x}}^{2}+\sigma_{\mathrm{y}}^{2}\right)=0$

Topic: Mathematics, Statistics, NDA, NDA 2014 (Phase 2)
The relation 'has the same father as' over the set of children is:

Options:
A only reflexive
B only symmetric
C only transitive
D an equivalence relation

The correct answer is:
only reflexive

Solution:
The relation 'has the same father as' over the set of children is onlyreflexive. $(\because$ In reflexivity an element isrelated to itself)

Topic: Mathematics, Sets and Relations, NDA, NDA 2012 (Phase 2)
The relation $R=\{(1,1),$,
$(2,2),(3,3)(1,2),(2,3)$
$(1,3)\}$ on a set $\mathrm{A}=\{1,2,3\}$ is

Options:
A reflexive, transitive but not symmetric
B reflexive, symmetric but not transitive
C symmetric, transitive but not reflexive
D reflexive but neither symmetric nor transitive

The correct answer is:
reflexive, transitive but not symmetric

Solution:
Let $\mathrm{R}=\{(1,1),(2,2),(3,3),(1,2),(2,3),(1,3)\}$
Reflexive Since, IR $1,2 \mathrm{R} 2,3 \mathrm{R} 3$ in the set $\mathrm{R}$ $\mathrm{R}$ is reflexive relation Symmetric Since $1 \mathrm{R} 2$ but 2 is not related to 1 in $\mathrm{R}$ $\mathrm{R}$ isnot symmetric relation. Transitive $\mathbb{1 R} 2,2 \mathrm{R} 3 \Rightarrow 1 \mathrm{R}^{3}$
$\mathrm{R}$ is transitive relation Hence, $\mathrm{R}$ isreflexive and transitive only.

Topic: Mathematics, Sets and Relations, NDA, NDA 2010 (Phase 2)
The relation $R$ in the set $Z$ of integers given by $R=\{(a, b)$ $a-b$ is divisible by 5 \}is

Options:
A reflexive
B reflexive but not symmetric
C symmetric and transitive
D an equivalence relation

The correct answer is:
an equivalence relation

Solution:
For reflexive:
$(a, a)=a-a=0$ is divisible by 5 . For symmetric:
If $(a-b)$ is divisible by 5 , then $b-a=-(a-b)$ is also divisible by 5 . Thus relation is symmetric.
For transitive If $(a-b)$ and $(b-c)$ is divisible by 5 Then $(\mathrm{a}-\mathrm{c})$ is also divisible by 5 .
Thusrelation is transitive $\mathrm{R}$ is an equivalence relation.

Topic: Mathematics, Sets and Relations, NDA, NDA 2013 (Phase 2)
The relation $S$ is defined on the set of integers $Z$ as $x S y$ if integer $x$ devides integer $y$. Then

Options:
A $s$ is an equivalence relation
B $s$ is onlyreflexiveand symmetric.
C $s$ is only reflexive and transitive
D $s$ is only symmetric and transitive

The correct answer is:
$s$ is only reflexive and transitive

Solution:
The relation $\mathrm{S}$ is defined on the set of integers $\mathrm{Z}$ and xSy, if integer $x$ divides integer $y$ Reflexive: Since, every integer divides itself integer $\mathrm{x}$ divides integer $\mathrm{x}$ $\Rightarrow x$ Sx
Hence, $\mathrm{S}$ is reflexive.
Symmetric : Let $\mathrm{x}, \mathrm{y} \in \mathrm{Z}$ such that $\mathrm{xSy}$
L.e., integer $\mathrm{x}$ divides integer $\mathrm{y}$ Now, this does notimplies that integer ydivides integer $\mathrm{x}$
e.g. Take $x=2$ and $y=4$ Then, 2 divides 4 but 4 does not divides 2 .
Thus, $\mathrm{S}$ is not symmetric.
Transitive : Let $x, y, z \in Z$ such that $x$ Sy and $y S z$. $\Rightarrow$ integer $x$ divides integer $y$ and integer y divides integer $\Rightarrow$ integer $\mathrm{x}$ divides integer $\mathrm{z}$ $\Rightarrow \mathrm{xSz}$
Hence, $\mathrm{S}$ is transitive

Topic: Mathematics, Sets and Relations, NDA, NDA 2014 (Phase 2)
The relationship between the pair of compounds shown above are respectively,

Options:
A enantiomer, diastereomer, diastereomer
B enantiomer, enantiomer, diastereomer
C enantiomer, homomer (identical), diastereomer
D homomer (identical), diastereomer, geometrical isomer

The correct answer is:
enantiomer, homomer (identical), diastereomer

Solution:
Hint :

Topic: Chemistry, General Organic Chemistry, WBJEE, WBJEE 2023
The relative abundance of two isotopes of an element with atomic weight 85 and 87 is 75% and 25% respectively. Then calculate the average atomic weight of element

Options:
A 86.0
B 85.5
C 75.5
D 40.0

The correct answer is:
85.5

Solution:
Average atomic weight of element =85×3+87×13+1=85.5

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
The right option for the statement "Tyndall effect is exhibited by", is:

Options:
A NaCl solution
B Glucose solution
C Starch solution
D Urea solution

The correct answer is:
Starch solution

Solution:
Tyndall effect is exhibited by colloidal solutions and not true solutions. Starch solution is colloid, so it will exhibit the Tyndall effect.

Topic: Chemistry, Surface Chemistry, NEET, NEET 2021
The roots of $A x^{2}+B x+C=0$ are $r$ and $s$. For the roots of $x^{2}+p x+q=0$ to be $r^{2}$ and $s^{2}$, what must be the value of $p$

Options:
A $\left(B^{2}-4 A C\right) / A^{2}$
B $\left(B^{2}-2 A C\right) / A^{2}$
C $\left(2 A C-B^{2}\right) / A^{2}$
D $B^{2}-2 C$

The correct answer is:
$\left(2 A C-B^{2}\right) / A^{2}$

Solution:
Since, $r$ and $s$ are the roots of $A x^{2}+B x+C=0$, then
$r+s=-\frac{B}{A}$ and $r s=\frac{C}{A}$
Now, Given roots of $x^{2}+p x+q=0$ be $r^{2}$ and $s^{2}$
$\Rightarrow r^{2}+s^{2}=-p$ and $r^{2} s^{2}=q$
$\Rightarrow \frac{B^{2}}{A^{2}}-\frac{2 C}{A}=-p$
$\Rightarrow \frac{B^{2}-2 A C}{A^{2}}=-p$
$\Rightarrow \quad p=\frac{2 A C-B^{2}}{A^{2}}$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2009 (Phase 1)
The roots of the equation $\left|\begin{array}{ccc}1 & \mathrm{t}-1 & 1 \\ \mathrm{t}-1 & 1 & 1 \\ 1 & 1 & \mathrm{t}-1\end{array}\right|=0$ are

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

The correct answer is:
$-1,2$

Solution:
$\begin{array}{ll} & 1(\mathrm{t}-2)-(\mathrm{t}-1)\left[(\mathrm{t}-1)^{2}-1\right]+1(\mathrm{t}-2)=0 \\ \Rightarrow & (\mathrm{t}-2)-\mathrm{t}(\mathrm{t}-1)(\mathrm{t}-2)+(\mathrm{t}-2)=0 \\ & \Rightarrow(\mathrm{t}-2)[1-(\mathrm{t}-1)(\mathrm{t})+1]=0 \\ & \Rightarrow(\mathrm{t}-2)\left(\mathrm{t}^{2}-\mathrm{t}-2\right)=0 \\ & \Rightarrow(\mathrm{t}-2)(\mathrm{t}-2)(\mathrm{t}+1)=0 \\ & \Rightarrow \mathrm{t}=2, \mathrm{t}=-1\end{array}$
Hence, required roots are $-1,2 .$

Topic: Mathematics, Determinants, NDA, NDA 2013 (Phase 1)
The roots of the equation $2 a^{2} x^{2}-2 a b x+b^{2}=0$ when. and $b>0$ are

Options:
A Some times complex
B Always irrational
C Always complex
D Always real

The correct answer is:
Always complex

Solution:
We have, $2 a^{2} x^{2}-2 a b x+b^{2}=0$ Discriminent, $\mathrm{D}=(-2 \mathrm{ab})^{2}-4\left(2 \mathrm{a}^{2}\right)\left(\mathrm{b}^{2}\right)$
$=4 a^{2} b^{2}-8 a^{2} b^{2}=-4 a^{2} b^{2} < 0$
Roots are always complex.

Topic: Mathematics, Quadratic Equation, NDA, NDA 2014 (Phase 1)
The roots of the equation $(q-r) x^{2}+(r-p) x+(p-q)=0$ are

Options:
A $(\mathrm{r}-\mathrm{p}) /(\mathrm{q}-\mathrm{r}), 1 / 2$
B $(\mathrm{p}-\mathrm{q}) /(\mathrm{q}-\mathrm{r}), \mathrm{j}$
C $(q-r) /(p-q)$,
D $(r-p) /(p-q), 1 / 2$

The correct answer is:
$(\mathrm{p}-\mathrm{q}) /(\mathrm{q}-\mathrm{r}), \mathrm{j}$

Solution:
Given equation, $(q-r) x^{2}+(r-p) x+(p-q)=0$
On observing the equation, it is clear that 1 is root of equation. If $x=1$, then $q-r+r-p+p-q=0$
1 is one root of given equation. Since, the given equation is quadratic equation, we
know that product of roots is $\frac{\mathrm{c}}{\mathrm{a}}$
Let the second root be $\alpha$.
$\therefore \quad(1)(\alpha)=\frac{\mathrm{p}-\mathrm{q}}{\mathrm{q}-\mathrm{r}} \Rightarrow \alpha=\frac{\mathrm{p}-\mathrm{q}}{\mathrm{q}-\mathrm{r}}$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2017 (Phase 2)
The roots of the equation $x^{2}-8 x+16=0$

Options:
A are imaginary
B are distinct and real
C are equal and real
D cannot be ascertained

The correct answer is:
are equal and real

Solution:
Discriminant, $\mathrm{D}=(-8)^{2}-4 \times 16=0$ Roots are real and equal

Topic: Mathematics, Quadratic Equation, NDA, NDA 2013 (Phase 2)
The roots of the equation $\left|\begin{array}{lll}x & \alpha & 1 \\ \beta & x & 1 \\ \beta & \gamma & 1\end{array}\right|=0$ are independent of

Options:
A $\alpha$
B $\beta$
C $\gamma$
D $\alpha, \beta$ and $\gamma$

The correct answer is:
$\alpha$

Solution:
Given $\left|\begin{array}{lll}x & \alpha & 1 \\ \beta & x & 1 \\ \beta & \gamma & 1\end{array}\right|=0$
$\Rightarrow x(x-\gamma)-\alpha(0)+1(\gamma \beta-x \beta)=0$
$\Rightarrow x^{2}-x \gamma+\gamma \beta-x \beta=0$
$\Rightarrow x^{2}-(\gamma+\beta) x+\gamma \beta=0$
$\Rightarrow x=\frac{(\gamma+\beta) \pm \sqrt{\gamma^{2}+\beta^{2}-2 \gamma \beta}}{2}$
( $\because$ roots of Quad. eqn. $a x^{2}+b x+c=0$ are
$\left.\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\right)$
$\Rightarrow x=\frac{(\gamma+\beta) \pm(\gamma-\beta)}{2}$
$\Rightarrow x=\frac{\gamma+\beta+\gamma-\beta}{2}, \frac{\gamma+\beta-\gamma+\beta}{2}$
$\Rightarrow x=\gamma, \beta .$
Hence, roots of the equation $\left|\begin{array}{lll}x & \alpha & 1 \\ \beta & x & 1 \\ \beta & \gamma & 1\end{array}\right|=0$ are
independent of $\alpha$.

Topic: Mathematics, Determinants, NDA, NDA 2011 (Phase 2)
The roots ofthe equation $(x-p)(x-q)=r^{2}$, where $p, q, r$ are
real, are

Options:
A always complex
B always real
C always purely imaginary
D None of these

The correct answer is:
always real

Solution:
Given, equation is $(x-p)(x-q)=$ Now, $\begin{aligned} D &=\sqrt{(p+q)^{2}-4\left(p q-r^{2}\right)} \\ &=\sqrt{(p-q)^{2}+4 r^{2}} \geq 0 \end{aligned}$
Hence, roots are always real

Topic: Mathematics, Quadratic Equation, NDA, NDA 2009 (Phase 2)
The scalar triple product $(\vec{A} \times \vec{B}) \cdot \vec{C}$ of three vectors
$\vec{A}, \vec{B}, \vec{C}$ determines

Options:
A Volume of a parallelopiped
B Volume of a tetrahedron
C Volume of an ellipsoid
D None of the above

The correct answer is:
Volume of a parallelopiped

Solution:
The scalar triple product $(\overrightarrow{\mathbf{A}} \times \overrightarrow{\mathbf{B}}) \cdot \overrightarrow{\mathbf{C}}$ of three vectors
$\overrightarrow{\mathbf{A}}, \overrightarrow{\mathbf{B}}, \overrightarrow{\mathbf{C}}$ determines volume of a parallelopiped.

Topic: Mathematics, Vector Algebra, NDA, NDA 2008 (Phase 2)
The scores of 15 students in an examination were recorde as $10,5,8,16,18,20,8,10,16,20,18,11,16,14$ and 12 . Afte calculating the mean, median and mode, an error is found One of the values is wrongly written as 16 instead of 18 Which of the following measures of central tendency w1 change?

Options:
A Mean and median
B Median and mode
C Mode only
D Mean and mode

The correct answer is:
Mean and mode

Solution:
Mean of the scores $=\frac{202}{15}$ Mean of the correct scores $=\frac{200}{15}$
i.e., Mean changes. Median is same for both cases i.e., 14 Mode is proportional to mean.

Topic: Mathematics, Statistics, NDA, NDA 2016 (Phase 2)
The second degree equation \(x^{2}+4 y^{2}-2 x-4 y+2=0\) represents

Options:
A A point
B An ellipse of semi-major axis 1
C An ellipse with eccentricity $\frac{\sqrt{3}}{2}$
D None of the above

The correct answer is:
A point

Solution:
$\left(x^{2}-2 x+1\right)+\left(4 y^{2}-4 y+1\right)=0$
$\Rightarrow(x-1)^{2}+(2 y-1)^{2}=0$
$\Rightarrow \quad x=1, y=\frac{1}{2}$
It is a point.

Topic: Mathematics, Straight Lines, NDA, NDA 2018 (Phase 2)
The second maturation division of the mammalian ovum occurs

Options:
A Shortly after ovulation before the ovum makes entry into the Fallopian tube
B Until the nucleus of the sperm has fused with that of the ovum
C In the Graffian follicle following the first maturation division.
D After the ovum has been penetrated by a sperm

The correct answer is:
After the ovum has been penetrated by a sperm

Solution:
Oogenesis starts with the division of oogonia (gamete mother cells) giving rise to primary oocyte which enters into prophase - I of the meiotic division and gets temporarily arrested at this stage. These primary oocyte gets surrounded by primary, secondary, and tertiary follicles respectively. The tertiary follicle grows in size and completes its first meiotic division to give rise to the haploid secondary oocyte. This secondary oocyte forms a new membrane called zona pellucida surrounding it. During fertilization, the sperm enters into the cytoplasm of the ovum through the zona pellucida and the plasma membrane. This induces the completion of the meiotic division (2nd division) of the secondary oocyte. The second meiotic division is also unequal and results in the formation of a second polar body and a haploid ovum (ootid).

Topic: Biology, Human Reproduction, NEET
The second meiotic division of ________P________ is unequal and results in the formation of ______Q________.

Select the appropriate option for P and Q.

Options:
A P-secondary spermatocytes; Q-spermatids
B P-primary oocyte; Q-ovum
C P-secondary oocyte; Q-polar body
D P-primary spermatocytes; Q-spermatids

The correct answer is:
P-secondary oocyte; Q-polar body

Solution:
The entry of sperm into the cytoplasm of secondary oocytes brings the signal to induce the completion of the meiotic division. The second meiotic division of secondary oocyte is unequal which results in the formation of a large size haploid ovum and a tiny haploid cell called a polar body.

Topic: Biology, Human Reproduction, NEET
The second overtone of an open pipe has the same frequency as the first overtone of a closed pipe $2 \mathrm{~m}$ long. The length of the open pipe is

Options:
A $8 \mathrm{~m}$
B $4 \mathrm{~m}$
C $2 \mathrm{~m}$
D $1 \mathrm{~m}$

The correct answer is:
$4 \mathrm{~m}$

Solution:
Let $L$ be the length of the open pipe. The fundamental frequency of the open pipe is given by
$v=\frac{v}{2 L}, v=$ velocity of sound in air
The second overtone of the open pipe has a
frequency $3 v_O=\frac{3 v}{2 L_O} \mathrm{~Hz}$
The length of the closed pipe $L_C=2 \mathrm{~m}$.
The fundamental frequency of the closed pipe is given by $v=\frac{v}{4 L}$
The first overtone of the closed pipe has a frequency
$$
3 v_C=\frac{3 v}{4 L_C}=\frac{3 v}{4 \times 2}=\frac{3 v}{8} \mathrm{~Hz}
$$
Now $3 v_O=3 v_C$ or $\frac{3 v}{2 L_O}=\frac{3 v}{8}$
or $2 L_O=8$ or $L_O=4 \mathrm{~m}$

Topic: Physics, Waves and Sound, AIIMS, AIIMS 2010
The second trophic level in a lake is ___.

Options:
A Phytoplankton
B Zooplankton
C Benthos
D Fishes

The correct answer is:
Zooplankton

Solution:
Zooplankton forms the secondary trophic level in a lake as they are the primary consumers in food chain that is aquatic, feeding upon Phytoplankton.

Topic: Biology, Ecosystem, NEET
The secondary metabolites belonging to same category are
a. Abrin
b. Vinblastin
c. Rubber
d. Ricin
e. Carotenoids

Options:
A a, b only
B a, b, c only
C a, d only
D a, c, e only

The correct answer is:
a, d only

Solution:
In animal tissues, one notices the presence of all such
categories of compounds like nitrogen bases, nucleosides or nucleotides, etc. These are called primary metabolites. However, when one analyses plant, fungal, and microbial cells, one would see thousands of compounds other than these called primary metabolites, e.g. alkaloids, flavonoids, rubber, essential oils, antibiotics, colored pigments, scents, gums, spices. These are called secondary metabolites. A list of secondary metabolites is given in the table below.

Pigments	Carotenoids, Anthocyanins,
etc.

Alkaloids	Morphine, Codeine, etc.

Terpenoides	Monoterpenes, Diterpenes etc.

Essential oils	Lemon grass oil, etc.

Toxins	Abrin, Ricin

Lectins	Concanavalin A

Drugs	Vinblastin, curcumin, etc.

Polymeric
substances	Rubber, gums, cellulose

Topic: Chemistry, Biomolecules, NEET
The secondary precursors of photochemical smog are

Options:
A $\mathrm{SO}_2$ and $\mathrm{NO}_2$
B $\mathrm{NO}_2$ and hydrocarbons
C $\mathrm{NO}_2$ and PAN
D $\mathrm{O}_3$ and PAN

The correct answer is:
$\mathrm{O}_3$ and PAN

Solution:
The primary precursors of photochemical smog are $\mathrm{NO}_2$ and hydrocarbons. The secondary precursors are ozone and PAN (peroxyacyl nitrates).

Topic: Chemistry, Environmental Chemistry, AP EAMCET, AP EAMCET 2021 (23 Aug Shift 2)
The secondary productivity means

Options:
A Rate of increase in the biomass of autotrophs
B Rate of increase in biomass of heterotrophs
C The rate at which the organic molecules are formed in an autotrophs.
D The rate at which the organic molecules are used in an autotrophs

The correct answer is:
Rate of increase in biomass of heterotrophs

Solution:
Secondary productivity is the rate at which organic matter or biomass is stored at the level of consumers.

Topic: Biology, Ecosystem, NEET
The secondary structure of a protein refers to

Options:
A $\alpha$-helical backbone
B hydrophobic interactions
C sequence of $\alpha$-amino acids
D fixed configuration of the polypeptide backbone

The correct answer is:
$\alpha$-helical backbone

Solution:
Secondary structure involves $\alpha$-helical and $\beta$-pleated sheet like structure. Where as Primary strucure involves sequence of $\alpha$-amino acids polypeptide chain.

Topic: Chemistry, Biomolecules, VITEEE, VITEEE 2011
The secondary structure of proteins is derived from

Options:
A Hydrogen bonding
B Peptide linkage
C Folding of chains
D Disulphide linkage

The correct answer is:
Hydrogen bonding

Solution:
Peptide linkages give rise to primary structure of protein and the arrangement of polypeptide chains assumed as a result of hydrogen bonding is called secondary structure of proteins. Folding of chains and disulphide linkage etc. give rise to tertiary structure of protein.

Topic: Chemistry, Biomolecules, NEET
The secondary valency of platinum in tetra amminedichloroplatinum (IV) chloride is

Options:
A $+2$
B 3
C 6
D $+4$

The correct answer is:
6

Solution:
Secondary valency corresponds to the coordination number. The formula of the given complex is $\left[\mathrm{Pt} \mathrm{Cl}_{2}\left(\mathrm{NH}_{3}\right)_{4}\right] \mathrm{Cl}_{2}$.
In it coordination number of $\mathrm{Pt}=2+4=6$
$\therefore$ Secondary valency of Pt $=6$.

Topic: Chemistry, Coordination Compounds, KCET, KCET 2012
The self-inductance of two coils of a transformer is $20 \mathrm{mH}$ and $30 \mathrm{mH}$. What is the resistance between them?

Options:
A 0
B $1.5 \Omega$
C $600 \Omega$
D Infinite

The correct answer is:
Infinite

Solution:
Transformers are built in such a way that both primary and secondary coils are well insulated from each other. Hence, resistance between them is considered to be infinite.

Topic: Physics, Electromagnetic Induction, AP EAMCET, AP EAMCET 2021 (25 Aug Shift 1)
The sequence of basic strength of alky1 amines follows the order

Options:
A $\mathrm{RNH}_{2} < \mathrm{R}_{2} \mathrm{NH}>\mathrm{R}_{3} \mathrm{~N}$
B $\mathrm{R}_{2} \mathrm{NH}_{2} < \mathrm{R}_{2} \mathrm{NH} < \mathrm{R}_{3} \mathrm{~N}$
C $\mathrm{R}_{2} \mathrm{NH} < \mathrm{RNH}_{2} < \mathrm{R}_{3} \mathrm{~N}$
D $\mathrm{RNH}_{2} < \mathrm{R}_{2} \mathrm{NH} < \mathrm{R}_{3} \mathrm{~N}$

The correct answer is:
$\mathrm{RNH}_{2} < \mathrm{R}_{2} \mathrm{NH}>\mathrm{R}_{3} \mathrm{~N}$

Solution:
It is expected that the basic nature of amines should be in order tertiary > secondary $>$ primary but the observed order in the case of lower members is found to be as secondary > Primary $>$ tetriary. This anomalous behaviour of tetriary amines is due to steric factors i.e crowding of alkyl groups cover nitrogen atom from all sides thus makes the approach and bonding by a proton relatively difficult which results the maximum steric strain in tetiary amines. The electrons are there but the path is blocked, resulting the reduction in basicity. Thus the correct order is
$\mathrm{R}_{2} \mathrm{NH}>\mathrm{RNH}_{2}>\mathrm{R}_{3} \mathrm{~N}$

Topic: Chemistry, Amines, VITEEE, VITEEE 2006
The set of all points, where the function $f(\mathrm{x})=\sqrt{1-\mathrm{e}^{-\mathrm{x}^{2}}}$ is differentiable, is

Options:
A $(0, \infty)$
B $(-\infty, \infty)$
C $(-\infty, 0) \cup(0, \infty)$
D $(-1, \infty)$

The correct answer is:
$(-\infty, 0) \cup(0, \infty)$

Solution:
$\mathrm{f}(\mathrm{x})=\sqrt{1-\mathrm{e}^{-\mathrm{x}^{2}}}$
$\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{2 \sqrt{1-\mathrm{e}^{-\mathrm{x}^{2}}}} \cdot\left(-\mathrm{e}^{-\mathrm{x}^{2}}\right)(-2 \mathrm{x})$
$f^{\prime}(x)=\frac{x e^{-x^{2}}}{\sqrt{1-e^{-x^{2}}}}$
$\mathrm{f}^{\prime}(\mathrm{x})$ is defined for all values of $\mathrm{x}$, except $0 .$ $\therefore \mathrm{f}(\mathrm{x})$ is differentiable on $(-\infty, 0) \cup(0, \infty)$

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2017 (Phase 2)
The set of bivariate observation $\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right),\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right), \ldots . .(\mathrm{xn}, \mathrm{yn})$
are such that all the values are distinct and all the observations fall on a straight line with non-zero slope. Then the possible values of the correlation coefficient between and $\mathrm{y}$ are $\quad\mathrm{}$

Options:
A 0 and 1 only
B 0 and $-1$ only
C 0,1 and $-1$
D $-1$ and 1 only

The correct answer is:
$-1$ and 1 only

Solution:
By definition.

Topic: Mathematics, Statistics, NDA, NDA 2018 (Phase 2)