datavorous/entrance-exam-dataset huggingface dataset:
Nitrosoamines $\left(\mathrm{R}_{2} \mathrm{~N}-\mathrm{N}=\mathrm{O}\right)$ are soluble in water. On heating them with concentrated $\mathrm{H}_{2} \mathrm{SO}_{4}$, they give secondary amines. This reaction is called

Options:
A Perkin reaction
B Sandmeyer's reaction
C Fitting reaction
D Liebermann nitroso reaction

The correct answer is:
Liebermann nitroso reaction

Solution:
The given reaction is known as Liebermann Nitroso reaction.

Topic: Chemistry, Amines, VITEEE, VITEEE 2018
Number of chromosomes present in secondary spermatocyte is

Options:
A 22
B 23
C 24
D 25

The correct answer is:
23

Solution:
The number of chromosomes present in secondary spermatocyte is 23 . It is because the secondary spermatocytes are formed after the lst meiotic division (reductional division) of diploid primary spermatocytes, hence they have haploid set of chromosomes,
i.e. 23 in count.

Topic: Biology, Human Reproduction, JIPMER, JIPMER 2019
Number of secondary carbon atoms present in the compounds is respectively:

Options:
A 6,4,5
B 4,5,6
C 5,4,6
D 6,2,1

The correct answer is:
6,4,5

Solution:
Secondary carbons are those which attached with two other carbon atoms.

Topic: Chemistry, General Organic Chemistry, NEET
Number of telephone calls received in 245 succesive one minute intervals at an exchange is given below in the following frequency distribution. $\quad[2014-I]$ \begin{array}{|c|c|c|c|c|c|c|c|c|}
\hline Number of calls & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\
\hline Frequency & 14 & 21 & 25 & 43 & 51 & 40 & 39 & 12 \\
\hline
\end{array}
What is the mode of the distribution?

Options:
A 3
B 4
C 5
D 6

The correct answer is:
4

Solution:
The higher frequency is 51 mode $=$ value of the variable corresponding to the highe frequency $154 \equiv 4$

Topic: Mathematics, Statistics, NDA, NDA 2014 (Phase 1)
Number of telephone calls received in 245 succesive one minute intervals at an exchange is given below in the following frequency distribution. $\quad[2014-I]$ \begin{array}{|c|c|c|c|c|c|c|c|c|}
\hline Number of calls & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\
\hline Frequency & 14 & 21 & 25 & 43 & 51 & 40 & 39 & 12 \\
\hline
\end{array}
What is the median of the distribution?

Options:
A $3.5$
B 4
C $4.5$
D 5

The correct answer is:
4

Solution:
$\frac{\mathrm{N}}{2}=\frac{245}{2}=122.5$
Required mean $=4$

Topic: Mathematics, Statistics, NDA, NDA 2014 (Phase 1)
Number of telephone calls received in 245 succesive one minute intervals at an exchange is given below in the following frequency distribution. $\quad[2014-I]$ \begin{array}{|c|c|c|c|c|c|c|c|c|}
\hline Number of calls & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\
\hline Frequency & 14 & 21 & 25 & 43 & 51 & 40 & 39 & 12 \\
\hline
\end{array}
What is the mean of the distribution?

Options:
A $3.76$
B $3.84$
C $3.96$
D $4.05$

The correct answer is:
$3.76$

Solution:
$\frac{\Sigma \mathrm{fx}}{\mathrm{N}}=\frac{922}{245}=3.76$

Topic: Mathematics, Statistics, NDA, NDA 2014 (Phase 1)
Number of $\mathrm{X}$ is randomly selected from the set of odd numbers and $\mathrm{Y}$ is randomly selected from the set of even numbers of the set $\{1,2,3,4,5,6,7\}$. Let $Z=(\mathrm{X}+\mathrm{Y}) . .$
What is $P(Z=10)$ equal to ?

Options:
A 0
B $1 / 2$
C $1 / 3$
D $1 / 5$

The correct answer is:
0

Solution:
Set $A=\{1,2,3,4,5,6,7\}$ and $z=x+y$
$\mathrm{x}=$ set of odd numbers
$\mathrm{y}=$ set of even numbers
$\mathrm{P}(\mathrm{Z}=10)=\frac{\mathrm{n}\left(\mathrm{E}_{2}\right)}{\mathrm{n}(\mathrm{S})}=\frac{0}{12}=0$

Topic: Mathematics, Probability, NDA, NDA 2014 (Phase 1)
Number of $\mathrm{X}$ is randomly selected from the set of odd numbers and $\mathrm{Y}$ is randomly selected from the set of even numbers of the set $\{1,2,3,4,5,6,7\}$. Let $Z=(\mathrm{X}+\mathrm{Y}) . .$
What is $P(Z$ is the product of two prime numbers $)$ equal to ?

Options:
A 0
B $1 / 2$
C $1 / 4$
D None of these

The correct answer is:
$1 / 4$

Solution:
Set $A=\{1,2,3,4,5,6,7\}$ and $z=x+y$
$\mathrm{x}=$ set of odd numbers
$\mathrm{y}=$ set of even numbers
$\mathrm{Z}=$ product of two prime numbers $\mathrm{Z}=\mathrm{x}+\mathrm{y}=7+6=13$
$\mathrm{n}\left(\mathrm{E}_{4}\right)=3$
$\mathrm{P}(\mathrm{Z}=9)=\frac{\mathrm{n}\left(\mathrm{E}_{4}\right)}{\mathrm{n}(\mathrm{S})}=\frac{3}{12}=\frac{1}{4}$

Topic: Mathematics, Probability, NDA, NDA 2014 (Phase 1)
Number of $\mathrm{X}$ is randomly selected from the set of odd numbers and $\mathrm{Y}$ is randomly selected from the set of even numbers of the set $\{1,2,3,4,5,6,7\}$. Let $Z=(\mathrm{X}+\mathrm{Y}) . .$
What is $P(Z=11)$ equal to ?

Options:
A 0
B $1 / 4$
C $1 / 6$
D $1 / 12$

The correct answer is:
$1 / 12$

Solution:
Set $A=\{1,2,3,4,5,6,7\}$ and $z=x+y$
$\mathrm{x}=$ set of odd numbers
$\mathrm{y}=$ set of even numbers
$Z>11$ is only possible when $x=7$ and $y=6$
$P(>11)=\frac{n\left(E_{3}\right)}{n(S)}=\frac{1}{12}$

Topic: Mathematics, Probability, NDA, NDA 2014 (Phase 1)
Number of $\mathrm{X}$ is randomly selected from the set of odd numbers and $\mathrm{Y}$ is randomly selected from the set of even numbers of the set $\{1,2,3,4,5,6,7\}$. Let $Z=(\mathrm{X}+\mathrm{Y}) . .$
Suppose A and $\mathrm{B}$ are two events. Event $B$ has occurred and it is known that $\mathrm{P}(\mathrm{B}) < 1$. What is $\mathrm{P}\left(\mathrm{A} \mid \mathrm{B}^{\mathrm{c}}\right)$ equal to ?

Options:
A $\frac{\mathrm{P}(\mathrm{A})-\mathrm{P}(\mathrm{B})}{1-\mathrm{P}(\mathrm{B})}$
B $\frac{\mathrm{P}(\mathrm{A})-\mathrm{P}(\mathrm{AB})}{1-\mathrm{P}(\mathrm{B})}$
C $\frac{\mathrm{P}(\mathrm{A})+\mathrm{P}\left(\mathrm{B}^{\mathrm{c}}\right)}{1-\mathrm{P}(\mathrm{B})}$
D None of these

The correct answer is:
$\frac{\mathrm{P}(\mathrm{A})-\mathrm{P}(\mathrm{AB})}{1-\mathrm{P}(\mathrm{B})}$

Solution:
$\mathrm{P}\left(\mathrm{A} / \mathrm{B}^{\prime}\right)=\frac{\mathrm{P}\left(\mathrm{A} \cap \mathrm{B}^{\prime}\right)}{\mathrm{P}\left(\mathrm{B}^{\prime}\right)}=\frac{\mathrm{P}(\mathrm{A})-\mathrm{P}(\mathrm{AB})}{1-\mathrm{P}(\mathrm{B})}$

Topic: Mathematics, Probability, NDA, NDA 2014 (Phase 2)
Number of $\mathrm{X}$ is randomly selected from the set of odd numbers and $\mathrm{Y}$ is randomly selected from the set of even numbers of the set $\{1,2,3,4,5,6,7\}$. Let $Z=(\mathrm{X}+\mathrm{Y}) . .$
What is $P(Z=5)$ equal to ?

Options:
A $1 / 2$
B $1 / 3$
C $1 / 4$
D $1 / 6$

The correct answer is:
$1 / 6$

Solution:
Set $A=\{1,2,3,4,5,6,7\}$ and $z=x+y$
$\mathrm{x}=$ set of odd numbers
$\mathrm{y}=$ set of even numbers
$\mathrm{n}(\mathrm{S})=12$
$n\left(E_{1}\right)=2$
$\mathrm{P}(\mathrm{Z}=5)=\frac{\mathrm{n}\left(\mathrm{E}_{1}\right)}{\mathrm{n}(\mathrm{S})}=\frac{2}{12}=\frac{1}{6}$

Topic: Mathematics, Probability, NDA, NDA 2014 (Phase 1)
$\mathrm{O}(0,0), \mathrm{A}(0,3), \mathrm{B}(4,0)$ are the vertices of triangle $\mathrm{OAB}$.
A force $10 \hat{\mathrm{i}}$ acts at $\mathrm{B}$. What is the magnitude of moment of force about the vertex $\mathrm{A}$ ?

Options:
A 0
B 30 unit
C 40 unit
D 50 unit

The correct answer is:
30 unit

Solution:
As given:
$\mathrm{O}(0,0), \mathrm{A}(0,3), \mathrm{B}(4,0)$ are the vertices of triangle
OAB
$\therefore \overrightarrow{\mathrm{OA}}=\overrightarrow{\mathrm{r}}=3 \hat{\mathrm{j}}$ and $\overrightarrow{\mathrm{F}}=10 \hat{\mathrm{i}}$
Movement of force about the vertex
$\mathrm{A}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{F}}=3 \hat{\mathrm{j}} \times 10 \hat{\mathrm{i}}=-30 \hat{\mathrm{k}}$
$\therefore \quad$ Magnitude of moment $=|30 \hat{\mathrm{k}}|=30$ unit

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2007 (Phase 2)
$\mathrm{OAB}$ is a given triangle such that $\overrightarrow{\mathrm{OA}}=\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{OB}}=\overrightarrow{\mathrm{b}}$. Also $\mathrm{C}$
is a point on $\overrightarrow{\mathrm{AB}}$ such that $\overrightarrow{\mathrm{AB}}=2 \overrightarrow{\mathrm{BC}}$. What is $\overrightarrow{\mathrm{AC}}$ equal
to ?

Options:
A $\frac{1}{2}(\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}})$
B $\frac{1}{2}(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{a}})$
C $\frac{3}{2}(\vec{a}-\vec{b}) \quad$
D $\frac{3}{2}(\vec{b}-\vec{a})$

The correct answer is:
$\frac{1}{2}(\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}})$

Solution:
In $\begin{aligned} \Delta & \mathrm{OAB}, \\ & \text { Let } \overrightarrow{\mathrm{OA}}=\overrightarrow{\mathrm{a}} \\ \overrightarrow{\mathrm{OB}} &=\overrightarrow{\mathrm{b}} \\ \overrightarrow{\mathrm{OA}} &+\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{OB}} \\ \Rightarrow \overrightarrow{\mathrm{AB}} &=\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}} \\ &=\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}} \\ \overrightarrow{\mathrm{AB}} &=2 \overrightarrow{\mathrm{BC}} \end{aligned}$
$\Rightarrow \quad C$ is mid point of $\overrightarrow{\mathrm{AB}}$
$\therefore \quad \overrightarrow{\mathrm{AC}}=\frac{1}{2} \overrightarrow{\mathrm{AB}}=\frac{1}{2}(\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}})$

Topic: Mathematics, Vector Algebra, NDA, NDA 2006 (Phase 1)
Observe the diagram given below and identify a and b.

Options:
A a:Amphimixis, b: Parthenocarpy
B a: Amphimixis, b: Parthenogenesis
C a: Parthenocarpy, b: Amphimixis
D a: Parthenogenesis, b: sexual reproduction

The correct answer is:
a: Amphimixis, b: Parthenogenesis

Solution:
Amphimixis is the sexual reproduction, where the two gametes from two parents fuse to produce the offspring. Parthenogenesis (Gk. Parthenos-virgin, genesis-descent). It is the development of a new individual from a single gamete without fusion with another gamete. Parthenogenesis is the process of production of offsprings from the unfertilized eggs and is commonly observed among invertebrates.

Topic: Biology, Reproduction in Organisms, NEET
Observe the graphs given below and choose the options correctly describing the color of Pigments 1, 2, and 3 in a chromatogram.

Options:
A Pigment 1: yellow green, Pigment 2: yellow-orange and Pigment 3: blue green
B Pigment 1: yellow - orange, Pigment 2: blue green and Pigment 3: yellow green
C Pigment 1: blue green, Pigment 2: yellow green and Pigment 3: yellow-orange
D Pigment 1: yellow green, Pigment 2: blue green and Pigment 3: yellow-orange

The correct answer is:
Pigment 1: yellow green, Pigment 2: yellow-orange and Pigment 3: blue green

Solution:
The above graph represents the absorption spectrum. It is a graph that depicts the absorption of various wavelengths of light by photosynthetic pigments. In this graph pigment, 1 is chlorophyll b, pigment 2 is carotenoids and pigment 3 is chlorophyll a. On separation of the leaf pigments of any green plant through paper chromatography we find that chlorophyll a appears bright or blue-green, chlorophyll b appears yellow-green, xanthophylls appear yellow and carotenoids appear yellow to yellow-orange.

Topic: Biology, Photosynthesis in Higher Plants, NEET
Observe the picture given below and identify the disease-causing nematode that causes a great reduction in yield of tobacco plants.

Options:
A Ancylostoma duodenale
B Wuchereria bancrofti
C Meloidogyne incognita
D Ascaris lumbricoides

The correct answer is:
Meloidogyne incognita

Solution:
A nematode Meloidogyne incognita infects the roots of tobacco plants and causes a great reduction in yield, by causing root-knot diseases.

Topic: Biology, Biotechnology and Its Applications, NEET
Occurrence of triploid ( $3 n)$ primary endosperm nucleus is a characteristic feature of

Options:
A Algae
B Gymnosperms
C Angiosperms
D Bryophytes

The correct answer is:
Angiosperms

Solution:
Endosperm is the food laden tissue which is meant for nourishing the embryo in seed plants. In gymnosperms, it represents the female gametophyte and thus is haploid(n). In angiosperms, the endosperm is a special tissue which is formed as a result of fusion of a male gamete with diploid secondary nucleus of the central cell (vegetative fertilization or triple fusion). The fusion product is primary endosperm cell having a triploid (3n) endosperm nucleus.

Topic: Biology, Sexual Reproduction in Flowering Plants, AIIMS, AIIMS 2013
"Omnis-cellula-e-cellula" was given by

Options:
A Virchow
B Hooke
C Leeuwenhoek
D Brown.

The correct answer is:
Virchow

Solution:
Rudolf Virchow was the first to suggest that new cells are formed from the division of the preexisting cells - omnis-cellula-e-cellula (every cell is derived from a previous cell). Robert Hooke was the first to coin the term "cell" for small structures in a piece of cork under a microscope. His observations were published in a book named micrographia. Leeuwenhoek was the first person to observe and describe microscopic organisms and living cell. He observed nucleus in RBC of salmon fish and used simple lens and observed nuclei and unicellular organisms including bacteria. In 1676, he described the bacteria and gave the term animalcules. His observations laid the foundations for the science of bacteriology and microbiology. Robert Brown (1831) described and named nucleus.

Topic: Biology, Biological Classification, AIIMS, AIIMS 2007
On a railway route there are 20 stations. What is the number of different tickets required in order that it may be possible to travel from every station to every other station?

Options:
A 40
B 380
C 400
D 420

The correct answer is:
380

Solution:
From each railway station, there are 19 different tickets to be issued. There are 20 railway station So, total number of tickets $=20 \times 19=380$.

Topic: Mathematics, Permutation Combination, NDA, NDA 2007 (Phase 2)
On analysis of a protein chain, it was found that it is made up of just a peptide bond. The possible structure of the protein is its

Options:
A primary structure.
B secondary structure.
C tertiary structure.
D quaternary structure.

The correct answer is:
primary structure.

Solution:
Proteins contain amino acids joined by peptide bonds. This protein structure can be primary, secondary, tertiary or quaternary. The primary structure is the simplest structure in which position of amino acids can be checked. Thus these structures contain only peptide bonds. The secondary structure is coiling of primary structure i.e. polypeptide chain. These can be alpha helix or beta pleated sheets that show presence of hydrogen bonding apart from the peptide bond. Tertiary structures are woollen ball appearance of polypeptide chains and thus show the presence of many different bonds like ionic, disulphide, etc. Quaternary is a mixture of primary, secondary and tertiary structures.

Topic: Chemistry, Biomolecules, NEET
On closing an open organ pipe from one end, it is noticed that the frequency of third harmonic is $50 \mathrm{~Hz}$ more than the fundamental frequency of vibration in open organ pipe. The fundamental frequency of open organ pipe is

Options:
A $100 \mathrm{~Hz}$
B $50 \mathrm{~Hz}$
C $250 \mathrm{~Hz}$
D $200 \mathrm{~Hz}$

The correct answer is:
$100 \mathrm{~Hz}$

Solution:
Frequency of $\mathrm{m}^{\text {th }}$ harmonic in an open organ pipe is $f_m=\frac{m v}{2 L}$
Fundamental frequency in an open organ pipe, for $m=1 f_1=\frac{v}{2 L}$
Frequency of nth harmonic in a closed organ pipe, $f_n^{\prime}=\frac{n v}{4 L}$
Frequency of 3rd harmonic in a closed organ pipe, $f^{\prime}{ }_3=\frac{3 v}{4 L}$
Now, $f_3^{\prime}-f_1=50$
$\frac{3 v}{4 L}-\frac{v}{2 L}=50 \Rightarrow \frac{v}{L}=200$
Thus, fundamental frequency in an open organ pipe,
$f_1=\frac{v}{2 L}=\frac{200}{2} \mathrm{~Hz}=100 \mathrm{~Hz}$

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2022 (08 Aug Shift 1)
On closing an open organ pipe from one end, it is noticed that the frequency of third harmonic is $50 \mathrm{~Hz}$ more than the fundamental frequency of vibration in open organ
pipe. The fundamental frequency of open organ pipe is

Options:
A $250 \mathrm{~Hz}$
B $100 \mathrm{~Hz}$
C $50 \mathrm{~Hz}$
D $200 \mathrm{~Hz}$

The correct answer is:
$100 \mathrm{~Hz}$

Solution:
Fundamental frequency of closed pipe $=\mathrm{n}_{1}=\frac{\mathrm{V}}{4 \ell}$
Frequency of third harmonic $n_{3}=\frac{3 \mathrm{~V}}{4 \ell}$
Fundamental frequency of open pipe $=n_{1}{ }^{\prime}=\frac{V}{2 \ell}$
$\frac{3 \mathrm{~V}}{4 \ell}-\frac{\mathrm{V}}{2 \ell}=50$
$\therefore \frac{V}{4 \ell}=50 \quad \therefore \frac{V}{2 \ell}=100 \mathrm{~Hz}$

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2020 (12 Oct Shift 1)
On simplifying $\frac{\sin ^{3} A+\sin 3 \mathrm{~A}}{\sin \mathrm{A}}+\frac{\cos ^{3} \mathrm{~A}-\cos 3 \mathrm{~A}}{\cos \mathrm{A}}$, we get

Options:
A $\sin 3 \mathrm{~A}$
B $\cos 3 \mathrm{~A}$
C $\sin \mathrm{A}+\cos \mathrm{A}$
D 3

The correct answer is:
3

Solution:
$\frac{\sin ^{3} \mathrm{~A}+\sin 3 \mathrm{~A}}{\sin \mathrm{A}}+\frac{\cos ^{3} \mathrm{~A}-\cos 3 \mathrm{~A}}{\cos \mathrm{A}}$
$\Rightarrow \frac{\sin ^{3} \mathrm{~A}+3 \sin \mathrm{A}-4 \sin ^{3} \mathrm{~A}}{\sin \mathrm{A}}$
$+\frac{\cos ^{3} A-\left[4 \cos ^{3} A-3 \cos A\right]}{\cos A}$
$\Rightarrow \frac{3 \sin A-3 \sin ^{3} A}{\sin A}+\frac{\left(-3 \cos ^{3} A+3 \cos A\right)}{\cos A}$
$=3-3 \sin ^{2} \mathrm{~A}-3 \cos ^{2} \mathrm{~A}+3$
$=6-3\left(\cos ^{2} \mathrm{~A}+\sin ^{2} \mathrm{~A}\right)=6-3(1)=3$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2015 (Phase 2)
On the basis of above pedigree chart, select the correct option

Options:
A Autosomal recessive trait
B Sex-linked recessive trait
C Sex-linked dominant trait
D Autosomal dominant trait

The correct answer is:
Autosomal dominant trait

Solution:
The given pedigree chart is a representation of an autosomal dominant trait. Such a trait is carried by the autosomes and is irrelevant to the sex of the person. As given in the chart, it is inherited by both males and females equally and does not skip generations. The latter is ensured by the dominant nature of the trait, meaning only a single copy of the gene can cause it. A striking contrast of such a trait is observed with an autosomal recessive trait, in which both copies of the gene are required to inherit the trait. Examples of autosomal dominant diseases include myotonic dystrophy, Huntington disease, polycystic kidney disease.

Topic: Biology, Principles of Inheritance and Variation, NEET
On the basis of standard electrode potential of redox couples given below, find out which of the following is the strongest oxidising agent. $\left(E^{\circ}\right.$ values $: \mathrm{Fe}^{3+}\left|\mathrm{Fe}^{2+}=+0.77 \mathrm{~V} ; \mathrm{I}_{2(s)}\right| \mathrm{I}^{-}=$ $\left.+0.54 \mathrm{~V} ; \mathrm{Cu}^{2+}\left|\mathrm{Cu}=+0.34 \mathrm{~V} ; \mathrm{Ag}^{+}\right| \mathrm{Ag}=+0.80 \mathrm{~V}\right)$

Options:
A $\mathrm{Fe}^{3+}$
B $\mathrm{I}_{2(s)}$
C $\mathrm{Cu}^{2+}$
D $\mathrm{Ag}^{+}$

The correct answer is:
$\mathrm{Ag}^{+}$

Solution:
Higher the electrode potential, better is the oxidising agent. Since, the electrode
potential decreases in the order :
$\mathrm{Ag}^{+}\left|\mathrm{Ag}(+0.80 \mathrm{~V})>\mathrm{Fe}^{3+}\right| \mathrm{Fe}^{2+}(+0.77 \mathrm{~V})>$
$\mathrm{I}_{2(s)}\left|\mathrm{I}^{-}(+0.54 \mathrm{~V})>\mathrm{Cu}^{2+}\right| \mathrm{Cu}(+0.34 \mathrm{~V})$
Hence, $\mathrm{Ag}^{+}$is the strongest oxidising agent.

Topic: Chemistry, Electrochemistry, AIIMS, AIIMS 2017
On the basis of standard electrode potential values, suggest which of the following reactions would not take place? .

Options:
A Cu + Zn2+→ Cu2++ Zn
B Mg + Fe2+→ Mg2++ Fe
C Br2+ 2Cl–→ Cl2 + 2Br–
D Fe + Cd2+→ Cd + Fe2+

The correct answer is:
Cu + Zn2+→ Cu2++ Zn

Solution:
Correct Option is : (A)
Cu + Zn2+→ Cu2++ Zn

Topic: Chemistry, Redox Reactions, NEET
On the basis of the following observations made with aqueous solutions, What are the secondary valences of Pd in PdCl2.4NH3, Ni in NiCl2.6H2O and Pt in PtCl2.2NH3respectively?

Options:
A Secondary 2, Secondary 6, Secondary 3
B Secondary 4, Secondary 1, Secondary 4
C Secondary 4, Secondary 6, Secondary 4
D Secondary 7, Secondary 3, Secondary 5

The correct answer is:
Secondary 4, Secondary 6, Secondary 4

Solution:
Correct Option is : (C)
Secondary 4, Secondary 6, Secondary 4

Topic: Chemistry, Coordination Compounds, NEET
On the basis of the following observations made with aqueous solutions, What are the secondary valences of Pt in PtCl4.2HCl and Co in CoCl3.4NH3respectively?

Options:
A Secondary 2 and Secondary 8
B Secondary 1 and Secondary 3
C Secondary 6 and Secondary 4
D Secondary 6 and Secondary 6

The correct answer is:
Secondary 6 and Secondary 6

Solution:
Correct Option is : (D)
Secondary 6 and Secondary 6

Topic: Chemistry, Coordination Compounds, NEET
On the set $Z$ of integers, relation $\mathrm{R}$ is defined as "a $\mathrm{R} \mathrm{b} \Leftrightarrow$ $\mathrm{a}+2 \mathrm{~b}$ is an integral multiple of $3^{\prime \prime}$. Which oneofthe following statements is correct for $\mathrm{R}$ ?

Options:
A $\mathrm{R}$ is only reflexive
B $\mathrm{R}$ is only symmetric
C $\mathrm{R}$ is only transitive
D $\mathrm{R}$ is an equivalence relation

The correct answer is:
$\mathrm{R}$ is an equivalence relation

Solution:
The given relation is $\mathrm{aRb} \Leftrightarrow \mathrm{a}+2 \mathrm{~b}$, is an integral multiple of 3
In this relation $\mathrm{aRa} \Leftrightarrow \mathrm{a}+2 \mathrm{a}=3 \mathrm{a}$, an integral multiple of $3 . \mathrm{So}$, it is
reflexive $\mathrm{aRb} \Leftrightarrow \mathrm{a}+2 \mathrm{~b}$ and
$b R a=b+2 a+4 b-4 b$
$=2(a+2 b)-3 b$ is also an integral multiple of 3 . So, it issymmetric Let there be another value c, $\mathrm{bRc}=\mathrm{b}+2 \mathrm{c}$, be an integral multiple of 3 . Then $a \mathrm{Rc}=\mathrm{a}+2 \mathrm{c}$
So, $a R b+b R c=a+2 b+b+2 c=a+2 c+3 b$ is integral
multiple of 3 , hence, $\mathrm{a}+2 \mathrm{c}$ is also integral multiple $\mathrm{o}$ :
$=0, \mathrm{aRb}$ and $\mathrm{bRc} \Rightarrow \mathrm{aRc} . \mathrm{So}$, it is transitive
Therefore, relation is reflexive, symmetric as well transitive. Hence, $\mathrm{R}$ is an equivalence relation.

Topic: Mathematics, Sets and Relations, NDA, NDA 2006 (Phase 2)
On what basis the elements are arranged in the modern periodic table

Options:
A Increasing mass
B Increasing volume
C Increasing atomic number
D Alphabetically

The correct answer is:
Increasing atomic number

Solution:
Increasing atomic number. Mosley found that atomic no. was better fundamental property than atomic weight.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET
Once BOD from the sewage is reduced significantly, the 'flocs' are allowed to sediment and it is known as _________.

Options:
A primary sludge
B secondary sludge
C activated sludge
D inactivated sludge

The correct answer is:
activated sludge

Solution:
Once BOD of sewage water is reduced significantly, the effluent is then passed into the settling tank where the bacterial ' flocs' are allowed to sediment. This sediment is called as activated sludge.

Topic: Biology, Microbes in Human Welfare, NEET
One bag contains 3 white and 2 black balls, another bag contains 5 white and 3 black balls. If a bag is chosen at random and a ball is drawn from it, what is the chance that it is white?

Options:
A $\frac{3}{8}$
B $\frac{49}{80}$
C $\frac{8}{13}$
D $\frac{1}{2}$

The correct answer is:
$\frac{49}{80}$

Solution:
Probability of choosing one bag from two bags $=\frac{1}{2}$
Probability of choosing white ball from first bag $=\frac{3}{5}$
Probability of choosing white ball from second bag $=\frac{5}{8}$
$\therefore$ Required probability $=\frac{1}{2}\left(\frac{3}{5}+\frac{5}{8}\right)$
$=\frac{1}{2}\left(\frac{24+25}{40}\right)=\frac{1}{2} \times \frac{49}{40}=\frac{49}{80}$

Topic: Mathematics, Probability, NDA, NDA 2018 (Phase 1)