datavorous/entrance-exam-dataset huggingface dataset:
The probability of a non-leap year having 53 Mondays is .........

Options:
A $\frac{2}{7}$
B $\frac{1}{7}$
C $\frac{5}{7}$
D $\frac{6}{7}$

The correct answer is:
$\frac{1}{7}$

Solution:
Number of days in non Leap year $=365$
$$
\begin{aligned}
& =52 \times 7+1 \\
& =52 \text { weeks }+1 \text { day }
\end{aligned}
$$
The one additional day may be any one of the day
$$
\begin{aligned}
\therefore \quad n(A) & =1 \\
n(s) & =7 \\
\mathrm{P}(\text { Getting } 53 \text { sundays }) & =\frac{n(A)}{n(S)}=\frac{1}{7}
\end{aligned}
$$
Hence, option (2) is correct.

Topic: Mathematics, Probability, AP EAMCET, AP EAMCET 2020 (22 Sep Shift 2)
The probability of guessing a correct answer is $\frac{x}{12}$. If the probability of not guessing the correct answer is $\frac{2}{3}$, then what is $x$ equal to?

Options:
A 2
B 3
C 4
D 6

The correct answer is:
4

Solution:
Given Probability of guessing a correct answer $=\frac{x}{12}$
and probability of not guessing the correct answer $=\frac{2}{3}$
As we know
$P$ (occurence of an event) $+P$ (non-occurence of an event $)=1$
$\therefore \quad \frac{x}{12}+\frac{2}{3}=1 \Rightarrow \frac{x+8}{12}=1 \Rightarrow x=12-8=4$

Topic: Mathematics, Probability, NDA, NDA 2010 (Phase 1)
The probability that a non-leap year contains 53 Sunday is

Options:
A $\frac{1}{7}$
B $\frac{1}{9}$
C $\frac{2}{7}$
D $\frac{1}{5}$

The correct answer is:
$\frac{1}{7}$

Solution:
A non-leap year contains 365 days and there are total 52 weeks and 1 day in a non-leap year. Hence there should be 52 Sunday also.
But the 1 extra day apart from those 52 weeks may be either Monday, Tuesday, Wednesday, Thursday, Friday, Saturday or a Sunday.
Clearly, total possible events are $=7$
and Number of favourable events $=1$
Hence probability $=\frac{1}{7}$

Topic: Mathematics, Probability, AP EAMCET, AP EAMCET 2023 (19 May Shift 1)
The probability that a ship safely reaches a port is $\frac{1}{3}$. The probability that out of 5 ships, at least 4 ships would arrive safely is

Options:
A $\frac{1}{243}$
B $\frac{10}{243}$
C $\frac{11}{243}$
D $\frac{13}{243}$

The correct answer is:
$\frac{11}{243}$

Solution:
Probability that a ship reaches port $=\frac{1}{3}$
Probability that a ship not reaching port $=1-\frac{1}{3}=\frac{2}{3}$ Number of ships $(\mathrm{n})=5$. $\mathrm{r}=4,5$
$\mathrm{p}(4)+\mathrm{p}(5)={ }^{5} \mathrm{c}_{4}\left(\frac{1}{3}\right)^{4}\left(\frac{2}{3}\right)^{5-4}+{ }^{5} \mathrm{c}_{5}\left(\frac{1}{3}\right)^{5}\left(\frac{2}{3}\right)^{5-5}$
$=5\left(\frac{2}{3^{5}}\right)+\frac{1}{3^{5}}$
$=\frac{10}{243}+\frac{1}{243}=\frac{11}{243}$

Topic: Mathematics, Probability, NDA, NDA 2017 (Phase 2)
The probability that a student passes in mathematics is $4 / 9$ and that he passes in physics is $2 / 5$. Assuming that passing in mathematics and physics are independent of each other, what is the probability that he passes in mathematics but fails in physics?

Options:
A $\frac{4}{15}$
B $\frac{8}{45}$
C $\frac{26}{45}$
D $\frac{19}{45}$

The correct answer is:
$\frac{4}{15}$

Solution:
Probability of passing in mathematics $=\frac{4}{9}$
Probability of passing in physics $=\frac{2}{5}$
Probability of failure in physics $=1-\frac{2}{5}=\frac{3}{5}$ Given that both the events are independent. Required probability $=\frac{4}{9} \times \frac{3}{5}=\frac{4}{15}$

Topic: Mathematics, Probability, NDA, NDA 2006 (Phase 2)
The probability that in a random arrangement of the letters of the word 'UNIVERSITY', the two I's do not come together is

Options:
A $4 / 5$
B $1 / 5$
C $1 / 10$
D $9 / 10$

The correct answer is:
$4 / 5$

Solution:
$$
\begin{array}{l}
\because 1 \mathrm{U}, 1 \mathrm{~N}, 2 \mathrm{I}, 1 \mathrm{~V}, 1 \mathrm{E}, 1 \mathrm{R}, 1 \mathrm{~S}, 1 \mathrm{~T}, 1 \mathrm{Y} \\
\therefore \text { Total number of possible arrangements }=\frac{10 !}{2 !} \\
\text { and favourable arrangements }=\frac{10 !}{2 !}-9 ! \\
\therefore \text { Required probability }=\frac{\frac{10 !}{2 !}-9 !}{\frac{10 !}{2 !}} \\
=\frac{9 !(5-1)}{9 ! \times 10} \times 2=\frac{4}{5}
\end{array}
$$

Topic: Mathematics, Probability, NDA, NDA 2014 (Phase 2)
The process during which the differentiation occurs without any cell division

Options:
A Spermiogenesis
B Spermatocytogenesis
C Spermacytogenesis
D Spermatidogenesis

The correct answer is:
Spermiogenesis

Solution:
Having completed the first meiotic division, these cells give rise to daughter cells called secondary spermatocytes, which divide to complete meiosis to form round spermatids.
The round spermatids do not divide but undergo a complex metamorphosis, called spermiogenesis, to become spermatozoa that are released into the lumen of the seminiferous tubule by a process called spermiation.

Topic: Biology, Human Reproduction, NEET
The process in which colloids, when subjected to DC electric field move towards an electrode is

Options:
A Brownian movement
B Tyndall effect
C Peptization
D Electrophoresis

The correct answer is:
Electrophoresis

Solution:
Electrophoresis

Topic: Chemistry, Surface Chemistry, TS EAMCET, TS EAMCET 2022 (20 Jul Shift 2)
The product of an amine ' $\mathrm{X}$ ' with benzene sulphonyl chloride produces the product which is insoluble in alkali. The product of ' $\mathrm{X}$ ' with ethanoyl chloride is

Options:
A $\mathrm{C}_6 \mathrm{H}_5 \mathrm{NHCOCH}_3$
B $\mathrm{C}_6 \mathrm{H}_5 \mathrm{~N}\left(\mathrm{CH}_3\right) \mathrm{COCH}_3$
C $\mathrm{C}_6 \mathrm{H}_5 \mathrm{~N}\left(\mathrm{CH}_3\right) \mathrm{CH}_2 \mathrm{CH}_3$
D $\mathrm{C}_6 \mathrm{H}_5 \mathrm{NHCH}_2 \mathrm{CH}_3$

The correct answer is:
$\mathrm{C}_6 \mathrm{H}_5 \mathrm{~N}\left(\mathrm{CH}_3\right) \mathrm{COCH}_3$

Solution:
A secondary amine gives an insoluble compound upon reaction with Hinsberg reagent.
Thus, ' $\mathrm{X}$ ' will be

Topic: Chemistry, Amines, AP EAMCET, AP EAMCET 2023 (18 May Shift 2)
The product of first nine terms of a GP is, in general, equal to which one of the following?

Options:
A The 9 th power of the 4 th term
B The 4 th power of the 9 th term
C The 5 th power of the 9 th term
D The 9 th power of the 5 th term

The correct answer is:
The 9 th power of the 5 th term

Solution:
Let a be the first term and $\mathrm{r}$, the common ratio First nine terms of a GP are a, ar, ar $^{2}, \ldots .$ ar $^{8}$.
$$
\begin{aligned}
\therefore \quad \mathrm{P} &=\mathrm{a} . \mathrm{ar} \cdot \mathrm{ar}^{2} \ldots . \mathrm{ar}^{8}=\mathrm{a}^{9} \cdot \mathrm{r}^{1+2+\ldots+8} \\
&=\mathrm{a}^{9} \cdot \mathrm{r}^{\frac{8.9}{2}}=\mathrm{a}^{9} \mathrm{r}^{36}=\left(\mathrm{ar}^{4}\right)^{9}=\left(\mathrm{T}_{5}\right)^{9}
\end{aligned}
$$
$=9$ th power of the 5 th term

Topic: Mathematics, Sequences and Series, NDA, NDA 2008 (Phase 1)
The product of $\mathrm{y}$ the perpendiculars from the two points $(\pm 4,0)$ to the line $3 x \cos \phi+5 y \sin \phi=15$ is

Options:
A 25
B 16
C 9
D 8

The correct answer is:
9

Solution:
If length of perpendicular be $\mathrm{p}_{1}$ from the point $(4,0)$ $\begin{aligned} \mathrm{p}_{1} &=\left|\frac{12 \cos \phi-15}{\sqrt{(3 \cos \phi)^{2}+(5 \cos \phi)^{2}}}\right| \\ &=\frac{15-12 \cos \phi}{\sqrt{(3 \cos \phi)^{2}+(5 \cos \phi)^{2}}} \end{aligned}$
If length of perpendicular be $\mathrm{p}_{2}$ from the point $(-4,0)$
$p_{2}=\left|\frac{-12 \cos \phi-15}{\sqrt{(3 \cos \phi)^{2}+(5 \cos \phi)^{2}}}\right|$
$\begin{aligned}=& \frac{(12 \cos \phi+15)}{\sqrt{(3 \cos \phi)^{2}+(5 \sin \phi)^{2}}} \\ \mathrm{p}_{1} \cdot \mathrm{p}_{2} &=\frac{(15-12 \cos \phi)(12 \cos \phi+15)}{(3 \cos \phi)^{2}+(5 \sin \phi)^{2}} \\ &=\frac{\left(225-144 \cos ^{2} \phi\right)}{9 \cos ^{2} \phi+25 \sin ^{2} \phi}=\frac{9\left(25-16 \cos ^{2} \phi\right)}{25-16 \cos ^{2} \phi} \\ &=9 \end{aligned}$

Topic: Mathematics, Straight Lines, NDA, NDA 2015 (Phase 2)
The production of food grains in Maharashtra is given for the 12 years from 1992 to 2003 . Which one of the following representations is most suitable to depict the data?

Options:
A A simple bar diagram
B A pie diagram
C A component bar diagram with the components arranged in chronological order
D A broken line graph $\quad$

The correct answer is:
A simple bar diagram

Solution:
A simple bar diagram is most suitable for this.

Topic: Mathematics, Statistics, NDA, NDA 2006 (Phase 1)
The products resulting from atmospheric reactions of hydrocarbons and nitrogen oxides in the presence of sunlight are called:

Options:
A Primary pollutants
B Secondary pollutants
C Tertiary pollutants
D Non-pollutants

The correct answer is:
Secondary pollutants

Solution:
An air pollutant is a substance in the air that can have adverse effects on humans and the ecosystem. The substance can be solid particles, liquid droplets, or gases. A pollutant can be of natural origin or man-made. Secondary pollutants are not emitted directly. Rather, they form in the air when primary pollutants react or interact. Ground-level ozone is a prominent example of a secondary pollutant.

Topic: Biology, Environmental Issues, NEET
The profit function, in rupees, of a firm selling $x$ items $(x \geq 0)$ per week is given by $\mathrm{P}(\mathrm{x})=-3500+(400-\mathrm{x}) \mathrm{x} . \mathrm{How}$
many items should the firm sell so that the firm has maximum profit?

Options:
A 400
B 300
C 200
D 100

The correct answer is:
200

Solution:
$\quad P(x)=-3500+(400-x) x=-3500+400 x-x^{2}$
On differentiating w.r.t. $x$, we get $P^{\prime}(x)=400-2 x$
Put $P^{\prime}(x)=0$ for maxima or minima $\Rightarrow 400-2 x=0$
$\Rightarrow x=200$
Now $P^{\prime \prime}(x)=-2 x$
$\Rightarrow P^{\prime \prime}(200)=-400 < 0$
$\therefore P(x)$ is maximum at $x=200$ Hence 200 items should the firm sell so that the firm has maximum profit.

Topic: Mathematics, Application of Derivatives, NDA, NDA 2009 (Phase 1)
The projections of a directed line segment on the coordinate axes are $12,4,3$ respectively. What are the direction cosines of the line segment?

Options:
A $\left\langle\frac{12}{13}, \frac{4}{13}, \frac{3}{13}\right\rangle$
B $\left\langle\frac{12}{13},-\frac{4}{13}, \frac{3}{13}\right\rangle$
C $\left\langle\frac{12}{13},-\frac{4}{13},-\frac{3}{13}\right\rangle$
D $\left\langle-\frac{12}{13},-\frac{4}{13}, \frac{3}{13}\right\rangle$

The correct answer is:
$\left\langle\frac{12}{13}, \frac{4}{13}, \frac{3}{13}\right\rangle$

Solution:
Direction cosine of line segment $=\left(\frac{12}{13}, \frac{4}{13}, \frac{3}{13}\right)$
Option (a) is correct.

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2015 (Phase 1)
The projections of a directed line segment on the coordinate axes are $12,4,3$ respectively. What is the length of the line segment?

Options:
A 19 units
B 17 units
C 15 units
D 13 units

The correct answer is:
13 units

Solution:
The projection of a directed line segment on the co-ordinate axes are $12,4,3$, respectively. 4e Length of the line segment $=\sqrt{12^{2}+4^{2}+3^{2}}$
$=\sqrt{144+16+9}=\sqrt{169}=13$ units
$\therefore \quad$ Option (d) is correct.

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2015 (Phase 1)
The property of colloid is

Options:
A Scattering of light
B settling under gravity
C Dialysis
D Emulsion

The correct answer is:
Scattering of light

Solution:
Tyndal effect is scattering of light by colloidal soln.

Topic: Chemistry, Surface Chemistry, NEET
The protective power of a lyophilic colloidal sol is expressed in terms of

Options:
A

Critical micelle concentration

B

Standard reduction potential

C Gold number
D Oxidation number

The correct answer is:
Gold number

Solution:
The protective power of lyophilic colloidal sol is expressed in terms of gold number. Gold number is the number of milligrams of the protective colloid which prevent the coagulation of 10 ml of red gold sol. Smaller the gold number higher the protective power of a colloid.

Topic: Chemistry, Surface Chemistry, AP EAMCET, AP EAMCET 2021 (19 Aug Shift 2)
The protein haemoglobin found in humans can be described having a ___________ structure.

Options:
A primary
B secondary
C tertiary
D quaternary

The correct answer is:
quaternary

Solution:
When a protein is made up of two or more than two polypeptide held together variously than it form quaternary structure. Protein haemoglobin has quaternary structure as it is made up of 4 polypeptide or subunits. Two of these are identical to each other. Hence two subunits of alpha type and two subunits of beta type together constitutes the human haemoglobin molecule.

Topic: Chemistry, Biomolecules, NEET
The proteins with a prosthetic group are known as-

Options:
A Complex proteins
B Conjugated proteins
C Secondary proteins
D Essential proteins

The correct answer is:
Conjugated proteins

Solution:
Conjugated protein = alpha amino acid + non protein material (prosthetic group)

Topic: Chemistry, Biomolecules, NEET
The QRS complex in a standard ECG represents:

Options:
A Depolarisation of auricles
B Depolarisation of ventricles
C Repolarisation of ventricles
D Repolarisation of auricles

The correct answer is:
Depolarisation of ventricles

Solution:
An electrocardiogram (ECG), is a graphical representation of the electrical activity of the heart during a cardiac cycle. The machine used to obtain an electrocardiogram is known as electrocardiography and this technique is called electrocardiography. The interpretation of ECG is as follows:
• The P wave is a small upward wave that represents electrical excitation (or depolarization) of the atria which leads to contraction of both the atria.
• The QRS (wave) complex represents the depolarization of the ventricles, which initiates the ventricular contraction (ventricular systole). The contraction of the ventricles start shortly after Q and marks the beginning of the systole.
• The T wave represents the return of the ventricles from excited (depolarized) to a normal state (i.e., repolarisation). The end of the T wave marks the end of the systole.
• Thus, by counting the number of QRS complexes that occur in a given time period, the heartbeat rate of an individual can be determined.

Topic: Biology, Body Fluids and Circulation, NEET, NEET 2020 (Phase 1)
The quadratic equation $x^{2}+b x+4=0$ will have real roots if

Options:
A $\quad b \leq-4$ only
B $b \geq 4$ only
C $-4 < b < 4$
D $\quad b \leq-4, b \geq 4$

The correct answer is:
$\quad b \leq-4, b \geq 4$

Solution:
If root are real $b^{2}-4 \times 4 \geq$
$b^{2} \geq 16$
$b \leq-4, b \geq 4$
$\alpha$ and $\beta$ are roots of given equation. $(a \alpha+b)(a \beta+b)=a^{2} \alpha \beta+a b \alpha+a b \beta+b^{2}$
$=a^{2} \alpha \beta+a b(\alpha+\beta)+b^{2}$
From the given quadratic equation
$\alpha+\beta=\frac{-b}{a}, \alpha \beta=\frac{c}{a}$
$a^{2} \times \frac{c}{a}+a b \times-\frac{b}{a}+b^{2}=a c$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2013 (Phase 2)
The Quantum Hall Resistance $\mathrm{R}_{\mathrm{H}}$ is a fundamental constant with dimensions of resistance. If $\mathrm{h}$ is Planck's constant and e the electron charge, then the dimension of $\mathrm{R}_{\mathrm{H}}$ is the same as -

Options:
A $\mathrm{e}^{2} / \mathrm{h}$
B $\mathrm{h} / \mathrm{e}^{2}$
C $\mathrm{h}^{2} / \mathrm{e}$
D $\mathrm{e} / \mathrm{h}^{2}$

The correct answer is:
$\mathrm{h} / \mathrm{e}^{2}$

Solution:
$\mathrm{R}=\frac{\mathrm{V}}{\mathrm{i}}=\frac{\mathrm{V} \times \mathrm{i}}{\mathrm{i}^{2}}=\frac{\mathrm{P}}{\mathrm{i}^{2}}$
energy $=\mathrm{h} \mathrm{v}=\frac{\mathrm{h}}{\mathrm{t}}$
Power $=\frac{\text { energy }}{\mathrm{t}}$
$\mathrm{P}=\frac{\mathrm{h}}{\mathrm{t}^{2}}$
$\mathrm{i}=\frac{\mathrm{e}}{\mathrm{t}}$
$\frac{\mathrm{P}}{\mathrm{i}^{2}}=\frac{\mathrm{h}}{\mathrm{e}^{2}}$

Topic: Physics, Units and Dimensions, KVPY, KVPY 2011 (SB/SX)
The quiescent centre in root meristem serves as a

Options:
A site for storage of food which is utilized during maturation
B reservoir of growth hormones
C reserve for replenishment of damaged cells of the meristem
D region for absorption of water.

The correct answer is:
reserve for replenishment of damaged cells of the meristem

Solution:
A zone of inactive cells is present in the central part of the root apex called quiscent centre. It is a hemispherical region of inactive cells present in between the root cap and the active meristematic region of root apex. This zone contains several hundred cells which are characterized by having lower concentration of DNA, RNA and protein. The quiescent centre acts as reservoir of cells and becomes active whenever the previously active initials get damaged or during the establishment of the secondary roots.

Topic: Biology, Anatomy of Flowering Plants, AIIMS, AIIMS 2003
The radius of a circle is uniformly increasing at the rate of $3 \mathrm{~cm} / \mathrm{s}$. What is the rate of increase in area, when the radius is $10 \mathrm{~cm} ?$

Options:
A $6 \pi \mathrm{cm}^{2} / \mathrm{s}$
B $10 \pi \mathrm{cm}^{2} / \mathrm{s}$
C $30 \pi \mathrm{cm}^{2} / \mathrm{s}$
D $60 \pi \mathrm{cm}^{2} / \mathrm{s}$

The correct answer is:
$60 \pi \mathrm{cm}^{2} / \mathrm{s}$

Solution:
Given $\frac{d r}{d t}=3$
Let $\mathrm{A}=$ Area of circle $=\pi \mathrm{r}^{2}$
$\therefore \frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}}=2 \pi \mathrm{r} \cdot \frac{\mathrm{d} \mathrm{r}}{\mathrm{dt}}=6 \pi \mathrm{r}$
Now, $\left.\frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}}\right|_{\mathrm{r}=10}=6 \times 10 \times \pi=60 \pi \mathrm{cm}^{2} / \mathrm{s}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2012 (Phase 2)
The radius of the circle $x^{2}+y^{2}+x+c=0$ passing through the origin is

Options:
A $\frac{1}{4}$
B $\frac{1}{2}$
C 1
D 2

The correct answer is:
$\frac{1}{2}$

Solution:
Circle is passing through origin then $\mathrm{C}=0$ Now, $x^{2}+y^{2}+x=0$
$\mathrm{x}^{2}+\mathrm{x}+\frac{1}{4}-\frac{1}{4}+\mathrm{y}^{2}=0$
$\left(\mathrm{x}+\frac{1}{2}\right)^{2}+\mathrm{y}^{2}=\left(\frac{1}{2}\right)^{2}$

Topic: Mathematics, Circle, NDA, NDA 2013 (Phase 2)
The radius of the sphere $\quad[2015-I I]$ $3 x^{2}+3 y^{2}+3 z^{2}-8 x+4 y+8 z-15=0$ is

Options:
A 2
B 3
C 4
D 5

The correct answer is:
3

Solution:
$3 x^{2}+3 y^{2}+3 z^{2}-8 y+4 y+8 z-15=0$
$\Rightarrow x^{2}+y^{2}+z^{2}-\frac{8}{3} x+\frac{4}{3} y+\frac{8}{3} z-5=0$
$\Rightarrow\left(x-\frac{4}{3}\right)^{2}-\frac{16}{9}+\left(y+\frac{2}{3}\right)^{2}-\frac{4}{9}+\left(z+\frac{4}{3}\right)^{2}-\frac{16}{9}-5=0$
$\Rightarrow\left(x-\frac{4}{3}\right)^{2}+\left(y+\frac{2}{3}\right)^{2}+\left(z+\frac{4}{3}\right)^{2}=(3)^{2}$
So radius is 3 .

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2015 (Phase 2)
The range of frequency bands used for standard AM
broadcast is

Options:
A 540 – 1600 KHz
B 88 – 108 MHz
C 800 – 900 MHz
D 3.7 – 4.2 GHz

The correct answer is:
540 – 1600 KHz

Solution:
No solution. Refer to answer key.

Topic: Physics, Communication System, TS EAMCET, TS EAMCET 2022 (19 Jul Shift 1)
The rank of \(\left[\begin{array}{ccc}2 & 1 & 1 \\ 0 & 3 & -1 \\ 1 & -1 & 1\end{array}\right]\) is

Options:
A 1
B 2
C 3
D 4

The correct answer is:
2

Solution:
\(\begin{aligned}
& \begin{aligned}
A & =\left[\begin{array}{ccc}
2 & 1 & 1 \\
0 & 3 & -1 \\
1 & -1 & 1
\end{array}\right] \\
R_3 \rightarrow & 2 R_3-R_1 \\
& =\left[\begin{array}{ccc}
2 & 1 & 1 \\
0 & 3 & -1 \\
0 & -3 & 1
\end{array}\right] \\
R_3 \rightarrow & R_3+R_2 \\
& =\left[\begin{array}{ccc}
2 & 1 & 1 \\
0 & 3 & -1 \\
0 & 0 & 0
\end{array}\right]
\end{aligned}
\end{aligned}\)
\(\therefore\) Rank of \(A=2\)
Hence, option (b) is correct.

Topic: Mathematics, Matrices, AP EAMCET, AP EAMCET 2020 (18 Sep Shift 2)
The rate at which light energy is converted into chemical energy of organic molecules in the ecosystem is:

Options:
A Net primary productivity
B Gross secondary productivity
C Net secondary productivity
D Gross primary productivity

The correct answer is:
Gross primary productivity

Solution:
Gross primary productivity is the total rate of photosynthesis. Light energy is converted into chemical energy including the organic matter used up in respiration during the measurement period.

Topic: Biology, Ecosystem, NEET
The rate at which light energy is converted to the chemical energy of organic molecules is:

Options:
A Net primary productivity
B Gross primary productivity
C Net secondary productivity
D Gross secondary productivity

The correct answer is:
Gross primary productivity

Solution:
The rate at which organic molecules are formed in a green plant is called gross productivity. Gross primary productivity of an ecosystem is the rate of production of organic matter during photosynthesis. A considerable amount of GPP is utilised by plants in respiration.

Topic: Biology, Ecosystem, NEET
The rate constant of a reaction is given by $k=P Z e^{-E_{a} / R T}$ under standard notation. In order to speed up the reaction, which of the following factors has to be decreased?

Options:
A $Z$
B Both $Z$ and $T$
C $E_{a}$
D $T$

The correct answer is:
$E_{a}$

Solution:
$\therefore k=P Z e^{-E a / R T}$
The above reaction expression is given in terms of exponents. We have to increase the speed of reaction. So, value of $k$ has to be more and as per the given expression.
$k \propto \frac{1}{e^{E_{a} / R T}}$
if $E a$ decreases, than $\left[e^{E_{a} / R T}\right]$ also decreases and $\left[e^{\frac{1}{E_{a} / R T}}\right]$ increases.
$\therefore k$ also increases

Topic: Chemistry, Chemical Kinetics, KCET, KCET 2020
The rate of conversion of light energy into the chemical energy of organic molecules in an ecosystem is called

Options:
A Gross primary productivity
B Net primary productivity
C Net secondary productivity
D Gross secondary productivity

The correct answer is:
Gross primary productivity

Solution:
The amount of matter found in the organism is called biomass.
Gross primary production (GPP) is the rate at which an ecosystem's producers capture and store a given amount of chemical energy as biomass in a given length of time. (Definition). It is also the rate of conversion of light energy to chemical energy by producers in an ecosystem.
Total production of organic matter is called gross primary productivity and the energy left after utilization of producers is termed net primary productivity. Community productivity is the difference between GPP and the community respiration. The rate of assimilation of food energy by consumers is called secondary productivity and this energy is available to carnivores as they are the consumers.

Topic: Biology, Ecosystem, NEET
The rate of organic matter produced by producers in their bodies per unit time and area is called as

Options:
A GPP
B NPP
C Secondary productivity
D Ecological efficiency

The correct answer is:
GPP

Solution:
The amount of energy accumulation in green plants as biomass or organic matter per unit area over a time period is known as primary productivity. The rate of total capture of energy or the rate of the total production of organic material (biomass) during photosynthesis, is known as gross primary productivity.

Topic: Biology, Ecosystem, NEET
The rate of synthesis of organic matter by consumers is

Options:
A net primary productivity
B gross primary productivity
C standing state
D secondary productivity

The correct answer is:
secondary productivity

Solution:
Carnivores are organisms which obtain nutrient by killing and eating other animals. These form the secondary consumers preceded by the herbivore in the food chain. The rate of assimilation of food energy by consumers is called secondary productivity and this energy is available to carnivores as they are the consumers.
Total production of organic matter is called gross primary productivity and the energy left after utilization of producers is termed net primary productivity.

Topic: Biology, Ecosystem, NEET
The ratio EW of relative strength of electromagnetic force E to weak nuclear force W is,

Options:
A 10-11
B 1011
C 1020
D 10-20

The correct answer is:
1011

Solution:
The fundamental forces in decreasing order of their magnitude are, strong nuclear force (1)>the electromagnetic force (10−2 )> weak nuclear force (10−13)> gravitational force (10-49). So, the relation between electromagnetic force and weak nuclear force is,
EW=10-210-13=1011.

Topic: Physics, Motion In One Dimension, TS EAMCET, TS EAMCET 2021 (04 Aug Shift 2)