datavorous/entrance-exam-dataset huggingface dataset:
Study the following statements and select the option with correct statements.

(i) A lion eating a deer and a sparrow feeding on grains are ecologically similar in being consumers.

(ii) Predator star fish Pisaster helps in maintaining species diversity of some invertebrates.

(iii) Predators ultimately lead to the extinction of prey species.

(iv) Production of chemicals such as nicotine by the plants are metabolic disorders.

Options:
A (ii) and (iii)
B (iii) and (iv)
C (i) and (iv)
D (i) and (ii)

The correct answer is:
(i) and (ii)

Solution:
Predator and prey evolve together. Predators are prudent in nature, they never over exploits there prey as in absence of the prey predator will also become extinct. The prey is part of the predator's environment. Nicotine is an alkaloid found in the night shade family of plants (Solanaceae) that constitutes approximately 0.6–3.0% of dry weight of tobacco. These are not metabolic disorder products but are secondary metabolites or metabolic wastes.

Topic: Biology, Biodiversity and its Conservation, NEET
Study the four statements (1-4) given below and select the two correct ones out of them.
(1) A lion eating a deer and a sparrow feeding on grain are ecologically similar in being consumers
(2) Predator star fish Pisaster helps in maintaining species diversity of some invertebrates
(3) Predators ultimately lead to the extinction of prey species
(4) Production of chemicals such as nicotine, strychnine by the plants are metabolic disorders
The two correct statements are

Options:
A (2) and (3)
B (3) and (4)
C (1) and (4)
D (1) and (2)

The correct answer is:
(1) and (2)

Solution:
Both lion (carnivore) and sparrow (herbivore) are consumers. The Asteroidea occupy several important roles through out ecology and biology. Sea stars, such as Ochre star (Pisaster achraceus) have became widely known as the example of the keystone species concept in ecology. Most species are generalist predators, eating molluscs such as clams, oysters, etc.

Topic: Biology, Organisms and Populations, NEET, NEET 2010 (Screening)
Study the given diagram below of the stirred tank fermenter. What is the function of part labeled A?

Options:
A To maintain optimum pH
B To control the foam within the reactor
C To maintain optimum oxygen concentration
D To facilitate even mixing.

The correct answer is:
To control the foam within the reactor

Solution:
A stirred-tank reactor is usually cylindrical or with a curved base to facilitate the mixing of the reactor contents. The stirrer facilitates even mixing and oxygen availability throughout the bioreactor. The bioreactor has an agitator system, an oxygen delivery system, and a foam control system, a temperature control system, a pH control system, and sampling ports.

Topic: Biology, Biotechnology - Principles and Processes, NEET
Study the orbital diagrams of two atoms 'X' and 'Y'. Which subshell will be more stable and why?

Options:
A X, exchange energy is maximum, so is stability
B Y, exchange energy is maximum, so is stability
C X, exchange energy is minimum, so stability is maximum
D Y, exchange energy is minimum, so stability is maximum

The correct answer is:
X, exchange energy is maximum, so is stability

Solution:
In 'X' one electron can exchange energy with 5 orbitals while in 'Y' energy exchange is between 3 orbitals hence half-filled orbitals are more stable.

Topic: Chemistry, Structure of Atom, JEE Main
Study the pathway given below
In which of the following options, correct words for all the three blanks A, B and C are indicated?

Options:
A (a)
B (b)
C (c)
D (d)

The correct answer is:
(c)

Solution:
A-Fixation (of $\mathrm{CO}_2$ by PEP
carboxylase/oxygenase)
B-Decarboxylation
C-Regeneration

Topic: Biology, Respiration in Plants, NEET, NEET 2010 (Mains)
Study the pedigree chart given below :
What does it show?

Options:
A Inheritance of a recessive sex-linked disease like haemophilia
B Inheritance of a sex-linked inborn error of metabolism like phenylketonuria
C Inheritance of a condition like phenylketonuria as an autosomal recessive trait
D The pedigree chart is wrong as this is not possible

The correct answer is:
Inheritance of a condition like phenylketonuria as an autosomal recessive trait

Solution:
No Solution Available

Topic: Biology, Principles of Inheritance and Variation, NEET, NEET 2009 (Mains)
Study the pedigree chart of a certain family given below and select the correct conclusion which can be drawn for the character.

Options:
A The female parent is heterozygous
B The parents could not have had a normal daughter for this character
C The trait under study could not be colour blindness
D The male parent is homozygous dominant

The correct answer is:
The female parent is heterozygous

Solution:
The given pedigree chart shows that both the daughters received the gene from the parents, while son may be normal or affected. It shows that the female parent is heterozygous.

Topic: Biology, Principles of Inheritance and Variation, NEET, NEET 2010 (Mains)
Study the structure of maltose and mark the incorrect statement.

Options:
A Maltose is composed of two α -D-glucose units
B C-1 of one glucose is linked to C-4 of other unit
C It is a non-reducing sugar
D It is a disaccharide

The correct answer is:
It is a non-reducing sugar

Solution:
It is a reducing sugar. The free aldehyde group can be produced at C-1 of second glucose in solution and it shows reducing properties.

Topic: Chemistry, Biomolecules, JEE Main
Substance A2B(g) can undergo decomposition to form two set of products

If the molar ratio of A2gto A(g) is 5 : 3 in a set of product gases , then the energy involved in the decomposition of 1 mole of A2B(g) is:

Options:
A 48.75 kJ/mol
B 43.75 kJ/mol
C 46.25 kJ/mol
D 64.2 kJ/mol

The correct answer is:
43.75 kJ/mol

Solution:
If the molar ratio of A2gto A(g) is 5 : 3 in a set of product gases , then the energy involved in the decomposition of 1 mole of A2B(g) is:
ârH=58Ã40+38Ã50=43.75 kJ/ mol

Topic: Chemistry, Thermodynamics (C), JEE Main
Substance A has atomic mass number 16 and half life of 1 day. Another substance B has atomic mass number 32 and half life of 12 day. If both A and B simultaneously start undergo radio activity at the same time with initial mass 320 g each, how many total atoms of A and B combined would be left after 2 days

Options:
A 3.38 × 1024
B 6.76 × 1024
C 6.76 × 1023
D 1.69 × 1024

The correct answer is:
3.38 × 1024

Solution:
Number of nuclei left after n half-lives is given by,

N=N012tt12.

As, N0A=32016=20 moles and

N0B=32032=10 moles

Then, NA=N0A221=204=5 and

NB=N0B220.5=1024=0.625

So total N = 5.625

and number of atoms =5.625×6.023×1023

=3.38×1024

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2023 (29 Jan Shift 2)
Substance which is weakly repelled by a magnetic field is

Options:
A \( \mathrm{O}_{2} \)
B \( \mathrm{H}_{2} \mathrm{O} \)
C \( \mathrm{CrO}_{2} \)
D \( \mathrm{Fe}_{3} \mathrm{O}_{4} \)

The correct answer is:
\( \mathrm{H}_{2} \mathrm{O} \)

Solution:
Diamagnetic substances are weakly repelled by a magnetic field. H2O, NaCl and C6H6 are some examples of such substances. They are weakly magnetised in a magnetic field in opposite direction. Diamagnetism is shown by those substances in which all the electrons are paired and there are no unpaired electrons. Pairing of electrons cancels their magnetic moments, and they lose their magnetic character.

Topic: Chemistry, Solid State, JEE Main
Succulents are known to keep their stomata closed during the day to check transpiration. How do they meet their photosynthetic $\mathrm{CO}_2$ requirements?

Options:

None

Solution:
Succulent plants grow in dry and xeric conditions so, to prevent water loss through transpiration the stomata remains closed during day time. So that the gaseous exchange does not take place.
Thus plants have developed the mechanism to fix $\mathrm{CO}_2$ during night in the form of malic acid, which is a 4 carbon compound and are released during the day, inside the photosynthetic cells.

Topic: Biology, Photosynthesis in Higher Plants
Sucrose contains which of the following groups

Options:
A $-\mathrm{CHO}$
B $\gt C=0$
C Both (a) and (b)
D None of these

The correct answer is:
None of these

Solution:
Sucrose is a non-reducing sugar as both reducing groups (aldehyde and ketone) are bonded together.

Topic: Chemistry, Biomolecules, JEE Main
Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with t1/2=3.00 hours. What fraction of a sample of sucrose remains after 8 hours ?

Options:
A 0.811 M
B 0.675 M
C 0.465 M
D 0.158 M

The correct answer is:
0.158 M

Solution:
Correct Option is : (D)
0.158 M

Topic: Chemistry, Chemical Kinetics, NEET
Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with \(t_{1 / 2}=\) \(3 \cdot 00\) hours. What fraction of sample of sucrose remains after 8 hours ?

Options:

None

Solution:
Decomposition of sucrose is first order reaction,
\(k=\frac{2 \cdot 303}{t} \log \frac{\left[\mathrm{A}_0\right]}{[\mathrm{A}]}\) ...(i)
For time, \(t_{1 / 2}=3.0 \mathrm{hr}\)
\(k=\frac{0.693}{t_{1 / 2}}=\frac{0 \cdot 693}{3 \mathrm{hr}}=0.231 \mathrm{hr}^{-1}\)
Substituting \(k=0.231 \mathrm{hr}^{-1}\) in \((i)\) we get, \(0.231 \mathrm{hr}^{-1}=\frac{2 \cdot 303}{8 \mathrm{hr}} \log \frac{\left[\mathrm{A}_0\right]}{[\mathrm{A}]}\)
\(\begin{aligned}
&\log \frac{\left[\mathrm{A}_0\right]}{[\mathrm{A}]}=0.8024 \\
&\frac{\left[\mathrm{A}_0\right]}{[\mathrm{A}]}=\text { Antilog } 0.8024=6.345
\end{aligned}\)
\(\therefore\) fraction of sample remains
\(\frac{[\mathrm{A}]}{\left[\mathrm{A}_0\right]}=\frac{1}{6 \cdot 345}=0.158\)

Topic: Chemistry, Chemical Kinetics
Sucrose hydrolyses in acid solution into glucose and fructose following first order rate law with a half-life of 3.33 h at 25°C. After 9 h, the fraction of sucrose remaining is f. The value of log101f is ________ ×10-2. (Rounded off to the nearest integer)

[Assume: In10=2.303, ln 2=0.693]

Options:

The correct answer is:
81

Solution:
C12H22O11+H2O→t1/2=103hrorderC6H12O6Glucose+C6H12O6Fructose

t=0 a=A0

t=9hr a-x=At

from I order kinetic: k×t2.303=logA0At

⇒ln2×9103×2.303=log1f

⇒0.693×9×323.03=log1f

⇒log1f=0.81246=81.24×10-2

⇒x=81

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2021 (24 Feb Shift 2)
Sucrose is a

Options:
A Monosaccharide
B Disaccharide
C Trisaccharide
D Polysaccharide

The correct answer is:
Disaccharide

Solution:
Sucrose is a disaccharide. One molecule of sucrose on hydrolysis gives one molecule of glucose and one molecule of fructose.

Topic: Chemistry, Biomolecules, JEE Main
Sucrose on hydrolysis gives

Options:
A Two molecules of glucose
B Two molecules of fructose
C One molecule each of glucose and fructose
D One molecule each of glucose and mannose

The correct answer is:
One molecule each of glucose and fructose

Solution:
Sucrose is a disaccharide. One molecule of sucrose on hydrolysis gives one molecule of glucose and one molecule of fructose

Topic: Chemistry, Biomolecules, JEE Main
Sugar is the main source of energy as it produces energy on metabolic decomposition. But these days low calorie drinks are more popular, why?

Options:
A Low-calorie drinks are more affordable than sugary beverages.
B Low-calorie drinks are easier to find in stores.
C Because in such drinks artificial sweetening agents are present which do not metabolize hence do not produce any energy therefore it does not cause diseases like diabetes.
D Low-calorie drinks provide more sustained and long-lasting energy compared to sugar.

The correct answer is:
Because in such drinks artificial sweetening agents are present which do not metabolize hence do not produce any energy therefore it does not cause diseases like diabetes.

Solution:
Correct Option is : (C)
Because in such drinks artificial sweetening agents are present which do not metabolize hence do not produce any energy therefore it does not cause diseases like diabetes.

Topic: Chemistry, Chemistry in Everyday Life, NEET
Sugar moiety in DNA and RNA molecules respectively are

Options:
A β-D-2-deoxyribose, β-D-deoxyribose
B β-D-ribose, β-D-2deoxyribose
C β-D-2-deoxyribose, β-D-ribose
D β-Ddeoxyribose, β-D-2deoxyribose

The correct answer is:
β-D-2-deoxyribose, β-D-ribose

Solution:
Deoxyribo nucleic acid or DNA contains 2-deoxy-D-ribose sugar unit, whereas ribo nucleic acid or RNA contains D-ribose sugar unit.

β-D-2 deoxyribose in DNA

β-D-ribose in RNA

Topic: Chemistry, Biomolecules, JEE Main
Sugar moiety in DNA and RNA molecules respectively are

Options:
A β-D-2-deoxyribose, β-D-deoxyribose
B β-D-ribose, β-D-2deoxyribose
C β-D-2-deoxyribose, β-D-ribose
D β-Ddeoxyribose, β-D-2deoxyribose

The correct answer is:
β-D-2-deoxyribose, β-D-ribose

Solution:
Deoxyribo nucleic acid or DNA contains 2-deoxy-D-ribose sugar unit, whereas ribo nucleic acid or RNA contains D-ribose sugar unit.

β-D-2 deoxyribose in DNA

β-D-ribose in RNA

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2022 (29 Jun Shift 1)
Sugar which does not give reddish brown precipitate with Fehling's reagent is:

Options:
A Sucrose
B Lactose
C Glucose
D Maltose

The correct answer is:
Sucrose

Solution:
Sucrose does not give Fehling's test positive, and it does not have any aldehyde or ketone group free to undergo oxidation.

It is used to distinguish between aldehyde and ketone. Aldehydes are easily oxidized and give positive Fehling’s test. Ketones apart from alpha-hydroxy-ketones do not give positive Fehling’s test as they are not oxidized easily. Aldose monosaccharides and ketose monosaccharides both give positive Fehling’s test. Presence of glucose in urine can be determined using Fehling’s test.

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2024 (30 Jan Shift 1)
Sugars are

Options:
A Optically active polyhydroxy aldehydes
B Optically active polyhydroxy ketones
C Optically active polyhydroxy aldehydes or ketones
D Polyhydroxy aldehydes or ketones which may or may not be optically active

The correct answer is:
Polyhydroxy aldehydes or ketones which may or may not be optically active

Solution:
sugar may be defined as optically active polyhydroxy aldehydes or ketones as it contains at least one asymmetrical (chiral) carbo and is, therefore, optically active

Topic: Chemistry, Biomolecules, JEE Main
Suitable conditions for melting of ice are:

Options:
A High temperature and high pressure
B High temperature and low pressure
C Low temperature and low pressure
D Low temperature and high pressure

The correct answer is:
High temperature and high pressure

Solution:
Ice⇌water .
According to Le-Chatelier's principle, at equilibrium, if we change temperature or pressure of a reaction it will shift in such a way so as to undo the change.
Melting of ice is endothermic process, so supply of heat shifts equilibrium in forward direction.
Volume of ice is more than volume of water, so increase in pressure shifts equilibrium to right.
High temperature and high pressure are favourable conditions of melting of ice.

Topic: Chemistry, Chemical Equilibrium, JEE Main
Suitable reagents $\mathrm{X}$ and $\mathrm{Y}$, respectively, in the following reactions are

Options:
A aq. $\mathrm{KOH}, \mathrm{CrO}_3$
B alc. $\mathrm{KOH}, \mathrm{Cu} / \Delta$
C aq. $\mathrm{NaHCO}_3, \mathrm{KMnO}_4$
D alc. $\mathrm{KOH}, \mathrm{NaOI}$

The correct answer is:
alc. $\mathrm{KOH}, \mathrm{NaOI}$

Solution:

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Sulphan' is

Options:
A a mixture of $\mathrm{SO}_{3}$ and $\mathrm{H}_{2} \mathrm{SO}_{3}$
B $100 \%$ conc. $\mathrm{H}_{2} \mathrm{SO}_{4}$
C a mixture of gypsum and conc. $\mathrm{H}_{2} \mathrm{SO}_{4}$
D $100 \%$ oleum (a mixture of $100 \% \mathrm{SO}_{3}$ in $100 \%$ $\left.\mathrm{H}_{2} \mathrm{SO}_{4}\right)$

The correct answer is:
$100 \%$ oleum (a mixture of $100 \% \mathrm{SO}_{3}$ in $100 \%$ $\left.\mathrm{H}_{2} \mathrm{SO}_{4}\right)$

Solution:
$\mathrm{H}_{2} \mathrm{SO}_{4}$ saturated with $\mathrm{SO}_{3}$ is called oleum of sulphan.
$$
\mathrm{H}_{2} \mathrm{SO}_{4}+\mathrm{SO}_{3} \longrightarrow \mathrm{H}_{2} \mathrm{S}_{2} \mathrm{O}_{7}
$$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Sulphide ion is soft base and its ores are common for metals

a Pb b Al c Ag d Mg

Choose the correct answer from the options given below

Options:
A a and c only
B a and d only
C a and b only
D c and d only

The correct answer is:
a and c only

Solution:
Pb and Ag commonly exist in the form of sulphide ore like PbS (galena) and Ag2S (Argentite) Al is mainly found in the form of oxide ore whereas Mg is found in the form of halide ore.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2021 (22 Jul Shift 1)
Sulphide ores are common for the metals

Options:
A Ag, Cu and Pb
B Ag, Cu and Sn
C Ag, Mg and Pb
D Al, Cu and Pb

The correct answer is:
Ag, Cu and Pb

Solution:
Silver, copper and lead mainly found in their sulphides ores like Argentite Ag2S, Copperpyrite CuFeS2 and Galena (PbS)

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
Sulphide, which is insoluble in dilute HNO3:-

Options:
A PbS
B CuS
C CdS
D HgS

The correct answer is:
HgS

Solution:
HgS is insoluble in HNO3, it is soluble only in aqua regia.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Sulphonamides act as

Options:
A
Antiseptic

B
Analgesic

C
Antimicrobials

D
Antipyretic

The correct answer is:
Antimicrobials

Solution:
Sulphonamides act as antimicrobials.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main, JEE Main 2012 (26 May Online)
Sulphur dioxide and oxygen were allowedâ to diffuse through a porous partition. 20 dm3 of SO2 diffuses through the porous partition in 60 seconds. The volume of O2 in dm3 which diffuses under the similar condition in 30 seconds will be (atomic mass of sulphur= 32 u);

Options:
A 28.2
B 10.0
C 14.1
D 7.09

The correct answer is:
14.1

Solution:
As per Graham's law,
V SO 2 / t SO 2 V O 2 / t O 2 = p SO 2 p O 2 M O 2 M SO 2 = M O 2 M SO 2
2 0 dm 3 / 6 0 sec V O 2 / 3 0 sec = 3 2 6 4 = 1 1 · 4 1 4
2 0 × 3 0 6 0 × V O 2 = 1 1 · 4 1 4 ⇒ V O 2 = 1 0 × 1 · 4 1 4
= 14.14 dm3

Topic: Chemistry, States of Matter, JEE Main, JEE Main 2014 (19 Apr Online)
Sulphur dioxide is one of the components of polluted air. SO2 is also a major contributor to acid rain. The correct and complete reaction to represent acid rain caused by SO2 is :

Options:
A 2SO2+O2→2SO3

B SO2+O3→SO3+O2

C SO2+H2O2→H2SO4

D 2SO2+O2+2H2O→2H2SO4

The correct answer is:
2SO2+O2+2H2O→2H2SO4

Solution:
Acid rain is a by product of a variety of human activities that emit the oxides of sulphur and nitrogen in the atmosphere. As mentioned earlier, burning of fossil fuels (which contain sulphur and nitrogenous matter) such as coal and oil in power stations and furnaces or petrol and diesel in motor engines produce sulphur dioxide and nitrogen oxides. SO2 and NO2 after oxidation and reaction with water are major contributors to acid rain, because polluted air usually contains particulate matter that catalyse the oxidation.

2SO2g+O2g+2H2Ol→2H2SO4aq

4NO2g+O2g+2H2Ol→4HNO3aq

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2022 (29 Jun Shift 2)
Sulphur (S) containing amino acids from the following are:

(a) isoleucine

(b) cysteine

(c) lysine

(d) methionine

(e) glutamic acid

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

The correct answer is:
b, d

Solution:
Isoleucine -

Isoleucine is a neutral amino acid.

Cysteine -

Cysteine is a sulphur containing amino acid. It contain thiol group.

Lysine -

Lysine is a basic amino acid.

Methionine -

Methionine is a sulphur containing amino acid contain thioether group.

Glutamic acid -

Glutamic acid is an acidic amino acid.

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2023 (08 Apr Shift 1)
Sulphur sol contains

Options:
A Discrete S-atoms
B Discrete S-molecules
C Large aggregates of S-molecules
D Water dispersed in Solid Sulphur

The correct answer is:
Large aggregates of S-molecules

Solution:
Sulphur sol contains large aggregates of \( \mathrm{S} \) molecules.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Sulphuric acid provides a example of

Options:
A Co-ordinate bonds
B Non-covalent compound
C Covalent and co-ordinate bond
D Non-covalent ion

The correct answer is:
Covalent and co-ordinate bond

Solution:
Sulphuric acid contain, covalent and co-ordinate bond.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Sum of bond order of CO and NO+ is

Options:

The correct answer is:
6

Solution:
Bond order is the number of chemical bonds between a pair of atoms and indicates the stability of a bond.

Bond order=Number of bonding electrons-number of antibonding electrons2

The given two molecules have the same number of electrons(14) like nitrogen molecule.

Hence, CO⇒ C¯≡O+ :BO=3

NO+⇒N≡O+ :BO=3

The sum of the bond order of CO and NO+

=3+3=6

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2024 (27 Jan Shift 1)
Sum of eccentric angles $(\alpha, \beta, \gamma, \delta)$ of co-normal points of ellipse equals to an odd multiple of $\pi$.

Options:
A true
B false
C insufficient data
D Contradictory

The correct answer is:
true

Solution:
Let $P, Q, R, s$ be four concyclic points on an ellipse, whose eccentric angles $\alpha, \beta, \gamma, \delta$ respectively.
Then equation of the chords $P Q$ and $R S$ are
$\begin{aligned} & \Rightarrow \frac{x}{a} \cos \left(\frac{\alpha+\beta}{2}\right)+\frac{y}{b} \sin \left(\frac{\alpha+\beta}{2}\right)-\cos \left(\frac{\alpha-\beta}{2}\right)=0 \text { and } \\ & \Rightarrow \frac{x}{a} \cos \left(\frac{\gamma+\delta}{2}\right)+\frac{y}{b} \sin \left(\frac{\gamma+\delta}{2}\right)-\cos \left(\frac{\gamma-\delta}{2}\right)=0\end{aligned}$
Now, the equation of any were passing through $P, Q, R$ and $S$ is given by.
$\begin{aligned} &\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}-1\right)+\lambda\left(\frac{x}{a} \cos \left(\frac{\alpha+\beta}{2}\right)+\frac{y}{b} \sin \left(\frac{\alpha+\beta}{2}\right)-\cos \left(\frac{\alpha-\beta}{2}\right)\right) \\ & \times\left(\frac{x}{a} \cos \left(\frac{\gamma+\delta}{2}\right)+\frac{y}{b} \sin \left(\frac{\gamma+\delta}{2}\right)-\cos \left(\frac{\gamma-\delta}{2}\right)\right)=0\end{aligned}$
But the given points are concyclic. Hence, this equation will represent a circle, if co-efficient of $x^2=$co-efficient $y^2$.
and co-efficient of $x y=0$
Now equation of the coefficient of $x y=0$
$\begin{aligned} & \Rightarrow \cos \left(\frac{\alpha+\beta}{2}\right) \sin \left(\frac{\gamma+\delta}{2}\right)+\sin \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\gamma+\delta}{2}\right)=0 \\ & \Rightarrow \sin \left(\frac{\alpha+\beta+\gamma+\delta}{2}\right)=0=\sin n \pi \\ & \therefore \frac{1}{2}(\alpha+\beta+\gamma+\delta)=n \pi \text { or } \alpha+\beta+\gamma+\delta=2 n \pi\end{aligned}$
where, $n$ is any integer.
Hence. the sum of eccentric angles of four cyclic points on an ellipse is always an even multiple of $\pi$.

Topic: Mathematics, Ellipse, JEE Main
Sum of infinite number of terms of GP is 20 and sum of their square is 100. The common ratio of GP is

Options:
A
5

B
3/5

C
8/5

D
1/5

The correct answer is:
8/5

Solution:

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2002
Sum of magnitude of two forces is $25 \mathrm{~N}$. The resultant of these forces is normal to the smaller force and has a magnitude of $10 \mathrm{~N}$. Then the two forces are

Options:
A $14.5 \mathrm{~N}, 10.5 \mathrm{~N}$
B $16 \mathrm{~N}, 9 \mathrm{~N}$
C $13 \mathrm{~N}, 12 \mathrm{~N}$
D $20 \mathrm{~N}, 5 \mathrm{~N}$

The correct answer is:
$14.5 \mathrm{~N}, 10.5 \mathrm{~N}$

Solution:
Sum of two forces
$F_1+F_2=25$
Resultant of these forces
$(10)^2=F_1^2+F_2^2+2 F_1 F_2 \cos \theta$
Since, resultant is normal to the smaller force
$\therefore \quad F_1+F_2 \cos \theta=0$
From Eqs. (i), (ii) and (iii), we get
$\begin{gathered}
F_1=14.5 \mathrm{~N} \\
F_2=10.5 \mathrm{~N}
\end{gathered}$

Topic: Physics, Mathematics in Physics, JEE Main