datavorous/entrance-exam-dataset huggingface dataset:
The gravitational field, due to the 'left over part' of a uniform sphere (from which a part as shown, has been 'removed out'), at a very far off point, $\mathrm{P}$, located as shown, would be (nearly) :

Options:
A
$\frac{5}{6} \frac{G M}{x^2}$

B
$\frac{8}{9} \frac{G M}{x^2}$

C
$\frac{7}{8} \frac{G M}{x^2}$

D
$\frac{6}{7} \frac{G M}{x^2}$

The correct answer is:
$\frac{7}{8} \frac{G M}{x^2}$

Solution:
Let mass of smaller sphere (which has to be removed) is $\mathrm{m}$
Radius $=\frac{\mathrm{R}}{2}($ from figure $)$
$$
\begin{aligned}
& \frac{\mathrm{M}}{\frac{4}{3} \pi \mathrm{R}^3}=\frac{\mathrm{m}}{\frac{4}{3} \pi\left(\frac{\mathrm{R}}{2}\right)^3} \\
& \Rightarrow \mathrm{m}=\frac{\mathrm{M}}{8}
\end{aligned}
$$
Mass of the left over part of the sphere
$$
\mathrm{M}^{\prime}=\mathrm{M}-\frac{\mathrm{M}}{8}=\frac{7}{8} \mathrm{M}
$$
Therefore gravitational field due to the left over part of the sphere
$$
=\frac{\mathrm{GM}^{\prime}}{\mathrm{x}^2}=\frac{7}{8} \frac{\mathrm{GM}}{\mathrm{x}^2}
$$

Topic: Physics, Gravitation, JEE Main, JEE Main 2013 (09 Apr Online)
The gravitational field in a region is given by: $\vec{E}=(5 N / k g) \hat{i}+(12 N / k g) \hat{j}$
If the potential at the origin is taken to be zero, then the ratio of the potential at the points $(12 \mathrm{~m}, 0)$ and $(0,5 \mathrm{~m})$ is :

Options:
A
Zero

B
1

C
$\frac{144}{25}$

D
$\frac{25}{144}$

The correct answer is:
1

Solution:
From question, $\mathrm{E}_{\mathrm{x}}=5 \mathrm{~N} / \mathrm{kg}$ and $\mathrm{E}_{\mathrm{y}}=12 \mathrm{~N} / \mathrm{kg}$
Gravitational potential
$=$ Gravitational field $\times$ distance
$\therefore \mathrm{V}_{(12 \mathrm{~m}, 0)}=\mathrm{E}_{\mathrm{x}} \times 12 \mathrm{~J} / \mathrm{kg}$
and $\mathrm{V}_{(0,5 \mathrm{~m})}=\mathrm{E}_{\mathrm{y}} \times 5 \mathrm{~J} / \mathrm{kg}$
(Given : potential at the origin is zero)
$$
\therefore \frac{V_{(12 \mathrm{~m}, 0)}}{\mathrm{V}_{(0,5 \mathrm{~m})}}=\frac{\mathrm{E}_{\mathrm{x}} \times 12}{\mathrm{E}_{\mathrm{y}} \times 5}=\frac{5 \times 12}{12 \times 5}=1
$$

Topic: Physics, Electrostatics, JEE Main, JEE Main 2013 (25 Apr Online)
The gravitational field in a region is given by g→=5i^+12j^ N kg-1. The change in the gravitational potential energy of a particle of mass 2 kg when it is taken from the origin to a point 7 m,-3 m is

Options:
A 71ââJ
B 1 3 5 8 J
C 2ââJ
D 1  J

The correct answer is:
2  J

Solution:
dV=-g→.dr→

⇒∆V=-g→.∆r→
=-5i^+12j^·7i^-3j^
= - 3 5 - 3 6 = 1
⇒ΔU=mΔV=2J

Topic: Physics, Gravitation, JEE Main, JEE Main 2014 (19 Apr Online)
The gravitational field in a region is given by gâ=5i^+12j^ N kg-1. The change in the gravitational potential energy of a particle of mass 2 kg when it is taken from the origin to a point 7 m, -3 m is

Options:
A 71ââJ
B 1 3 5 8 J
C 2ââJ
D 1  J

The correct answer is:
2  J

Solution:
dV=-Eâ.drâ

ââV=-Eâ.ârâ
=-5i^+12j^Â·7i^-3j^
= - 3 5 - 3 6 = 1
âÎU=mÎV=2J

Topic: Physics, Gravitation, JEE Main
The gravitational force acting on a particle, due to a solid sphere of uniform density and radius $R$, at a distance of $3 R$ from the centre of the sphere is $F_1$. A spherical hole of radius $(R / 2)$ is now made in the sphere as shown in the figure. The sphere with hole now exerts a force $F_2$ on the same particle. Ratio of $F_1$ and $F_2$ is

Options:
A $\frac{50}{41}$
B $\frac{41}{50}$
C $\frac{41}{42}$
D $\frac{25}{41}$

The correct answer is:
$\frac{50}{41}$

Solution:
Gravitational force due to solid sphere is
$$
F_1=\frac{G M m}{(3 R)^2}=\frac{G M m}{9 R^2}
$$
where, $M$ and $m$ are mass of solid sphere and particle respectively. Gravitational force on particle due to sphere with cavity
$$
\begin{aligned}
F_2 & =\frac{G M m}{9 R^2}-\frac{G\left(\frac{M}{8}\right) m}{(5 R / 2)^2} \\
& =\frac{G M m}{R^2}\left[\frac{1}{9}-\frac{4}{8 \times 25}\right] \\
& =\frac{G M m}{R^2}\left[\frac{41}{50 \times 9}\right] \\
\therefore \quad \frac{F_1}{F_2} & =\frac{50}{41}
\end{aligned}
$$

Topic: Physics, Gravitation, JEE Main
The gravitational force between two point masses $m_1$ and $m_2$ at separation $r$ is given by $F=k \frac{m_1 m_2}{r^2}$. The constant $k$

Options:
A Depends on system of units only
B Depends on medium between masses only
C Depends on both (a) and (b)
D Is independent of both (a) and (b)

The correct answer is:
Depends on system of units only

Solution:
$k$ represents gravitational constant which depends only on the system of units.

Topic: Physics, Gravitation, JEE Main
The gravitational force between two stones of mass $1 \mathrm{~kg}$ each separated by a distance of 1 metre in vacuum is

Options:
A Zero
B $6.675 \times 10^{-5}$ newton
C $6.675 \times 10^{-11}$ newton
D $6.675 \times 10^8$ newton

The correct answer is:
$6.675 \times 10^{-11}$ newton

Solution:
$F=G \frac{m_1 m_2}{r^2}=6.675 \times \frac{1 \times 1}{1^2} \times 10^{-11}=6.675 \times 10^{-11} \mathrm{~N}$

Topic: Physics, Gravitation, JEE Main
The gravitational force $F_g$ between two objects does not depend on

Options:
A Sum of the masses
B Product of the masses
C Gravitational constant
D Distance between the masses

The correct answer is:
Sum of the masses

Solution:
The correct option is option $1$.
Sum of the masses
1. Gravitational force is defined as the force of attraction between two bodies that are separated by a distance from each other.
2. According to Newton's law of gravitation, this force of attraction is proportional to the product of their masses and inversely proportional to the square of the distance between them.
3. Mathematically it is represented as
$F_g \propto \frac{m_1 m_2}{r^2}$
4. $F_g=\frac{G m_1 m_2}{r^2}$ where $G, m_1, m_2, r$ are gravitational constant, masses of bodies 1 and 2, and separation distance between them respectively.
5. From the above formula, it is clear that gravitational force depends on the product of the masses, the Gravitational constant, and the distance between the masses.
6. But it does not depend on the sum of the masses.

Topic: Physics, Gravitation, JEE Main
The gravitational potential at a point above the surface of earth is -5.12×107 J kg-1 and the acceleration due to gravity at that point is 6.4 m s-2. Assume that the mean radius of earth to be 6400 km. The height of this point above the earth's surface is:

Options:
A 1600 km
B 540 km
C 1200 km
D 1000 km

The correct answer is:
1600 km

Solution:
Gravitational potential at a distance r from the centre of the earth is given by, V=-GMr.

Therefore, we can write

-GMERE+h=-5.12×107 …i

also acceleration due to gravity at that point will be

GMERE+h2=6.4 …ii

By (i) and (ii), we get

-GMERE+hGMERE+h2=5.12×1076.4

⇒RE+h=80×105 m=8000 km⇒h=8000 km-6400 km=1600 km

Topic: Physics, Gravitation, JEE Main, JEE Main 2024 (30 Jan Shift 1)
The gravitational potential of two homogeneous spherical shells A and B, having the same surface mass density, at their respective centres are in the ratio 3 : 4 . If the two shells coalesce into single one such that the surface mass density remains the same, then the ratio of potential at an internal point of the new shell and A is equal to

Options:
A 3Â :Â 2
B 4Â :Â 3
C 5 : 3
D 5Â :Â 6

The correct answer is:
5 : 3

Solution:
MA=σ4πRA2, MB=σ4π RB2
where σ is surface density
VA=-GMARA,   VB=-GMBRB
VAVB=MAMBRBRA=σ4πRA2σ4πRB2RBRA=RARB
Given VAVB=RARB=34
then RB=43 RA
For new shell of mass M and Radius R -
VVA=RRA=RA2+RB2RA=53

Topic: Physics, Gravitation, JEE Main
The greatest and least value of $\sin x \cos x$ are

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

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

Solution:
Let $f(x)=\sin x \cos x=\frac{1}{2} \sin 2 x$
We know $-1 \leq \sin 2 x \leq 1 \Rightarrow-\frac{1}{2} \leq \frac{1}{2} \sin 2 x \leq \frac{1}{2}$
Thus the greatest and least value of f(x) are $\frac{1}{2}$ and $-\frac{1}{2}$ respectively.

Topic: Mathematics, Trigonometric Equations, JEE Main
The greatest and the least value of $\left(\sin ^{-1} x\right)^3+\left(\cos ^{-1} x\right)^3$ are

Options:
A $-\frac{\pi}{2}, \frac{\pi}{2}$
B $-\frac{\pi^3}{8}, \frac{\pi^3}{8}$
C $\frac{7 \pi^3}{8}, \frac{\pi^3}{32}$
D None of these

The correct answer is:
$\frac{7 \pi^3}{8}, \frac{\pi^3}{32}$

Solution:
$\begin{aligned} & \text { We have }\left(\sin ^{-1} x\right)^3+\left(\cos ^{-1} x\right)^3 \\ & =\left(\sin ^{-1} x+\cos ^{-1} x\right)^3-3 \sin ^{-1} x \cos ^{-1} x\left(\sin ^{-1} x+\cos ^{-1} x\right) \\ & =\frac{\pi^3}{8}-3\left(\sin ^{-1} x \cos ^{-1} x\right) \frac{\pi}{2} \\ & =\frac{\pi^3}{8}-\frac{3 \pi}{2} \sin ^{-1} x\left(\frac{\pi}{2}-\sin ^{-1} x\right)\end{aligned}$
$\begin{aligned}
& =\frac{\pi^3}{8}-\frac{3 \pi^2}{4} \sin ^{-1} x+\frac{3 \pi}{2}\left(\sin ^{-1} x\right)^2 \\
& =\frac{\pi^3}{8}+\frac{3 \pi}{2}\left[\left(\sin ^{-1} x\right)^2-\frac{\pi}{2} \sin ^{-1} x\right] \\
& =\frac{\pi^3}{8}+\frac{3 \pi}{2}\left[\left(\sin ^{-1} x-\frac{\pi}{4}\right)^2\right]-\frac{3 \pi^3}{32} \\
& =\frac{\pi^3}{32}+\frac{3 \pi}{2}\left(\sin ^{-1} x-\frac{\pi}{4}\right)^2
\end{aligned}$
$\therefore$ The least value is $\frac{\pi^3}{32}$
and since
$\left(\sin ^{-1} x-\frac{\pi}{4}\right)^2 \leq\left(\frac{3 \pi}{4}\right)^2$
$\therefore$ The greatest value is $\frac{\pi^3}{32}+\frac{9 \pi^2}{16} \times \frac{3 \pi}{2}=\frac{7 \pi^3}{8}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The greatest and the least value of $\left(\sin ^{-1} x\right)^3+\left(\cos ^{-1} x\right)^3$ are

Options:
A $-\frac{\pi}{2}, \frac{\pi}{2}$
B $-\frac{\pi^3}{8}, \frac{\pi^3}{8}$
C $\frac{7 \pi^3}{8}, \frac{\pi^3}{32}$
D None of these

The correct answer is:
$\frac{7 \pi^3}{8}, \frac{\pi^3}{32}$

Solution:
$\begin{aligned} & \text { We have }\left(\sin ^{-1} x\right)^3+\left(\cos ^{-1} x\right)^3 \\ & =\left(\sin ^{-1} x+\cos ^{-1} x\right)^3-3 \sin ^{-1} x \cos ^{-1} x\left(\sin ^{-1} x+\cos ^{-1} x\right) \\ & =\frac{\pi^3}{8}-3\left(\sin ^{-1} x \cos ^{-1} x\right) \frac{\pi}{2} \\ & =\frac{\pi^3}{8}-\frac{3 \pi}{2} \sin ^{-1} x\left(\frac{\pi}{2}-\sin ^{-1} x\right) \\ & =\frac{\pi^3}{8}-\frac{3 \pi^2}{4} \sin ^{-1} x+\frac{3 \pi}{2}\left(\sin ^{-1} x\right)^2 \\ & =\frac{\pi^3}{8}+\frac{3 \pi}{2}\left[\left(\sin ^{-1} x\right)^2-\frac{\pi}{2} \sin ^{-1} x\right] \\ & =\frac{\pi^3}{8}+\frac{3 \pi}{2}\left[\left(\sin ^{-1} x-\frac{\pi}{4}\right)^2\right]-\frac{3 \pi^3}{32} \\ & =\frac{\pi^3}{32}+\frac{3 \pi}{2}\left(\sin ^{-1} x-\frac{\pi}{4}\right)^2 \\ & \therefore \text { The least value is } \frac{\pi^3}{32} \\ & \text { and since }\left(\sin ^{-1} x-\frac{\pi}{4}\right)^2 \leq\left(\frac{3 \pi}{4}\right)^2 \\ & \therefore \text { The greatest value is } \frac{\pi^3}{32}+\frac{9 \pi^2}{16} \times \frac{3 \pi}{2}=\frac{7 \pi^3}{8} \\ & \end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The greatest biodiversity on Earth is found in:

Options:
A Tropical Amazonian rain forest in South Africa
B Temperate Amazonian rain forest in North America
C Tropical Amazonian rain forest in South America
D Temperate Amazonian rain forest in South America

The correct answer is:
Tropical Amazonian rain forest in South America

Solution:
Tropical Amazonian rain forest in South America has the greatest diversity.
Tropical habitats are relatively more constant and less seasonal. Such environments tend to promote niche specialization that gives more diversity. Also, such environments remain undisturbed for millions of years leading to evolution and species diversification.

Topic: Biology, Biodiversity and its Conservation, NEET
The greatest coefficient in the expansion of $(1+x)^{2 n+2}$ is

Options:
A $\frac{(2 n)!}{(n!)^2}$
B $\frac{(2 n+2)!}{\{(n+1)!\}^2}$
C $\frac{(2 n+2)!}{n!(n+1)!}$
D $\frac{(2 n)!}{n!(n+1)!}$

The correct answer is:
$\frac{(2 n+2)!}{\{(n+1)!\}^2}$

Solution:
Greatest coefficient of $(1+x)^{2 n+2}$ is $={ }^{(2 n+2)} C_{n+1}=\frac{(2 n+2)!}{\{(n+1)!\}^2}$

Topic: Mathematics, Binomial Theorem, JEE Main
The greatest contribution of root pressure is

Options:
A to re-establish the continuous chains of water molecules in the xylem which often break under the enormous tensions created by transpiration.
B to play a major role in water movement up tall trees over 50 feet in height.
C to act as a main driving force to push water from the roots in the xylem towards the leaves.
D to give water high tensile strength and high capillarity action.

The correct answer is:
to re-establish the continuous chains of water molecules in the xylem which often break under the enormous tensions created by transpiration.

Solution:
A positive pressure called root pressure is developed as various ions from the soil are actively transported into the vascular tissues of the roots, water follows its potential gradient and increases the pressure inside the xylem. It is responsible for pushing up water to small heights in the stem. It can't push water over long distances. The greatest contribution of root pressure may be to re-establish the continuous chains of water molecules in the xylem which often break under the enormous tensions created by transpiration. Root pressure does not account for the majority of water transport; most plants meet their need by transpiratory pull. Cohesion, adhesion and surface tension give water high tensile strength, i.e., an ability to resist a pulling force, and high capillarity, i.e., the ability to rise in thin tubes.

Topic: Biology, Transport in Plants, NEET
The greatest height to which a man can throw a stone is $h$. The greatest distance to which he can throw it, will be

Options:
A $\frac{h}{2}$
B h
C 2h
D 3h

The correct answer is:
2h

Solution:
For greatest height $\theta=90^{\circ}$
$\begin{aligned} & H_{\max }=\frac{u^2 \sin ^2\left(90^{\circ}\right)}{2 g}=\frac{u^2}{2 g}=h \text { (given) } \\ & R_{\max }=\frac{u^2 \sin ^2 2\left(45^{\circ}\right)}{g}=\frac{u^2}{g}=2 h\end{aligned}$

Topic: Physics, Motion In Two Dimensions, JEE Main
The greatest integer less than 2+16 is equal to

Options:
A 196
B 197
C 198
D 199

The correct answer is:
197

Solution:
2+16+2-16=26C023+6C222+6C421+6C620
2+16+2-16=28+60+30+1=198
2+16=198-2-16
as 2-1â0.414⇒2-16â0,1
⇒2+16=198-f,fâ0,1
⇒ integral part of 2+16=197

Topic: Mathematics, Binomial Theorem, JEE Main
The greatest integer less than or equal to $(\sqrt{3}+2)^5$ is

Options:
A 721
B 722
C 723
D 724

The correct answer is:
723

Solution:
Let $(2+\sqrt{3})^5=Y+F$
$[\because Y$ is integer, $F$ is fraction]
$$
\because \quad 2-\sqrt{3} < 1 \Rightarrow(2-\sqrt{3})^5 < 1
$$
Let $\quad(2-\sqrt{3})^5=F^{\prime}$
$$
\begin{aligned}
\therefore \quad & Y+F+F^{\prime}=(2+\sqrt{3})^5+(2-\sqrt{3})^5 \\
& =2\left(2^5+{ }^5 C_2 2^3(3)+{ }^5 C_4 2 \cdot 3^2\right) \\
& Y+F+F^{\prime}=2(32+240+90) \\
& Y+F+F^{\prime}=724 \\
& Y=724-\left(F+F^{\prime}\right) \\
& Y=724-1=723 \quad\left[\because F+F^{\prime}=1\right]
\end{aligned}
$$

Topic: Mathematics, Binomial Theorem, JEE Main
The greatest integer less than or equal to the sum of first 100 terms of the sequence 13,59,1927,6581,… is equal to ______

Options:

The correct answer is:
98

Solution:
Let S=13+59+1927+⋯ upto 100 terms

S=3-231+32-2232+33-2333+.... upto 100 terms

So, Tn=1−23n⇒S100=100−231−231001−23

=98+223100 ⇒ S100=98

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2022 (25 Jun Shift 1)
The greatest positive integer k, for which 49k+1 is a factor of the sum 49125+49124+…+492+49+1, is

Options:
A 32
B 63
C 60
D 35

The correct answer is:
63

Solution:
This series is a G.P. with first term 49125 and common ratio 149.

So, its sum =49126-148=4963+14963-148

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2020 (07 Jan Shift 1)
The greatest term in the expansion of $\sqrt{3}\left(1+\frac{1}{\sqrt{3}}\right)^{20}$ is

Options:
A $\frac{25840}{9}$
B $\frac{24840}{9}$
C $\frac{26840}{9}$
D None of these

The correct answer is:
$\frac{25840}{9}$

Solution:
Let $(r+1)^{\text {th }}$ term be the greatest term. Then
$\begin{aligned} & T_{r+1}=\sqrt{3} \cdot{ }^{20} C_r\left(\frac{1}{\sqrt{3}}\right)^r \text { and } T_r=\sqrt{3} \cdot{ }^{20} C_{r-1}\left(\frac{1}{\sqrt{3}}\right)^{r-1} \\ & \text { Now } \frac{T_{r+1}}{T_r}=\frac{20-r+1}{r}\left(\frac{1}{\sqrt{3}}\right)\end{aligned}$
$\begin{aligned} & \therefore T_{r+1} \geq T_r \Rightarrow 20-r+1 \geq \sqrt{3} r \\ & \Rightarrow 21 \geq r(\sqrt{3}+1) \Rightarrow r \leq \frac{21}{\sqrt{3}+1} \Rightarrow r \leq 7.686 \Rightarrow r=7\end{aligned}$
Hence the greatest term is $T_8=\sqrt{3}{ }^{20} C_7\left(\frac{1}{\sqrt{3}}\right)^7=\frac{25840}{9}$

Topic: Mathematics, Binomial Theorem, JEE Main
The greatest term in the expansion of 3+2x51, where x=15, is

Options:
A 5th term
B 6th term
C 7th term
D 9th term

The correct answer is:
7th term

Solution:
The greatest term in the expansion of x+yn is the [k]+1th term, if kâZ, where k=n+1yx+y, â is GIF
Here,
k=51+12x3+2x
=52253+25
=10415+2=6
Thus, 7th term is the greatest term.

Topic: Mathematics, Binomial Theorem, JEE Main
The greatest value of \( c \in R \) for which the system of linear equations \( x-c y-c z=0, c x-y+c z=0, c x+c y-z=0 \) has a non-trivial solution, is

Options:
A \( -1 \)
B \( 2 \)
C \( \frac{1}{2} \)
D \( 0 \)

The correct answer is:
\( \frac{1}{2} \)

Solution:
∵ System of equation has non trivial solution. It is possible only when ∆=0.
∴∆ =1-c-cc-1ccc-1=0
⇒11-c2+c-c-c2-cc2+c=0
⇒2c3+3c2-1=0
⇒c=-1, 12

Hence, the greatest value of c=12.

Topic: Mathematics, Determinants, JEE Main
The greatest value of c∈R for which the system of linear equations x-cy-cz=0, cx-y+cz=0, cx+cy-z=0 has a non-trivial solution, is

Options:
A -1
B 2
C 12
D 0

The correct answer is:
12

Solution:
∵ System of equation has non trivial solution. It is possible only when ∆=0.
∴∆ =1-c-cc-1ccc-1=0
⇒11-c2+c-c-c2-cc2+c=0
⇒2c3+3c2-1=0
⇒c=-1, 12

Hence, the greatest value of c=12.

Topic: Mathematics, Determinants, JEE Main, JEE Main 2019 (08 Apr Shift 1)
The green alga used by T.W Engelmann in his prism experiment is

Options:
A Chlorella
B Spirogyra
C Cladophora
D Ulothrix

The correct answer is:
Cladophora

Solution:
T.W Engelmann by using a prism split light into its spectral components and then illuminated a green alga, Cladophora, placed in a suspension of aerobic bacteria. The bacteria were used to detect the sites of O2 evolution. He observed that the bacteria accumulated mainly in the region of blue and red light of the split spectrum.

Topic: Biology, Photosynthesis in Higher Plants, NEET
The green house gas/es is (are):

(A) Carbon dioxide

(B) Oxygen

(C) Water vapour

(D) Methane

Choose the most appropriate answer from the options given below:

Options:
A (A) and (C) only
B (A) only
C (A), (C) and (D) only
D (A) and (B) only

The correct answer is:
(A), (C) and (D) only

Solution:
Greenhouse effect, a warming of Earth's surface and troposphere (the lowest layer of the atmosphere) caused by the presence of water vapour, carbon dioxide, methane, and certain other gases in the air. Of those gases, known as greenhouse gases, water vapour has the largest effect.

The green house gases are CO2, H2Ovapour and CH4.

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2021 (16 Mar Shift 2)
The green scum seen in the freshwater bodies is

Options:
A
Blue green algae

B
Red algae

C
Green algae

D
Both (a) and (b)

The correct answer is:
Both (a) and (b)

Solution:
The green scum seen in the fresh water bodies mainly consists of green algae and blue-green algae. While the red algae are mostly marine.

Topic: Biology, Environmental Issues
The grid represents a region in space containing a uniform electric field ( each square of size 1mÃ1m ) . If potentials at point O , A , B , C , D , E , F , G , H are respectively 0 , â1 , â2 , 1 , 2 , 0 , â1 , 1 and 0 volts. find the electric field intensity.

Options:
A i^+j^ V/m
B i^âj^ V/m
C âi^+j^ V/m
D −i^+j^ V/m

The correct answer is:
i^âj^ V/m

Solution:
OEH is an equipotential surface, the uniform electric field must be perpendicular to it pointing from higher to lower potential as shown.
Hence E=(i^–j^2)
E=(vE–vB)EB=0–(–2)2=2
∴E→=2 (i^–j^)2=i^–j^

Topic: Physics, Electrostatics, JEE Main
The ground state energy of a hydrogen atom is -13.6 eV. The energy of second excited state of He+ ion in eV is:

Options:
A -27.2
B -6.04
C -3.4
D -54.4

The correct answer is:
-6.04

Solution:
E=-13.6z2n2=-13.62232=-6.04 eV

(Second exited state=Third shell, n=3)

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2019 (10 Jan Shift 2)
The ground state energy of hydrogen atom is â13.6 eV. If the electron jumps to the ground state from the 3rd excited state, the wavelength of the emitted photon is

Options:
A

875 Aâ

B

1052 Aâ

C

752 Aâ

D

974 Aâ

The correct answer is:

974 Aâ

Solution:
The energy of an electron in nth orbit is given by E n = −13.6 n 2   eV

The required energy to jump electron to the ground state from the 3rd excited state

E= E 4 − E 1

= −13.6 ( 4 ) 2 −[ −13.6 ( 1 ) 2 ] =−0.85+13.6=12.75 eV

∴ The wavelength of the photon emitted is λ= hc E [ ∵E= hc λ ]

=6.626×10−34×3×10812.75×1.6×10−19=19.878×10−720.4=0.974×10−7=974A∘

Topic: Physics, Atomic Physics, JEE Main
The ground state energy of hydrogen atom is $-13.6 \mathrm{eV}$. When its electron is in the first excited state, its excitation energy is

Options:
A 0
B $3.4 \mathrm{eV}$
C $6.8 \mathrm{eV}$
D $10.2 \mathrm{eV}$

The correct answer is:
$10.2 \mathrm{eV}$

Solution:
Excitation energy $=\mathrm{E}_{\mathrm{f}}-\mathrm{E}_{\mathrm{i}}$
$=-3.4-(-13.6)=10.2 \mathrm{eV}$

Topic: Physics, Atomic Physics, NEET, NEET 2008 (Mains)
The group having triangular planar structure is:

Options:
A BF3, NF3, CO32-
B CO32-, NO3-, SO3
C NH3, SO3, CO32-
D NCl3, BCl3, SO3

The correct answer is:
CO32-, NO3-, SO3

Solution:
CO3-2, NO3-, SO3:plane triangular shape. Central atom in all these three species is sp2 hybridized, without any lone pairs on the central atom.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2017 (09 Apr Online)
The group number, number of valence electrons, and valency of an element with atomic number 15, respectively, are:

Options:
A 15,6 and 2
B 16,6 and 3
C 16,5 and 2
D 15,5 and 3

The correct answer is:
15,5 and 3

Solution:
Atomic number 15 electronic configuration is
15→1s2,2s2,2p6,3s2,3p3
and this element is Phosphorous.
Group no =15
Valence e-⇒e- present in valence shell
=2+3=5
Valency =3 , it needs 3 electron to complete its octet.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2019 (12 Apr Shift 1)
The group of chemicals used as pesticide is

Options:
A Aldrin, Sodium Chlorate, Sodium arsinite
B DDT, Aldrin
C Sodium chlorate, DDT, PAN
D Dieldrin, Sodium arsinite, Tetrachloroethene

The correct answer is:
DDT, Aldrin

Solution:
DDT(dichlorodiphenyltrichloroethane) and Aldrin are two synthetic pesticides that were widely used for insect control. Pesticides are groups of chemicals used for destruction of insects, weeds, fungi, bacteria etc. They are generally called insecticides, fungicides, bactericides, herbicides and rodenticides.

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2023 (06 Apr Shift 2)
The group of molecules having identical shape is:

Options:
A PCl5,IF5,Â XeO2F2
B BF3,Â PCl3,Â XeO3
C SF4,Â XeF4,Â CCl4
D ClF3,XeOF2, XeF3+

The correct answer is:
ClF3,XeOF2, XeF3+

Solution:
All molecules have same hybridisation i.e sp3d and have similar structure, the T-shaped. So,they are isostructural

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2016 (09 Apr Online)
The H amount of thermal energy is developed by a resistor in 10 s when a current of 4 A is passed through it. If the current is increased to 16 A, the thermal energy developed by the resistor in 10 s will be

Options:
A H
B 16H
C H4
D 4H

The correct answer is:
16H

Solution:
Heat energy developed in the resistor due to current is given by, H=i2Rt. Therefore

⇒H∝i2 (for t= constant)

Now required ratio, HH'=4162

⇒H'=16H

Topic: Physics, Current Electricity, JEE Main, JEE Main 2023 (31 Jan Shift 2)