datavorous/entrance-exam-dataset huggingface dataset:
Ecello for reaction, 4Als+3O2s+6H2O+4OH-→4AlOH4- is 2.73 V. If Gfo of OH- and H2O are -157 kJ mole-1 and -237.2 kJ mol-1 determine Gfo for AlOH4-

Options:
A -1580 kJ
B -1303 kJ
C -1260 kJ
D -1380 kJ

The correct answer is:
-1303 kJ

Solution:
For given cell reaction,
ÎG o â=ââ nE o F
So, ÎG o â=ââ12âÃâ2.73âÃâ96500âJ
=â3.1613âÃâ 10 3 âkJ
Now for given reactions,
ÎG o â=â4âÃâ G f o Al OH 4 â â6âÃâ G f o H 2 O â4âÃâ G f o OH â
(Also note that Gfo for elements is zero)
-3.1613Ã103=4GfoAlOH4--6Ã-237.2-4Ã-157
So, G f o Al OH 4 −  = −1303 kJ

Topic: Chemistry, Electrochemistry, JEE Main
$\mathrm{E}_{\text {cell }}^0$ of the reaction $\mathrm{Mg}_{(\mathrm{s})}+2 \mathrm{Ag}^{+}(0.0001 \mathrm{M}) \rightleftharpoons \mathrm{Mg}^{2+}(0.01 \mathrm{M})+2 \mathrm{Ag}_{(\mathrm{s})}$ is $3.17 \mathrm{~V}$. The $\mathrm{E}_{\text {cell }}$ of the reaction and its cell notation respectively are :

Options:
A $2.993 \mathrm{~V}, \mathrm{Ag}\left|\mathrm{Ag}^{+}(0.0001 \mathrm{M}) \| \mathrm{Mg}^2+(0.01 \mathrm{M})\right| \mathrm{Mg}$
B $\text { 3.993 V, } \mathrm{Mg}\left|\mathrm{Mg}^{2+}(0.0001 \mathrm{M}) \| \mathrm{Ag}^{+}(0.01 \mathrm{M})\right| \mathrm{Ag}$
C $2.993 \mathrm{~V}, \mathrm{Mg}\left|\mathrm{Mg}^{2+}(0.01 \mathrm{M}) \| \mathrm{Ag}^{+}(0.0001 \mathrm{M})\right| \mathrm{Ag}$
D $\text { 3.993 V, Ag| } \mathrm{Ag}^{+}(0.01 \mathrm{M}) \| \mathrm{Mg}^{2+}(0.0001 \mathrm{M}) \mid \mathrm{Mg}$

The correct answer is:
$2.993 \mathrm{~V}, \mathrm{Mg}\left|\mathrm{Mg}^{2+}(0.01 \mathrm{M}) \| \mathrm{Ag}^{+}(0.0001 \mathrm{M})\right| \mathrm{Ag}$

Solution:
No solution. Refer to answer key.

Topic: Chemistry, Electrochemistry, JEE Main
ECu2+|Cu0=+0.34 V
EZn2+|Zn0=−0.76 V
Identify the incorrect statement from the options below for the above cell:

Options:
A If Eext>1.1 V, e– flow from Cu to Zn
B If Eext>1.1 V, Zn dissolves at Zn electrode and Cu deposits at Cu electrode
C If Eext<1.1 V, Zn dissolves at anode and Cu deposits at cathode
D If Eext=1.1 V, no flow of e– or current occurs

The correct answer is:
If Eext>1.1 V, Zn dissolves at Zn electrode and Cu deposits at Cu electrode

Solution:
Ecell∘=ECu2+Cu∘−EZn2+Zn∘=1.1 V

So If E=1.1 V no electron will flow

At E>1.1 V cell act as electrolytic cell and electron will flow from Cu to Zn.

At E<1.1 V cell act as electrochemical so Zn dissolve and Cu deposit.

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2020 (04 Sep Shift 1)
$\int \mathrm{e}^{\left(\mathrm{e}^{\mathrm{x}}+\mathrm{x}\right)} \mathrm{dx}=$

Options:
A $\mathrm{e}^{\mathrm{x}}+\mathrm{x}+\mathrm{c}$
B $\mathrm{e}^{\left(\mathrm{e}^{\mathrm{x}}\right)} \cdot \mathrm{x}+\mathrm{cd}$
C $\mathrm{e}^{\left(\mathrm{e}^{\mathrm{x}}\right)}+\mathrm{c}$
D $\mathrm{e}^{\left(\mathrm{e}^{\mathrm{x}}\right)}\left(\mathrm{e}^{\mathrm{x}}-1\right)+\mathrm{c}$

The correct answer is:
$\mathrm{e}^{\left(\mathrm{e}^{\mathrm{x}}\right)}+\mathrm{c}$

Solution:
$$
I=\int e^{e^x+x} d x=\int e^{e^x} \cdot e^x d x
$$
Put $\mathrm{e}^{\mathrm{x}}=\mathrm{t} \Rightarrow \mathrm{e}^{\mathrm{x}} \mathrm{dx}=\mathrm{dt}$
$$
\therefore \mathrm{I}=\int \mathrm{e}^{\mathrm{t}} \mathrm{dt}=\mathrm{e}^{\mathrm{t}}+\mathrm{c}=\mathrm{e}^{\mathrm{e}^{\mathrm{x}}}+\mathrm{c}
$$

Topic: Mathematics, Indefinite Integration, JEE Main
$\int e^{\log (\sin x)} d x=$

Options:
A $\sin x+c$
B $-\cos x+c$
C $e^{\log (\cos x)}+c$
D None of these

The correct answer is:
$-\cos x+c$

Solution:
$\int e^{\log (\sin x)} d x=\int \sin x d x=-\cos x+c$

Topic: Mathematics, Indefinite Integration, JEE Main
$E^{\circ}$ of a cell $a A+b B \rightarrow c C+d D$ is

Options:
A $-\frac{R T}{n F} \log \frac{[C]^c[D]^d}{[A]^a[B]^b}$
B $-R T \log \frac{[a]^A[b]^B}{[a]^C[d]^D}$
C $-\frac{R T}{n F} \log \frac{[C]^c[d]^D}{[A]^a[B]^b}$
D $-\frac{R T}{n F} \log \frac{[C]^c[d]^D}{[a]^A[B]^b}$

The correct answer is:
$-\frac{R T}{n F} \log \frac{[C]^c[D]^d}{[A]^a[B]^b}$

Solution:
The equation for the cell potential of a galvanic cell is (1) $E^{\circ} c e l l=E^{\circ}$ cathode - $E^{\circ}$ anode.
This equation reflects the difference between the standard reduction potentials of the cathode and anode in a galvanic cell. Here's a breakdown of the terms:
- Cell potential ($E^{\circ}$ cell): This is the voltage produced by the cell under standard conditions (specific concentration, temperature, and pressure). A positive cell potential indicates a spontaneous reaction, while a negative cell potential indicates a non-spontaneous reaction.
- Standard reduction potential $\left(E^{\circ}\right)$ : This is the potential of a half-cell reaction measured against a standard hydrogen electrode (SHE) under standard conditions. It represents the tendency of a species in a half-reaction to gain electrons and be reduced. More positive reduction potentials indicate a greater tendency to undergo reduction (gain electrons).
In a galvanic cell, the cathode is the electrode where reduction occurs (electrons are gained) and the anode is the electrode where oxidation occurs (electrons are lost). The standard reduction potential of the cathode is always more positive than the standard reduction potential of the anode. Subtracting the two potentials gives the overall cell potential, $E^{\circ}$ cell.
Therefore, option (1) accurately describes the relationship between the standard reduction potentials of the cathode and anode in determining the cell potential of a galvanic cell.

Topic: Chemistry, Electrochemistry, JEE Main
â«ex2+sinâ¡2x1+cosâ¡2x dx=

Options:
A extan⁡x+C
B ex+tan⁡x+C
C 2extan⁡x+C
D extan⁡2x+C

The correct answer is:
extan⁡x+C

Solution:
I= ∫ e x ( 2+2sinxcosx 2 cos 2 x )dx
I= ∫ e x ( tanx+ sec 2 x )dx
I= e x tanx+C
∵ ∫ e x f x + f ′ x dx = e x · f + C

Topic: Mathematics, Indefinite Integration, JEE Main
$\int e^{x} \cdot \sec x(1+\tan x) d x=$

Options:
A $e^{x} \operatorname{cosec} x+c$
B $e^{x} \sec x+c$
C $e^{x} \cot x+c$
D $e^{x} \tan x+c$

The correct answer is:
$e^{x} \sec x+c$

Solution:
$\int e^{x} \sec x(1+\tan x) d x$
$=\int e^{x}(\sec x+\sec x \tan x)=e^{x} \sec x+c$

Topic: Mathematics, Indefinite Integration, JEE Main
$\int e^x \tan ^2\left(e^x\right) d x=$

Options:
A $\tan \left(e^x\right)-x+c$
B $e^x\left(\tan e^x-1\right)+c$
C $\sec \left(e^x\right)+c$
D $\tan \left(e^x\right)-e^x+c$

The correct answer is:
$\tan \left(e^x\right)-e^x+c$

Solution:
Put $e^x=t \Rightarrow e^x d x=d t$, then
$\begin{aligned}
\int e^x \tan ^2\left(e^x\right) d x & =\int \tan ^2 t d t=\int\left(\sec ^2 t-1\right) d t \\
& =\tan t-t+c=\tan \left(e^x\right)-e^x+c .
\end{aligned}$

Topic: Mathematics, Indefinite Integration, JEE Main
→ Δ E1 mechanism (A)
Major product A is

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Each of A and B tosses a fair coin simultaneously 10 times. The probability that both of them will not get tail at the same toss is equal to

Options:
A 3410
B 2710
C 1410
D 1210

The correct answer is:
3410

Solution:
For any toss, total cases: HH,HT,TH,TT ( 4 cases)
∴ both of them will not get tail at same toss, then the favorable cases: HH,HT,TH ( 3 cases)
Hence, the required probability =3410

Topic: Mathematics, Probability, JEE Main
Each of the compounds PtNH36Cl4,CrNH36Cl3,CoNH34Cl3 and K 2 PtCl 6 were dissolved in water to make its 0.01M solution. The correct order of their increasing conductivity in solution is

Options:
A K2PtCl6<CoNH34·Cl3<CrNH36Cl3<PtNH36Cl4
B CrNH36Cl3<CoNH34·Cl3<K2PtCl6<PtNH36Cl4
C CoNH34·Cl3<K2PtCl6<CrNH36Cl3<PtNH36Cl4
D PtNH36Cl4<CoNH36·Cl3<CrNH36Cl3<K2PtCl6

The correct answer is:
CoNH34·Cl3<K2PtCl6<CrNH36Cl3<PtNH36Cl4

Solution:
In aqueous solution 0.001âM , then complexes with dissociate to give the ions :
[Pt ( NH 3 ) 6 Cl 4 ]â [Pt ( NH 3 ) 6 ] 4+ â+â 4Cl â ;â5âions
[Cr(NH 3 ) 6 ]Cl 3 â [Cr ( NH 3 ) 6 ] 3+ +â 3Cl â ;â4âions
[Co(NH 3 ) 4 Cl 2 ]Clâ [Co(NH 3 ) 4 Cl 2 ] + +â Cl - ;â2âions
K 2 [PtCl 6 ]⇌ PtCl 6 2− +2K +  ; 3 ions

Topic: Chemistry, Coordination Compounds, JEE Main
Each of the persons A and B independently tosses three fair coins. The probability that both of them get the same number of heads is:

Options:
A 58
B 18
C 516
D 1

The correct answer is:
516

Solution:
Each of the persons A and B independently tosses three fair coins.

Let X be the number of heads obtained by A and Y be the number of heads obtained by B.

Both X and Y are binomial variate with parameters, n=3 & p=12

Probability that both A and B obtain the same number of heads

=PX=0PY=0+PX=1PY=1+PX=2PY=2+PX=3PY=3

=C03 1232+C13 1232+C23 1232+C33 1232

=1× 1232+3× 1232+3× 1232+1× 1232

=1261+9+9+1

=2064

=516

Topic: Mathematics, Probability, JEE Main, JEE Main 2021 (27 Aug Shift 2)
Each of the properties of sound listed in column A primarily depends on one of the quantities in column B. Choose the
matching pairs from two columns
$\begin{array}{l|l}
\text{Column A} & \text{Column B} \\
\hline \text{Pitch} & \text{Waveform} \\
\text{Quality} & \text{Frequency} \\
\text{Loudness} & \text{Intensity}
\end{array}$

Options:
A Pitch-waveform; Quality-frequency; Loudness-intensity
B Pitch-frequency; Quality-waveform; Loudness-intensity
C Pitch-intensity; Quality-waveform; Loudness- frequency
D Pitch-waveform; Quality-intensity; Loudness-frequency

The correct answer is:
Pitch-frequency; Quality-waveform; Loudness-intensity

Solution:
The correct option is 2 Pitch-frequency; Quality-waveform; Loudness-intensity Pitch depends on frequency, loudness depends on intensity and the quality depends on waveform.

Topic: Physics, Waves and Sound, JEE Main
Each of the roots of the equation $x^3-6 x^2+6 x-5=0$ are increased by $h$. So that the new transformed equation does not contain $x^2$ term, then $h$ is equal to

Options:
A $1$
B $2$
C $\frac{1}{2}$
D $\frac{1}{3}$

The correct answer is:
$2$

Solution:
Given equation is
$x^3-6 x^2+6 x-5=0$
$x$ of this equation replaced by $x+h$.
$\begin{gathered}\therefore(x+h)^3-6(x+h)^2+6(x+h)-5=0 \\ x^3+h^3+3 x^2 h+3 x h^2-6\left(x^2+h^2+2 x h\right) \\ +6 x+6 h-5=0\end{gathered}$
Since, coefficient of $x^2=0$
$\therefore \quad 3 h-6=0 \Rightarrow h=2$

Topic: Mathematics, Quadratic Equation, JEE Main
Each of the six ideal batteries of emf $20 \mathrm{~V}$ is comnected to an external resistance of $4 \Omega$ as shown in the figure. The current through the resistance is

Options:
A $6 \mathrm{~A}$
B $3 \mathrm{~A}$
C $4 \mathrm{~A}$
D $15 \mathrm{~A}$

The correct answer is:
$15 \mathrm{~A}$

Solution:
Resistance $=4 \Omega$
The equivalent emf of the combination
$$
E_{e q}=n E=3 \times 20=60
$$
The cell have no internal resistance.
$\therefore$ Main current flowing through the load
$$
i=\frac{n E}{R}=\frac{60}{4}=15 \mathrm{~A}
$$

Topic: Physics, Current Electricity, TS EAMCET, TS EAMCET 2017
Each of the two strings of length $51.6 \mathrm{~cm}$ and $49.1 \mathrm{~cm}$ are tensioned separately by $20 \mathrm{~N}$ force. Mass per unit length of both the strings is same and equal to $1 \mathrm{~g} / \mathrm{m}$. When both the strings vibrate simultaneously the number of beats is :

Options:
A 3
B 5
C 7
D 8

The correct answer is:
7

Solution:
$\Delta v=\frac{\mathrm{V}}{2 \ell_1}-\frac{\mathrm{V}}{2 \ell_2}=\frac{\mathrm{V}}{2}\left[\frac{1}{\ell_1}-\frac{1}{\ell_2}\right]$
$$
=\frac{1}{2} \sqrt{\frac{\mathrm{T}}{\mu}}\left\{\frac{1}{\ell_1}-\frac{1}{\ell_2}\right\}
$$

Topic: Physics, Waves and Sound, NEET, NEET 2009 (Mains)
Each of these circles represents an animal group. Which one of these options has animal groups which can be depicted by this diagram?

I	II	III	IV

i	Animals with three germ layers	Animals having bilateral symmetry	Animals with no coelom	Animals with a chitinous exoskeleton.

ii	Animals with radial symmetry	Animals with calcareous ossicles	Animals with rasping organs called a radula	Animals with three pairs of legs

iii	Animals with completely extracellular digestion	Animals with a chitinous exoskeleton.	Animals with three pairs of legs and two pairs of wings	Animals with book gills as their respiratory organs

iv	Animals with notochord during embryonic stages	Animals with well- defined jaws	Animals with 12 pairs of cranial nerves	Animals which can maintain their internal temperature constant

Options:
A i
B ii
C iii
D iv

The correct answer is:
iv

Solution:
In this diagram I includes II which includes III which includes IV. Thus group IV should be a sub group of group III which should be a sub group of II which should be a sub group of I.
In option i, Animals with three germ layers (Triploblastic) are present in all phyla from Platyhelminthes to Chordata. Animals having bilateral symmetry are present in all phyla from Platyhelminthes to Chordata except Echinodermata. Animals with no coelom (triploblastic + bilateral symmetry) are present in Phylum Platyheminthes. But members of Platyhelminthes don't have chitinous exoskeleton. This is seen in Arthropoda which is not included under Platyhelminthes.
In option ii, Animals with radial symmetry are present in Phylum Cnidaria, Phylum Ctenophora and Phylum Echinodermata. Animals with calcareous ossicles are present in Phylum Echinodermata. Animals with rasping organs called radula are present in Mollusca which is not included in Echinodermata.
In option iii, Animals with complete extracellular digestion are present in all phyla from Platyhelminthes to Chordata. Animals with chitinous exoskeleton are present in Arthropoda. Animals with three pairs of legs and two pairs of wings are Insects which are included under Arthropoda. Animals with book gills as their respiratory organs are crustacans which are not insects.
In option iv. Animals with notochord during embryonic stages are included in Phylum Chordata. Animals with well- defined jaws are present in Gnathostomata which is division under Phylum Chordata. Animals with 12 pairs of cranial nerves are amniotes includes in Class Reptilia, Aves and Mammalia included under division Gnathostomata. Animals which can maintain their internal temperature constant (homeotherms) are birds and mammals. So this is the correct option.

Topic: Biology, Animal Kingdom, NEET
Each quarter of a vessel of depth $H$ is filled with liquids of the refractive indices $n_1, n_2, n_3$ and $n_4$ from the bottom respectively. The apparent depth of the vessel when looked normally is

Options:
A $\frac{H\left(n_1+n_2+n_3+n_4\right)}{4}$
B $\frac{\left(n_1+n_2+n_3+n_4\right)}{4 H}$
C $\frac{H\left(\frac{1}{n_1}+\frac{1}{n_2}+\frac{1}{n_3}+\frac{1}{n_4}\right)}{4}$
D $\frac{H\left(\frac{1}{n_1}+\frac{1}{n_2}+\frac{1}{n_3}+\frac{1}{n_4}\right)}{2}$

The correct answer is:
$\frac{H\left(\frac{1}{n_1}+\frac{1}{n_2}+\frac{1}{n_3}+\frac{1}{n_4}\right)}{4}$

Solution:
Correct option is 3. $\frac{H\left(\frac{1}{n_1}+\frac{1}{n_2}+\frac{1}{n_3}+\frac{1}{n_4}\right)}{4}$
Apparent depth of bottom
$\begin{aligned} & =\frac{H / 4}{\mu_1}+\frac{H / 4}{\mu_2}+\frac{H / 4}{\mu_3}+\frac{H / 4}{\mu_4} \\ & =\frac{H}{4}\left(\frac{1}{\mu_1}+\frac{1}{\mu_2}+\frac{1}{\mu_3}+\frac{1}{\mu_4}\right)\end{aligned}$

Topic: Physics, Ray Optics, JEE Main
Each side of a box made of metal sheet in cubic shape is a at room temperature T, the coefficient of linear expansion of the metal sheet is α. The metal sheet is heated uniformly, by a small temperature ΔT, so that its new temperature is T+ΔT. Calculate the increase in the volume of the metal box.

Options:
A 4a3αΔT
B 3a3αΔT
C 4πa3αΔT
D 43πa3αΔT

The correct answer is:
3a3αΔT

Solution:
ΔV=VγΔT

ΔV=3a3αΔT

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2021 (24 Feb Shift 1)
Each side of a square subtends an angle of 60Â° at the top of a tower 5 meters high standing at the center of the square. If a meters is the length of each side of the square, then a is equal to ( use 2=1.41)

Options:

The correct answer is:
7.05

Solution:
Let, ABCD be a square with each side of length a.

It is given that â BPC=60Â°
Let, M be the midpoint of BC.

Then, â BPM=â CPM=30Â°

In âBMP, right angled at M,
we have, tanâ¡â BPM=BMPM
In âOPM
3a24 = a24+52
â´a2=50âa=52m

Topic: Mathematics, Heights and Distances, JEE Main
Each side of a square subtends an angle of 60o at the top of a tower 5 meters high standing in the center of the square. If a meters is the length of each side of the square, then a is equal to

Options:
A 32
B 42
C 52
D 62

The correct answer is:
52

Solution:
Let ABCD be a square of each side of length a meters.

It is given that ∠BPC=60o.
Let M be the midpoint of BC.
Then, ∠BPM=∠CPM=30o
In ∆BMP, right angled at M,
We have
tan⁡∠BPM=BMPM⇒tan⁡30°=a/2PM=13

⇒PM=3a2
In ∆OPM,

PM2=OM2+OP2
3a24=a24+52
∴a2=50⇒a=52 m

Topic: Mathematics, Heights and Distances, JEE Main
Earth is moving around the sun in elliptical orbit as shown. The ratio of $O B$ and $O A$ is $R$. Then the ratio of Earth's velocities at $A$ and $B$ is

Options:
A $R^{-1}$
B $\sqrt{R}$
C $R$
D $R^{2 / 3}$

The correct answer is:
$R$

Solution:
In central force motion, the angular momentum of body remains conserved.
i.e., Angular momentum,
$L=r \times p=m v r$
[at minor and major axis $\mathbf{v}$ and $\vec{r}$ are perpendicular]
So, if $v_{1}$ and $v_{2}$ be the speed of earth at $A$ and $B$, respectively
Then, $\quad m v_{1} r_{1}=m v_{2} r_{2}$
$\Rightarrow \quad \frac{v_{1}}{v_{2}}=\frac{r_{2}}{r_{1}}=\frac{O B}{O A} \Rightarrow \frac{v_{1}}{v_{2}}=R$
So, option (c) is correct.

Topic: Physics, Gravitation, KCET, KCET 2013
Earth receives 1400 W m-2 of solar power. If all the solar energy falling on a lens of area 0.2 m-2 is focussed on to a block of ice of mass 280 g, then what is the time (in min) taken by the ice to melt completely? [Lfusion=3.3x105 J kg-1]

Options:

The correct answer is:
5.5

Solution:
Heat required to melt the block
= (0.28kg) (3.3 x 105 J/kg)= 9. 24 x 104J
Heat received per second = (1400J/ m2 ) (0. 2m2 ) = 280 J
Time taken (in minutes) =9.24Ã104280Ã60 = 5. 5

Topic: Physics, Thermodynamics, JEE Main
Earth revolves round the sun in a circular orbit of radius 'R'. The angular
momentum of the revolving earth is directly proportional to

Options:
A $\mathrm{R}^{2}$
B $\mathrm{R}^{3}$
C $\mathrm{R}$
D $\sqrt{\mathrm{R}}$

The correct answer is:
$\sqrt{\mathrm{R}}$

Solution:
Correct option is A)
We know that angular momentum of Earth
$$
\begin{array}{l}
\mathrm{L}=\mathrm{mvR} \\
\Rightarrow \mathrm{L}=\mathrm{m} \sqrt{\frac{\mathrm{GM}}{\mathrm{R}}} \times \mathrm{R} \\
\Rightarrow \mathrm{L} \propto \mathrm{R}^{1 / 2}
\end{array}
$$

Topic: Physics, Motion In Two Dimensions, JEE Main
Earth's magnetism is due to:

Options:
A Dynamo effect
B Doppler effect
C Solar effect
D Magnus effect

The correct answer is:
Dynamo effect

Solution:
The Dynamo effect:
- It is a theory that explains the origin of the Earth's main magnetism in terms of a self-sustaining dynamo.
- In this dynamo mechanism, fluid motion in the Earth's outer core moves conducting material (liquid iron) across an already existing, weak magnetic field and generates an electric current.
- The electric current, in turn, produces a magnetic field that also interacts with the fluid motion to create a secondary magnetic field.
- Together, the two fields are stronger than the original and lie essentially along the axis of the Earth's rotation.

Topic: Physics, Magnetic Properties of Matter, JEE Main
Earthworms have no skeleton but during burrowing, the anterior end becomes turgid and acts as a hydraulic skeleton. It is due to

Options:
A Setae
B Coelomic fluid
C Blood
D Gut peristalsis

The correct answer is:
Coelomic fluid

Solution:
Setae help in attachment, but turgidity and hydraulic skeleton is due to coelomic fluid.

Topic: Biology, Structural Organisation in Animals, NEET, NEET 2008 (Mains)
Eccentricity of the conic $16 x^2+7 y^2=112$ is

Options:
A $3 / \sqrt{7}$
B $7 / 16$
C $3 / 4$
D $4 / 3$

The correct answer is:
$3 / 4$

Solution:
$\frac{x^2}{\frac{112}{16}}+\frac{y^2}{\frac{112}{7}}=1$. Therefore, $e=\sqrt{1-\frac{112}{16} \cdot \frac{7}{112}}=\frac{3}{4}$.

Topic: Mathematics, Ellipse, JEE Main
Eccentricity of the parabola $x^2-4 x-4 y+4=0$ is

Options:
A $e=0$
B $e=1$
C $e\gt4$
D $e=4$

The correct answer is:
$e=1$

Solution:
Always eccentricity of parabola is $\mathrm{e}=1$.

Topic: Mathematics, Parabola, JEE Main
Ecosan is

Options:
A Use of organisms to neutralize, absorb, and assimilate the pollutants present in sewage.
B A sustainable system for handling excreta.
C Recycling mechanism for the removal of waste.
D The strategy to remove heavy metals from sewage.

The correct answer is:
A sustainable system for handling excreta.

Solution:
Ecological sanitation, commonly abbreviated as ecosan (also spelled eco-san), is an approach to sanitation provision which aims to safely reuse excreta in agriculture as manure. It desires to 'close the loop' mainly for the nutrients and organic matter between sanitation and agriculture.

Topic: Biology, Environmental Issues, NEET
Eddy currents are produced when

Options:
A a metal is kept in varying magnetic field
B a metal is kept in steady magnetic field
C a circular coil is placed in a magnetic field
D through a circular coil, current is passed

The correct answer is:
a metal is kept in varying magnetic field

Solution:
Eddy can be created when a conductor is moving through a magnetic field, or when the magnetic field surrounding a stationary conductor is varying. Option A

Topic: Physics, Electromagnetic Induction, JEE Main
Edge length of face centred cubic unit cell of an ionic solid AB is 508 pm. If the radius of A+ ion is 110 pm, the radius of B- ion in pm will be

Options:

The correct answer is:
144

Solution:
Edge length =2rA++2rB-
508=2×110+2rB-
2rB-=508-220=288
rB-=144 pm

Topic: Chemistry, Solid State, JEE Main
Effect of a couple is

Options:
A purely linear motion
B purely rotational motion
C linear and rotational motion
D no motion

The correct answer is:
purely rotational motion

Solution:
A couple mainly comprises of two paralleI forces which are both equal in magnitude but opposite in direction. There is no line of action along which both forces act and hence it produces only rotational motion.

Topic: Physics, Rotational Motion, AP EAMCET, AP EAMCET 2020 (22 Sep Shift 1)
Effective capacitance between $A$ and $B$ in the figure shown is (all capacitance are in $\mu \mathrm{F}$ )

Options:
A $21 \mu \mathrm{F}$
B $23 \mu \mathrm{F}$
C $\frac{3}{14} \mu F$
D $\frac{14}{3} \mu F$

The correct answer is:
$\frac{14}{3} \mu F$

Solution:
Given circuit is balanced Whetstone bridge. So capacitor of $2 \mu F$-can be dropped from the circuit
$\Rightarrow C_{A B}=2+\frac{8}{3}=\frac{14}{3} \mu F$

Topic: Physics, Capacitance, JEE Main
Effective capacitance of parallel combination of two capacitors C1 and C2 is 10μF. When these capacitors are individually connected to a voltage source of 1V, the energy stored in the capacitor C2 is 4 times that of C1. If these capacitors are connected in series, their effective capacitance will be:

Options:
A 4.2μF
B 3.2μF
C 1.6μF
D 8.4μF

The correct answer is:
1.6μF

Solution:
Given C1+C2=10μF    ...(i)
412C1V2=12C2V2
⇒  4C1=C2...(ii)
From equation (i) and (ii)
C1=2μF
C2=8μF
If they are in series
Ceq=C1C2C1+2=1.6μF

Topic: Physics, Capacitance, JEE Main, JEE Main 2020 (08 Jan Shift 1)
Effective filtration pressure in glomerulus is maintained due to

Options:
A powerful pumping action of the heart.
B secretion of adrenaline.
C afferent arteriole being slightly larger than efferent arteriole.
D vacuum develops in collecting duct and sucks the blood.

The correct answer is:
afferent arteriole being slightly larger than efferent arteriole.

Solution:
The glomerulus is a capillary network within the Bowman's capsules. Blood enters glomerular capillaries through afferent arterioles and leaves through efferent arterioles. The diameter of the afferent arteriole is much more than that of efferent arteriole. This increases the volume of blood in the glomerulus and increases the filtration rate.

Topic: Biology, Excretory Products and their Elimination, NEET
Effective resistance between $A$ and $B$ is

Options:
A $15 \Omega$
B $5 \Omega$
C $\frac{5}{2} \Omega$
D $20 \Omega$

The correct answer is:
$5 \Omega$

Solution:
The given figure is balance wheat stone bridge.

Topic: Physics, Current Electricity, JEE Main