datavorous/entrance-exam-dataset huggingface dataset:
The general solution of the differential equation
$x \cos \frac{y}{x}(y d x+x d y)=y \sin \frac{y}{x}(x d y-y d x)$ is

Options:
A $\log (x y)=\log \cos \frac{x}{y}+C$
B $\cos \left(\frac{y}{x}\right)=\frac{C}{x y}$
C $\log (x y)=\log \sec \frac{x}{y}+C$
D $x+y+C=0$

The correct answer is:
$\cos \left(\frac{y}{x}\right)=\frac{C}{x y}$

Solution:
We have, $x \cos \left(\frac{y}{x}\right)(y d x+x d y)$ $=y \sin \frac{y}{x}(x d y-y d x)$
$\Rightarrow \quad x y \sin \left(\frac{y}{x}\right) d y-y^2 \sin \left(\frac{y}{x}\right) d x$ $=x y \cos \left(\frac{y}{x}\right) d x+x^2 \cos \left(\frac{y}{x}\right) d y$
$\Rightarrow\left[x y \sin \left(\frac{y}{x}\right)-x^2 \cos \left(\frac{y}{x}\right)\right] d y$ $=\left[x y \cos \left(\frac{y}{x}\right)+y^2 \sin \left(\frac{y}{x}\right)\right] d x$
$\Rightarrow \frac{d y}{d x}=\frac{x y \cos \left(\frac{y}{x}\right)+y^2 \sin \left(\frac{y}{x}\right)}{x y \sin \left(\frac{y}{x}\right)-x^2 \cos \left(\frac{y}{x}\right)}$
$\Rightarrow \frac{d y}{d x}=\frac{\frac{y}{x} \cos \left(\frac{y}{x}\right)+\left(\frac{y}{x}\right)^2 \sin \left(\frac{y}{x}\right)}{\frac{y}{x} \sin \left(\frac{y}{x}\right)-\cos \left(\frac{y}{x}\right)}$
Put, $y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$
$\therefore \quad v+x \frac{d v}{d x}=\frac{v \cos v+v^2 \sin v}{v \sin v-\cos v}$
$\Rightarrow \quad x \frac{d v}{d x}=\frac{2 v \cos v}{v \sin v-\cos v}$
$\Rightarrow \quad \frac{v \sin v-\cos v}{v \cos v} d v=2 \frac{d x}{x}$
$\Rightarrow \quad \int\left(\tan v-\frac{1}{v}\right) d v=2 \int \frac{d x}{x}$
$\Rightarrow \quad \log |\sec v|-\log |v|=2 \log |x|+\log \left|C_1\right|$
$\Rightarrow \quad \log \left|\frac{\sec v}{v}\right|-\log |x|^2=\log \left|C_1\right|$
$\Rightarrow \quad \log \left|\frac{\sec v}{v x^2}\right|=\log \left|C_1\right|$
$\Rightarrow \quad \frac{\sec v}{v x^2}=C_1 \Rightarrow \frac{\sec \left(\frac{y}{x}\right)}{\frac{y}{x}\left(x^2\right)}=C_1$
$\Rightarrow \quad \frac{\sec \left(\frac{y}{x}\right)}{x y}=C_1 \Rightarrow \sec \left(\frac{y}{x}\right)=C_1 x y$
$\Rightarrow \quad \frac{1}{\cos \left(\frac{y}{x}\right)}=\frac{C}{x y} \quad\left(\because\right.$ where $\left.\frac{1}{C_1}=C\right)$

Topic: Mathematics, Differential Equations, JEE Main
The general solution of the differential equation y2-x3dx-xydy=0, x≠0 is (where c is a constant of integration)

Options:
A y2+2x2+cx3=0
B y2-2x2+cx3=0
C y2-2x3+cx2=0
D y2+2x3+cx2=0

The correct answer is:
y2+2x3+cx2=0

Solution:
Given differential equation can be written as:
y2 dx-xy dy=x3 dx
⇒y dx-x dy=x3y dx
⇒ y dx-x dyy2=xy3 dx
⇒ ∫dxyxy3= ∫dx
⇒ xy-3+1 -3+1=x+c

⇒ y2=-2x3-cx2

⇒y2+2x3+cx2=0

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2019 (12 Apr Shift 2)
The general solution of the equation \( \tan 2 \theta \cdot \tan \theta=1 \) for \( n \in Z \) is

Options:
A \( (2 n+1) \frac{\pi}{4} \)
B \( (2 n+1) \frac{\pi}{6} \)
C \( (2 n+1) \frac{\pi}{2} \)
D \( \frac{1}{1}(2 n+1) \frac{\pi}{3} \)

The correct answer is:
\( (2 n+1) \frac{\pi}{6} \)

Solution:
Given, tan⁡2θ=1tan⁡θ
⇒ tan⁡2θ=tan⁡π2-θ
⇒ 2θ=nπ+π2-θ
⇒ θ=π6(2n+1)

Topic: Mathematics, Trigonometric Equations, JEE Main
The general solution of the system of equations sin 3 x + sin 3 2 Ï 3 + x + sin 3 4 Ï 3 + x + 3 4 cos 2 x = 0 and cosâ¡xâ 0 is

Options:
A x=2k+1π10, k∈Z
B x=2k+1π5, k∈Z
C x=4k+1Ï10, kâZ
D x=4k+15π, k∈Z

The correct answer is:
x=4k+1Ï10, kâZ

Solution:
∵sin3x=3sinx-4 sin3x ⇒sin3x=143sinx-sin3x
The given equation reduces to
143sinx-sin3x+143sin2π3+x- sin2π+3x+14(3sin4π3+x-sin4π+3x+34cos2x=0
3sin⁡x+sin2π3+x+sin⁡4π3+x-sin⁡3x+sin⁡2π+3x+sin⁡4π+3x+3cos⁡2x=0
3 sinx+2sinx cos⁡ 2π3-3sin3x+3cos2x=0
⇒ sin 3x=cos 2x ⇒ cosπ2-3x=cos 2x⇒2x=2kπ±π2-3x
⇒ 5x=2kπ+π2=4k+1π2 ∀k∈Z or x=-2kπ+π2, ∀k∈Z
⇒ x=4k+1π10, ∀k∈Z cos⁡x≠0

Topic: Mathematics, Trigonometric Equations, JEE Main
The general solution of $x^2 \frac{d y}{d x}=2$ is

Options:
A $y=c+\frac{2}{x}$
B $y=c-\frac{2}{x}$
C $y=2 c x$
D $y=c-\frac{3}{x^2}$

The correct answer is:
$y=c-\frac{2}{x}$

Solution:
$\frac{d y}{d x}=\frac{2}{x^2} \Rightarrow d y=\frac{2}{x^2} d x$, Now integrate it.

Topic: Mathematics, Differential Equations, JEE Main
The general solution of $x d y-y d x=y d y$ is

Options:
A $y=A e^{-x / y}$
B $y=A e^x$
C $\frac{y}{x}=A e^x$
D $\frac{x}{y}+\frac{y}{x}=c$

The correct answer is:
$y=A e^{-x / y}$

Solution:
$x d y-y d x=y d y$
$\begin{aligned} & \Rightarrow \quad-(y d x-x d y)=y d y \\ & \Rightarrow \quad-\left(\frac{y d x-x d y}{y^2}\right)=\frac{y d y}{y^2}\end{aligned}$
$\Rightarrow \quad-\int d\left(\frac{x}{y}\right)=\int \frac{d y}{y}$
$\Rightarrow \quad-\frac{x}{y}=\ln |y|+C$
$\Rightarrow \quad y=A e^{-x / y}$

Topic: Mathematics, Differential Equations, JEE Main
The general solution of $y^2 d x+\left(x^2-x y+y^2\right) d y=0$ is :

Options:
A $\tan ^{-1}\left(\frac{y}{x}\right)=\log y+C$
B $2 \tan ^{-1}\left(\frac{x}{y}\right)+\log x+C=0$
C $\log \left(y+\sqrt{x^2+y^2}\right)+\log y+C=0$
D $\sinh ^{-1}\left(\frac{x}{y}\right)+\log y+C=0$

The correct answer is:
$\tan ^{-1}\left(\frac{y}{x}\right)=\log y+C$

Solution:
We have,
$y^2 d x+\left(x^2-x y+y^2\right) d y=0$
$\Rightarrow \quad \frac{d y}{d x}=\frac{-y^2}{x^2-x y+y^2}$
It is a homogeneous linear differential equation
Put $y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$
$\therefore v+x \frac{d v}{d x}=\frac{-v^2 x^2}{x^2-v x^2+x^2 v^2}=\frac{-v^2}{v^2-v+1}$
$x \frac{d v}{d x}=\frac{-v^2-v^3+v^2-v}{v^2-v+1}=\frac{-v^3-v}{v^2-v+1}$
$\Rightarrow \quad \frac{\left(v^2-v+1\right)}{-v^3-v} d v=\frac{1}{x} d x$
$\Rightarrow \quad \frac{-\left(v^2+1\right)+v}{v\left(v^2+1\right)} d v=\frac{1}{x} d x$
$\therefore \quad-\frac{1}{v} d v+\frac{1}{v^2+1} d v=\frac{1}{x} d x$
On integrating both sides, we get
$-\log v+\tan ^{-1} v=\log x+C$
$\Rightarrow \quad \tan ^{-1} v=\log x v+C$
$\therefore \quad \tan ^{-1}\left(\frac{y}{x}\right)=\log y+C$

Topic: Mathematics, Differential Equations, JEE Main
The general trend in the properties of elements of carbon family shows that with increase in atomic number

Options:
A the tendency towards catenation increases
B the tendency to show +2 oxidation state increases
C metallic character decreases
D the tendency to form complexes with covalency higher than four decreases

The correct answer is:
the tendency to show +2 oxidation state increases

Solution:
Here statement B is correct as with increase in atomic number down the group the tendency to show +2 oxidation state increases due to inert pair effect.
In carbon family, the tendency towards catenation decreases on moving down the group, because the strength of M-M bond decreases.
Metallic character always increases on moving down the group. Carbon is strictly non-metallic; silicon is essentially a non-metal in its chemical property; germanium is a semimetal with pronounced metallic properties; tin and lead are truly metallic.
Carbon atom does not form complexes, because it does not have d-orbitals in its valence shell. The other elements of group have d-orbitals in their valence shell, therefore, the tendency to form complexes with covalency higher than four increases on moving down the group.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
The genes, which remain confined to non homologous region of Y chromosome, are called

Options:
A autosomal genes.
B holandric genes.
C completely sex-linked genes.
D mutant genes.

The correct answer is:
holandric genes.

Solution:
The genes located on the non-homologous region of the Y chromosome are called as Y linked or holandric gene and their inheritance as Y linked inheritance. Since the Y chromosome is present only in males these genes are inherited to all male progeny thus called a holandric gene.

Topic: Biology, Principles of Inheritance and Variation, NEET
The genetic defect-adenosine deaminase (ADA) deficiency may be cured permanently by

Options:
A Enzyme replacement therapy
B Periodic infusion of genetically engineered lymphocytes having functional $\mathrm{ADA} \mathrm{cDNA}$
C Administering adenosine deaminase activators
D Introducing bone marrow cells producing ADA into cells at early embryonic stages

The correct answer is:
Introducing bone marrow cells producing ADA into cells at early embryonic stages

Solution:
No Solution Available

Topic: Biology, Biotechnology and Its Applications, NEET, NEET 2009 (Mains)
The geometric mean of 6 observations was calculated as 13. It was later observed that one of the observations was recorded as 28 instead of 36. The correct geometric mean is

Options:
A 9716
B 3Â 9716
C 13 9716
D 13 7 9 1 6

The correct answer is:
13 9716

Solution:
Let x1, x2â¦, x6 are the observations and x1=28
â 28. x2â¦x6=136
â x2â¦x6=13628
Now correct observation is 36
â 36.x2â¦x6=13628Ã36
So, now correct geometric mean =13 9716

Topic: Mathematics, Sequences and Series, JEE Main
The geometry around boron in the product 'B' formed from the following reaction is

BF3+NaH⟶450 KA+NaF

A+NMe3→B

Options:
A trigonal planar
B tetrahedral
C pyramidal
D square planar

The correct answer is:
tetrahedral

Solution:
2BF3+6NaH→B2H6A+6NaF

In this reaction hydride ions replace the fluoride ions. and diborane is formed, which is a Lewis acid due to incomplete octet. It breaks symmetrically with bases like tertiary amines.

B2H6+2Me3N→2Me3N+BBH3-

In this product geometry around Boron will be tetrahedral. And it will have sp3 hybridisation.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main, JEE Main 2022 (25 Jul Shift 1)
The geometry of ClO 3 − ion according to valence shell electron pair repulsion (VSEPR) theory will be:

Options:
A Planar triangular
B Trigonal Pyramidal
C Tetrahedral
D Square planar

The correct answer is:
Trigonal Pyramidal

Solution:
Hybridization is sp 3 and shape is trigonal pyramidal.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The geometry of sulphur trioxide molecule is

Options:
A Tetrahedral
B Trigonal planar
C Pyramidal
D Square planar

The correct answer is:
Trigonal planar

Solution:
In $SO_3$ sulphur is $s p^2$ hybridized so its shape will be trigonal planar.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The geometry of XeOF4 by VSEPR theory is:

Options:
A Trigonal bipyramidal
B Square pyramidal
C Pentagonal planar
D Octahedral

The correct answer is:
Square pyramidal

Solution:
H=12V+M-C+A
=128+4=6
sp3d2 hybridization
4 B.P + 1 B.P (Doubly bonded) + 1 L.P
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on axial position.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The germ pores on the pollen grains are found on the

Options:
A Exine only
B Intine only
C Both exine and intine
D Vegetative cell

The correct answer is:
Exine only

Solution:
At certain places, exine remains unthickened or missing; these places are known as germ pores.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
The Gibbs energy change (in J) for the given reaction at Cu2+=Sn2+=1 M and 298 K is :

Cus+Sn2+aq.→Cu2+aq.+Sns;

ESn2+Sn0=−0.16 V, ECu2+Cu0=0.34 V,Take F=96500 C mol−1

Options:

The correct answer is:
96500

Solution:
Ecell 0=ESn2+/Sn0-ECu2+/Cu0

=-0.16-0.34

=-0.50V

ΔG0=-nFEcell0

=-2×96500×(-0.5)

=96500 J

=96.5 KJ=96500 J

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2020 (02 Sep Shift 1)
The Gibbs' energy for the decomposition of $\mathrm{Al}_2 \mathrm{O}_3$ at $500^{\circ} \mathrm{C}$ is as follows
$$
\begin{gathered}
\frac{2}{3} \mathrm{Al}_2 \mathrm{O}_3 \longrightarrow \frac{4}{3} \mathrm{Al}+\mathrm{O}_2 ; \\
\Delta_r G=+960 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{gathered}
$$
The potential difference needed for the electrolytic reduction of aluminium oxide $\left(\mathrm{Al}_2 \mathrm{O}_3\right)$ at $500^{\circ} \mathrm{C}$ is at least

Options:
A $4.5 \mathrm{~V}$
B $3.0 \mathrm{~V}$
C $2.5 \mathrm{~V}$
D $5.0 \mathrm{~V}$

The correct answer is:
$2.5 \mathrm{~V}$

Solution:
The half cell reaction are at anode
$\left.2 \mathrm{O}^{2-}+4 e^{-} \longrightarrow \mathrm{O}_2\right] \times 3$
at cathode
$\left.\mathrm{Al}^{3+} \longrightarrow \mathrm{Al}+3 e^{-}\right] 4$
Net reaction
$4 \mathrm{Al}+6 \mathrm{O}^{2-} \longrightarrow 3 \mathrm{O}_2+4 \mathrm{Al}$
$\begin{aligned}
& \text {or } \frac{4}{3} \mathrm{Al}+2 \mathrm{O}^{2-} \longrightarrow \mathrm{O}_2+\frac{4}{3} \mathrm{Al} \\
& \therefore \quad n=\frac{12}{3}=4 \\
& \Delta G^{\circ}=-n F E^{\circ} \\
& \text {Here, } \Delta G^{\circ}=+960 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& =960 \times 1000 \mathrm{~J} \mathrm{~mol}^{-1} \\
& n=4 \\
& F=96500 \text {coulomb } \\
& \therefore \quad 960 \times 1000=-4 \times 96500 \times E^{\circ} \\
& E^{\circ}=-\frac{960000}{4 \times 96500} \\
& =-2.48 \mathrm{~V} \\
& \text {Potential difference } \approx 2.5 \mathrm{~V}
\end{aligned}$

Topic: Chemistry, Electrochemistry, NEET, NEET 2012 (Mains)
The Gibbs energy for the decomposition of $\mathrm{Al}_2 \mathrm{O}_3$ at $500^{\circ} \mathrm{C}$ is as follows :
$$
\frac{2}{3} \mathrm{Al}_2 \mathrm{O}_3 \rightarrow \frac{4}{3} \mathrm{Al}+\mathrm{O}_2, \Delta_{\mathrm{r}} \mathrm{G}=+966 \mathrm{~kJ} \mathrm{~mol}^{-1}
$$
The potential difference needed for electrolytic reduction of $\mathrm{Al}_2 \mathrm{O}_3$ at $500^{\circ} \mathrm{C}$ is at least

Options:
A
$4.5 \mathrm{~V}$

B
$3.0 \mathrm{~V}$

C
$2.5 \mathrm{~V}$

D
$5.0 \mathrm{~V}$

The correct answer is:
$2.5 \mathrm{~V}$

Solution:
$\Delta G=-n F E \quad \Rightarrow E=\frac{-\Delta G}{n F}$
$E=-\frac{966 \times 10^3}{4 \times 96500}$
$=-2.5 \mathrm{~V}$
$\therefore$ The potential difference needed for the reduction $=2.5 \mathrm{~V}$

Topic: Chemistry, Electrochemistry, JEE Main
The Gibbs energy for the decomposition of $\mathrm{Al}_2 \mathrm{O}_3$ at $500^{\circ} \mathrm{C}$ is as follows :
$$
\frac{2}{3} \mathrm{Al}_2 \mathrm{O}_3 \rightarrow \frac{4}{3} \mathrm{Al}+\mathrm{O}_2, \Delta_{\mathrm{r}} \mathrm{G}=+966 \mathrm{~kJ} \mathrm{~mol}^{-1}
$$
The potential difference needed for electrolytic reduction of $\mathrm{Al}_2 \mathrm{O}_3$ at $500^{\circ} \mathrm{C}$ is at least

Options:
A
$4.5 \mathrm{~V}$

B
$3.0 \mathrm{~V}$

C
$2.5 \mathrm{~V}$

D
$5.0 \mathrm{~V}$

The correct answer is:
$2.5 \mathrm{~V}$

Solution:
$\Delta G=-n F E \quad \Rightarrow E=\frac{-\Delta G}{n F}$
$E=-\frac{966 \times 10^3}{4 \times 96500}$
$=-2.5 \mathrm{~V}$
$\therefore$ The potential difference needed for the reduction $=2.5 \mathrm{~V}$

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2010
The Gibbs energy for the decomposition of $\mathrm{Al}_2 \mathrm{O}_3$ at $500 \mathrm{~C}$ is as follows :
$\frac{2}{3} \mathrm{Al}_2 \mathrm{O}_3 \rightarrow \frac{4}{3} \mathrm{Al}+\mathrm{O}_2, \Delta_r G=+940 \mathrm{~kJ} \mathrm{~mol}^{-1}$
The potential difference needed for the electrolytic reduction of aluminium oxide at $500^{\circ} \mathrm{C}$ should be at least :

Options:
A
$4.5 \mathrm{~V}$

B
$3.0 \mathrm{~V}$

C
$5.0 \mathrm{~V}$

D
$2.5 \mathrm{~V}$

The correct answer is:
$3.0 \mathrm{~V}$

Solution:
In the reaction
$\frac{2}{3} \mathrm{Al}_2 \mathrm{O}_3 \longrightarrow \frac{4}{3} \mathrm{Al}+\mathrm{O}_2$
For the oxidation half-reaction
$\mathrm{Al}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Al}$
no. of electron transfered $(n)=3$
$\begin{aligned}
& \Delta \mathrm{G}^{\circ}=-n \mathrm{FE}^{\circ} \\
& 940=3 \times 96500 \times \mathrm{E}^{\circ}
\end{aligned}$
$\begin{aligned}
\mathrm{E}^{\circ} & =\frac{940 \times 10^3 \mathrm{~J}}{3 \times 96500} \\
& =3.24 \approx 3 \mathrm{~V}
\end{aligned}$

Topic: Chemistry, Electrochemistry, JEE Main
The Gibbs energy for the decomposition of $\mathrm{Al}_2 \mathrm{O}_3$ at $500 \mathrm{~C}$ is as follows :
$\frac{2}{3} \mathrm{Al}_2 \mathrm{O}_3 \rightarrow \frac{4}{3} \mathrm{Al}+\mathrm{O}_2, \Delta_r G=+940 \mathrm{~kJ} \mathrm{~mol}^{-1}$
The potential difference needed for the electrolytic reduction of aluminium oxide at $500^{\circ} \mathrm{C}$ should be at least :

Options:
A
$4.5 \mathrm{~V}$

B
$3.0 \mathrm{~V}$

C
$5.0 \mathrm{~V}$

D
$2.5 \mathrm{~V}$

The correct answer is:
$3.0 \mathrm{~V}$

Solution:
In the reaction
$\frac{2}{3} \mathrm{Al}_2 \mathrm{O}_3 \longrightarrow \frac{4}{3} \mathrm{Al}+\mathrm{O}_2$
For the oxidation half-reaction
$\mathrm{Al}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Al}$
no. of electron transfered $(n)=3$
$\begin{aligned}
& \Delta \mathrm{G}^{\circ}=-n \mathrm{FE}^{\circ} \\
& 940=3 \times 96500 \times \mathrm{E}^{\circ}
\end{aligned}$
$\begin{aligned}
\mathrm{E}^{\circ} & =\frac{940 \times 10^3 \mathrm{~J}}{3 \times 96500} \\
& =3.24 \approx 3 \mathrm{~V}
\end{aligned}$

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2013 (25 Apr Online)
The given compound exists in polar form in which there is a close loop of Huckle's number of electrons. The number of electrons in the outer loop is.

Options:

The correct answer is:
14

Solution:
The molecule is aromatic because it follows Huckle's rule. The outer periphery has 14 electrons.

Topic: Chemistry, General Organic Chemistry, JEE Main
The given diagram is of adenine. What group should be present in the spot marked X?

Options:
A -CH3
B -NH2
C -COOH
D -CH2OH

The correct answer is:
-NH2

Solution:
Adenine is a nitrogen base present in nucleic acids. This belongs to the purine group. Purine groups are nitrogen bases with two rings. A four-membered ring is attached to a six-membered ring. Purines are guanine and adenine and these are present in both DNA's and RNA's. The main difference between adenine and guanine is that guanine shows an extra = O group. Both the nitrogen bases have –NH2 groups but present on different carbon atoms. Thus the 'x' in the figure represents the amino group. Adenine is called as 6-amino purine.

Topic: Chemistry, Biomolecules, NEET
The given diagram shows an undisturbed sedimentary strata at the bottom of the ocean. The fossils found in layer 'B' resemble the fossils found in layer 'A'. This similarity suggests that

Options:
A the fossils in layer 'B' were formed before fossils in layer 'A'
B modern forms of the life may have evolved from earlier forms of life
C vertebrate fossils are only found in sediments
D the fossils in layer 'A' must be more complex than layer 'B'

The correct answer is:
modern forms of the life may have evolved from earlier forms of life

Solution:
Preserved remains or impressions of once living organisms from the past on Earth's crust are called fossils. Cross-section of earth's crust will provide information on the arrangement of sediments formed by rock, one over the other during the geological period. Different-aged rock sediments contain fossils of different life-forms. In the given diagram, fossil of layer B followed the fossil of layer A. It also suggests that they belong to the same region of different timeline and display similarity. From the similarity between the fossils of layer A and B, it can be deduced that modern life forms have evolved from earlier life form.

Topic: Biology, Evolution, NEET
The given diagram shows four processes i.e., isochoric, isobaric, isothermal and adiabatic. The correct assignment of the processes, in the same order is given by:

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

The correct answer is:
d a b c

Solution:
a→ Isobaric process
b→ Isothermal process
c→ Adiabatic process
d→ Isochoric process

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2019 (08 Apr Shift 2)
The given ecological pyramid is

Options:
A Pyramid of energy for grassland and pyramid of biomass for a lake
B Pyramid of biomass for grassland and pyramid of number for tree
C Pyramid of number for sea and pyramid of biomass for grassland
D Pyramid of energy and biomass for a pond

The correct answer is:
Pyramid of number for sea and pyramid of biomass for grassland

Solution:
In the grassland ecosystem, the pyramid of numbers is upright or straight. The producers are mainly grasses are always more in number. The primary consumers as grasshoppers, rabbits, mice etc. Are Lesser than the number of grasses. The primary consumers are eaten by a relatively smaller number of secondary consumers as a frog, lizard, snake etc. These secondary consumers are consumed by only a few tertiary or top consumers as the hawk, peacock or other birds.

Topic: Biology, Ecosystem, NEET
The given figure shows the elliptical orbit of a planet m about the sun S. If t1 is the time taken by the planet to move from C to D and t2 is the time taken by the planet to move from A to B, then the relation between t1 and t2 is [Given, area SCD=2area SAB]

Options:
A t1=4t2
B t1=2t2
C t1=t2
D 2t1=t2

The correct answer is:
t1=2t2

Solution:
According to kepler's second law
Equal areas are swept in equal time. t1, the time taken to go from C to D = 2t2 where t2 is the time taken to go from A to B. As it is given that area SCD = 2SAB.​

Topic: Physics, Gravitation, JEE Main
The given graph shows the variation of velocity with displacement. Which, one of the graphs given below correctly represents the variation of acceleration with displacement

Options:
A
B
C
D

The correct answer is:

Solution:
The v - x equation from the given graph can be written as
v=(-v0x0)x+v0... (i)

a=dvdt=(-v0x0)dxdt

=(-v0x0)v

Substituting v from Eq. (i) we get,
a=(-v0x0)&lsqb;(-v0x0)x+v0&rsqb;

a=(v0x0)2x-v02x0

Thus, a-x graph is a straight line with positive slope and negative intercept.
Correct answer is

Topic: Physics, Motion In Two Dimensions, JEE Main
The given graph shows variation (with distancer from centre) of:

Options:
A Electric field of a uniformly charged sphere
B Potential of a uniformly charged spherical shell
C Potential of a uniformly charged sphere
D Electric field of a uniformly charged spherical shell

The correct answer is:
Potential of a uniformly charged spherical shell

Solution:
Electric potential is constant inside a charged spherical shell.

Topic: Physics, Electrostatics, JEE Main, JEE Main 2019 (11 Jan Shift 1)
The given illustration shows the skull of two different mammals. Which of the following accurately describes the differences between these skulls?

Options:
A Skull A has less brain capacity than skull B
B Skull A is of primate and skull B is not of primate
C Skull A is the skull of human and skull B is the skull of an ape
D Skull A and skull B are of baby chimpanzee and adult chimpanzee respectively

The correct answer is:
Skull A is the skull of human and skull B is the skull of an ape

Solution:
The main differences between human and ape skull in the above diagram are as follows":
1 Ape skull is more protruding
2 Apes have large brow ridges
3 In apes the forehead is receding
4 Zygomatic arch is much larger in apes.
All the above features are seen in B and thus it represents Skull of Apes.

Topic: Biology, Evolution, NEET
The given paragraph describes the resting state of a nerve fiber.

When a neuron is resting, the axonal membrane is comparatively _____P_____ to potassium ions ( K+ ) and _____Q_____ to sodium ions ( Na+ ). Similarly, the membrane is ______R_____to negatively charged proteins present in the axoplasm. Consequently, the axoplasm inside the axon contains high concentration of _____S_____ and _____T______. P, Q, R, S and T are respectively

Options:
A More permeable, nearly impermeable, impermeable, K+, proteins
B Less permeable, impermeable, impermeable, K+, proteins
C More permeable, nearly impermeable, less permeable, Na+, K+
D Less permeable, highly permeable, impermeable, K+, Ca2+

The correct answer is:
More permeable, nearly impermeable, impermeable, K+, proteins

Solution:
During the resting state, the axonal membrane is
A. More permeable to K+.
B. Nearly impermeable to Na+.
C. Impermeable to negatively charged proteins present in the axoplasm.
Axoplasm has high concentration of K+ and negatively charged proteins and low concentration of Na+. In contrast, the fluid outside the axon contains a low concentration of K+, a high concentration of Na+, and thus forms a concentration gradient. These ionic gradients across the resting membrane are maintained by the Na–K pump which actively transports 3Na+ outwards for 2K+ into the cell.
The outer surface of the axonal membrane is positively charged. The inner surface of the axonal membrane is negatively charged. The axonal membrane is polarised and the potential difference is termed as resting potential.

Topic: Biology, Neural Control and Coordination, NEET