datavorous/entrance-exam-dataset huggingface dataset:
Using very little soap while washing clothes, does not serve the purpose of cleaning of clothes, because

Options:
A soap particles remain floating in water as ions.
B colloidal structure of soap in water is completely disturbed.
C the hydrophobic part of soap is not able to take away grease.
D the micelles are not formed due to concentration of soap, below its CMC value.

The correct answer is:
the micelles are not formed due to concentration of soap, below its CMC value.

Solution:
Micelle or associate colloid formation occurs above a certain concentration known as CMC. So, if the concentration of soap is less that means it hasn't reached the CMC value. That is why it does not serve the purpose of cleaning of clothes.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main, JEE Main 2022 (25 Jun Shift 1)
Using Young's double slit experiment, a monochromatic light of wavelength 5000Å produces fringes of fringe width 0.5 mm. If another monochromatic light of wavelength 6000Å is used and the separation between the slits is doubled, then the new fringe width will be

Options:
A 0.5 mm
B 1.0 mm
C 0.6 mm
D 0.3 mm

The correct answer is:
0.3 mm

Solution:
We know that fringe width is given by, β=λDd

⇒β'β=λ'λ×dd'=60005000×d2d

⇒β'=65×2×0.5=0.3 mm

Topic: Physics, Wave Optics, JEE Main, JEE Main 2022 (29 Jun Shift 1)
U-tube moves with a constant speed v parallel to the surface of a stationary liquid. The cross-section area of the lower part of the tube lowered into the liquid, is equal to S1 and that of the top part located over the liquid is S2. Friction and formation of waves should be neglected. The density of the fluid is ρ. Neglect difference in heights at both the openings of the tube.

The velocity of the liquid coming out of the top part as seen by an observer on the ground will be

Options:
A vS1S2
B v1+S1S2
C zero
D none of these

The correct answer is:
v1+S1S2

Solution:
Using the equation of continuity, we get
S1v1=S2v2
v1=v (liquid is entering in pipe with a speed v with respect to the pipe)
âv2=S1S2v
v2 : velocity of liquid at S2 with respect to pipe
â´ With respect to ground, the speed of liquid will be,
v+S1S2v=v1+S1S2

Topic: Physics, Mechanical Properties of Fluids, JEE Main
UV light of $4.13 \mathrm{eV}$ is incident on a photosensitive metal surface having work function $3.13 \mathrm{eV}$. The maximum kinetic energy of ejected photoelectrons will be:

Options:
A $4.13 \mathrm{eV}$
B $3.13 \mathrm{eV}$
C $1 \mathrm{eV}$
D $7.26 \mathrm{eV}$

The correct answer is:
$1 \mathrm{eV}$

Solution:
$\begin{aligned} & \mathrm{E}_{\text {photon }}=(\text { work function })+\mathrm{K} \cdot \mathrm{E}_{\max } \\ & \therefore 4.13=3.13+\mathrm{K} \cdot \mathrm{E}_{\max } \\ & \therefore \mathrm{K}_{\mathrm{max}}=1 \mathrm{eV}\end{aligned}$

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2024 (09 Apr Shift 2)
V1mL of NaOH of molarity X and V2mL of Ba(OH)2 of molarity Y are mixed together. The mixture is completely neutralized by 100 mL of 0.1N HCl. If V1/V2=1/4 and X/Y = 4 then what fraction of acid is neutralized by Ba(OH)2

Options:
A 0.5
B 0.25
C 0.33
D 0.67

The correct answer is:
0.67

Solution:
Meq. of NaOH + Meq. of Ba(OH)2=Meq. of HCl
X.V1+2Y.V2=100×0.1=10
or 4Y.V24+2Y.V2=10
or 3YV2=10
2Y.V2=10×23=6.66
Fraction used by Ba(OH)2=6.6610=0.67

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
$\alpha, v$ and $u$ represent most probable velocity, average velocity and root mean square velocity respectively of a gas at a particular temperature. The correct order among the following is

Options:
A
$u>v>\alpha$

B
$v>u>\alpha$

C
$\alpha>u>v$

D
$u>\alpha>v$

The correct answer is:
$u>v>\alpha$

Solution:
$$
\begin{aligned}
u: v: \alpha & =\sqrt{\frac{3 R T}{M}}: \sqrt{\frac{8 R T}{\pi M}}: \sqrt{\frac{2 R T}{M}} \\
& =\sqrt{3}: \sqrt{\frac{8}{\pi}}: \sqrt{2}
\end{aligned}
$$
$$
=1.224: 1.128: 1
$$

Topic: Chemistry, States of Matter, JEE Main, JEE Main 2012 (12 May Online)
VBB can vary between 0 and 5 V. Find the minimum value of base current and VBB, so that transistor works in saturation mode. [Given Î²=200 and VBE=1 V]

Options:
A 20 μA, 2.8 V
B 25 Î¼A, 3.5 V
C 20 Î¼A, 3.5 V
D 25 Î¼A, 2.8 V

The correct answer is:
20 Î¼A, 3.5 V

Solution:
When switched on,
VCE=0
VCC-RC iC=0
ic=VCCRC=51Ã103=5Ã10-3 A
ic=Î²iB
iB=iCÎ²=5Ã10-3200=2.5Ã10-6 A =2.5 Î¼A
Using KVL at input side,
VBB-iB RB-VBE=0
VBB=VBE+iBRB
=1+100Ã103Ã25Ã10-6
=1+2.5
=3.5 V

Topic: Physics, Semiconductors, JEE Main
$v_e$ and $v_p$ denotes the escape velocity from the earth and another planet having twice the radius and the same mean density as the earth. Then

Options:
A $v_e=v_p$
B $v_e=v_p / 2$
C $v_e=2 v_p$
D $v_e=v_p / 4$

The correct answer is:
$v_e=v_p / 2$

Solution:
$v_e=\sqrt{\frac{2 G M}{R}}=R \sqrt{\frac{8}{3} \pi G \rho}$
If mean density is constant then $v_{\mathrm{e}} \propto R$
$\frac{v_e}{v_p}=\frac{R_e}{R_p}=\frac{1}{2} \Rightarrow v_e=\frac{v_p}{2}$

Topic: Physics, Gravitation, JEE Main
Vacuole in a plant cell

Options:
A is membrane-bound and contains storage proteins and lipids
B is membrane-bound and contains water and excretory substances
C lacks membrane and contains air
D lacks membrane and contains water and excretory substances

The correct answer is:
is membrane-bound and contains water and excretory substances

Solution:
The cytoplasm of many plant and some animal cells contains numerous small or large sized, hollow, liquid filled structures, the vacuoles. The vacuoles of plant cells are bounded by a single semipermeable membrane known as tonoplast. These vacuoles contain water, phenol, flavonols, anthocyanins, alkaloids and storage products such as sugars and proteins. The vacuoles of animal cells are bounded by a lipoproteinaceous membrane and their function is storage, transmission of materials and maintenance of internal pressure of cell.

Topic: Biology, Cell - The Unit of Life, NEET, NEET 2008 (Screening)
Vacuole in a plant cell

Options:
A Lacks membrane and contains water and excretory substances
B Is membrane-bound and contains storage proteins and lipids
C Is membrane-bound and contains water and excretory substances
D Lacks membrane and contains air

The correct answer is:
Is membrane-bound and contains water and excretory substances

Solution:
It is having tonoplast as membrane and stores water and excretory substances.

Topic: Biology, Cell - The Unit of Life, NEET, NEET 2008 (Mains)
Value of gas constant R is

Options:
A
$0.082$ litre atm

B
$0.987 \mathrm{cal} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$

C
$8.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$

D
$83 \mathrm{~erg} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$

The correct answer is:
$8.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$

Solution:
No Solution Available

Topic: Chemistry, States of Matter, JEE Main, JEE Main 2002
Value of Henry's constant \( \mathrm{K}_{\mathrm{H}} \)

Options:
A increases with increase in temperature
B decreases with increase in temperature
C remains constant.
D first increases then decreases.

The correct answer is:
increases with increase in temperature

Solution:
Henry's law is a gas law which states that at the amount of gas that is dissolved in a liquid is directly proportional to the partial pressure of that gas above the liquid when the temperature is kept constant.

Value of Henry’s constant increases with increase in temperature and solubility decreases.

Topic: Chemistry, Solutions, JEE Main
Value of KP for the equilibrium reaction N2O4( g)⇌2NO2( g) at 288 K is 47.9. The KC for this reaction at same temperature is (Nearest integer)
R=0.083 L bar K-1 mol-1

Options:

The correct answer is:
2

Solution:
Kp=KcRT∆n

∆n=1

KC=KpRT=47.90.083×288=2

Topic: Chemistry, Chemical Equilibrium, JEE Main, JEE Main 2021 (22 Jul Shift 1)
Value of $K_p$ in the reaction
$\mathrm{MgCO}_{3(s)} \rightleftharpoons \mathrm{MgO}_{(s)}+\mathrm{CO}_{2(g)}$ is

Options:
A $K_p=P_{\mathrm{CO}_2}$
B $K_p=P C O_2 \times \frac{P_{\mathrm{CO}_2} \times P_{\mathrm{MgO}}}{P_{\mathrm{MgCO}_3}}$
C $K_p=\frac{P_{\mathrm{CO}_2} \times P_{\mathrm{Mg0}}}{P_{\mathrm{MgCO}_3}}$
D $K_p=\frac{P_{\mathrm{MgCO}_3}}{P_{\mathrm{CO}_2} \times P_{\mathrm{MgO}}}$

The correct answer is:
$K_p=P_{\mathrm{CO}_2}$

Solution:
In this reaction gaseous molecule count
$\mathrm{MgCO}_3 \rightarrow \mathrm{MgO}_{(\mathrm{s})}+\mathrm{CO}_{2(\mathrm{~g})}$
$K_p=P_{\mathrm{CO}_2}$

Topic: Chemistry, Chemical Equilibrium, JEE Main
Values of dissociation constant, $K_a$ are given as follows :
$\begin{array}{lc}\text { Acid } & \boldsymbol{K}_{\boldsymbol{a}} \\ \mathrm{HCN} & 6.2 \times 10^{-10} \\ \mathrm{HF} & 7.2 \times 10^{-4} \\ \mathrm{HNO}_2 & 4.0 \times 10^{-4}\end{array}$
Correct order of increasing base strength of the base $\mathrm{CN}^{-}, \mathrm{F}^{-}$and $\mathrm{NO}_2^{-}$will be :

Options:
A
$\mathrm{F}^{-} < \mathrm{CN}^{-} < \mathrm{NO}_2^{-}$
B
$\mathrm{NO}_2^{-} < \mathrm{CN}^{-} < \mathrm{F}^{-}$
C
$\mathrm{F}^{-} < \mathrm{NO}_2^{-} < \mathrm{CN}^{-}$
D
$\mathrm{NO}_2^{-} < \mathrm{F}^{-} < \mathrm{CN}^{-}$

The correct answer is:
$\mathrm{F}^{-} < \mathrm{NO}_2^{-} < \mathrm{CN}^{-}$

Solution:
Higher the value of $\mathrm{K}_a$ lower will be the value of $\mathrm{pK}_{\mathrm{a}}$ i.e. higher will be the acidic nature. Further since $\mathrm{CN}^{-}, \mathrm{F}^{-}$and $\mathrm{NO}_2^{-}$are conjugate base of the acids $\mathrm{HCN}, \mathrm{HF}$ and $\mathrm{HNO}_2$ respectively hence the correct order of base strength will be
$$
\mathrm{F}^{-} < \mathrm{NO}_2^{-} < \mathrm{CN}^{-}
$$
$(\because$ stronger the acid weaker will be its conjugate base)

Topic: Chemistry, Ionic Equilibrium, JEE Main
Values of dissociation constant, $K_a$ are given as follows :
$\begin{array}{lc}\text { Acid } & \boldsymbol{K}_{\boldsymbol{a}} \\ \mathrm{HCN} & 6.2 \times 10^{-10} \\ \mathrm{HF} & 7.2 \times 10^{-4} \\ \mathrm{HNO}_2 & 4.0 \times 10^{-4}\end{array}$
Correct order of increasing base strength of the base $\mathrm{CN}^{-}, \mathrm{F}^{-}$and $\mathrm{NO}_2^{-}$will be :

Options:
A
$\mathrm{F}^{-} < \mathrm{CN}^{-} < \mathrm{NO}_2^{-}$
B
$\mathrm{NO}_2^{-} < \mathrm{CN}^{-} < \mathrm{F}^{-}$
C
$\mathrm{F}^{-} < \mathrm{NO}_2^{-} < \mathrm{CN}^{-}$
D
$\mathrm{NO}_2^{-} < \mathrm{F}^{-} < \mathrm{CN}^{-}$

The correct answer is:
$\mathrm{F}^{-} < \mathrm{NO}_2^{-} < \mathrm{CN}^{-}$

Solution:
Higher the value of $\mathrm{K}_a$ lower will be the value of $\mathrm{pK}_{\mathrm{a}}$ i.e. higher will be the acidic nature. Further since $\mathrm{CN}^{-}, \mathrm{F}^{-}$and $\mathrm{NO}_2^{-}$are conjugate base of the acids $\mathrm{HCN}, \mathrm{HF}$ and $\mathrm{HNO}_2$ respectively hence the correct order of base strength will be
$$
\mathrm{F}^{-} < \mathrm{NO}_2^{-} < \mathrm{CN}^{-}
$$
$(\because$ stronger the acid weaker will be its conjugate base)

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2013 (22 Apr Online)
Values of m, for which the line y=mx+25 is a tangent to the hyperbola 16x2-9y2=144, are the roots of the equation x2-a+bx-4=0, then the value of a+b is equal to

Options:

The correct answer is:
0

Solution:
Equation of hyperbola is x29-y216=1.
Equation of tangent is y=mx+9m2-16

⇒9m2-16=25⇒m=±2
⇒a+b= sum of roots =0.

Topic: Mathematics, Hyperbola, JEE Main
Values of work function (W0) for a few metals are given below

Metal	Li	Na	K	Mg	Cu	Ag

W0eV	2.42	2.3	2.25	3.7	4.8	4.3

The number of metals which will show photoelectric effect when light of wavelength 400 nm falls on it is _____

Given: h=6.6×10–34 J s

c=3×108 ms–1

e=1.6×10–19 C

Options:

The correct answer is:
3

Solution:
The minimum amount of energy necessary to start the emission of electrons from a metal's surface is known as its work function. The work function is the minimum amount of energy required to remove an electron from a metal's surface to infinity.

Energy of incident photon =hcλ

=6.6×10-34×3×108400×10-9 J

=6.6×10-34×3×108400×10-9×1.6×10-9 eV

=3.1 eV

If radiation energy is greater than work function of metal, then metal show photo electric effect.

∴ Li, Na and K will show photoelectric effect.

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2023 (12 Apr Shift 1)
Van der Waal's equation for a gas is stated as,
P=nRTV-nb-anV2.
This equation reduces to the perfect gas equation, P=nRTV When,

Options:
A temperature is sufficiently high and pressure is low.
B both temperature and pressure are very low.
C both temperature and pressure are very high.
D temperature is sufficiently low and pressure is high.

The correct answer is:
temperature is sufficiently high and pressure is low.

Solution:
At very high temperatures & low pressures, the characteristic constant behaviour can be understood from virial expansion
Z = PV m RT = 1 + b - a RT 1 V m + b 2 V m 2 + ....
Vm, molar volume large i.e b 2 V m 2 → o
i.e Third term in virial expansion can be ignored.

In qualitative terms, the real gas will show ideal behaviour when the intermolecular forces of attraction will be negligible which is possible at high temperature and low pressure.
Also b - a RT → o
i.e. Z ≈ 1 and gas behaves ideally.

Topic: Chemistry, States of Matter, JEE Main, JEE Main 2014 (09 Apr Online)
Van-Arkel process and Mond's process are respectively used for refining of:

Options:
A Zr and Ti
B Ni and Zr
C Ti and Ni
D Ni and Fe

The correct answer is:
Ti and Ni

Solution:
Van-Arkel process ⇒ Ti, Hf, Zr,B

Mond's Process ⇒Ni

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
Vander Waal's equation of state is obeyed by real gases. For $n$ moles of a real gas, the expression will be

Options:
A $\left(\frac{P}{n}+\frac{n a}{V^2}\right)\left(\frac{V}{n-b}\right)=R T$
B $\left(P+\frac{a}{V^2}\right)(V-b)=n R T$
C $\left(P+\frac{n a}{V^2}\right)(n V-b)=n R T$
D $\left(P+\frac{n^2 a}{V^2}\right)(V-n b)=n R T$

The correct answer is:
$\left(P+\frac{n^2 a}{V^2}\right)(V-n b)=n R T$

Solution:
$\left(P+\frac{n^2 a}{V^2}\right)(V-n b)=n R T$

Topic: Chemistry, States of Matter, JEE Main
Vanillin compound obtained from vanilla beans, has total sum of oxygen atoms and \(\pi\) electrons is ______

Options:

The correct answer is:
11

Solution:
Vanillin compound is an organic compound molecular formula $\mathrm{C}_8 \mathrm{H}_8 \mathrm{O}_3$. It is a phenolic aldehyde. Its functional compounds include aldehyde, hydroxyl and ether. It is the primary component of the extract of the vanilla beans.
Total sum of oxygen atoms and $\pi$-electrons is $3+8=11$
Total number of oxygen atoms $=3$
Total number of $\pi$-bonds $=4$
$\therefore$ Total number of $\pi$-electrons $=8$

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2024 (04 Apr Shift 2)
van't Hoff factors of equimolal solutions of sodium chloride, barium chloride and glucose in water are

Options:
A $2,3,0$ respectively
B $2,3,6$ respectively
C $2,3,4$ respectively
D $2,3,1$ respectively

The correct answer is:
$2,3,1$ respectively

Solution:
$\mathrm{NaCl}(a q) \longrightarrow \mathrm{Na}^{+}(a q)+\mathrm{Cl}^{-}(a q)$
van't Hoff factor, $i=2$
$\mathrm{BaCl}_{2}(a q) \longrightarrow \mathrm{Ba}^{2+}(a q)+2 \mathrm{Cl}^{-}(a q), i=3$
Glucose does not undergo dissociation or association hence, $i=1$

Topic: Chemistry, Solutions, JEE Main
Van't Hoff's equation for a chemical reaction under equilibrium is given by standard reaction enthalpy at temperature T and K is the equilibrium constant. Predict how K will vary with temperature for an exothermic reaction.

Options:
A K decreases as the temperature rises
B K remain unchanged
C K increases as the temperature rise
D Information is not sufficient

The correct answer is:
K decreases as the temperature rises

Solution:
Here K decreases as the temperature rises as K ∝1T .

Topic: Chemistry, Chemical Kinetics, NEET
Vapour pressure of a solution of 5g of non- electrolyte in 100g of water at a particular temperature is 2985N/m2 . The vapour pressure of pure water is 3000N/m2 . The molecular weight of the solute is

Options:
A 60
B 120
C 180
D 380

The correct answer is:
180

Solution:
Po-PsPo=W2M2W1M1
=3000-29853000=5M210018 or M2=180

Topic: Chemistry, Solutions, JEE Main
Vapour pressure of chloroform $\left(\mathrm{CHCl}_3\right)$ and dichloromethane $\left(\mathrm{CH}_2 \mathrm{Cl}_2\right)$ at $25^{\circ} \mathrm{C}$ are $200 \mathrm{~mm} \mathrm{Hg}$ and $415 \mathrm{~mm} \mathrm{Hg}$ respectively. Vapour pressure of the solution obtained by mixing $25.5 \mathrm{~g}$ of $\mathrm{CHCl}_3$ and $4 \mathrm{O} \mathrm{g}$ of $\mathrm{CH}_2 \mathrm{Cl}_2$ at the same temperature will be:
[Molecular mass of $\mathrm{CHCl}_3=119.5 \mathrm{~g} / \mathrm{mol}$ and molecular mass of $\mathrm{CH}_2 \mathrm{Cl}_2=85 \mathrm{~g} / \mathrm{mol}$ ]

Options:
A $173.9 \mathrm{mmHg}$
B $615.0 \mathrm{mmHg}$
C $347.9 \mathrm{mmHg}$
D $28.5 \mathrm{mmHg}$

The correct answer is:
$347.9 \mathrm{mmHg}$

Solution:
Molar mass of $\mathrm{CH}_2 \mathrm{Cl}_2=12 \times 1+1 \times 2+35.5 \times 2=85 \mathrm{~g} \mathrm{~mol}^{-1}$
Molar mass of $\mathrm{CHCl}_3=12 \times 1+1 \times 1+35.5 \times 3=119.5 \mathrm{~g} \mathrm{~mol}^{-1}$
Moles of $\mathrm{CH}_2 \mathrm{Cl}_2=40 \mathrm{~g} / 85 \mathrm{~g} \mathrm{~mol}^{-1}=0.47 \mathrm{~mol}$
Moles of $\mathrm{CHCl}_3=25.5 \mathrm{~g} / 119.5 \mathrm{~g} \mathrm{~mol}^{-1}=0.213 \mathrm{~mol}$
Total number of moles $=0.47+0.213=0.683 \mathrm{~mol}$
Mole fraction of component 2
$$
=0.47 \mathrm{~mol} / 0.683 \mathrm{~mol}=0.688
$$
Mole fraction of component 1
$$
=1.00-0.688=0.312
$$
We know that:
$$
\begin{aligned}
& \mathrm{P}_{\mathrm{T}}=\mathrm{p}_1^0+\left(\mathrm{p}_2^0-\mathrm{p}_1^0\right) \mathrm{x}_2 \\
& =200+(415-200) \times 0.688 \\
& =200+147.9 \\
& =347.9 \mathrm{~mm} \mathrm{Hg}
\end{aligned}
$$

Topic: Chemistry, Solutions, NEET, NEET 2012 (Mains)
Vapour pressure of pure benzene is 119 torr and that of toluene is $37.0$ torr at the same temperature. Mole fraction of toluene in vapour phase which is in equilibrium with a solution of benzene and toluene having a mole fraction of toluene $0.50$, will be :

Options:
A
$0.137$

B
$0.237$

C
$0.435$

D
$0.205$

The correct answer is:
$0.237$

Solution:
$\mathrm{P}_{\mathrm{A}}=\mathrm{P}_{\mathrm{A}}^0 \times x_{\mathrm{A}}=$ total pressure $\times \mathrm{y}_{\mathrm{A}}$
$$
\mathrm{P}_{\mathrm{B}}=\mathrm{P}_{\mathrm{B}}^0 \times x_{\mathrm{B}}=\text { total pressure } \times \mathrm{y}_{\mathrm{B}}
$$
where $\mathrm{x}$ and $\mathrm{y}$ represents mole fraction in liquid and vapour phase respectively.
$$
\frac{\mathrm{P}_{\mathrm{B}}^0 x_{\mathrm{B}}}{\mathrm{P}_{\mathrm{A}}^0 x_{\mathrm{A}}}=\frac{\mathrm{y}_{\mathrm{B}}}{\mathrm{y}_{\mathrm{A}}} ; \frac{\mathrm{P}_{\mathrm{B}}^0\left(1-x_{\mathrm{A}}\right)}{\mathrm{P}_{\mathrm{A}}^0 x_{\mathrm{A}}}=\frac{1-\mathrm{y}_{\mathrm{A}}}{\mathrm{y}_{\mathrm{A}}}
$$
on putting values $\frac{119(1-0.50)}{37 \times 0.50}=\frac{1-y_{\mathrm{A}}}{\mathrm{y}_{\mathrm{A}}}$ on solving $\mathrm{y}_{\mathrm{A}}=0.237$

Topic: Chemistry, Solutions, JEE Main, JEE Main 2013 (23 Apr Online)
Variable ellipses are drawn with x=-4 as directrix and the origin as the corresponding focus. The eccentricity of the locus of the extremities of the minor axes of these ellipses is

Options:
A 1
B 2
C 2
D 12

The correct answer is:
1

Solution:
Now k=b and h=ae
Also ae-ae=4⇒a1-e2=4e
⇒a21-e2=4ae
⇒b2=4h
⇒k2=4h⇒y2=4x
⇒ eccentricity =1

Topic: Mathematics, Ellipse, JEE Main
Vasa recta are absent or highly reduced in

Options:
A juxtamedullary nephrons
B cortical nephrons
C both juxtamedullary and cortical nephrons
D medullary nephrons

The correct answer is:
cortical nephrons

Solution:
The cortical nephrons are mainly responsible for the concentration of the urine. The glomeruli, PCT, DCT, loop of Henle, are limited to the cortex in the cortical nephrons, whereas in the juxtamedullary nephrons they extend into the medullary portion of the kidney. The vasa recta is highly reduced or absent in the cortical nephrons because they have reduced loop of Henle.

Topic: Biology, Excretory Products and their Elimination, NEET
Vascular bundles of roots are

Options:
A Conjoint
B Concentric
C Bicollateral
D Radial

The correct answer is:
Radial

Solution:
The xylem and phloem strands alternate with each other separated by parenchymatous cells. Such kinds of vascular bundles are called radial and found mainly in roots.

Topic: Biology, Anatomy of Flowering Plants, NEET
Vascular cambium and cork cambium are the examples of ______.

Options:
A apical meristem
B lateral meristem
C intercalary meristem
D promeristem

The correct answer is:
lateral meristem

Solution:
Growth in the plant is of two types i.e.primary and secondary growth. Primary growth mainly occurs in the primary meristem. They help in increasing the length of the plant because they present on the axis of it. On the lateral side of it, axillary buds are present that form branches. Secondary meristem is of two types i.e. vascular cambium and cork cambium. Vascular cambium is present in between vascular tissue that helps in the secondary growth of the plants and increases the girth of the plants by the addition of cells in the lateral direction. Similarly, cork cambium also helps in the addition of cells in the cortex. So the correct answer is B .

Topic: Biology, Anatomy of Flowering Plants, NEET
Vascular tissues in flowering plants develop from

Options:
A Dermatogen
B Phellogen
C Plerome
D Periblem

The correct answer is:
Plerome

Solution:
Periblem - Ground tissue
Phellogen -Cork cambium
Dermatogen - Epidermis

Topic: Biology, Anatomy of Flowering Plants, NEET, NEET 2008 (Mains)
Vasopressin is produced and secreted by

Options:
A

Produced	Secreted

Hypothalamus	Hypothalamus

B

Produced	Secreted

Neurohypophysis	Hypothalamus

C

Produced	Secreted

Hypothalamus	Neurohypophysis

D

Produced	Secreted

Neurohypophysis	Neurohypophysis

The correct answer is:

Produced	Secreted

Hypothalamus	Neurohypophysis

Solution:
Neurohypophysis (pars nervosa) also known as posterior pituitary, stores, and releases two hormones called oxytocin and vasopressin, which are actually synthesized by the hypothalamus and are transported axonally to the neurohypophysis. Vasopressin acts mainly at the kidney and stimulates the reabsorption of water and electrolytes by the distal tubules and thereby reduces the loss of water through urine (diuresis). Hence, it is also called an anti-diuretic hormone (ADH).

Topic: Biology, Chemical Coordination and Integration, NEET
Vector $A$ has a magnitude of 10 units and makes an angle of $30^{\circ}$ with the positive $X$-axis. Vector $B$ has a magnitude of 20 units and makes an angle of $30^{\circ}$ with the negative $X$-axis. What is the magnitude of the resultant between these two vectors?

Options:
A $20 \sqrt{3}$
B $35$
C $15 \sqrt{3}$
D $10 \sqrt{3}$

The correct answer is:
$10 \sqrt{3}$

Solution:
Given,
$$
\begin{gathered}
A=10 \text { units } \\
B=20 \text { units } \\
\theta=30^{\circ}
\end{gathered}
$$
The given situation is shown in the figure. The angle between $\mathbf{A}$ and $\mathbf{B}$,
$$
\theta=180^{\circ}-30^{\circ}-30^{\circ}=120^{\circ}
$$
$\therefore$ Magnitude of the resultant vector,
$$
\begin{aligned}
R &=\sqrt{A^{2}+B^{2}+2 A B \cos \theta} \\
&=\sqrt{10^{2}+20^{2}+2 \times 10 \times 20 \cos 120^{\circ}} \\
&=\sqrt{100+400-200} \\
&=10 \sqrt{3}
\end{aligned}
$$

Topic: Physics, Mathematics in Physics, JEE Main
Vector $\vec{A}$ makes equal angles with $x, y$ and $z$ axis. Value of its components (in terms of magnitude of $\vec{A}$ ) will be

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

The correct answer is:
$\frac{A}{\sqrt{3}}$

Solution:
Let the components of $\dot{A}$ makes angles $\alpha, \beta$ and $\gamma$ with $x, y$ and $z$ axis respectively then $\alpha=\beta=\gamma$
$\cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1$
$\Rightarrow 3 \cos ^2 \alpha=1 \Rightarrow \cos \alpha=\frac{1}{\sqrt{3}}$
$\therefore A_x=A_y=A_z=A \cos \alpha=\frac{A}{\sqrt{3}}$

Topic: Physics, Mathematics in Physics, JEE Main