datavorous/entrance-exam-dataset huggingface dataset:
Uncertainty in position of a $0.25 \mathrm{~g}$ particle is $10^{-5}$. Uncertainty of velocity is $\left(h=6.6 \times 10^{-34} \mathrm{Js}\right)$

Options:
A $1.2 \times 10^{34}$
B $2.1 \times 10^{-29}$
C $1.6 \times 10^{-20}$
D $1.7 \times 10^{-9}$

The correct answer is:
$2.1 \times 10^{-29}$

Solution:
According to $\Delta x \times m \times \Delta v=\frac{h}{4 \pi} ; \quad \Delta v=\frac{h}{\Delta x \times m \times 4 \pi}$
$=\frac{6.6 \times 10^{-34}}{10^{-5} \times 0.25 \times 3.14 \times 4}=2.1 \times 10^{-29} \mathrm{~m} / \mathrm{s}$

Topic: Chemistry, Structure of Atom, JEE Main
Uncertainty in the position of an electron (mass $=9.1 \times 10^{-31} \mathrm{~kg}$ ) moving with a velocity $300 \mathrm{~ms}^{-1}$, accurate upto $0.001 \%$, will be

Options:
A
$19.2 \times 10^{-2} \mathrm{~m}$

B
$5.76 \times 10^{-2} \mathrm{~m}$

C
$1.92 \times 10^{-2} \mathrm{~m}$

D
$3.84 \times 10^{-2} \mathrm{~m}$
(h $\left.=6.63 \times 10^{-34} \mathrm{Js}\right)$

The correct answer is:
$1.92 \times 10^{-2} \mathrm{~m}$

Solution:
$$
\begin{aligned}
& \Delta \mathrm{x} . \Delta \mathrm{V} \geq \frac{\mathrm{h}}{4 \pi \mathrm{m}} \\
& \Delta \mathrm{x} \geq \frac{\mathrm{h}}{4 \pi \mathrm{m} \Delta \mathrm{V}}=\frac{6.63 \times 10^{-34}}{4 \times 3.14 \times 9.1 \times 10^{-31} \times 300 \times \frac{0.001}{100}} \\
&=\frac{6.63 \times 10^{-34}}{4 \times 3.14 \times 9.1 \times 3 \times 10^{-31} \times 10^{-3}} \\
&=0.01933 \\
&=1.93 \times 10^{-2}
\end{aligned}
$$

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2006
Under ambient conditions, the total number of gases released as products in the final step of the reaction scheme shown below is

XeF6âCompletehydrolysisP+ Other product âOH-/H2O QâSlow disproportionationin OH-/H2O Products

Options:
A 0
B 1
C 2
D 3

The correct answer is:
2

Solution:
XeF6âComplete HydrolysisXeO3P+HFâOH-/H2OHXeO4-Q
âOHâ/H2OâSlowâ disproportionationXeO64â+Xeg+H2O+O2g

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Under ambient conditions, which among the following surfactants will form micelles in aqueous solution at lowest molar concentration?

Options:
A CH3CH211N⊕CH33Br_
B CH3-CH213-OSO3-Na+
C CH3CH215N⊕CH33Br⊝
D CH3—CH28—COO⊝Na+

The correct answer is:
CH3CH215N⊕CH33Br⊝

Solution:
CMC is inversely related to the length of hydrophobic chain.
C16H33 N+CH33Br- forms cationic micelle with longest hydrophobic chain and at lowest CMC value.

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2015 (11 Apr Online)
Under an adiabatic process, the volume of an ideal gas gets doubled. Consequently, the mean collision time between the gas molecule changes from τ1 to τ2 . If CPCv=γ for this gas then a good estimate for τ2τ1 is given by

Options:
A 2
B 12
C 12γ
D 12γ+12

The correct answer is:
12γ+12

Solution:
Relaxation time τ∝VT
And T∝1Vγ-1
τ∝V1+γ-12
τ∝V1+γ2
τfτi=2VVγ+12
τfτi=2γ+12

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2020 (07 Jan Shift 2)
Under hydrolytic conditions, the compounds used for preparation of linear polymer and for chain termination, respectively, are

Options:
A CH3SiCl3 and SiCH34
B CH32SiCl2 and CH33SiCl
C CH32SiCl2 and CH3SiCl3
D SiCl4 and CH33SiCl

The correct answer is:
CH32SiCl2 and CH33SiCl

Solution:
Under hydrolytic conditions, the compound used for the preparation of linear polymer and for chain termination are (CH3)2SiCl2 and (CH3)3SiCl respectively.
The hydrolysis of dimethyldichlorosilane gives dimethyl silane diol which on condensation polymerization forms a linear polymer.
The hydrolysis of trimethylchlorosilane gives trimethyl silanol which can not undergo polymerization as it contains only one -OH group so it is used for chain termination.

Topic: Chemistry, Polymers, JEE Main
Under identical conditions, the $\mathrm{S}_{\mathrm{N}} 1$ reaction will occur most efficiently with

Options:
A tert-butyl chloride
B 1 -chlorobutane
C 2 -methyl- 1 -chloropropane
D 2 -chlorobutane

The correct answer is:
tert-butyl chloride

Solution:
$\mathrm{S}_{\mathrm{N}^{1}}$ reaction is most effective in $3^{\circ}$ carbon.

Topic: Chemistry, General Organic Chemistry, JEE Main
Under isothermal conditions, two soap bubbles of radii ' $r_{1}$ ' and ' $r_{2}$ ' coalesce to form a big drop. The radius of the big drop is

Options:
A $\left(r_{1}-r_{2}\right)^{\frac{1}{2}}$
B $\left(r_{1}+r_{2}\right)^{\frac{1}{2}}$
C $\left(r_{1}^{2}+r_{2}^{2}\right)^{\frac{1}{2}}$
D $\left(r_{1}^{2}-r_{2}^{2}\right)^{\frac{1}{2}}$

The correct answer is:
$\left(r_{1}^{2}+r_{2}^{2}\right)^{\frac{1}{2}}$

Solution:
Under isothermal conditions the surface tension remains constant. If $r$ is the radius of the digger drop, then Final surface energy $=$ Initial surface energy $\therefore 8 \pi r^{2} T=8 \pi r_{1}^{2} T+8 \pi r_{2}^{2} T$
$\therefore r^{2}=r_{1}^{2}+r_{2}^{2}$
$\therefore r=\left(r_{1}^{2}+r_{2}^{2}\right)^{\frac{1}{2}}$

Topic: Physics, Mechanical Properties of Fluids, MHT CET, MHT CET 2020 (12 Oct Shift 2)
Under isothermal conditions, two soap bubbles of radii ' $r_1$ ' and ' $r_2$ ' combine to forms a single soap bubble of radius ' $R$ '. The surface tension of soap solution is ( $\mathrm{P}=$ outside pressure)

Options:
A $\frac{\mathrm{P}\left(\mathrm{R}^3+\mathrm{r}_1^3+\mathrm{r}_2^3\right)}{4\left(\mathrm{r}_1^2-\mathrm{r}_2^2+\mathrm{R}^2\right)}$
B $\frac{\mathrm{P}\left(\mathrm{R}^2+\mathrm{r}_1^2+\mathrm{r}_2^2\right)}{4\left(\mathrm{r}_1^2-\mathrm{r}_2^2+\mathrm{R}^2\right)}$
C $\frac{P\left(R^3-r_1^3-r_2^3\right)}{4\left(r_1^2+r_2^2-R^2\right)}$
D $\frac{\mathrm{P}\left(\mathrm{R}^2-\mathrm{r}_1^2-\mathrm{r}_2^2\right)}{4\left(\mathrm{r}_1^3+\mathrm{r}_2^3-\mathrm{R}^3\right)}$

The correct answer is:
$\frac{P\left(R^3-r_1^3-r_2^3\right)}{4\left(r_1^2+r_2^2-R^2\right)}$

Solution:
Pressure inside the first bubble $=\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_1}$
Pressure inside the second bubble $=\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_2}$
Using the formula $\mathrm{PV}=\mathrm{nR} \theta \quad(\theta=$ absolute temp $)$
$\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_1}\right) \cdot \frac{4 \pi}{3} \mathrm{r}_1^3=\mathrm{n}_1 \mathrm{R} \theta \quad$ ( $\mathrm{R}$ is molar gas constant)
$\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_2}\right) \cdot \frac{4 \pi}{3} \mathrm{r}_2^3=\mathrm{n}_2 \mathrm{R} \theta$
and $\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{R}}\right) \cdot \frac{4 \pi}{3} \mathrm{R}^3=\left(\mathrm{n}_1+\mathrm{n}_2\right) \mathrm{R} \theta$
$\therefore\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{R}}\right) \cdot \frac{4 \pi}{3} \mathrm{R}^3=\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_1}\right) \cdot \frac{4 \pi}{3} \mathrm{r}_1^3+\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_2}\right) \cdot \frac{4 \pi \mathrm{r}_2^3}{3}$
$\therefore\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{R}}\right) \mathrm{R}^3=\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_1}\right) \mathrm{r}_1^3+\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_2}\right) \mathrm{r}_2^3$
on solving: $\mathrm{T}=\frac{\mathrm{P}\left(\mathrm{R}^3-\mathrm{r}_1^3-\mathrm{r}_2^3\right)}{4\left(\mathrm{r}_1^2+\mathrm{r}_2^2-\mathrm{R}^2\right)}$

Topic: Physics, Mechanical Properties of Fluids, JEE Main
Under similar conditions of pressure and temperature, 40 ml of slightly moist hydrogen chloride gas is mixed with 20 ml of ammonia gas, the final volume of gas at the same temperature and pressure will be

Options:
A 100 ml
B 20 ml
C 40 ml
D 60 ml

The correct answer is:
20 ml

Solution:
\(\mathrm{NH}_3+\mathrm{HCl} \rightarrow \mathrm{NH}_4 \mathrm{Cl}\)
According to the given question, \(1 \mathrm{ml} \mathrm{NH}_3\) reacts with 1 ml HCl to produce 1 ml of \(\mathrm{NH}_4 \mathrm{Cl}\).
So, 20 ml of ammonia reacts with 20 ml of HCl to produce 20 ml of resultant \(\mathrm{NH}_4 \mathrm{Cl}\). The rest of HCl will stay unreacted.
So, the correct option is '20 ml'.

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
Under standard conditions, the density of a gas is 1.3âmg cm-3 and the velocity of propagation of sound, in it, is 330 m s-1. The number of degrees of freedom of gas is

Options:

The correct answer is:
5

Solution:
Vsound=γPρ and γ=1+2f
∴ 1+2f=vsound2×ρP

1+2f=(330)2×(1.3)1.01×105

1+2f=1.40

f=5

Topic: Physics, Kinetic Theory of Gases, JEE Main
Under strong reaction conditions, what happens to tertiary alcohols during oxidation?

Options:
A They form carboxylic acids.
B They form ketones.
C They undergo dehydration.
D They remain unchanged.

The correct answer is:
They form carboxylic acids.

Solution:
Under strong reaction conditions, tertiary alcohols can undergo cleavage of various C-C bonds and form a mixture of carboxylic acids.

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
Under the influence of a uniform magnetic field, a charged particle moves with constant speed $\mathrm{V}$ in a circle of radius $R$. The time period of rotation of the particle :

Options:
A Depends on both $v$ and $R$
B Depends on $v$ and not on $R$
C Depends on R and not on $v$
D Is independent of both $v$ and $R$

The correct answer is:
Is independent of both $v$ and $R$

Solution:
$\mathrm{T}=\frac{2 \pi \mathrm{m}}{2 \mathrm{~B}}$
$\mathrm{T}$ is time period

Topic: Physics, Magnetic Effects of Current, JEE Main
Under the influence of a uniform magnetic field, a charged particle moves with constant speed $\mathrm{V}$ in a circle of radius $R$. The time period of rotation of the particle :

Options:
A Depends on both $v$ and $R$
B Depends on $v$ and not on $R$
C Depends on R and not on $v$
D Is independent of both $v$ and $R$

The correct answer is:
Is independent of both $v$ and $R$

Solution:
$\mathrm{T}=\frac{2 \pi \mathrm{m}}{2 \mathrm{~B}}$
$\mathrm{T}$ is time period

Topic: Physics, Magnetic Effects of Current, NEET, NEET 2009 (Mains)
Under the influence of force ' $F_1$ ' the body oscillates with a period ' $\mathrm{T}_1$ ' and due to another force ' $\mathrm{F}_2$ ' body oscillates with period ' $\mathrm{T}_2$ '. If both forces acts simultaneously then the resultant period is
(consider displacement is same in all three cases)

Options:
A $\mathrm{T}=\sqrt{\frac{\mathrm{T}_1^2+\mathrm{T}_2^2}{\mathrm{~T}_1^2 \mathrm{~T}_2^2}}$
B $\mathrm{T}=\sqrt{\frac{\mathrm{T}_1^2 \mathrm{~T}_2^2}{\mathrm{~T}_1^2+\mathrm{T}_2^2}}$
C $\mathrm{T}=\sqrt{\frac{\mathrm{T}_1^2}{\mathrm{~T}_2^2}}$
D $\mathrm{T}=\sqrt{\mathrm{T}_1^2+\mathrm{T}_2^2}$

The correct answer is:
$\mathrm{T}=\sqrt{\frac{\mathrm{T}_1^2 \mathrm{~T}_2^2}{\mathrm{~T}_1^2+\mathrm{T}_2^2}}$

Solution:
$\mathrm{F}=\mathrm{kx}$ and $\mathrm{k}=\mathrm{m} \omega^2$
$\therefore \quad F-m \omega^2 x$
$\therefore \quad \omega^2=\frac{F}{m x}$
For First Force, $\omega_1^2=\frac{\mathrm{F}_1}{\mathrm{mx}}$
For Second Force, $\omega_2^2=\frac{\mathrm{F}_2}{\mathrm{mx}}$
$\therefore \quad$ Resultant Force will be
$\omega_3^2=\frac{\mathrm{F}_1+\mathrm{F}_2}{\mathrm{mx}}$
$\omega_3^2=\omega_1^2+\omega_2^2$
$\frac{\omega_3^2}{4 \pi^2}=\frac{\omega_1^2}{4 \pi^2}+\frac{\omega_2^2}{4 \pi^2}$ ....(dividing by $4 \pi^2$ )
$\frac{1}{\mathrm{~T}_3^2}=\frac{1}{\mathrm{~T}_1^2}+\frac{1}{\mathrm{~T}_2^2}$
$\mathrm{T}_3^2=\sqrt{\frac{\mathrm{T}_1^2 \mathrm{~T}_2^2}{\mathrm{~T}_1^2+\mathrm{T}_{2 \ell}^2}}$

Topic: Physics, Oscillations, JEE Main
Under the same conditions, how many mL of 1 M KOH and 0.5 M H2SO4 solutions, respectively, when mixed to form at a total volume of 100 mL produces the highest rise in temperature?

Options:
A 67, 33
B 33, 67
C 40, 60
D 50, 50

The correct answer is:
50, 50

Solution:
2KOH(aq)2 mol+H2SO4(aq)1 molâK2SO4+H2O+Heat
Thus, maximum heat is liberated if 2 mol of KOH reacts with 1 mol H2SO4 (or if the ratio of the amounts of KOH to H2SO4 is 2 : 1).
Let x = volume of KOH, then 100 mL â x = volume of H2SO4
Number of mmol KOH =xÃ1 mmol of KOH
Number of mmol of H2SO4=(100-x)Ã0.5 mmol of H2SO4
Maximum heat is liberated if
x m mol of KOH(100-x)Ã0.5 m mol H2SO4=21
So x = 50 mL

Topic: Chemistry, Ionic Equilibrium, JEE Main
Under the same load, wire A having length 5.0 m and cross section 2.5×10-5 m2 stretches uniformly by the same amount as another wire B of length 6.0 m and a cross section of 3.0×10-5 m2 stretches. The ratio of the Young's modulus of wire A to that of wire B will be:

Options:
A 1:4
B 1:1
C 1:10
D 1:2

The correct answer is:
1:1

Solution:
Elongation in uniform wire under load is given by, Δl=FlSY, where S is area of cross-section.

As F is same for both wire and Δl is also same, we can write

ΔlF=lSY⇒lASAYA=lBSBYB

⇒52.5×10-5×YA=63×10-5×YB

⇒YAYB=1

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2023 (31 Jan Shift 2)
Under unfavourable conditions, many zooplankton species in lakes and ponds are known to enter

Options:
A Aestivation
B Migration
C Diapause
D Hibernation

The correct answer is:
Diapause

Solution:
Organism response to abiotic factors in different ways. If the stressful external conditions are localized or remains only for a short duration, than many zooplanktons species in lakes and ponds are known to enter diapauses, a stage of suspended development.

Topic: Biology, Organisms and Populations, NEET
Under what conditions is the tangent to a given curve at a point perpendicular to $\mathrm{x}$ -axis ?

Options:
A $\frac{d y}{d x}=0$
B $\frac{d y}{d x}=1$
C $\frac{d x}{d y}=0$
D $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=1 \quad$

The correct answer is:
$\frac{d x}{d y}=0$

Solution:
Tangent to a given curve at a point is perpendicular to
$\mathrm{x}$ -axis gives $\frac{\mathrm{dy}}{\mathrm{dx}}=\tan \frac{\pi}{2}$
$\Rightarrow \quad \frac{\mathrm{dx}}{\mathrm{dy}}=\cot \frac{\pi}{2}=0$

Topic: Mathematics, Application of Derivatives, JEE Main
Under which condition the gene ratio remains constant for any species?

Options:
A Sexual selection
B Random mating
C Mutation
D Gene flow

The correct answer is:
Random mating

Solution:
Genetic equilibrium means that the frequency of occurrence of alleles of a gene is supposed to remain fixed and even remain the same through generations. Gene frequencies tend to remain constant from generation to generation when disturbing factors are not present. Factors that disturb the natural equilibrium of gene frequencies include mutation, migration (or gene flow), random genetic drift, and natural selection.
Related Theory
Random mating involves the mating of individuals regardless of any physical, genetic or social preference. In other words, the mating between two organisms is not influenced by any environmental, hereditary or social interaction. Hence, potential mates have an equal chance of being selected.

Topic: Biology, Organisms and Populations, NEET, NEET 2002
Under which of the following condition is the relation $\Delta \mathrm{H}=\Delta \mathrm{E}+\mathrm{P} \Delta \mathrm{V}$ valid for a closed system?

Options:
A Constant Pressure
B Constant temperature
C Constant temperature and pressure
D Constant temperature, pressure and composition

The correct answer is:
Constant Pressure

Solution:
This is applicable when pressure remains constant.

Topic: Chemistry, Thermodynamics (C), WBJEE, WBJEE 2009
Under which of the following conditions will there be no change in the reading frame of the following mRNA?
5'AACAGCGGUGCUAUU 3'

Options:
A Insertion of G at 5th position
B Deletion of G from 5th position
C Insertion of A and G at 4th and 5th positions respectively
D Deletion of GGU from 7th 8th and 9th positions

The correct answer is:
Deletion of GGU from 7th 8th and 9th positions

Solution:
Genetic codes are read in the sequence of 3. Insertion or deletion of a single base changes the reading frame of codon and is called frameshift mutation. However, insertion or deletion of three or its multiple bases insert or delete in one or multiple codon hence one or multiple amino acids and reading frame remains unaltered from that point onwards.

Topic: Biology, Molecular Basis of Inheritance, NEET, NEET 2019 (All India)
Under which one of the following condition does the system of equations
$\begin{array}{l}
k x+y+z=k-1 \\
x+k y+z=k-1 \\
x+y+k z=k-1
\end{array}$
have no solution?

Options:
A $k=1$
B $k \neq-2$
C $k=1$ or $k=-2$
D $k=-2$

The correct answer is:
$k=1$ or $k=-2$

Solution:
The given system of equations are $\mathrm{kx}+\mathrm{y}+\mathrm{z}=\mathrm{k}-1$
$x+k y+z=k-1$
$\mathrm{x}+\mathrm{y}+\mathrm{kz}=\mathrm{k}-1$
$A=\left[\begin{array}{lll}k & 1 & 1 \\ 1 & k & 1 \\ 1 & 1 & k\end{array}\right], B=\left[\begin{array}{l}k-1 \\ k-1 \\ k-1\end{array}\right]$ and $\mathrm{x}=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$
Now, $|A|=\left|\begin{array}{lll}k & 1 & 1 \\ 1 & k & 1 \\ 1 & 1 & k\end{array}\right|$
$=k\left(k^{2}-1\right)-1(k-1)+1(1-k)$
$=k^{3}-k-k+1+1-k$
$=k^{3}-3 k+2$
The given system of equations has no solution, if
$|A|=0$
$\Rightarrow k^{3}-3 k+2=0$
$\Rightarrow(k-1)^{2}(k+2)=0$
$\Rightarrow k=1$ or $k=-2$

Topic: Mathematics, Determinants, JEE Main
Underground stems of some plants spread to new niches, and when older parts die new plants are formed. An example of such a modification is

Options:
A Banana and pineapple
B Jasmine and mint
C Grass and strawberry
D Chrysanthemum and Pistia

The correct answer is:
Grass and strawberry

Solution:
Underground stems of some plants such as grass and strawberry, etc., spread to new niches, and when older parts die new plants are formed. In plants like mint and jasmine, a slender lateral branch arises from the base of the main axis and after growing aerially for some time arch downwards to touch the ground. A lateral branch with short internodes and each node bearing a rosette of leaves and a tuft of roots are found in aquatic plants like Pistia and Eichhornia. In banana, pineapple, and Chrysanthemum, the lateral branches originate from the basal and underground portion of the main stem, grow horizontally beneath the soil and then come out obliquely upward giving rise to leafy shoots.

Topic: Biology, Morphology of Flowering Plants, NEET
Underline the correct alternative:
(a) When a conservative force does positive work on a body, the potential energy of the body increases/ decreases/remains unaltered.
(b) Work done by a body against friction always results in a loss of its kinetic/potential energy.
(c) The rate of change of total momentum of a many particle system is proportional to the external force/ sum of the internal forces on the system.
(d) In an inelastic collision of two bodies, the quantities which don't change after the collision are the total kinetic energy/total linear momentum/total energy of the system of two bodies.

Options:

None

Solution:
(a) Potential energy of the body decreases. The conservative force does positive work on a body, when it displaces the body in the direction of the force. The body therefore approaches the centre of force thus, decreasing $x$. Hence P.E. decreases.
(b) Work is done by a body against friction at the expense of its kinetic energy. Hence work done by a body against friction results in a loss of its kinetic energy.
(c) Internal forces cannot change the total or net momentum of a system. Hence the rate of change of total momentum of many particle system is proportional to the external force on the system.
(d) In an elastic collision of two bodies the quantities which do not change after the collision are total linear momentum and total energy of the system of two bodies (if the system is isolated). The total K.E. of the system, is not conserved as it may change to an equivalent amount of energy in some other form.

Topic: Physics, Work Power Energy
Uniform ring of radius $\mathrm{R}$ is moving on a horizontal surface with speed $\mathrm{v}$ and then climbs up a ramp of inclination $30^{\circ}$ to a hcight $h$. Thcre is no slipping in the cntirc motion. Then h is

Options:
A $\mathrm{v}^{2} / 2 \mathrm{~g}$
B $\mathrm{v}^{2} / \mathrm{g}$
C $3 \mathrm{v}^{2} / 2 \mathrm{~g}$
D $2 \mathrm{v}^{2} / \mathrm{g}$

The correct answer is:
$\mathrm{v}^{2} / \mathrm{g}$

Solution:
\{Using conservation of energy)
$$
\begin{array}{l}
\frac{1}{2} \mathrm{mV}^{2}+\frac{1}{2} \mathrm{mR}^{2}\left(\frac{\mathrm{V}}{\mathrm{R}}\right)^{2}=\mathrm{mgh} \\
\mathrm{m}\left(\frac{\mathrm{V}^{2}}{2}+\frac{\mathrm{V}^{2}}{2}\right)=\mathrm{mgh} \\
\mathrm{h}=\frac{\mathrm{V}^{2}}{\mathrm{~g}}
\end{array}
$$

Topic: Physics, Work Power Energy, JEE Main
Unique features of Bryophytes is that they:

Options:
A Produce spores
B Have sporophyte attached to gametophyte
C Lack roots
D Lack vascular tissues

The correct answer is:
Have sporophyte attached to gametophyte

Solution:
In Bryophytes, the plant body is gametophyte and sporophyte remain attached to the gametophyte and produces spores. The sporophyte is dependent on the gametophyte.

Topic: Biology, Plant Kingdom, NEET
Unisexuality of flowers prevents

Options:
A Autogamy and geitonogamy
B Autogamy, but not geitonogamy
C Both geitonogamy and xenogamy
D Geitonogamy, but not xenogamy

The correct answer is:
Autogamy, but not geitonogamy

Solution:
Autogamy but not geitnogamy : Unisexuality promotes or favours the cross pollination (xenogamy) and prevent self pollination (Autogamy). Geitnogamy involve transfer of pollen grain between two flowers of the same plant so unisexuality will not prevent geitnogamy.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET, NEET 2008 (Mains)
Unit mass of a liquid with volume $V_1$ is completely changed into a gas of volume $V_2$ at a constant external pressure $P$ and temperature $T$. If the latent heat of evaporation for the given mass is $L$, then the increase in the internal energy of the system is

Options:
A Zero
B $P\left(V_2-V_1\right)$
C $L-P\left(V_2-V_1\right)$
D $L$

The correct answer is:
$L-P\left(V_2-V_1\right)$

Solution:
$\begin{aligned} & \Delta Q=\Delta V+P \Delta V \Rightarrow m L=\Delta U+P\left(V_2-V_1\right) \\ & \Rightarrow \Delta U=L-P\left(V_2-V_1\right)~(\therefore m=1)\end{aligned}$

Topic: Physics, Thermodynamics, JEE Main
Units of \( a \) in Van der Waal's equation of state is

Options:
A \( \mathrm{N} \mathrm{m}^{4} \mathrm{~mol}^{-1} \)
B \( \mathrm{N} \mathrm{m}^{2} \mathrm{~mol}^{-1} \)
C \( \mathrm{N}^{2} \mathrm{~m} \mathrm{~mol}^{-1} \)
D \( \mathrm{N} \mathrm{m}^{4} \mathrm{~mol}^{-2} \)

The correct answer is:
\( \mathrm{N} \mathrm{m}^{4} \mathrm{~mol}^{-2} \)

Solution:
Van der Waals equation is given as P+an2V2V−nb=nRT, here P is pressure, V is volume, T is temperature, n is the number of moles, and R is the universal gas constant.

The dimensional formula of pressure isP=ForceArea=MLT−2L2=ML-1T−2

The dimensional formula of the volume is L3

As we know, the quantities with the same dimensions are added or subtracted. Thus, the dimension of P=the dimension of an2V2.

Therefore, a=PV2n2=ML-1T-2L32mol2=ML5T-2mol-2

Thus, the unit of constant a is N m4 mol-2.

Topic: Physics, Units and Dimensions, JEE Main
Units of rate constant of first and zero order reactions in terms of molarity M unit are respectively

Options:
A
$\mathrm{sec}^{-1}, \mathrm{Msec}^{-1}$

B
$\mathrm{sec}^{-1}, \mathrm{M}$

C
Msec $^{-1}, \sec ^{-1}$

D
$\mathrm{M}, \sec ^{-1}$

The correct answer is:
$\mathrm{sec}^{-1}, \mathrm{Msec}^{-1}$

Solution:
$\mathrm{K}=\left(\mathrm{mol} \mathrm{~L}^{-1}\right)^{1-\mathrm{n}} \mathrm{sec}^{-1}, \mathrm{n}=0,1$

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2002
Unlike dry cells, the mercury cell has a constant cell potential throughout its useful life. Why?

Options:
A The mercury cell uses a solid electrolyte, which maintains constant ionic conductivity.
B The mercury cell has a high initial voltage, which remains unchanged over time.
C Because Ions are not involved in the overall cell reaction of mercury cells.
D The mercury cell contains a constant supply of mercury, ensuring a consistent potential.

The correct answer is:
Because Ions are not involved in the overall cell reaction of mercury cells.

Solution:
Correct Option is : (C)
Because Ions are not involved in the overall cell reaction of mercury cells.

Topic: Chemistry, Electrochemistry, NEET