datavorous/entrance-exam-dataset huggingface dataset:
As shown in the figure, a network of resistors is connected to a battery of 24 V with an internal resistance of 3 Ω. The currents through the resistors R4 and R5 are I4 and I5 respectively. The values of I4 and I5 are:

Options:
A I4=85 A and I5=25 A
B I4=245 A and I5=65 A
C I4=65 A and I5=245 A
D I4=25 A and I5=85 A

The correct answer is:
I4=25 A and I5=85 A

Solution:
Equivalent resistance of circuit
Req=3+2×22+2+2+20×520+5+2=12 Ω

Current through battery i=2412=2 A

Now, I4=R5R4+R5×2=520+5×2=25 A

and I5=i-I4=2-25=85 A

Topic: Physics, Current Electricity, JEE Main, JEE Main 2023 (24 Jan Shift 1)
As shown in the figure, a particle is moving with constant speed π m s-1. Considering its motion from A to B, the magnitude of the average velocity is:

Options:
A 3 m s-1
B π m s-1
C 1.53 m s-1
D 23 m s-1

The correct answer is:
1.53 m s-1

Solution:
The given data is

v=π m s-1

By using trigonometric ratios in the triangle shown above the displacement can be found out by

sin60°=AB2R⇒AB=R3

The formula for distance is

d=vt⇒t=dv⇒t=120°360°×2πRπ=2R3

So, the average velocity is

vavg=3R2R3=332

=1.53 m s-1

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2023 (06 Apr Shift 2)
As shown in the figure, a particle of mass m=100 gm is attached with four identical springs, each of length l=10 cm. Initial tension in each spring is F0=25 N. Neglecting gravity, calculate the angular frequency (in rad s-1) of small oscillations of the particle along a line perpendicular to the plane of the figure.

Options:

The correct answer is:
100

Solution:
For small x we can assume the tension in each spring to remain constant

F=-4F0sin⁡θ≈-4F0xl

a=-4F0mlx

ω=4F0ml=100 rad s-1

Topic: Physics, Oscillations, JEE Main
As shown in the figure, a plane mirror is fixed at a height of 50 cm from the bottom of tank containing water μ=43. The height of water in the tank is 8 cm. A small bulb is placed at the bottom of the water tank. The distance of image of the bulb formed by mirror from the bottom of the tank is ______ cm.

Options:

The correct answer is:
98

Solution:
The data given is

d=8 cmμ=43

Apparent depth is

h'=dμ=843=6 cm
Distance of object from mirror =42+6=48 m
Distance of image from bottom of tank =48+50=98 m

Topic: Physics, Ray Optics, JEE Main, JEE Main 2023 (11 Apr Shift 2)
As shown in the figure, a point charge Q is placed at the centre of conducting spherical shell of inner radius a and outer radius b. The electric field due to charge Q in three different regions I, II and III is given by : (I : r < a, II : a < r < b, III : r > b)

Options:
A EI= 0, EII=0, EIII≠0
B EI≠0, EII=0, EIII≠0
C EI≠0, EII=0, EIII=0
D EI=0, EII=0, EIII=0

The correct answer is:
EI≠0, EII=0, EIII≠0

Solution:
In region I, due to charge at centre, electric field will be present.

But in region II, electric field will be zero as electric field inside material of conductor is zero.

Again for outside region, net charges enclosed is not zero and therefore electric field will exist.

Topic: Physics, Electrostatics, JEE Main, JEE Main 2023 (30 Jan Shift 2)
As shown in the figure, a potentiometer wire of resistance 20 Ω and length 300 cm is connected with resistance box (R.B.) and a standard cell of emf 4 V. For a resistance 'R' of resistance box introduced into the circuit, the null point for a cell of 20 mV is found to be 60 cm. The value of 'R' is _____ Ω.

Options:

The correct answer is:
780

Solution:
Given here, l=300 cm, Rw=20 Ω and ε=4 V.

Using the relation, E2=l1l2E1, we have

E=ACABVA-VB

∴20×10-3=60300×4×20R+20⇒5×10-3=4R+20

∴R=780 Ω

Topic: Physics, Current Electricity, JEE Main, JEE Main 2022 (28 Jul Shift 1)
As shown in the figure, a single conducting wire is bent to form a loop in the form of a circle of radius 'r' concentrically inside a square of side ' $a$ ', where $a: r=8: \pi$. A battery B drives a current through the wire. If the battery B and the gap G are of negligible sizes, determine the strength of magnetic field at the common centre $O$.

Options:
A $\cdot \frac{\mu_{0} \mid}{2 \pi a} \sqrt{2}(\sqrt{2}-1)$
B $\frac{\mu_{0} I}{2 \pi a}(\sqrt{2}+1)$
C $\frac{\mu_{0} I}{\pi a} 2 \sqrt{2}(\sqrt{2}+1)$
D $\frac{\mu_{0} I}{\pi \mathrm{a}} 2 \sqrt{2}(\sqrt{2}-1)$

The correct answer is:
$\frac{\mu_{0} I}{\pi \mathrm{a}} 2 \sqrt{2}(\sqrt{2}-1)$

Solution:
Hint:
$\frac{a}{r}=\frac{8}{\pi}, r=\frac{\pi a}{8}$
$B=\frac{\mu_{0} \mid}{2 r}-\frac{4 \mu_{0} I}{4 \pi a / 2} \times \sqrt{2}=\frac{\mu_{0} I}{2 r}-\frac{2 \sqrt{2} \mu_{0} I}{\pi a}=\frac{\mu_{0} \mid \times 8}{2 \pi a}-\frac{2 \sqrt{2} \mu_{0} I}{\pi a}$
$=\frac{\mu_{0} \mid}{\pi a}(4-2 \sqrt{2})=\frac{\mu_{0} I}{\pi a} 2 \sqrt{2}(\sqrt{2}-1)$

Topic: Physics, Magnetic Effects of Current, JEE Main
As shown in the figure, after passing through the medium 1. The speed of light v2 in medium 2 will be :
(Given c=3×108 m s-1 )

Options:
A 1.0×108 m s-1
B 0.5×108 m s-1
C 1.5×108 m s-1
D 3.0×108 m s-1

The correct answer is:
1.0×108 m s-1

Solution:
The refractive index of a medium is given by, n=μrεr.

For medium 2, n2=1×9=3

Now speed of light in a medium will be,

cv=n⇒v=3×1083=1×108 m s-1

Topic: Physics, Ray Optics, JEE Main, JEE Main 2022 (28 Jul Shift 1)
As shown in the figure, an equilateral triangle $A B C$ is formed by joining three rods of equal lengths and $D$ is the midpoint of $A B$.
Coefficient of linear expansion of the material of $A B$ is $\alpha_1$ and that of $A C$ and $B C$ is $\alpha_2$. If the length $D C$ remains constant for small changes in temperature, then

Options:
A $\alpha_1=\alpha_2$
B $\alpha_1=4 \alpha_2$
C $\alpha_2=4 \alpha_1$
D $\alpha_1=\frac{\alpha_2}{2}$

The correct answer is:
$\alpha_1=4 \alpha_2$

Solution:
For a change of temperature by $t$, change in length $D C$ is $\triangle D C$, then $\triangle D C^2=\triangle A C^2-\triangle A D^2$
$$
\begin{aligned}
& =l\left(1+\alpha_2 t\right)^2-\left(\frac{l}{2}\left(1+\alpha_2 t\right)\right)^2 \\
& =l^2\left(2 \alpha_2 t\right)-\frac{l^2}{4}\left(2 \alpha_1 t\right)
\end{aligned}
$$
(After neglecting terms $\alpha_2^2 t^2$ and $\alpha_1^2 t^2$, being very small)
As, $\quad \Delta D C=0 \Rightarrow l^2 2 \alpha_2 t-\frac{l^2}{4}\left(2 \alpha_1 t\right)=0$
$$
\Rightarrow \quad \alpha_1=4 \alpha_2
$$

Topic: Physics, Thermal Properties of Matter, AP EAMCET, AP EAMCET 2018 (22 Apr Shift 2)
As shown in the figure an inductor of inductance 200 mH is connected to an AC source of emf 220 Vand frequency 50 Hz. The instantaneous voltage of the source is 0 V when the peak value of current is aπ A. The value of a is _____ .

Options:

The correct answer is:
242

Solution:
The reactance of the given inductor is,

XL=ωL=2πfL=2π×50×0.2=20π Ω.

The maximum current in the given circuit will be,

i0=2VrmsXL=2×22020π=242π A

Therefore, a=242

Topic: Physics, Alternating Current, JEE Main, JEE Main 2022 (24 Jun Shift 1)
As shown in the figure, forces of $10^5 \mathrm{~N}$ each are applied in opposite directions, on the upper and lower faces of a cube of side $10 \mathrm{~cm}$, shifting the upper face parallel to itself by $0.5 \mathrm{~cm}$. If the side of another cube of the same material is, $20 \mathrm{~cm}$ then under similar conditions as above, the displacement will be:

Options:
A
$1.00 \mathrm{~cm}$

B
$0.25 \mathrm{~cm}$

C
$0.37 \mathrm{~cm}$

D
$0.75 \mathrm{~cm}$

The correct answer is:
$0.25 \mathrm{~cm}$

Solution:
For same material the ratio of stress to strain is same For first cube
$$
\text { Stress }_1=\frac{\text { force }_1}{\text { area }_1}=\frac{10^5}{\left(0.1^2\right)}
$$
For second block,
$$
\begin{aligned}
&\text { stress }_2=\frac{\text { force }_2}{\text { area }_2}=\frac{10^5}{\left(0.2^2\right)} \\
&\text { strain }_2=\frac{\text { change in length }{ }_2}{\text { original length }}=\frac{x}{0.2}
\end{aligned}
$$
$x$ is the displacement for second block.
For same material, $\frac{\text { stress }_1}{\text { strain }_1}=\frac{\text { stress }_2}{\text { strain }_2}$
$$
\text { or, } \frac{\frac{10.5}{(0.1)^2}}{\frac{0.5 \times 10^{-2}}{0.1}}=\frac{\frac{10^5}{(0.2)^2}}{\frac{x}{0.2}}
$$
$$
\text { Solving we get, } x=0.25 \mathrm{~cm}
$$

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
As shown in the figure the voltmeter reads 2 V across 5 Ω resistor. The resistance of the voltmeter is _____ Ω

Options:

The correct answer is:
20

Solution:
Let the resistance of the voltmeter be R Ω. The voltmeter is in parallel with the 5 Ω. So the total resistance in the circuit is

5R5+R+2 Ω

Applying Ohm's law in the whole circuit, V=IR,

I=3 2+5R5+R A

It is given that the potential difference across 5 Ω is 2 V.

Thus, by Ohm's law,

32+5R5+R×5RR+5=2⇒15+3R10+7R×5R5+R=2⇒15R=20+14R⇒R=20 Ω

Topic: Physics, Current Electricity, JEE Main, JEE Main 2023 (06 Apr Shift 2)
As shown in the figure, two infinitely long, identical wires are bent by 90o and placed in such a way that the segments LP and QM are along the x - axis, while segments PS and QN are parallel to the y - axis. If OP=OQ=4cm, and the magnitude of the magnetic field at O is 10-4 T, and the two wires carry equal currents (see figure), the magnitude of the current in each wire and the direction of the magnetic field at O will be μ0=4π×10-7NA-2:

Options:
A 20 A, perpendicular into the page
B 20 A, perpendicular out of the page
C 40 A, perpendicualr into the page
D 40 A, perpendicular out of the page

The correct answer is:
20 A, perpendicular into the page

Solution:
As we know that the magnetic field due to a infinite wire at a distance of d is given as

B=μ0i4πd

but there are two wires so that the net magnetic field is given as

B=2μ0i4πd
10-4=24π×10-7×i4π×4100
i=20A

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2019 (12 Jan Shift 1)
As shown in the figure, two particles, each of mass 'm' tied at the ends of a light string of length 2a are kept on a frictionless horizontal surface. When the mid point (P) of the string is pulled vertically upwards with a small but constant force F, the particles move towards each other on the surface. Magnitude of acceleration of each particle, when the separation between them becomes 2x is

Options:
A F2maa2-x2
B F2mxa2-x2
C F2mxa
D F2ma2-x2x

The correct answer is:
F2mxa2-x2

Solution:
From the above diagram,

tanθ=a2-x2x, 2Tsinθ=F and Tcosθ=ma

from the above equations,

a=Fcotθ2m

⇒a=F2m1tanθ=F2mxa2-x2

Topic: Physics, Laws of Motion, JEE Main
As shown in the figure, wheel A of radius rA=10 cm is coupled by a belt B to the wheel C of radius rC=25 cm. The angular speed of wheel A is increased from rest at a constant rate of 1.6 rad s-2. Time after which wheel C reaches a rotational speed of 100 rpm, assuming the belt does not slip, is nearly

Options:
A 4 s
B 8 s
C 12 s
D 16 s

The correct answer is:
16 s

Solution:
at=αA rA=αCrC
αC= αA rArC= 1.6×1025=0.64 rad s-2
t=ωCαC=100×2π600.64=16.35 s

Topic: Physics, Rotational Motion, JEE Main
As shown in the following figures, block of mass 2 kg is in equilibrium in the vertical plane.

At a certain instant right spring in case-1 and right string in case-2 are cut. Then the ratio of instantaneous acceleration of block in case-1 to case-2 just after the cutting is (strings and springs are ideal)

Options:
A 0
B 1
C 2524
D none of these

The correct answer is:
2524

Solution:
2Tsin⁡37°=20
2T×35=20
T=503 N
Now when spring is cut in case-1 then net force F1 acting on block is,
F1=202+5032-2×20×503×35 (resultant force of the remaining T=503 N and mg=20 N forces)
=503 N
Similarly, when the string is cut in case-2 then net force F2 acting on block is F2=20×45=16 N
∴a1a2=503×116=5048=2524

Topic: Physics, Laws of Motion, JEE Main
As shown in the schematic below, a rod of uniform cross-sectional area A and length l is carrying a constant current i through it and voltage across the rod is measured using an ideal voltmeter. The rod is stretched by the application of a force F.

Which of the following graphs would show the variation in the voltage across the rod as function of the strain ε when the strain is small. Neglect Joule heating.

Options:
A
B
C
D

The correct answer is:

Solution:
When rod is stretched, its length increases. Potential drop across the rod also increases due to increase in
resistance of rod.

Resistance of rod, R=ρlA=ρl2X
(∵ volume of rod, X=Al)

Change in resistance of rod, ΔR=ρ.2lΔlX

Change in potential drop across rod,

ΔV=iΔR=i.ρ.2lΔlX=2iρl2X.Δll=2iρl2X.ε

or ΔV∝ε

As change in potential drop is directly proportional to strain, voltage as a function of strain is as shown below;

Topic: Physics, Current Electricity, JEE Main
As the atomic number of the alkaline earth metal ion increases, what happens to the solubility of their carbonates in water?

Options:
A Increases
B Decreases
C Remains constant
D Fluctuates

The correct answer is:
Decreases

Solution:
The solubility of alkaline earth metal carbonates in water decreases as the atomic number of the metal ion increases. This trend is due to the increasing cationic size and decreasing charge density as we move down the group.

Topic: Chemistry, s Block Elements, NEET
As the electron in Bohr's orbit of hydrogen atom passes from state $n=2$ to, $n=1$, the $\mathrm{KE}(\mathrm{K})$ and the potential energy (U) changes as

Options:
A $K$ four fold, $U$ also four fold
B $K$ two fold, $V$ also two fold
C $K$ four fold, $U$ two fold
D $K$ two fold, $U$ four fold

The correct answer is:
$K$ four fold, $U$ also four fold

Solution:
$\mathrm{KE}$ of an electron in nth orbit : $\mathrm{K}_{\mathrm{n}} \propto \frac{1}{\mathrm{n}^{2}}$ and PE of an electron in nth orbit :
$\mathrm{U}_{\mathrm{n}} \propto \frac{1}{\mathrm{n}^{2}}$
$\therefore$ When an electron passes from state $\mathrm{n}=2$ to $\mathrm{n}=1$ $\frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}=\frac{1^{2}}{2^{2}}=\frac{1}{4}$
$\begin{array}{l}
\text { or } \mathrm{K}_{1}=4 \mathrm{~K}_{2} \\
\frac{\mathrm{U}_{2}}{\mathrm{U}_{1}}=\frac{\mathrm{l}^{2}}{2^{2}}=\frac{1}{4} \\
\text { or } \mathrm{U}_{1}=4 \mathrm{U}_{2}
\end{array}$

Topic: Physics, Atomic Physics, JEE Main
As the quantum number increases, the difference in energy between consecutive energy levels

Options:
A Remains the same
B Increases
C Decreases
D Sometimes increases and sometimes decreases

The correct answer is:
Decreases

Solution:
$E_{n+1}-E_n=13.6\left[\frac{1}{n^2}-\frac{1}{(n+1)^2}\right]$
$\begin{aligned} & \Delta \mathrm{E}_{\mathrm{n}}=13.6\left[\frac{(\mathrm{n}+1)^2-\mathrm{n}^2}{\mathrm{n}^2(\mathrm{n}+1)^2}\right] \\ & =13.6\left[\frac{1+2 \mathrm{n}}{\mathrm{n}^2(\mathrm{n}+1)^2}\right]\end{aligned}$
For large value of $n$
$\Delta \mathrm{E}_{\mathrm{n}}=13.6\left[\frac{2 \mathrm{n}}{\mathrm{n}^4}\right]$
$\begin{aligned} & \propto \frac{1}{\mathrm{n}^3} \\ & \text { So, } \mathrm{n} \uparrow \uparrow \Rightarrow \Delta \mathrm{E}_{\mathrm{n}} \downarrow \downarrow\end{aligned}$

Topic: Physics, Atomic Physics, AP EAMCET, AP EAMCET 2022 (08 Jul Shift 1)
As the size of alkyl/aryl groups in alcohols increases, their solubility in water generally:

Options:
A Increases
B Decreases
C Remains unchanged
D Becomes unpredictable

The correct answer is:
Decreases

Solution:
Solubility decreases with the increase in the size of hydrophobic alkyl/aryl groups.

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
As the temperature increases average kinetic energy of molecules increases. What would be the effect of increase of temperature on pressure provided the volume is constant?

Options:
A
Increases

B
Decreases

C
Remains same

D
Becomes half

The correct answer is:
Increases

Solution:
From Gay-Lussac's law, at constant volume, as the temperature is increased, pressure increases.

Topic: Chemistry, States of Matter
As the temperature is increased, the period of a pendulum

Options:
A
increases as its effective length increases even though its centre of mass still remains at the centre of the bob

B
decreases as its effective length increases even though its centre of mass still remains at the centre of the bob

C
increases as its effective length increases due to shifting to centre of mass below the centre of the bob

D
decreases as its effective length remains same but the centre of mass shifts above the centre of the bob

The correct answer is:
increases as its effective length increases even though its centre of mass still remains at the centre of the bob

Solution:
As the temperature increased the length $(L)$ of the pendulum increases due to expansion is linear.
So, Time period of pendulum
$$
\begin{aligned}
&T=2 \pi \sqrt{\frac{L}{g}} \\
&T \propto \sqrt{L}
\end{aligned}
$$
Hence on increasing temperature, time period $(T)$ also increases.

Topic: Physics, Thermal Properties of Matter
As the temperature is raised from $20^{\circ} \mathrm{C}$ to $40^{\circ} \mathrm{C}$, the average kinetic energy of neon atoms changes by a factor of which of the following?

Options:
A
$1 / 2$

B
2

C
$\frac{313}{293}$

D
$\sqrt{\frac{313}{293}}$

The correct answer is:
$\frac{313}{293}$

Solution:
$\frac{E_1}{E_2}=\frac{T_1}{T_2}$
$\frac{E_1}{E_2}=\frac{293}{313}$
$\because$ factor $=\frac{313}{293}$

Topic: Chemistry, States of Matter, JEE Main, JEE Main 2004
As the temperature of water increases, its viscosity

Options:
A Remains unchanged
B Decreases
C Increases
D Increases or decreases depending on the external pressure

The correct answer is:
Decreases

Solution:
Correct option is 2. Decreases
The energy that is caused by increased temperature makes the molecules move at a faster rate to a level where they overcome the bonds or the binding forces of the molecules. This makes the liquid more fluid decreasing its viscosity.

Topic: Physics, Mechanical Properties of Fluids, JEE Main
As we go from species to kingdom in a taxonomic hierarchy, the number of common characteristics:

Options:
A
will decrease

B
will increase

C
remain same

D
may increase or decrease

The correct answer is:
will decrease

Solution:
Lower the taxa, more are the number of shared characteristics within the members of the taxon. So, the lowest taxon shares the maximum number of morphological similarities. As we move towards the higher hierarchy, i.e., class, kingdom, similarities decrease.

Topic: Biology, The Living World
As we go from species to kingdom in a taxonomic hierarchy, the number of common characteristics

Options:
A will increase
B will decrease
C remains the same
D may increase or decrease

The correct answer is:
will decrease

Solution:
On moving up the taxonomical hierarchy number of common characteristics between organism decreases but number on organism goes on increasing. Kingdom has maximum number of organism but similarities between these organism is minimum.

Topic: Biology, The Living World, NEET
As we go from the equator of the earth to pole of the earth, the value of acceleration due to gravity

Options:
A decreases
B remains same.
C decreases up to latitude of $45^{\circ}$ and increases thereafter.
D increases.

The correct answer is:
increases.

Solution:
$(A)$
As centrifugal force decreases.

Topic: Physics, Gravitation, MHT CET, MHT CET 2020 (16 Oct Shift 2)
As we go from the equator to the poles, the value of acceleration due to gravity (g)

Options:
A remains the same
B decreases
C increases
D decreases upto a latitude of 45°

The correct answer is:
increases

Solution:
The radius of the earth at poles is less than equator therefore as we go from the equator to the poles, the value of g increases.
Also as we go from the equator to the poles, the effect of rotation decreases and becomes zero at pole so the value of g increases.

Topic: Physics, Gravitation, NEET
As you move down the group of alkaline earth metals, what happens to the tendency to form halide hydrates?

Options:
A It increases
B It decreases
C It remains constant
D It varies unpredictably

The correct answer is:
It decreases

Solution:
The tendency to form halide hydrates gradually decreases down the group of alkaline earth metals.

Topic: Chemistry, s Block Elements, NEET
As you move from lithium (Li) to cesium (Cs) within the alkali metal group, what happens to the atomic size?

Options:
A It decreases
B It remains constant
C It increases
D It fluctuates randomly

The correct answer is:
It increases

Solution:
The atomic and ionic radii of alkali metals increase on moving down the group, from lithium (Li) to cesium (Cs). This trend occurs because additional energy levels are added with each successive element, leading to larger atomic sizes.

Topic: Chemistry, s Block Elements, NEET
Ascaris is characterized by

Options:
A Presence of true coclom and metamerism (metamerisation)
B Absence of true coclom but presence of metamerism
C Presence of neither true coclom nor metamerism
D Presence of true coclom but absence of metamerism

The correct answer is:
Presence of neither true coclom nor metamerism

Solution:
No Solution Available

Topic: Biology, Animal Kingdom, NEET, NEET 2008 (Mains)
Ascaris is characterized by the

Options:
A presence of neither true coelom nor metamerism.
B presence of true coelom but absence of metamerism.
C presence of both true coelom and metamerism.
D absence of true coelom but presence of metamerism.

The correct answer is:
presence of neither true coelom nor metamerism.

Solution:
Ascaris belong to Phylum Aschelminthes. The main distinguishing feature is that sexes are separate. These are known as roundworms as they show presence of body cavity. These show well presence of organ system and hence digestive system is also complete. These can be free living or parasitic in nature. The parasites can be in ectoparasites or endoparasites depending on whether they are on the host or inside the host. These show presence of pseudocoloem as mesoderm is in the form of pouches. Hence true coloem is absent and these organisms also lack metamerism.

Topic: Biology, Animal Kingdom, NEET
Ascent of xylem sap, caused due to transpiration, depends mainly on three physical properties of water. These properties result in

Options:
A low tensile strength but high capillarity.
B low tensile strength as well as low capillarity.
C high tensile strength as well as high capillarity.
D high tensile strength but low capillarity.

The correct answer is:
high tensile strength as well as high capillarity.

Solution:
The transpiration driven ascent of xylem sap depends mainly on the following physical properties of water:
• Cohesion – mutual attraction between water molecules.
• Adhesion – the attraction of water molecules to polar surfaces (such as the surface of tracheary elements).
• Surface Tension – water molecules are attracted to each other in the liquid phase more than to water in the gas phase.
These properties give water high tensile strength, i.e., an ability to resist a pulling force, and high capillarity, i.e., the ability to rise in thin tubes. In plants, capillarity is aided by the small diameter of the tracheary elements – the tracheids and vessel elements.

Topic: Biology, Transport in Plants, NEET
Asexual reproduction mainly takes place in

Options:
A Higher animals
B Lower animals
C Plants
D All the above

The correct answer is:
Lower animals

Solution:
Asexual reproduction basically occurs in prokaryotes like bacteria and in some eukaryotes which are single-celled and multi-celled organisms. Some of the asexual methods that are observed in animals are binary fission (e.g. Amoeba, bacteria), budding (e.g. Hydra), fragmentation (e.g. Planaria) etc.

Topic: Biology, Reproduction in Organisms, NEET
Aspartame is one of the good artificial sweeteners whose use is limited to cold foods and soft drinks because

Options:
A It has very low boiling point
B It gets dissociated at cooking temperature
C It is sweetener at low temperature
D It is insoluble at higher temperatures

The correct answer is:
It gets dissociated at cooking temperature

Solution:
Aspartame gets dissociated at cooking temperature

Topic: Chemistry, Chemistry in Everyday Life, JEE Main