datavorous/entrance-exam-dataset huggingface dataset:
The magnetic potential due to a magnetic dipole at a point on its axis situated at a distance of 20 cm from its center is 1.5×10-5 T m. The magnetic moment of the dipole is _______ A m2. (Given : μ04π=10-7 T m A-1)

Options:

The correct answer is:
6

Solution:
The formula to calculate the magnetic potential due to a magnetic dipole can be written as

V=μ04πMr2 ...1

Substitute the values of the known parameters into equation (1) and solve to calculate the required magnetic moment of the dipole.

1.5×10-5=10-7×M20×10-22⇒M=1.5×10-5×20×20×10-410-7⇒M=1.5×4=6 A m2

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2024 (29 Jan Shift 1)
The magnetic properties of a magnet is lost at its

Options:
A melting point
B boiling point
C Curie point
D triple point

The correct answer is:
Curie point

Solution:
The magnetic properties of a magnet is lost at its Curie point because above this temperature. Magnetic domains to be disrupted permanently.

Topic: Physics, Magnetic Properties of Matter, COMEDK, COMEDK 2014
The magnetic susceptibility is negative for:

Options:
A diamagnetic material only
B paramagnetic material only
C ferromagnetic material only
D paramagnetic and ferromagnetic materials

The correct answer is:
diamagnetic material only

Solution:
As we know the relation between the magnetic permeability and susceptibility of material is, μr=1+χm …(i)

∵ For diamagnetic substances, μr<1

So, according to Eq. (i), the magnetic susceptibility χm of diamagnetic substance will be negative. While in case of para and ferromagnetic substances subsceptibility is positive.

Topic: Physics, Magnetic Properties of Matter, JEE Main
The magnetic susceptibility of a material of a rod is 499. Permeability in vacuum is 4π×10-7Hm-1. Absolute permeability of the material of the rod is

Options:
A 4π×10-4 Hm-1
B 2π×10-4 Hm-1
C 3π×10-4 Hm-1
D π×10-4 Hm-1

The correct answer is:
2π×10-4 Hm-1

Solution:
μ=μ01+xm

=4π×10-7×500

=2π×10-4 Hm-1

Topic: Physics, Magnetic Properties of Matter, JEE Main, JEE Main 2021 (20 Jul Shift 2)
The magnetic susceptibility of a material of a rod is 499. The absolute permeability of vacuum is 4ÏÃ10-7 H m-1. The absolute permeability of the material of a rod is

Options:
A ÏÃ10-4 H m-1
B 2ÏÃ10-4 H m-1
C 3ÏÃ10-4 H m-1
D 4ÏÃ10-4 H m-1

The correct answer is:
2ÏÃ10-4 H m-1

Solution:
Absolute permeability of material of rod
Î¼=Î¼rÎ¼0=1+XmÎ¼0
â´ Î¼=1+499Ã4ÏÃ10-7=2ÏÃ10-4 H m-1

Topic: Physics, Magnetic Properties of Matter, JEE Main
The magnetic susceptibility of a paramagnetic substance at -173 °C is 1.5×10-2. To have the susceptibility 0.5×10-2, the change in temperature in °C is

Options:
A 100
B 180
C 200
D 220

The correct answer is:
200

Solution:
Given:

xm=1.5×10-2

T1=173 °C+273=100 K

From Curie's Law,

xm∝1T

xm1xm2=T2T1

Substitute the values in the above equation

1.5×10-20.5×10-2=T2100

T2=300 K

T2=300 K-273

T2=27 °C

The change in temperature is,

ΔT=T1-T

=27 °C--173 °C

=200 °C

Topic: Physics, Magnetic Properties of Matter, JEE Main
The magnetised wire of moment $M$ and length $l$ is bent in the form of semicircle of radius $r$. Then its magnetic moment is

Options:
A $\frac{2 M}{\pi}$
B $2 M$
C $\frac{M}{\pi}$
D zero

The correct answer is:
$\frac{2 M}{\pi}$

Solution:
On bending a wire its pole strength remains unchanged whereas its magnetic moment changes.
New magnetic moment,
$$
M^{\prime}=m(2 r)=m\left(\frac{2 l}{\pi}\right)=\frac{3 M}{\pi}
$$

Topic: Physics, Magnetic Effects of Current, AP EAMCET, AP EAMCET 2008
The magnetised wire of moment $M$ and length $l$ is bent in the form of semicircle of radius $r$. Then its magnetic moment is

Options:
A $\frac{2 M}{\pi}$
B $2 M$
C $\frac{M}{\pi}$
D zero

The correct answer is:
$\frac{2 M}{\pi}$

Solution:
On bending a wire its pole strength remains unchanged whereas its magnetic moment changes.
New magnetic moment,
$$
M^{\prime}=m(2 r)=m\left(\frac{2 l}{\pi}\right)=\frac{3 M}{\pi}
$$

Topic: Physics, Magnetic Effects of Current, TS EAMCET, TS EAMCET 2008
The magnetization of bar magnet of length 5 cm, cross sectional area 2 cm2 and net magnetic moment 1 Am2 is

Options:
A 3×105Am
B 4×105Am
C 2×105Am
D 1×105Am

The correct answer is:
1×105Am

Solution:
Given, l=5cm=5×10-2m,
a=2cm2=2×10-4m2
And M=1Am-2
The magnetisation of the bar magnet is given by
l=MV
where, V = volume of bar magnet = a×l
⇒l=Ma×l
Substituting given values, we get
l=12×10-4×5×10-2=110-5
=1×105Am

Topic: Physics, Magnetic Properties of Matter, JEE Main
The magnets of two suspended coll galvanometers are of the same strength so that they produce identical uniform magnetic fields in the region of the coils. The coil of the first one is in the shape of a square of side a and that of the second one is circular of radius $\frac{a}{\sqrt{\pi}},$ When the same current is passed through the coils, the ratio of the torque experienced by the first coil to that experienced by the second one is

Options:
A $1: \frac{1}{\sqrt{\pi}}$
B 1: 1
C $\pi: 1$
D $1: \pi$

The correct answer is:
1: 1

Solution:
In galvanometer $I \sim \theta=G \theta$ Where, $\quad G=\frac{k}{N A B}$
When coil is in square of shape,
$$
G=\frac{k}{N A B}=\frac{k}{N a^{2} B}
$$
When coil is in circular shape of radius $a / \sqrt{\pi}$
$$
\begin{aligned}
G &=\frac{k}{N A B}=\frac{k}{N \pi r^{2} \cdot B} \\
&=\frac{k}{N \pi\left(\frac{a}{\sqrt{\pi}}\right)^{2} \cdot B}=\frac{k}{N a^{2} B}
\end{aligned}
$$
So, value of $G$ is same for both type of shape of coil. So, the torque $e=N B L A \sin \theta$ will be same in both the case
$$
e: e=1: 1
$$

Topic: Physics, Current Electricity, JEE Main
The magnification produced by the objective lens and the eye lens of a compound microscope are \( 25 \) and \( 6 \) respectively.
The magnification of this microscope is

Options:
A \( 25 \)
B \( 50 \)
C \( 150 \)
D \( 200 \)

The correct answer is:
\( 150 \)

Solution:
Given M0=6⇒25Me
∴ magnification of this microscope is
M=M0 ×Me

=25×6=150

Topic: Physics, Ray Optics, JEE Main
The magnifying power of a telescope can be increased by

Options:
A Increasing focal length of the system
B Fitting eye piece of high power
C Fitting eye piece of low power
D Increasing the distance of objects

The correct answer is:
Fitting eye piece of high power

Solution:
Magnifying power of telescope is $\frac{f_o}{f_e}$, so as $\frac{1}{f_e}$ increases, magnifying power increases

Topic: Physics, Ray Optics, JEE Main
The magnifying power of a telescope is 9. When it is adjusted for parallel rays the distance between the objective and eyepiece is $20 \mathrm{~cm}$. The focal length of lenses are

Options:
A $10 \mathrm{~cm}, 10 \mathrm{~cm}$
B $15 \mathrm{~cm}, 5 \mathrm{~cm}$
C $18 \mathrm{~cm}, 2 \mathrm{~cm}$
D $11 \mathrm{~cm}, 9 \mathrm{~cm}$

The correct answer is:
$18 \mathrm{~cm}, 2 \mathrm{~cm}$

Solution:
Given, $M=\frac{f_0}{f_e}=9$ and $f_0+f_e=20$
$f_o=9 f_e$
So, $9 f_e+f_e=20$
$\begin{aligned}
& \therefore f_e=2 \mathrm{~cm} \\
& f_o=9 \times 2 \\
& f_0=18 \mathrm{~cm}
\end{aligned}$

Topic: Physics, Ray Optics, JEE Main
The magnifying power of a telescope with tube length 60cm is 5. What is the focal length of its eye piece?

Options:
A 20cm
B 40cm
C 30cm
D 10cm

The correct answer is:
10cm

Solution:
m=f0fe
5=f0fe
fo=5fe
fo+fe=60
6fe=60
fe=10

Topic: Physics, Ray Optics, JEE Main, JEE Main 2020 (08 Jan Shift 1)
The magnifying power of an astronomical telescope for relaxed vision is 16 . On adjusting, the distance between the objective and eye lens is $34 \mathrm{~cm}$. Then the focal length of objective and eye lens will be respectively

Options:
A $17 \mathrm{~cm}, 17 \mathrm{~cm}$
B $20 \mathrm{~cm}, 14 \mathrm{~cm}$
C $32 \mathrm{~cm}, 2 \mathrm{~cm}$
D $30 \mathrm{~cm}, 4 \mathrm{~cm}$

The correct answer is:
$32 \mathrm{~cm}, 2 \mathrm{~cm}$

Solution:
Correct option is 2. $20 \mathrm{~cm}, 14 \mathrm{~cm}$
In this case $|m|=\frac{f_0}{f_e}=16$....(i)
and length of telescope $=f_0+f_e=34 \ldots$...(ii)
Solving (i) and (ii), we get $f_{\mathrm{k}}=2 \mathrm{~cm}, \mathrm{Fo}_0=32 \mathrm{~cm}$

Topic: Physics, Ray Optics, JEE Main
The magnitude and direction of the current in the circuit shown will be

Options:
A \( \frac{7}{3} A \) from \( a \) to \( b \) through \( e \).
B \( \frac{7}{3} A \) from \( b \) to \( a \) through \( e \).
C \( 1 A \) from \( b \) to \( a \) through \( e \).
D \( 1 A \) from \( a \) to \( b \) through \( e \).

The correct answer is:
\( 1 A \) from \( a \) to \( b \) through \( e \).

Solution:
Suppose the current i is flowing from a to b through e.

Applying Kirchhoff's voltage law,

10-4-2i-3i-1i=0

⇒6-6i=0

⇒i=1 A

Since value of current is positive means our proposed direction is correct.

Hence, the current will flow from a to b through e.

Topic: Physics, Current Electricity, JEE Main
The magnitude of electric field intensity $E$ is such that, an electron placed in it would experience an electrical force equal to its weight is given by

Options:
A $mge$
B $\frac{m g}{e}$
C $\frac{e}{m g}$
D $\frac{e^2}{m^2} g$

The correct answer is:
$\frac{m g}{e}$

Solution:
According to the question, $e E=m g \Rightarrow E=\frac{m g}{e}$

Topic: Physics, Electrostatics, JEE Main
The magnitude of magnetic fields at a distacne 'r' from the centre of a short bar magnet, in longitudinal position to transverse position is in the ratio

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

The correct answer is:
2:1

Solution:
For longitudenal positions magnetic field will be $B_{1} \alpha \frac{2 M}{d^{3}}$
For transverse positions magnetic field will be $B_{2} \alpha \frac{M}{d^{3}}$
$\therefore \frac{B_{1}}{B_{2}}=2: 1$

Topic: Physics, Magnetic Properties of Matter, JEE Main
The magnitude of magnetic induction at mid-point O due to current arrangement as shown in figure will be

.

Options:
A μ0I2πa
B 0
C μ0I4πa
D μ0Iπa

The correct answer is:
μ0Iπa

Solution:
The magnetic field at point O due to the two semi-infinite portions of both wires lying along line BE will be 0.

Using right-hand rule, it can be determined that the magnetic field due to both the semi-infinite portions of wires BC and ET which are perpendicular to BE will be outwards from the plane.

The magnetic field due to one such portion will be:

B=μ0I4πa2

Since the magnetic field of both the portions is in same direction, the net magnetic field at O will be:

Bnet=2×μ0I4πa2=μ0Iπa

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2023 (29 Jan Shift 1)
The magnitude of the average electric field normally present in the atmosphere just above the surface of the Earth is about 150 N/C, directed inward towards the center of the Earth. This gives the total net surface charge carried by the Earth to be : [Given : ∈ O = 8.85 × 1 0 - 1 2 C 2 / N-m 2 , R E = 6.37 × 1 0 6 m ]

Options:
A -680 kC
B + 670 kC
C + 680 kC
D - 670 kC

The correct answer is:
-680 kC

Solution:
E=150 N/C
We know that Ï = E → · A →
= E × 4 π r 2
= 1 5 0 × 4 π × R e 2
Now, Ï = Q in ∈ o
Q in = Ï ∈ o = 1 5 0 × 4 π R e 2 × 8.85 × 1 0 - 1 2 C
= 1 5 0 × 4 π 6.37 × 1 0 6 2 × 8.85 × 1 0 - 1 2
= 6 0 0 π × 6.37 × 6.37 × 8.85 C
= 6 8 0 kC
As Electric field is inward, Q will be negative

Topic: Physics, Electrostatics, JEE Main, JEE Main 2014 (09 Apr Online)
The magnitude of the change in oxidising power of the MnO4-/Mn2+ couple is x×10-4 V if the H+ concentration is decreased from 1 M to 10-4M at 25°C. (Assume concentration of MnO4- and Mn2+ to be same on change in H+ concentration). The value of x is __________.

(Rounded off to the nearest integer)

[Given: 2.303RTF=0.059]

Options:

The correct answer is:
3776

Solution:
Eqn is-

MnO4-+H⊕+5e-→Mn+2+4H2O

Nernst equation:

Ecell =ECell 0-0.0595logMn+2MnO4-1H+8

(I) Given H⊕=1M

E1=E0-0.0595logMn+2MnO4-

(II) Now: H⊕=10-4M

E2=E0-0.0595logMn+2MnO4-×110-48

=E0-0.0595logMn+2MnO4-+0.0595log10-32

therefore: E1-E2=0.0595×32

=0.3776 V=3776×10-4

x=3776

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2021 (24 Feb Shift 2)
The magnitude of the energy gap for an insulator is

Options:
A less than $1 \mathrm{eV}$
B between $2 \mathrm{eV}$ to $3 \mathrm{eV}$
C more than $3 \mathrm{eV}$
D between $1 \mathrm{eV}$ to $2 \mathrm{eV}$

The correct answer is:
more than $3 \mathrm{eV}$

Solution:
In an insulator, there exists a large band gap between conduction band and valence band $\mathrm{Eg}(\mathrm{Eg}\gt3 \mathrm{eV}$ ), that results in no free electrons in the conduction band, and therefore no electrical conduction is possible.

Topic: Physics, Semiconductors, JEE Main
The magnitude of the magnetic field at the centre of an equilateral triangular loop of side 1 m which is carrying a current of 10 A is:
[Take Î¼0=4ÏÃ10-7 N A-2 ]

Options:
A 3Â Î¼T
B 1Â Î¼T
C 18 Î¼T
D 9Â Î¼T

The correct answer is:
18 Î¼T

Solution:
The perpendicular distance of centroid from any side, 123 m
Magnetic field due to one side.
B1=ÂµoI4Ïr[cosÎ¸1+cosÎ¸2]
=Î¼0(10)4Ïr123cos30Â°+cos30Â°
=53Ï Î¼0 32+32=15Î¼0Ï
Net magnetic field B=3B1=45ÏÎ¼0=45ÏÃ4ÏÃ10-7
=180Ã10-7 T
=18 Î¼T

Topic: Physics, Magnetic Effects of Current, JEE Main
The magnitude of the magnetic field $(B)$ due to loop $A B C D$ at the origin $(O)$ is

Options:
A
zero

B
$\frac{\mu_0(b-a)}{24 a b}$

C
$\frac{\mu_0 \mathrm{I}}{4 \pi}\left[\frac{b-a}{a b}\right]$

D
$\frac{\mu_o \mid}{4 \pi}\left[2(b-a)+\frac{\pi}{3}(a+b)\right]$

The correct answer is:
$\frac{\mu_0(b-a)}{24 a b}$

Solution:
Net magnetic field due to loop $A B C D$ at $O$ is
$$
\begin{aligned}
& B=B_{A B}+B_{B C}+B_{C D}+B_{D A} \\
& =0+\frac{\mu_o I}{4 \pi a} \times \frac{\pi}{6}+0-\frac{\mu_o I}{4 \pi b} \times \frac{\pi}{6}=\frac{\mu_o I}{24 a}-\frac{\mu_o I}{24 b}=\frac{\mu_o I}{24 a b}(b-a)
\end{aligned}
$$

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2009
The magnitude of the projection of the vector 2i^+3j^+k^ on the vector perpendicular to the plane containing the vectors i^+j^+k^ and i^+2j^+3k^, is:

Options:
A 36
B 32
C 6
D 32

The correct answer is:
32

Solution:
Vector perpendicular to i^+j^+k^ and i^+2j^+3k^

=i^j^k^111123=i^-2j^+k^

∴ Required projection

=|(2i^+3j^+k^)⋅ (i^-2j^+k^)||i^-2j^+k^|

=|2-6+1|6=36=32

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2019 (08 Apr Shift 1)
The magnitude of the projection of the vector $\mathbf{a}=4 \mathbf{i}-3 \mathbf{j}+2 \mathbf{k}$ on the line which makes equal angles with the coordinate axes is

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

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

Solution:
Let the vector $\mathbf{v}$ make an angle $\alpha$ with each of the three axes, then direction cosine of $\mathbf{v}$ are $ < \cos \alpha, \cos \alpha, \cos \alpha>$
Also, $\quad \cos ^2 \alpha+\cos ^2 \alpha+\cos ^2 \alpha=1$
Hence, direction cosine of $\mathbf{v}$ are $ < \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}>$ or $ < -\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}>$
So, the required line which makes equal angle with the coordinate axes is
$\mathbf{v}= \pm \frac{1}{\sqrt{3}} \mathbf{i} \pm \frac{1}{\sqrt{3}} \mathbf{j} \pm \frac{1}{\sqrt{3}} \mathbf{k}$
Now, the magnitude of the projection of the vector $\mathbf{a}=4 \mathbf{i}-3 \mathbf{j}+2 \mathbf{j}$ on line $\mathbf{v}$.
$\therefore$ Projection of a along r
$\begin{aligned}
\mathbf{v} & =\frac{\mathbf{a} \cdot \mathbf{v}}{|\mathbf{v}|} \\
& =\frac{3 / \sqrt{3}}{1}=\sqrt{3}
\end{aligned}$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The magnitude of torque on a particle of mass $1 \mathrm{~kg}$ is 2.5 Nm about the origin. If the force acting on it is $1 \mathrm{~N},$ and the distance of the particle from the origin is $5 \mathrm{~m},$ the angle between the force and the position vector is (in radians):

Options:
A $\frac{\pi}{6}$
B $\frac{\pi}{3}$
C $\frac{\pi}{8}$
D $\frac{\pi}{4}$

The correct answer is:
$\frac{\pi}{6}$

Solution:
Torque about the origin $=\vec{\tau}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{F}}$

$=r F \sin \theta \Rightarrow 2.5=1 \times 5 \sin \theta$

$\sin \theta=0.5=\frac{1}{2}$

$\Rightarrow \theta=\frac{\pi}{6}$

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2019 (11 Jan Shift 2)
The magnitude of vectors OA→, OB→ and OC→ in the given figure are equal. The direction of OA→+OB→-OC→ with x-axis will be:

Options:
A tan−1⁡(3−1+2)(1+3−2)
B tan−1⁡(1−3−2)(1+3+2)
C tan−1⁡(3−1+2)(1−3+2)
D tan−1⁡(1+3−2)(1−3−2)

The correct answer is:
tan−1⁡(1−3−2)(1+3+2)

Solution:
From the above diagram, the equations are,
OA→=Rcos30∘i^+sin30∘j^

OB→=Rcos60∘i^-sin60°j^

OC→=R-cos45°i^+sin45°j^

⇒OA→+OB→−OC→=R12+32+12i^+R12−32−12j^, angle with the x-axis.

⇒tanα=12−32−1212+32+12=1−3−21+3+2

⇒α=tan−11−3−21+3+2.

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2021 (26 Aug Shift 1)
The magnitude of work done by a gas that undergoes a reversible expansion along the path \( \mathrm{ABC} \) shown in the figure is

Options:

The correct answer is:
48

Solution:
W=126+10×6=48J

Topic: Chemistry, Thermodynamics (C), JEE Main
The main arena of various types of activities of a cell is

Options:
A plasma membrane
B mitochondrion
C cytoplasm
D nucleus

The correct answer is:
cytoplasm

Solution:
The main arena of various types of activities of a cell is cytoplasm. Cytoplasm is an aqueous substance containing a variety of cell organelle along with non-living inclusions. The soluble part of cytoplasm forms the background material or ground substance between the cell organelles.

Topic: Biology, Cell - The Unit of Life, NEET, NEET 2010 (Screening)
The main cause of extinction of species from tropical areas is:

Options:
A Over-exploitation
B Deforestation
C Alien species invasion
D More than one option is correct

The correct answer is:
Deforestation

Solution:
Habitat loss and fragmentation: Deforestation is the most important cause driving animals and plants to extinction. The most dramatic examples of habitat loss come from tropical rain forests. Once covering more than 14 percent of the earthâs land surface, these rain forests now cover no more than 6 percent. They are being destroyed fast.

Topic: Biology, Biodiversity and its Conservation, NEET
The main cause of the disintegration of the endometrial lining is

Options:
A LH surge
B degeneration of corpus luteum
C ovulation during mid-cycle
D implantation leads to pregnancy

The correct answer is:
degeneration of corpus luteum

Solution:
Menstrual cycle shows changes in ovary and uterus simultaneously. It begins with menstruation i.e. bleeding phase. It continues for five days. Menstruation means shedding of endometrial lining from uterus. This occurs due to decrease in levels of ovarian hormones: Estrogen and Progesterone. These hormones decrease because in the ovary the corpus luteum of previous cycle start disintegrating. Thus degeneration of corpus luteum causes the ovarian hormones to decrease and thus endometrium degenerates leading to bleeding phase.

Topic: Biology, Human Reproduction, NEET
The main cause of the population explosion in the World is

Options:
A Excellent job facilities
B Increase in agricultural production
C Excellent health care
D Fewer battles and wars

The correct answer is:
Excellent health care

Solution:
The main cause of population explosion in the World is excellent health care as the better medical facilities controls the incidence and spread of the disease. Many of the fatal diseases, epidemics and proper care of new born children and their mothers decreases the death rate and hence enhances the population growth.

Topic: Biology, Organisms and Populations, NEET
The main cloning vectors used for DNA sequencing in the Human Genome Project are ________________.

Options:
A T-DNA.
B BAC and YAC.
C Expression vectors.
D pBR322 cloning vectors.

The correct answer is:
BAC and YAC.

Solution:
For sequence annotation in HGP, the total DNA from a cell is isolated and converted into random fragments of relatively smaller sizes (recall DNA is a very long polymer, and there are technical limitations in sequencing very long pieces of DNA) and cloned in suitable host using specialised vectors. The cloning results in the amplification of each piece of DNA fragment so that it could be subsequently sequenced with ease. The commonly used hosts were bacteria and yeast, and the vectors were called BAC (bacterial artificial chromosomes) and YAC (yeast artificial chromosomes), respectively.

Topic: Biology, Molecular Basis of Inheritance, NEET
The main collecting duct of the lymphatic system, the thoracic duct, receives lymph from the

Options:
A left side of the head, neck, chest, left upper extremity and the left side of the body below the ribs
B right side of the head, neck, chest, left upper extremity and the right side of body below the ribs
C right side of head, neck, chest, left upper extremity and the entire body below the ribs
D left side of the head, neck, chest, left upper extremity and the entire body below the ribs

The correct answer is:
left side of the head, neck, chest, left upper extremity and the entire body below the ribs

Solution:
As the blood passes through the capillaries some amount of it oozes out in the surrounding tissues. This is known as tissue fluid or interstitial fluid. When this fluid enters the lymphatic system it is known as lymph. Lymph is a colorless fluid as it lacks haemoglobin. It contains cells known as Lymphocytes which is important for immune system. The Lymphatic system consists of lymphatic capillaries, nodes, vessels and ducts. The lymphatic capillaries join to form lymphatic vessels. Large lymphatic vessels finally pour into one of the two principal vessels: Thoracic duct or Right Lymphatic duct. The thoracic duct receives lymph from the entire left side of the body.

Topic: Biology, Body Fluids and Circulation, NEET
The main constituent of enamel on the surface of teeth is:

Options:
A 3Ca3PO42·CaF2
B Ca3PO42·H2O
C 3Ca3PO42·Ca(OH)2
D CaF2

The correct answer is:
3Ca3PO42·Ca(OH)2

Solution:
Enamel provides the hard surface to the teeth. It has the carbonate substituted hydroxyapatite crystallites. It is formed from the minerals. It has the calcium phosphate and calcium hydroxide.

Hence, the main constituent of enamel on the surface of teeth is 3Ca3PO42·Ca(OH)2.

Topic: Chemistry, Environmental Chemistry, AP EAMCET, AP EAMCET 2021 (20 Aug Shift 2)
The main constituent of vinegar is ____

Options:
A formic acid
B oxalic acid
C nitric acid
D acetic acid

The correct answer is:
acetic acid

Solution:
The main constituent of vinegar is acetic acid \(\left(\mathrm{CH}_3 \mathrm{COOH}\right)\). It contains \(5-8 \%\) acetic acid by volume. Vinegar is mainly used in culinary arts.

Topic: Chemistry, Carboxylic Acid Derivatives, AP EAMCET, AP EAMCET 2020 (18 Sep Shift 1)
The main constituents of Dettol are:

Options:
A Chloroxylenol and $\alpha$-terpineol.
B Toluene and phenol.
C Aniline and benzene.
D None of the above.

The correct answer is:
Chloroxylenol and $\alpha$-terpineol.

Solution:
Chloroxylenol and $\alpha$-terpineol are the main constituents of dettol.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main
The main difference between active and passive transport across cell membrane is

Options:
A passive transport is non-selective whereas active transport is selective
B passive transport requires a concentration gradient across a biological membrane whereas active transport requires energy to move solutes
C passive transport is confined to anionic carrier proteins whereas active transport is confined to cationic channel proteins
D active transport occurs more rapidly than passive transport

The correct answer is:
passive transport requires a concentration gradient across a biological membrane whereas active transport requires energy to move solutes

Solution:
The main difference between active and passive transport across the cell is that passive transport requires concentration gradient across biological membrane whereas during active transport, the movement of molecules is from high concentration to low concentration that means they move against the concentration gradient by using ATP.

Topic: Biology, Transport in Plants, NEET, NEET 2019 (Odisha)