datavorous/entrance-exam-dataset huggingface dataset:
In human male testes, the regions outside the seminiferous tubules are called interstitial spaces, and they contain

Options:
A Immunologically competent cells
B Leydig cells
C Sertoli cells
D Both A & B

The correct answer is:
Both A & B

Solution:
Testes consist of structural units known as seminiferous tubules that are present inside the lobules of testes. Each lobule consists of one to three tubules in a highly coiled manner. Seminiferous tubules comprised of two cells, the main spermatogonia i.e. mother cells that form the sperms and the Sertoli cells or nurse cells where the sperms mature. In between the seminiferous tubules is interstitial tissue that contains Leydig cells, blood vessels, immune cells etc.

Topic: Biology, Human Reproduction, NEET
In humans, at the end of the first meiotic division, the male germ cells differentiate into the

Options:
A Spermatozonia
B Primary spermatocytes
C Secondary spermatocytes
D Spermatids

The correct answer is:
Secondary spermatocytes

Solution:
After the first meiotic division primary spermatocytes divide to form secondary spermatocytes. Whereas, after second meiotic division secondary spermatocytes will divide to form spermatids.

Topic: Biology, Human Reproduction, NEET, NEET 2008 (Mains)
In humans, blood passes from the post caval to the diastolic right atrium of heart due to

Options:
A Pressure difference between the post caval and atrium
B Pushing open of the venous valves
C Suction pull
D Stimulation of the sino auricular node

The correct answer is:
Pressure difference between the post caval and atrium

Solution:
Blood always flows from higher pressure towards the lower pressure.

Topic: Biology, Body Fluids and Circulation, NEET, NEET 2008 (Mains)
In hydrogen atom, the de-Broglie wavelength of an electron in the second Bohr orbit is [Given that, Bohr radius, $\mathrm{a}_0=52.9 \mathrm{pm}$ ]

Options:
A $211.6 \mathrm{pm}$
B $211.6 \pi \mathrm{pm}$
C $52.9 \pi \mathrm{pm}$
D $105.8 \mathrm{pm}$

The correct answer is:
$211.6 \pi \mathrm{pm}$

Solution:
According to Bohr,
$$
\begin{aligned}
& \mathrm{mvr}=\frac{\mathrm{nh}}{2 \pi} \\
& 2 \pi \mathrm{r}=\frac{\mathrm{nh}}{\mathrm{mv}}=\mathrm{n} \lambda
\end{aligned}
$$
..(i) $\left[\because \lambda=\frac{\mathrm{h}}{\mathrm{mv}}\right]$
where, $\mathrm{r}=$ radius, $\lambda=$ wavelength
$\mathrm{n}=$ number of orbit
Also,
$$
r=\frac{a_0 n^2}{\mathrm{Z}}
$$
where, $\mathrm{a}_0=$ Bohr radius $=52.9 \mathrm{pm}$
$\mathrm{Z}=$ atomic number
On substituting the value of ' $r$ ' from Eq. (ii) to Eq. (i), we get
$$
\begin{aligned}
\mathrm{n} \lambda & =\frac{2 \pi \mathrm{n}^2 \mathrm{a}_0}{\mathrm{Z}} \Rightarrow \lambda=\frac{2 \pi \mathrm{na}_0}{\mathrm{Z}} \\
\lambda & =2 \pi \times 2 \times 52.9 \quad[\because \mathrm{n}=2, \mathrm{Z}=1] \\
& =211.6 \pi \mathrm{pm}
\end{aligned}
$$

Topic: Chemistry, Structure of Atom, JEE Main
In hydrogen atoms, radius of the smallest orbit of the electron is $a_0$, the radius of the third orbit is

Options:
A $\frac{a_0}{9}$
B $3 \mathrm{a}_0$
C $6 a_0$
D $9 \mathrm{a}_0$

The correct answer is:
$9 \mathrm{a}_0$

Solution:
Concept: Bohr orbit radius
The electron undergoes a uniform circular motion in Bohr's orbit. The coulombic attraction of the nucleus balancing the centrifugal force:
$\frac{\mathrm{mv}^2}{\mathrm{r}}=\frac{\mathrm{eZ}}{4 \pi \varepsilon_0 \mathrm{r}^2}$
Further, the angular momentum of electron in Bohr's orbit is:
$\mathrm{mvr}=\frac{\mathrm{nh}}{2 \pi}$
On squaring the second equation and dividing it by the first equation:
$r \propto \frac{n^2}{Z}$
The smallest Bohr orbit has radius $\mathrm{a}_0$ as $\mathrm{n}=1$. Using the above functional dependence of the Bohr orbit, the third orbit is has radius $9 \mathrm{a}_0$.

Topic: Physics, Atomic Physics, JEE Main
In hydrogen azide, the bond orders of bonds (I) and (II) are ______.

Options:
A
I $ < 2$, II $>2$
B
I $>2$, II $>2$

C
I$>2$, II $ < 2$
D
I $ < 2$, II $ < 2$

The correct answer is:
I $ < 2$, II $>2$

Solution:
As in the resonance structure of hydrogen azide, it can be seen that number of $\mathrm{N}-\mathrm{N}$ bonds is $\leq 2$
Hence for bond (I), bond order will be $ < 2$ whereas for bond (II), number of bond $\geq 2$.
Thus its bond order will be $>2$.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
In hydrogen discharge tube, it is observed that through a given cross-section 3.13Ã1015 electrons are moving from right to left and 3.12Ã1015 protons are moving from left to right per second. The current in the discharge tube (in mA) will be

Options:

The correct answer is:
1

Solution:
Number of electrons ne=3.13×1015
Number of protons np=3.12×1015
Now, the current is given by the relation
i=neqe+npqp
=3.13×1015×1.6×10-19+3.12×1015×1.6×10-19
=1×10-3=1 mA
Now, due to excess the charge on electrons, the direction of the current will be towards the right.

Topic: Physics, Current Electricity, JEE Main
In hydrogen - oxygen fuel cell, combustion of hydrogen occurs to

Options:
A
generate heat

B
remove adsorbed oxygen from electrode surfaces

C
produce high purity water

D
create potential difference between the two electrodes

The correct answer is:
create potential difference between the two electrodes

Solution:
No Solution Available

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2004
In hydrogen spectrum, if the shortest wavelength in Balmer series is $\lambda$, the shortest wavelength in Brackett series is

Options:
A $\lambda$
B $\lambda / 2$
C $4 \lambda$
D $9 \lambda$

The correct answer is:
$4 \lambda$

Solution:
For Balmer series,
$\frac{1}{\lambda_{\text {Balmer }}}=R\left(\frac{1}{2^2}-\frac{1}{n^2}\right)$, where $n=3,4,5 \ldots$
For shortest wavelength, $n=\infty$
$\begin{array}{rlrl}\Rightarrow & \frac{1}{\lambda} & =R\left(\frac{1}{2^2}\right)=\frac{R}{4} \\ \text { or } & \lambda=\frac{4}{R}\end{array}$
For Brackett series,
$\frac{1}{\lambda_{\text {Brackett }}}=R\left(\frac{1}{4^2}-\frac{1}{n^2}\right)$, where $n=5,6,7, \ldots$
Again, for shortest wavelength, $n=\infty$
$\begin{aligned} & \Rightarrow \frac{1}{\lambda_{\text {Brackett }}}=R\left(\frac{1}{4^2}\right)=\frac{R}{16} \\ & \text { or } \lambda_{\text {Brackett }}=\frac{16}{R}=4\left(\frac{4}{R}\right)=4 \lambda\end{aligned}$
[From Eq. (i)]

Topic: Physics, Atomic Physics, JEE Main
In hydrogen spectrum, the shortest wavelengths of Lyman and Balmer series are $\lambda_1$ and $\lambda_2$ respectively. The Rydberg constant of hydrogen is

Options:
A $\frac{\lambda_1+\lambda_2}{2}$
B $\frac{4\left(\lambda_2-\lambda_1\right)}{3 \lambda_1 \lambda_2}$
C $\frac{3\left(\lambda_2-\lambda_1\right)}{4 \lambda_1 \lambda_2}$
D $\frac{2\left(\lambda_2-\lambda_1\right)}{3 \lambda_1 \lambda_2}$

The correct answer is:
$\frac{4\left(\lambda_2-\lambda_1\right)}{3 \lambda_1 \lambda_2}$

Solution:
For Lymen series
$\begin{aligned} & \frac{1}{\lambda_1}=\mathrm{R}_{\mathrm{H}}\left[\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right] \\ & \mathrm{n}_1=1 \text { and } \mathrm{n}_2=\infty\end{aligned}$
$\frac{1}{\lambda_1}=\mathrm{R}_{\mathrm{H}}$ ...(1)
For balmer series
$\begin{aligned} & \frac{1}{\lambda_2}=\mathrm{R}_{\mathrm{H}}\left[\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right] \\ & \mathrm{n}_1=2 \text { and } \mathrm{n}_2=\infty\end{aligned}$
$\frac{1}{\lambda_2}=\frac{\mathrm{R}_{\mathrm{H}}}{4}$ ...(2)
$\begin{aligned} & \frac{1}{\lambda_2}-\frac{1}{\lambda_1}=\frac{\mathrm{R}_{\mathrm{H}}}{4}-\mathrm{R}_{\mathrm{H}} \\ & \frac{\lambda_1-\lambda_2}{\lambda_1 \lambda_2}=\frac{-3 \mathrm{R}_{\mathrm{H}}}{4}\end{aligned}$
Rydberg constant, $\mathrm{R}_{\mathrm{H}}=\frac{4\left(\lambda_2-\lambda_1\right)}{3 \lambda_1 \lambda_2}$

Topic: Physics, Atomic Physics, JEE Main
In hydrolysis of a salt of weak acid and strong base, $A^{-}+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{HA+} \mathrm{OH}^{-}$, the hydrolysis constant $\left(K_h\right)$ is equal to ________.

Options:
A $\frac{K_w}{K_a}$
B $\frac{K_w}{K_b}$
C $\sqrt{\frac{K_a}{C}}$
D $\frac{K_w}{K_a \times K_b}$

The correct answer is:
$\frac{K_w}{K_a}$

Solution:
Hydrolysis constant $h=\frac{K_w}{K_a}$

Topic: Chemistry, Ionic Equilibrium, JEE Main
In India, 'ecosan' toilets are found in

Options:
A West Bengal
B Sikkim
C Andaman and Nicobar
D Kerala

The correct answer is:
Kerala

Solution:
Ecological sanitation, commonly abbreviated as ecosan, is an approach to sanitation provision which aims to safely reuse excreta in agriculture. It desires to 'close the loop' mainly for the nutrients and organic matter between sanitation and agriculture. It is a process where human excreta can be recycled into a resource as a natural fertilizer, which reduces the need of chemical fertilizers. In India, Kerala was the first state to use it on a large scale.

Topic: Biology, Environmental Issues, NEET
In India electricity is supplied for domestic use at $220 \mathrm{~V}$. It is supplied at $110 \mathrm{~V}$ in USA. If the resistance of a $60 \mathrm{~W}$ bulb for use in India is $R$, the resistance of a $60 \mathrm{~W}$ bulb for use in USA will be:

Options:
A $R$
B $2 R$
C $\frac{R}{4}$
D $\frac{R}{2}$

The correct answer is:
$\frac{R}{4}$

Solution:
In India, $P_1=\frac{(220)^2}{R}$
In USA $\quad P_V=\frac{(110)^2}{R_0}$
As $\quad P_1=P_V$
$$
\begin{aligned}
& \Rightarrow \quad \frac{(220)^2}{R}=\frac{(110)^2}{R_V} \\
& \therefore R_V=\frac{110 \times 110}{220 \times 220} \Rightarrow f v=\frac{R}{4}
\end{aligned}
$$

Topic: Physics, Current Electricity, JEE Main
In India, forests constitute about

Options:
A 19.4% of the land area
B 33.7% of the land area
C 22% of the land area
D 67% of the and area

The correct answer is:
22% of the land area

Solution:
India aims at maintaining a minimum of 33% of the country's geographical area under forest and tree cover. The total forest cover of the country as per the India State of Forest Report 2009 is 690899 sq.km which constitutes 21.02% of the geographical area of the country.

Topic: Biology, Biodiversity and its Conservation, NEET
In infinite dilutions, the equivalent conductances of $\mathrm{Ba}^{2+}$ and $\mathrm{Cl}^{-}$are 127 and $76 \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}$ eqvt ${ }^{-1}$. The equivalent conductivity of $\mathrm{BaCl}_2$ at indefinite dilution is

Options:
A $101.5$
B $139.5$
C $203.5$
D $279.5$

The correct answer is:
$139.5$

Solution:
$\lambda^{\infty} \mathrm{BaCl}_2=\frac{1}{2} \lambda^{\infty} \mathrm{Ba}^{2+}+\lambda^{\infty} \mathrm{Cl}^{-}$
$=\frac{127}{2}+76=139.5 \mathrm{ohm}^{-1} \mathrm{~cm}^{-1} \mathrm{eq}^{-1}$

Topic: Chemistry, Electrochemistry, JEE Main
In isolation of which one of the following metals from their ores, the use of cyanide salt is commonly not involved?

Options:
A Zinc
B Silver
C Gold
D Copper

The correct answer is:
Copper

Solution:
For ZnS,KCN is used as depressant.

In the metallurgy of silver and gold, the respective metal is leached with a dilute solution of NaCN or KCN in the presence of air, which supplies O2. The metal is obtained later by replacement reaction.

4Ms+8CN-aq+2H2Oaq+O2g→4MCN2-aq+4OH-aq M=Ag or Au

2MCN2-aq+Zns→ZnCN42-aq+2Ms

Copper metal is not involved in this process.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2022 (29 Jun Shift 1)
In isothermal process, which of the following is not true?

Options:
A Temperature remains constant
B Internal energy does not change
C No heat enters or leaves the system
D none of the above

The correct answer is:
No heat enters or leaves the system

Solution:
In isothermal process, heat enters or leaves the system, to keep the temperature constant, so statement (c) is wrong.

Topic: Physics, Thermodynamics, JIPMER, JIPMER 2006
In kinetic theory of gases, a molecule of mass $m$ of an ideal gas collides with a wall of vessel with velocity $V$. The change in the linear momentum of the molecule is

Options:
A $2 m V$
B $m V$
C $-m V$
D Zero

The correct answer is:
$2 m V$

Solution:
$\Delta p=m V-(-m V)=2 m V$

Topic: Physics, Kinetic Theory of Gases, JEE Main
In Kjeldahl's method for estimation of nitrogen, CuSO4 acts as :

Options:
A

Reducing agent

B Catalytic agent
C Hydrolysis agent
D Oxidising agent

The correct answer is:
Catalytic agent

Solution:
Kjeldahl's method is used for the estimation of nitrogen. The organic compound is heated with conc. H2SO4 in presence of K2SO4 and a little CuSO4 to convert all the nitrogen into NH42SO4.K2SO4 raises the boiling point of H2SO4 while CuSO4 catalyses the reaction.

Hence, B is the answer.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2024 (01 Feb Shift 1)
In $\mathrm{KO}_2$, the nature of oxygen species and the oxidation state of oxygen atom are, respectively:

Options:
A
Superoxide and $-1$

B
Superoxide and $-1 / 2$

C
Peroxide and $-1 / 2$

D
Oxide and $-2$

The correct answer is:
Superoxide and $-1 / 2$

Solution:
In $\mathrm{KO}_2$, the nature of oxygen species and the oxidation state of oxygen atom are superoxide (superoxide ion is $\mathrm{O}_2^{-}$) and $-1 / 2$ respectively.
Let $x$ be oxidation state of oxygen. The oxidation state of $\mathrm{K}$ is $+1$. Hence
$$
\begin{gathered}
+1+2(x)=0 \\
2 x=-1 \\
x=-\frac{1}{2}
\end{gathered}
$$

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
In Kranz anatomy, the bundle sheath cells have

Options:
A thin walls, many intercellular spaces and no chloroplasts
B thick walls, no intercellular spaces and large number of chloroplasts
C thin walls, no intercellular spaces and several chloroplasts
D thick walls, many intercellular space and few chloroplasts

The correct answer is:
thick walls, no intercellular spaces and large number of chloroplasts

Solution:
$\mathrm{C}_4$ plants have a characteristic leaf anatomy, called 'Kranz' anatomy. Kranz means 'wreath' (ring) and is a reflection of the arrangement of cells. The large sized bundle sheath cells may form several layers around the vascular bundles. These cells are characterised by having a large number of agranal chloroplasts, thick walls impervious to gaseous exchange and no intercellular spaces.

Topic: Biology, Photosynthesis in Higher Plants, NEET, NEET 2011 (Mains)
In LC circuit the inductance L=40 mH and capacitance C=100 μF. If a voltage Vt=10sin314t is applied to the circuit, the current in the circuit is given as:

Options:
A 0.52cos⁡314t
B 10cos⁡314t
C 5.2cos⁡314t
D 0.52sin⁡314t

The correct answer is:
0.52cos⁡314t

Solution:
z=R2+XC-XL2
R=0
Z=XC-XL
=1ωC-ωL
=1314×100×10-6-314×40×10-3
=31.84-12.56
=19.28 Ω
i=V0Zsin⁡314t+π2
∴i=V0Zcos⁡314t⇒i=1019.28cos⁡314t
⇒i=0.52cos⁡314t

Topic: Physics, Alternating Current, JEE Main, JEE Main 2020 (09 Jan Shift 2)
In lab $\mathrm{H}_2 \mathrm{O}_2$ is prepared by

Options:
A Cold $\mathrm{H}_2 \mathrm{SO}_4+\mathrm{BaO}_2$
B $\mathrm{HCl}+\mathrm{BaO}_2$
C Conc. $\mathrm{H}_2 \mathrm{SO}_4+\mathrm{Na}_2 \mathrm{O}_2$
D Conc. $\mathrm{H}_2 \mathrm{SO}_4+\mathrm{Na}_2 \mathrm{O}_2$

The correct answer is:
Cold $\mathrm{H}_2 \mathrm{SO}_4+\mathrm{BaO}_2$

Solution:
$\mathrm{H}_2 \mathrm{SO}_4+\mathrm{BaO}_2 \rightarrow \mathrm{BaSO}_4+\mathrm{H}_2 \mathrm{O}_2$

Topic: Chemistry, Hydrogen, JEE Main
In laboratory silicon can be prepared by the reaction

Options:
A By heating potassium fluosilicate with potassium
B By heating carbon in electric furnace
C Silica with magnesium
D None of these

The correct answer is:
Silica with magnesium

Solution:
In laboratory, Silicon is prepared by the reaction of Silica with Magnesium.
Reaction:
$\mathrm{SiO}_2+2 \mathrm{Mg} \rightarrow \mathrm{Si}+2 \mathrm{MgO}$

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
In laminar flow, how does the velocity change from one layer to the next as you move away from the fixed surface?

Options:
A It decreases
B It remains constant
C It becomes turbulent
D It increases

The correct answer is:
It increases

Solution:
In laminar flow, as you move away from the fixed surface.

Topic: Chemistry, States of Matter, NEET
In Langmuir’s model of adsorption of a gas on a solid surface

Options:
A
the rate of dissociation of adsorbed molecules from the surface does not depend on the surface covered

B
the adsorption at a single site on the surface may involve multiple molecules at the same time

C
the mass of gas striking a given area of surface is proportional to the pressure of the gas

D
the mass of gas striking a given area of surface is independent of the pressure of the gas

The correct answer is:
the mass of gas striking a given area of surface is proportional to the pressure of the gas

Solution:
No Solution Available

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2006
In Lassaigne's extract of organic compound both nitrogen and sulphur are present, which gives blood-red colour with Fe3+ due to the formation of

Options:
A FeSCN2+
B Fe4FeCN63.xH2O
C NaSCN
D FeCN5NOS4-

The correct answer is:
FeSCN2+

Solution:
When the organic sample containing nitrogen and sulphur is strongly heated with sodium metal the ionic compound sodium thiocyanate is formed. Sodium thiocyanate gives blood-red colour with neutral FeCl3.

Na +N+C+S→NaSCNSCN- + Fe3+→FeSCN2+

Topic: Chemistry, General Organic Chemistry, JEE Main
In \( L-C-R \) circuit, in the condition of resonance, the value of power factor is

Options:
A zero
B \( 1 / 2 \)
C \( 1 \)
D infinite

The correct answer is:
\( 1 \)

Solution:
It's only possible at resonance. i.e. P.F=1 , XL=XC⇒ωL=1/ωC

Power factor in L-C-R circuit is defined as the ratio of resistance and its total impedance i.e.

Power factor=RZ.

For power factor is maximum, Total impedance of circuit is must be equal to resistance of circuit. i.e. R=Z.

Topic: Physics, Alternating Current, JEE Main
In leaves of $\mathrm{C}_4$ plants malic acid synthesis during $\mathrm{CO}_2$ fixation occurs in

Options:
A Guard cells
B Epidermal cells
C Mesophyll cells
D Bundle sheath

The correct answer is:
Mesophyll cells

Solution:
Bundle sheath malic acid is transported from mesophylls.

Topic: Biology, Photosynthesis in Higher Plants, NEET, NEET 2008 (Mains)
In LED visible light is produced by

Options:
A gallium phosphide
B gallium arsenide
C germanium phosphide
D silicon phosphide

The correct answer is:
gallium arsenide

Solution:
In materials like gallium arsenide the number of photons of light energy is sufficient to produce quite intense visible light.

Topic: Physics, Semiconductors, JEE Main
In lichens, sexual reproduction is performed by

Options:
A Fungal partner only
B Algal partner only
C Fungal and algal partners (both)
D Either fungal partner or algal partner (not both)

The correct answer is:
Fungal partner only

Solution:
Sexual reproduction in lichens is performed only or mainly by its fungal component.

Topic: Biology, Biological Classification, NEET
In light reaction for the synthesis of ATP, the site is

Options:
A Cytoplasm
B Stromal side of the thylakoid membrane
C Lumen side of the thylakoid membrane
D Inter-membrane space

The correct answer is:
Stromal side of the thylakoid membrane

Solution:
During photosynthesis in plants, ATP is synthesized by ATP synthase with the help of a proton gradient created in the lumen through the membrane of the thylakoid and into the chloroplast stroma. ATP synthase consists of two main subunits: FO and F1. This has a rotational motor mechanism allowing for ATP production. The F1 part that is involved in ATP synthesis is located at the stromal side of the thylakoid membrane.

Topic: Biology, Photosynthesis in Higher Plants, NEET
In linear programming problems when does feasibility change ?

Options:
A Change in objective functions coefficient
B Change in right hand side of feasible region
C Addition of new variable
D Feasibility doesn't change

The correct answer is:
Change in right hand side of feasible region

Solution:
Linear programming feasibility is if a solution meets all restrictions. Feasible problems have solutions. Infeasible solution meets all requirements.
Changes in limitations or objective function can affect feasibility. If a new requirement makes it impossible to develop a solution t constraints, the problem becomes infeasible. However, removing a constraint may make the challenge doable.
Changing the objective function might also affect feasibility. The task is viable if the new objective function solves all restrictions. infeasible if the new objective function yields a solution that violates all restrictions.

Topic: Mathematics, Linear Programming, JEE Main
In linkage isomerism, how do the spatial arrangements of ligands around the metal center change?

Options:
A The spatial arrangement remains the same.
B The spatial arrangement varies depending on the ligand.
C The spatial arrangement flips completely.
D The spatial arrangement is reversed.

The correct answer is:
The spatial arrangement flips completely.

Solution:
Linkage isomerism involves a complete flip or inversion in the spatial arrangement of ligands around the metal center.

Topic: Chemistry, Coordination Compounds, NEET
In liquation process used for tin Sn, the metal

Options:
A is reacted with acid
B is dissolved in water
C is brought to molten form which is made to flow on a slope
D is fused with NaOH,

The correct answer is:
is brought to molten form which is made to flow on a slope

Solution:
Liquation is a metallurgical method for separating metals from an ore or alloy.

This method was largely used to remove lead containing silver from copper, but it can also be used to remove antimony minerals from ore, and refine tin.

Liquation process is used for metal having low melting point such as tin in which they are heated and brought to molten state and made to flow down the slope while impurities with higher melting point left on the top.

Liquation is similar to distillation, except instead of vaporizing the impure metal, it is melted.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2022 (29 Jul Shift 2)
In lysosomes,

a. The acidic condition is maintained by pumping protons to the exterior part.
b. Intercellular digestion occurs.
c. All types of macromolecules are digested.
d. The enzymes present in vesicles belong to acid hydrolases.

Select those statements which are correct.

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

The correct answer is:
c and d

Solution:
Lysosomes are formed by budding off vesicles from the Golgi apparatus They are bound by a single membrane about 500 nm. They are involved in intracellular digestion and contain the enzymes needed to digest all types of macromolecules. The enzymes present in such vesicles belong to acid hydrolases: The acidic condition is maintained by pumping protons into the interior of lysosomes. These are the organelles that are involved in the process of phagocytosis.

Topic: Biology, Cell - The Unit of Life, NEET
In majority of nephrons in humans, the loop of Henle is

Options:
A too short and extends only very little into the cortex.
B too long and extends very deep into the cortex.
C too short and extends only very little into the medulla.
D too long and extends very deep into the medulla.

The correct answer is:
too short and extends only very little into the medulla.

Solution:
The nephron is the structural and functional unit of the kidney. Each nephron consists of a glomerulus and the renal tubule. Renal tubule consists of Bowman's capsule and PCT, DCT and Henle's loop. Based on the length of the loop of Henle, two types of nephrons are seen, cortical and juxta-medullary nephrons. Cortical nephrons have a short loop of Henle as it extends in the cortex only and the majority of nephrons are of this type. This type of nephron is mainly responsible for excretion function. Juxta medullary nephrons have a long loop and extend in the medulla. This is primarily responsible for the concentration of urine so that water loss is prevented.

Topic: Biology, Excretory Products and their Elimination, NEET