datavorous/entrance-exam-dataset huggingface dataset:
Maximum number of electrons in a subshell or an atom is determined by the following :

Options:
A $2 n^2$
B $4 l+2$
C $2 l+2$
D $4 l-2$

The correct answer is:
$4 l+2$

Solution:
Maximum no. of electrons in any subshell
$$
=4 l+2
$$

Topic: Chemistry, Structure of Atom, JEE Main
Maximum number of electrons in a subshell or an atom is determined by the following :

Options:
A $2 n^2$
B $4 l+2$
C $2 l+2$
D $4 l-2$

The correct answer is:
$4 l+2$

Solution:
Maximum no. of electrons in any subshell
$$
=4 l+2
$$

Topic: Chemistry, Redox Reactions, NEET, NEET 2009 (Mains)
Maximum number of electrons in a subshell with $l=3$ and $n=4$ is

Options:
A 14
B 16
C 10
D 12

The correct answer is:
14

Solution:
$n$ represents the main energy level and $l$ represents the subshell.
If $n=4$ and $l=3$, the subshell is $4 f$.
In $f$ subshell, there are 7 orbitals and each orbital can accommodate a maximum of two electrons, so, maximum number of electrons in 4
$\begin{aligned}
4 f \text { subshell } & =7 \times 2 \\
& =14
\end{aligned}$

Topic: Chemistry, Structure of Atom, JEE Main
Maximum number of electrons in a subshell with $l=3$ and $n=4$ is

Options:
A 14
B 16
C 10
D 12

The correct answer is:
14

Solution:
$n$ represents the main energy level and $l$ represents the subshell.
If $n=4$ and $l=3$, the subshell is $4 f$.
In $f$ subshell, there are 7 orbitals and each orbital can accommodate a maximum of two electrons, so, maximum number of electrons in 4
$\begin{aligned}
4 f \text { subshell } & =7 \times 2 \\
& =14
\end{aligned}$

Topic: Chemistry, Structure of Atom, NEET, NEET 2012 (Screening)
Maximum number of electrons in an orbit of an atom is determined by
( $\mathbf{n}$ = Orbit Number)

Options:
A $2 \mathrm{n}$
B $2 n^2$
C $(2 n+1)$
D $n$

The correct answer is:
$2 n^2$

Solution:
The number of electrons in an orbit is obtained by $2 \times \mathrm{n}^2$, where $\mathrm{n}$ is the number of orbits.
For example: Number of electrons in K shell i.e., in $1^{\text {st }}$ orbit can be obtained as follows :-
Here $n=1$, by using $2 \times n^2$ we get maximum number of electrons in K-shell i.e., $1^{\text {st }}$ shell $=2$.

Topic: Physics, Atomic Physics, JEE Main
Maximum number of electrons that can be accommodated in shell with n=4 are:

Options:
A 16
B 32
C 50
D 72

The correct answer is:
32

Solution:
The maximum number of electrons that could be present in a shell is given by the rule 2n2.

Where n is the shell number.

The first shell K shell can accommodate a maximum of 2 electrons (2 × 12 = 2).

The second shell L shell can accommodate a maximum of 8 electrons (2 × 22 = 8).

Likewise, the 4th shell named N can accommodate a maximum of 32 electrons (2 × 42 = 2 × 16 = 32).

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2023 (30 Jan Shift 2)
Maximum number of electrons that can be accommodated in the subshell with azimuthal quantum number $1=4$, is

Options:
A 10
B 8
C 16
D 18

The correct answer is:
18

Solution:
Total number of electrons $=2(2 l+1)=18$

Topic: Chemistry, Structure of Atom, JEE Main
Maximum number of isomeric monochloro derivatives which can be obtained from 2,2,5,5- tetramethylhexane by chlorination is _________

Options:

The correct answer is:
3

Solution:
The given alkane 2,2,5,5- tetramethylhexane has two types hydrogen, hence, it forms two different structural isomers on chlorination.

Total numbers of isomers are equal to three as second molecule is optically active.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2023 (24 Jan Shift 2)
Maximum number of molecules of $\mathrm{CH}_{3} \mathrm{I}$ that can react with a molecule of $\mathrm{CH}_{3} \mathrm{NH}_{2}$ are

Options:
A 3
B 4
C 2
D 1

The correct answer is:
3

Solution:
$\mathrm{CH}_{3} \mathrm{NH}_{2} \underset{\substack{\text { Methly } \\ \text { amine }}}{\stackrel{\mathrm{CH}_{3} \mathrm{I}}{\longrightarrow}} \underset{-\mathrm{HI}}{\longrightarrow}\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NHH} \underset{\begin{array}{c}\text { Dimethyl } \\ \text { amine }\end{array}}{\stackrel{\mathrm{CH}{ }_{3} \mathrm{I}}{\longrightarrow}} \underset{-\mathrm{HI}}{\longrightarrow}$
$\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N}$ Trimethyl amine $\downarrow \mathrm{CH}_{3} \mathrm{I}$
$\left(\mathrm{CH}_{3}\right)_{4} \mathrm{~N}^{+} \mathrm{I}^{-}$
Tetramethyl ammonium iodide
Hence, three molecules of $\mathrm{CH}_{3} \mathrm{I}$ is used.

Topic: Chemistry, Amines, JEE Main
Maximum number of molecules of methyl iodide that can react with one molecule of methyl amine is

Options:
A 4
B 3
C 2
D 1

The correct answer is:
4

Solution:
The reaction of methyl iodide and methyl amine is as follows
$\therefore 4$ molecules of methyl iodide can react with one molecule of methyl amine.

Topic: Chemistry, Amines, JEE Main
Maximum $\mathrm{p} \pi-\mathrm{p} \pi$ back bonding exists in

Options:
A $\mathrm{BCl}_3$
B $\mathrm{BI}_3$
C $\mathrm{BBr}_3$
D $\mathrm{BF}_3$

The correct answer is:
$\mathrm{BF}_3$

Solution:
In $\mathrm{BF}_3$, boron atom has vacant $2 p_z$ A.O. and fluorine atom has electron pair in $2 p \mathrm{~A} . \mathrm{O}$. So, maximum overlapping is possible between vacant $2 p_z$ A.O. of boron and fully filled $2 p$ A.O. of fluorine due to similar size of A.O. of boron and fully filled 2 p A.O. of fluorine due to similar size of AO's.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Maximum percent rectifier efficiency of half wave rectifier is :

Options:
A 40.6
B 80.2
C 20.3
D More than one of the above

The correct answer is:
40.6

Solution:
Concept:
Half wave rectifier
The efficiency of a rectifier is defined as the ratio of dc output power to input power.
The efficiency of a half-wave rectifier will be:
$\begin{aligned}
& \eta=\frac{P_{d c}}{P_{a c}} \\
& \eta=\frac{\frac{V_{d c}^2}{R_L}}{\frac{V_{\text {Tms }}^2}{R_L}} \\
& \mathrm{~V}_{\mathrm{DC}}=\mathrm{DC} \text { or average output voltage } \\
& \mathrm{R}_{\mathrm{L}}=\text { Load Resistance }
\end{aligned}$
For a half-wave rectifier, the output DC voltage or the average voltage is given by:
$V_{D C}=\frac{V_m}{\pi}$
Also, the RMS voltage for a half-wave rectifier is given by:
$V_{r m s}=\frac{V_m}{2}$
Calculation: $\quad$
The efficiency for a half-wave rectifier will be:
$\eta=\frac{\left(\frac{V_m}{\pi}\right)^2}{\left(\frac{v_m}{2}\right)^2}=40.6 \%$
For Half wave rectifier maximum efficiency $=40.6 \%$
Note: For Full wave rectifier maximum efficiency $=81.2 \%$

Topic: Physics, Semiconductors, JEE Main
Maximum volume of air that can be exhaled after forceful inhalation, is termed as

Options:
A Total lung capacity
B Vital capacity
C Expiratory capacity
D Functional Residual capacity.

The correct answer is:
Vital capacity

Solution:
Vital Capacity (VC = ERV + TV + IRV) - It is the maximum volume of air that a person can inspire after forced expiration.
Functional Residual Capacity (FRC = ERV + RV) - It refers to the volume of air that remains in the lungs after a normal expiration.
Expiratory Capacity (EC = TV + ERV) - It is the total volume of air that a person can expire after a normal inspiration.
Total Lung Capacity (TLC = ERV+TV+IRV+RV = VC+RV) - It refers to the total volume of air that can be accommodated in the lungs at the end of forced inspiration.

Topic: Biology, Breathing and Exchange of Gases, NEET
Maxwell's equations relate to

Options:
A law of gravitation
B basic laws of electromagnetism
C laws of electrostatics
D nuclear model of an atom

The correct answer is:
basic laws of electromagnetism

Solution:
Maxwell's equations are a group of partial differential equations that, together along with the Lorentz force law, form the inspiration of classical electromagnetism, classical optics, and electric circuits.

Maxwell's equations describe how electric and magnetic fields are generated by charges, currents, and changes of the fields. One important consequence of the equations is that they demonstrate how fluctuating electric and magnetic fields propagate at the speed of light.

Topic: Physics, Electromagnetic Waves, JEE Main
m-chlorobenzaldehyde on treatment with 50%KOH solution yields

Options:
A
B
C
D

The correct answer is:

Solution:
Meta–chlorobenzaldehyde will undergo Cannizzaro reaction with 50% KOH to give mchlorobenzoate ion and m-chlorobenzyl alcohol.

Since m-chlorobenzaldehyde does not contain α hydrogen thus with 50% KOH it undergoes the Cannizzaro reaction. The Cannizzaro reaction for the m-chlorobenzaldehyde is given as:

Since it is a disproportionation reaction so one molecule of m-chlorobenzaldehyde is reduced to m-chlorobenzyl alcohol at the cost of others which is oxidized potassium m-chlorobenzoate.

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2024 (30 Jan Shift 2)
Mean deviation of \( 6,8,12,15,10,9 \) through mean is

Options:
A \( 10 \)
B \( 2.33 \)
C \( 2.5 \)
D none of these.

The correct answer is:
\( 2.33 \)

Solution:
Mean is the ratio of sum of all the observations to total number of observations.

Here, mean, x¯=6+8+12+15+10+96=10

The arithmetic average of the deviations (all taking positive) from the mean, median or mode is known as mean deviation.

∴ Mean deviation from mean =Σxi-x¯n

=6-10+8-10+12-10+15-10+10-10+9-106

=4+2+2+5+0+16=146=2.33

Topic: Mathematics, Statistics, JEE Main
Mean deviation of the series a2,a2+d,a2+2d,........,a2+2nd from its mean is

Options:
A n+1d2n+1
B nd2n+1
C nn+1d2n+1
D 2n+1dnn+1

The correct answer is:
nn+1d2n+1

Solution:
mean x ¯ = 2n+1 2 · a 2 + a 2 +2nd 2n+1 = a 2 +nd
Σ x— x ¯ =2d 1+2+……+n =n n+1 d
â´MÂ·DÂ·=nn+1d2n+1

Topic: Mathematics, Statistics, JEE Main
Mean free path of molecules in a polyatomic gas is independent of

Options:
A number density of the molecules
B volume of the molecule
C temperature of the gas
D gas constant R

The correct answer is:
gas constant R

Solution:
$\begin{aligned} \text { Mean free path } & =\frac{1}{\sqrt{2} \pi d^2 n_V} \\ & =\frac{\mathbf{v} t}{\sqrt{2}\left(\pi d^2 \mathbf{v} t\right) n_V}=\frac{\mathbf{v} t}{\sqrt{2}(V) n_V}\end{aligned}$
where, $d=$ size (diameter) of a molecule,
$n_V=$ number of molecules per unit volume,
$=$ number density of the molecules,
$\mathbf{v}=$ average velocity of gas molecules,
$t=$ time interval between two successive collisions
and $\quad V=$ volume of the molecule.
So clearly, mean free path depends on number density of the molecules, volume of the molecule, temperature of the gas (as average velocity of gas molecules depend on temperature of the gas).
But mean free path doesn't depend on gas constant $R$.

Topic: Physics, Kinetic Theory of Gases, JEE Main
Mean kinetic energy (or average energy) per gm molecule of a monoatomic gas is given by

Options:
A $\frac{3}{2} R T$
B $\frac{1}{2} K T$
C $\frac{1}{2} R T$
D $\frac{3}{2} K T$

The correct answer is:
$\frac{3}{2} R T$

Solution:
For one gm mole; average kinetic energy $=\frac{3}{2} R T$

Topic: Physics, Kinetic Theory of Gases, JEE Main
Mean of 5 observations is 7 . If four of these observations are $6,7,8,10$ and one is missing then the variance of all the five observations is :

Options:
A
4

B
6

C
8

D
2

The correct answer is:
2

Solution:
Let 5 th observation be $x$.
Given mean $=7$
$$
\begin{aligned}
& \therefore 7=\frac{6+7+8+10+x}{5} \\
& \Rightarrow x=4
\end{aligned}
$$
Now, Variance
$$
\begin{gathered}
=\sqrt{\frac{(6-7)^2+(7-7)^2+(8-7)^2+(10-7)^2+(4-7)^2}{5}} \\
=\sqrt{\frac{1^2+0^2+1^2+3^2+3^2}{5}}=\sqrt{\frac{20}{5}}=\sqrt{4}=2
\end{gathered}
$$

Topic: Mathematics, Statistics, JEE Main, JEE Main 2013 (22 Apr Online)
Mean of 100 items is 49 . It was discovered that three items which should have been $60,70,80$ were wrongly read as $40,20,50$ respectively. The correct mean is

Options:
A $48$
B $82 \frac{1}{2}$
C $50$
D $80$

The correct answer is:
$50$

Solution:
Sum of 100 items $=49 \times 100=4900$
Sum of items added $=60+70+80=210$
Sum of items replaced $=40+20+50=110$
New sum $=4900+210-110=5000$
$\therefore$ Correct mean $=\frac{5000}{100}=50$.

Topic: Mathematics, Statistics, JEE Main
Mean of $n$ observations $x_{1}, x_{2}, \ldots, x_{n}$ is $\bar{x}$. If an observation $x_{q}$, is replaced by $x_{q}^{\prime}$ then the new mean is

Options:
A $\bar{x}-x_{q}+x_{q}^{\prime}$
B $\frac{(n-1) \bar{x}+x_{q}^{\prime}}{n}$
C $\frac{(n-1) \bar{x}-x_{q}^{\prime}}{n}$
D $\frac{n x-x_{q}+x_{q}^{\prime}}{n}$

The correct answer is:
$\frac{n x-x_{q}+x_{q}^{\prime}}{n}$

Solution:
We have,
$$
\bar{x}=\frac{x_{1}+x_{2}+\ldots+x_{q}+\ldots+x_{2}}{n}
$$
$\Rightarrow \quad \Sigma x=n \bar{x}$
If $x_{q}$, is replaced by $x_{q}^{\prime}$, then new total will be $\Sigma x'=\Sigma x-x_{4}-x_{4}^{\prime}$
New mean will be
$$
\begin{array}{l}
\bar{x}^{\prime}=\frac{\Sigma x^{\prime}}{n} \\
\bar{x}^{\prime}=\frac{\sum x-x_{q}+x_{i}^{\prime}}{n} \\
\bar{x}^{\prime}=\frac{n \bar{x}-x_{q}+x_{i}^{\prime}}{n}
\end{array}
$$

Topic: Mathematics, Statistics, JEE Main
Mean of the numbers 1,2,3,…………,9 with respective weights 12+1,22+2,32+3,…,92+9 is

Options:

The correct answer is:
7

Solution:
Here, for each xi=i, weight =wi=i2+i
Hence, the required mean =ΣwixiΣwi=∑i=19ii2+i∑i=19i2+i
=∑i=19i3+∑i=19i2∑i=19i2+∑i=19i=92n+124+99+118+1699+118+16+99+12=7

Topic: Mathematics, Statistics, JEE Main
Mechanism of a hypothetical reaction
X2+Y2→2XY is given below
(1) X2→X+X (fast)
(2) X+Y2⇌XY+Y (slow)
(3) X+Y→XY (fast)
The overall order of the reaction will be

Options:
A 2
B 0
C 1.5
D 1

The correct answer is:
1.5

Solution:
X2→X+X (Fast)
X+Y2⇌XY+Y (Slow)
As we know that slow step is
Rate determine step
∴Rate=KX[Y2] (i)
From last step
K=X2X2
X2=K1[X2] (ii)
From equation (i)&(ii)
Rate =KK1X212Y2
Rate =K'X212Y2
Order =12+1=1.5

Topic: Chemistry, Chemical Kinetics, JEE Main
Medicines used in the treatment of cancer patients involves, which of the following complex compound?

Options:
A
B
C
D

The correct answer is:

Solution:
The Cis isomer of [PtNH32Cl2] is used as an anticancer drug for treating several types of malignant tumours.
â

Topic: Chemistry, Coordination Compounds, JEE Main
Megaspores are produced from the megaspore mother cells after:

Options:
A mitotic division
B formation of thick wall
C differentiation
D meiotic division.

The correct answer is:
formation of thick wall

Solution:
A hypodermal cell of the nucellus enlarges to form an archesporia cell which transformed into a megaspore mother cell. Each diploid megaspore mother cell undergoes meiosis to form linear tetrad of megaspores. Three megaspores degenerate and remaining functional one enlarges at the chalazal end.
Related Theory
The functional megaspore undergoes three subsequent mitotic divisions and forms the eight nuclei. The eight nuclei arrange themselves into three groups. Three nuclei migrate at the micropylar end and forms the egg apparatus, consisting of two synergids and a round egg cell. The other three nuclei arrange themselves at the chalazal end and form the antipodal cells. The remaining two nuclei come together as polar nuclei in the centre of the embryo sac. All the cells of the embryo sac are haploid.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET, NEET 2013 (Karnataka)
Megaspores are produced from the megaspore mother cells after :

Options:
A Mitotic division
B Formation of a thick wall
C Differentiation
D Meiotic division

The correct answer is:
Meiotic division

Solution:
The process of formation of megaspore from megaspore mother cell is called megasporogenesis. Ovules generally differentiates single megaspore mother cell in the micropylar region. It is large cell with prominent nucleus and dense cytoplasm, it undergoes meiotic division to form four megaspore, three of which degenerates and only one remains functional.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
Meiosis and mitosis are found to be the two types of cell division. A cell is observed to have 50 chromosomes in its nucleus during interphase before entering S-phase. How many chromosomes will it have after undergoing division by the two-division process?

Meiosis	Mitosis

1	100	25

2	50	50

3	50	100

4	25	50

Options:
A 1
B 2
C 3
D 4

The correct answer is:
4

Solution:
Meiosis is a type of reductional division process where the chromosome number is halved in the daughter cell compared to its parent cell. It is significant in the process of reproduction for making gametes. Mitosis is a type of cell division where the chromosome number between parent cell and daughter cell remains the same. It is also called as equational division. Therefore, if parent cell has 50 chromosomes then after mitosis the daughter cell will have 50 chromosomes as well but after meiosis, the daughter cell formed will have 25 chromosomes only. Thus, row 4th is the correct answer.

Topic: Biology, Cell Cycle and Cell Division, NEET
Meiosis occurs in organisms during

Options:
A
sexual reproduction

B
vegetative reproduction

C
Both (a) and (b)

D
None of these

The correct answer is:
sexual reproduction

Solution:
Meiosis occurs in sexually reproducing organisms to reduce the chromosome number to half, before their gametes unite, so as to maintain the constant chromosome number $(2 n)$ in the progeny.
Vegetative reproduction is a kind of asexual reproduction occurring in plants and does not involve the formation and fusion of gametes.

Topic: Biology, Cell Cycle and Cell Division
Melamine is used in formation of melamine formaldehyde resin. How many nitrogen atoms are present in melamine unit.

Options:

The correct answer is:
6

Solution:

Topic: Chemistry, Biomolecules, JEE Main
Melamine polymer is a copolymer of

Options:
A melamine and acetaldehyde
B melamine and formaldehyde
C phenol and formaldehyde
D non of the above

The correct answer is:
melamine and formaldehyde

Solution:
Melamine polymer is a copolymer of melamine and formaldehyde.

Topic: Chemistry, Polymers, JEE Main
Melamine polymer is formed by the condensation of

Options:
A

B

C

D

The correct answer is:

Solution:
Melamine is a nitrogen-based compound used by many manufacturers to create a number of products, especially plastic dish ware. It's also used in plastic products, Dry-erase boards. It is formed by condensation polymerisation of formaldehyde and melamine.

Topic: Chemistry, Polymers, JEE Main, JEE Main 2022 (25 Jul Shift 1)
Melatonin is secreted by __________ which is located at __________ .

Options:
A

(i)	(ii)

Pineal	Dorsal side of the forebrain

B

(i)	(ii)

Pineal	Middle side of the forebrain

C

(i)	(ii)

Pineal	Base of the diencephalon

D

(i)	(ii)

Pineal	Ventral side of the forebrain

The correct answer is:

(i)	(ii)

Pineal	Dorsal side of the forebrain

Solution:
The Pineal Gland
Location : Dorsal side of forebrain.
Hormone secreted : Melatonin.
Functions of melatonin :
1. Melatonin is very important for the regulation of a 24-hour (diurnal) rhythm of our body.
Example : It helps in maintaining the normal rhythms of sleep-wake cycle, body temperature, etc.
2. It also affects - metabolism, pigmentation, menstrual cycle and our immune system.

Topic: Biology, Chemical Coordination and Integration, NEET
Mendeleef's periodic law is based on

Options:
A Atomic weight
B Atomic number
C Number of neutrons
D None of the above

The correct answer is:
Atomic weight

Solution:
Mendeleev's periodic law is based on atomic mass. He proposed that properties of elements varied regularly with the atomic mass.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
Menstrual cycle is ceased during pregnancy is due to negative feedback of:

Options:
A FSH
B ADH
C Oxytocin
D Progesterone

The correct answer is:
Progesterone

Solution:
After fertilization the corpus luteum and later on placenta also secretes large amount of progesterone which is essential to maintain endometrium for implantation and other pregnancy conditions.

Topic: Biology, Human Reproduction, NEET
Mention briefly about the circulatory system of earthworm.

Options:

None

Solution:
Circulatory system of earthworm-
- Earthworm consists of a closed circulatory system, which consists of heart, blood vessels and blood.
$-\quad$ Blood glands are present into the $4^{\text {th }}, 5^{\text {th }}$ and $6^{\text {th }}$ segments of the body, they produce blood cells and haemoglobin which remains dissolved in plasma.
$-7^{\text {th }}, 9^{\text {th }}, 12^{\text {th }}$ and $13^{\text {th }}$ segments contains four pairs of pulsatile hearts.
- Blood circulates in one direction, which is maintained by contractions of the heart.
- Alimentary canal, nerve cord and bodywall gets blood by smaller blood vessels.
- Blood cells are phagocytic in nature.

Topic: Biology, Structural Organisation in Animals
Mention the importance of LH surge during menstrua cycle.

Options:

None

Solution:
During the mid menstrual cycle at about 14 th day, rapid secretion of LH leads to its maximum level called LH surge, that induces rupture of Graafian follicle and the release of ovum (ovulation).
Later ovulation (ovulatory phase) is followed by the luteal phase during which the remaining parts of the Graafian follicle transform in the form of corpus luteum. The corpus luteum secretes large amounts of progesterone that is essential for maintenance of the endometrium. Such an endometrium is necessary for the implantation of the fertilised ovum and other events of pregnancy.

Topic: Biology, Human Reproduction
Mention the main hormone involved in parturition.

Options:
A Oestrogen
B Progesterone
C Oxytocin
D LH

The correct answer is:
Oxytocin

Solution:
The main hormone involved in parturition is oxytocin.

Topic: Biology, Human Reproduction, NEET
Mention two similarities between
(a) Aves and mammals
(b) A frog and crocodile
(c) A turtle and Pila

Options:

None

Solution:
(a) Following are the similarities between aves and mammals
(i) Presence of four chambered heart.
(ii) The members of both the groups are homeotherms, i.e., warm blooded. They are able to maintain constant body temperature.
(b) Similarities between frog and crocodile include:
(i) They are cold blooded animals. The members of both the groups are poikilotherms, i.e., they lack the capacity to regulate their body temperature.
(ii) Frogs and crocodiles are oviparous animals.
(c) Similarities between turtle and Pila include
(i) Bodyis covered with dryand cornified skin in both animals. In turtle, the epidermal covering is known as scales whereas in case of Pila, it is known as calcareous shell.
(ii) Both animals are oviparous.

Topic: Biology, Animal Kingdom
Mercury boils at $367^{\circ} \mathrm{C}$. However, mercury thermometers are made such that they can measure temperature upto $500^{\circ} \mathrm{C}$. This is done by

Options:
A maintaining vacuum aboves mercury column in the stem of the thermometer.
B filling nitrogen gas at high pressure above the mercury column.
C filling oxygen gas at high pressure above the mercury column.
D filling nitrogen gas at low pressure above the mercury column.

The correct answer is:
filling nitrogen gas at high pressure above the mercury column.

Solution:
If we fill nitrogen gas at high pressure above mercury level, the boiling point of mercury is increased which can extend to the range upto $500^{\circ}(C)$.

Topic: Physics, Kinetic Theory of Gases, JIPMER, JIPMER 2017
Mercury boils at $367^{\circ} \mathrm{C}$. However, mercury thermometers are made such that they can measure temperature upto $500^{\circ} \mathrm{C}$. This is done by

Options:
A maintaining vacuum above mercury column in the stem of the thermometer
B filling nitrogen gas at high pressure above the mercury column
C filling oxygen gas at high pressure above the mercury column
D filling nitrogen gas at low pressure above the mercury column

The correct answer is:
filling nitrogen gas at high pressure above the mercury column

Solution:
If we fill nitrogen gas at high pressure above mercury level, the boiling point of mercury is increased which can extend the range upto $500^{\circ} \mathrm{C}$.

Topic: Physics, Kinetic Theory of Gases, JIPMER, JIPMER 2009