datavorous/entrance-exam-dataset huggingface dataset:
The decreasing order of stability of oxides of halogen is

Options:
A fluorine $>$ chlorine $>$ iodine $>$ bromine
B fluorine $>$ chlorine $>$ bromine $>$ iodine
C iodine $>$ bromine $>$ chlorine $>$ fluorine
D iodine $>$ chlorine $>$ bromine $>$ fluorine

The correct answer is:
iodine $>$ chlorine $>$ bromine $>$ fluorine

Solution:
In oxides of halogen, the bonds are mainly covalent due to small difference in electronegativity between the halogens and oxygen. The bond polarity, however, increases as we move from $F$ to $I$.
The stability of oxides of iodine is greater than those of chlorine, while bromine oxides are the least stable. Iodine-oxygen bond is stable due to greater polarity of the bond, while the stability of the chlorine-oxygen bond is due to multiple bond formation involving $d$-orbitals of the chlorine atom. Bromine being in between lacks both these characteristics. Thus, the stability of oxides of halogens decreases in the following order
$$
\mathrm{I}>\mathrm{Cl}>\mathrm{Br}>\mathrm{F}
$$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
The decreasing order of stability of oxides of halogen is

Options:
A fluorine $>$ chlorine $>$ iodine $>$ bromine
B fluorine $>$ chlorine $>$ bromine $>$ iodine
C iodine $>$ bromine $>$ chlorine $>$ fluorine
D iodine $>$ chlorine $>$ bromine $>$ fluorine

The correct answer is:
iodine $>$ chlorine $>$ bromine $>$ fluorine

Solution:
In oxides of halogen, the bonds are mainly covalent due to small difference in electronegativity between the halogens and oxygen. The bond polarity, however, increases as we move from $F$ to $I$.
The stability of oxides of iodine is greater than those of chlorine, while bromine oxides are the least stable. Iodine-oxygen bond is stable due to greater polarity of the bond, while the stability of the chlorine-oxygen bond is due to multiple bond formation involving $d$-orbitals of the chlorine atom. Bromine being in between lacks both these characteristics. Thus, the stability of oxides of halogens decreases in the following order
$$
\mathrm{I}>\mathrm{Cl}>\mathrm{Br}>\mathrm{F}
$$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), COMEDK, COMEDK 2015
The decreasing order of the bond moment of E-H bond in NH3 , PH3 , AsH3 and SbH3 is given by

Options:
A NH3 > PH3 > AsH3 > SbH3
B SbH3 > AsH3 > PH3 > NH3
C NH3 > AsH3 > PH3 > SbH3
D NH3 > SbH3 > AsH3 > PH3

The correct answer is:
NH3 > SbH3 > AsH3 > PH3

Solution:
Electronegativities of N(3.0), P (2.1), As (2.0), Sb (1.9) and H (2.1) point out clearly that the correct order of decreasing polarity is :
NH3 > SbH3 > AsH3 > PH3

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
The decreasing values of bond angles from $\mathrm{NH}_3\left(106^{\circ}\right)$ to $\mathrm{SbH}_3\left(101^{\circ}\right)$ down group-15 of the periodic table is due to

Options:
A
increasing bp-bp repulsion

B
increasing p-orbital character in $\mathrm{sp}^3$

C
decreasing Ip-bp repulsion

D
decreasing electronegativity

The correct answer is:
decreasing electronegativity

Solution:
No Solution Available

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2006
The deficiencies of micronutrients, not only affect the growth of plants but also vital functions such as photosynthetic and mitochondrial electron flow. Among the list given below, which group of three elements shall affect most, both photosynthetic and mitochondrial electron transport?

Options:
A Co, Ni, Mo
B Ca, K, Na
C Mn, Co, Ca
D Cu, Mn, Fe

The correct answer is:
Cu, Mn, Fe

Solution:
Iron is mainly available in the ferrous form and it is absorbed in the ferric form, also. It is a part of catalases, peroxidases, cytochromes etc. and plays a role in the electron transport system in photosynthesis. Manganese is absorbed by the plants when it is in the bivalent form. Manganese participates in the photolysis of water in photosystem II during photosynthesis and thus it helps in the electron transport of water to chlorophyll. Copper is absorbed on the clay particles as divalent cations, from where it can be absorbed by the plants by exchange mechanism. It is constituent of plastocyanin which takes part in electron transport during photosynthetic phosphorylation.

Topic: Biology, Mineral Nutrition, NEET
The deficiencies of micro-nutrients, not only affects growth of plants but also vital functions such as photosynthetic and mitochondrial electron flow. Among the list given below, which group of three elements shall affect most, both photosynthetic and mitochondrial electron transport?

Options:
A $\mathrm{Cu}, \mathrm{Mn}, \mathrm{Fe}$
B Co, Ni, Mo
C $\mathrm{Mn}, \mathrm{Co}, \mathrm{Ca}$
D $\mathrm{Ca}, \mathrm{K}, \mathrm{Na}$

The correct answer is:
$\mathrm{Cu}, \mathrm{Mn}, \mathrm{Fe}$

Solution:
Micronutrients are minerals obtained from the soil and present in plant tissues at concentrations usually less than $3 \mu$ $\mathrm{mol} \mathrm{g}^{-1}$ dry matter. Cu (copper), Mn (manganese) and Fe (iron) are those micronutrients, which affect both photosynthesis and mitochondrial electron transport because they are the main constituents of various electron carriers.

Topic: Biology, Mineral Nutrition, JIPMER, JIPMER 2007
The deficiencies of micronutrients, not only affects the growth of plants but also vital functions such as photosynthetic and mitochondrial electron flow. Among the list given below, which group of three elements shall affect most, both photosynthetic and mitochondrial electron transport.

Options:
A Co, Ni, Mo
B Ca, K, Na
C Mn, Co, Ca
D Cu, Mn, Fe

The correct answer is:
Cu, Mn, Fe

Solution:
Iron is mainly available in the ferrous form and it is absorbed in the ferric form, also. It plays a role in the electron transport system in photosynthesis. Manganese is absorbed by the plants when it is in the bivalent form. Manganese participates in the photolysis of water in pigment system II during photosynthesis and thus it helps in the electron transport from water to chlorophyll. Copper is absorbed on the clay particles as divalent cations. It is constituent of plastocyanin which takes part in electron transport during photosynthetic phosphorylation

Topic: Biology, Mineral Nutrition, NEET
The deflection in a galvanometer falls to 14th of its initial value when it is shunted by 3 Î©. If an additional shunt of 2 Î© is connected in parallel to earlier shunt, the deflection in the galvanometer falls to 1nth. What is the value of n?

Options:

The correct answer is:
8.5

Solution:
Iâ I 4 = 3I 4
RI 4 = 3I 4 Ã3
R=9 Î©
I= I 1 + I 2 ...(1)
9 I 1 = 6 5 I 2 ...(2)
â 15 2 I 1 = I 2
Using equation (1)
I= I 1 + 15 2 I 1 = 17 2 I 1
I 1 = 2I 17
I 1 = I 17/2 = I 8.5

Topic: Physics, Current Electricity, JEE Main
The deflection in a moving coil galvanometer is reduced to half when it is shunted with a \( 40 \Omega \) coil. The resistance of the
galvanometer is

Options:
A \( 15 \Omega \)
B \( 20 \Omega \)
C \( 40 \Omega \)
D \( 80 \Omega \)

The correct answer is:
\( 40 \Omega \)

Solution:
I∝Q
IgI=12
SG+S=12
40G+40=12⟹G=40Ω

Topic: Physics, Magnetic Effects of Current, JEE Main
The deflection in moving coil galvanometer falls from 25 divisions to 5 division when a shunt of 24 Ω is applied. The resistance of galvanometer coil will be:

Options:
A 12 Ω
B 96 Ω
C 48 Ω
D 100 Ω

The correct answer is:
96 Ω

Solution:
Let x represents the current per division though the galvanometer.

After applying shunt

Now, if G is the galvanometer resistance, from the above figure, it can be written that

5x×G=20x×24⇒G=4×24⇒G=96 Ω

Topic: Physics, Current Electricity, JEE Main, JEE Main 2024 (29 Jan Shift 1)
The deflection produced in a tangent galvanometer, whose coil has a resistance of \(9 \Omega\) is \(30^{\circ}\). The potential difference across the coil is \(4.5 \mathrm{~V}\). If the number of turns in the coil is 10 , the radius of the coil is (Given, \(B_{\mathrm{H}}=3.14 \times 10^{-5} \mathrm{~T}\) )

Options:
A \(2 \sqrt{3} \times 10^{-2} \mathrm{~m}\)
B \(10 \sqrt{3} \times 10^{-2} \mathrm{~m}\)
C \(6 \times 10^{-2} \mathrm{~m}\)
D \(3.5 \times 10^{-2} \mathrm{~m}\)

The correct answer is:
\(10 \sqrt{3} \times 10^{-2} \mathrm{~m}\)

Solution:
For tangent galvanometer,
Deflection, \(\theta=30^{\circ}\)
Resistance of coil, \(R=9 \Omega\)
Potential difference across coil,
\(V=4.5 \mathrm{~V}\)
Number of turns in the coil
\(\begin{aligned}
N & =10 \\
B_H & =3.14 \times 10^{-5} \mathrm{~T}
\end{aligned}\)
Current flowing through the coil,
\(I=\frac{V}{R}=\frac{4.5}{9}=0.5 \mathrm{~A}\)
We know that, current flowing through the galvanometer is given as
\(\begin{aligned}
I & =\frac{2 r B_H \tan \theta}{\mu_0 N} \\
\Rightarrow \quad r & =\frac{\mu_0 N I}{2 B_H \tan \theta} \\
& =\frac{4 \pi \times 10^{-7} \times 10 \times 0.5}{2 \times 3.14 \times 10^{-5} \times \tan 30^{\circ}} \\
& =10 \sqrt{3} \times 10^{-2} \mathrm{~m}
\end{aligned}\)

Topic: Physics, Magnetic Properties of Matter, JEE Main
The degenerate orbitals of CrH2O63+ are

Options:
A dx2-y2 and dxy.
B dz2 and dxz.
C dyz and dz2.
D dxz and dyz.

The correct answer is:
dxz and dyz.

Solution:
Degenerate orbitals of CrH2O63+:

Orbitals having same energy call degenerate orbital, And here water acts as weak ligands which split d-orbital into two parts.

Hence, according to the options given, degenerate orbitals are dxz and dyz.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2019 (09 Apr Shift 1)
The degree and order of the differential equation of the family of all parabolas whose axis is X-axis, are respectively.

Options:
A
2,3

B
2,1

C
1,2

D
3,2

The correct answer is:
3,2

Solution:
$y^2=4 a(x-h), 2 y_1=4 a \Rightarrow y_1=2 a \Rightarrow y_1{ }^2+y_1=0$
Degree $=1$, order $=2$

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2003
The degree of adsorption of solution on solid surface depends on concentration of solution
xm=KC1n
In which of the conditions, we get following type of graph?

Options:
A C = 0
B 1n=0
C C= constant
D C = 2 M

The correct answer is:
1n=0

Solution:
Horizontal straight line will be possible when slope is zero.
i.e., 1n=0 .

Topic: Chemistry, Surface Chemistry, JEE Main
The degree of dissociation is 0.4 at 420 K and 1.0 atm for the gaseous reaction

PCl5g⟶PCl3g+Cl2(g) . Assuming ideal behavior of all gases, calculate the density of equilibrium mixture of equilibrium mixture at 420 K and 1.0 atm. (P = 31, Cl = 35.5)

Options:
A 4.30
B 3.54
C 2.64
D 1.56

The correct answer is:
4.30

Solution:
PCl 5 g â PCl 3 g + Cl 2 g t = 0 a 0 0 t eq 0.6a 0.4a 0.4a
nTotal=1.4a
∴   M av = 208 a 1.4 a              =148.57 ∴   P mix = P M av RT               = 1×148.57 0.0821×420 =4.30

Topic: Chemistry, Chemical Equilibrium, JEE Main
The degree of dissociation of $0.1 \mathrm{M} \mathrm{HCN}$ solution is $0.01 \%$. Its ionisation constant would be

Options:
A $10^{-3}$
B $10^{-5}$
C $10^{-7}$
D $10^{-9}$

The correct answer is:
$10^{-9}$

Solution:
$H A=H^{+}+A^{-}$
$K=C a^2=0.1 \times\left(10^{-4}\right)^2=10^{-9}$

Topic: Chemistry, Ionic Equilibrium, JEE Main
The degree of dissociation $(\alpha)$ of a weak electrolyte, $A_x B_y$ is related to van't Hoff factor (i) by the expression:

Options:
A
$\alpha=\frac{i-1}{x+y+1}$

B
$\alpha=\frac{x+y-1}{i-1}$

C
$\alpha=\frac{x+y+1}{i-1}$

D
$\alpha=\frac{i-1}{(x+y-1)}$

The correct answer is:
$\alpha=\frac{i-1}{(x+y-1)}$

Solution:
$$
\begin{aligned}
& i=1-\alpha+n \alpha=1+\alpha(n-1) \\
& \frac{i-1}{n-1}=\alpha \\
& A_x B_y \rightarrow x A^{+y}+y B^{-x} \\
& n=x+y
\end{aligned}
$$
So $\alpha=\frac{i-1}{x+y-1}$

Topic: Chemistry, Solutions, JEE Main, JEE Main 2011
The degree of dissociation of a weak monoprotic acid of concentration 1.2 ×  10 –3  M having Ka=1.0Ã10â4 is

Options:
A 1
B 10
C 15
D 25

The correct answer is:
25

Solution:
By Î±=KaC=10-41.2Ã10-3=112
% Î±=28.86% , as value of Î± is greater than 5% so we can not take assumptions and will have to use the exact formula.
So Î± will be given by:
Î±=-10-4+10-8+4Ã10-3Ã1.2Ã10-42Ã1.2Ã10-3
So the correct value of Î±=25 %

Topic: Chemistry, Ionic Equilibrium, JEE Main
The degree of dissociation of acetic acid in a 0.1âM solution is 1.0 Ã 10â2 . The pKa of acetic acid value.

Options:
A 3
B 4
C 5
D 6

The correct answer is:
5

Solution:
Ka=Cα21-α=0.1×10-221-0.01=1.01×10-5
pKa=-log⁡1×10-5=5

Topic: Chemistry, Ionic Equilibrium, JEE Main
The degree of dissociation of $\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}$ in dilute aq. solution containing $7.0 \mathrm{~g}$ of salt per $100 \mathrm{~g}$ of water at $100^{\circ} \mathrm{C}$ is $70 \%$. If vapour pressure of water at $100^{\circ} \mathrm{C}$ is $760 \mathrm{~mm} \mathrm{Hg}$. The vapour pressure of solution is

Options:
A 735
B $\quad 730$
C 760
D $\quad 746$

The correct answer is:
$\quad 746$

Solution:
$\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2} \rightleftharpoons \mathrm{Ca}^{2+}+2 \mathrm{NO}_{3}^{-}$
$1-0.71 \quad 0.7 \quad 2 \times 0.7$
$\quad\left(\because \alpha=\frac{70}{100}=0.7\right)$
$\quad \mathrm{n}_{1}=\frac{100}{18}=5.55$
$\frac{\mathrm{p}^{0}-\mathrm{p}_{\mathrm{s}}}{\mathrm{p}^{0}}=\frac{\mathrm{n}_{2} \times \mathrm{i}}{\mathrm{n}_{1}+\mathrm{n}_{2}}, \frac{760-\mathrm{p}_{\mathrm{s}}}{760}$
$=\frac{0.042 \times 2.4}{5.55+0.042}$
$\therefore \mathrm{p}_{\mathrm{s}}-746 \mathrm{~mm} \mathrm{Hg} .$

Topic: Chemistry, Solutions, JEE Main
The degree of dissociation of PCl5(g) for the equilibrium PCl5g⇌PCl3g+Cl2(g) is approximately related to the pressure at equilibrium (P) by the relation [α≪1]

Options:
A α∝P
B α∝1P
C α∝1P2
D α∝1P4

The correct answer is:
α∝1P

Solution:
PCl 5 â PCl 3 + Cl 2 Total pressure = P 1 0 0 1 - Î± Î± Î±
â´KP=Î±P 1+Î±Î±1+Î±P1-Î±1+Î±P
=Î±21-Î±2PâÎ±2PÎ±âª1
Neglecting 'Î±2'
KP=Î±2P
â´Î±=KPP

Topic: Chemistry, Chemical Equilibrium, JEE Main
The degree of dissociation of PCl5(α) obeying the equilibrium,
PCl5(g)⇌PCl3(g)+Cl2(g) , is approximately related to the pressure at equilibrium by

Options:
A α∝P
B α∝1P
C α∝1P2
D α∝1P4

The correct answer is:
α∝1P

Solution:
So, Kp=Î±2(1-Î±)P1+Î±=Î±2P1-Î±2
or Î±=KPP (if 1-Î±2=1.)

Topic: Chemistry, Chemical Equilibrium, JEE Main
The degree of dissociation 'α' of the reaction

N2O4g⇌2NO2g

Can be related to Kp as [Given: Total pressure at equilibrium = P)

Options:
A α=KPP4+KPP
B α=KP4+KP
C α=KP/P4+KP/P1/2
D α=KP4+KP1/2

The correct answer is:
α=KP/P4+KP/P1/2

Solution:
N2O4g⇌2NO2g
KP=2α21–α×P1+α1
KP=4α21–α2P
α=KP/P4+KP/P1/2

Topic: Chemistry, Chemical Equilibrium, JEE Main
The degree of ionization of $0.4 \mathrm{M}$ acetic acid will be $\left(K_{a}=1.8 \times 10^{-5}\right)$

Options:
A $6.71 \times 10^{-3}$
B $1.6 \times 10^{-3}$
C $0.4 \times 1.8 \times 10^{-5}$
D $1.8 \times 10^{-5}$

The correct answer is:
$6.71 \times 10^{-3}$

Solution:
For a weak acid (acetic acid),
$$
\begin{aligned}
K_{a} &=C \alpha^{2} \\
1.8 \times 10^{-5} &=0.4 \times \alpha^{2} \\
\alpha &=\sqrt{\frac{1.8 \times 10^{-5}}{0.4}} \\
&=6.71 \times 10^{-3}
\end{aligned}
$$

Topic: Chemistry, Ionic Equilibrium, JEE Main
The degree of ionization of $0.10 \mathrm{M}$ lactic acid is $4.0 \%$
The value of $K_c$ is

Options:
A $1.66 \times 10^{-5}$
B $1.66 \times 10^{-4}$
C $1.66 \times 10^{-3}$
D $1.66 \times 10^{-2}$

The correct answer is:
$1.66 \times 10^{-4}$

Solution:
$\%$ dissociation $=4 \%$
degree of dissociation $(\alpha)=\frac{4}{100}=0.04$
For lactic acid
$$
\mathrm{CH}_3 \mathrm{CH}(\mathrm{OH}) \mathrm{COOH} \rightleftharpoons \mathrm{CH}_3 \mathrm{CH}(\mathrm{OH}) \mathrm{COO}^{-}+\mathrm{H}^{+}
$$
Initial concentration
$\begin{array}{cll}C \text { mol L } & 0 & 0 \\ \text { At. equilibrium } C(1-\alpha) & C \alpha & C \alpha\end{array}$
$\begin{aligned} \therefore \quad K_c & =\frac{C \alpha \cdot C \alpha}{C(1-\alpha)}=\frac{C \alpha^2}{(1-\alpha)} \\ & =\frac{0.1 \times 0.04 \times 0.04}{(1-0.04)} \\ & =\frac{1.6 \times 10^{-4}}{0.96}=1.66 \times 10^{-4}\end{aligned}$

Topic: Chemistry, Solutions, JEE Main
The degree of ionization of a compound depends on

Options:
A Size of solute molecules
B Nature of solute molecules
C Nature of vessel used
D Quantity of electricity passed

The correct answer is:
Nature of solute molecules

Solution:
The degree of ionization of a solute depends upon its nature, concentration, and temperature.

Topic: Chemistry, Ionic Equilibrium, JEE Main
The degree of the differential equation $\frac{d^2 y}{d x^2}+\sqrt{1+\left(\frac{d y}{d x}\right)^3}=0$

Options:
A 1
B 2
C 3
D 6

The correct answer is:
2

Solution:
$\frac{d^2 y}{d x^2}=-\sqrt{1+\left(\frac{d y}{d x}\right)^3}$
On squaring, we get $\left(\frac{d^2 y}{d x^2}\right)^2=1+\left(\frac{d y}{d x}\right)^3$
Obviously the degree is 2 .

Topic: Mathematics, Differential Equations, JEE Main
The degree of the differential equation $\frac{d^2 y}{d x^2}-\sqrt{\frac{d y}{d x}-3}=x$ is

Options:
A $2$
B $1$
C $1 / 2$
D $3$

The correct answer is:
$2$

Solution:
$\frac{d^2 y}{d x^2}-\sqrt{\frac{d y}{d x}-3}=x \Rightarrow \frac{d^2 y}{d x^2}-x=\sqrt{\frac{d y}{d x}-3}$
Squaring both sides, we get$\left(\frac{d^2 y}{d x^2}-x\right)^2=\left(\frac{d y}{d x}-3\right)$
$\Rightarrow\left(\frac{d^2 y}{d x^2}\right)^2+x^2-2 x \frac{d^2 y}{d x^2}=\frac{d y}{d x}-3$ Clearly, degree $=2$.

Topic: Mathematics, Differential Equations, JEE Main
The degree of the differential equation, of which \( y^{2}=4 a(x+a) \) is a solution, is

Options:
A \( 1 \)
B \( 2 \)
C \( 3 \)
D None of these

The correct answer is:
\( 2 \)

Solution:
We have, y2=4a x+a… 1
On differentiating w.r.t. x , we get
2y dydx=4a ⇒ a=y2dydx.
On substituting the value of a in equation (1), we get
y2=2ydydxx+y2dydx
⇒y=2xdydx+y dydx2
⇒y 1-dydx2=2x . dydx, which is the required differential equation.

The degree of the differential equation is 2
Hence, B is the correct answer.

Topic: Mathematics, Differential Equations, JEE Main
The degrees of freedom of a triatomic gas is

Options:
A $2$
B $4$
C $6$
D $8$

The correct answer is:
$6$

Solution:
Degree of Freedom:
- The degree of freedom refers to the number of independent ways a gas molecule can travel.
- It is an independent physical parameter in the formal definition of a physical system's state.
- The number of degrees of freedom refers to how many different ways a molecule in the gas phase may move, spin, or vibrate in space.
- The degrees of freedom of the system is given by the formula
$\mathrm{f}=3 \mathrm{~N}-\mathrm{K}$
Where,
1. $\mathrm{f}=$ degrees of freedom
2. $\mathrm{N}=$ Number of Particles in the system.
3. $\mathrm{K}$ = Independent relation among the particles
- For a tri-atomic gas
$\begin{aligned}& \mathrm{N}=3 \\& \mathrm{~K}=3 \\& \mathrm{f}=3 \times 3-3 \\& \mathrm{f}=9-3 \\& \mathrm{f}=6\end{aligned}$
Therefore, the degree of freedom of triatomic gas is 6.

Topic: Physics, Kinetic Theory of Gases, JEE Main
The d-electronic configuration of CoCl42- in tetrahedral crystal field is emt2n. The sum of m and number of unpaired electrons is

Options:

The correct answer is:
7

Solution:
The oxidation state of Co in complex CoCl42- is

x-4=-2

x=+2

Electronic configuration of Co is C27o=Ar4s23d7

So, Co+2=[Ar]3d7

Since Cl- is a weak field ligand, so the given complex will have a tetrahedral geometry with following Crystal-field splitting

So, value of (m)+(unpaired electrons) is (4)+(3)=7.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2023 (24 Jan Shift 1)
The delicate balance of CO2 and O2 is NOT disturbed by

Options:
A Respiration
B Burning of coal
C Deforestation
D Burning of petroleum

The correct answer is:
Respiration

Solution:
Plants use up carbon dioxide during photosynthesis and release oxygen into the atmosphere. This maintains the delicate balance between oxygen and carbon dioxide in the atmosphere.

The burning of fuels and cutting of trees is disturbing this balance in nature. As a result of such activities, the amount of carbon dioxide in the air is going up and the level of oxygen is coming down. Continuous and excessive use of fuels and deforestation would increase this imbalance. This is bringing about a warming of the atmosphere, called global warming. Respiration, is a natural process, So balance of CO2 and O2 not disturbed by respiration.

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2023 (10 Apr Shift 2)
The denaturation process in Polymerase Chain Reaction (PCR) does not involve

Options:
A breaking of hydrogen bonds of the original DNA double helix
B heating of starting solution to 94oC
C addition of nucleotides to the primer by DNA polymerase
D creation of single-stranded DNA templates

The correct answer is:
addition of nucleotides to the primer by DNA polymerase

Solution:
PCR stands for Polymerase Chain Reaction. In this reaction, gene (segment of DNA) of interest is synthesised in vitro and multiple copies are generated. The main steps of PCR-
Denaturation - where the double-stranded template DNA is heated to separate it into two single strands.
Annealing - where the temperature is lowered to enable the DNA primers to attach to the template DNA.
Extension - where the temperature is raised and the new strand of DNA is made by the Taq polymerase
enzyme. Thus, option C is correct as it is part of extension step and not denaturation.

Topic: Biology, Biotechnology - Principles and Processes, NEET
The denominator of a fraction exceeds the square of the numerator by 16, then the least value of the fraction is

Options:
A -14
B -18
C 112
D 116

The correct answer is:
-18

Solution:
Let the numerator be x, then f x = x x 2 + 1 6
On differentiating w.r.t. x, we get
f ′ x = x 2 + 1 6 · 1 - x 2 x x 2 + 1 6 2
= x 2 + 1 6 - 2 x 2 x 2 + 1 6 2 = 1 6 - x 2 x 2 + 1 6 2 ...(i)
Putting f'(x)=0 for maxima or minima, we get,
16-x2=0 ⇒ x=4,-4
Again, on differentiating w.r.t. x, we get
f ″ x = x 2 + 1 6 2 - - 2 x - 1 6 - x 2 2 x 2 + 1 6 2 x x 2 + 1 6 4
At x=4, f"(x)<0
∴ f(x) is maximum at x=4
And at x=-4, f"(x)>0⇒ f(x) is minimum
∴ Least value of f x = - 4 1 6 + 1 6 = - 1 8

Topic: Mathematics, Application of Derivatives, JEE Main
The densities of a liquid at $0^{\circ} \mathrm{C}$ and $100^{\circ} \mathrm{C}$ are respectively 1.0127 and 1 . A specific gravity bottle is filled with $300 \mathrm{~g}$ of the liquid at $0^{\circ} \mathrm{C}$ upto the brim and it is heated to $100^{\circ} \mathrm{C}$. Then, the mass of the liquid expelled in grams is: (Coefficient of linear expansion of glass $\left.=9 \times 10^{-6} /{ }^{\circ} \mathrm{C}\right)$

Options:
A $\frac{3}{10.1}$
B $\frac{3}{1.01}$
C $\frac{3.81}{1.0127}$
D $\frac{3.81}{0.0127}$

The correct answer is:
$\frac{3}{1.01}$

Solution:
Density at $0^{\circ} \mathrm{C}, \rho_0=1.0127$
Density at $100^{\circ} \mathrm{C}, \rho_{100}=1$
Coefficient of real expansion of liquid,
$\gamma_{\text {real }}=\frac{\rho_0-\rho_{100}}{\rho_{100} \times \Delta t}$
$=\frac{1.0127-1}{1 \times 100}=0.0127 \times 10^{-2}$
$=1.27 \times 10^{-4}$
$\gamma_{\text {real }}=\gamma_{\text {app }}+\gamma_g$
$\gamma_g=$ coefficient of volume expansion of glass $=3 \alpha$
$\therefore \quad 1.27 \times 10^{-4}=\gamma_{\mathrm{app}}+3 \alpha$
$1.27 \times 10^{-4}=\gamma_{\text {app }}+3 \times 9 \times 10^{-6}$
$\gamma_{\mathrm{app}}=1.27 \times 10^{-4}-27 \times 10^{-6}$
$=1.27 \times 10^{-4}-0.27 \times 10^{-4}$
$=1 \times 10^{-4}$
$\therefore \quad \gamma_{\text {app }}=\frac{\text { mass expelled }}{\text { remaining mass } \times \Delta t}$
$1 \times 10^{-4}=\frac{m_1-m_2}{m_2 \times 100}$
$\left(\frac{m_1}{m_2}-1\right)=1 \times 10^{-4} \times 100=10^{-2}$
$\frac{m_1}{m_2}=1+10^{-2}=1.01$
$m_2=\frac{m_1}{1.01}=\frac{300}{1.01}$
Mass expelled $=m_1-m_2$
$=300-\frac{300}{1.01}$
$=\frac{3}{1.01}$

Topic: Physics, Thermal Properties of Matter, AP EAMCET, AP EAMCET 2003