datavorous/entrance-exam-dataset huggingface dataset:
The steps involved in genetically modifying an organism is:

Options:
A

Identification of DNA with desirable genes

B

Maintenance of introduced DNA in the host

C

Introduction of the Identified DNA into the host

D All of these

The correct answer is:
All of these

Solution:
This technology called cloning of antibiotic resistance gene, in E. coil. There are three basic steps in genetically modifying an organism:
(i) Identification of DNA with desirable genes.
(ii) Introduction of the Identified DNA into the host.
(iii) Maintenance of introduced DNA in the host and transfer of the DNA to its progeny.
The cut piece of DNA was then linked with the plasmid DNA. There plasmid DNA act as vector to transfer the piece of DNA attached to it.
A plasmid can be used as vectors deliver an alien piece of DNA into the lost organism. DNA molecules join the ends and make a new combination of circular autonomously replicating DNA created in Vitro and is known as recombination DNA.

Topic: Biology, Biotechnology - Principles and Processes, NEET
The stepwise stability constants of a complex are given below. What is its overall reaction stability constant \(\left(\beta_4\right)\) ?
\(\begin{aligned}
& M+L \rightleftharpoons M L ; K_1=1.0 \times 10^4 \\
& M L+L \rightleftharpoons M L_2 ; K_2=1.0 \times 10^3 \\
& M L_2+L \rightleftharpoons M L_3 ; K_3=1.0 \times 10^3 \\
& M L_3+L \rightleftharpoons M L_4 ; K_4=1.0 \times 10^2
\end{aligned}\)
(Overall reaction : \(M+4 L \rightleftharpoons M L_4\) )

Options:
A \(1.0 \times 10^{12}\)
B \(12.1 \times 10^3\)
C \(1.0 \times 10^6\)
D \(1.0 \times 10^8\)

The correct answer is:
\(1.0 \times 10^{12}\)

Solution:
Given,
\(\begin{gathered}
M+L \rightleftharpoons M L ; K_1=1.0 \times 10^4 \\
M L+L \rightleftharpoons M L_2 ; K_2=1.0 \times 10^3 \\
M L_2+L \rightleftharpoons M L_3 ; K_3=1.0 \times 10^3 \\
M L_3+L \rightleftharpoons M L_4 ; K_4=1.0 \times 10^2
\end{gathered}\)
Overall reaction,
\(M+4 L \rightleftharpoons M L_4\)
The overall stability constant,
\(\begin{aligned}
\beta_4 & =K_1 \times K_2 \times K_3 \times K_4 \\
\beta_4 & =1.0 \times 10^4 \times 1.0 \times 10^3 \times 1 \times 10^3 \times 1 \times 10^2 \\
& =1 \times 10^{12}
\end{aligned}\)

Topic: Chemistry, Chemical Equilibrium, JEE Main
The stereoisomers that are formed by electrophilic addition of bromine to trans-but-2-ene is/are :

Options:
A 2 identical mesomers
B 2 enantiomers
C 2 enantiomers and 2 mesomers
D 1 racemic and 2 enantiomers

The correct answer is:
2 identical mesomers

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2021 (01 Sep Shift 2)
The steroid responsible for balance of water and electrolytes in our body is

Options:
A
insulin

B
melatonin

C
testosterone

D
aldosterone

The correct answer is:
aldosterone

Solution:
Aldosterone acts at the renal tubules stimulating the reabsorption of $\mathrm{Na}^{+}$, water and excretion of $\mathrm{K}^{+}$and phosphate ions. Thus, it helps in maintenance of electrolytes, body fluid volume, osmotic pressure and blood pressure. Insulin is a peptide hormone, that plays an important role in maintaining the normal blood glucose levels.
Testosterone is a male sex hormone that stimulates male features like muscular growth, facial and axial hair, aggressiveness, low pitch of voice, etc. and spermatogenesis. Melatonin Hormone has a very important role in regulating 24 hours (diurnal) rhythm of our body.

Topic: Biology, Chemical Coordination and Integration
The stoichiometry and solubility product of a salt with the solubility curve given below is, respectively:

Options:
A X2Y,2×10-9M3
B XY2,4×10-9M3
C XY2,1×10-9M3
D XY,2×10-6M3

The correct answer is:
XY2,4×10-9M3

Solution:
X Y 2 s → X + aq 10 −3 + 2 Y − aq 2× 10 −3
Ksp=X+Y−2
Ksp=10−32×10−32
Ksp=4×10−9M3

Topic: Chemistry, Ionic Equilibrium, JEE Main
The stoichiometry and solubility product of a salt with the solubility curve given below is, respectively:

Options:
A X2Y,2×10-9M3
B XY2,4×10-9M3
C XY2,1×10-9M3
D XY,2×10-6M3

The correct answer is:
XY2,4×10-9M3

Solution:
X Y 2 s → X + aq 10 −3 + 2 Y − aq 2× 10 −3
Ksp=X+Y−2
Ksp=10−32×10−32
Ksp=4×10−9M3

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2020 (08 Jan Shift 1)
The stopping potential for a photoelectric emission process is $10 \mathrm{~V}$. Find the maximum kinetic energy of the electrons ejected in the process.

Options:
A $3.2 \times 10^{-19} \mathrm{~J}$
B $1.6 \times 10^{-19} \mathrm{~J}$
C $1.6 \times 10^{-18} \mathrm{~J}$
D $0$

The correct answer is:
$1.6 \times 10^{-18} \mathrm{~J}$

Solution:
The relation between maximum kinetic energy and stopping potential is given by:
$K E_{\operatorname{Max}}=\mathrm{e} V_S$
Where $e$ is charge on an electron and $V_S$ is stopping potential $\left(e=1.6 \times 10^{-19} \mathrm{C}\right)$
EXPLANATION:
Given that: Stopping potential $\left(V_S\right)=10 \mathrm{~V}$
Maximum kinetic energy of electrons $\left(\mathrm{KE}_{\mathrm{MaX}}\right)=e V_S=1.6 \times 10^{-19} \times 10=1.6 \times$ $10^{-18} \mathrm{~J}$

Topic: Physics, Dual Nature of Matter, JEE Main
The stopping potential for electrons emitted from a photosensitive surface illuminated by light of wavelength 491 nm is 0.710 V. When the incident wavelength is changed to a new value, the stopping potential is 1.43 V. The new wavelength is :

Options:
A 382 nm
B 309 nm
C 329 nm
D 400 nm

The correct answer is:
382 nm

Solution:
hcλ=ϕ+eVs

1240491=ϕ+0.71 ...1

1240λ=ϕ+1.43 ...2

∴λ=382 nm

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2021 (25 Feb Shift 2)
The stopping potential for photoelectric emission from a metal surface is plotted along Y-axis and frequency ν of incident light along X-axis. A straight line is obtained as shown. Planck's constant is given by

Options:
A slope of the line
B product of slope of the line and charge on the electron
C intercept along Y-axis divided by charge on the electron
D product of intercept along X-axis and mass of the electron

The correct answer is:
product of slope of the line and charge on the electron

Solution:
Einstein's equation is
V=hνe-hν0e
Here, the slope of the line =he
⇒ h= Planck's constant = slope ×e

Topic: Physics, Dual Nature of Matter, JEE Main
The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength 6630Å is 0.42 V. If the threshold frequency is x×1013 s, where x is (nearest integer): (Given, speed light =3×108 m s-1. Planck's constant =6.63×10-34 J s)

Options:

The correct answer is:
35

Solution:
We know that, KEmax=hf-ϕ=hcλ-ϕ

Or, hcλ-ϕ=eV0, where, ϕ=hνth and V0 is stopping potential.

So, threshold frequency is

νth=cλ-eV0h

=3×10866330×10-10-1.6×10-19×0.426.63×10-34

=36.630×1015-1.6×0.426.63×1015

=101536.630-1.6×0.426.63=0.4524-0.1013

νth=35.11×1013

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2022 (26 Jun Shift 2)
The stopping potential for photoelectrons from a metal surface is $V_{1}$ when monochromatic light of frequency $v_1$ is incident on it. The stopping potential becomes $V_{2}$ when monochromatic light of another frequency is incident on the same metal surface. If $h$ be the Planck's constant and $e$ be the charge of an electron, then the frequency of light in the second case is

Options:
A $v_{1}-\frac{e}{h}\left(V_{2}+V_{1}\right)$
B $v_{1}+v \frac{e}{h}\left(V_{2}+V_{1}\right)$
C $v_{1}-\frac{e}{h}\left(V_{2}-V_{1}\right)$
D $v_{1}+\frac{e}{h}\left(V_{2}-V_{1}\right)$

The correct answer is:
$v_{1}+\frac{e}{h}\left(V_{2}-V_{1}\right)$

Solution:
Here, $h v_{1}=\phi_{0}=e V_{1}$ ...(i)
and $h v_{2}=\phi_{0}+e V_{2}$...(ii)
From Eqs. (i) and (ii), we have
$\begin{aligned}
& h\left(v_{2}-v_{1}\right)=e\left(V_{2}-v_{1}\right) \\
\Rightarrow \quad & v_{2}=\frac{e}{h}\left(V_{2}-v_{1}\right)+v_{1}
\end{aligned}$

Topic: Physics, Dual Nature of Matter, JEE Main
The stopping potential in the context of photoelectric effect depends on the following property of incident electromagnetic radiation:

Options:
A Phase
B Intensity
C Amplitude
D Frequency

The correct answer is:
Frequency

Solution:
According to Einstein’s photoelectric equation, stopping potential depends on frequency as

hν-hν∘=eVV=hν-hν∘e

Hence stopping potential depends on frequency.

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2021 (16 Mar Shift 1)
The stopping potential Vo (in volt) as a function of frequency (v) for a sodium emitter, is shown in the figure. The work function of sodium, form the data plotted in the figure, will be:
(Given: Planck’s constant h=6.63×10-34 J s , electron charge ( e ) =1.6×10-19 C)

Options:
A 2.12 eV
B 1.82 eV
C 1.95 eV
D 1.66 eV

The correct answer is:
1.66 eV

Solution:
From Einstein's photoelectric equation, hν=ϕ+eV0 .

⇒V0=hνe-ϕe ......1

At ν=4×1014 Hz, V0=0.

Putting the value in equation 1, we get

0=hve-ϕe

So, Work function (ϕ)=hνe  in eV
=6.63×10-34×4×10141.6×10-19
=1.66 eV

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2019 (12 Apr Shift 1)
The straight chain polymer is formed by

Options:
A hydrolysis of $\left(\mathrm{CH}_3\right)_3 \mathrm{SiCl}$ followed by condensation polymerisation
B hydrolysis of $\mathrm{CH}_3 \mathrm{SiCl}_3$ followed by condensation polymerisation
C hydrolysis of $\left(\mathrm{CH}_3\right)_4 \mathrm{Si}$ by addition polymerisation
D hydrolysis of $\left(\mathrm{CH}_3\right)_2 \mathrm{SiCl}_2$ followed by condensation polymerisation

The correct answer is:
hydrolysis of $\left(\mathrm{CH}_3\right)_2 \mathrm{SiCl}_2$ followed by condensation polymerisation

Solution:
$\mathrm{RSiCl}_3$ (trichloro derivatives) on hydrolysis followed by condensation polymerisation yields branched chain silicones.
Straight chain silanes are silicone oils. These are more stable at high temperature than mineral oils and have less tendency to thicken at low temperature.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
The straight chain polymer is formed by -

Options:
A hydrolysis of $\left(\mathrm{CH}_3\right)_2 \mathrm{SiCl}_2$ followed by condensation polymerization
B hydrolysis of $\left(\mathrm{CH}_3\right)_3 \mathrm{SiCl}$ followed by condensation polymerization
C hydrolysis of $\mathrm{CH}_3 \mathrm{SiCl}_3$ followed by condensation polymerization
D hydrolysis of $\left(\mathrm{CH}_3\right)_4 \mathrm{Si}$ by addition polymerization

The correct answer is:
hydrolysis of $\left(\mathrm{CH}_3\right)_2 \mathrm{SiCl}_2$ followed by condensation polymerization

Solution:

Topic: Chemistry, Polymers, NEET, NEET 2009 (Mains)
The straight line passing through $(-1,1)$ and remaining parallel to the line common to the pairs of lines provided by $6 x^2-x y-12 y^2=0$ and $15 x^2+14 x y-8 y^2=0$, is

Options:
A $5 \mathrm{x}-2 \mathrm{y}+7=0$
B $3 x+4 y-1=0$
C $3 x-4 y+7=0$
D $2 x-3 y+5=0$

The correct answer is:
$3 x+4 y-1=0$

Solution:
$$
\begin{aligned}
& \text { } 6 x^2-x y-12 y^2=0 \\
& \Rightarrow 6 x^2-3 x y+8 x y-12 y^2=0 \\
& \Rightarrow(2 x-3 y)(3 x+4 y)=0 \ldots \text { (i) } \\
& \Rightarrow 15 x^2+14 x y-8 y^2=0 \\
& \Rightarrow 15 x^2+20 x y-6 x y-8 y^2=0 \\
& \Rightarrow(5 x-2 y)(3 x+4 y)=0 \ldots \text { (ii) }
\end{aligned}
$$
Equation of common line from (i) and (ii) is
$$
\begin{aligned}
& 3 \mathrm{x}+4 \mathrm{y}=0 \\
& \Rightarrow \mathrm{y}=\frac{-3}{4} \mathrm{x} \Rightarrow \operatorname{slope}(\mathrm{m})=\frac{-3}{4}
\end{aligned}
$$
Equation of straight line passes through $(-1,1)$ and parallel to $3 \mathrm{x}+4 \mathrm{y}=0$
$$
y-1=\frac{-3}{4}(x+1) \Rightarrow 3 x+4 y-=0
$$

Topic: Mathematics, Straight Lines, AP EAMCET, AP EAMCET 2022 (06 Jul Shift 1)
The straight line touching the circle $x^2+y^2-2 x-3=0$ and remaining normal to the circle $x^2+y^2-4 y-6=0$ is

Options:
A $4 x-3 y+6=0$
B $y+2=0$
C $4 x+3 y-6=0$
D $2 x+3=0$

The correct answer is:
$4 x-3 y+6=0$

Solution:
Line is normal to circle $x^2+y^2-4 y-6=0$
$\therefore$ It passes through centre $(0,2)$.
Let $m$ be slope of line.
$\therefore$ Equation of line is $y-2=m x$
$$
\Rightarrow \quad m x-y+2=0
$$
This line is tangent to $x^2+y^2-2 x-3=0$
centre is $(1,0)$, radius is 2
$$
\begin{aligned}
& \therefore \quad 2=\left|\frac{m+2}{\sqrt{1+m^2}}\right| \\
& \Rightarrow \quad 4\left(1+m^2\right)=(m+2)^2 \\
& \therefore \quad 3 m^2-4 m=0 \\
& \Rightarrow \quad m=0 \\
& \text { or } \quad m=\frac{4}{3} \\
&
\end{aligned}
$$
$\therefore \quad$ Line can be $y=2$
or $\quad 4 x-3 y+6=0$

Topic: Mathematics, Circle, AP EAMCET, AP EAMCET 2022 (07 Jul Shift 2)
The straight line $x+2 y=1$ meets the coordinate axes at $A$ and $\mathrm{B}$. A circle is drawn through $\mathrm{A}, \mathrm{B}$ and the origin. Then the sum of perpendicular distances from $\mathrm{A}$ and $\mathrm{B}$ on the tangent to the circle at the origin is:

Options:
A $\frac{\sqrt{5}}{2}$
B $2 \sqrt{5}$
C $\frac{\sqrt{5}}{4}$
D $4 \sqrt{5}$

The correct answer is:
$\frac{\sqrt{5}}{2}$

Solution:
Let equation of circle be $x^{2}+y^{2}+2 g x+2 f y=0$

As length of intercept on $x$ axis is $1=2 \sqrt{g^{2}-c}$

$\Rightarrow|g|=\frac{1}{2}$

length of intercept on $y$ -axis $=\frac{1}{2}=2 \sqrt{f^{2}-c}$

$\Rightarrow|f|=\frac{1}{4}$

Equation of circle that passes through given points is

$x^{2}+y^{2}-x-\frac{y}{2}=0$

Tangent at (0,0) is,

$(y-0)=\left(\frac{d y}{d x}\right)_{(0,0)}^{(x-0)} \cdot(\mathrm{x}-0)$

$\Rightarrow 2 x+y=0$

Perpendicular distance from $B(1,0)$ on the tangent to the circle $=\frac{\frac{1}{2}}{\sqrt{5}}$

Perpendicular distance from $B\left(0, \frac{1}{2}\right)$ on the tangent to

the circle $=\frac{2}{\sqrt{5}}$

Sum of perpendicular distance $=\frac{\frac{1}{2}+2}{\sqrt{5}}=\frac{\sqrt{5}}{2}$.

Topic: Mathematics, Circle, JEE Main, JEE Main 2019 (11 Jan Shift 1)
The straight line $y=2 x+\lambda$ does not meet the parabola $y^2=2 x$, if

Options:
A $\lambda \lt \frac{1}{4}$
B $\lambda\gt\frac{1}{4}$
C $\lambda=4$
D $\lambda=1$

The correct answer is:
$\lambda\gt\frac{1}{4}$

Solution:
$y=2 x+\lambda$ does not meet, if $\lambda\gt\frac{a}{m}=\frac{1}{2.2}=\frac{1}{4} \Rightarrow \lambda\gt\frac{1}{4}$.

Topic: Mathematics, Parabola, JEE Main
The straight line y=2x meets y=fx at P, where fx is a solution of the differential equation dydx=x2+xyx2+y2 such that f1=3, then f ′ x at point P is

Options:
A 35
B 53
C 2
D 1

The correct answer is:
35

Solution:
Note f ′ x = dy dx = 1+ y x 1+ y x 2
At point P,yx=2
So, dydxP=1+21+4=35

Topic: Mathematics, Differential Equations, JEE Main
The straight lines $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}$ and $\frac{x-1}{2}=\frac{y-2}{2}=\frac{z-3}{-2}$ are

Options:
A Parallel lines
B Intersecting at $60^{\circ}$
C Skew lines
D Intersecting at right angle

The correct answer is:
Intersecting at right angle

Solution:
Since $2(1)+2(2)+(-2)(3)=0$.
Hence lines are intersecting at right angles.

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The straight lines l1 and l2 pass through the origin and trisect the line segment of the line L:9x+5y=45 between the axes. If m1 and m2 are the slopes of the lines l1 and l2, then the point of intersection of the line y=(m1+m2)x with L lies on

Options:
A y–2x=5
B 6x+y=10
C y–x=5
D 6x–y=15

The correct answer is:
y–x=5

Solution:
Given,

The straight lines l1 and l2 pass through the origin and trisect the line segment of the line L:9x+5y=45 between the axes,

And m1 and m2 are the slopes of the lines l1 and l2,

Now on plotting the diagram we get,

Given equation of line, L:9x+5y=45

⇒x5+y9=1

Now using the section formula between point A5,0 & B0,9 we get the value of point C & D

⇒C≡103,3 and D≡53,6

Now finding the slope m1 & m2 we get,

m1=3-0103-0 =910 & m2=6×35=185

So, equation of line y=m1+m2x will be,

y=910+3610x=92x

So, intersection point with L will be,

7y=45⇒y=457, x=107

Hence, y-x=45-107=5

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2023 (06 Apr Shift 1)
The straight lines $x+2 y-9=0, \quad 3 x+5 y-5=0$ and $a x+b y-1=0$ are concurrent, if the straight line $35 x-22 y+1=0$ passes through the point

Options:
A $(a, b)$
B $(b, a)$
C $(-a,-b)$
D None of these

The correct answer is:
$(a, b)$

Solution:
The three lines are concurrent. if $\left|\begin{array}{ccc}1 & 2 & -9 \\ 3 & 5 & -5 \\ a & b & -1\end{array}\right|=0$
$\Rightarrow 35 a-22 b+1=0$
which is true if the line $35 x-22 y+1=0$ passes through $(a, b)$.

Topic: Mathematics, Straight Lines, JEE Main
The strain-stress curves of three wires of different materials are shown in the figure. $P, Q$ and $R$ are the elastic limits of the wires. The figure shows that

Options:
A Elasticity of wire $P$ is maximum
B Elasticity of wire $Q$ is maximum
C Tensile strength of $R$ is maximum
D None of the above is true

The correct answer is:
None of the above is true

Solution:
As stress is shown on $x$-axis and strain on $y$-axis
So we can say that $Y=\cot \theta=\frac{1}{\tan \theta}=\frac{1}{\text { slope }}$
So elasticity of wire $P$ is minimum and of wire $R$ is maximum
So elasticity of wire $P$ is minimum and of wire $R$ is maximum

Topic: Physics, Mechanical Properties of Solids, JEE Main
The stream of a river is flowing with a speed of 2 km h-1. A swimmer can swim at a speed of 4 km h-1. The direction of the swimmer with respect to the flow of the river, to cross the river straight, is

Options:
A 150Â°
B 90°
C 120Â°
D 60Â°

The correct answer is:
120°

Solution:
To reach the point directly opposite to the starting point, the resultant velocity of swimmer should be as shown in figure.

From figure
sinâ¡Î¸=24=12
Î¸=30Â°
Angle with flow =90Â°+30Â°=120Â°

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2019 (09 Apr Shift 1)
The strength of 50 volume solution of hydrogen peroxide is _____ g/L (Nearest integer).

Given:
Molar mass of H2O2 is 34 g mol-1
Molar volume of gas at STP=22.7 L.

Options:

The correct answer is:
150

Solution:
Molarity =Volume strength of H2O2Molar volume of oxygen gas at STP×2

Molarity =5011.35

The strength of the solution=Molarity×Molecular weight of H2O2

∴ Strength in gm/L=5011.35×34.

Topic: Chemistry, Hydrogen, JEE Main, JEE Main 2023 (30 Jan Shift 2)
The strength of 11.2 volume solution of H2O2 is

[Given that, the molar mass of H=1 g mol-1 and O=16 g mol-1]

Options:
A 1.7 %
B 13.6 %
C 34 %
D 3.4 %

The correct answer is:
3.4 %

Solution:
Molarity of H2O=Volume strength11.2

M=11.211.2=1

∴ 1 molar means 1000 mL solution contains 1 mol H2O2=1×34 g

Since, the strength is represented as weight of the solute in 1 L of the solution. The strength of the solution will be 34 g L-1.

Note: The question should've mentioned that we need to find the mass by volume percentage.

Percentage w/v =341000×100=3.4%

Topic: Chemistry, Hydrogen, JEE Main, JEE Main 2019 (08 Apr Shift 2)
The strength of an aqueous NaOH solution is most accurately determined by titrating: (Note: consider that an appropriate indicator is used)

Options:
A Aq.NaOH in a pipette and aqueous oxalic acid in a burette
B Aq.NaOH in a burette and aqueous oxalic acid in a conical flask
C Aq.NaOH in a burette and concentrate H2SO4 in a conical flask
D Aq.NaOH in a volumetric flask and concentrated H2SO4 in a conical flask

The correct answer is:
Aq.NaOH in a burette and aqueous oxalic acid in a conical flask

Solution:
Oxalic acid is a primary standard solution while H2SO4 is a secondary standard solution.

Titration is a common laboratory method of quantitative chemical analysis to determine the concentration of an identified analyte. A reagent, termed the titrant or titrator, is prepared as a standard solution of known concentration and volume.

Topic: Chemistry, Redox Reactions, JEE Main, JEE Main 2020 (08 Jan Shift 1)
The strength of the magnetic field at a point r near a long straight current-carrying wire is B. The field at a distance r2 will be

Options:
A B2
B B4
C 2B
D 4B

The correct answer is:
2B

Solution:
B∝1r⇒B1B2=r2r1⇒BB2=r2r⇒B2=2B

Topic: Physics, Magnetic Effects of Current, JEE Main
The stress along the length of a rod (with rectangular cross-section) is $1 \%$ of the Young's modulus of its material. What is the approximate percentage of change of its volume? (Poisson's ratio of the material of the rod is $0.3 .$ )

Options:
A $3 \%$
B $1 \%$
C $0.7 \%$
D $0.4 \%$

The correct answer is:
$0.4 \%$

Solution:
$\because$ Stress, $\frac{F}{\Delta A}=1 \%$ of $Y=\frac{Y}{100}$
But Young's modulus, $Y=\frac{\text { stress }}{\text { strain }}=\frac{\frac{F}{\Delta A}}{\frac{\Delta l}{l}}$
$\therefore \quad Y=\frac{\frac{Y}{100}}{\frac{\Delta}{l}} \quad\left(\right.$ putting $\left.\frac{F}{\Delta A}=\frac{Y}{100}\right)$
$\therefore \quad \frac{\Delta l}{l}=\frac{1}{100}$
Poisson's ratio, $\sigma=\frac{-\frac{\Delta r}{r}}{\frac{\Delta l}{l}}$
$\therefore \quad \frac{\Delta r}{r}=-\sigma \cdot \frac{\Delta l}{l}=\frac{-0.3}{100}$
$$
\frac{\Delta r}{r}=\frac{-0.3}{100}
$$
$\therefore$ Change in volume, $\frac{\Delta V}{V}=\frac{2 \Delta r}{r}+\frac{\Delta l}{l}$
$$
\begin{array}{l}
=\frac{2 \times(-0.3)}{100}+\frac{1}{100} \\
=\frac{1-0.6}{100}=\frac{0.4}{100}
\end{array}
$$
$\Delta V \%=0.4 \%$

Topic: Physics, Mechanical Properties of Solids, JEE Main
The stress along the length of a rod (with rectangular cross-section) is \( 1 \% \) of the Young's modulus of its material. What is the approximate percentage of change of its volume? (Poisson's ratio of the material of the rod is \( 0.3 \) )

Options:
A \( 3 \% \)
B \( 1 \% \)
C \( 0.7 \% \)
D \( 0.4 \% \)

The correct answer is:
\( 0.4 \% \)

Solution:
Given,

The stress along the length of a rod (with rectangular cross-section) is 1% of the Young's modulusY of its material and

Poisson's ratio,γ=-lateral strainlongitudinal strain=0.3

⇒lateral strain=-γ∆ll=-0.3∆ll.

By Hooke's law,

stress=Y×∆ll

⇒0.01Y=Y∆ll

⇒∆ll=0.01.

If length, breadth, and width of the rod are represented by l,b and w respectively then

Volume, V=lbw

⇒∆VV=∆ll+∆bb+∆ww

⇒∆VV=∆ll-γ∆ll-γ∆ll

⇒∆VV=∆ll-2γ∆ll

⇒∆VV=0.011-20.3

⇒∆VV=0.004

∴∆VV×100=0.4%.

Topic: Physics, Mechanical Properties of Solids, JEE Main
The stress versus strain graphs for wires of two materials $A$ and $B$ are as shown in the figure. If $Y_A$ and $Y_B$ are the Young 's modulii of the materials, then

Options:
A $Y_B=2 Y_A$
B $Y_A=Y_B$
C $Y_B=3 Y_A$
D $Y_A=3 Y_B$

The correct answer is:
$Y_A=3 Y_B$

Solution:
$\frac{Y_A}{Y_B}=\frac{\tan \theta_A}{\tan \theta_B}=\frac{\tan 60}{\tan 30}=\frac{\sqrt{3}}{1 / \sqrt{3}}=3 \Rightarrow Y_A=3 Y_B$

Topic: Physics, Mechanical Properties of Solids, JEE Main
The strongest acid amongst the following compounds is :

Options:
A
$\mathrm{HCOOH}$

B
$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}(\mathrm{Cl}) \mathrm{CO}_2 \mathrm{H}$

C
$\mathrm{ClCH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{COOH}$

D
$\mathrm{CH}_3 \mathrm{COOH}$

The correct answer is:
$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}(\mathrm{Cl}) \mathrm{CO}_2 \mathrm{H}$

Solution:
Electron releasing groups (Alkyl groups) de stabilizes conjugate base.
The $+$ I effect of $\mathrm{C}_3 \mathrm{H}_7$ is less than $-1$ effect of $\mathrm{Cl}$
$\mathrm{K}_{\mathrm{a}}$ of $\mathrm{HCOOH}$ is $17.9 \times 10^{-5}$
$\mathrm{K}_{\mathrm{a}}$ of $\mathrm{CH}_3 \mathrm{CH}_2 \underset{\mathrm{Cl}}{\mathrm{CH}}-\mathrm{COOH}$ is $139 \times 10^{-5}$

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2011
The strongest acid from the following is

Options:
A

B

C

D

The correct answer is:

Solution:
In the case of substituted phenols, the acidity of phenols increases in the presence of the electron-withdrawing group. This is due to the stability of the phenoxide ion generated. In case of meta substituted phenols, the acidic nature can be explained on the basis of inductive effect. Among the given groups -NO2 & -Cl groups are electron withdrawing groups and methyl group is electron releasing group.

Acidic strength order is :

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2023 (06 Apr Shift 2)
The strongest base amongst the following is (In aqueous state):

Options:
A NH3
B C2H5NH2
C C2H52N
D C2H53N

The correct answer is:
C2H52N

Solution:
The basic order depends upon the following three factors:
1. +I effect, basicity increases
2. steric factor - When R group around N increases, then its ability to accept proton decreases.
3. Solvation of ammonium ion - When R group around N increases, then its solvation in water is less.
In aqueous state, 2o C2H52N is most stable

Topic: Chemistry, General Organic Chemistry, JEE Main