datavorous/entrance-exam-dataset huggingface dataset:
The negative $\mathrm{Zn}$ pole of a Daniell cell, sending a constant current through a circuit, decreases in mass by $0.13 \mathrm{~g}$ in 30 minutes. If the electrochemical equivalent of $\mathrm{Zn}$ and $\mathrm{Cu}$ are $32.5$ and $31.5$ respectively, the increase in the mass of the positive Cu pole in this time is

Options:
A
$0.180 \mathrm{~g}$

B
$0.141 \mathrm{~g}$

C
$0.126 \mathrm{~g}$

D
$0.242 \mathrm{~g}$

The correct answer is:
$0.126 \mathrm{~g}$

Solution:
No Solution Available

Topic: Physics, Current Electricity, JEE Main, JEE Main 2003
The net magnetic flux of a magnetic field passing through a closed surface is

Options:
A \( \frac{q}{\mu_{0}} \)
B Zero
C \( \mu_{0} I \)
D \( \oiint \vec{B} \cdot \overrightarrow{d s} \)

The correct answer is:
Zero

Solution:
According to Maxwell's second equation, the total magnetic flux passing through a closed surface is zero.

∯B⇀.ds⇀ = 0

where ∯B⇀.ds⇀ is the surface integral of the magnetic field over a closed surface. This law confirms that the magnetic field is generated due to the dipole nature of the magnet as the number of magnetic lines entering the surface is equal to the number of magnetic lines leaving it. Thus, magnetic lines form closed loops and no isolated magnetic monopole can exist.

Topic: Physics, Electromagnetic Waves, JEE Main
The net work done on the gas in the cycle ABCDA is

Options:
A
Zero

B
$276 \mathrm{R}$

C
$1076 \mathrm{R}$

D
$1904 \mathrm{R}$

The correct answer is:
$276 \mathrm{R}$

Solution:
$$
\begin{aligned}
\text { Net work done in a cycle } & =W_{A B}+W_{B C}+W_{C B}+W_{B A} \\
& =400 R+2 \times 2.303 \times 500 R \ln 2-400 R-414 R \\
& =1000 R \times \ln 2-600 R \times \ln 2=400 R \times \ln 2=276 R
\end{aligned}
$$

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2009
The net work done through a series of changes reported is figure at the end of cycle for an ideal gas is equal to

Options:
A zero
B -5PV
C -3PV
D -2PV

The correct answer is:
-3PV

Solution:
Net work done = Area covered by the cycle (Clockwise)
=-124P-P×3V-V
=-12×6PV
=-3PV

Topic: Chemistry, Thermodynamics (C), JEE Main
The neutralization occurs when 10 mL of 0.1 M acid 'A' is allowed to react with 30 mL of 0.05 M base MOH2. The basicity of the acid 'A' is
[M is a metal]

Options:

The correct answer is:
3

Solution:
Let basicity of acid is $x$

A+MOH2⟶Salt+Water

At complete neturization.

Milli eq. of acid = Milli eq. of base

n0.1×10=20.05×30

n=3

Topic: Chemistry, Redox Reactions, JEE Main, JEE Main 2022 (25 Jun Shift 2)
The new varieties of plants are produced by

Options:
A Selection and hybridisation
B Mutation and selection
C Introduction and mutation
D Selection and introduction

The correct answer is:
Selection and hybridisation

Solution:
The main purpose of plant breeding is the manipulation of plant species to create desired plant types that are better suited for cultivation. The manupulated plants are disease resistant and gives better yields. Such plants are selected and cross hybridised to bring their traits together in the progeny.

Topic: Biology, Strategies for Enhancement in Food Production, NEET
The nitride ion in lithium nitride is composed of

Options:
A 7 protons +10 electrons
B 10 protons +10 electrons
C 7 protons +7 protons
D 10 protons +7 electrons

The correct answer is:
7 protons +10 electrons

Solution:
In case of $N^{3-}, p=7$ and $c=10$

Topic: Chemistry, Structure of Atom, JEE Main
The nitrosation of N, N-dimethylaniline takes place through the attack of electrophile

Options:
A nitronium ion
B protonated nitrous acid
C nitrous acid
D nitrosonium ion

The correct answer is:
nitrosonium ion

Solution:
The actual nitrosation agent is the nitrosyl cation, NO+ which is formed as follows

Aromatic tertiary amine i.e., N, N-dimethylaniline, undergo electrophilic substitution with nitrosonium ion at p-position of phenyl ring.

Topic: Chemistry, Amines, JEE Main
The noble gas that does not occur in the atmosphere is;

Options:
A Kr
B Ne
C Ra
D He

The correct answer is:
Ra

Solution:
In the atmosphere, all the noble gases occur except radon. Out of all noble gases, argon is the major constituent. A decay product of radium is radon.

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2019 (10 Apr Shift 2)
The noble gas which forms maximum number of compounds is

Options:
A $A r$
B $\mathrm{He}$
C $\mathrm{Xe}$
D $\mathrm{Ne}$

The correct answer is:
$\mathrm{Xe}$

Solution:
$\mathrm{XeF}_2, \mathrm{XeOF}_2, \mathrm{XeF}_4, \mathrm{XeOF}_4, \mathrm{XeF}_6, \mathrm{XeO}_3$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
The nodal plane in the Ï - bond of ethene is located

Options:
A A plane parallel to the molecular plane
B A plane perpendicular to the molecular plane which bisects the carbon - carbon Ï -bond at right angle .
C A plane perpendicular to the molecular plane which contains the carbon - carbon σ - bond .
D Along the molecular plane

The correct answer is:
Along the molecular plane

Solution:
π Bond's Nodal Plane along & Passing through Internuclear Axis

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The non-metal that does not exhibit positive oxidation state is:

Options:
A Chlorine
B Iodine
C Fluorine
D Oxygen

The correct answer is:
Fluorine

Solution:
Fluorine is most electronegative in the long form of periodic table. Due to its highest electronegativity, it will only accept the electron so, it exhibits O.S. value of -1 only.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2016 (09 Apr Online)
The non-metal used in the purple coloured solution applied on wounds as an antiseptic is

Options:
A Chlorine
B Iodine
C Bromine
D Sulphur

The correct answer is:
Iodine

Solution:
Iodine is a non-metal that is used to make a purple-colored antiseptic solution called tincture of iodine that is applied to wounds. Tincture of iodine is made by dissolving iodine, sodium or potassium iodide, and plant extract in ethanol and water. The solution contains \(2-7 \%\) iodine and has antiseptic properties that prevent bacterial growth. When applied to a wound, the solution sterilizes it and protects it from microbial infection.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main
The non-zero verctors $\vec{a}, \vec{b}$ and $\vec{c}$ are related by $\vec{a}=8 \vec{b}$ and $\vec{c}=-7 \vec{b}$. Then the angle between $\vec{a}$ and $\overrightarrow{\mathrm{c}}$ is

Options:
A
0

B
$\pi / 4$

C
$\pi / 2$

D
$\pi$

The correct answer is:
$\pi$

Solution:
Since $\vec{a}=8 \vec{b}$
$$
\overrightarrow{\mathrm{c}}=-7 \overrightarrow{\mathrm{b}}
$$
$\therefore \vec{a}$ and $\vec{b}$ are like vectors and $\vec{b}$ and $\vec{c}$ are unlike.
$\Rightarrow \vec{a}$ and $\vec{c}$ will be unlike
Hence, angle between $\vec{a}$ and $\vec{c}=\pi$.

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2008
The normal activity of living carbon-containing matter is found to be about 15 decays per minute for every gram of carbon. This activity arises from the small proportion
of radioactive ${ }_6^{14} \mathrm{C}$ present with the stable carbon isotope ${ }_6^{12} \mathrm{C}$ when the organism is dead, its interaction with the atmosphere (which maintains the above equilibrium activity) ceases and its activity begins to drop. From the known half lives of ${ }_6^{14} \mathrm{C}$, and the measured activity, the age of the specimen can be approximately estimated. This is the principle of ${ }_6^{14} \mathrm{C}$ dating used in archaeology. Suppose a specimen from Mohenjodaro gives an activity of 9 decays per minute per gram of carbon. Estimate the approximate age of the Indus-Valley civilisation.

Options:

None

Solution:
Normal activity $R_0=15$ decays $/ \mathrm{min}$;
Present activity $R=9$ decays $/ \mathrm{min} ; \mathrm{T}=5730$ yrs, $\mathrm{t}=$ ? $\frac{\mathrm{N}}{\mathrm{N}_0}=\frac{\mathrm{R}}{\mathrm{R}_0}=\frac{9}{15}$ also, $\frac{\mathrm{N}}{\mathrm{N}_0}=\mathrm{e}^{-\lambda \mathrm{t}}=\frac{9}{15}$
$$
\Rightarrow \mathrm{e}^{-\lambda . \mathrm{t}}=\frac{3}{5} \Rightarrow \mathrm{e}^{\lambda \mathrm{t}}=\frac{5}{3}
$$
Taking log on both sides,
$\Rightarrow \lambda t \log _{\mathrm{e}} \mathrm{e}=2.303 \log _{\mathrm{e}} 1.667 \Rightarrow \lambda \mathrm{t}=2.303 \times 0.228$
$\Rightarrow \mathrm{t}=\frac{0.5109}{\lambda}$ also $\lambda=\frac{0.693}{\mathrm{~T}}=\frac{0.693}{5730} \mathrm{yr}^{-1}$
$$
\Rightarrow \mathrm{t}=0.5109 \times \frac{5730}{0.693} \Rightarrow \mathrm{t}=4224.3 \mathrm{yr}
$$

Topic: Physics, Nuclear Physics
The normal at $\left(2, \frac{3}{2}\right)$ to the ellipse, $\frac{x^2}{16}+\frac{y^2}{3}=1$ touches a parabola, whose equation is

Options:
A
$y^2=-104 x$

B
$y^2=14 x$

C
$y^2=26 x$

D
$y^2=-14 x$

The correct answer is:
$y^2=-104 x$

Solution:
Ellipse is $\frac{x^2}{16}+\frac{y^2}{3}=1$
Now, equation of normal at $(2,3 / 2)$ is
$$
\begin{aligned}
& \frac{16 x}{2}-\frac{3 y}{3 / 2}=16-3 \\
& \Rightarrow \quad 8 x-2 y=13 \\
& \Rightarrow \quad y=4 x-\frac{13}{2}
\end{aligned}
$$
Let $y=4 x-\frac{13}{2}$ touches a parabola $y^2=4 a x$.
We know, a straight line $y=m x+c$ touches a parabola $y^2=4 a x$ if $a-m c=0$
$$
\therefore \quad a-(4)\left(-\frac{13}{2}\right)=0 \Rightarrow a=-26
$$
Hence, required equation of parabola is $y^2=4(-26) x=-104 x$

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2012 (26 May Online)
The normal at the point $\left(b t_1^2, 2 b t_1\right)$ on a parabola meets the parabola again in the point $\left(b t_2^2, 2 b t_2\right)$, then

Options:
A $t_2=-t_1-\frac{2}{t_1}$
B $t_2=-t_1+\frac{2}{t_1}$
C $t_2=t_1-\frac{2}{t_1}$
D $t_2=t_1+\frac{2}{t_1}$

The correct answer is:
$t_2=-t_1-\frac{2}{t_1}$

Solution:
\(\begin{aligned}
& y^2=4 b x \\
& \Rightarrow \frac{d y}{d x}=\frac{2 b}{y}
\end{aligned}\)
Therefore, at the point \(\left(\mathrm{bt}_1^2, 2 b \mathrm{t}_1\right), \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\mathrm{t}_1}\)
So, the slope of normal at \(\left(b t_1^2, 2 b t_1\right)=\frac{-1}{\left(\frac{d y}{d x}\right)}=-t_1\)
Therefore, the equation of normal at \(\left(b t_1^2, 2 b t_1\right)\) is \(\left(y-2 b t_1\right)=-t_1\left(x-b t_1^2\right)\)
The point \(\left(\mathrm{bt}_2^2, 2 \mathrm{bt}_2\right)\) also lies on the normal
\(\begin{aligned}
& \Rightarrow\left(2 b t_2-2 b t_1\right)=-t_1\left(b t_2^2-b t_1^2\right) \\
& \Rightarrow 2 b\left(t_2-t_1\right)=-t_1 b\left(t_2-t_1\right)\left(t_2+t_1\right) \\
& \Rightarrow 2 b\left(t_2-t_1\right)+t_1 b\left(t_2-t_1\right)\left(t_2+t_1\right)=0 \\
& \Rightarrow b\left(t_2-t_1\right)\left(2+t_1\left(t_2+t_1\right)\right)=0 \\
& \Rightarrow 2+t_1\left(t_2+t_1\right)=0 \ldots .\left[t_2 \neq t_1\right] \text { and }[b \neq 0] \\
& \Rightarrow t_2=-t_1-\frac{2}{t_1}
\end{aligned}\)

Topic: Mathematics, Parabola, JEE Main
The normal at the point $\left(\mathrm{bt}_1{ }^2, 2 \mathrm{bt}_1\right)$ on a parabola meets the parabola again in the point $\left(\mathrm{bt}_2{ }^2, 2 \mathrm{bt}_2\right)$, then

Options:
A
$\mathrm{t}_2=\mathrm{t}_1+\frac{2}{\mathrm{t}_1}$

B
$t_2=-t_1-\frac{2}{t_1}$

C
$\mathrm{t}_2=-\mathrm{t}_1+\frac{2}{\mathrm{t}_1}$

D
$\mathrm{t}_2=\mathrm{t}_1-\frac{2}{\mathrm{t}_1}$

The correct answer is:
$t_2=-t_1-\frac{2}{t_1}$

Solution:
Fundamental theorem (fact) $t_2=-t_1-\frac{2}{t_1}$

Topic: Mathematics, Parabola, JEE Main, JEE Main 2003
The normal at the point \( P\left(a p^{2}, 2 a p\right) \) meets the parabola \( y^{2}=4 a x \) again at \( Q\left(a q^{2}, 2 a q\right) \) such that the lines joining the origin to \( P \) and \( Q \) are at right angle. Then

Options:
A \( p^{2}=2 \)
B \( q^{2}=2 \)
C \( p=2 q \)
D \( q=2 p \)

The correct answer is:
\( p^{2}=2 \)

Solution:
Since the normal at Pap2,2ap to y2=4ax meets the parabola at Qaq2, 2aq, therefore PQ is the focal chord. Hence

q=-p-2p ...i

Since OP⊥OQ,

∴ 2ap-0ap2-0×2aq-0aq2-0=-1

⇒ pq=-4

⇒ p-p-2p=-4 [Using (i)]

⇒ p2=2

Topic: Mathematics, Parabola, JEE Main
The normal density of a material is ρ and its bulk modulus of elasticity is K. The magnitude of increase in density of material, when a pressure P is applied uniformly on all sides, will be :

Options:
A PKρ
B ρPK
C ρKP
D KρP

The correct answer is:
ρPK

Solution:
ρ=MV

dρρ=-dVV

k=-PdVV

-dVV=Pk

dρρ=Pk⇒dρ=ρPk

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2021 (26 Feb Shift 1)
The normal rain water is slightly acidic and its pH value is 5.6 because of which one of the following ?

Options:
A CO2+H2O→H2CO3
B N2O5+H2O→2HNO3
C 2SO2+O2+2H2O→2H2SO4
D 4NO2+O2+2H2O→4HNO3

The correct answer is:
CO2+H2O→H2CO3

Solution:
Rain water has a pH of 5-6 due to presence of H+ions formed by the reaction of rain water with carbon dioxide present in the atmosphere.

H2O(l)+CO2( g)⇌H2CO3

H2CO3⇌H++HCO3

So, the correct option is A.

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2023 (31 Jan Shift 2)
The normal reaction N for a vehicle of 800 kg mass, negotiating a turn on a 30° banked road at maximum possible speed without skidding is ____ ×103 kg m s-2.

Options:
A 10.2
B 7.2
C 12.4
D 6.96

The correct answer is:
10.2

Solution:
At vmax, f will be limiting in nature.

∴ Balancing force in vertical direction, Ncos30°-mg-μNcos60°=0
⇒Ncos30°-μcos60°=mg
∴ N=800×10(0.87-0.1)≈10.2×103 kg m s-2
Hence, option 1.

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2021 (20 Jul Shift 1)
The normal to a curve at $P(x, y)$ meets the $x$-axis at $G$. If the distance of $G$ from the origin is twice the abscissa of $P$, then the curve is a

Options:
A
ellipse

B
parabola

C
circle

D
pair of straight lines

The correct answer is:
ellipse

Solution:
Equation of normal is $Y-y=-\frac{d x}{d y}(X-x)$
$\begin{aligned}
& \Rightarrow G \equiv\left(x+y \frac{d y}{d x}, 0\right) \\
& \left|x+y \frac{d y}{d x}\right|=|2 x| \\
& \Rightarrow y \frac{d y}{d x}=x \text { or } y \frac{d y}{d x}=-3 x \\
& y d y=x d x \text { or } y d y=-3 x d x \\
& \frac{y^2}{2}=\frac{x^2}{2}+c \text { or } \frac{y^2}{2}=-\frac{3 x^2}{2}+c \\
& x^2-y^2=-2 c \text { or } 3 x^2+y^2=2 c.
\end{aligned}$

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2007
The normal to the curve x2+2xy-3y2=0, at 1, 1

Options:
A Meets the curve again in the fourth quadrant
B Does not meet the curve again
C Meets the curve again in the second quadrant
D Meets the curve again in the third quadrant

The correct answer is:
Meets the curve again in the fourth quadrant

Solution:
Given x2+2xy-3y2=0
x+3yx-y=0
Pair of straight lines passing through the origin.
∵ x+3y=0 or x-y=0
Normal exists at 1, 1 which is on x-y=0
⇒ The slope of normal at 1, 1=-1
∴ Equation of normal will be
y-1=-x-1 y-y1=mx-x1
y-1=-x+1
x+y=2
Now, find the point of intersection with x+3y=0.
x+y=2x+3y=0
----------------
-2y=2 ⇒y=-1, x=3
∵ 3, -1 lies in the fourth quadrant.

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2015 (04 Apr)
The normal to the curve $x=a(1+\cos \theta), y=a \sin \theta$ at ' $\theta$ ' always passes through the fixed point

Options:
A
$(a, 0)$

B
$(0, a)$

C
$(0,0)$

D
$(a, a)$

The correct answer is:
$(a, 0)$

Solution:
Eliminating $\theta$, we get $(x-a)^2+y^2=a^2$.
Hence normal always pass through $(a, 0)$

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2004
The normal to the curve $x=a(\cos \theta+\theta \sin \theta), y=a(\sin \theta-\theta \cos \theta)$ at any point ' $\theta$ ' is such that

Options:
A
it passes through the origin

B
it makes angle $\frac{\pi}{2}+\theta$ with the $x$-axis

C
it passes through $\left(\mathrm{a} \frac{\pi}{2},-\mathrm{a}\right)$

D
it is at a constant distance from the origin

The correct answer is:
it is at a constant distance from the origin

Solution:
Clearly $\frac{\mathrm{dy}}{\mathrm{dx}}=\tan \theta \Rightarrow$ slope of normal $=-\cot \theta$
Equation of normal at ' $\theta$ ' is
$y-a(\sin \theta-\theta \cos \theta)=-\cot \theta(x-a(\cos \theta+\theta \sin \theta)$
$\Rightarrow y \sin \theta-a \sin ^2 \theta+a \theta \cos \theta \sin \theta=-x \cos \theta+a \cos ^2 \theta+a \theta \sin \theta \cos \theta$
$\Rightarrow x \cos \theta+y \sin \theta=a$
Clearly this is an equation of straight line which is at a constant distance 'a' from origin.

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2005
The normal to the curve yx-2 x-3=x+6 at the point where the curve intersects the y-axis passes through the point:

Options:
A -12, -12
B 12,12
C 12, -13
D 12,13

The correct answer is:
12,12

Solution:
To find the point where curve intersects the y-axis
Put, x=0, we get

6y=6⇒y=1
Hence, the point is (0, 1).
On differentiating
⇒dydx x-2 x-3+y2x-5=1
Put, x=0, y=1, we get
⇒ dydx+6+1-5=1
⇒dydx=1
⇒ Slope of normal =-1
Equation of normal at 0, 1

⇒y-1=-1x-0
⇒x+y-1=0
Point 12,12 satisfy the normal.

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2017 (02 Apr)
The normal to the hyperbola x2a2-y29=1 at the point 8,33 on it passes through the point

Options:
A 15,-23
B 9,23
C -1,93
D -1,63

The correct answer is:
-1,93

Solution:
Given,

hyperbola: x2a2-y29=1

it passes through 8,33

∵64a2-279=1⇒a2=16

Now, equation of normal to hyperbola in point form

∵16x8+9y33=16+9

⇒2x+3y=25

-1,93 satisfies the equation.

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2022 (26 Jun Shift 2)
The normal to the parabola y2=4x at P9,6 meets the parabola again at Q. If the tangent at Q meets the directrix at R, then the slope of another tangent drawn from point R to this parabola is

Options:
A 11
B 113
C 311
D 3

The correct answer is:
113

Solution:
Parameter corresponding to P is t1=3
Hence, the parameter corresponding to Q is t2=-t1-2t1=-3-23=-113

Let, St3 be the point where another tangent from R touches the parabola
Now, if the tangent at Qt2 and St3 meet on the directrix at R then t2t3=-1

⇒t3=311

So the equation of the tangent at S is 311y=x+9121
Hence, the slope of the tangent at S is 113

Topic: Mathematics, Parabola, JEE Main
The normal to the parabola y 2 =8x at the point (2,4) meets the parabola again at the point

Options:
A –18,–12
B –18,12
C 18,12
D 18,-12

The correct answer is:
18,-12

Solution:
We have, t 2 =− t 1 − 2 t 1

Since a=2, t1=1
∴t2=−3
Hence, the other end will be (a t 2 2 , 2a t 2 ) i.e. (18,-12).

Topic: Mathematics, Parabola, JEE Main
The normality of 30 volume \(\mathrm{H}_2 \mathrm{O}_2\) is

Options:
A \(2.678 \mathrm{~N}\)
B \(5.336 \mathrm{~N}\)
C \(8.034 \mathrm{~N}\)
D \(6.685 \mathrm{~N}\)

The correct answer is:
\(5.336 \mathrm{~N}\)

Solution:
Hints: Volume strength \(=5.6 \times\) normality
\(\begin{aligned}
& 30=5.6 \times \mathrm{N} \\
& \Rightarrow \mathrm{N}=\frac{30}{5.6}=5.3
\end{aligned}\)

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
The north pole of a long horizontal bar magnet is being brought closer to a vertical conducting plane along the perpendicular direction. The direction of the induced current in the conducting plane will be

Options:
A Horizontal
B Vertical
C Clockwise
D Anticlockwise

The correct answer is:
Anticlockwise

Solution:
According to Lenz's law
$e=-N\left(\frac{\Delta B}{\Delta t}\right) \cdot A \cos \theta$

Topic: Physics, Electromagnetic Induction, JEE Main
The Novolac polymer has mass of 963 g. The number of monomer units present in it are

Options:

The correct answer is:
9

Solution:
The molar mass of the monomer unit in the box =107

Assume there are n number of monomer units,

So, total molar mass n×107 = 963

n=9

Topic: Chemistry, Polymers, JEE Main, JEE Main 2022 (25 Jun Shift 2)
The nuclear activity of a radioactive element becomes 18th of its initial value in 30 years. The half-life of radioactive element is ________ years.

Options:

The correct answer is:
10

Solution:
A=A0e-λt

A08=A0e-λt⇒λt=ln8

λt=3ln2

ln2λ=t3=303=10 years

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2021 (25 Jul Shift 2)
The nuclei of which one of the following pairs of nuclei are isotones?

Options:
A ${ }_{34} \mathrm{Se}^{74},{ }_{31} \mathrm{Ga}^{71}$
B ${ }_{38} \mathrm{Sr}^{84},{ }_{38} \mathrm{Ga}^{86}$
C ${ }_{42} \mathrm{Se}^{92},{ }_{40} \mathrm{Ga}^{92}$
D ${ }_{20} \mathrm{Se}^{40},{ }_{16} \mathrm{Ga}^{12}$

The correct answer is:
${ }_{34} \mathrm{Se}^{74},{ }_{31} \mathrm{Ga}^{71}$

Solution:
Isotones means number of neutron remains same.

Topic: Physics, Nuclear Physics, NEET, NEET 2005
The nucleus is a significant organelle found in the cell. Read the following statements about the nucleus and select the option that correctly determines the number of false statements.

(i) The nuclear envelope consists of two parallel membranes having space between them called pericentriolar space.

(ii) The outer membrane of the nucleus usually remains continuous with the Golgi apparatus.

(iii) The nuclear matrix or the nucleoplasm contains nucleolus and chromatin.

(iv) Larger and more numerous nucleoli are present in cells actively carrying out lipid synthesis

Options:
A One
B Two
C Three
D Four

The correct answer is:
Three

Solution:
Nucleus first discovered by Brown found in eukaryotic cell has an envelope made of two parallel membranes. The space between membranes is called as pernuclear space. The outer membrane of the nucleus has been observed to run continuous with endoplasmic reticulum rather than Golgi bodies. Nucleolus and chromatin are found within the matrix of nucleus. Cells actively carrying out protein synthesis generally show larger and multiple nucleoli.

Topic: Biology, Cell - The Unit of Life, NEET
The nucleus ${ }_n^m X$ emits one $\alpha$ particle and $2 \beta$-particles. The resulting nucleus is

Options:
A $\begin{aligned} & m-4 \\ & n-2\end{aligned} Y$
B $\begin{aligned} & m-6 \\ & n-4\end{aligned} Z$
C ${ }_n^{m-6} Z$
D ${ }_n^{m-4} X$

The correct answer is:
${ }_n^{m-4} X$

Solution:
Emission of $\alpha$-particle decreases the mass number and the atomic number by 4 and 2 respectively. Emission of $\beta$-particle increases the atomic number by 1 while the mass number remains unchanged.
After the emission of the one $\alpha$-particle and two $\beta$-particles
Decreases in mass number $=4-0=4$
Decreases in atomic number $=2-2=0$
$\therefore$ The resulting the nucleus is $m-4, X$.
i.e. the isotope of the given nucleus.

Topic: Physics, Nuclear Physics, AIIMS, AIIMS 2012
The nucleus of an atom is spherical. The relation between radius of the nucleus and mass number A is given by 1.25Ã10-13ÃA13 cm . If radius of atom is one Ã and the mass number is 64, then the fraction of the atomic volume that is occupied by the nucleus is xÃ10-13 . Calculate x

Options:

The correct answer is:
1.25

Solution:
Radius of nucleus =1.25âÃ10-13ÃA13âcm
=1.25âÃ10-13Ã6413=5Ã10-13âcm
Radius of atom = 1Ãâ=10-8cm.
Volume of nucleusVolume of atom=43Ïâ(5Ã10-13)343Ïâ(10-8)3
=1.25âÃ10-13 .

Topic: Chemistry, Chemical Kinetics, JEE Main
The nucleus of the element having atomic number 25 and atomic weight 55 will contain

Options:
A 25 protons and 30 neutrons
B 25 neutrons and 30 protons
C 55 protons
D 55 neutrons

The correct answer is:
25 protons and 30 neutrons

Solution:
No. of protons $=$ Atomic no. $=25$ and no. of neutron $=55-$ $25=30$.

Topic: Chemistry, Structure of Atom, JEE Main
The number 292 in decimal system is expressed in binary system by

Options:
A 100001010
B 10001000 .
C 100100100
D 101010000

The correct answer is:
100100100

Solution:
Remainder
Required answer $=100100$

Topic: Mathematics, Sets and Relations, NDA, NDA 2012 (Phase 1)
The number \((101)^{100}-1\) is divisible by

Options:
A \(10^4\)
B \(10^6\)
C \(10^8\)
D \(10^{12}\)

The correct answer is:
\(10^4\)

Solution:
\(\begin{aligned}\text {Hints: }
& (101)^{100}-1={ }^{100} \mathrm{C}_1 100+{ }^{100} \mathrm{C}_2 100^2+{ }^{100} \mathrm{C}_3 100^3+\ldots \ldots +{ }^{100} \mathrm{C}_{100} 100^{100} \\
& =100^2\left[1+{ }^{100} \mathrm{C}_2+{ }^{100} \mathrm{C}_3 100+\ldots \ldots .\right] \\
& =\left(10^4\right)
\end{aligned}\)

Topic: Mathematics, Binomial Theorem, JEE Main
The number and type of bonds between two carbon atoms in CaC2 are:

Options:
A One sigma and one pi bonds
B One sigma and two pi bonds
C One sigma and half pi bond
D One sigma bond

The correct answer is:
One sigma and two pi bonds

Solution:
The structure of CaC2 is Ca2+ -Câ¡C- so it has one sigma and 2 pi bonds.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main