datavorous/entrance-exam-dataset huggingface dataset:
The \(\mathrm{F}^{-}\)ions make the enamel on teeth much harder by converting hydroxyapatite (the enamel on the surface of teeth) into much harder fluoroapatite having the formula.

Options:
A \(\left[3\left(\mathrm{Ca}_3\left(\mathrm{PO}_4\right)_2\right) \cdot \mathrm{Ca}(\mathrm{OH})_2\right]\)
B \(\left[3\left(\mathrm{Ca}_3\left(\mathrm{PO}_4\right)_3\right) \cdot \mathrm{CaF}_2\right]\)
C \(\left[3\left(\mathrm{Ca}_2\left(\mathrm{PO}_4\right)_2\right) \cdot \mathrm{Ca}(\mathrm{OH})_2\right]\)
D \(\left[3\left(\mathrm{Ca}_3\left(\mathrm{PO}_4\right)_2\right) \cdot \mathrm{CaF}_2\right]\)

The correct answer is:
\(\left[3\left(\mathrm{Ca}_3\left(\mathrm{PO}_4\right)_2\right) \cdot \mathrm{CaF}_2\right]\)

Solution:
Fluoroapatite $\Rightarrow\left[3 \mathrm{Ca}_3\left(\mathrm{PO}_4\right)_2 \cdot \mathrm{CaF}_2\right]$

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2024 (09 Apr Shift 1)
The \( \mathrm{pK}_{\mathrm{a}} \) values of ionisable groups in lysine are \( 2.18,8.95 \) and \( 10.79 \) respectively. Find isoelectric point of lysine

Options:

The correct answer is:
9.87

Solution:
Isoelectric point= 8.95+10.79 2 =9.87

Topic: Chemistry, Biomolecules, JEE Main
The matrix $\left[\begin{array}{ccc}0 & 5 & -7 \\ -5 & 0 & 11 \\ 7 & -11 & 0\end{array}\right]$ is known as

Options:
A Upper triangular matrix
B Skew symmetric matrix
C Symmetric matrix
D Diagonal matrix

The correct answer is:
Skew symmetric matrix

Solution:
The given matrix is a skew-symmetric matrix
$\left[\because A^{\prime}=-A\right] \text {. }$

Topic: Mathematics, Matrices, JEE Main
The matrix $A^2+4 A-5 I$, where $I$ is identity matrix and $A=\left[\begin{array}{cc}1 & 2 \\ 4 & -3\end{array}\right]$, equals :

Options:
A
$4\left[\begin{array}{ll}2 & 1 \\ 2 & 0\end{array}\right]$

B
$4\left[\begin{array}{cc}0 & -1 \\ 2 & 2\end{array}\right]$

C
$32\left[\begin{array}{ll}2 & 1 \\ 2 & 0\end{array}\right]$

D
$32\left[\begin{array}{ll}1 & 1 \\ 1 & 0\end{array}\right]$

The correct answer is:
$4\left[\begin{array}{ll}2 & 1 \\ 2 & 0\end{array}\right]$

Solution:
$\begin{aligned} & \mathrm{A}^2+4 \mathrm{~A}-5 \mathrm{I}=\mathrm{A} \times \mathrm{A}+4 \mathrm{~A}-5 \mathrm{I} \\ = & {\left[\begin{array}{cc}1 & 2 \\ 4 & -3\end{array}\right] \times\left[\begin{array}{cc}1 & 2 \\ 4 & -3\end{array}\right]+4\left[\begin{array}{cc}1 & 2 \\ 4 & -3\end{array}\right]-5\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] } \\ & =\left[\begin{array}{cc}9 & -4 \\ -8 & 17\end{array}\right]+\left[\begin{array}{cc}4 & 8 \\ 16 & -12\end{array}\right]-\left[\begin{array}{cc}5 & 0 \\ 0 & 5\end{array}\right] \\ & =\left[\begin{array}{cc}9+4-5 & -4+8-0 \\ -8+16-0 & 17-12-5\end{array}\right]=\left[\begin{array}{ll}8 & 4 \\ 8 & 0\end{array}\right] \\ & =4\left[\begin{array}{ll}2 & 1 \\ 2 & 0\end{array}\right]\end{aligned}$

Topic: Mathematics, Matrices, JEE Main, JEE Main 2013 (09 Apr Online)
The matrix $A=\left[\begin{array}{cc}i & 1-2 i \\ -1-2 i & 0\end{array}\right]$ is which of the following

Options:
A Symmetric
B Skew-symmetric
C Hermitian
D Skew-hermitian

The correct answer is:
Skew-hermitian

Solution:
Since for $A=\left[\begin{array}{cc}i & 1-2 i \\ -1-2 i & 0\end{array}\right]$. So, $(\bar{A})^T=-A$. Thus, $A$ is skew hermitian.

Topic: Mathematics, Matrices, JEE Main
The matrix \(A^2+4 A-5 I\), where \(I\) is identity matrix and \(A=\left[\begin{array}{cc}1 & 2 \\ 4 & -3\end{array}\right]\), equals :

Options:
A \(4\left[\begin{array}{ll}2 & 1 \\ 2 & 0\end{array}\right]\)
B \(4\left[\begin{array}{cc}0 & -1 \\ 2 & 2\end{array}\right]\)
C \(32\left[\begin{array}{ll}2 & 1 \\ 2 & 0\end{array}\right]\)
D \(32\left[\begin{array}{ll}1 & 1 \\ 1 & 0\end{array}\right]\)

The correct answer is:
\(4\left[\begin{array}{ll}2 & 1 \\ 2 & 0\end{array}\right]\)

Solution:
\(\begin{aligned}
& A^2+4 A-5 I=A \times A+4 A-5 I \\
= & {\left[\begin{array}{cc}
1 & 2 \\
4 & -3
\end{array}\right] \times\left[\begin{array}{cc}
1 & 2 \\
4 & -3
\end{array}\right]+4\left[\begin{array}{cc}
1 & 2 \\
4 & -3
\end{array}\right]-5\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] } \\
= & {\left[\begin{array}{cc}
9 & -4 \\
-8 & 17
\end{array}\right]+\left[\begin{array}{cc}
4 & 8 \\
16 & -12
\end{array}\right]-\left[\begin{array}{cc}
5 & 0 \\
0 & 5
\end{array}\right] } \\
= & {\left[\begin{array}{cc}
9+4-5 & -4+8-0 \\
-8+16-0 & 17-12-5
\end{array}\right]=\left[\begin{array}{cc}
8 & 4 \\
8 & 0
\end{array}\right] } \\
= & 4\left[\begin{array}{cc}
2 & 1 \\
2 & 0
\end{array}\right]
\end{aligned}\)

Topic: Mathematics, Matrices, JEE Main
The maximum amount of oxygen is lost from the blood in

Options:
A arteries
B capillaries of body
C left atrium of the heart
D capillaries surrounding alveoli

The correct answer is:
capillaries of body

Solution:
Oxygen is transported in various ways: about 97% is transported as oxyhaemoglobin, a combined state of oxygen and haemoglobin.
The remaining 3% is carried in the form of dissolved oxygen in the blood plasma. The difference in the partial pressure of oxygen in the blood vessels (capillaries of the body) and tissues triggers the diffusion of oxygen to the tissues and carbon dioxide to the blood.

Topic: Biology, Breathing and Exchange of Gases, NEET
The maximum amplitude for an amplitude modulated wave is found to be 12 V while the minimum amplitude is found to be 3 V. The modulation index is 0.6x where x is _____.

Options:

The correct answer is:
1

Solution:
Amax=Ac+Am=12

Amin=Ac-Am=3

⇒Ac=152 and Am=92

Modulation index =AmAc=92152=0.6

⇒x=1

Topic: Physics, Communication System, JEE Main, JEE Main 2021 (27 Jul Shift 2)
The maximum and minimum amplitude of an amplitude modulated wave is 16 V and 8 V respectively. The modulation index for this amplitude modulated wave is x×10-2. The value of x is _______. (Round off your answer to the nearest integer)

Options:

The correct answer is:
33

Solution:
Modulation index =Amax-AminAmax+Amin

=16-816+8=824=13=0.33

x×10-2=0.33

So, $x=33$

Topic: Physics, Communication System, JEE Main, JEE Main 2021 (26 Feb Shift 1)
The maximum and minimum distances of a comet from the Sun are 1.6×1012 m and 8.0×1010 m respectively. If the speed of the comet at the nearest point is 6×104 m s-1, the speed at the farthest point is

Options:
A 1.5×103 m s-1
B 6.6×103 m s-1
C 3.0×103 m s-1
D 4.5×103 m s-1

The correct answer is:
3.0×103 m s-1

Solution:
By angular momentum conservation :

mv1r1=mv2r2

v1=48×10141.6×1012=3000 m s-1

=3×103 m s-1.

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2021 (16 Mar Shift 1)
The maximum and minimum distances of a comet from the sun are $1.4 \times 10^{12} \mathrm{~m}$ and $7 \times 10^{10} \mathrm{~m}$. If its velocity nearest to the sun is $6 \times 10^{14} \mathrm{~ms}^{-1}$, what is its velocity in the farthest position?
(Assume the comet to be spherical)

Options:
A $3 \times \mathrm{10}^{-18} \quad$
B $2 \times \mathrm{10}^{-18} \quad$
C $\mathrm{10}^{-18} \quad$
D $4 \times \mathrm{10}^{-18} \quad$

The correct answer is:
$3 \times \mathrm{10}^{-18} \quad$

Solution:
The given situation is shown below
$a_{\max }=1.4 \times 10^{12} \mathrm{~m}$
$a_{\min }=7 \times 10^{10} \mathrm{~m}$
The velocity of the comet is maximum, when it is nearest to the sun and minimum when it is farthest from the sun.
$\therefore \quad v_{\max }=6 \times 10^{14} \mathrm{~m} / \mathrm{s}$
Applying the law of conservation of angular momentum at points $A$ and $B$, we get
$$
\begin{aligned}
&m v_{\min } a_{\max }=m v_{\max } a_{\min } \quad(\quad L=m v r) \\
&\Rightarrow \quad v_{\min }=\frac{v_{\max } \times a_{\min }}{a_{\max }} \\
&=\frac{6 \times 10^{14} \times 7 \times 10^{10}}{1.4 \times 10^{12}} \\
&=3 \times 10^{13} \mathrm{~ms}^{-1}
\end{aligned}
$$
No option is correct.

Topic: Physics, Gravitation, JEE Main
The maximum and minimum tension in the string whirling in a vertical circle of the radius 2.5 m with constant speed are in the ratio 5 : 3 then its velocity is (g=9.8 m s-2)

Options:
A 98 m s-1
B 7 m s-1
C 490 m s-1
D 4.9 m s-1

The correct answer is:
98 m s-1

Solution:
In this problem it is assumed that particle although moving in a vertical loop but it speed remain constant
Tension at lowest point Tmax=mv2r+mg
Tension at highest point Tmin=mv2r-mg
TmaxTmin=mv2r+mgmv2r-mg=53
3mv2r+3mg=5mv2r-5mg
2mv2r=8mg
v=4rg
=4×2.5×9.8
=98 m s-1

Topic: Physics, Laws of Motion, NEET
The maximum and minimum values of the function $f:[R \rightarrow[R$ defined by $f(x)=5 \cos x+3 \cos \left(x+\frac{\pi}{3}\right)+8$ for all $x \in[R$, are respectively.

Options:
A 15, 1
B 8, - 8
C -7, - 15
D 1, - 15

The correct answer is:
15, 1

Solution:
$$
\begin{aligned}
& f(x)=5 \cos x+3 \cos \left(x+\frac{\pi}{3}\right)+8 \\
& =5 \cos x+3\left[\cos x \cdot \cos \frac{\pi}{3}-\sin x \sin \frac{\pi}{3}\right]+8 \\
& =5 \cos x+3\left[\frac{1}{2} \cos x-\frac{\sqrt{3}}{2} \sin x\right]+8 \\
& =\frac{13}{2} \cos x-\frac{3 \sqrt{3}}{2} \sin x+8
\end{aligned}
$$
We know that, $a \cos x-b \sin x$ is lie between $\left[-\sqrt{a^2+b^2}, \sqrt{a^2+b^2}\right]$
$$
\begin{gathered}
-\sqrt{\left(\frac{13}{2}\right)^2+\left(\frac{-3 \sqrt{3}}{2}\right)^2}+8 \leq f(x) \\
\leq \sqrt{\left(\frac{13}{2}\right)^2+\left(\frac{-3 \sqrt{3}}{2}\right)^2}+8 \\
-7+8 \leq f(x) \leq 7+8 \\
1 \leq f(x) \leq 15
\end{gathered}
$$
So, maximum value of $f(x)$ is 15 and minimum value is 1 .

Topic: Mathematics, Functions, JEE Main
The maximum and minimum values of $x^3-18 x^2+96 x$ in interval $(0,9)$ are

Options:
A 160,0
B 60,0
C 160
D 120, 28

The correct answer is:
160

Solution:
Let $y=x^3-18 x^2+96 x \Rightarrow \frac{d y}{d x}=3 x^2-36 x+96=0$
$\therefore \quad x^2-12 x+32=0 \Rightarrow(x-4)(x-8)=0 \quad, \quad x=4,8$
Now, $\quad \frac{d^2 y}{d x^2}=6 x-36$
At $x=4 \frac{d^2 y}{d x^2}=24-36=-12 \lt 0$
$\therefore$ At $x=4$ function will be maximum
and $[f(4)]_{\text {max. }}=64-288+384=160$
At $x=8, \quad \frac{d^2 y}{d x^2}=48-36=12\gt0$
$\therefore$ At $x=8$, function will be minimum and $[f(8)]_{\text {min. }}=128$.

Topic: Mathematics, Application of Derivatives, JEE Main
The maximum and the minimum equivalent resistance obtained by combining n identical resistors of resistance R, are Rmax and Rmin respectively. The ratio RmaxRmin is equal to

Options:
A n
B n2
C n2-1
D n3

The correct answer is:
n2

Solution:
To get maximum equivalent resistance all resistance must be connected in series
∴Reqmax=R+R+R+…nR.=nR
To get minimum equivalent resistance all resistances must be connected in parallel.
∴1Reqmin=1R+1R+1R+…1n,1Reqmin =nR
⇒Reqmin=Rn∴ReqmaxReqmin=nRR/n=n2

Topic: Physics, Current Electricity, JEE Main
The maximum area (in sq. units) bounded by y=sin⁡x, y=ax ∀a∈1,4 and the line π-2x=0 is

Options:
A π2
B π22-1
C π+2
D π2-4

The correct answer is:
π22-1

Solution:
The area bounded is maximum if slope of the line y=ax is greatest, i.e. 4 .
Hence, the required area =â«0Ï24x-sinâ¡xdx
=2x2+cosâ¡x0Ï2
=Ï22+0-0+1
=Ï22-1 sq. units

Topic: Mathematics, Area Under Curves, JEE Main
The maximum area (in sq. units) of a rectangle having its base on the x- axis and its other two vertices on the parabola, y=12-x2 such that the rectangle lies inside the parabola, is :

Options:
A 202
B 32
C 36
D 183

The correct answer is:
32

Solution:
Since, given parabola is symmetric about the y- axis, hence rectangle will also be symmetric about y- axis.

Let one vertex of the rectangle on the x- axis be α,0, then

Area of rectangle A=2α.12-α2

Differentiating both sides with respect to α, we get

⇒dAdα=24-6α2=0⇒α=2, -2

For area to be maximum, put α=2 in the equation for area of the rectangle, we get

Amax=2×2×12-22=32

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2019 (12 Jan Shift 1)
The maximum area of a right angled triangle with hypotenuse $h$ is :

Options:
A
$\frac{h^2}{2 \sqrt{2}}$

B
$\frac{h^2}{2}$

C
$\frac{h^2}{\sqrt{2}}$

D
$\frac{h^2}{4}$

The correct answer is:
$\frac{h^2}{4}$

Solution:
Let base $=b$
Altitude (or perpendicular) $=\sqrt{h^2-b^2}$ Area, $\mathrm{A}=\frac{1}{2} \times$ base $\times$ altitude $=\frac{1}{2} \times b \times \sqrt{h^2-b^2}$ $\Rightarrow \frac{d \mathrm{~A}}{d b}=\frac{1}{2}\left[\sqrt{h^2-b^2}+b \cdot \frac{-2 b}{2 \sqrt{h^2-b^2}}\right]$ $=\frac{1}{2}\left[\frac{h^2-2 b^2}{\sqrt{h^2-b^2}}\right]$
Put $\frac{d \mathrm{~A}}{d b}=0, \Rightarrow b=\frac{h}{\sqrt{2}}$
Maximumarea $=\frac{1}{2} \times \frac{h}{\sqrt{2}} \times \sqrt{h^2-\frac{h^2}{2}}=\frac{h^2}{4}$

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2013 (22 Apr Online)
The maximum distance from origin of a point on the curve $x=a \sin t-b \sin \left(\frac{a t}{b}\right)$
$y=a \cos t-b \cos \left(\frac{a t}{b}\right)$, both $a, b>0$ is

Options:
A
a - b

B
a + b

C
$\sqrt{a^2+b^2}$

D
$\sqrt{a^2-b^2}$

The correct answer is:
a - b

Solution:
Distance of origin from $(x, y)=\sqrt{x^2+y^2}$
$$
=\sqrt{a^2+b^2-2 a b \cos \left(t-\frac{a t}{b}\right)}=\sqrt{a^2+b^2-2 a b}\left[\because \max \cdot \cos \left(t-\frac{a t}{b}\right)=1\right]=a-b
$$

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2002
The maximum electron density in the ionosphere in the morning is 1010 m-3. At noon time, it increases to 2Ã1010 m-3. Find the ratio of critical frequency at noon and the critical frequency in the morning.

Options:
A 2.0
B 2.82
C 4.00
D 1.414

The correct answer is:
1.414

Solution:
Given, the maximum electron density in the morning
Nmax=1010 m-3
fc'=9Nmax1/2=910101/2
Critical frequency fc′=9×105
The maximum electron density at noon
Nmax=2×1010 m-3
Critical frequency
fc′′=9×Nmax′12=9×2×101012=9×2×105
The ratio of critical frequency at noon and critical frequency in the morning.
fc′′fc′=9×2×1059×105
Or fc′′fc′=2=1.414

Topic: Physics, Communication System, JEE Main
The maximum frequency for reflection of sky waves from a certain layer of the ionosphere is found to be fmax = 9 (Nmax)1/2, where Nmax is the maximum electron density at that layer of the ionosphere. On a certain day, it is observed that signals of frequencies higher than 8 MHz are not received by reflection from the F2 layer of the ionosphere. Estimate the maximum electron densities of the F2 layers on that day.

Options:
A 7.9Ã1011 m-3
B 8.9×1011 m-3
C 1.9×1011 m-3
D 7.1×1011 m-3

The correct answer is:
7.9Ã1011 m-3

Solution:
The maximum frequency for reflection of sky waves
fmax = 9(Nmax)1/2
where Nmax is a maximum electron density.
For F2 layer,

fmax=8 MHz

So, 8Ã106=9(Nmax)1/2

maximum electron density,Nmax=8Ã10692=7Â·9Ã1011 m-3

Topic: Physics, Communication System, JEE Main
The maximum height reached by a projectile is $64 \mathrm{~m}$. If the initial velocity is halved, the new maximum height of the projectile is _____ $\mathrm{m}$.

Options:

The correct answer is:
16

Solution:
$\begin{aligned} & \mathrm{H}_{\max }=\frac{\mathrm{u}^2 \sin ^2 \theta}{2 \mathrm{~g}} \\ & \frac{\mathrm{H}_{1 \text { max }}}{\mathrm{H}_{2 \max }}=\frac{\mathrm{u}_1^2}{\mathrm{u}_2^2} \\ & \frac{64}{\mathrm{H}_{2 \max }}=\frac{\mathrm{u}^2}{(\mathrm{u} / 2)^2} \\ & \mathrm{H}_{2 \max }=16 \mathrm{~m}\end{aligned}$

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2024 (05 Apr Shift 2)
The maximum intensity of fringes in Young's experiment is I. If one of the slit is closed, then the intensity at that place becomes $I_0$. Which of the following relation is true?

Options:
A $I=I_0$
B $I=2 I_0$
C $I=4 I_0$
D There is no relation between $/$ and $I_{\circ}$

The correct answer is:
$I=4 I_0$

Solution:
Suppose slit width's are equal, so they produces waves of equal intensity say $I^{\prime}$. Resultant intensity at any point $I_R=4 / \cos ^2 \phi$ where $\phi$ is the phase difference between the waves at the point of observation.
For maximum intensity $\phi=0^{\circ} \Rightarrow I_{\max }=4 I^{\prime}=I \ldots$ (i)
If one of slit is closed, Resultant intensity at the same point will be $I^{\prime}$ only i.e. $I^{\prime}=I_0$ (ii)
Comparing equation (i) and (ii) we get
$I=4 I_0$

Topic: Physics, Wave Optics, JEE Main
The maximum kinetic energy of a photoelectron liberated from the surface of lithium with work function $2.35 \mathrm{eV}$ by electromagnetic radiation whose electric component varies with time as :
$E=a\left[1+\cos \left(2 \pi f_1 t\right)\right] \cos 2 \pi f_2 t$ (where $a$ is a constant $)$ is $\left(f_1=3.6 \times 10^{15} \mathrm{~Hz}\right.$,
and $f_2=1.2 \times 10^{15} \mathrm{~Hz}$ and Planck's constant $\left.h=6.6 \times 10^{-34} \mathrm{Js}\right)$

Options:
A 2.64 eV
B 7.55 eV
C 12.52 eV
D 17.45 eV

The correct answer is:
17.45 eV

Solution:
Here, work function, $W_0=2.35 \mathrm{eV}$ and the electric component of electromagnetic radiation
$$
\begin{aligned}
& E=a\left[1+\cos \left(2 \pi f_1 t\right)\right] \cos \left(2 \pi f_2 t\right) \\
& \Rightarrow E=\left[a \cos \left(2 \pi f_2 t\right)+a \cos \left(2 \pi f_1 t\right) \cos \left(2 \pi f_2 t\right)\right] \\
& \left(\because \cos A \cos B=\frac{1}{2}[\cos (A+B)-\cos (A-B))\right. \\
& \Rightarrow E=a \cos \left(2 \pi f_2 t\right)+\frac{a}{2} \cos 2 \pi\left(f_1+f_2\right) t-\frac{a}{2} \cos 2 \pi\left(f_1-f_2\right) t
\end{aligned}
$$
So, the electric component has 3 sub-components with frequencies are,
$$
f_2,\left(f_1+f_2\right) \text { and }\left(f_1-f_2\right)
$$
So, for maximum kinetic energy of photoelectron, we take photon of maximum frequency. Hence,
$$
\begin{aligned}
E_{\max } & =\frac{h v_{\max }}{e}=\frac{6.6 \times 10^{-34} \times\left(3.6 \times 10^{15}+1.2 \times 10^{15}\right)}{1.6 \times 10^{-19}} \\
& =19.8 \mathrm{eV}
\end{aligned}
$$
Hence, the maximum kinetic energy,
$$
\mathrm{KE}_{\text {max }}=E_{\text {max }}-W_0=19.8-2.35=17.45 \mathrm{eV}
$$
Hence, the correct option is (d).

Topic: Physics, Dual Nature of Matter, JEE Main
The maximum kinetic energy of photoelectrons emitted from a metal surface increases from 0.4 eV to 1.2 eV when the frequency of the incident radiation is increased by 40%. What is the work function (in eV) of the metal surface?

Options:

The correct answer is:
1.6

Solution:
According to Einstein's photoelectric equation,

Kmax=hυ-ϕ0

where Kmax is the maximum kinetic energy of photoelectrons and ϕ0 is the work function of the metal surface

Initially,

hυ=0.4+ϕ0

When the frequency is increased by 40%, then

1.4hυ=1.2+ϕ0

On solving we get

ϕ0=1.6 eV

Topic: Physics, Dual Nature of Matter, JEE Main
The maximum load a wire can withstand without breaking, when its length is reduced to half of its original length, will

Options:
A
be double

B
be half

C
be four times

D
remain same

The correct answer is:
remain same

Solution:
As we know that,
$$
\text { Breaking stress }=\frac{\text { Breaking force }}{\text { Area of cross-section }} \text {..(i) }
$$
When length of the wire changes (or by reducing half) area of cross-section remains same.
Hence, breaking force will be same because breaking stress does not depend on length.

Topic: Physics, Mechanical Properties of Solids
The maximum mass of an atom is concentrated in which of these?

Options:
A Nucleus.
B Neutrons.
C Protons.
D Electrons.

The correct answer is:
Nucleus.

Solution:
Since the nucleus contains protons and neutrons, most of the mass of an atom is concentrated in its nucleus.

Topic: Physics, Atomic Physics, JEE Main
The maximum negative integral value of b for which the point 2b+3,b2 lies above the line 3x-4y-aa-2=0, ∀a∈R is

Options:
A -1
B -3
C -2
D -4

The correct answer is:
-2

Solution:
Let x1,y2 lie on the line
∴ 3x1-4y2-aa-2=0⇒4y2=3x1-aa-2
Now, y2<y1⇒3x1-aa-24<y1
Putting x1=2b+3, y1=b2, we get,

⇒32b+3-aa-2<4b2.
⇒a2-2a+4b2-6b-9>0

Now ∀a∈R, D<0

⇒4-44b2-6b-9<0
⇒1-4b2+6b+9<0⇒4b2-6b+10>0
⇒2b2-3b-5>0⇒2b-5b+1>0
⇒b∈-∞,-1∪52,∞

Hence, the maximum negative integer value of b is -2

Topic: Mathematics, Straight Lines, JEE Main
The maximum number of $90^{\circ}$ angles between bond pair of electrons is observed in

Options:
A
dsp ${ }^3$ hybridization

B
$\mathrm{sp}^3 \mathrm{~d}^2$ hybridization

C
$\mathrm{dsp}^2$ hybridization

D
$\mathrm{sp}^3 \mathrm{~d}$ hybridization

The correct answer is:
$\mathrm{sp}^3 \mathrm{~d}^2$ hybridization

Solution:
$\mathrm{sp}^3 \mathrm{~d}^2$ hybridisation confirms to octahedral or square bipyramidal configuration $\therefore$ all the bond angles are $90^{\circ}$ in the structure

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2004
The maximum number of atoms that may will in the same plane in the following molecule are x .

Then the value of x is.

Options:

The correct answer is:
8

Solution:
All atoms of this molecule can lie in a single plane

Topic: Chemistry, General Organic Chemistry, JEE Main
The maximum number of electrons that can be accommodated in $\mathrm{M}$ shell is:

Options:
A $8$
B $2$
C $18$
D $32$

The correct answer is:
$18$

Solution:
The general formula of filling electron in the shell is $2\left(n^2\right)$; where $n$ is the number of the shell.
\(\begin{array}{|l|l|l|l|} \hline \mathrm{M} & n=3 & =2\left(3^2\right) & 18 e^- \\ \hline \end{array}\)

Topic: Physics, Atomic Physics, JEE Main
The maximum number of electrons that can be accommodated in the L-shell of an atom is:

Options:
A \( 2 \)
B \( 4 \)
C \( 8 \)
D \( 16 \)

The correct answer is:
\( 8 \)

Solution:
L-shell contains 2s and 2p sub-shells. There are total four orbitals in these sub-shells and each orbital can accommodate two electrons. Therefore, it contains a total of eight electrons.

Topic: Chemistry, Structure of Atom, JEE Main
The maximum number of isomeric ethers with the molecular formula $\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}$ is

Options:
A 2
B 3
C 4
D 5

The correct answer is:
3

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Main
The maximum number of lone pairs of electron on the central atom from the following species is ......
ClO3-, XeF4, SF4 and l3-

Options:

The correct answer is:
3

Solution:
ClO3-: The central atom is Cl, there is one lone pair of electrons on the central atom.

XeF4: The central atom is Xe. It has two lone pair on central atom.

SF4: The central atom is S. It contains one lone pair on central atom.

I3-: It contains three lone pair on central atom.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2023 (11 Apr Shift 2)
The maximum number of moles of Ba3PO42 that can be formed if 2 mole BaCl2 is mixed with 1 mole Na3PO4 is

Options:
A 0.7
B 0.25
C 0.33
D 0.50

The correct answer is:
0.50

Solution:
The balanced chemical reaction is

3BaCl2+2Na3PO4→Ba3PO42+6NaCl

According to the reaction,

3 mole BaCl2 reacts with 2 mole Na3PO42

Thus, 2 mole BaCl2 will require 23×2=43=1.33 mole Na3PO4
Since sodium phosphate is taken in less quantity then the required quantity, here, Na3PO4 is the limiting reactant. The stoichiometric calculations are done based on the limiting reagent.

According to the reaction.
2 mole Na3PO4 gives 1 mole Ba3PO42

So, 1 mole Na3PO4 will give 0.5 mole Ba3PO42.

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
The maximum number of normal that can be drawn from a point to a parabola is

Options:
A $0$
B $1$
C $2$
D $3$

The correct answer is:
$3$

Solution:
The correct option is D 3
equation of normal is given by $y=m x-2 a m-a m^3$
where equation of parabola is $y^2=4 a x$
clearly, it is a cubic equation in $\mathrm{m}$ and hence three values of possible $m$ for normal hence three normal's can be drawn from any point(maximum)

Topic: Mathematics, Parabola, JEE Main
The maximum number of normal that can be drawn from a point to a parabola is

Options:
A $0$
B $1$
C $2$
D $3$

The correct answer is:
$3$

Solution:
equation of normal is given by $y=m x-2 a m-a^3$
where equation of parabola is $y^2=4 a x$
clearly, it is a cubic equation in $\mathrm{m}$ and hence three values of possible $m$ for normal hence three normal's can be drawn from any point(maximum)

Topic: Mathematics, Parabola, JEE Main
The maximum number of orbitals which can be identified with \(\mathrm{n}=4\) and \(m_l=0\) is ______

Options:

The correct answer is:
4

Solution:
So answer is 4 .

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2024 (04 Apr Shift 2)