datavorous/entrance-exam-dataset huggingface dataset:
The percentage hydrolysis of $0.15 \mathrm{M}$ solution of ammonium acetate, $\mathrm{K}_{\mathrm{a}}$ for $\mathrm{CH}_{3} \mathrm{COOH}$ is $1.8 \times 10^{-5}$

and $\mathrm{K}_{\mathrm{b}}$ for $\mathrm{NH}_{3}$ is $1.8 \times 10^{-5}$

Options:
A 0.556
B 4.72
C 9.38
D 5.56

The correct answer is:
0.556

Solution:
$\alpha=\sqrt{\frac{\mathrm{K}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{a}} \times \mathrm{K}_{\mathrm{b}}}}=\sqrt{\frac{1 \times 10^{-14}}{1.8 \times 10^{-5} \times 1.8 \times 10^{-5}}}=0.55$

Topic: Chemistry, Ionic Equilibrium, JEE Main
The percentage increase in the speed of transverse waves produced in a stretched string if the tension is increased by 4%, will be ___ %.

Options:

The correct answer is:
2

Solution:
v=Tμ

ΔVV=12ΔTT

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2021 (25 Feb Shift 2)
The percentage of adenine in DNA isolated from human liver is observed to be 30.7%. What is the expected percentage of thymine, guanine, and cytosine?

Options:
A T = 19.3%, G = 19.3%, C = 30.7%
B T = 19.3%, G = 30.7%, C = 19.3%
C T = 30.7%, G = 19.3%, C = 30.7%
D T = 30.7%, G = 19.3%, C = 19.3%

The correct answer is:
T = 30.7%, G = 19.3%, C = 19.3%

Solution:
According to Chargaff's rule ratio of adenine to thymine is one and the ratio of guanine to cytosine is one. Thus, percentage of adenine is equal to thymine. If 30.7 % is adenine than 30.7% is thymine, and remaining 38.6% is guanine and cytosine (G = 19.3%, C = 19.3%).

Topic: Biology, Molecular Basis of Inheritance, NEET
The percentage of oxygen in CH2O is

Options:
A 40%
B 6.6%
C 53.33%
D 49%

The correct answer is:
53.33%

Solution:
(i) As per molecular formula
CH2O=12+2+16=30
(ii) As we know in case of percentage, the total mass =100
(iii) When total mass =30
the mass of oxygen =16
So, when total mass =100
Percentage of oxygen =16×10030=53.33%

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
The percentage of $\mathrm{P}_2 \mathrm{O}_5$ in diammonium hydrogen phosphate $\left(\mathrm{NH}_4\right)_2 \mathrm{HPO}_4$ is

Options:
A $23.48$
B $46.96$
C $53.78$
D $71.00$

The correct answer is:
$53.78$

Solution:
$\underset{2(36+1+31+64)=264}{2\left(\mathrm{NH}_4\right)_2 \mathrm{HPO}_4} \equiv \underset{62+80=142}{\mathrm{P}_2 \mathrm{O}_5}$
$\%$ of $\mathrm{P}_2 \mathrm{O}_5=\frac{\text { wt. of } \mathrm{P}_2 \mathrm{O}_5}{\text { wt of salt }} \times 100$
$=\frac{142}{264} \times 100=53.78 \%$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
The percentage of p-character in the orbitals forming P – P bonds in P4 is

Options:

The correct answer is:
75

Solution:
P is sp3 hybridised in P4. so, p character is 75% in it.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
The perimeter of a parallelogram whose sides are represented by the lines x+2y+3=0, 3x+4y-5=0, 2x+4y+5=0 and 3x+4y-10=0 is equal to

Options:
A 52+55 units
B 5+55 units
C 5+525 units
D 5+552 units

The correct answer is:
52+55 units

Solution:
Let, the parallelogram is ABCD as shown in the figure.
Let, the distance between AB & CD is d1 and the distance between AD & BC is d2 and
∠DAB=∠DCB=θ then,

AD=d1cosecθ and DC=d2cosecθ
⇒ Perimeter of the parallelogram ABCD is 2AD+DC=2cosecθd1+d2
Now, d1=-32+541+14=125 & d2=52-541+916=1
and tan⁡θ=-12+341+38=211 or cot⁡θ=112
or cosecθ=1+1214=552
⇒ Perimeter =2552125+1=52+55

Topic: Mathematics, Straight Lines, JEE Main
The perimeter of a sector is $p$. The area of the sector is maximum when its radius is

Options:
A $\sqrt{p}$
B $\frac{1}{\sqrt{p}}$
C $\frac{p}{2}$
D $\frac{p}{4}$

The correct answer is:
$\frac{p}{4}$

Solution:
Since $x\gt0$, so no maximum value can be found.
Perimeter of a sector $=p$. Let $A O B$ be the sector with radius $r$. If angle of the sector be $\theta$ radians, then area
of sector
$\text { (A) }=\frac{1}{2} r^2 \theta...(i)$
Length of $\operatorname{arc}(s)=r \theta$ or $\quad \theta=\frac{s}{r}$.
Therefore perimeter of the sector $=(p)=r+s+r=2 r+s...(ii)$
Substituting $\theta=\frac{s}{r}$ in (i), $A=\left(\frac{1}{2} r^2\right)\left(\frac{s}{r}\right)=\frac{1}{2} r s$
$\Rightarrow \quad s=\frac{2 A}{r}$
Now substituting the value of $s$ in (ii), we get $p=2 r+\left(\frac{2 A}{r}\right)$ or $2 A=p r-2 r^2$.
Differentiating with respect to $r$, we get $2 \frac{d A}{d r}=p-4 r$. We know that for the maximum value of area $\frac{d A}{d r}=0$ or $p-4 r=0$ or $r=\frac{p}{4}$.

Topic: Mathematics, Application of Derivatives, JEE Main
The perimeter of a triangle is $10 \mathrm{~cm}$. If one of its side is $4 \mathrm{~cm}$, then remaining sides of the triangle, when area of triangle is maximum are

Options:
A $5 \mathrm{~cm}, 1 \mathrm{~cm}$
B $3 \cdot 6 \mathrm{~cm}, 2 \cdot 4 \mathrm{~cm}$
C $3 \mathrm{~cm}, 3 \mathrm{~cm}$
D $2 \mathrm{~cm}, 4 \mathrm{~cm}$

The correct answer is:
$3 \mathrm{~cm}, 3 \mathrm{~cm}$

Solution:
Let $\mathrm{a}, \mathrm{b}, \mathrm{c}$ be the sides of the triangle.
The perimeter of triangle $(2 \mathrm{~s})=\mathrm{a}+\mathrm{b}+\mathrm{c}$
Let $\mathrm{a}=4$ and $2 \mathrm{~s}=10$ i.e. $\mathrm{s}=5$
$\therefore 10=4+b+c \Rightarrow b=6-c$
Now, area of triangle $=\Delta=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{5(5-4)(5-6+c)(5-c)}$
$\Delta=\sqrt{5(1)(\mathrm{c}-1)(5-\mathrm{c})} \Rightarrow \Delta^{2}=5(\mathrm{c}-1)(5-\mathrm{c})$
$=5\left(5 c-c^{2}-5+c\right)=5\left(-c^{2}+6 c-5\right)$
Let $f(c)=5\left(-c^{2}+6 c-5\right) \Rightarrow f^{\prime}(c)=5(-2 c+6) \Rightarrow f^{\prime \prime}(c)=5(-2)=-10$
For extreme value, $f^{\prime}(c)=0$ i.e. $5(-2 c+6)=0 \Rightarrow c=3$
Thus $f^{\prime \prime}(3)=-10 < 0 \Rightarrow f(c)$ has maximum value at $c=3$
$\therefore \mathrm{b}=6-3=3$ i.e. the lengths of the remaining sides are $3 \mathrm{~cm}$ and $3 \mathrm{~cm}$.

Topic: Mathematics, Application of Derivatives, MHT CET, MHT CET 2020 (12 Oct Shift 2)
The perimeter of the locus of the point at which the two circles x2+y2=1 and x-42+y2=4 subtend equal angles is

Options:
A 43π
B 83π
C 23π
D 163π

The correct answer is:
163π

Solution:
sin⁡θ=1h2+k2=2h-42+k2
⇒h2-8h+16+k2=4h2+4k2
⇒3x2+3y2+8x-16=0
Radius of the circle is 432+163=16+489
=83
⇒ Perimeter is equal to 16π3

Topic: Mathematics, Circle, JEE Main
The period of a simple pendulum is doubled, when

Options:
A Its length is doubled
B The mass of the bob is doubled
C Its length is made four times
D The mass of the bob and the length of the pendulum are doubled

The correct answer is:
Its length is made four times

Solution:
$T=2 \pi \sqrt{\frac{L}{g}} \Rightarrow T \propto \sqrt{L}$

Topic: Physics, Oscillations, JEE Main
The period of $|\cos x|$ will be

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

The correct answer is:
$\pi$

Solution:
Let $f(x)=|\cos x|$
From figure, it is clear that the period of $|\cos x|$ is $\pi$.

Topic: Mathematics, Functions, JEE Main
The period of $f(x)=x-[x]$, if it is periodic, is

Options:
A $f(x)$ is not periodic
B $\frac{1}{2}$
C $1$
D $2$

The correct answer is:
$1$

Solution:
Let $f(x)$ be periodic with period $T$.
Then, $f(x+T)=f(x)$ for all $x \in R$
$\Rightarrow x+T-[x+T]=x-[x]$, for all $x \in R$
$\Rightarrow x+T-x=[x+T]-[x]$
$\Rightarrow[x+T]-[x]=T$ for all $x \in R \Rightarrow T=1,2,3,4, \ldots \ldots .$.
The smallest value of $T$ satisfying
$f(x+T)=f(x)$ for all $x \in R$ is 1 .
Hence $f(x)=x-[x]$ has period 1 .

Topic: Mathematics, Functions, JEE Main
The period of oscillation of a bar magnet at a place is 2 s. At the same place, the period of oscillation of another identical bar magnet whose magnetic moment is 4 times to that of first magnet is

Options:
A $4 \mathrm{~s}$
B $1 \mathrm{~s}$
C $2 \mathrm{~s}$
D $0.5 \mathrm{~s}$

The correct answer is:
$1 \mathrm{~s}$

Solution:
Period of oscillation of a bar, $\mathrm{T}_1=2 \mathrm{~s}$
Let magnetic moment, $M_1=M$ then $M_2=4 M$
$$
\begin{aligned}
& \mathrm{T}=2 \pi \sqrt{\frac{\mathrm{I}}{\mathrm{MH}}} \\
& \frac{\mathrm{T}_1}{\mathrm{~T}_2}=\sqrt{\frac{\mathrm{I}}{\mathrm{M}_1 \mathrm{H}} \times \frac{\mathrm{M}_2 \mathrm{H}}{\mathrm{I}}} \\
& =\sqrt{\frac{4 \mathrm{M}}{\mathrm{M}}} \\
& \Rightarrow \frac{2}{\mathrm{~T}_2}=2 \\
& \Rightarrow \mathrm{T}_2=1 \mathrm{~s} .
\end{aligned}
$$

Topic: Physics, Magnetic Effects of Current, JEE Main
The period of oscillation of a simple pendulum is given by T=2ÏLg, where L is the length of the pendulum and g is the acceleration due to gravity. The length is measured using a meter scale which has 500 divisions. If the measured value L is 50 cm, the accuracy in the determination of g is 1.1% and the time taken for 100 oscillations is 100 seconds, what should be the possible error in measurement of the clock in one minute (in milliseconds)?

Options:
A 1
B 2
C 5
D 0.25

The correct answer is:
5

Solution:
Given,
T=2πlg
or T2=4π2lg
∴ 2∆TT=∆ll+∆gg ...(i)
Now, l=50 cm, ∆l=2 mm=0.2 cm
∆gg=1.1%=1.1100
Put these values in Equation. (i), then we get
∆TT=120.250+1.1100
=7.5×10-3 s or 7.5 ms
∵ In 100 s, resolution of the clock is 7.5 ms .
∴ In 60 s resolution of the clock is 7.5×60100≈5 ms

Topic: Physics, Units and Dimensions, JEE Main
The period of oscillation of a simple pendulum is T=2πLg. Measured value of L is 1.0 m from meter scale having a minimum division of 1 mm and time of one complete oscillation is 1.95 s measured from stopwatch of 0.01 s resolution. The percentage error in the determination of 'g' will be:

Options:
A 1.03%
B 1.33%
C 1.30%
D 1.13%

The correct answer is:
1.13%

Solution:
T=2πℓg

g=4π2ℓT2

Δgg=Δℓℓ+2ΔTT

Δgg=1×10-31+2×0.011.95

Δgg=0.0113 or 1.13%

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2021 (24 Feb Shift 2)
The period of oscillation of a simple pendulum is T=2πlg. Measured value of l is 20.0 cm, known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using a wristwatch of 1 s resolution. The accuracy in the determination of g is

Options:
A 5%
B 4%
C 3%
D 1%

The correct answer is:
3%

Solution:
∴T=2πlg⇒g=4π2lT2
∴ Error in g can be calculated as
Δgg=Δll+2ΔTT.
∴ Total time for n oscillation is t=nT where T= time for oscillation.
⇒Δtt=ΔTT
⇒Δgg=Δll+2Δtt
Given that Δl=1 mm=10-3 m, l=20×10-2 m
Δt=1 s, t=90 s.
% error in g is
Δgg×100=Δll+2Δtt×100
=10-320×10-2+2×190×100=12+209=0.5+2.22=2.72%
≅3%

Topic: Physics, Mathematics in Physics, JEE Main
The period of oscillation of a simple pendulum is T=2πlg. Measured value of l is 20.0 cm, known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using a wristwatch of 1 s resolution. The accuracy in the determination of g is

Options:
A 5%
B 4%
C 3%
D 1%

The correct answer is:
3%

Solution:
∴T=2πlg⇒g=4π2lT2
∴ Error in g can be calculated as
Δgg=Δll+2ΔTT.
∴ Total time for n oscillation is t=nT where T= time for oscillation.
⇒Δtt=ΔTT
⇒Δgg=Δll+2Δtt
Given that Δl=1 mm=10-3 m, l=20×10-2 m
Δt=1 s, t=90 s.
% error in g is
Δgg×100=Δll+2Δtt×100
=10-320×10-2+2×190×100=12+209=0.5+2.22=2.72%
≅3%

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2015 (04 Apr)
The period of revolution of planet A around the sun is eight times that of B. How many times is the distance of A from the sun, greater than that of B?

Options:
A 2
B 3
C 4
D 5

The correct answer is:
3

Solution:
As T2∝r3,
so, TA2TB2=rA3rB3
or rA=4rB
so, rA-rB=4rB-rB=3rB

Topic: Physics, Gravitation, JEE Main
The period of $\sin ^2 \theta$ is

Options:
A
$\pi^2$

B
$\pi$

C
$2 \pi$

D
$\pi / 2$

The correct answer is:
$\pi$

Solution:
$\sin ^2 \theta=\frac{1-\cos 2 \theta}{2} ;$ Period $=\frac{2 \pi}{2}=\pi$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main, JEE Main 2002
The period of the function $f(\theta)=\sin \frac{\theta}{3}+\cos \frac{\theta}{2}$ is

Options:
A $3 \pi$
B $6 \pi$
C $9 \pi$
D $12 \pi$

The correct answer is:
$12 \pi$

Solution:
We have, $f(\theta)=\sin \frac{\theta}{3}+\cos \frac{\theta}{2}$
LCM of 3 and 2 is 6
$\Rightarrow \quad \frac{\theta}{6}=2 \pi \Rightarrow \theta=12 \pi$
$\therefore$ Period is $12 \pi$

Topic: Mathematics, Trigonometric Equations, JEE Main
The period of the function \( f(\theta)=\sin \frac{\theta}{3}+\cos \frac{\theta}{2} \) is

Options:
A \( 3 \pi \)
B \( 6 \pi \)
C \( 9 \pi \)
D \( 12 \pi \)

The correct answer is:
\( 12 \pi \)

Solution:
Period of sin⁡θ3=6π
And period of cos⁡θ2=4π
∴ Period of fx=LCM6π, 4π=12π

Topic: Mathematics, Functions, JEE Main
The periodic table consists of 18 groups. An isotope of copper, on bombardment with protons, undergoes a nuclear reaction yielding element $X$ as shown below. To which group, element $X$ belongs in the periodic table?

${ }_{29}^{63} \mathrm{Cu}+{ }_{1}^{1} \mathrm{H} \rightarrow 6_{0}^{1} n+{ }_{2}^{4} \alpha+2{ }_{1}^{1} \mathrm{H}+\mathrm{X}$

Options:

The correct answer is:
8

Solution:
${ }_{29}^{63} \mathrm{Cu}+{ }_{1}^{1} \mathrm{H} \rightarrow 6{ }_{0}^{1} n+{ }_{2}^{4} \mathrm{He}+2{ }_{1}^{1} \mathrm{H}+{ }_{Z}^{A} X$

Balancing the atomic mass and atomic number

$63+1=(6 \times 1)+4+2+A \Rightarrow A=52$

$29+1=(6 \times 0)+2+2+Z \Rightarrow Z=26$

Thus ${ }_{Z}^{A} X={ }_{26}^{52} X$ or ${ }_{26}^{52} \mathrm{Fe}$

Hence, $X$ belongs to group 8 in the periodic table.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
The periodic time of a body executing simple harmonic motion is $3 \mathrm{sec}$. After how much interval from time $t=0$, its displacement will be half of its amplitude

Options:
A $\frac{1}{8} \mathrm{sec}$
B $\frac{1}{6} \mathrm{sec}$
C $\frac{1}{4} \mathrm{sec}$
D $\frac{1}{3} \mathrm{sec}$

The correct answer is:
$\frac{1}{4} \mathrm{sec}$

Solution:
$\begin{aligned} & y=a \sin \frac{2 \pi}{T} t \Rightarrow \frac{a}{2}=a \sin \frac{2 \pi t}{3} \Rightarrow \frac{1}{2}=\sin \frac{2 \pi t}{3} \\ & \Rightarrow \sin \frac{2 \pi t}{3}=\sin \frac{\pi}{6} \Rightarrow \frac{2 \pi t}{3}=\frac{\pi}{6} \Rightarrow t=\frac{1}{4} \mathrm{sec}\end{aligned}$

Topic: Physics, Oscillations, JEE Main
The perpendicular bisector of a line segment with end points 1,2,6 and -3,6,2 passes through -6,2,4 and has the equation of the form x+6l=y-2m=z-4n (Where l,m,n are integers, l is a prime number and l>0), then the value of lmn-l+m+n equals to

Options:
A -3
B -5
C -7
D -9

The correct answer is:
-7

Solution:
Midpoint of the line segment is 1-31,2+62,6+22≡-1,4,4
Parallel vector to the required line =-1+6i^+4-2j^+4-4k^
=5i^+2j^+0k^
Hence, equation of the line is
x+65-y-22=z-40
⇒l=5, m=2, n=0

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The perpendicular bisector of the line segment joining A1,4 and Bt,3 has y-intercept equal to -4. Then, the product of all possible values of t is equal to

Options:
A 1
B 2
C -16
D -4

The correct answer is:
-16

Solution:
The slope of the perpendicular bisector =t-1
The mid-point of the line segment =t+12,72
∴ Equation of the bisector is y-72=t-1x-t+12
Putting x=0 and y=-4, we get,

t=±4

Hence, the required product =4-4=-16

Topic: Mathematics, Straight Lines, JEE Main
The perpendicular bisector of the line segment joining $P(1,4)$ and $Q(k, 3)$ has $y$-intercept - 4. Then a possible value of $k$ is

Options:
A
1

B
2

C
$-2$

D
$-4$

The correct answer is:
$-4$

Solution:
Slope of bisector $=k-1$
Middle point $=\left(\frac{\mathrm{k}+1}{2}, \frac{7}{2}\right)$
Equation of bisector is
$y-\frac{7}{2}=(k-1)\left(x-\frac{(k+1)}{2}\right)$
Put $\mathrm{x}=0$ and $\mathrm{y}=-4$
$$
\Rightarrow k=\pm 4
$$

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2008
The perpendicular distance from the origin to the plane containing the two lines, x + 23=y - 25=z + 57 and x - 11=y - 44=z + 47, is

Options:
A 116
B 116
C 11
D 611

The correct answer is:
116

Solution:
Direction vector of normal to the plane =i^j^k^357147=7i^-14j^+7k^

Hence, direction ratio of normal to the plane =<1, -2, 1>

Thus, equation of plane is x-2y+z+11=0

Hence, perpendicular distance from origin =1112+-22+12=116

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2019 (12 Jan Shift 1)
The perpendicular distance from the point $(1,2)$ to common chord of the circles $x^2+y^2-2 x+4 y-4=0$ and $x^2+y^2+4 x-6 y-3=0$ is........ units.

Options:
A $\frac{13}{\sqrt{123}}$
B $\frac{13}{\sqrt{136}}$
C $\frac{13}{\sqrt{63}}$
D $\frac{13}{\sqrt{132}}$

The correct answer is:
$\frac{13}{\sqrt{136}}$

Solution:
Given circle,
$x^2+y^2-2 x+4 y-4=0$
and $x^2+y^2+4 x-6 y-3=0$
Equation of common chord of circle is $S_1-S_2=0$
$\therefore \quad\left(x^2+y^2-2 x+4 y-4\right)$
$-\left(x^2+y^2+4 x-6 y-3\right)=0$
$\begin{array}{ll}\Rightarrow & -6 x+10 y-1=0 \\ \Rightarrow & 6 x-10 y+1=0\end{array}$
Perpendicular distance from the point $(1,2)$ to the line $6 x-10 y+1=0$ is $\left|\frac{6-20+1}{\sqrt{6^2+10^2}}\right|$
$=\frac{13}{\sqrt{136}}$

Topic: Mathematics, Circle, JEE Main
The perpendicular distance from the point $(1, \pi)$ to the line joining $\left(1,0^{\circ}\right)$ and $\left(1, \frac{\pi}{2}\right)$, (in polar coordinates) is

Options:
A $2$
B $\sqrt{3}$
C $1$
D $\sqrt{2}$

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

Solution:
Given points $(1, \pi),\left(1,0^{\circ}\right)$ and $\left(1, \frac{\pi}{2}\right)$ are in polar form.
Now, change in cartesian form,
$$
\begin{aligned}
& (1, \pi) \rightarrow(1 \cdot \cos \pi, 1 \cdot \sin \pi) \rightarrow(-1,0) \\
& \left(1,0^{\circ}\right) \rightarrow\left(1 \cdot \cos 0^{\circ}, 1 \cdot \sin 0^{\circ}\right) \rightarrow(1,0) \\
& \text { and }\left(1, \frac{\pi}{2}\right) \rightarrow\left(1 \cdot \cos \frac{\pi}{2}, 1 \cdot \sin \frac{\pi}{2}\right) \rightarrow(0,1) \\
&
\end{aligned}
$$
Now, equation of the line passing through $(1,0)$ and $(0,1)$ is
$$
\begin{aligned}
& (y-0)=\frac{1-0}{0-1}(x-1) \\
& \Rightarrow \quad y=-x+1 \\
& \Rightarrow \quad x+y-1=0 \\
&
\end{aligned}
$$
So, the perpendicular distance from the point $(-1,0)$ to the line (i) is
$$
=\frac{|-1+0-1|}{\sqrt{1+1}}=\frac{2}{\sqrt{2}}=\sqrt{2}
$$

Topic: Mathematics, Straight Lines, JEE Main
The perpendicular distance of the straight line $12 x+5 y=7$ from the origin is given by

Options:
A $\frac{7}{13}$
B $\frac{12}{13}$
C $\frac{5}{13}$
D $\frac{1}{13}$

The correct answer is:
$\frac{7}{13}$

Solution:
Required distance $=\frac{7}{\sqrt{(12)^2+5^2}}=\frac{7}{13}$

Topic: Mathematics, Straight Lines, JEE Main
The persistent nucellus is a

Options:
A thin remaining layer of nucellus
B thick functional layer of integument
C multilayer of pollen sac
D thin tissue form by endosperm

The correct answer is:
thin remaining layer of nucellus

Solution:
In some seeds, remnants of nucellus are also persistent, which is termed as perisperm.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
The person shown in the pictures is affected by a certain disease. Identify the CORRECT disease.

Options:
A a-caused by Wuchereria bancrofti and W. Malayi, b-caused by Epidermophyton and Microsporum
B a-Elephantiasis, b-ringworm
C a-caused by a nematode, b-caused by fungi
D All of the above

The correct answer is:
All of the above

Solution:
Lymphatic filariasis which is also commonly known as elephantiasis is a rare condition that occurs when the parasite is transmitted to humans through mosquitoes. It causes swelling and enlargement of legs, sex organs, arms, and breasts. This disease is caused by three types of roundworms classified as nematodes-Wuchereria bancrofti-causes 90% of cases, Brugia malayi and Brugia timori cause remainder cases.

Topic: Biology, Human Health and Diseases, NEET
The pH at which Mg(OH)2 Ksp=1×10-11 begins to precipitate from a solution containing 0.10M Mg2+ ions is ______.

Options:

The correct answer is:
09

Solution:
Precipitation when Q=Ksp

Q=Reaction quotient, it is the ratio of products to reactants at any time of reaction.

Ksp=Solubility product.

MgOH2⇋Mg2++2OH-

Mg2+OH-2=10-11

0.1×OH-2=10-11⇒OH-=10-5

⇒pOH=5 ⇒pH=14-5=9

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2024 (30 Jan Shift 1)
The pH of 0.1M solution of a weak acid HA is 4.50. It is neutralised with certain ammount of NaOH solution to decrease the acid content to half of initial value. Calculate the pH of the resulting solution.

Options:

The correct answer is:
8

Solution:
pH=pKa+log⁡[Salt][Acid]
pH=pKa
when weak acid is half neutralized by strong base.
For weak acid = pH =  1 2 pKa − log C
4.5 =  1 2 [pKa− log10 −1 ]
9−1 = pKa⇒pKa = 8

Topic: Chemistry, Ionic Equilibrium, JEE Main
The pH of 0.1 (M) solution of the following salts increases in the order

Options:
A NaCl<NH4Cl<NaCN<HCl
B HCl<NH4Cl<NaCl<NaCN
C NaCN<NH4Cl<NaCl<HCl
D HCl<NaCl<NaCN<NH4Cl

The correct answer is:
HCl<NH4Cl<NaCl<NaCN

Solution:
pH pf NaCl=7 ;pH of NH4Cl<7 ;
pH of NaCN>7and pH of 0.1 MHCl=1

Topic: Chemistry, Ionic Equilibrium, JEE Main
The pH of 0.1 M solution of the following salts increases in the order

Options:
A NaCl<NH4Cl<NaCN<HCl
B HCl<NH4Cl<NaCl<NaCN
C NaCN<NH4Cl<NaCl<HCl
D HCl<NaCl<NaCN<NH4Cl

The correct answer is:
HCl<NH4Cl<NaCl<NaCN

Solution:
HCl: Strong acid (low pH)
NH4Cl : Salt of weak base + strong acid → it gives acidic solution. (pH less than 7)
NaCl: Salt of strong base + strong base → it gives neutral solution with pH = 7.
NaCl: Salt of strong base + weak acid → it gives basic solution with pH > 7.
Thus the increasing order of pH is
HCl<NH4Cl<NaCl<NaCN

Topic: Chemistry, Ionic Equilibrium, JEE Main
The pH of a $0.1$ molar solution of the acid $\mathrm{HQ}$ is $3$ . The value of the ionization constant, Ka of this acid is :

Options:
A
$3 \times 10^{-1}$

B
$1 \times 10^{-3}$

C
$1 \times 10^{-5}$

D
$1 \times 10^{-7}$

The correct answer is:
$1 \times 10^{-5}$

Solution:
$\left[\mathrm{H}^{+}\right]=\sqrt{\mathrm{K}_{\mathrm{a}} \cdot \mathrm{C}} \Rightarrow 10^{-3}=\sqrt{\mathrm{K}_{\mathrm{a}} \cdot 10^{-1}}$
$\Rightarrow \mathrm{K}_{\mathrm{a}}=10^{-5}$

Topic: Chemistry, Ionic Equilibrium, JEE Main
The pH of a $0.1$ molar solution of the acid $\mathrm{HQ}$ is $3$ . The value of the ionization constant, Ka of this acid is :

Options:
A
$3 \times 10^{-1}$

B
$1 \times 10^{-3}$

C
$1 \times 10^{-5}$

D
$1 \times 10^{-7}$

The correct answer is:
$1 \times 10^{-5}$

Solution:
$\left[\mathrm{H}^{+}\right]=\sqrt{\mathrm{K}_{\mathrm{a}} \cdot \mathrm{C}} \Rightarrow 10^{-3}=\sqrt{\mathrm{K}_{\mathrm{a}} \cdot 10^{-1}}$
$\Rightarrow \mathrm{K}_{\mathrm{a}}=10^{-5}$

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2012 (Offline)
The pH of a solution is 5.0. To this solution sufficient acid is added to decreases the pH to 2.0. The no. of times the concentration of H+ increased is

Options:

The correct answer is:
1000

Solution:
H3O+ before the addition of acid is
[H2O+]initial=10-pH=10-5M
[H3O+]final=10-2M
So [H3O+]final[H3O+]initial=10-2M10-5M
=1000 times

Topic: Chemistry, Ionic Equilibrium, JEE Main
The $\mathrm{pH}$ of a solution is increased from 3 to 6 . Its $\mathrm{H}^{+}$ion concentration will be

Options:
A Reduced to half
B Doubled
C Reduced by 1000 times
D Increased by 1000 times

The correct answer is:
Reduced by 1000 times

Solution:
When $p H=3$, then $\left[H^{+}\right]=10^{-3} M$
after that we increased the $p H$ from 3 to 6 then $\left[\mathrm{H}^{+}\right]=10^{-6} M$ means reduced 1000 times

Topic: Chemistry, Ionic Equilibrium, JEE Main
The pH of a solution obtained by mixing 50 mL of 1 M HCl and 30 mL of 1 M NaOH is x×10-4. The value of x is (Nearest integer)
log 2.5=0.3979

Options:

The correct answer is:
6021

Solution:
Milli equivalents of HClNaVa=50×1=50
Milli equivalents of NaOHNbVb=30×1=30
Since NaVa>NbVb
H+=NaVa-NbVbVa+Vb=50-3080=2080=0.25=2.5×10-1
pH=-logH+=-log2.5×10-1=1-0.3979=0.6021
pH×104=0.6021×104=6021

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2021 (31 Aug Shift 2)
The $\mathrm{pH}$ of a solution obtained by mixing $50 \mathrm{ml}$ of $0.4 \mathrm{~N} \mathrm{HCl}$ and $50 \mathrm{ml}$ of $0.2 \mathrm{~N} \mathrm{NaOH}$ is

Options:
A $-\log 2$
B $-\log 0.2$
C 1.0
D 2.0

The correct answer is:
1.0

Solution:
N.eq. for $\mathrm{HCl}=\frac{0.4}{1000} \times 50=0.02$
$\begin{aligned} & \text { N.eq. for } \mathrm{NaOH}=\frac{0.2}{1000} \times 50=0.1 \\ & \text { Now }\left[\mathrm{OH}^{-}\right] \text {left }=0.1-0.02 \\ & {\left[\mathrm{OH}^{-}\right]=.08=8 \times 10^{-2} \mathrm{M}} \\ & p \mathrm{OH}=-\log 8 \times 10^{-2} \mathrm{M} ; \mathrm{pOH}=1.0\end{aligned}$

Topic: Chemistry, Ionic Equilibrium, JEE Main