datavorous/entrance-exam-dataset huggingface dataset:
If the potential energy of a gas molecule is $U=\frac{M}{r^{6}}-\frac{N}{r^{12}}, M$ and $N$ being positive constants, then the potential energy at equilibrium must be

Options:
A zero
B $N M^{2} / 4$
C $M N^{2} / 4$
D $M^{2} / 4 N$

The correct answer is:
$M^{2} / 4 N$

Solution:
Given, $U=\frac{M}{r^{6}}-\frac{N}{r^{12}}$
$\begin{array}{l}
\therefore \mathrm{F}=\frac{-\mathrm{du}}{\mathrm{dr}}=\frac{-\mathrm{d}}{\mathrm{dr}}\left(\frac{\mathrm{M}}{\mathrm{r}^{6}}-\frac{\mathrm{N}}{\mathrm{r}^{12}}\right) \\
=-\left(\frac{-6 \mathrm{M}}{\mathrm{r}^{7}}+\frac{12 \mathrm{~N}}{\mathrm{r}^{13}}\right)=\left(\frac{6 \mathrm{M}}{\mathrm{r}^{7}}-\frac{12 \mathrm{~N}}{\mathrm{r}^{13}}\right)
\end{array}$
For equilibrium position, $\mathrm{F}=0$
$\therefore \frac{6 \mathrm{M}}{\mathrm{r}^{7}}=\frac{12 \mathrm{~N}}{\mathrm{r}^{13}} \text { or } \mathrm{r}^{6}=\frac{2 \mathrm{~N}}{\mathrm{M}}$
Hence, $U=\frac{M}{(2 N / M)}-\frac{N}{(2 N / M)^{2}}=\frac{M^{2}}{4 N}$

Topic: Physics, Work Power Energy, JEE Main
If the potential energy of a spring is $V$ on stretching it by $2 \mathrm{~cm}$, then its potential energy when it is stretched by $10 \mathrm{~cm}$ will be

Options:
A $V / 25$
B $5 \mathrm{~V}$
C $V / 5$
D $25 \mathrm{~V}$

The correct answer is:
$25 \mathrm{~V}$

Solution:
$U=\frac{1}{2}\left(\frac{Y A}{L}\right) l^2 \quad \therefore U \propto l^2$
$\frac{U_2}{U_1}=\left(\frac{l_2}{l_1}\right)^2=\left(\frac{10}{2}\right)^2=25 \Rightarrow U_2=25 U_1$

Topic: Physics, Mechanical Properties of Solids, JEE Main
If the potential of a capacitor having capacity $6 \mu \mathrm{F}$ is increased from $10 \mathrm{~V}$ to $20 \mathrm{~V}$, then increase in its energywill be

Options:
A $4 \times 10^{-4} \mathrm{~J}$
B $4 \times 10^{-6} \mathrm{~J}$
C $9 \times 10^{-4} \mathrm{~J}$
D $12 \times 10^{-6} \mathrm{~J}$

The correct answer is:
$9 \times 10^{-4} \mathrm{~J}$

Solution:
Capacitance of capacitor (C) $=6 \mu \mathrm{F}$ $=6 \times 10^{-6} \mathrm{~F}$; Initial potential $\left(\mathrm{V}_{1}\right)=10 \mathrm{~V}$ and
final potential $\left(\mathrm{V}_{2}\right)=20 \mathrm{~V}$.
The increase in energy $(\Delta \mathrm{U})$
$$
\begin{array}{l}
=\frac{1}{2} \mathrm{C}\left(\mathrm{V}_{2}^{2}-\mathrm{V}_{1}^{2}\right) \\
=\frac{1}{2} \times\left(6 \times 10^{-6}\right) \times\left[(20)^{2}-(10)^{2}\right] \\
=\left(3 \times 10^{-6}\right) \times 300=9 \times 10^{-4} \mathrm{~J} .
\end{array}
$$

Topic: Physics, Capacitance, JEE Main
If the pressure amplitude in a sound wave is tripled, then the intensity of sound is increased by a factor of

Options:
A 9
B 3
C 6
D $\sqrt{3}$

The correct answer is:
9

Solution:
Intensity $\propto(\text { Amplitude })^2$

Topic: Physics, Waves and Sound, JEE Main
If the principal quantum number n=6, the correct sequence of filling of electrons will be:

Options:
A ns→n-1 d→n-2 f→np
B ns→np→n-1d→n-2f
C ns→n-2f→np→n-1d
D ns→n-2f→n-1d→np

The correct answer is:
ns→n-2f→n-1d→np

Solution:
As per n+l rule electrons fill first in that orbital which have least n+l value.
When n+l values are same, then electron fills that orbital which have lowest n value.
when n=6 as per n+l rule when n=6
ns subshell â6+0=6
nâ1 d subshell â5+2=7
nâ2 f subshell â4+3=7
np subshell â6+1=7
ns , n-2f, n-1d,np
n+l values â7, 7, 7
n value â4, 5, 6

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2015 (10 Apr Online)
If the probability of a horse A winning a race is $1 / 4$ and the probability of a horse $B$ winning the same race is $1 / 5$, then the probability that either of them will win the race is

Options:
A $\frac{1}{20}$
B $\frac{9}{20}$
C $\frac{11}{20}$
D $\frac{19}{20}$

The correct answer is:
$\frac{9}{20}$

Solution:
$P(A \cup B)=P(A)+P(B)=\frac{1}{4}+\frac{1}{5}=\frac{9}{20}$
$\{$ Since events are mutually exclusive, so $P(A \cap B)=0\}$

Topic: Mathematics, Probability, JEE Main
If the probability of hitting a target by a shooter, in any shot is 13, then the minimum number of independent shots at the target required by him so that the probability of hitting the target at least once is greater than 56, is:

Options:
A 4
B 5
C 6
D 3

The correct answer is:
5

Solution:
Phitting target=13
Let number of trials =n
Phitting  target  atleast once
=1-P not hit
=1-23n>56
⇒16>23n
So minimum number of trials =5

Topic: Mathematics, Probability, JEE Main, JEE Main 2019 (10 Jan Shift 2)
If the probability of $X$ to fail in the examination is 0.3 and that for $Y$ is 0.2 , then the probability that either $X$ or $Y$ fail in the examination is

Options:
A $0.5$
B $0.44$
C $0.6$
D None of these

The correct answer is:
$0.44$

Solution:
Here $P(X)=0.3 ; \quad P(Y)=0.2$
Now $P(X \cup Y)=P(X)+P(Y)-P(X \cap Y)$
Since these are independent events, so
$P(X \cap Y)=P(X) \cdot P(Y)$
Thus required probability $=0.3+0.2-0.06=0.44$

Topic: Mathematics, Probability, JEE Main
If the probability that a student is not a swimmer is $1 / 5$, then the probability that out of 5 students one is swimmer is

Options:
A ${ }^5 C_1\left(\frac{4}{5}\right)^4\left(\frac{1}{5}\right)$
B ${ }^5 C_1 \frac{4}{5}\left(\frac{1}{5}\right)^4$
C $\frac{4}{5}\left(\frac{1}{5}\right)^4$
D None of these

The correct answer is:
${ }^5 C_1 \frac{4}{5}\left(\frac{1}{5}\right)^4$

Solution:
Required probability $={ }^5 C_1\left(\frac{4}{5}\right)\left(\frac{1}{5}\right)^4$

Topic: Mathematics, Probability, JEE Main
If the probability that the random variable X takes values x is given by P(X=x)=k(x+1)3-x, x=0,1,2,3,……, where k is a constant, then P(X≥2) is equal to

Options:
A 727
B 718
C 1118
D 2027

The correct answer is:
727

Solution:
Given,

The probability that the random variable X takes values x is given by P(X=x)=k(x+1)3-x, x=0,1,2,3,……, where k is a constant,

Now we know that,

P(X=0)+P(x=1)+P(x=2)+………=1

⇒k30+2k31+3k32+…..=1

⇒k1+23+332+….=1

Now finding,

S=1+23+332+433+… ....1

⇒S3=13+232+332+…… .....2

Now on subtracting above two equation we get,

⇒2S3=1+13+132+……

⇒S=94

Hence, k1+23+332+….=1

⇒k=49

Now finding,

P(X≥2)=P(2)+P(3)+……

⇒P(X≥2)=1-P(0)-P(1)

⇒P(X≥2)=1-k1+2k3=1-2027=727

Topic: Mathematics, Probability, JEE Main, JEE Main 2023 (08 Apr Shift 2)
If the product of height and square of the radius of the greatest cone obtained by rotating a right-angled triangle of hypotenuse 2 meters about a side is k33, then k is equal to

Options:

The correct answer is:
16

Solution:
Let, the cone is obtained by rotating about BC

From the diagram, the radius of the cone is AB and the height is BC
The volume generated, V=13πr2h=13π4-x2x
For maximum volume, dVdx=0
⇒4-3x2=0⇒x=23

∴Vmax=π34-4323=16π93
Also, d2Vdx2<0

Hence, x4-x2=234-43=1633

Topic: Mathematics, Application of Derivatives, JEE Main
If the product of the matrix $B=\left[\begin{array}{ccc}2 & 6 & 4 \\ 1 & 0 & 1 \\ -1 & 1 & -1\end{array}\right]$
with a matrix $A$ has the inverse $C=\left[\begin{array}{ccc}-1 & 0 & 1 \\ 1 & 1 & 3 \\ 2 & 0 & 2\end{array}\right]$, then $A^{-1}$ equals

Options:
A $\left[\begin{array}{ccc}-3 & -5 & 5 \\ 0 & 9 & 14 \\ 2 & 2 & 9\end{array}\right]$
B $\left[\begin{array}{ccc}-3 & 5 & 5 \\ 0 & 0 & 9 \\ 2 & 14 & 16\end{array}\right]$
C $\left[\begin{array}{ccc}-3 & -5 & -5 \\ 0 & 0 & 2 \\ 2 & 14 & 6\end{array}\right]$
D $\left[\begin{array}{ccc}-3 & -5 & -5 \\ 0 & 9 & 2 \\ 2 & 14 & 6\end{array}\right]$

The correct answer is:
$\left[\begin{array}{ccc}-3 & -5 & -5 \\ 0 & 9 & 2 \\ 2 & 14 & 6\end{array}\right]$

Solution:
$(B A)^{-1}=C$ (given)
or $A^{-1} B^{-1}=C$
or $A^{-1}\left[\begin{array}{ccc}2 & 6 & 4 \\ 1 & 0 & 1 \\ -1 & 1 & -1\end{array}\right]^{-1}=\left[\begin{array}{ccc}-1 & 0 & 1 \\ 1 & 1 & 3 \\ 2 & 0 & 2\end{array}\right]$
Multiply by $B$ on both sides, we get
$A^{-1}\left(B^{-1} B\right)=\left[\begin{array}{ccc}-1 & 0 & 1 \\ 1 & 1 & 3 \\ 2 & 0 & 2\end{array}\right]\left[\begin{array}{ccc}2 & 6 & 4 \\ 1 & 0 & 1 \\ -1 & 1 & -1\end{array}\right]$
or
$$
A^{-1}=\left[\begin{array}{ccc}
-3 & -5 & -5 \\
0 & 9 & 2 \\
2 & 14 & 6
\end{array}\right]_{3 \times 3}
$$

Topic: Mathematics, Matrices, JEE Main
If the product of three terms of G.P. is 512 . If 8 added to first and 6 added to second term, so that number may be in A.P., then the numbers are

Options:
A $2,4,8$
B $4,8,16$
C $3,6,12$
D None of these

The correct answer is:
$4,8,16$

Solution:
Let three terms of a G.P. are $\frac{a}{r}, a, a r$
so, $\frac{a}{r} \cdot a \cdot a r=512 \Rightarrow a^3=8^3 \Rightarrow a=8$
From second condition, we get $\frac{a}{r}+8, a+6, a r$ will be in A.P.
$\Rightarrow \quad 2(a+6)=\frac{a}{r}+8+a r$ $\Rightarrow \quad 28=8\left\{\frac{1}{r}+1+r\right\}$
$\Rightarrow \frac{1}{r}+r+1=\frac{7}{2} \Rightarrow \frac{1}{r}+r-\frac{5}{2}=0$
$\Rightarrow \quad r^2-\frac{5}{2} r-1=0$ $\Rightarrow 2 r^2-5 r+2=0$
$\Rightarrow \quad(2 r-1)(r-2)=0 \Rightarrow$ $r=\frac{1}{2}, r=2 \quad(r\gt1)$
$\Rightarrow r=2$. Hence required numbers are $4,8,16$
Trick : Check for (1) $2+8,4+6,8$ are not in A.P.
(2) $4+8,8+6,16$ i.e. $12,14,16$ are in A.P.

Topic: Mathematics, Sequences and Series, JEE Main
If the projection of 2i^+4j^-2k^ on i^+2j^+αk^ is zero. Then, the value of α will be

Options:

The correct answer is:
5

Solution:
Given here:

a→=2i^+4j^-2k^ and b→=i^+2j^+αk^

Now, a→·b→=0, as a→ should be perpendicular to b→.

∴ 2×1+4×2-2×α=0

∴ α=5

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2022 (28 Jul Shift 1)
If the projection of the vector i^+2j^+k^ on the sum of the two vectors 2i^+4j^-5k^ and -λi^+2j^+3k^ is 1, then λ is equal to _______.

Options:

The correct answer is:
5

Solution:
Sum of two vectors =2i^+4j^-5k^+(-λi^+2j^+3k^)=(2-λ)i^+6j^-2k^
Projection =1=1(2-λ)+2(6)+1×(-2)(2-λ)2+62+22

⇒12-λλ2-4λ+44=1
⇒(12-λ)2=λ2-4λ+44
⇒144+λ2-24λ=λ2-4λ+44
⇒100=20λ
⇒λ=5

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2021 (26 Aug Shift 2)
If the projections of a line segment on the $x, y$ and $z$-axes in 3-dimensional space are 2, 3 and 6 respectively, then the length of the line segment is :

Options:
A
12

B
7

C
9

D
6

The correct answer is:
7

Solution:
Length of the line segment
$$
=\sqrt{(2)^2+(3)^2+(6)^2}=7
$$

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2013 (23 Apr Online)
If the proton in the hydrogen atom had a charge of (+4/3)e and the electron had a charge of (−3/4)e, where e = 1.6 × 10–19C, would the Bohr formula for the H-atom remain unchanged? Provide reasons for your answer.

Options:
A Yes, because the Bohr formula involves only the product of the charges.
B No, because the Bohr formula directly depends on the individual charges of the electron and the proton.
C Yes, because the Bohr formula is independent of the charges of the particles.
D No, because the Bohr formula is sensitive to the ratio of the charges of the particles.

The correct answer is:
Yes, because the Bohr formula involves only the product of the charges.

Solution:
Correct Option is : (A)
Yes, because the Bohr formula involves only the product of the charges.

Topic: Chemistry, Atomic Physics, NEET
If the radii of nuclei of 13Al27 and 30Zn64 are R1 and R2 respectively, then R1R2=

Options:
A 6427
B 34
C 2764
D 43

The correct answer is:
34

Solution:
Given R1 and R2 are the radii of atomic nuclei of mass number 27 and 64 respectively, then the ratio

R1R2=A1A213=2764=34

Topic: Physics, Nuclear Physics, JEE Main
If the radius and acceleration due to gravity both are doubled, escape velocity of earth will become

Options:
A $11.2 \mathrm{~km} / \mathrm{s}$
B $22.4 \mathrm{~km} / \mathrm{s}$
C $5.6 \mathrm{~km} / \mathrm{s}$
D $44.8 \mathrm{~km} / \mathrm{s}$

The correct answer is:
$22.4 \mathrm{~km} / \mathrm{s}$

Solution:
$v=\sqrt{2 g R}$. If $g$ and $R$ both are doubled then $v$ will becomes two times i.e. $11.2 \times 2=22.4 \mathrm{~km} / \mathrm{s}$

Topic: Physics, Gravitation, JEE Main
If the radius of curvature of the path of two particles of same mass are in the ratio 3: 4, then in order to have constant centripetal force, their velocities will be in the ratio of:

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

The correct answer is:
3:2

Solution:
Given m1=m2 and r1r2=34

As centripetal force is given by F=mv2r, in order to have the same centripetal force, we can write

F1=F2⇒m1v12r1=m2v22r2⇒v1v2=r1r2=32

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2024 (29 Jan Shift 1)
If the radius of earth shrinks by 2% while its mass remains same. The acceleration due to gravity on the earth's surface will approximately

Options:
A decrease by 2%
B decrease by 4%
C increase by 2%
D increase by 4%

The correct answer is:
increase by 4%

Solution:
The acceleration due to gravity on the surface of the earth is given by, g=GMR2.

The radius shrunk by 2%. Therefore, ∆RR×100=-2%.

As the mass is constant and the change in the radius of the earth is relatively small, we can write

∆gg=-2∆RR⇒∆gg×100=4%

Hence, the acceleration due to gravity increases by 4%.

Topic: Physics, Gravitation, JEE Main, JEE Main 2022 (28 Jul Shift 1)
If the radius of first Bohr's orbit is x, then de Broglie wavelength of an electron in 3rd orbit is nearly

Options:
A 2πx
B 6Ïx
C 9x
D x/3

The correct answer is:
6πx

Solution:
If r3=Radius of 3rd orbit

and r1=Radius of 1st orbit

r3=r1Ãnâ¡2=xÃ32=9x
âµ 2Ïr3=nÎ»
â´Î»=2Ïr3nâ¡=2ÏÃ9x3=6Ïx

Topic: Physics, Atomic Physics, JEE Main
If the radius of first Bohr's orbit of hydrogen is x , then de-Broglie wavelength of electron in its 3rd orbit is

Options:
A 2πx
B 6πx
C 9x
D x3

The correct answer is:
6πx

Solution:
r n ∝ n 2 z
for Hydrogen atom, r n ∝ n 2
r 1 =x
r 3 =9x
Now, ⇒ 2πr 3 =nλ
⇒λ= 2 πr 3 3
⇒λ= 2 3 π×9x
∴ λ=6πx

Topic: Chemistry, Structure of Atom, JEE Main
If the radius of first orbit of $\mathrm{H}$ atom is $a_0$, the deBroglie wavelength of an electron in the third orbit is

Options:
A
$4 \pi a_0$

B
$8 \pi a_0$

C
$6 \pi a_0$

D
$2 \pi a_0$

The correct answer is:
$6 \pi a_0$

Solution:
$r_n=a_0 n^2$
$$
\begin{aligned}
& r=a_0 \times(3)^2=9 a_0 \\
& m v r=\frac{n h}{2 \pi} \\
& m v=\frac{n h}{2 \pi \mathrm{r}}=\frac{3 h}{2 \pi \times 9 a_0}=\frac{h}{6 \pi a_0} \\
& \lambda=\frac{h}{m v}=\frac{h}{h} \times 6 \pi a_0=6 \pi a_0
\end{aligned}
$$

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2012 (12 May Online)
If the radius of $\mathrm{H}$ is $0.53 Ã…$, then what will be the radius of ${ }_{3} \mathrm{Li}^{2+}$?

Options:
A $0.17 Ã…$
B $0.36 Ã…$
C $0.53 Ã…$
D $0.59 Ã…$

The correct answer is:
$0.17 Ã…$

Solution:
Radius of orbit $=\frac{\mathrm{n}^{2} \mathrm{a}_{\mathrm{o}}}{\mathrm{z}}\left(\mathrm{a}_{\mathrm{o}}=0.529 \mathrm{~A}\right)$
Radius of $\mathrm{H}=\frac{(1)^{2} \times 0.529 \mathrm{~A}}{1}=0.53 \mathrm{~A}$
Thus, the radius of ${ }_{3} \mathrm{Li}^{2+}$ will be $=\frac{(1)^{2} \times 0.529}{3}=0.17 Ã…$

Topic: Chemistry, Structure of Atom, JEE Main
If the radius of the 1327Al nucleus is estimated to be 3.6 Fermi, then the radius of 52125Te nucleus be nearly :

Options:
A 6 Fermi
B 8 Fermi
C 4 Fermi
D 5 Fermi

The correct answer is:
6 Fermi

Solution:
R R 0 1 1 / 3
R AI R Te = R 0 A AI 1 / 3 R 0 A Te 1 / 3 = 3 5
â´ R Te = 5 3 Ã 3 . 6
RTe = 6 Fermi

Topic: Physics, Nuclear Physics, JEE Main
If the radius of the 3nd Bohr's orbit of hydrogen atom is r3 and the radius of 4th Bohr's orbit is r4. Then

Options:
A r4=916r3
B r4=169r3
C r4=34r3
D r4=43r3

The correct answer is:
r4=169r3

Solution:
For H-atom, Z=1

rn∝n2z

r3r4=3242=916

r4=169r3

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2022 (26 Jun Shift 1)
If the radius of the circle passing through the origin and touching the line x+y=2 at 1,1 is r units, then the value of 32r is

Options:

The correct answer is:
3

Solution:
Family of circles touching the given line at the given point is
x-12+y-12+λx+y-2=0
It passes through 0,0, hence 2-2λ=0⇒λ=1
So, the equation of the circle is
x2+y2-x-y=0
Radius =12

Topic: Mathematics, Circle, JEE Main
If the radius of the circle $x^2+y^2-18 x+12 y+k=0$ be 11 , then $k=$

Options:
A $347$
B $4$
C $-4$
D $49$

The correct answer is:
$-4$

Solution:
$(\text { Radius) })^2=g^2+f^2-c$ or $121=81+36-k$
or $k=-4$.

Topic: Mathematics, Circle, JEE Main
If the radius of the earth becomes half of its present value (mass remaining the same), the new length of the day would be

Options:
A 6 hours
B 12 hours
C 48 hours
D 96 hours

The correct answer is:
6 hours

Solution:
The angular momentum of the earth about its axis will remain constant

L=Iω=25MR22πT

⇒T=4πM5LR2

∴T∝R2

If the radius of earth is reduced to half of its present value then time period will become one-fourth of present value

Topic: Physics, Gravitation, NEET
If the radius of the earth becomes $x$ times its present value, the new period of rotation in hours is

Options:
A $6x^2$
B $12x^2$
C $24x^2$
D $48x^2$

The correct answer is:
$24x^2$

Solution:
Angular momentum will remain conserved
$\begin{aligned}
& \mathrm{L}=\mathrm{I} \omega \\
& \therefore \mathrm{I}_1 \omega_1=\mathrm{I}_2 \omega_2 \\
& \Rightarrow \frac{2}{5} \mathrm{MR}^2 \times \frac{2 \pi}{\mathrm{T}_1}=\frac{2}{5} \mathrm{M}(\mathrm{xR})^2 \times \frac{2 \pi}{\mathrm{T}_2}
\end{aligned}$
$\begin{aligned} & \Rightarrow T_2=T_1 x^2 \\ & \Rightarrow T_2=24 x^2\end{aligned}$

Topic: Physics, Rotational Motion, TS EAMCET, TS EAMCET 2023 (12 May Shift 1)
If the radius of the earth shrinks by $1 \%$, its mass remaining the same, then the acceleration due to gravity on the earth surface would

Options:
A increase by $1 \%$
B remain unchanged
C increase by $2 \%$
D decrease by $9.8 \%$

The correct answer is:
increase by $2 \%$

Solution:
Acceleration due to gravity,
$g_1=\frac{G M}{R_1^2}$
As, $M$ remain constant and $R$ becomes
$R_2=R_1-1 \% \text { of } R_1=R_1-\frac{R_1}{100}=\frac{99}{100} R_1$
Therefore, the new gravitational acceleration becomes,
$g_2=\frac{G M}{R_2^2}=\frac{G M \times(100)^2}{(99)^2 R_1^2}=\frac{(100)^2}{(99)^2} g_1=1.02 g_1$
Change in the gravitational acceleration,
$g_2-g_1=1.02 g_1-g_1=0.02 g_1$
Hence, the percentage increase is
$\frac{g_2-g_1}{g_1} \times 100=0.02 \times 100=2 \%$
So, if the radius of earth shrinks by $1 \%$, the value of $g$ would increase by $2 \%$.

Topic: Physics, Gravitation, TS EAMCET, TS EAMCET 2020 (11 Sep Shift 2)
If the radius of the earth were to shrink by 1% its mass remaining same, the acceleration due to gravity on the earth's surface would

Options:
A Decrease by 2%
B Remain unchanged
C Increase by 2%
D Become zero

The correct answer is:
Increase by 2%

Solution:
g=GMR2; g′=GMR′2=GM1002992R2
% increase in g=(g′-g)×100g
=g′g-1×100=100992-1×100
1+1992-1×100≈2%

Topic: Physics, Gravitation, NEET
If the radius of the earth were to shrink by 1% its mass remaining same, the acceleration due to gravity on the earth's surface would

Options:
A Decrease by 2%
B Remain unchanged
C Increase by 2%
D Become zero

The correct answer is:
Increase by 2%

Solution:
g=GMR2; g′=GMR′2=GM1002992R2
% increase in g=(g′-g)×100g
=g′g-1×100=100992-1×100
1+1992-1×100≈2%

Topic: Physics, Gravitation, JEE Main
If the radius of the earth were to shrink by one percent, its mass remaining the same, the value of g on the earth's surface would

Options:
A increase by 0.5%
B increase by 2%​
C decrease by 0.5%​
D decrease by 2%​

The correct answer is:
increase by 2%​

Solution:
g = GM R 2
Δ g g × 100 = 2 Δ R R × 100
= 2 × 1 % = 2 %
Therefore value of g will increase by 2%

Topic: Physics, Gravitation, NEET
If the radius of the first orbit of hydrogen atom is a0, then de Broglie’s wavelength of electron in 3rd orbit is

Options:
A πa06
B πa03
C 6πa0
D 3πa0

The correct answer is:
6πa0

Solution:
According to Bohr's angular momentum equation,

mvr=nh2π

The angular momentum equation for third orbit is

mvr=3h2π

2πr3=hmv

According de Broglie's equation

λ=hmv

∴ λ=2π9a03=6πa0

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2023 (06 Apr Shift 2)
If the radius of the hydrogen atom is $53 \mathrm{pm}$, the radius of the $\mathrm{He}^{+}$ion is closest to

Options:
A $108 \mathrm{pm}$
B $81 \mathrm{pm}$
C $27 \mathrm{pm}$
D $13 \mathrm{pm}$

The correct answer is:
$27 \mathrm{pm}$

Solution:
$r_{n}=\frac{R_{H} n^{2}}{Z}$
$r_{H_{e+}}=\frac{53 n^{2}}{Z}$ $=\frac{53 \times 1^{2}}{2}=27$ approx.

Topic: Chemistry, Structure of Atom, JEE Main
If the radius of the spherical Gaussian surface is increased then the electric flux due to a point charge enclosed by the surface

Options:
A remains unchanged
B zero
C increases
D decreases

The correct answer is:
remains unchanged

Solution:
Flux depends on the charge enclosed. Therefore, size of the Gaussian sphere would not affect the net flux.

Topic: Physics, Electrostatics, MHT CET, MHT CET 2022 (10 Aug Shift 2)
If the radius of the spherical Gaussian surface is increased then the electric flux
due to a pooint charge enclosed by the surface

Options:
A decreases.
B remains unchanged.
C increases.
D is zero.

The correct answer is:
remains unchanged.

Solution:
(A)
Flux depends on the charge enclosed.

Topic: Physics, Electrostatics, MHT CET, MHT CET 2020 (19 Oct Shift 1)
If the radius ratio is in the range of $0.731-1$, then the coordination number will be

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

The correct answer is:
$8$

Solution:
The radius ratio for co-ordination number 4,6 and 8 lies in between the ranges $[0.225-0.414],[0.414-0.732]$ and $[0.732-1]$ respectively.

Topic: Chemistry, Solid State, JEE Main
If the range of $f(\theta)=\frac{\sin ^4 \theta+3 \cos ^2 \theta}{\sin ^4 \theta+\cos ^2 \theta}, \theta \in \mathbb{R}$ is $[\alpha, \beta]$, then the sum of the infinite G.P., whose first term is 64 and the common ratio is $\frac{\alpha}{\beta}$, is equal to________

Options:

The correct answer is:
96

Solution:
$\begin{aligned} & f(\theta)=\frac{\sin ^4 \theta+3 \cos ^2 \theta}{\sin ^4 \theta+\cos ^2 \theta} \\ & f(\theta)=1+\frac{2 \cos ^2 \theta}{\sin ^4 \theta+\cos ^2 \theta} \\ & f(\theta)=\frac{2 \cos ^2 \theta}{\cos ^4 \theta-\cos ^2 \theta+1}+1 \\ & f(\theta)=\frac{2}{\cos ^2 \theta+\sec ^2 \theta-1}+1 \\ & \left.f(\theta)\right|_{\text {min. }}=1 \\ & f(\theta)_{\text {max. }}=3 \\ & S=\frac{64}{1-1 / 3}=96\end{aligned}$

Topic: Mathematics, Functions, JEE Main, JEE Main 2024 (08 Apr Shift 1)
If the range of fx=tan-1x+2sin-1x+cos-1x is a,b, then

Options:
A a=π4
B a=-π2
C b=5π4
D b=3π2

The correct answer is:
b=5π4

Solution:
fx=tan-1x+sin-1x+sin-1x+cos-1x
fx=tan-1x+sin-1x+π2;x∈-1,1 ∵sin-1x+cos-1x=π2
∵fx is an increasing function
∴ Range of fx∈f-1,f1
f-1=tan-1-1+sin-1-1+π2=-π4
f1=tan-11+sin-11+π2=5π4
∴ Range of fx∈-π4,5π4

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If the range of $\operatorname{sech}^{-1} x+\operatorname{cosech}^{-1} x$ is $[a, b]$, then

Options:
A $a=0, b=1$
B $a=\sqrt{2}, b=\infty$
C $a=\log (1+\sqrt{2}), b=\infty$
D $a=0, b=\log (1+\sqrt{2})$

The correct answer is:
$a=\log (1+\sqrt{2}), b=\infty$

Solution:
$\begin{aligned} & \text { Range of } \operatorname{sech}^{-1} x=\operatorname{cosech}^{-1} x \text { is }[a, b] \\ & \because \operatorname{sech}^{-1} x=\cosh ^{-1}\left(\frac{1}{x}\right)=\ln \left(\frac{1}{x}+\sqrt{\frac{1}{x^2}-1}\right) \\ & \therefore \text { Domain } x \in(0,1] \\ & \operatorname{cosech}^{-1}(x)=\sinh ^{-1}\left(\frac{1}{x}\right)=\ln \left(\frac{1}{x}+\sqrt{\frac{1}{x^2}+1}\right) \\ & \therefore \text { Domain } x \in R-\{0\} \\ & \therefore \text { Domain of } \operatorname{sech}^{-1}(x)+\operatorname{cosech}^{-1}(x) \text { is Eqs. (i) and (ii) } \\ & \Rightarrow \quad x \in(0,1] \\ & \therefore \quad f(0)=\operatorname{sech}^{-1}(0)+\operatorname{cosech}^{-1}(0)=\infty \\ & \quad f(1)=\operatorname{sech}^{-1}(1)+\operatorname{cosech}^{-1}(1)=0+\ln (1+\sqrt{2}) \\ & \therefore \text { Range }=[(\ln (1+\sqrt{2}), \infty) \\ & \text { Hence, } a=\ln (1+\sqrt{2}), b=\infty\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, TS EAMCET, TS EAMCET 2021 (06 Aug Shift 2)