datavorous/entrance-exam-dataset huggingface dataset:
The radiosotope tritium $\left({ }_1^3 \mathrm{H}\right)$ has a lalf life of 12.3 years. If the initial amount of tritium is $32 \mathrm{mg}$, how many milligrams of it would remain after 49.2 years

Options:
A $1 \mathrm{mg}$
B $2 \mathrm{mg}$
C $4 \mathrm{mg}$
D $8 \mathrm{mg}$

The correct answer is:
$2 \mathrm{mg}$

Solution:
$$
\begin{aligned}
\mathrm{T} & =\mathrm{t}^{1 / 2} \times n \\
n & =\frac{49.2}{12.3}=4 \\
\mathrm{~N} & =\mathrm{N}_0(1 / 2)^n \\
& =32(1 / 2)^4 \\
& =2 \mathrm{mg}
\end{aligned}
$$

Topic: Chemistry, Chemical Kinetics, JEE Main
The radiosotope tritium $\left({ }_1^3 \mathrm{H}\right)$ has a lalf life of 12.3 years. If the initial amount of tritium is $32 \mathrm{mg}$, how many milligrams of it would remain after 49.2 years

Options:
A $1 \mathrm{mg}$
B $2 \mathrm{mg}$
C $4 \mathrm{mg}$
D $8 \mathrm{mg}$

The correct answer is:
$2 \mathrm{mg}$

Solution:
$$
\begin{aligned}
\mathrm{T} & =\mathrm{t}^{1 / 2} \times n \\
n & =\frac{49.2}{12.3}=4 \\
\mathrm{~N} & =\mathrm{N}_0(1 / 2)^n \\
& =32(1 / 2)^4 \\
& =2 \mathrm{mg}
\end{aligned}
$$

Topic: Chemistry, Chemical Kinetics, NEET, NEET 2003
The radius in kilometer to which the present radius of earth (R=6400 km) to be compressed so that the escape velocity is increased 10 time is _______.

Options:

The correct answer is:
64

Solution:
Ve=2GmR …1

10 Ve=2GmR' …2

∴10=RR'

⇒R'=R100=6400100=64 km

Topic: Physics, Gravitation, JEE Main, JEE Main 2021 (17 Mar Shift 1)
The radius of 2nd orbit of He+ of Bohr's model is r1 and that of fourth orbit of Be3+ is represented as r2. Now the ratio r2r1 is x:1. The value of x is _____.

Options:

The correct answer is:
2

Solution:
The formula to calculate the radius r of any nth Bohr's orbit is given by

r=r0n2Z ...1

where, r0 is Bohr's radius, n is the principal quantum number and Z is the atomic number.

Thus, for Helium ion,

r1=r0222= 2r0 ...2

And, for the Beryllium ion,

r2=r0424=4r0 ...3

Divide equation (3) by equation (2) to obtain the required ratio.

r2r1=4r02r0=2

Hence, x=2.

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2023 (13 Apr Shift 1)
The radius of a body moving in a circle with constant angular velocity is given by $r=$ $4 \mathrm{t} 2$, with respect to time. What is the magnitude of the tangential velocity at $t=2 \mathrm{~s}$, if the angular velocity is $7 \mathrm{rad} / \mathrm{s}$ ?

Options:
A $112$
B $113$
C $56$
D $28$

The correct answer is:
$112$

Solution:
Given, $r=4 \mathrm{t}^2$, and angular velocity, $\boldsymbol{\omega}=7$ $\mathrm{rad} / \mathrm{s}$
We know that tangential velocity, $v=r \omega$
At $t=2 s, r=4(2)^2=16 \mathrm{~m}$
Substituting the given values, $v=16 \times 7=$ $112 \mathrm{~m} / \mathrm{s}$
Therefore, the magnitude of the tangential velocity at $\mathrm{t}=2 \mathrm{~s}$ is $112 \mathrm{~m} / \mathrm{s}$.

Topic: Physics, Laws of Motion, JEE Main
The radius of a calcium ion is $94 \mathrm{pm}$ and of the oxide ion is $146 \mathrm{pm}$. The possible crystal structure of calcium oxide will be

Options:
A
tetrahedral

B
trigonal

C
octahedral

D
pyramidal

The correct answer is:
octahedral

Solution:
As per formula,
$$
\begin{aligned}
\text { radius ratio } & =\frac{\text { radius of cation }}{\text { radius of anion }} \\
& =\frac{94}{146}=0.643
\end{aligned}
$$
Since the value is between $0.414-0.732$ hence the coordination no. will be 6 and geometry will be octahedral.

Topic: Chemistry, Solid State, JEE Main, JEE Main 2012 (12 May Online)
The radius of a circle, having minimum area, which touches the curve y=4-x2 and the lines, y=x is:

Options:
A 22+1
B 22-1
C 42-1
D 42+1

The correct answer is:
42-1

Solution:
Length of perpendicular from centre of circle to tangent line r=k2
And r=4-k
⇒ r=4-k=k2
⇒4=k+2k2

⇒ k=422+1
⇒r=422+12
⇒r=42+1

⇒ r=42-1

Topic: Mathematics, Circle, JEE Main, JEE Main 2017 (02 Apr)
The radius of a circle is increasing at a rate of $0.1 \mathrm{~cm} \mathrm{~s}^{-1}$. Then the rate of change of area, when its radius is $5 \mathrm{~cm}$, is .........

Options:
A $\pi^2 \mathrm{~cm}^2 \mathrm{~s}^{-1}$
B $\pi \mathrm{cm}^2 \mathrm{~s}^{-1}$
C $2 \pi \mathrm{cm}^2 \mathrm{~s}^{-1}$
D $\frac{\pi}{2} \mathrm{~cm}^2 \mathrm{~s}^{-1}$

The correct answer is:
$\pi \mathrm{cm}^2 \mathrm{~s}^{-1}$

Solution:
Let the circle have radius $r \mathrm{~cm}$ and area $A \mathrm{~cm}^2$ and it is given that $\frac{d r}{d t}=0.1 \mathrm{~cm} . \mathrm{s}^{-1}$ and as we know area $A=\pi r^2 \Rightarrow \frac{d A}{d t}=2 \pi r \frac{d r}{d t}$
$\therefore$ Rate of change of area, when radius is $5 \mathrm{~cm}$ is $\left.\frac{d A}{d t}\right|_{r=5 \mathrm{~cm}}=2 \pi(5)(0.1)=\pi \mathrm{cm}^2 \cdot \mathrm{s}^{-1}$

Topic: Mathematics, Differential Equations, JEE Main
The radius of a circle which touches $y$-axis at $(0,3)$ and cuts intercept of 8 units with $x$-axis, is

Options:
A $3$
B $2$
C $5$
D $8$

The correct answer is:
$5$

Solution:
Obviously from figure,
Radius is $r=\sqrt{4^2+3^2}=5$.

Topic: Mathematics, Circle, JEE Main
The radius of a nucleus of mass number 64 is 4.8 fermi. Then the mass number of another nucleus having radius of 4 fermi is 1000x, where x is _________.

Options:

The correct answer is:
27

Solution:
Radius of a nucleus is given by, R=R0 A1/3.

Therefore, R3∝A

⇒4.843=64A

⇒64 A=(1.2)3

⇒64A=1.44×1.2

⇒A=641.44×1.2=1000x

⇒x=144×1264=27

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2024 (01 Feb Shift 1)
The radius of a right circular cylinder increases at the rate of 0.2 cm/sec and the height decreases at the rate of 0.1 cm/sec. The rate of change of the volume of the cylinder when the radius is 1cm and the height is 2cm is

Options:
A π10cm3/sec
B 9π10cm3/sec
C 7π10cm3/sec
D π5cm3/sec

The correct answer is:
7π10cm3/sec

Solution:
Volume of cylinder, V=πr2h
dVdt=πr2dhdt+2rhdrdt
Given, dhdt=-0.1, drdt=0.2
dVdt=π-0.112+2120.2
dVdt=π-110+45=7π10cm3/sec

Topic: Mathematics, Application of Derivatives, JEE Main
The radius of a sphere is measured to be (7.50±0.85) cm. Suppose the percentage error in its volume is x. The value of x, to the nearest x, is ___ .

Options:

The correct answer is:
34

Solution:
∵v=43πr3

taking log & then differentiate

dVV=3drr=3×0.857.5×100%=34%

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2021 (18 Mar Shift 2)
The radius of an atomic nucleus of mass number 64 is 4.8 fermi. Then the mass number of another atomic nucleus of radius 6 fermi is

Options:
A 64
B 81
C 100
D 125

The correct answer is:
125

Solution:
Density of nucleus is constant. Therefore,

A1A2=r1r23⇒A2=64.83×64=125

Topic: Physics, Nuclear Physics, JEE Main
The radius of circle, touching the parabola y2=8x at 2,4 and passing through 0,4, is

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

The correct answer is:
2 units

Solution:
Equation of the tangent at 2,4 on the parabola y2=8x is y4=8x+22⇒y=x+2
Let equation of circle touching line y=x+2 at 2,4 is x-22+y-42+λx-y+2=0 which passes through 0,4 ⇒4+0+λ0-4+2⇒λ=2
⇒Equation of the required circle is x2+y2-2x-10y+24=0
⇒ The radius of the circle is 12+52-24=2

Topic: Mathematics, Parabola, JEE Main
The radius of curvature for a convex lens is $40 \mathrm{~cm}$, for each surface. Its refractive index is 1.5 . The focal length will be

Options:
A 40 cm
B 20 cm
C 80 cm
D 30 cm

The correct answer is:
40 cm

Solution:
By formula $\frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
$\begin{aligned} & =(1.5-1)\left(\frac{1}{40}+\frac{1}{40}\right)=0.5 \times \frac{1}{20}=\frac{1}{40} \\ \therefore f & =40 \mathrm{~cm}\end{aligned}$

Topic: Physics, Ray Optics, JEE Main
The radius of curvature of each surface of a convex lens having refractive index 1.8 is 20 cm. The lens is now immersed in a liquid of refractive index 1.5. The ratio of power of lens in air to its power in the liquid will be x: 1. The value of x is

Options:

The correct answer is:
4

Solution:
The power of a lens is given by

P1=1f

Using Lens Maker's formula,

P1=(1.8−1)1R1−1R2 and

P2=1fimmersed =1.81.5−11R1−1R2

Hence, the ratio is,

P1P2=(1.8−1)1.81.5−1=0.8×1.50.3=4

Topic: Physics, Ray Optics, JEE Main, JEE Main 2023 (11 Apr Shift 1)
The radius of Earth is about $6400 \mathrm{~km}$ and that of Mars is $3200 \mathrm{~km}$, and mass of the Earth is about 10 times mass of Mars. An object weight $200 \mathrm{~N}$ on the surface of Earth. Then, its weight on the surface of Mars will be

Options:
A $80 \mathrm{~N}$
B $40 \mathrm{~N}$
C $20 \mathrm{~N}$
D $8 \mathrm{~N}$

The correct answer is:
$80 \mathrm{~N}$

Solution:
Given, radius of Earth and Mars are $6400 \mathrm{~km}$ and $3200 \mathrm{~km}$, respectively.
Mass of Earth $=10$ (Mass of Mars)
$M_e=10 M_m$
We know that, acceleration due to gravity at surface of a planet is
$g=\frac{G M}{R^2},$
where, $M=$ mass of planet
and $R=$ radius of planet.
Taking the ratio of acceleration due to gravity of Mars to Earth,
$\frac{g_m}{g_e}=\frac{\frac{G M_m}{R_m^2}}{\frac{G M_e}{R_e^2}}=\frac{M_m}{M_e} \times\left(\frac{R_e}{R_m}\right)^2$
By substituting the values, we get
$\begin{aligned} & \frac{g_m}{g_e}=\frac{1}{10} \times\left(\frac{6400}{3200}\right)^2=\frac{1}{10} \times 4 \\ & g_m=\frac{2}{5} g_e\end{aligned}$
Given, the weight on earth's surface,
$w_e=m g_e=200 \mathrm{~N}$
$\therefore$ Weight on the Mars surface,
$w_m=m g_m=\frac{2}{5}\left(m g_e\right)=\frac{2}{5} \times 200 \mathrm{~N}$
$=80 \mathrm{~N}$

Topic: Physics, Gravitation, JEE Main
The radius of electron's second stationary orbit in Bohr's atom is R. The radius of 3rd orbit will be

Options:
A R3
B 2.25 R
C 3R
D 9R

The correct answer is:
2.25 R

Solution:
As we know, radius of hydrogen like atom r∝n2Z.

Therefore,

r2r3=n2n32

⇒Rr3=232

⇒ r3=94R

=2.25R

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2023 (31 Jan Shift 2)
The radius of fifth orbit of Li++ is_______×10-12 m. Take: radius of hydrogen atom=0.51 A∘

Options:

The correct answer is:
425

Solution:
The formula to calculate the radius of Lithium-ion is given by

RLi=RH×n2Z ...(1)

Substitute the values of the known parameters into equation (1) to calculate the required radius of the ion.

RLi=0.51×253×10-10 m=425×10-12 m

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2023 (06 Apr Shift 1)
The radius of germanium (Ge) nuclide is measured to be twice the radius of Be49. The number of nucleons in Ge are

Options:
A 73
B 74
C 75
D 72

The correct answer is:
72

Solution:
Let the radius of Be49 nucleus be r. Then, radius of germanium (Ge) nucleus will be 2r.
Radius of nucleus is given by
R=R0A1/3
∴ R1R2=A1A21/3
⇒ r2r=9A21/3 (∵A1=9)
⇒ 123=9A2
Hence, A2=9 × 23 =9×8=72
Thus, in germanium (Ge) nucleus number of nucleons is 72.

Topic: Physics, Nuclear Physics, JEE Main
The radius of germanium (Ge) nuclide is measured to be twice the radius of 4 9 Be. Calculate the number of nucleons in Ge.

Options:

The correct answer is:
72

Solution:
Nuclear radius, R=R0(A)13,
or Râ(A)13
Râ¡ Be Râ¡ Ge = 9 1 / 3 Aâ¡ 1 / 3 or, Râ¡ Beâ¡ 2 Râ¡ Be = 9 1 / 3 Aâ¡ 1 / 3 âµ Given Râ¡ Geâ¡ = 2 Râ¡ Beâ¡
or (A)13= 2 Ã (9)13
or A=23Ã9=8Ã9=72
â´ The number of nucleons in Ge is 72.

Topic: Physics, Atomic Physics, JEE Main
The radius of gyration of a body about an axis at a distance \( 6 \mathrm{~cm} \) from its centre of mass is \( 10 \mathrm{~cm} \). Then its radius of gyration about a parallel axis through its centre of mass will be

Options:
A \( 80 \mathrm{~cm} \)
B \( 8 \mathrm{~cm} \)
C \( 0.8 \mathrm{~cm} \)
D \( 8.0 \mathrm{~cm} \)

The correct answer is:
\( 8 \mathrm{~cm} \)

Solution:
From the theorem of parallel axis, the moment of inertia I is equal to
I=ICM+Ma2
where ICM moment of inertia is about centre of mass and a the distance of axis from centre.
∴ I=MK2+M×62
MK12=MK2+36M
⇒ K12=K2+36
⇒ 102=K2+36
⇒ K2=100-36=64
⇒ K=8 cm

Topic: Physics, Rotational Motion, JEE Main
The radius of gyration of a cylindrical rod about an axis of rotation perpendicular to its length and passing through the center will be _____ m. Given, the length of the rod is 103 m.

Options:

The correct answer is:
5

Solution:
Moment of inertia of the rod about an axis perpendicular to its length and passing through the centre is I=ml212.

Using the relation, I=mk2, we have

⇒k2=l212⇒k=l12=l23=10323=5 m

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2022 (26 Jul Shift 2)
The radius of $\mathrm{La}^{3+}$ (Atomic number of $\mathrm{La}=57$ ) is $1.06 Å$. Which one of the following given values will be closest to the radius of $\mathrm{Lu}^{3+}$ (Atomic number of $\mathrm{Lu}=71$ ) ?

Options:
A $1.40 Å$
B $1.06 Å$
C $0.85 Å$
D $1.60 Å$

The correct answer is:
$0.85 Å$

Solution:
Ionic radii $\propto \frac{1}{z}$
Thus, $\frac{z_{2}}{z_{1}} \Rightarrow \frac{1.06}{\left(\text { Ionic radii of } \mathrm{Lu}^{3+}\right)}=\frac{71}{57}$
$\Rightarrow$ Ionic radii of $\mathrm{Lu}^{3+}=0.85 Å$

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
The radius of $\mathrm{La}^{3+}$ (Atomic number of $\mathrm{La}=57$ ) is $1.06 Å$. Which one of the following given values will be closest to the radius of $\mathrm{Lu}^{3+}$ (Atomic number of $\mathrm{Lu}=71$ )?

Options:
A
$1.40 Å$

B
$1.06 Å$

C
$0.85 Å$

D
$1.60 Å$

The correct answer is:
$0.85 Å$

Solution:
$\Omega \propto \frac{1}{\mathrm{Z}}$
$\frac{\Omega_1}{\Omega_2}=\frac{z_2}{z_1} \Rightarrow \frac{1.06}{\Omega_2}=\frac{71}{57} \Rightarrow \Omega_2=0.85 Å$

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2003
The radius of the 2nd orbit of Li2+ is x. The expected radius of the 3rd orbit of Be3+ is

Options:
A 94x
B 49x
C 2716x
D 1627x

The correct answer is:
2716x

Solution:
Radius of orbit (r) = 0.529 × n2Z

Where, n = principal quantum number of orbit

Z = Atomic number

(rLi2+)n=2 = 0.529 × (2)23 = x -----(1)

rBe3+n=3 = 0.529 ×(3)24 -----(2)

Divide equation (1) by equation (2)

rLi2+n=2rBe3+n=3 = 0.529 ×(2)230.529 ×(3)24

⇒x(rBe3+)n=3 = 1627

∴rBe3+n=3 = 27x16

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2023 (25 Jan Shift 1)
The radius of the \( 4^{\text {th }} \) Bohr orbit is \( 0.864 \mathrm{~nm} \). The de Broglie wavelength of the electron in that orbit is:

Options:
A \( 13.297 Å \)
B \( 1.3565 \mathrm{~nm} \)
C \( 1.3291 \mathrm{pm} \)
D \( 0.1329 \mathrm{~nm} \)

The correct answer is:
\( 1.3565 \mathrm{~nm} \)

Solution:
Circumference of Bhor's atomic orbit is related to De-Broglie wave length (λ) is as given below2π rn=nλ

n= Orbit number

n=4

rn = radius of the nth orbit

rn = 0.864 nm⇒ λ=2π rnn=2×3.14×0.8644 nm
= 1.3565 nm

Topic: Physics, Dual Nature of Matter, JEE Main
The radius of the circle in which the sphere
$x^2+y^2+z^2+2 x-2 y-4 z-19=0$ is cut by the plane $x+2 y+2 z+7=0$ is

Options:
A
4

B
1

C
2

D
3

The correct answer is:
3

Solution:
centre of sphere $=(-1,1,2)$
Radius of sphere $\sqrt{1+1+4+19}=5$
Perpendicular distance from centre to the plane
$\mathrm{OC}=\mathrm{d}=\left|\frac{-1+2+4+7}{\sqrt{1+4+4}}\right|=\frac{12}{3}=4$
$\mathrm{AC}^2=\mathrm{AO}^2-\mathrm{OC}^2=5^2-4^2=9 \Rightarrow \mathrm{AC}=3$

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2003
The radius of the circle touching the line x+y=4 at 1,3 and intersecting x2+y2=4 orthogonally is

Options:
A 324 units
B 34 units
C 32 units
D 423 units

The correct answer is:
324 units

Solution:
Let the equation of the circle is x-12+y-32+λx+y-4=0
⇒x2+y2+λ-2x+λ-6y+10-4λ=0
which is orthogonal to x2+y2=4
Now, 2g1g2+2f1f2=c1+c2 implies 0+0=10-4λ-4⇒λ=32
Hence, the equation of the circle is
x2+y2-x2-9y2+4=0
Radius=116+8116-4=1816
=324

Topic: Mathematics, Circle, JEE Main
The radius of the circle with centre at 3,2 and whose common chord with the circle C: x2+y2-4x-8y+16=0 is also a diameter of the circle C, is

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

The correct answer is:
3 units

Solution:
Let the equation of the circle be x2+y2-6x-4y+c=0
So, the common chord of these 2 circles is
2x-4y+16-c=0
It passes through the centre 2,4 of the given circle, so
4-16+16-c=0⇒c=4
Hence, the radius of the circle =9+4-4=3 units

Topic: Mathematics, Circle, JEE Main
The radius of the director circle of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, is

Options:
A $a-b$
B $\sqrt{a-b}$
C $\sqrt{a^2-b^2}$
D $\sqrt{a^2+b^2}$

The correct answer is:
$\sqrt{a^2-b^2}$

Solution:
Equation of director-circle of the hyperbola
$\begin{aligned}
& \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \text { is } x^2+y^2=a^2-b^2 \\
& \text { So, radius }=\sqrt{a^2-b^2}
\end{aligned}$

Topic: Mathematics, Hyperbola, JEE Main
The radius of the earth is $6400 \mathrm{~km}$ and $g=10 \mathrm{~m} / \mathrm{sec}^2$. In order that a body of $5 \mathrm{~kg}$ weighs zero at the equator, the angular speed of the earth is

Options:
A $1 / 80$ radian/sec
B $1 / 400$ radian/sec
C $1 / 800$ radian/sec
D $1 / 1600$ radian/sec

The correct answer is:
$1 / 800$ radian/sec

Solution:
For the condition of weightlessness at equator
$\omega=\sqrt{\frac{g}{R}} \quad \therefore \quad \omega=\sqrt{\frac{1}{640 \times 10^3}}=\frac{1}{800} \mathrm{rad} / \mathrm{s}$

Topic: Physics, Gravitation, JEE Main
The radius of the Earth is 6400 km. If the height of an antenna is 500 m, then its range is

Options:
A 800 km

B 100 km
C 80 km
D 10 km

The correct answer is:
80 km

Solution:
Given, R=6400 km=6.4Ã106 m,h=500 m
â´ Range, d=2hR
=2Ã500Ã6.4Ã106
=64Ã108=8Ã104=80 km

Topic: Physics, Communication System, JEE Main
The radius of the first orbit of hydrogen atom is 0.52×10-8 cm. The radius of the first orbit of helium atom is

Options:
A 0.26×10-8 cm
B 0.52×10-8 cm
C 1.04×10-8 cm
D 2.08×10-8 cm

The correct answer is:
0.26×10-8 cm

Solution:
rn=n2Z0.52×10-8cm
For H-atom Z=1 , n = 1, r1=0.52×10-8cm
For He atom, n = 1, Z = 2,
r1=12×0.5×10-8 cm
=0.26×10-8 cm

Topic: Chemistry, Structure of Atom, JEE Main
The radius of the first orbit of hydrogen is rH, and the energy in the ground state is -13.6 eV. Considering a Î¼-- particle with a mass 207me revolving around a proton as in hydrogen atom, the energy and radius of proton and Î¼--combination respectively in the first orbit are (assume nucleus to be stationary)

Options:
A -13.6Ã207 eV, rH207
B -207×13.6 eV, 207rH
C -13.6207 eV, rH207
D -13.6207 eV, 207rH

The correct answer is:
-13.6Ã207 eV, rH207

Solution:
The total energy of nth orbit,
En= -me48Îµ02h2Â·1n2
Obviously Enâm
â´ EÎ¼Ee=mÎ¼me
â EÎ¼=mÎ¼meÃEe
Ground state energy of a proton in a hydrogen atom,
EÎ¼=-13.6Ã207meme eV
=-13.6Ã207 eV
( âµ mÎ¼=207me, where me is the mass of an electron)
We know that,
r=Îµ0h2n2207Ïmee2
For ground state n=1 for proton, we have
rÎ¼=Îµ0h2207ÏmeÂ·e2
But Îµ0h2ÏmeÂ·e2 = ground state radius of a hydrogen atom
Îµ0h2ÏmeÂ·e2=rH
â´ rÎ¼=rH207

Topic: Physics, Atomic Physics, JEE Main
The radius of the first orbit of hydrogen is $\mathrm{r}_{\mathrm{H}}$, and the energy in the ground state is - $13.6 \mathrm{eV}$. Considering a $\mu^{-}$-particle with a mass $207 m_e$ revolving round a proton as in hydrogen atom, the energy and radius of proton and $\mu^{-}$- combination respectively in the first orbit are (assume nucleus to be stationary)

Options:
A $-13.6 \times 207 \mathrm{eV}, \frac{\mathrm{C}^{\mathrm{H}}}{207}$
B $-207 \times 13.6 \mathrm{eV}, 207 r_{\mathrm{H}}$
C $-\frac{13.6}{207} \mathrm{eV}, \frac{\mathrm{F}}{207}$
D $-\frac{13.6}{207} \mathrm{eV}, 207 \mathrm{r}_{\mathrm{H}}$

The correct answer is:
$-13.6 \times 207 \mathrm{eV}, \frac{\mathrm{C}^{\mathrm{H}}}{207}$

Solution:
The total energy of $n$th orbit
$E_n=-\frac{m e^4}{8 \varepsilon_0^2 h^2} \cdot \frac{1}{n^2}$
Obviously $E_n \propto m$
$\begin{array}{llll}
& \therefore & \frac{E_\mu}{E_e} & =\frac{m_\mu}{m_e} \\
\Rightarrow & E_\mu & =\frac{m_\mu}{m_e} \times E_e
\end{array}$
Ground state energy of a proton in hydrogen atom,
$\begin{aligned}
E_\mu & =-13.6 \times \frac{207 m_e}{m_e} \mathrm{eV} \\
& =-13.6 \times 207 \mathrm{eV}
\end{aligned}$
$\left(\because m_\mu=207 m_e\right.$, where $m_e$ is the mass of electron $)$
We know that
$r=\frac{\varepsilon_0 h^2 n^2}{207 \pi m_{\mathrm{e}} \mathrm{e}^2}$
For ground state $(n-1)$ for proton, we have
$r_\mu=\frac{\varepsilon_0 h^2}{207 \pi m_e \cdot e^2}$
But $\frac{\varepsilon_0 h^2}{\pi m_{\mathrm{e}} \cdot e^2}=$ ground state radius of hydrogen atom
$\begin{aligned}
\frac{\varepsilon_0 h^2}{\pi m_e \cdot e^2} & =r_{\mathrm{H}} \\
\therefore \quad r_\mu & =\frac{r_{\mathrm{H}}}{207}
\end{aligned}$

Topic: Physics, Atomic Physics, JEE Main
The radius of the largest sphere which fits properly at the centre of the edge of a body centred cubic unit cell is :

(Edge length is represented by 'a')

Options:
A 0.027 a
B 0.047 a
C $0.134 \mathrm{a}$
D 0.067 a

The correct answer is:
0.067 a

Solution:
For bcc

$$

\Rightarrow \mathrm{R}=\frac{\sqrt{3} \mathrm{a}}{4}

$$

$$

\therefore \text { Empty space at edge }=a-2 R=a-\frac{\sqrt{3} a}{2}

$$

$=$ diameter of sphere.

$$

\therefore \quad \mathrm{r}_{\text {sphere }}=\frac{a-\frac{\sqrt{3}}{2} \mathrm{a}}{2}=\left(\frac{2-\sqrt{3}}{4}\right) \mathrm{a}=0.067 \mathrm{a}

$$

Topic: Chemistry, Solid State, JEE Main, JEE Main 2019 (11 Jan Shift 2)
The radius of the orbit of a geostationary satellite is (mean radius of earth is ' $R$ ', angular velocity about own axis is ' $\omega$ ' and acceleration due to gravity on earth's surface is ' $\mathrm{g}$ ')

Options:
A $\left(\frac{\mathrm{gR}^2}{\omega^2}\right)^{\frac{1}{3}}$
B $\left(\frac{\mathrm{gR}^2}{\omega^2}\right)^{\frac{2}{3}}$
C $\left(\frac{\mathrm{gR}^2}{\omega^2}\right)^{\frac{1}{2}}$
D $\frac{\mathrm{gR}^2}{\omega^2}$

The correct answer is:
$\left(\frac{\mathrm{gR}^2}{\omega^2}\right)^{\frac{1}{3}}$

Solution:
$\mathrm{mr} \omega^2=\frac{\mathrm{GMm}}{\mathrm{r}^2}$
$\omega^2=\frac{G M}{r^3}=\frac{g R^2}{r^3} \quad \ldots\left(\because g=\frac{G M}{R^2}\right)$
$\therefore \quad$ Radius of the orbit of the satellite is:
$\mathrm{r}=\left(\frac{\mathrm{gR}^2}{\omega^2}\right)^{\frac{1}{3}}$

Topic: Physics, Gravitation, JEE Main
The radius of the second Bohr orbit for hydrogen atom is
(Planck's constant, (h)=6.6262×10−34Js; mass of electron =9.1091×10-31kg; charge of electron =1.60210×10-19C; permittivity of vacuum, (∈0)=8.854185×10-12kg-1m-3A2)

Options:
A 4.76 Å
B 0.529 Å
C 2.12 Å
D 1.65 Å

The correct answer is:
2.12 Å

Solution:
Radius of the Bohr's orbit is calculated by the following formula rn=0.529×n2zÅ

Radius is directly proportional to square of nth orbit and inversely proportional to atomic number. This formula works for single electron species

here rn=radius of the nth orbit

Z=1 and n=orbit or shell number
r2=0.529×221 Å
=2.12 Å

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2017 (02 Apr)
The radius of the second Bohr orbit, in terms of the Bohr radius (a 0 ) , of Li2+ is given as xa0y . Find the sum of x+y here?

Options:

The correct answer is:
7

Solution:
We know
r=0.529Ãn2Zâ«
r=a0Ãn2Z
n=2
Z=3
r= a 0 ×4 3 = 4 a 0 3
Hence x=4,y=3
So, sum of x+y=4+3=7

Topic: Chemistry, Structure of Atom, JEE Main
The radius of third stationary orbit of electron for Bohr's atom is R. The radius of fourth stationary orbit will be:

Options:
A 43R
B 169R
C 34R
D 916R

The correct answer is:
169R

Solution:
From Bohr's atomic model, it follows that

rn=n2Zr1 ...1

Following equation (1), it can be written that

r4r3=4232⇒r4=169R

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2024 (27 Jan Shift 1)
The radius of two metallic spheres$A$ and $B$ are $r_1$ and $r_2$ respectively $\left(r_1 \gt r_2\right)$. They are connected by a thin wire and the system is given a certain charge. The charge will be greater

Options:
A On the surface of the sphere $B$
B On the surface of the sphere $A$
C Equal on both
D Zero on both

The correct answer is:
On the surface of the sphere $A$

Solution:
After connection of wire, potential becomes equal
$\therefore \frac{Q_1}{r_1}=\frac{Q_2}{r_2} \Rightarrow \frac{Q_1}{Q_2}=\frac{r_1}{r_2}$ when $r_1 \gt r_2$, then $Q_1 \gt Q_2$

Topic: Physics, Capacitance, JEE Main
The radius of which of the following orbit is same as that of the first Bohr's orbit of hydrogen atom

Options:
A $\mathrm{He}^{+}(n=2)$
B $\mathrm{Li}^{2+}(n=2)$
C $\mathrm{Li}^{2+}(\mathrm{n}=3)$
D $B e^{3+}(n=2)$

The correct answer is:
$B e^{3+}(n=2)$

Solution:
$r_H=0.529 \frac{n^2}{z} Å$
For hydrogen; $n=1$ and $z=1$ therefore
$r_H=0.529 Å$
For $B e^{3+}: Z=4$ and $n=2$ Therefore
$r_{B e^{3+}}=\frac{0.529 \times 2^2}{4}=0.529 Å$

Topic: Chemistry, Structure of Atom, JEE Main