datavorous/entrance-exam-dataset huggingface dataset:
A solid sphere is in rolling motion. In rolling motion a body possesses translational kinetic energy \( \left(K_{t}\right) \) as well as rotational kinetic energy \( \left(K_{r}\right) \) simultaneously. The ratio \( K_{t}:\left(K_{t}+K_{r}\right) \) for the sphere is

Options:
A \( 10: 7 \)
B \( 5: 7 \)
C \( 7: 10 \)
D \( 2: 5 \)

The correct answer is:
\( 5: 7 \)

Solution:
Transnational KE = 1 2 m v 2
Translational KE+
Rotational KE
= 1 2 m v 2 + 1 2 I ω 2
= 1 2 m v 2 + 1 2 ( 2 5 m r 2 ) ω 2
∵ for rolling motion v=rω
= 1 2 m v 2 + 1 2 × 2 5 m r 2 ( v r ) 2 = 7 10 m v 2
So, K t K t + K r = 1 2 m v 2 7 10 m v 2 = 5 7

Topic: Physics, Rotational Motion, JEE Main
A solid sphere is rolling on a horizontal plane without slipping. If the ratio of angular momentum about axis of rotation of the sphere to the total energy of moving sphere is π:22 then, the value of its angular speed will be_________rad s-1.

Options:

The correct answer is:
4

Solution:
The angular momentum is given by

L=ICMω=2MR2ω5

The total kinetic energy is

E=12(Mv2+2MR2ω25)

The ratio of angular momentum and kinetic energy is

LE=ICMω12Mv2+12·25MR2ω2=25MR2ω710MR2ω2

⇒π22=47·1ω⇒ω=4 rad s-1

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2023 (13 Apr Shift 1)
A solid sphere is rolling on a surface as shown in figure, with a translational velocity $v \mathrm{~m} \mathrm{~s}^{-1}$. If it is to climb the inclined surface continuing to roll without slipping, then minimum velocity for this to happen is

Options:
A
$\sqrt{2 g h}$

B
$\sqrt{\frac{7}{5} g h}$

C
$\sqrt{\frac{7}{2} g h}$

D
$\sqrt{\frac{10}{7} g h}$

The correct answer is:
$\sqrt{\frac{10}{7} g h}$

Solution:
Minimum velocity for a body rolling without slipping
$$
v=\sqrt{\frac{2 g h}{1+\frac{K^2}{R^2}}}
$$
For solid sphere, $\frac{K^2}{R^2}=\frac{2}{5}$
$$
\therefore \quad v=\sqrt{\frac{2 g h}{1+\frac{K^2}{R^2}}}=\sqrt{\frac{10}{7} g h}
$$

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2012 (12 May Online)
A solid sphere is rotating freely about its symmetric axis in free space. The radius of the sphere is increased keeping its mass same. Which of the following physical quantities would remain constant for the sphere?

Options:
A Rotational kinetic energy
B Moment of inertia
C Angular velocity
D Angular momentum

The correct answer is:
Angular momentum

Solution:
∵τext =dL→dt
So, when τ ext  =0⇒  d L → dt =0
⇒  L → = constant
So angular momentum remains constant.

Topic: Physics, Rotational Motion, NEET, NEET 2018
A solid sphere is rotating in free space. If the radius of the sphere is increased keeping mass same which one of the following will not be affected?

Options:
A
moment of inertia

B
angular momentum

C
angular velocity

D
rotational kinetic energy.

The correct answer is:
angular momentum

Solution:
Let it be assume that in "free space" not only the acceleration due to gravity it acting but al there are no external torque acting but also there are no external torque acting on the sphere. If due to internal changes in the system, the radius has increased, then the law of the conservation of angular momentum holds good.

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2004
A solid sphere made out of metal having a bulk modulus of 2Ã1011 Nm-2 is charged uniformly to a surface charge density Ï=Î±Îµ0 Cm-2, where Îµ0 is the permittivity of free space and Î± (magnitude = 1) is a positive constant of appropriate dimensions. If the volume strain produced in the sphere after charging is xÃ10-11, then the value of x is

Options:

The correct answer is:
0.25

Solution:
Bulk modulus is given by

B=-ΔPΔVV

The outward electrostatic pressure is

Pelectrostatic=σ22ε0=-ΔP

⇒ΔVV=σ22ε0B

⇒ΔVV=αε022ε02×1011=0.25×10-11

Topic: Physics, Mechanical Properties of Solids, JEE Main
A solid sphere of density $\eta(\gt1)$ times lighter than water is suspended in a water tank by a string tied to its base as shown in fig. If the mass of the sphere is $m$ then the tension in the string is given by

Options:
A $\left(\frac{\eta-1}{\eta}\right) m g$
B $\eta m g$
C $\frac{m g}{\eta-1}$
D $(\eta-1) m g$

The correct answer is:
$(\eta-1) m g$

Solution:
Tension in spring $T=$ upthrust - weight of sphere
$=V \sigma g-V \rho g=V \eta \rho g-V \rho g \quad(\mathrm{~A} s \sigma=\eta \rho)$
$=(\eta-1) V \rho g=(\eta-1) m g$.

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A solid sphere of mass 1 kg rolls without slipping on a plane surface. Its kinetic energy is 7×10-3 J. The speed of the centre of mass of the sphere is ______ cm s-1.

Options:

The correct answer is:
10

Solution:
Let velocity of centre of mass of sphere be v.

Total kinetic energy of solid sphere is KE=12mv2+12Iω2, where, I is moment of inertia of the solid sphere about the axis passing through it's centre of mass and ω=vR pure rolling condition.

Therefore,

7×10-3=12×1×(v)2+1225(1)R2vR2

⇒12v275=7×10-3

⇒v2=10-2

⇒v=10-1=0.1 m s-1=10 cm s-1

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2023 (31 Jan Shift 1)
A solid sphere of mass 2 kg is making pure rolling on a horizontal surface with kinetic energy 2240 J. The velocity of centre of mass of the sphere will be ______ m s-1.

Options:

The correct answer is:
40

Solution:
Let velocity of centre of mass of sphere be v.

Total kinetic energy of solid sphere is KE=12mv2+12Iω2, where, I is moment of inertia of the solid sphere about the axis passing through it's centre of mass and ω=vR pure rolling condition.

Therefore,

2240=122(v)2+1225(2)R2vR2

⇒2240=v2+25v2

⇒v=40 m s-1.

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2023 (29 Jan Shift 1)
A solid sphere of mass 500 g radius 5 cm is rotated about one of its diameter with angular speed of 10 rad s-1. If the moment of inertia of the sphere about its tangent is x×10-2 times its angular momentum about the diameter. Then the value of x will be

Options:

The correct answer is:
35

Solution:
The angular momentum about the diameter is

Ldiameter =25MR2ω;

The moment of inertia of a sphere is ICM=25MR2. The moment of inertia about the tangent is

Itangent =ICM+MR2=25MR2+MR2=75MR2.

It is given that ω=10 rad s-1

Using the given relation between the moment of inertia and the angular momentum,

75MR2=x×10-2×25MR2ω⇒x=3.5×102ω=3.5×10=35

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2023 (11 Apr Shift 1)
A solid sphere of mass M and radius a is surrounded by a uniform concentric spherical shell of thickness 2a and mass 2M. The gravitational field at distance 3a from the centre will be:

Options:
A GM9a2
B GM3a2
C 2GM3a2
D 2GM9a2

The correct answer is:
GM3a2

Solution:
Gravitational field at point P,
=GM3a2+G2M3a2
=GM9a2+2GM9a2
=GM3a2

Topic: Physics, Gravitation, JEE Main, JEE Main 2019 (09 Apr Shift 1)
A solid sphere of mass 'M' and radius 'R' has moment of inertia 'l' about its diameter.
It is recast into a disc of thickness 't' whose moment of inertia about an axis passing
through its edge and perpendicular to its plane, remains 'I'. Radius of the disc will
be

Options:
A $\mathrm{R}/\sqrt{19}$
B $R / \sqrt{15}$
C $2 \mathrm{R} / \sqrt{15}$
D $2 R / \sqrt{19}$

The correct answer is:
$2 \mathrm{R} / \sqrt{15}$

Solution:
Moment of inertia of solid sphere is
$\mathrm{I}_{\mathrm{s}}=\frac{2}{5} \mathrm{MR}^{2}$
Moment of inertia of the disc is given by about the given axis is
$\begin{aligned}
\mathrm{I}_{d} &=\frac{3}{2} \mathrm{Mr}^{2} \\
\therefore \mathrm{r}^{2} &=\frac{4}{15} \mathrm{R}^{2} \quad \therefore \frac{3}{2} \mathrm{Mr}^{2}=\frac{2}{5} \mathrm{MR}^{2} \\
\therefore \mathrm{r} &=\frac{2 \mathrm{R}}{\sqrt{15}}
\end{aligned}$

Topic: Physics, Rotational Motion, MHT CET, MHT CET 2020 (14 Oct Shift 2)
A solid sphere of mass $M$, radius $R$ has moment of inertia ' $I$ ' about its diameter. It is recast into a disc of thickness ' $t$ ' whose moment of inertia about an axis passing through its edge and perpendicular to its plane remains ' $I$ '. Radius of the disc will be

Options:
A $\frac{4 \mathrm{R}}{\sqrt{11}}$
B $\frac{3 R}{4}$
C $\frac{2 \mathrm{R}}{\sqrt{15}}$
D $\frac{2 R}{3}$

The correct answer is:
$\frac{2 \mathrm{R}}{\sqrt{15}}$

Solution:
For solid sphere, $\mathrm{I}=\frac{2}{5} \mathrm{MR}^2$
For disc, $\mathrm{I}^{\prime}=\frac{\mathrm{MR}^{\prime 2}}{2}+\mathrm{MR}^{\prime 2}=\frac{3}{2} \mathrm{MR}^{\prime 2}$
$$
\begin{aligned}
& \because I^{\prime}=I \\
& \therefore \frac{3}{2} M R^{\prime 2}=\frac{2}{5} M^2 \\
& R^{\prime}=\frac{2}{\sqrt{15}} R
\end{aligned}
$$

Topic: Physics, Rotational Motion, MHT CET, MHT CET 2021 (20 Sep Shift 2)
A solid sphere of radius $r_1=1 \mathrm{~cm}$ carries charge distributed uniformly over it with density $\rho_1=-3 \mathrm{C} / \mathrm{cm}^3$. It is surrounded by a concentric spherical shell of radius $r_2=2 \mathrm{~cm}$ carrying uniform charge density $\rho_2=\frac{1}{2} \mathrm{C} / \mathrm{cm}^2$. If $E_d$ denotes the magnitude of the electric field at distance $d$ from the common centre of the spheres, then

Options:
A $E_d=\frac{1}{3 \varepsilon_0 d^2}, d \leq 1 \mathrm{~cm}$
B $E_d=\frac{1}{\varepsilon_0 d^2}, d \leq 1 \mathrm{~cm}$
C $E_d=\frac{d}{3 \varepsilon_0}, d \leq 1 \mathrm{~cm}$
D $E_d=\frac{d}{\varepsilon_0}, d \leq 1 \mathrm{~cm}$

The correct answer is:
$E_d=\frac{d}{\varepsilon_0}, d \leq 1 \mathrm{~cm}$

Solution:
Electric field at a point inside inner sphere is only due to charge enclosed.
We use Gauss' law to get,
$$
E=\left(\frac{q}{4 \pi \varepsilon_0 R^3}\right) r=\frac{\rho_R}{3 \varepsilon_0}
$$
Here,
$$
\rho=-3\left(\frac{\mathrm{C}}{\mathrm{cm}^3}\right)
$$
So,
$$
E=\frac{-d}{\varepsilon_0} \text { or }|\mathbf{E}|=\frac{d}{\varepsilon_0}
$$

Topic: Physics, Electrostatics, JEE Main
A solid sphere, of radius R acquires a terminal velocity v1 when falling (due to gravity) through a viscous fluid having a coefficient of viscosity η. The sphere is broken into 27 identical solid spheres. If each of these spheres acquires a terminal velocity, v2, when falling through the same fluid, the ratio v1v2 equals:

Options:
A 19
B 27
C 127
D 9

The correct answer is:
9

Solution:
If R is the radius of initial drop and r is the radius of new drop, then
43πR3=27×43πr3⇒r=R3
Terminal velocity,

vT∝r2
⇒ v1v2=R2R32=9

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2019 (12 Apr Shift 2)
A solid sphere of radius R and mass M is rolled by a force F acting at the top of the sphere as shown in the figure. The sphere starts from rest and rolls without slipping, then

Options:
A work done by force F when the center moves a distance S is 2FS
B work done by force F when the center moves a distance S is 3FS
C work done by force F when the center moves a distance S is FS
D work done by force F when the center moves a distance S is 4FS

The correct answer is:
work done by force F when the center moves a distance S is 2FS

Solution:
When the centre moves a distance S, the topmost point displaces by 2S. The work done is by the force is

W=2FS

Topic: Physics, Rotational Motion, JEE Main
A solid sphere of radius R carries a charge Q+q distributed uniformly over its volume. A very small point like piece of it of mass m gets detached from the bottom of the sphere and falls down vertically under gravity. This piece carries charge q. If it acquires a speed ν when it has fallen through a vertical height y (see figure), then (assume the remaining portion to be spherical)

Options:
A v2=yqQ4πϵ0R2ym+g
B v2=yqQ4πϵ0R(R+y)m+g
C v2=2yQqR4πϵ0(R+y)3m+g
D v2=2yqQ4πε0R(R+y)m+g

The correct answer is:
v2=2yqQ4πε0R(R+y)m+g

Solution:
By using total energy conservation
ΔKE+(ΔPE)electo +(ΔPE)gravitation =0

12mV2+kQqR+y-kQqR+(-mgy)=0

12mV2=mgy+kQq1R-1R+y; V2=2gy+2kQqmyR(R+y)

V2=2yqQ4πε0R(R+y)m+g

Topic: Physics, Electrostatics, JEE Main, JEE Main 2020 (05 Sep Shift 1)
A solid sphere of radius \( R \) gravitationally attracts a particle placed at \( 3 \mathrm{R} \) from its centre with a force \( F_{1} \). Now a spherical cavity of radius \( \left(\frac{R}{2}\right) \) is made in the sphere (as shown in figure) and the force becomes \( F_{2} \). The value of \( F_{1}: F_{2} \) is:

Options:
A \( 41: 50 \)
B \( 50: 41 \)
C \( 25: 36 \)
D \( 36: 25 \)

The correct answer is:
\( 50: 41 \)

Solution:
Let the initial mass of the sphere is m'. Hence, mass of
a removed portion will be m'/8,F1=m.E.=m.Gm'9R2

F2=mG·m'3R2-G·m'/85R/22=Gm'9R2-Gm'×48×25=19-150Gm'R2

F2=4150×9·Gm'R2⇒F1F2=19×50×941=5041

Topic: Physics, Gravitation, JEE Main
A solid sphere of radius R has a charge Q distributed in its volume with a charge density ρ = kra where k and a are constants and r is the distance from its centre. If the electric field at r = R 2 is 18 times that at r = R, find the value of a.

Options:

The correct answer is:
2

Solution:
From Gauss theorem,

E∝qr2
(q = charge enclosed)
Therefore
E 2 E 1 = q 2 q 1 = r 1 2 r 2 2
or
8 = ∫ 0 R 4 π r 2 kr deg dr ∫ 0 R / 2 4 π r 2 kr deg dr × R 2 R 2

Solving this equation we get, a = 2.

Topic: Physics, Electrostatics, JEE Main
A solid sphere of radius R has a charge Q distributed in its volume with a charge density Ï=kra, where k and a are constants and r is the distance from its centre. If the electric field at r=R2 is 18 times that at r=R, then find the value of a

Options:
A 2
B 3
C 2.5
D 0.2

The correct answer is:
2

Solution:
From Gauss' theorem,
E∝qr2 q=charge enclosed
E2E1=q2q1=r12r22
8=∫0R4πr2kradr∫0R/24πr2kradr×R22R2
Solving this equation we get, a=2.

Topic: Physics, Electrostatics, JEE Main
A solid sphere of radius r is gently placed on a rough horizontal surface with an initial angular speed Ï0 but no linear velocity. If the coefficient of friction is Î¼, then the time t when the slipping will stop is

Options:
A 2 7 r Ï 0 Î¼ g
B 3 7 r ω 0 μ g
C 4 7 r ω 0 μ g
D r ω 0 μ g

The correct answer is:
2 7 r ω 0 μ g

Solution:
Let m be the mass of the sphere.
Since, it is a case of backward slipping, force of friction is in forward direction. Limiting friction will act in this case.
Linear acceleration a = f m = Î¼ mg m = Î¼ g
Angular retardation Î± = Ï I = f Â· r 2 5 mr 2 = 5 2 Î¼ g r
Slipping will be stopped when v=rÏ
at = r(Ï0 - Î±t)
Î¼ gt = r ( Ï 0 - 5 2 Î¼ g t r )
7 2 Î¼ gt = r Ï 0
â´ t = 2 7 r Ï 0 Î¼ g

Topic: Physics, Rotational Motion, JEE Main
A solid sphere of radius $\mathrm{R}$ is set into motion on a rough horizontal surface with a linear speed $v_0$ in forward direction and an angular velocity $\omega_0=v_0 / 2 R$ in counter coefficient of friction is $\mu$, then find the time after which sphere starts pure rolling.

Options:
A $t_o=\frac{3 v_o}{7 \mu g}$
B $t_o=\frac{2 v_0}{7 \mu g}$
C $t_o=\frac{4 v_0}{7 \mu g}$
D $t_0=\frac{v_0}{7 \mu g}$

The correct answer is:
$t_o=\frac{3 v_o}{7 \mu g}$

Solution:
As the sphere is slipping, kinetic friction will act,
$\begin{aligned}& f_k=\mu m g, a_{c m}=\frac{f_k}{m}=\mu g, \\& a=\frac{f_k R}{I_{c m}}=\frac{\mu m g R}{\frac{2}{5} m R^2}=\frac{5 \mu g}{2 R}\end{aligned}$
when the sphere starts pure rolling, velocity $v$ of com will be $\omega R$.
$\begin{aligned}& v=v_0-\mu g \mathrm{t} \\& \omega=-\omega_0+a t\end{aligned}$
for pure rolling,
$\begin{aligned}& v=\omega R \\& \Rightarrow v_0-\mu g t=-\left(-\omega_0+a t\right) R \\& \Rightarrow v_0-\mu g t=\left(\frac{-v_0}{2 R}+\frac{5 \mu g}{2 R}\right) R \\& \Rightarrow \frac{3 v_0}{2}=\frac{7}{2} \mu g t \Rightarrow t_0=\frac{3 v_0}{7 \mu g}\end{aligned}$

Topic: Physics, Rotational Motion, JEE Main
A solid sphere of radius r made of a soft material of bulk modulus K is surrounded by a liquid in a cylindrical container. A massless piston of area a floats on the surface of the liquid, covering entire cross-section of cylindrical container. When a mass m is placed on the surface of the piston to compress the liquid, the fractional decrement in the radius of the sphere drr , is:

Options:
A mgKa
B Kamg
C Ka3mg
D mg3Ka

The correct answer is:
mg3Ka

Solution:
For sphere,∨=43πr3⇒∆∨∨=3∆rr⇒P=KΔVV
⇒P=3KΔRR⇒mg3Ka=ΔRR.

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2018 (08 Apr)
A solid sphere of radius R made up of a material of bulk modulus K is surrounded by a liquid in a cylindrical container. A massless piston of area A floats on the surface of the liquid. When a mass M is placed on the piston to compress the liquid, the fractional change in the radius of the sphere is

Options:
A MgAK
B Mg3AK
C 3MgAK
D Mg2AK

The correct answer is:
Mg3AK

Solution:
Change in pressure due to placing of mass on piston is,
âp=MgA
From Bulk modulus definition
K= âÎp ÎV
ââVV=âpK=MgAK
From V=43Ïr3
âVV=3âRR
â âRR=13âVV
=Mg3AK

Topic: Physics, Mechanical Properties of Solids, JEE Main
A solid sphere of radius R makes a perfect rolling down on a plane which is inclined to the horizontal axis at an angle θ. If the radius of gyration is k, then its acceleration is

Options:
A gsinθ1+k2R2
B gsinθR2+k2
C gsinθ2R2+k2
D gsinθ21+k2R2

The correct answer is:
gsinθ1+k2R2

Solution:
Referring to above diagram,

balancing forces along the inclined plane,

ma=mgsinθ-f ...1

torque is given by,

τ=Iα=fR but α=aR & l=mk2

∴ mk2×aR=fR⇒f=amk2R2

Putting in equation 1,

ma=mgsinθ-amk2R2

∴ a1+k2R2=gsinθ

∴ a=gsinθ1+k2R2

Topic: Physics, Rotational Motion, JEE Main
A solid sphere of uniform density and radius R applies a gravitational force of attraction equal to F1 on a particle placed at a distance 3R from the centre of the sphere. A spherical cavity of the radius R2 is now made in the sphere as shown in the figure. The sphere with the cavity now applies a gravitational force F2 on the same particle. The ratio F2F1 is

Options:
A 9 5 0
B 4 1 5 0
C 3 2 5
D 2 2 2 5

The correct answer is:
4 1 5 0

Solution:
From the superposition principle,
F1 = Fr + Fc
Here, Fr = force due to remaining part = F2
and Fc = force due to cavity
Now, F 1 = GMm 3 R 2 = GMm 9 R 2
F c = G M 8 m 5 2 R 2 = GMm 5 0 R 2
∴ F 2 = F 1 - F c = 4 1 GMm 4 5 0 R 2
∴ F 2 F 1 = 4 1 5 0

Topic: Physics, Gravitation, JEE Main
A solid sphere of uniform density and radius R applies a gravitational force of attraction equal to F1 on a particle placed at P, distance 2 R from the centre O of the sphere. A spherical cavity of the radius R2 is now made in the sphere as shown in the figure.
The sphere with the cavity now applies a gravitational force F2 on the same particle placed at P.
The ratio F2F1 will be

Options:
A 12
B 79
C 3
D 7

The correct answer is:
79

Solution:
Gravitational force due to solid sphere, F1=GMm2R2 where M and m are mass of the solid sphere and particle respectively and R is the radius of the sphere.
The gravitational force on the particle due to sphere with cavity = force due to a solid sphere - force due to sphere creating a cavity, assumed to be present above at that position.
i.e., F2=GMm4R2-GM/8m3R/22=736GMmR2
So F2F1=7GMm36R2GMm4R2=79

Topic: Physics, Gravitation, JEE Main
A solid sphere of uniform density and radius r applies a gravitational force of attraction equal to F1on a particle placed at P, distance 2R from the centre O of the sphere. A spherical cavity of the radius R2 is now made in the sphere as shown in the figure. The sphere with cavity now applied a gravitational force F2 on the same particle placed at P. The ratio F2F1 will be

Options:
A 12
B 79
C 3
D 7

The correct answer is:
79

Solution:
Gravitational force due to solid sphere, F⁡ 1 = G⁡ M⁡ m⁡ 2 R⁡ 2 ,
where M and m are mass of the solid sphere and particle respectively and R is the radius of the sphere. The gravitational force on particle due to sphere with cavity = force due to solid sphere creating cavity, assumed to be present above at that position
i.e., F2=GMm4R2=GM/8m3R/22=736GMmR2
So, F2F1=7GMm36R2/GMm4R2=79

Topic: Physics, Gravitation, JEE Main
A solid sphere rolls without slipping along the track shown in the figure. The sphere starts from rest from a height h above the bottom of the loop of radius R which is much larger than the radius of the sphere r. The minimum value of h for the sphere to complete the loop is

Options:
A 2.1 R

B 2.3 R

C 2.7 R

D 2.5 R

The correct answer is:
2.7 R

Solution:
At the topmost point of the loop minimum value of
the linear speed of the centre of the sphere should be :
v=gR
or translational kinetic energy
KT=12mv2=12mgR.
In case of pure rolling of a solid sphere the ratio
of rotational to translational kinetic energy is :
KRKT=25
∴ Total kinetic energy at topmost point should be :
K=5+25.KT=7512mgR=710mgR
Now from conservation of mechanical energy :
710mgR=mgh - 2R
∴ h = 2.7 R

Topic: Physics, Rotational Motion, JEE Main
A solid sphere rolls without slipping, first horizontal and then up to a point $X$ at height $h$ on an inclined plane before rolling down, as shown.
The initial horizontal speed of the sphere is

Options:
A $\sqrt{10 \mathrm{gh} / 7}$
B $\sqrt{7 g h / 5}$
C $\sqrt{5 g h / 7}$
D $\sqrt{2 g h}$

The correct answer is:
$\sqrt{10 \mathrm{gh} / 7}$

Solution:
$m g h=\frac{1}{2} m v^{2}+\frac{1}{2} \frac{2}{5} m R^{2} \frac{v^{2}}{R^{2}}$
$m g h=\frac{7}{10} m v^{2}$
$v=\sqrt{\frac{10 g h}{7}}$

Topic: Physics, Rotational Motion, JEE Main
A solid spherical ball and a hollow spherical ball of two different materials of densities $\rho_{1}$ and $\rho_{2}$ respectively have same outer radii and same mass. What will be the ratio, the moment of inertia (about an axis passing through the centre) of the hollow sphere to that of the solid sphere?

Options:
A $\frac{\rho_{2}}{\rho_{1}}\left(1-\frac{\rho_{2}}{\rho_{1}}\right)^{\frac{5}{3}}$
B $\frac{\rho_{2}}{\rho_{1}}\left[1-\left(1-\frac{\rho_{2}}{\rho_{1}}\right)^{\frac{5}{3}}\right]$
C $\frac{\rho_{2}}{\rho_{1}}\left(1-\frac{\rho_{1}}{\rho_{2}}\right)^{\frac{5}{3}}$
D $\frac{\rho_{2}}{\rho_{1}}\left[1-\left(1-\frac{\rho_{1}}{\rho_{2}}\right)^{\frac{5}{3}}\right]$

The correct answer is:
$\frac{\rho_{2}}{\rho_{1}}\left[1-\left(1-\frac{\rho_{1}}{\rho_{2}}\right)^{\frac{5}{3}}\right]$

Solution:
$\therefore \quad M_{1}=\rho_{1} \frac{4}{3} \pi R^{3}$ and $M_{2}=\rho_{2} \frac{4}{3} \pi\left(R^{3}-r^{3}\right)$
$\therefore \quad \rho_{1} \frac{4}{3} \pi R^{3}=\rho_{2} \frac{4}{3} \pi\left(R^{3}-r^{3}\right)$
$\Rightarrow \quad \rho_{1} R^{3}=\rho_{2}\left(R^{3}-r^{3}\right)$
$\Rightarrow \quad \frac{\rho_{1}}{\rho_{2}}=1-\frac{r^{3}}{R^{3}}$
$\therefore \frac{r}{R}=\left(1-\frac{\rho_{1}}{\rho_{2}}\right)^{\frac{1}{3}}$
Moment of inertia, $\begin{aligned} \frac{I_{M}}{I_{S}} &=\frac{\frac{2}{5} M_{2}\left[\frac{R^{5}-r^{3}}{R^{3}-r^{3}}\right]}{\frac{2}{5} M_{1} R^{2}} \\ &=\frac{\rho_{2} \frac{4}{3} \pi\left(R^{3}-r^{3}\right)\left[\frac{R^{5}-r^{5}}{\left(R^{3}-r^{3}\right)}\right]}{\rho_{1} \frac{4}{3} \pi R^{3} R^{2}} \\ &=\frac{\rho_{2}}{\rho_{1}}\left[\frac{R^{5}-r^{5}}{R^{5}}\right]=\frac{\rho_{2}}{\rho_{1}}\left[1-\left(\frac{r}{R}\right)^{3}\right] \end{aligned}$
$\therefore \quad \frac{I_{H}}{I_{s}}=\frac{\rho_{2}}{\rho_{1}}\left[1-\left(1-\frac{\rho_{1}}{\rho_{2}}\right)^{\frac{5}{3}}\right]$

Topic: Physics, Rotational Motion, JEE Main
A solid spherical ball is rolling on a frictionless horizontal plane surface about its axis of symmetry. The ratio of rotational kinetic energy of the ball to its total kinetic energy is

Options:
A 15
B 25
C 27
D 710

The correct answer is:
27

Solution:
Under pure rolling condition: v=ωR

Translational K.E=12mv2

Rotational K.E=12×25mR2vR2=15mv2

Total K.E=12mv2+15mv2=710mv2

So, Rotational K·ETotal K·E=15mv2710mv2=27

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2022 (26 Jun Shift 2)
A solid spherical ball of mass m is released from the topmost point of the shown semi-spherical shell. The track is sufficiently rough to enable pure rolling motion. The normal force between the ball and the shell at the lowest position is

Options:
A 1 2 7 mg
B 7 9 mg
C 1 7 7 mg
D 1 0 7 mg

The correct answer is:
1 7 7 mg

Solution:
From the law of conservation of mechanical energy,
Final energy = Translational kinetic energy + Rotational kinetic energy
Hence, Final energy = 12mv2+12Iω2
=12mv2+12×25mR2×vR2=710mv2
Hence, we finally obtain, mgR - r=710mv2 .......(1)
and N - mg = mv 2 R - r ..........(2)
Solving Eqs. (1) and (2), we get
N = 1 7 7 mg

Topic: Physics, Rotational Motion, JEE Main