datavorous/entrance-exam-dataset huggingface dataset:
A solid cube of the edge a is molten and moulded into eight identical small solid cubes and are placed on one other on a straight line with the edge of the bottom cube on the same horizontal plane on which big cube was placed, then the vertical shift in the centre of mass is

Options:
A 3a2
B 2a
C 5a2
D 3a

The correct answer is:
3a2

Solution:
Initial height of centre of mass hi=a2
As the volume remains same,
8l3=a3
l=a2
Final height of centre of mass
hf=8l2=4l
∴ Δh=4l-a2=2a-a2=3a2

Topic: Physics, Center of Mass Momentum and Collision, NEET
A solid cube of wood of side \( 2 a \) and mass \( M \) is resting on a horizontal surface as shown in the figure. The cube is free to rotate about a fixed axis \( A B \). A bullet of mass \( m(< < M) \) and speed \( v \) is shot horizontally at the face opposite to \( A B C D \) at a height of \( \frac{4 a}{3} \) from the surface to impart the cube and angular speed \( \omega \). It strikes the face and embeds in the cube. Then \( \omega \) is close to (note: the moment of inertia of the cube about an axis perpendicular to the face and passing through the center of mass is \( \frac{2 M a^{2}}{3} . \) )

Options:
A \( \frac{M v}{m a} \)
B \( \frac{M v}{2 m a} \)
C \( \frac{m v}{M a} \)
D \( \frac{m v}{2 M a} \)

The correct answer is:
\( \frac{m v}{2 M a} \)

Solution:
Applying the parallel axis theorem, I=Icm+Ma2, here Icm is the moment of inertia passing through centre of mass, a is the distance between two axes.

The moment of inertia of solid cube about AB is I=23Ma2+Ma22

I=83Ma2

The cube is free to rotate about AB axis. Thus, the angular momentum should be conserved.

Angular momentum mvr=Iω, here r=4a3.

So, mv×4a3=Iω

⇒ω=mv4a383Ma2=mv2Ma.

Topic: Physics, Rotational Motion, JEE Main
A solid cylinder and a solid sphere, having same mass M and radius R, roll down the same inclined plane from top without slipping. They start from rest. The ratio of velocity of the solid cylinder to that of the solid sphere, with which they reach the ground, will be

Options:
A 53
B 45
C 35
D 1415

The correct answer is:
1415

Solution:
During pure rolling, the work done by the friction is zero. Therefore, we can apply conservation of mechanical energy.

mgh=12mv2+12IvR2

⇒mgh=12m+IR2v2

⇒v=2gh1+ImR2

For a solid cylinder ImR2=12 and for a solid sphere, ImR2=25

vCylvSph=1+251+12=1415

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2022 (25 Jul Shift 1)
A solid cylinder is given a velocity v0 and an angular speed ω0 in the anticlockwise direction and then it is placed gently on a very long rough plank (initially at rest) as shown in the figure. Then, choose the incorrect statement:

Options:
A the friction force on the disc is in the backward direction till pure rolling starts
B the friction force between the disc and the plank is kinetic in nature till pure rolling starts
C the total momentum of system (disc and plank) is conserved
D the angular momentum of the disc about any point on the horizontal surface remains conserved

The correct answer is:
the angular momentum of the disc about any point on the horizontal surface remains conserved

Solution:
Initial velocity of point of contact of disc is in forward direction.

Hence friction is in backward direction and kinetic in nature.

Net force on the system is zero. So momentum is conserved.

Topic: Physics, Rotational Motion, JEE Main
A solid cylinder is kept on one edge of a plank of same mass and length 25 m placed on a smooth surface as shown in the figure. The coefficient of friction between the cylinder and the plank is 0.5. The plank is given a velocity of 20 m s-1 towards right. Find the time (in s) after which plank and cylinder will separate. [g=10 m s-2]

Options:

The correct answer is:
2

Solution:
Drawing free body diagram of the cylinder with respect to plank.
2Î¼mg=maâa=2Î¼g
Î¼mgR=12mR2Î±Â ;Î±-2Î¼gr
Acceleration of point of contact with respect to plank is 4Î¼g
Velocity of pure rolling starts,
-v+4Î¼gt=0
t=204Ã0.5Ã10=1s
Distance traveled by cylinder with respect to plank in 1s is
S′=-vt+122Î¼gt2=-15m
At t=1s , the velocity of cylinder with respect to plank is
t=1s
vrel=-v+2Î¼gt=-20+2Ã0.5Ã10=-10 m/s
Remaining 10 m will be travelled in time t′=1010=1s
â´ Total time = 2s

Topic: Physics, Rotational Motion, JEE Main
A solid cylinder is made of radius R and height 3R having mass density ρ . Now two half spheres of radius R are removed from both ends. The moment of inertia of remaining portion about axis ZZ' can be calculated as 296KπR5ρ . Find K.

Options:

The correct answer is:
5

Solution:
∵ Mcylinder=πR2×3Rρ.lcylinder=123πR3ρR2
Msphere=23πR3ρ,lsphere=2523πR3ρR2
∴ lremaining=lcylinder-lsphere=2930πR5ρ.

Topic: Physics, Rotational Motion, JEE Main
A solid cylinder of mass $M$ and radius $R$ rolls without slipping on an inclined plane of length $L$ and height $h$. What is the speed of its centre of mass when the cylinder reaches its bottom?

Options:
A $\sqrt{2 g h}$
B $\sqrt{\frac{3}{4} g h}$
C $\sqrt{\frac{4}{3} g h}$
D $\sqrt{4 g h}$

The correct answer is:
$\sqrt{\frac{4}{3} g h}$

Solution:
Here K.E. of centre of mass when cylinder reached the bottom $=\frac{1}{2} m v^2+\frac{1}{2} \mathrm{I} \omega^2 \ldots(\mathrm{i})$
Here $\mathrm{I}=\frac{1}{2} M K^2$ and $\omega=\frac{V}{R}$
$\begin{aligned}
& \therefore \mathrm{KE}=\frac{1}{2} m v^2+\frac{1}{2} M K^2 \cdot \frac{V^2}{r^2} \\
& =\frac{1}{2} m v^2\left[1+\frac{K^2}{R^2}\right]
\end{aligned}$
But for solid cylinder $K=\frac{R}{\sqrt{2}}$ or $\frac{K^2}{R^2}=\frac{1}{2}$
$\therefore K E=\frac{3}{4} m v^2$
PE of the solid cylinder at height $h$ is
$\mathrm{PE}=m g h$
From eq. (ii) and (iii)
$\begin{aligned}
& m g h=\frac{3}{4} m v^2 \\
& \Rightarrow v=\sqrt{\frac{4}{3} g h}
\end{aligned}$

Topic: Physics, Rotational Motion, JEE Main
A solid cylinder of mass m is wrapped with an inextensible light string and, is placed on a rough inclined plane as shown in the figure. The frictional force acting between the cylinder and the inclined plane is:

[The coefficient of static friction, μs, is 0.4 ]

Options:
A 72mg
B 5mg
C mg5
D 0

The correct answer is:
mg5

Solution:
Let's take solid cylinder is in equilibrium

T+f=mgsin60

TR-fR=0

Solving we get

T=freq=mgsinθ2

But limiting friction < required friction

μmgcos60°<mgsin60°2

∴ Hence cylinder will not remain in equilibrium

Hence f= kinetic

=μkN

=μkmgcos60°

=mg5

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2021 (18 Mar Shift 2)
A solid cylinder of radius $r_1=2.5 \mathrm{~cm}$, length $l_1=5.0 \mathrm{~cm}$ and temperature $40^{\circ} \mathrm{C}$ is suspended in an environment of temperature $60^{\circ} \mathrm{C}$. The thermal radiation transfer rate for cylinder is $1.0 \mathrm{~W}$. If the cylinder is stretched until its radius becomes $r_2=0.50 \mathrm{~cm}$, the thermal radiation transfer rate is changed to

Options:
A $3.35 \mathrm{~W}$
B $4.50 \mathrm{~W}$
C $0.75 \mathrm{~W}$
D $1.25 \mathrm{~W}$

The correct answer is:
$3.35 \mathrm{~W}$

Solution:
Keeping nature of surface and temperature of body and surroundings same, rate of heat transfer by radiation depends on area of body,
$\Rightarrow$ Rate of heat transfer $\propto$ Surface area of body
$\Rightarrow \quad \frac{Q_2}{Q_1}=\frac{S_2}{S_1} \Rightarrow Q_2=Q_1 \times \frac{S_2}{S_1}$ $\ldots(i)$
Now, volume remains same in stretching, so
$\pi r_1^2 l_1=\pi r_2^2 l_2$
$\Rightarrow \quad l_2=\frac{r_1^2}{r_2^2} \cdot l_1=\frac{(2.5)^2}{(0.5)^2} \times 5=125 \mathrm{~cm}$
So, area before stretching, $S_1=2 \pi r_1\left(l_1+r_1\right)$ and area after stretching, $S_2=2 \pi r_2\left(l_2+r_2\right)$ Hence by Eq. (i), we get
$Q_2=Q_1 \times \frac{S_2}{S_1}=Q_1 \times \frac{2 \pi r_2\left(l_2+r_2\right)}{2 \pi r_1\left(l_1+r_1\right)}$
$=Q_1 \times \frac{r_2\left(l_2+r_2\right)}{r_1\left(l_1+r_1\right)}=\frac{1 \times 0.5 \times 10^{-2}(125+0.5) \times 10^{-2}}{2.5 \times 10^{-2}(5+2.5) \times 10^{-2}}$
$=3.346=3.35 \mathrm{~W}$

Topic: Physics, Thermal Properties of Matter, JEE Main
A solid cylinder of radius $r_1=2.5 \mathrm{~cm}$, length $l_1=5.0 \mathrm{~cm}$ and temperature $40^{\circ} \mathrm{C}$ is suspended in an environment of temperature $60^{\circ} \mathrm{C}$. The thermal radiation transfer rate for cylinder is $1.0 \mathrm{~W}$. If the cylinder is stretched until its radius becomes $r_2=0.50 \mathrm{~cm}$, the thermal radiation transfer rate is changed to

Options:
A $3.35 \mathrm{~W}$
B $4.50 \mathrm{~W}$
C $0.75 \mathrm{~W}$
D $1.25 \mathrm{~W}$

The correct answer is:
$3.35 \mathrm{~W}$

Solution:
Keeping nature of surface and temperature of body and surroundings same, rate of heat transfer by radiation depends on area of body,
$\Rightarrow$ Rate of heat transfer $\propto$ Surface area of body
$\Rightarrow \quad \frac{Q_2}{Q_1}=\frac{S_2}{S_1} \Rightarrow Q_2=Q_1 \times \frac{S_2}{S_1}$ $\ldots(i)$
Now, volume remains same in stretching, so
$\pi r_1^2 l_1=\pi r_2^2 l_2$
$\Rightarrow \quad l_2=\frac{r_1^2}{r_2^2} \cdot l_1=\frac{(2.5)^2}{(0.5)^2} \times 5=125 \mathrm{~cm}$
So, area before stretching, $S_1=2 \pi r_1\left(l_1+r_1\right)$ and area after stretching, $S_2=2 \pi r_2\left(l_2+r_2\right)$ Hence by Eq. (i), we get
$Q_2=Q_1 \times \frac{S_2}{S_1}=Q_1 \times \frac{2 \pi r_2\left(l_2+r_2\right)}{2 \pi r_1\left(l_1+r_1\right)}$
$=Q_1 \times \frac{r_2\left(l_2+r_2\right)}{r_1\left(l_1+r_1\right)}=\frac{1 \times 0.5 \times 10^{-2}(125+0.5) \times 10^{-2}}{2.5 \times 10^{-2}(5+2.5) \times 10^{-2}}$
$=3.346=3.35 \mathrm{~W}$

Topic: Physics, Thermal Properties of Matter, TS EAMCET, TS EAMCET 2020 (10 Sep Shift 2)
A solid cylinder of radius $R$ made of a material of thermal conductivity $K_{1}$ is surrounded by a cylindrical shell of inner radius $R$ and outer radius $2 R$ made of a material of thermal conductivity $K_{2}$.
The two ends of the combined system are maintained at two different temperatures. Then there is no loss of heat across the cylindrical surface and the system is in steady state. The effective thermal conductivity of the system is

Options:
A $K_{1}+K_{2}$
B $\frac{K_{1} K_{2}}{K_{1}+K_{2}}$
C $\frac{2 K_{1}+K_{2}}{4}$
D $\frac{K_{1}+3 K_{2}}{4}$

The correct answer is:
$\frac{K_{1}+3 K_{2}}{4}$

Solution:
As there is no loss of heat, and the system is in steady state.
$\begin{array}{llrl}
& \therefore & Q_{\text {total }} & =Q_{1}+Q_{2} \\
\Rightarrow & & \frac{K\left(4 R^{2}\right) d \theta}{d x} & =\frac{K_{1} R^{2} d \theta}{d x}+\frac{K_{2}\left(3 R^{2}\right) d \theta}{d x} \\
\Rightarrow & & 4 K & =K_{1}+3 K_{2} \\
\Rightarrow & & K & =\frac{K_{1}+3 K_{2}}{4}
\end{array}$

Topic: Physics, Thermal Properties of Matter, KCET, KCET 2013
A solid disc of radius 20 cm and mass 10 kg is rotating with an angular velocity of 600 rpm, about an axis normal to its circular plane and passing through its centre of mass. The retarding torque required to bring the disc at rest in 10 s is _________π×10-1 N m

Options:

The correct answer is:
4

Solution:
τ=ΔLΔt=Iωf-ωiΔt

τ=mR22×[0-ω]Δt

=10×20×10-222×600×π30×10

=0.4π=4π×10-1

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2021 (25 Jul Shift 2)
A solid disc of radius \( 20 \mathrm{~cm} \) and mass \( 10 \mathrm{~kg} \) is rotating with an angular velocity of \( 600 \mathrm{rpm} \), about an axis normal to its circular plane and passing through its centre of mass. The retarding torque required to bring the disc at rest in \( 10 \mathrm{~s} \) is _____

Options:
A \(4 \pi \times 10^{-2}\)
B \(2 \pi\)
C \(4 \pi \times 10^{-1}\)
D \(2 \pi \times 10^{-2}\)

The correct answer is:
\(4 \pi \times 10^{-2}\)

Solution:
τ=ΔLΔt=Iωf-ωiΔt

τ=mR22×[0-ω]Δt

=10×20×10-222×600×π30×10

=0.4π=4π×10-2

Topic: Physics, Rotational Motion, JEE Main
A solid disc of radius a and mass m rolls down without slipping on an inclined plane making an angle θ with the horizontal. The acceleration of the disc will be 2 bgsinθ, where b is _______. (Round off to the Nearest Integer)

(g= acceleration due to gravity)

(θ= angle as shown in figure)

Options:

The correct answer is:
3

Solution:
a=gsinθ1+ImR2=gsinθ1+12=23 gsinθ

b=3

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2021 (16 Mar Shift 2)
A solid has a ' $b c c$ ' structure. If the distance of nearest approach between two atoms is $1.73 Å$, the edge length of the cell is

Options:
A
$314.20 \mathrm{pm}$

B
$1.41 \mathrm{pm}$

C
$200 \mathrm{pm}$

D
$216 \mathrm{pm}$

The correct answer is:
$200 \mathrm{pm}$

Solution:
For bcc structure
$$
d=\frac{\sqrt{3} a}{2}
$$
where $d=$ distance between two atoms $a=$ edge length
$$
\begin{aligned}
& 1.73=\frac{\sqrt{3}}{2} a \\
& a=\frac{2 \times 1.73}{\sqrt{3}}=2 Å=200 \mathrm{pm}
\end{aligned}
$$

Topic: Chemistry, Solid State, JEE Main, JEE Main 2012 (07 May Online)
A solid having density of $9 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{-3}$ forms face centred cubic crystals of edge length $200 \sqrt{2} \mathrm{pm} .$ What is the molar mass of the solid?

[Avogadro constant $\left.\approx 6 \times 10^{23} \mathrm{~mol}^{-1}, \pi \approx 3\right]$

Options:
A $0.0432 \mathrm{~kg} \mathrm{~mol}^{-1}$
B $0.0216 \mathrm{~kg} \mathrm{~mol}^{-1}$
C $0.0305 \mathrm{~kg} \mathrm{~mol}^{-1}$
D $0.4320 \mathrm{~kg} \mathrm{~mol}^{-1}$

The correct answer is:
$0.0305 \mathrm{~kg} \mathrm{~mol}^{-1}$

Solution:
Density $=\frac{Z \times M}{N_{A} \times a^{3}}$

$9 \times 10^{3}=\frac{4 \times \mathrm{M}}{\left(200 \times \sqrt{2} \times 10^{-12}\right)^{3} 6 \times 10^{23}}$

Topic: Chemistry, Solid State, JEE Main, JEE Main 2019 (11 Jan Shift 1)
A solid homogeneous sphere is moving on a rough horizontal surface, partly rolling and partly sliding. During the king of motion of the sphere.

Options:
A total kinetic energy is conserved
B angular momentum of the sphere about the point of contact with the plane is conserved
C only the rotational kinetic energy about centre of mass is conserved
D angular momentum about centre of mass is conserved

The correct answer is:
angular momentum of the sphere about the point of contact with the plane is conserved

Solution:
Since, the sphere is sliding, maximum force of friction will act passing through, the bottommost point. Torque of this force about point of contact is zero. Hence, angular momentum will remain conserved about the point of contact and not about the centre of mass. About centre of mass torque of friction will not be zero.

Topic: Physics, Rotational Motion, JEE Main
A solid is a compound of group \( 1 \) element and it gives a bright red colour in the flame test. The solid is

Options:
A \( \mathrm{LiBr} \)
B \( \mathrm{CsCl} \)
C \( \mathrm{KCl} \)
D \( \mathrm{NaCl} \)

The correct answer is:
\( \mathrm{LiBr} \)

Solution:
Lithium and strontium salts impart bright red colour to the flame and it says that it belongs to alkali metals which means it is salt of lithium.

Topic: Chemistry, s Block Elements, JEE Main
A solid is floating in a liquid with some fraction of its volume submerged. If for two different temperatures T1 and T2, the fractions of the volume of the solid submerged in the liquid are f1 and f2 respectively, then the coefficient of volume expansion of the liquid is [Assume that the expansion of the solid is negligible]

Options:
A f1-f2f2t1-f1t2
B f1-f2f1t1-f2t2
C f1+f2f2t1-f1t2
D f1+f2f1t1-f2t2

The correct answer is:
f1-f2f2t1-f1t2

Solution:
Let V be the volume of solid ; d be its density and m be its mass; if Î³ is the coefficient of volume expansion of the liquid, then
Density at temperature t1 is, d1=d01+Î³t1
Density at temperature t2 is, d2 = d01+Î³t2
According to Archimede's principle,
f1V d1=m=f2 Vd2
Or d1d2 =f2f1= d01+Î³t11+Î³t2d0
Or f1+f1Î³t2=f2+f2 Î³t1
f1-f2 = Î³f2t1-f1t2
Î³ = (f1-f2)f2t1-f1t2

Topic: Physics, Thermal Properties of Matter, JEE Main
A solid is formed and it has three types of atoms X, Y and Z. X forms a fcc lattice with Y atoms occupying all tetrahedral voids and Z atoms occupying half of octahedral voids. The formula of solid is

Options:
A X4YZ2
B X4Y2Z
C XY2Z4
D X2Y4Z

The correct answer is:
X2Y4Z

Solution:
X is in f.c.c. lattice
Number of atoms =atom at cornerÃ18+atoms at facesÃ12
X=8Ã18+6Ã12=4
Total tetrahedral voids =8
As y are at T.V.
So Y=8
Number of Octahedra voids =4
Z=12Ã4=2
So X=4,Y=8,Z=2
Hence the formula of solid is X4Y8Z2 or X2Y4Z .

Topic: Chemistry, Solid State, JEE Main
A solid is made of two elements $X$ and $Z$. The atoms $Z$ are in CCP arrangement while the atom $X$ occupy all the tetrahedral sites. What is the formula of the compound

Options:
A $X Z$
B $x Z_2$
C $X_2 Z$
D $X_2 Z_3$

The correct answer is:
$X_2 Z$

Solution:
Tetrahedral sites one double comparable to octahedral sites then ratio of $X$ and $Z$ respectively $2: 1$ since formula of the compound $x_2 zZ$.

Topic: Chemistry, Solid State, JEE Main
A solid maintained at $t_{1}^{\circ} \mathrm{C}$ is kept in an evacuated chamber at temperature $t_{2}^{\circ} \mathrm{C}$ $\left(t_{2} \gg t_{1}\right) .$ The rate of heat absorbed by the body is proportional to

Options:
A $t_{2}^{4}-t_{1}^{4}$
B $\left(f_{2}^{4}+273\right)-\left(t_{1}^{4}+273\right)$
C $t_{2}-t_{1}$
D $t_{2}^{2}-t_{1}^{2}$

The correct answer is:
$t_{2}-t_{1}$

Solution:
As we know that, the rate of cooling is
$$
\frac{d \theta}{d t}=b A\left(\theta-\theta_{0}\right)
$$
where, $b A=$ constant $\Rightarrow \quad \frac{d \theta}{d t} \propto \theta-\theta_{0}$
Also $-\frac{d T}{d t} \propto T-T_{0}$
where, $T=$ temperature and $\quad t=$ time
So, the rate of heat absorbed by cooled body is proportional to $t_{2}-t_{1}$

Topic: Physics, Thermal Properties of Matter, WBJEE, WBJEE 2015
A solid metal cube of edge length 2 cm is moving in the positive y-direction, at a constant speed of 6 m s-1. There is a uniform magnetic field of 0.1 T in the positive z-direction. The potential difference between the two faces of the cube, perpendicular to the x-axis, is

Options:
A 12 mV
B 1 mV
C 2 mV
D 6 mV

The correct answer is:
12 mV

Solution:
emf developed is given by, e=Bvl

∴e=0.12×10-2×6⇒e=12×10-3=12 mV

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2019 (10 Jan Shift 1)
A solid metallic cube having total surface area 24 m2 is uniformly heated. If its temperature is increased by 10°C, calculate the increase in volume of the cube. (Given α=5.0×10-4C∘-1).

Options:
A 2.4×106 cm3
B 1.2×105 cm3
C 6×104 cm3
D 4.8×105 cm3

The correct answer is:
1.2×105 cm3

Solution:
Total surface area of a cube is 6a2=24 m2⇒a2=4 ⇒a=2 m

Volume of a cube V=a3=8 m3

Change in volume is given by,

ΔV=V0γΔT, where γ=3α

ΔV=8×1.5×10-4×10 m3⇒ΔV=1.2×105 cm3

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2022 (25 Jun Shift 2)
A solid sphere, a hallow sphere and a ring are released from top of an inclined plane (frictionless)
so that they slide down the plane. Then maximum acceleration down the plane is for (no rolling)

Options:
A
solid sphere

B
hollow sphere

C
ring

D
all same

The correct answer is:
all same

Solution:
No Solution Available

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2002
A solid sphere and a hollow cylinder roll up without slipping on same inclined plane with same initial speed $v$. The sphere and the cylinder reaches upto maximum heights $h_1$ and $h_2$, respectively, above the initial level. The ratio $h_1: h_2$ is $\frac{n}{10}$. The value of $n$ is______.

Options:

The correct answer is:
7

Solution:
Gain in P.E. = Loss in K.E.
$\mathrm{mgh}=\frac{1}{2} \mathrm{mv}^2\left(1+\frac{\mathrm{K}^2}{\mathrm{R}^2}\right)$
$\mathrm{h} \propto 1+\frac{\mathrm{K}^2}{\mathrm{R}^2}$
$\frac{\mathrm{h}_1}{\mathrm{~h}_2}=\frac{1+\frac{2}{5}}{1+1}=\frac{7}{5 \times 2}=\frac{7}{10}$
$\mathrm{n}=7$

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2024 (04 Apr Shift 1)
A solid sphere and a solid cylinder of same mass and radius are rolling on a horizontal surface without slipping. The ratio of their radius of gyrations respectively ksph:kcyl is 2:x. The value of x is _______.

Options:

The correct answer is:
5

Solution:
Considering rotational axis as the diametrical axis for sphere and axis of cylinder. Then

K12=25R2 and K22=12R2

∴ K1K2=2/51/2=45

⇒K1K2=25

Comparing it with the given value in the question,

25=2x

∴ x=5

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2023 (15 Apr Shift 1)
A solid sphere and hollow sphere of the same mass and radius are given a spin about their centre of mass and then, they are placed on a rough horizontal surface. The spin angular velocity is the same for both the spheres and it is equal to Ï0. Once the pure rolling starts; let v1 and v2 be the linear speeds of their centres of mass, then

Options:
A v1=v2
B v1>v2
C v1<v2
D Data is insufficient

The correct answer is:
v1<v2

Solution:
From conservation of angular momentum about point of contact :
I Ï 0 = I Ï + mRv
or I Ï 0 = I v R + mRv
or v = I Ï 0 I R + mR
or
or v = Ï 0 1 R + mR I
Now I solid sphere < I hollow
â´ v solid < v hollow
â´ v 1 < v 2

Topic: Physics, Rotational Motion, JEE Main
A solid sphere and solid cylinder of identical radii approach an incline with the same linear velocity (see figure). Both roll without slipping all throughout. The two climb maximum heights hsph and hcyl on the inline. The ratio hsphhcyl is given by:

Options:
A 25
B 45
C 1415
D 1

The correct answer is:
1415

Solution:
From conservation of mechanical energy,
For solid sphere: 12msphV2+12Isphω2=msphghsph
V22+V25=ghsph
7V210=ghsph …(i)
For cylinder: 12mcylV2+12Icylω2=mcylghcyl
V22+V24=ghcyl
3V24=ghcyl …(ii)
From (i) and (ii) hsphhcyl=1415

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2019 (08 Apr Shift 2)
A solid sphere, disc and solid cylinder all of the same mass and made of the same material are allowed to roll down (from rest) on an inclined plane, then

Options:
A Solid sphere reaches the bottom first
B Solid sphere reaches the bottom last
C Disc will reach the bottom first
D All reach the bottom at the same time

The correct answer is:
Solid sphere reaches the bottom last

Solution:
Work-energy theorem: $\mathrm{W}_{\text {allforces }}=\Delta \mathrm{K} . \mathrm{E}$
$\mathrm{mgh}=\mathrm{K} . \mathrm{E} \Rightarrow$ all will have same $\mathrm{K}$.E. at the bottom.
$\begin{aligned}& \text { Also, } \mathrm{mgh}=\frac{1}{2} \mathrm{Iw}^2+\frac{1}{2} \mathrm{mv}^2 \\& \mathrm{mgh}=\frac{1}{2} \times \mathrm{Kmr}^2 \mathrm{w}^2+\frac{1}{2} \mathrm{mv}^2 \\& \mathrm{mgh}=\frac{1}{2} \times \mathrm{Kmv}^2+\frac{1}{2} \mathrm{mv}^2 \\& \Rightarrow \mathrm{v}^2=\frac{2 \mathrm{gh}}{1+\mathrm{K}}
\end{aligned}$
Acceleration of the body rolling purely on the inclines plane, $a=\frac{g \sin \theta}{1+\mathrm{K}}$ where
$\mathrm{K}$ is a constant.
Now, $v=$ at $\Rightarrow \mathrm{t}=\frac{\sqrt{2 g h(1+K)}}{g \sin \theta}$
As, $\mathrm{K}_{\mathrm{C}}=\mathrm{K}_{\mathrm{D}}\gt\mathrm{K}_{\mathrm{S}} \Rightarrow \mathrm{t}_{\mathrm{C}}=\mathrm{t}_{\mathrm{D}}\gt\mathrm{t}_{\mathrm{S}}$
Hence solid sphere will reach the bottom first.

Topic: Physics, Rotational Motion, JEE Main
A solid sphere having mass $m$ and radius $r$ rolls down an inclined plane. Then its kinetic energy is

Options:
A
$\frac{5}{7}$ rotational and $\frac{2}{7}$ translational

B
$\frac{2}{7}$ rotational and $\frac{5}{7}$ translational

C
$\frac{2}{5}$ rotational and $\frac{3}{5}$ translational

D
$\frac{1}{2}$ rotational and $\frac{1}{2}$ translational

The correct answer is:
$\frac{2}{7}$ rotational and $\frac{5}{7}$ translational

Solution:
$$
\begin{aligned}
& K . E_{\text {rotational }}=\frac{1}{2} I \omega^2 \\
& =\frac{1}{2} \frac{2}{5} \omega r^2 d^2\left(\because I_{\text {Solid sphere }}=\frac{2}{5} m r^2\right) \\
& K \cdot E_{\text {translational }}=\frac{1}{2} m v^2 \\
& \therefore \frac{K \cdot E_{\text {rotational }}}{K \cdot E_{\text {translational }}}=\frac{2}{5}
\end{aligned}
$$
Hence option (b) is correct

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2012 (07 May Online)