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|---|---|---|---|---|---|---|---|---|---|---|---|
lvb29ehp | physics | properties-of-matter | surface-tension,-excess-pressure-and-capillarity | <p>Pressure inside a soap bubble is greater than the pressure outside by an amount : (given : $$\mathrm{R}=$$ Radius of bubble, $$\mathrm{S}=$$ Surface tension of bubble)</p> | [{"identifier": "A", "content": "$$\\frac{S}{R}$$\n"}, {"identifier": "B", "content": "$$\\frac{4 \\mathrm{~S}}{\\mathrm{R}}$$\n"}, {"identifier": "C", "content": "$$\\frac{4 \\mathrm{R}}{\\mathrm{S}}$$\n"}, {"identifier": "D", "content": "$$\\frac{2 S}{R}$$"}] | ["B"] | null | <p>The difference in pressure inside a soap bubble as compared to the outside is due to the surface tension created by the soap film on the bubble. This difference in pressure can be calculated using the formula that relates the surface tension of the soap bubble to the radius of the bubble. The correct formula for the... | mcq | jee-main-2024-online-6th-april-evening-shift | 12,422 |
lvc57nse | physics | properties-of-matter | surface-tension,-excess-pressure-and-capillarity | <p>A big drop is formed by coalescing 1000 small droplets of water. The ratio of surface energy of 1000 droplets to that of energy of big drop is $$\frac{10}{x}$$. The value of $$x$$ is ________.</p> | [] | null | 1 | <p>$$R_{\text {big }}=10 R_{\text {small }}$$</p>
<p>$$ \Rightarrow {{{E_{1000}}} \over {{E_{big}}}} = {{1000 \times T \times 4\pi {{\left[ {{{{R_{big}}} \over {10}}} \right]}^2}} \over {T \times 4\pi R_{big}^2}} = {{10} \over 1}$$</p> | integer | jee-main-2024-online-6th-april-morning-shift | 12,423 |
74sEsWoySBVGwUIk | physics | rotational-motion | angular-momentum | Initial angular velocity of a circular disc of mass $$M$$ is $${\omega _1}.$$ Then two small spheres of mass $$m$$ are attached gently to diametrically opposite points on the edge of the disc. What is the final angular velocity of the disc? | [{"identifier": "A", "content": "$$\\left( {{{M + m} \\over M}} \\right)\\,\\,{\\omega _1}$$ "}, {"identifier": "B", "content": "$$\\left( {{{M + m} \\over m}} \\right)\\,\\,{\\omega _1}$$ "}, {"identifier": "C", "content": "$$\\left( {{M \\over {M + 4m}}} \\right)\\,\\,{\\omega _1}$$ "}, {"identifier": "D", "content":... | ["C"] | null | When two small spheres of mass $$m$$ are attached gently, the external torque, about the axis of rotation, is zero.
<br><br>So, $${{d\overrightarrow L } \over {dt}} = \overline z $$ = 0
<br><br>$$\overrightarrow L $$ = conserved
<br><br>So the angular momentum about the axis of rotation is conserved.
<p>$$\therefore$$ ... | mcq | aieee-2002 | 12,424 |
sjBaZtw9eMc4pxFP | physics | rotational-motion | angular-momentum | A particle of mass $$m$$ moves along line PC with velocity $$v$$ as shown. What is the angular momentum of the particle about P?
<img src="data:image/png;base64,UklGRiwMAABXRUJQVlA4ICAMAABwZgCdASrcAWUBP4G+1mY2LywnITH5usAwCWlu/EsYMutQ2/08/1/b//ueYP3VL9/z+6a7CfkRkp3YsAH5P/XPT7mU/oWiz0y73Y8p/6+O79x/9PA1FAeZ8GS84Ml5wZLzgyX... | [{"identifier": "A", "content": "$$mvL$$"}, {"identifier": "B", "content": "$$mvl$$ "}, {"identifier": "C", "content": "$$mvr$$ "}, {"identifier": "D", "content": "zero "}] | ["D"] | null | Angular momentum $$(L)$$
<br><br>$$=$$ ( linear momentum ) $$ \times $$ ( perpendicular distance of the line of action of momentum from the axis of rotation)
<br><br>$$ = mv \times r$$
<br><br>$$ = mv \times 0$$
<br><br>$$=0$$
<br><br>[ Here $$r=0$$ because the particle is
moving through the line PQ and
r is ... | mcq | aieee-2002 | 12,425 |
KZnaV6RylgRhrevz | physics | rotational-motion | angular-momentum | A particle performing uniform circular motion has angular frequency is doubled & its kinetic energy halved, then the new angular momentum is | [{"identifier": "A", "content": "$${L \\over 4}$$ "}, {"identifier": "B", "content": "$$2L$$ "}, {"identifier": "C", "content": "$$4L$$ "}, {"identifier": "D", "content": "$${L \\over 2}$$ "}] | ["A"] | null | We know Rotational Kinetic Energy$$={1 \over 2}I{\omega ^2},$$
<br><br>Angular Momentum $$L = I\omega \Rightarrow I = {L \over \omega }$$
<br><br>$$\therefore$$ Initial $$K.E. = {1 \over 2}{L \over \omega } \times {\omega ^2} = {1 \over 2}L\omega $$
<br><br>Final $$K.E'$$ = $${{K.E} \over 2}$$ = $${1 \over 2}{L'} \t... | mcq | aieee-2003 | 12,426 |
2yrP77y91dEoBdH3 | physics | rotational-motion | angular-momentum | A solid sphere is rotating in free space. If the radius of the sphere is increased keeping mass same which on of the following will not be affected ? | [{"identifier": "A", "content": "Angular velocity "}, {"identifier": "B", "content": "Angular momentum "}, {"identifier": "C", "content": "Moment of inertia "}, {"identifier": "D", "content": "Rotational kinetic energy "}] | ["B"] | null | Solid sphere is rotating in free space that means no external torque is operating on the sphere.
<br><br>Angular momentum will remain the same since external torque is zero. | mcq | aieee-2004 | 12,427 |
0P1oWpeepv5X83HE | physics | rotational-motion | angular-momentum | A thin horizontal circular disc is rotating about a vertical axis passing through its center. An insect is at rest at a point near the rim of the disc. The insect now moves along a diameter of the disc to reach its other end. During the journey of the insect, the angular speed of the disc. | [{"identifier": "A", "content": "continuously decreases "}, {"identifier": "B", "content": "continuously increases "}, {"identifier": "C", "content": "first increases and then decreases "}, {"identifier": "D", "content": "remains unchanged "}] | ["C"] | null | Here no external force is applied on the disc so Torque ($$\tau $$) = 0.
<br><br>So angular momentum is conserved.
<br><br>That means $${I_1}{\omega _1} = {I_2}{\omega _2}$$
<br><br>$$ \Rightarrow $$ $${\omega _2} = {{{I_1}{\omega _1}} \over {{I_2}}}$$
<br><br>$$\therefore$$ Angular speed is inversely proportional to M... | mcq | aieee-2011 | 12,431 |
pArPnFOcPLv5DIR5 | physics | rotational-motion | angular-momentum | A hoop of radius $$r$$ and mass $$m$$ rotating with an angular velocity $${\omega _0}$$ is placed on a rough horizontal surface. The initial velocity of the center of the hoop is zero. What will be the velocity of the center of the hoop when it cases to slip? | [{"identifier": "A", "content": "$${{r{\\omega _0}} \\over 4}$$ "}, {"identifier": "B", "content": "$${{r{\\omega _0}} \\over 3}$$"}, {"identifier": "C", "content": "$${{r{\\omega _0}} \\over 2}$$"}, {"identifier": "D", "content": "$${r{\\omega _0}}$$ "}] | ["C"] | null | <img class="question-image" src="https://imagex.cdn.examgoal.net/7SEWdgNFPTklUoJOB/ASdYXN5Qk0ZieNU7m9Cmhls5a20EM/hH4R5wyMjDYTTTwfN6RuWz/image.svg" loading="lazy" alt="JEE Main 2013 (Offline) Physics - Rotational Motion Question 191 English Explanation">
<br> From conservation of angular momentum at point of contact,
... | mcq | jee-main-2013-offline | 12,432 |
B4WVhMgVTWppG6pbXPQVF | physics | rotational-motion | angular-momentum | A cubical block of side 30 cm is moving with velocity 2 ms<sup>−1</sup> on a smooth horizontal surface. The surface has a bump at a point O as shown in figure. The angular
velocity (in rad/s) of the block immediately after it hits the bump, is :
<br/><br/><img src="data:image/png;base64,UklGRlYKAABXRUJQVlA4IEoKAABwoQCd... | [{"identifier": "A", "content": "5.0"}, {"identifier": "B", "content": "6.7"}, {"identifier": "C", "content": "9.4"}, {"identifier": "D", "content": "13.3"}] | ["A"] | null | Before hitting point 0,
<br><br>angular moment = mv $$ \times $$ $${a \over 2}$$
<br><br>After hitting point 0,
<br><br>Angular momentum = $${\rm I}\omega $$
<br><br>$$ \therefore $$ $${\rm I}\omega $$ = $${{mva} \over 2}$$
<br><br>$$ \Rightarrow $$ $$\omega $$ = $${{mva} \over {2{... | mcq | jee-main-2016-online-9th-april-morning-slot | 12,434 |
75ehPmtUXS2jAEDnNm3jF | physics | rotational-motion | angular-momentum | A particle of mass m is moving along the
side of a square of side ‘a’, with a uniform
speed v in the x-y plane as shown in the
figure :<br/><br/>
<img src="data:image/png;base64,UklGRhQRAABXRUJQVlA4IAgRAABwBAGdASoAA3ECP4G+2GW2L6ynIdGJWsAwCWlu4W2khmNwvV6l/3XqjuJuNO8ncX+8cbnCx1/8O/ndGTX//6uivqKVTG2opVMbailUxtqKVTG2opVMba... | [{"identifier": "A", "content": "$$\\overrightarrow L = mv\\left[ {{R \\over {\\sqrt 2 }} + a} \\right]\\widehat k$$\n<br>when the\nparticle is moving from B to C."}, {"identifier": "B", "content": "$$\\overrightarrow L = {{mv} \\over {\\sqrt 2 }}R\\widehat k$$\n<br> when the particle is moving from D to A."}, {"iden... | null | null | <p>The angular momentum is</p>
<p>$$\overrightarrow L = m(\overrightarrow r \times \overrightarrow v )$$</p>
<p>For the particle moving from $$D \to A$$, we have</p>
<p>$$\overrightarrow L = m\left[ {\left( {{R \over {\sqrt 2 }}\widehat i + {R \over {\sqrt 2 }}\widehat j} \right) \times v( - \widehat j)} \right]$$</... | mcqm | jee-main-2016-offline | 12,435 |
TXtHJA41WkRrlyAFFuWkc | physics | rotational-motion | angular-momentum | A particle of mass 20 g is released with an initial velocity 5 m/s along the curve from the point A, as shown in the figure. The point A is at height h from point B. The particle slides along the frictionless surface. when the particle reaches point b, its angular momentum about O will be :
<br/>(Take g = 10 m/s<sup>2<... | [{"identifier": "A", "content": "6 kg-m<sup>2</sup>/s"}, {"identifier": "B", "content": "8 kg-m<sup>2</sup>/s"}, {"identifier": "C", "content": "2 kg-m<sup>2</sup>/s"}, {"identifier": "D", "content": "3 kg-m<sup>2</sup>/s"}] | ["A"] | null | Work Energy Theorem from A to B
<br><br>Mgh = $${1 \over g}$$ mv$$_B^2$$ $$-$$ $${1 \over g}$$ mv$$_A^2$$
<br><br>2gh = $$v_B^2 - v_A^2$$
<br><br>2 $$ \times $$ 10 $$ \times $$ 10 = v$$_B^2$$ $$-$$ 5<sup>2</sup>
<br><br>v<sub>B</sub> = 15m/s
<br><br>Angular momentum about 0
<br><br>L<sub>0</sub> = mvr
<br><br>= 20 $$ \... | mcq | jee-main-2019-online-12th-january-evening-slot | 12,437 |
n2loNcUjIP1k2lGyy23rsa0w2w9jwzk8ih4 | physics | rotational-motion | angular-momentum | The time dependence of the position of a particle of mass m = 2 is given by $$\overrightarrow r \left( t \right) = 2t\widehat i - 3{t^2}\widehat j$$
. Its angular
momentum, with respect to the origin, at time t = 2 is | [{"identifier": "A", "content": "36 $$\\widehat k$$"}, {"identifier": "B", "content": "- 48 $$\\widehat k$$"}, {"identifier": "C", "content": "$$ - 34\\left( {\\widehat k - \\widehat i} \\right)$$"}, {"identifier": "D", "content": "$$48\\left( {\\widehat i + \\widehat j} \\right)$$"}] | ["B"] | null | $$\overrightarrow v = 2\widehat i - 6 + \widehat j$$<br><br>
At t = 2
<br>
$$\overrightarrow v = 2\widehat i - 12\widehat j$$<br><br>
$$\overrightarrow P = m\overrightarrow v = 4i - 24\widehat j$$<br><br>
At t = 2<br>
$$\overrightarrow r = 4\widehat i - 12\widehat j$$<br><br>
$$\overrightarrow L = \overrightarrow... | mcq | jee-main-2019-online-10th-april-evening-slot | 12,439 |
z1qdq6HCwPf3AM1IfK3rsa0w2w9jwzjh440 | physics | rotational-motion | angular-momentum | A solid sphere of mass M and radius R is divided into two unequal parts. The first part has a mass of $${{7M} \over 8}$$
and is converted into a uniform disc of radius 2R. The second part is converted into a uniform solid sphere.
Let I<sub>1</sub> be the moment of inertia of the disc about its axis and I<sub>2</sub> be... | [{"identifier": "A", "content": "65"}, {"identifier": "B", "content": "140"}, {"identifier": "C", "content": "185"}, {"identifier": "D", "content": "285"}] | ["B"] | null | $${I_1} = {{\left( {{{7M} \over 8}} \right){{\left( {2R} \right)}^2}} \over 2} = {{7M \times 4{R^2}} \over {2 \times 8}} = {{7M{R^2}} \over 4}$$<br><br>
$${I_2} = {2 \over 5}{M \over 8}{\left( {{R \over 2}} \right)^2} = {{2M} \over {5 \times 8}}{{{R^2}} \over 4} = {{M{R^2}} \over {80}}$$<br><br>
$${{{I_1}} \over {{I_2}... | mcq | jee-main-2019-online-10th-april-evening-slot | 12,440 |
OcTjHmAdk2nhPme9embPM | physics | rotational-motion | angular-momentum | A thin smooth rod of length L and mass M is
rotating freely with angular speed $$\omega $$<sub>0</sub> about an
axis perpendicular to the rod and passing
through its center. Two beads of mass m and
negligible size are at the center of the rod
initially. The beads are free to slide along the
rod. The angular speed of th... | [{"identifier": "A", "content": "$${{M{\\omega _0}} \\over {M + 3m}}$$"}, {"identifier": "B", "content": "$${{M{\\omega _0}} \\over {M + m}}$$"}, {"identifier": "C", "content": "$${{M{\\omega _0}} \\over {M + 6m}}$$"}, {"identifier": "D", "content": "$${{M{\\omega _0}} \\over {M + 2m}}$$"}] | ["C"] | null | Initial angular momentum = Final Angular
Momentum<br><br>
$${{M{L^2}} \over {12}}{\omega _0} = \left( {{{M{L^2}} \over {12}} + 2{{m{L^2}} \over 4}} \right)\omega $$<br><br>
$$ \Rightarrow \omega = {{M{\omega _0}} \over {M + 6m}}$$ | mcq | jee-main-2019-online-9th-april-evening-slot | 12,441 |
0jF0u1j47RGMUIQmtSBcR | physics | rotational-motion | angular-momentum | If the angular momentum of a planet of mass m, moving around the Sun in a circular orbit is L, about the center of the Sun, its areal velocity is : | [{"identifier": "A", "content": "$${L \\over m}$$"}, {"identifier": "B", "content": "$${4L \\over m}$$"}, {"identifier": "C", "content": "$${L \\over 2m}$$"}, {"identifier": "D", "content": "$${2L \\over m}$$"}] | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267094/exam_images/truruquwyu57sgraxwbz.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 9th January Morning Slot Physics - Rotational Motion Question 164 English Explanation">
<br><br>... | mcq | jee-main-2019-online-9th-january-morning-slot | 12,444 |
3we6V8BsX7efGEzb5rjgy2xukfrmar8l | physics | rotational-motion | angular-momentum | Four point masses, each of mass m, are fixed at the corners of a square of side $$l$$. The square is
rotating with angular frequency $$\omega $$, about an axis passing through one of the corners of the square
and parallel to its diagonal, as shown in the figure. The angular momentum of the square about this
axis is :
<... | [{"identifier": "A", "content": "3m$$l$$<sup>2</sup>$$\\omega $$"}, {"identifier": "B", "content": "4m$$l$$<sup>2</sup>$$\\omega $$"}, {"identifier": "C", "content": "m$$l$$<sup>2</sup>$$\\omega $$"}, {"identifier": "D", "content": "2m$$l$$<sup>2</sup>$$\\omega $$"}] | ["A"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266172/exam_images/tdo2snbecvkahx95xlug.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 6th September Morning Slot Physics - Rotational Motion Question 107 English Explanation">
<br><br>... | mcq | jee-main-2020-online-6th-september-morning-slot | 12,445 |
ORHXeUlrsLE7ptAwiHjgy2xukfosn6jf | physics | rotational-motion | angular-momentum | A thin rod of mass 0.9 kg and length 1 m is
suspended, at rest, from one end so that it can
freely oscillate in the vertical plane. A particle
of move 0.1 kg moving in a straight line with
velocity 80 m/s hits the rod at its bottom most
point and sticks to it (see figure). The angular
speed (in rad/s) of the rod immedi... | [] | null | 20 | <p>The given situation is shown in the following figure,</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l382zuq5/219ed7a0-1596-406b-a3ac-b74f9bf8be61/409a7fd0-d4bc-11ec-ad28-abe411cf4979/file-1l382zuq6.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l382zuq5/219ed7a0-159... | integer | jee-main-2020-online-5th-september-evening-slot | 12,446 |
bEtWbedAMiaQUiRiUYjgy2xukf258opx | physics | rotational-motion | angular-momentum | A person of 80 kg mass is standing on the rim
of a circular platform of mass 200 kg rotating
about its axis at 5 revolutions per minute (rpm).
The person now starts moving towards the
centre of the platform. What will be the
rotational speed (in rpm) of the platform when
the person reaches its centre _________. | [] | null | 9 | $${I_1}{\omega _1} = {I_2}{\omega _2}$$
<br><br>$$ \Rightarrow $$ $$\left( {{{M{R^2}} \over 2} + m{R^2}} \right){\omega _1} = {{M{R^2}} \over 2}{\omega _2}$$
<br><br>$$ \Rightarrow $$ $$\left( {1 + {{2m{R^2}} \over {M{R^2}}}} \right){\omega _1} = {\omega _2}$$
<br><br>$$ \Rightarrow $$ $$\left( {1 + {{2 \times 80} \ove... | integer | jee-main-2020-online-3rd-september-morning-slot | 12,448 |
Z1v8NC8U95YrjXGxT6jgy2xukexxod6m | physics | rotational-motion | angular-momentum | Two uniform circular discs are rotating
independently in the same direction around
their common axis passing through their
centres. The moment of inertia and angular
velocity of the first disc are 0.1 kg-m<sup>2</sup> and 10
rad s<sup>–1</sup> respectively while those for the second
one are 0.2 kg-m<sup>2</sup> and 5 r... | [{"identifier": "A", "content": "$${{20} \\over 3}J$$"}, {"identifier": "B", "content": "$${{5} \\over 3}J$$"}, {"identifier": "C", "content": "$${{10} \\over 3}J$$"}, {"identifier": "D", "content": "$${{2} \\over 3}J$$"}] | ["A"] | null | Angular momentum conserved for the system
<br><br>I<sub>1</sub>$${\omega _1}$$ + I<sub>2</sub>$${\omega _2}$$ = (I<sub>1</sub> + I<sub>2</sub>)$${\omega _f}$$
<br><br>$$ \Rightarrow $$ 0.1 × 10 + 0.2 × 5 = (0.1 + 0.2) × $${\omega _f}$$
<br><br>$$ \Rightarrow $$ $${\omega _f}$$ = $${{20} \over 3}$$
<br><br>Kinetic energ... | mcq | jee-main-2020-online-2nd-september-evening-slot | 12,449 |
oXr1x48VLDRKi13kts7k9k2k5lbu1m2 | physics | rotational-motion | angular-momentum | A uniformly thick wheel with moment of inertia
I and radius R is free to rotate about its centre
of mass (see fig). A massless string is wrapped
over its rim and two blocks of masses m<sub>1</sub> and
m<sub>2</sub> (m<sub>1</sub> $$ > $$ m<sub>2</sub>) are attached to the ends of the
string. The system is released ... | [{"identifier": "A", "content": "$${\\left[ {{{2\\left( {{m_1} + {m_2}} \\right)gh} \\over {\\left( {{m_1} + {m_2}} \\right){R^2} + I}}} \\right]^{{1 \\over 2}}}$$"}, {"identifier": "B", "content": "$${\\left[ {{{{m_1} + {m_2}} \\over {\\left( {{m_1} + {m_2}} \\right){R^2} + I}}} \\right]^{{1 \\over 2}}}gh$$"}, {"ident... | ["D"] | null | By conservation of energy :
<br><br>Loss in energy = Gain in enrgy
<br><br>m<sub>1</sub>gh = m<sub>2</sub>gh + $${1 \over 2}$$ m<sub>1</sub>V<sup>2</sup> + $${1 \over 2}$$ m<sub>2</sub>V<sup>2</sup> + $${1 \over 2}I{\omega ^2}$$
<br><br> $$ \Rightarrow $$ (m<sub>1</sub> – m<sub>2</sub>)gh = $${1 \over 2}$$ (m<sub>1</su... | mcq | jee-main-2020-online-9th-january-evening-slot | 12,450 |
xgguUJAR2MMkuQ4aev7k9k2k5gveqgd | physics | rotational-motion | angular-momentum | Consider a uniform rod of mass M = 4m and
length $$\ell $$ pivoted about its centre. A mass m
moving with velocity v making angle $$\theta = {\pi \over 4}$$ to
the rod's long axis collides with one end of the
rod and sticks to it. The angular speed of the
rod-mass system just after the collision is : | [{"identifier": "A", "content": "$${{3\\sqrt 2 } \\over 7}{v \\over \\ell }$$"}, {"identifier": "B", "content": "$${3 \\over 7}{v \\over \\ell }$$"}, {"identifier": "C", "content": "$${3 \\over {7\\sqrt 2 }}{v \\over \\ell }$$"}, {"identifier": "D", "content": "$${4 \\over 7}{v \\over \\ell }$$"}] | ["A"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263502/exam_images/ft4efq5vaelxd7r62j1c.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 8th January Morning Slot Physics - Rotational Motion Question 128 English Explanation">
<br><br>Ab... | mcq | jee-main-2020-online-8th-january-morning-slot | 12,451 |
BE7hgZymN0E2d9uhFNjgy2xukf6jrqfg | physics | rotational-motion | angular-momentum | A circular disc of mass M and radius R is rotating about its axis with angular speed $${\omega _1}$$
. If another
stationary disc having radius $${R \over 2}$$ and same mass M is droped co-axially on to the rotating disc.
Gradually both discs attain constant angular speed $${\omega _2}$$
the energy lost in the process... | [] | null | 20 | $${I_f}{\omega _f} = {I_i}{\omega _i}$$<br><br>$${I_i} = {{M{R^2}} \over 2}$$<br><br>$${I_f} = {{M{R^2}} \over 2} + {{M{{(R/2)}^2}} \over 2}$$<br><br>$$ = {5 \over 4}.{{M{R^2}} \over 2}$$<br><br>$$\left[ {{{M{R^2}} \over 2} + {M \over 2}{{\left( {{R \over 2}} \right)}^2}} \right]\omega ' = \left( {{{M{R^2}} \over 2}} \... | integer | jee-main-2020-online-4th-september-morning-slot | 12,452 |
WuHXW7gUgdMYj1BN9V1kluncx9i | physics | rotational-motion | angular-momentum | A cord is wound round the circumference of wheel of radius r. The axis of the wheel is horizontal and the moment of inertia about it is I. A weight mg is attached to the cord at the end. The weight falls from rest. After falling through a distance 'h', the square of angular velocity of wheel will be : | [{"identifier": "A", "content": "$${{2mgh} \\over {I + 2m{r^2}}}$$"}, {"identifier": "B", "content": "$${{2mgh} \\over {I + m{r^2}}}$$"}, {"identifier": "C", "content": "2gh"}, {"identifier": "D", "content": "$${{2gh} \\over {I + m{r^2}}}$$"}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266117/exam_images/svzalj1syrkpymtfs8gj.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 26th February Evening Shift Physics - Rotational Motion Question 100 English Explanation">
<br><br... | mcq | jee-main-2021-online-26th-february-evening-slot | 12,453 |
hUSqhRBf0fDVvJ6Kcr1kmj1x5wy | physics | rotational-motion | angular-momentum | A mass M hangs on a massless rod of length l which rotates at a constant angular frequency. The mass M moves with steady speed in a circular path of constant radius. Assume that the system is in steady circular motion with constant angular velocity $$\omega$$. The angular momentum of M about point A is L<sub>A</sub> wh... | [{"identifier": "A", "content": "L<sub>A</sub> is constant, both in magnitude and direction"}, {"identifier": "B", "content": "L<sub>B</sub> is constant in direction with varying magnitude"}, {"identifier": "C", "content": "L<sub>B</sub> is constant, both in magnitude and direction"}, {"identifier": "D", "content": "L<... | ["A"] | null | Net force on M is towards A, hence torque is zero
about A.
<br><br>$${\overrightarrow L _A} = \overrightarrow r \times \overrightarrow p $$<br><br>as $$r \bot p$$ so L<sub>A</sub> = constant | mcq | jee-main-2021-online-17th-march-morning-shift | 12,454 |
IerYEirovFn3oJiZwY1kmkr269n | physics | rotational-motion | angular-momentum | A thin circular ring of mass M and radius r is rotating about its axis with an angular speed $$\omega$$. Two particles having mass m each are now attached at diametrically opposite points. The angular speed of the ring will become : | [{"identifier": "A", "content": "$$\\omega {M \\over {M + m}}$$"}, {"identifier": "B", "content": "$$\\omega {{M + 2m} \\over M}$$"}, {"identifier": "C", "content": "$$\\omega {M \\over {M + 2m}}$$"}, {"identifier": "D", "content": "$$\\omega {{M - 2m} \\over {M + 2m}}$$"}] | ["C"] | null | $$\tau$$<sub>net</sub> = 0, so angular momentum is conserved<br><br>By angular momentum conservation<br><br>I<sub>i</sub>$$\omega$$<sub>i</sub> = I<sub>f</sub>$$\omega$$<sub>f</sub><br><br>(MR<sup>2</sup>)$$\omega$$ = (MR<sup>2</sup> + 2mR<sup>2</sup>)$$\omega$$<sub>f</sub><br><br>$$\omega$$<sub>f</sub> = $${{(M{R^2})\... | mcq | jee-main-2021-online-18th-march-morning-shift | 12,455 |
1krunysm5 | physics | rotational-motion | angular-momentum | A particle of mass 'm' is moving in time 't' on a trajectory given by<br/><br/>$$\overrightarrow r = 10\alpha {t^2}\widehat i + 5\beta (t - 5)\widehat j$$<br/><br/>Where $$\alpha$$ and $$\beta$$ are dimensional constants.<br/><br/>The angular momentum of the particle becomes the same as it was for t = 0 at time t = __... | [] | null | 10 | $$\overrightarrow r = 10\alpha {t^2}\widehat i + 5\beta (t - 5)\widehat j$$<br><br>$$\overrightarrow v = 20\alpha t\widehat i + 5\beta \widehat j$$<br><br>$$\overrightarrow L = m(\overrightarrow r \times \overrightarrow v )$$<br><br>$$ = m[10\alpha {t^2}\widehat i + 5\beta (t - 5)\widehat j] \times [20\alpha t\wide... | integer | jee-main-2021-online-25th-july-morning-shift | 12,456 |
1ktfjwrjv | physics | rotational-motion | angular-momentum | Two discs have moments of inertia I<sub>1</sub> and I<sub>2</sub> about their respective axes perpendicular to the plane and passing through the centre. They are rotating with angular speeds, $$\omega$$<sub>1</sub> and $$\omega$$<sub>2</sub> respectively and are brought into contact face to face with their axes of rota... | [{"identifier": "A", "content": "$${{{I_1}{I_2}} \\over {({I_1} + {I_2})}}{({\\omega _1} - {\\omega _2})^2}$$"}, {"identifier": "B", "content": "$${{{{({I_1} - {I_2})}^2}{\\omega _1}{\\omega _2}} \\over {2({I_1} + {I_2})}}$$"}, {"identifier": "C", "content": "$${{{I_1}{I_2}} \\over {2({I_1} + {I_2})}}{({\\omega _1} - {... | ["C"] | null | From conservation of angular momentum we get<br><br>$${I_1}{\omega _1} + {I_2}{\omega _2} = ({I_1} + {I_2})\omega $$<br><br>$$\omega = {{{I_1}{\omega _1} + {I_2}{\omega _2}} \over {{I_1} + {I_2}}}$$<br><br>$${k_i} = {1 \over 2}{I_1}\omega _1^2 + {1 \over 2}{I_2}\omega _2^2$$<br><br>$${k_f} = {1 \over 2}({I_1} + {I_2})... | mcq | jee-main-2021-online-27th-august-evening-shift | 12,457 |
1kth5aj8v | physics | rotational-motion | angular-momentum | Angular momentum of a single particle moving with constant speed along circular path : | [{"identifier": "A", "content": "changes in magnitude but remains same in the direction"}, {"identifier": "B", "content": "remains same in magnitude and direction"}, {"identifier": "C", "content": "remains same in magnitude but changes in the direction"}, {"identifier": "D", "content": "is zero"}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263868/exam_images/mm2wetdpg7wzm8abx2gd.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 31st August Morning Shift Physics - Rotational Motion Question 71 English Explanation"><br>$$\left... | mcq | jee-main-2021-online-31st-august-morning-shift | 12,458 |
1ktmwpjl2 | physics | rotational-motion | angular-momentum | A 2 kg steel rod of length 0.6 m is clamped on a table vertically at its lower end and is free to rotate in vertical plane. The upper end is pushed so that the rod falls under gravity, ignoring the friction due to clamping at its lower end, the speed of the free end of rod when it passes through its lowest position is ... | [] | null | 6 | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1kwosrtyn/5bdd1354-2542-4992-8ca1-578d08ae9c6d/15a2eff0-5358-11ec-b443-85f16d0c41b6/file-1kwosrtyo.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1kwosrtyn/5bdd1354-2542-4992-8ca1-578d08ae9c6d/15a2eff0-5358-11ec-b443-85f16d0c41b6/fi... | integer | jee-main-2021-online-1st-september-evening-shift | 12,459 |
1l58bgm1n | physics | rotational-motion | angular-momentum | <p>A thin circular ring of mass M and radius R is rotating with a constant angular velocity 2 rads<sup>$$-$$1</sup> in a horizontal plane about an axis vertical to its plane and passing through the center of the ring. If two
objects each of mass m be attached gently to the
opposite ends of a diameter of ring, the ring ... | [{"identifier": "A", "content": "$${M \\over {(M + m)}}$$"}, {"identifier": "B", "content": "$${{(M + 2m)} \\over {2M}}$$"}, {"identifier": "C", "content": "$${{2M} \\over {(M + 2m)}}$$"}, {"identifier": "D", "content": "$${{2(M + 2m)} \\over M}$$"}] | ["C"] | null | <p>$${I_1}{\omega _1} = {I_2}{\omega _2}$$</p>
<p>$$M{R^2}{\omega _1} = (M{R^2} + 2m{R^2}){\omega _2}$$</p>
<p>$${\omega _2} = \left( {{M \over {M + 2m}}} \right){\omega _1}$$</p>
<p>$${\omega _2} = 2\left( {{M \over {M + 2m}}} \right)$$</p> | mcq | jee-main-2022-online-26th-june-morning-shift | 12,461 |
1lgsxfqv2 | physics | rotational-motion | angular-momentum | <p>A circular plate is rotating in horizontal plane, about an axis passing through its center and perpendicular to the plate, with an angular velocity $$\omega$$. A person sits at the center having two dumbbells in his hands. When he stretches out his hands, the moment of inertia of the system becomes triple. If E be t... | [] | null | 3 |
The conservation of angular momentum states that the angular momentum (L) remains constant. The relation between kinetic energy (KE), angular momentum (L), and moment of inertia (I) is given by:
<br/><br/>
$$
\mathrm{KE}=\frac{\mathrm{L}^2}{2 \mathrm{I}}
$$
<br/><br/>
Using this relation, we can find the ratio of the ... | integer | jee-main-2023-online-11th-april-evening-shift | 12,463 |
1lguyo8ek | physics | rotational-motion | angular-momentum | <p>A solid sphere of mass $$500 \mathrm{~g}$$ and radius $$5 \mathrm{~cm}$$ is rotated about one of its diameter with angular speed of $$10 ~\mathrm{rad} ~\mathrm{s}^{-1}$$. If the moment of inertia of the sphere about its tangent is $$x \times 10^{-2}$$ times its angular momentum about the diameter. Then the value of ... | [] | null | 35 | $$
\begin{aligned}
& L_{\text {diameter }}=\frac{2}{5} M R^2 \omega ; \quad I_{\text {tangent }}=\frac{7}{5} M R^2 \\\\
& \begin{aligned}
\frac{I_{\text {tangent }}}{L_{\text {diameter }}} & =\frac{7 / 5}{2 / 5} \times \frac{1}{\omega}=\frac{7}{2 \omega} \\\\
& =\frac{7}{2 \times 10}=\frac{7}{20} \\\\
&= \frac{700}{20}... | integer | jee-main-2023-online-11th-april-morning-shift | 12,464 |
1lgvr89zh | physics | rotational-motion | angular-momentum | <p>Given below are two statements: one is labelled as Assertion $$\mathbf{A}$$ and the other is labelled as Reason $$\mathbf{R}$$</p>
<p>Assertion A : An electric fan continues to rotate for some time after the current is switched off.</p>
<p>Reason R : Fan continues to rotate due to inertia of motion.</p>
<p>In the li... | [{"identifier": "A", "content": "A is not correct but R is correct"}, {"identifier": "B", "content": "A is correct but R is not correct"}, {"identifier": "C", "content": "Both A and R are correct and R is the correct explanation of A"}, {"identifier": "D", "content": "Both A and R are correct but R is NOT the correct e... | ["C"] | null | <p>The correct answer is Both A and R are correct and R is the correct explanation of A.</p>
<p>Explanation:</p>
<p><b>Assertion A</b>: An electric fan continues to rotate for some time after the current is switched off. This is a correct statement. When you switch off the fan, it doesn't stop immediately but conti... | mcq | jee-main-2023-online-10th-april-evening-shift | 12,465 |
lsanle50 | physics | rotational-motion | angular-momentum | A uniform rod $A B$ of mass $2 \mathrm{~kg}$ and length $30 \mathrm{~cm}$ at rest on a smooth horizontal surface. An impulse of force $0.2 \mathrm{~Ns}$ is applied to end B. The time taken by the rod to turn through at right angles will be $\frac{\pi}{x} \mathrm{~s}$, where $x=$ _______ . | [] | null | 4 | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsbzsat7/6c6f871c-43eb-4810-bb22-d919a27dceee/3e3ce1a0-c5d4-11ee-927b-7ff3c4d3b955/file-6y3zli1lsbzsat8.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsbzsat7/6c6f871c-43eb-4810-bb22-d919a27dceee/3e3ce1a0-c5d4-11ee-92... | integer | jee-main-2024-online-1st-february-evening-shift | 12,466 |
jaoe38c1lsd8digo | physics | rotational-motion | angular-momentum | <p>A body of mass '$$m$$' is projected with a speed '$$u$$' making an angle of $$45^{\circ}$$ with the ground. The angular momentum of the body about the point of projection, at the highest point is expressed as $$\frac{\sqrt{2} m u^3}{X g}$$. The value of '$$X$$' is _________.</p> | [] | null | 8 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsiidqey/7faf5480-3b36-4be9-87f6-9e90f84e1ff3/aa8165a0-c969-11ee-b416-eff853096672/file-6y3zli1lsiidqez.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsiidqey/7faf5480-3b36-4be9-87f6-9e90f84e1ff3/aa8165a0-c969-11ee... | integer | jee-main-2024-online-31st-january-evening-shift | 12,467 |
jaoe38c1lsfm7qam | physics | rotational-motion | angular-momentum | <p>A body of mass $$5 \mathrm{~kg}$$ moving with a uniform speed $$3 \sqrt{2} \mathrm{~ms}^{-1}$$ in $$X-Y$$ plane along the line $$y=x+4$$. The angular momentum of the particle about the origin will be _________ $$\mathrm{kg} \mathrm{~m}^2 \mathrm{~s}^{-1}$$.</p> | [] | null | 60 | <p>$$y-x-4=0$$</p>
<p>$$d_1$$ is perpendicular distance of given line from origin.</p>
<p>$$\mathrm{d}_1=\left|\frac{-4}{\sqrt{1^2+1^2}}\right| \Rightarrow 2 \sqrt{2} \mathrm{~m}$$</p>
<p>So</p>
<p>$$\begin{aligned}
|\overrightarrow{\mathrm{L}}|=\mathrm{mvd}_1 & =5 \times 3 \sqrt{2} \times 2 \sqrt{2} \mathrm{~kg} \math... | integer | jee-main-2024-online-29th-january-evening-shift | 12,468 |
1lsg70wif | physics | rotational-motion | angular-momentum | <p>Two discs of moment of inertia $$I_1=4 \mathrm{~kg} \mathrm{~m}^2$$ and $$I_2=2 \mathrm{~kg} \mathrm{~m}^2$$, about their central axes & normal to their planes, rotating with angular speeds $$10 \mathrm{~rad} / \mathrm{s}$$ & $$4 \mathrm{~rad} / \mathrm{s}$$ respectively are brought into contact face to face... | [] | null | 24 | <p>To find the loss in kinetic energy when two spinning discs are brought together, we use the principle of conservation of angular momentum and the formula for kinetic energy. Here's how:</p><p>First, because angular momentum before and after they touch must be the same, we have:</p><p>$$I_1 \omega_1 + I_2 \omega_2 = ... | integer | jee-main-2024-online-30th-january-evening-shift | 12,469 |
1lsgd0f2p | physics | rotational-motion | angular-momentum | <p>A particle of mass $$\mathrm{m}$$ is projected with a velocity '$$\mathrm{u}$$' making an angle of $$30^{\circ}$$ with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height $$\mathrm{h}$$ is :</p> | [{"identifier": "A", "content": "$$\\frac{\\mathrm{mu}^3}{\\sqrt{2} \\mathrm{~g}}$$\n"}, {"identifier": "B", "content": "zero\n"}, {"identifier": "C", "content": "$$\\frac{\\sqrt{3}}{2} \\frac{\\mathrm{mu}^2}{\\mathrm{~g}}$$\n"}, {"identifier": "D", "content": "$$\\frac{\\sqrt{3}}{16} \\frac{\\mathrm{mu}^3}{\\mathrm{~g... | ["D"] | null | <p>$$\begin{aligned}
& \mathrm{L}=m u \cos \theta H \\
& =m u \cos \theta \times \frac{u^2 \sin ^2 \theta}{2 g} \\
& =\frac{m u^3}{2 g} \times \frac{\sqrt{3}}{2} \times\left(\frac{1}{2}\right)^2=\frac{\sqrt{3} m u^3}{16 g}
\end{aligned}$$</p> | mcq | jee-main-2024-online-30th-january-morning-shift | 12,470 |
lv3vefxa | physics | rotational-motion | angular-momentum | <p>A thin circular disc of mass $$\mathrm{M}$$ and radius $$\mathrm{R}$$ is rotating in a horizontal plane about an axis passing through its centre and perpendicular to its plane with angular velocity $$\omega$$. If another disc of same dimensions but of mass $$\mathrm{M} / 2$$ is placed gently on the first disc co-axi... | [{"identifier": "A", "content": "$$\\frac{4}{5} \\omega$$\n"}, {"identifier": "B", "content": "$$\\frac{5}{4} \\omega$$\n"}, {"identifier": "C", "content": "$$\\frac{3}{2} \\omega$$\n"}, {"identifier": "D", "content": "$$\\frac{2}{3} \\omega$$"}] | ["D"] | null | <p>To determine the new angular velocity of the system, we use the principle of conservation of angular momentum. When no external torque acts on a system, its angular momentum remains constant. Let's denote the initial angular momentum and the final angular momentum, respectively, as $$L_{\text{initial}}$$ and $$L_{\t... | mcq | jee-main-2024-online-8th-april-evening-shift | 12,471 |
atNumTaIjE3kMyJr | physics | rotational-motion | combined-translational-and-rotational-motion | A solid sphere, a hollow sphere and a ring are released from top of an inclined plane (frictionless) so that they slide down the plane. Then maximum acceleration down the plane is for (no rolling) | [{"identifier": "A", "content": "solid sphere "}, {"identifier": "B", "content": "hollow sphere "}, {"identifier": "C", "content": "ring "}, {"identifier": "D", "content": "all same "}] | ["D"] | null | Each bodies is sliding along the frictionless inclined plane and there is no rolling, therefore the acceleration of all the bodies is same $$\left( {g\,\sin \,\theta } \right).$$ | mcq | aieee-2002 | 12,472 |
nbiRKpbN7QLxHsp0 | physics | rotational-motion | combined-translational-and-rotational-motion | An annular ring with inner and outer radii $${R_1}$$ and $${R_2}$$ is rolling without slipping with a uniform angular speed. The ratio of the forces experienced by the two particles situated on the inner and outer parts of the ring, $${{{F_1}} \over {{F_2}}}\,$$ is | [{"identifier": "A", "content": "$${\\left( {{{{R_1}} \\over {{R_2}}}} \\right)^2}$$ "}, {"identifier": "B", "content": "$${{{{R_2}} \\over {{R_1}}}}$$"}, {"identifier": "C", "content": "$${{{{R_1}} \\over {{R_2}}}}$$ "}, {"identifier": "D", "content": "$$1$$"}] | ["C"] | null | Let the mass of each particle is m.
<br><br>Then force experienced by each particle, $$F = m{\omega ^2}R$$
<br><br>$$\therefore$$ $${{{F_1}} \over {{F_2}}} = {{m{\omega ^2}{R_1}} \over {m{\omega ^2}{R_2}}}$$
<br><br>$$ \Rightarrow $$ $${{{F_1}} \over {{F_2}}} = {{{R_1}} \over {{R_2}}}$$ | mcq | aieee-2005 | 12,473 |
KQJHwKrZeCnzHTDE | physics | rotational-motion | combined-translational-and-rotational-motion | A round uniform body of radius $$R,$$ mass $$M$$ and moment of inertia $$I$$ rolls down (without slipping) an inclined plane making an angle $$\theta $$ with the horizontal. Then its acceleration is | [{"identifier": "A", "content": "$${{g\\,\\sin \\theta } \\over {1 - M{R^2}/I}}$$ "}, {"identifier": "B", "content": "$${{g\\,\\sin \\theta } \\over {1 + I/M{R^2}}}$$ "}, {"identifier": "C", "content": "$${{g\\,\\sin \\theta } \\over {1 + M{R^2}/I}}$$ "}, {"identifier": "D", "content": "$${{g\\,\\sin \\theta } \\over {... | ["B"] | null | A uniform body of radius R, mass M and moment of inertia $$I$$
rolls down (without slipping) an inclined plane making an angle
θ with the horizontal. Then its acceleration is
<br><br>$$a = {{g\,\sin \,\theta } \over {1 + {I \over {M{R^2}}}}}$$ | mcq | aieee-2007 | 12,474 |
SkXe9OycfhSpUpyH | physics | rotational-motion | combined-translational-and-rotational-motion | A mass $$m$$ hangs with the help of a string wrapped around a pulley on a frictionless bearing. The pulley has mass $$m$$ and radius $$R.$$ Assuming pulley to be a perfect uniform circular disc, the acceleration of the mass $$m,$$ if the string does not slip on the pulley, is: | [{"identifier": "A", "content": "$$g$$ "}, {"identifier": "B", "content": "$${2 \\over 3}g$$ "}, {"identifier": "C", "content": "$${g \\over 3}$$ "}, {"identifier": "D", "content": "$${3 \\over 2}g$$ "}] | ["B"] | null | This is the free body diagram of pulley and mass
<img class="question-image" src="https://imagex.cdn.examgoal.net/YLozan7fjzZKlytst/ShmwGB0AEPkBmDOL837pXdxfXW0Nm/BXNtjtvWkUAO42ol24rvuO/image.svg" loading="lazy" alt="AIEEE 2011 Physics - Rotational Motion Question 194 English Explanation">
<br>For translation motion of ... | mcq | aieee-2011 | 12,475 |
5tyvIw22Ag2R0T7I22GXL | physics | rotational-motion | combined-translational-and-rotational-motion | The machine as shown has 2 rods of length1 m connected by a pivot at the top. The end of one rod is connected to the floor by a stationary pivot and the end of the other rod has a roller that rolls along the floor in a slot. As the roller goes back and forth, a 2 kg weight moves up and down. If the roller is moving to... | [{"identifier": "A", "content": "Constant speed "}, {"identifier": "B", "content": "decreasing speed "}, {"identifier": "C", "content": "increasing speed "}, {"identifier": "D", "content": "speed which is $${3 \\over 4}$$th of that of the roller when the weight is 0.4 m above the ground\n"}] | ["B"] | null | As the force F acts on the horizontal direction, the vertical force acting on the rod to raise the weight to move upwards happens with decreasing speed. | mcq | jee-main-2017-online-9th-april-morning-slot | 12,477 |
99IHjaK1GaolNtQJcbaLI | physics | rotational-motion | combined-translational-and-rotational-motion | A rod of length 50 cm is pivoted at one end. It is raised such that if makes an angle of 30<sup>o</sup> from the horizontal as shown and released from rest. Its angular speed when it passes through the horizontal (in rad s<sup>$$-$$1</sup>) will be (g = 10 ms<sup>$$-$$2</sup>)
<br/><br/><img src="data:image/png;base64... | [{"identifier": "A", "content": "$$\\sqrt {{{30} \\over 7}} $$"}, {"identifier": "B", "content": "$$\\sqrt {30} $$"}, {"identifier": "C", "content": "$${{\\sqrt {20} } \\over 3}$$"}, {"identifier": "D", "content": "$${{\\sqrt {30} } \\over 2}$$"}] | ["B"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l8p79v5n/51b3796a-b101-4142-b1f5-7e7a38ee632a/637a8cb0-4123-11ed-8a36-9f4a474c8481/file-1l8p79v5o.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l8p79v5n/51b3796a-b101-4142-b1f5-7e7a38ee632a/637a8cb0-4123-11ed-8a36-9f4a474c8481/fi... | mcq | jee-main-2019-online-9th-january-evening-slot | 12,478 |
ZOIYD1k9JaLRduLLgyo8B | physics | rotational-motion | combined-translational-and-rotational-motion | A homogeneous solid cylindrical roller of radius R and mass M is pulled on a cricket pitch by a horizontal force. Assuming rolling without slipping, angular acceleration of the cylinder is -
| [{"identifier": "A", "content": "$${F \\over {2mR}}$$"}, {"identifier": "B", "content": "$${2F \\over {3mR}}$$"}, {"identifier": "C", "content": "$${3F \\over {2mR}}$$"}, {"identifier": "D", "content": "$${F \\over {3mR}}$$"}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264318/exam_images/q3abepnq1xiuh5fhgfgm.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 10th January Morning Slot Physics - Rotational Motion Question 161 English Explanation">
<br><br... | mcq | jee-main-2019-online-10th-january-morning-slot | 12,479 |
TiZmOQnPzFnqfLMyGc5uA | physics | rotational-motion | combined-translational-and-rotational-motion | The following bodies are made to roll up
(without slipping) the same inclined plane from
a horizontal plane. : (i) a ring of radius R, (ii)
a solid cylinder of radius
R/2 and (iii) a solid
sphere of radius
R/4 . If in each case, the speed
of the centre of mass at the bottom of the incline
is same, the ratio of the maxi... | [{"identifier": "A", "content": "20 : 15 : 14"}, {"identifier": "B", "content": "4 : 3 : 2"}, {"identifier": "C", "content": "2 : 3 : 4"}, {"identifier": "D", "content": "10 : 15 : 7"}] | ["A"] | null | Total kinetic energy of a rolling body is given as
<br/><br/>$$
E_{\text {total }}=\frac{1}{2} m v^2\left[1+\frac{K^2}{R^2}\right]
$$
<br/><br/>where, $K$ is the radius of gyration.
<br/><br/>Using conservation law of energy,
<br/><br/>$$
\begin{array}{rlrl}
\frac{1}{2} m v^2\left[1+\frac{K^2}{R^2}\right] =m g h \\\\
... | mcq | jee-main-2019-online-9th-april-morning-slot | 12,482 |
lbTUQbol9VveYn7cut7k9k2k5dmi7d7 | physics | rotational-motion | combined-translational-and-rotational-motion | <img src="data:image/png;base64,UklGRnQIAABXRUJQVlA4IGgIAAAQRwCdASpIAQIBPm00mEkkIqKhIZD5kIANiWlu/HyZM70HZ1w/ox/U/5t2//3b+oftF1+Hjr2c9Mv4T6cL4o7nL5v+Jfz337/kn+f/lX9F/1/Ac/zn8W/rH6henf9N/6NsKv0L/sXji/2f8c/m/jL8RPzmf0A84eNz9K/Uc+Cv7R+pPvd+cP+//E/d1/p36J/3X9MPk4///8AEseM0oKegeM0oKegeM0oKegeM0oKeN4Ia/Illg+6q+czq5Ubs4aiziGm2... | [{"identifier": "A", "content": "$$r\\sqrt {{3 \\over {2gh}}} $$"}, {"identifier": "B", "content": "$$r\\sqrt {{3 \\over {4gh}}} $$"}, {"identifier": "C", "content": "$${1 \\over r}\\sqrt {{{4gh} \\over 3}} $$"}, {"identifier": "D", "content": "$${1 \\over r}\\sqrt {{{2gh} \\over 3}} $$"}] | ["C"] | null | mgh = $$\Delta $$KE
<br><br>= $${1 \over 2}m{v^2} + {1 \over 2}I{\omega ^2}$$
<br><br>For no slipping, v = $$\omega $$R
<br><br>$$ \therefore $$ mgh = $${1 \over 2}m{\omega ^2}{R^2} + {1 \over 2}{{m{R^2}} \over 2}{\omega ^2}$$
<br><br>$$ \Rightarrow $$ mgh = $${3 \over 4}m{\omega ^2}{R^2}$$
<br><br>$$ \Rightarrow $$ $$... | mcq | jee-main-2020-online-7th-january-morning-slot | 12,483 |
9jCyQhEAs6eWjkTr4Q7k9k2k5hhldlt | physics | rotational-motion | combined-translational-and-rotational-motion | A uniform sphere of mass 500 g rolls without
slipping on a plane horizontal surface with its
centre moving at a speed of 5.00 cm/s. Its
kinetic energy is : | [{"identifier": "A", "content": "8.75 \u00d7 10<sup>\u20133</sup> J"}, {"identifier": "B", "content": "1.13 \u00d7 10<sup>\u20133</sup> J"}, {"identifier": "C", "content": "8.75 \u00d7 10<sup>\u20134</sup> J"}, {"identifier": "D", "content": "6.25 \u00d7 10<sup>\u20134</sup> J"}] | ["C"] | null | K.E = $${1 \over 2}m{V^2} + {1 \over 2}{I_{cm}}{\omega ^2}$$
<br><br>= $${1 \over 2}m{V^2} + {1 \over 2} \times {2 \over 5}m{R^2} \times {{{V^2}} \over {{R^2}}}$$
<br><br>= $${1 \over 2}m{V^2} + {1 \over 5}m{V^2}$$
<br><br>= $${7 \over {10}}m{V^2}$$
<br><br>= $${7 \over {10}} \times 0.5 \times 25 \times {10^{ - 4}}$$
<... | mcq | jee-main-2020-online-8th-january-evening-slot | 12,484 |
JAFuoJFw7XQ8Wym02a1klt2buwh | physics | rotational-motion | combined-translational-and-rotational-motion | A sphere of radius 'a' and mass 'm' rolls along a horizontal plane with constant speed v<sub>0</sub>. It encounters an inclined plane at angle $$\theta$$ and climbs upward. Assuming that it rolls without slipping, how far up the sphere will travel?<br/><br/><img src="data:image/png;base64,UklGRugGAABXRUJQVlA4INwGAACwLQ... | [{"identifier": "A", "content": "$${{v_0^2} \\over {2g\\sin \\theta }}$$"}, {"identifier": "B", "content": "$${{7v_0^2} \\over {10g\\sin \\theta }}$$"}, {"identifier": "C", "content": "$${2 \\over 5}{{v_0^2} \\over {g\\sin \\theta }}$$"}, {"identifier": "D", "content": "$${{v_0^2} \\over {5g\\sin \\theta }}$$"}] | ["B"] | null | <picture><source media="(max-width: 907px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264712/exam_images/k8tujbunomsdtxrkoaza.webp"><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267069/exam_images/l2vqb7prslan4qwpptnn.webp"><source media="(max-wid... | mcq | jee-main-2021-online-25th-february-evening-slot | 12,485 |
Ib0wrCdscd7UvETY741kmipq0om | physics | rotational-motion | combined-translational-and-rotational-motion | A solid disc of radius 'a' and mass 'm' rolls down without slipping on an inclined plane making an angle $$\theta$$ with the horizontal. The acceleration of the disc will be $${2 \over b}$$g sin$$\theta$$ where b is ____________. (Round off to the Nearest Integer) (g = acceleration due to gravity, $$\theta$$ = angle as... | [] | null | 3 | <p>We know that, on an inclined plane</p>
<p>Acceleration, $$a = {{g\sin \theta } \over {1 + {I \over {m{R^2}}}}}$$</p>
<p>$$ \Rightarrow a = {{g\sin \theta } \over {1 + {1 \over 2}}}$$ [$$\because$$ For disc, $$I = {{m{R^2}} \over 2}$$]</p>
<p>$$ \Rightarrow a = {2 \over 3}g\sin \theta $$ ....... (i)</p>
<p>As per que... | integer | jee-main-2021-online-16th-march-evening-shift | 12,486 |
16kFcexlMjLkKgTlXA1kmj3ys5s | physics | rotational-motion | combined-translational-and-rotational-motion | The angular speed of truck wheel is increased from 900 rpm to 2460 rpm in 26 seconds. The number of revolutions by the truck engine during this time is _____________. (Assuming the acceleration to be uniform). | [] | null | 728 | $${\omega _f} = 2460 \times {{2\pi } \over {60}}$$<br><br>$$ = 82\pi $$<br><br>$${\omega _i} = {{900 \times 2\pi } \over {60}} = 30\pi $$<br><br>$$\alpha = {{{\omega _f} - {\omega _i}} \over t}$$<br><br>$$ = {{82\pi - 30\pi } \over {26}}$$<br><br>= 2 $$\pi$$ rad/sec<sup>2</sup><br><br>$$\theta = {{\omega _f^2 - \ome... | integer | jee-main-2021-online-17th-march-morning-shift | 12,487 |
S9wqyyoVVI3UqM0yF51kmj42zit | physics | rotational-motion | combined-translational-and-rotational-motion | The following bodies,<br/><br/>(1) a ring<br/><br/>(2) a disc<br/><br/>(3) a solid cylinder<br/><br/>(4) a solid sphere,<br/><br/>of same mass 'm' and radius 'R' are allowed to roll down without slipping simultaneously from the top of the inclined plane. The body which will reach first at the bottom of the inclined pla... | [] | null | 4 | $$a = {{g\sin \theta } \over {\left( {1 + {I \over {m{R^2}}}} \right)}}$$<br><br>I<sub>R</sub> = mR<sup>2</sup>, a<sub>R</sub> = g sin$$\theta$$/2<br><br>I<sub>D</sub> = $${{m{R^2}} \over 2}$$, a<sub>D</sub> = $${2 \over 3}$$ g sin$$\theta$$<br><br>I<sub>SC</sub> = $${{m{R^2}} \over 2}$$, a<sub>SC</sub> = $${2 \over 3}... | integer | jee-main-2021-online-17th-march-morning-shift | 12,488 |
1krpq8sv4 | physics | rotational-motion | combined-translational-and-rotational-motion | A circular disc reaches from top to bottom of an inclined plane of length 'L'. When it slips down the plane, it makes time 't<sub>1</sub>'. When it rolls down the plane, it takes time t<sub>2</sub>. The value of $${{{t_2}} \over {{t_1}}}$$ is $$\sqrt {{3 \over x}} $$. The value of x will be _______________. | [] | null | 2 | According to question, a circular disc reaches from top to bottom of an inclined plane of length L. This can be shown as <br><br> <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1kypnoi2v/a523ee22-f54c-4fa0-91c9-50dd1296647e/62ca9970-7b69-11ec-92f3-61d366101705/file-1kypnoi2w.png?format=png" data-orsrc=... | integer | jee-main-2021-online-20th-july-morning-shift | 12,490 |
1krqchuow | physics | rotational-motion | combined-translational-and-rotational-motion | A body rolls down an inclined plane without slipping. The kinetic energy of rotation is 50% of its translational kinetic energy. The body is : | [{"identifier": "A", "content": "Solid sphere"}, {"identifier": "B", "content": "Solid cylinder"}, {"identifier": "C", "content": "Hollow cylinder"}, {"identifier": "D", "content": "Ring"}] | ["B"] | null | $${1 \over 2}I{\omega ^2} = {1 \over 2} \times {1 \over 2}m{v^2}$$<br><br>$$I = {1 \over 2}m{R^2}$$<br><br>Body is solid cylinder | mcq | jee-main-2021-online-20th-july-evening-shift | 12,491 |
1krqfvxc1 | physics | rotational-motion | combined-translational-and-rotational-motion | A body rotating with an angular speed of 600 rpm is uniformly accelerated to 1800 rpm in 10 sec. The number of rotations made in the process is ___________. | [] | null | 200 | $${\omega _f} = {\omega _0} + \alpha t$$<br><br>$$\alpha = 1200 \times 6$$<br><br>$$\theta = {\omega _0}t + {1 \over 2}\alpha {t^2}$$<br><br>$$ = 600 \times {{10} \over {60}} + {1 \over 2} \times 1200 \times 6 \times {1 \over {36}}$$<br><br>$$\theta = 200$$ | integer | jee-main-2021-online-20th-july-evening-shift | 12,493 |
1krsu72qd | physics | rotational-motion | combined-translational-and-rotational-motion | Consider a situation in which a ring, a solid cylinder and a solid sphere roll down on the same inclined plane without slipping. Assume that they start rolling from rest and having identical diameter.<br/><br/>The correct statement for this situation is | [{"identifier": "A", "content": "All of them will have same velocity."}, {"identifier": "B", "content": "The ring has greatest and the cylinder has the least velocity of the centre of mass at the bottom of the inclined plane."}, {"identifier": "C", "content": "The sphere has the greatest and the ring has the least velo... | ["C"] | null | $${{{K_T}} \over {{K_R}}} = {{M{R^2}} \over {{I_{CM}}}}$$<br><br>I<sub>CM</sub> is maximum for ring.<br><br>$$\Rightarrow$$ v is least for ring. | mcq | jee-main-2021-online-22th-july-evening-shift | 12,494 |
1l546xdsu | physics | rotational-motion | combined-translational-and-rotational-motion | <p>A spherical shell of 1 kg mass and radius R is rolling with angular speed $$\omega$$ on horizontal plane (as shown in figure). The magnitude of angular momentum of the shell about the origin O is $${a \over 3}$$ R<sup>2</sup>$$\omega$$. The value of a will be :</p>
<p><img src="data:image/png;base64,UklGRnAJAABXRUJQ... | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "4"}] | ["C"] | null | <p>$${\overrightarrow L _0} = {\overrightarrow L _{of\,cm}} + {\overrightarrow L _{about\,cm}}$$</p>
<p>$$ \Rightarrow {a \over 3}{R^2}\omega = mvR + {2 \over 3}m{R^2}\omega = {5 \over 3}m{R^2}\omega $$</p>
<p>$$ \Rightarrow a = 5$$</p> | mcq | jee-main-2022-online-29th-june-morning-shift | 12,496 |
1l55joigw | physics | rotational-motion | combined-translational-and-rotational-motion | <p>A ball is spun with angular acceleration $$\alpha$$ = 6t<sup>2</sup> $$-$$ 2t where t is in second and $$\alpha$$ is in rads<sup>$$-$$2</sup>. At t = 0, the ball has angular velocity of 10 rads<sup>$$-$$1</sup> and angular position of 4 rad. The most appropriate expression for the angular position of the ball is :</... | [{"identifier": "A", "content": "$${3 \\over 2}{t^4} - {t^2} + 10t$$"}, {"identifier": "B", "content": "$${{{t^4}} \\over 2} - {{{t^3}} \\over 3} + 10t + 4$$"}, {"identifier": "C", "content": "$${{2{t^4}} \\over 3} - {{{t^3}} \\over 6} + 10t + 12$$"}, {"identifier": "D", "content": "$$2{t^4} - {{{t^3}} \\over 2} + 5t +... | ["B"] | null | <p>$$\alpha = {{d\omega } \over {dt}} = 6{t^2} - 2t$$</p>
<p>$$\int_0^\omega {d\omega = \int_0^t {(6{t^2} - 2t)dt} } $$</p>
<p>so $$\omega = 2{t^3} - {t^2} + 10$$</p>
<p>and $${{d\theta } \over {dt}} = 2{t^3} - {t^2} + 10$$</p>
<p>so $$\int_4^\theta {d\theta = \int_0^t {(2{t^3} - {t^2} + 10)dt} } $$</p>
<p>$$\the... | mcq | jee-main-2022-online-28th-june-evening-shift | 12,497 |
1l56w3ypg | physics | rotational-motion | combined-translational-and-rotational-motion | <p>A rolling wheel of 12 kg is on an inclined plane at position P and connected to a mass of 3 kg through a string of fixed length and pulley as shown in figure. Consider PR as friction free surface. The velocity of centre of mass of the wheel when it reaches at the bottom Q of the inclined plane PQ will be $${1 \over ... | [] | null | 3 | <p>For rolling wheel</p>
<p>$$[12g\sin \alpha - 3g\sin \alpha ] \times R = (2 \times 12{R^2} + 3{R^2}) \times {a \over R}$$</p>
<p>$$ \Rightarrow {{9g\sin \alpha } \over {27}} = a$$</p>
<p>$$ \Rightarrow a = {{g\sin \alpha } \over 3}$$</p>
<p>$$\therefore$$ $$v = \sqrt {2 \times {{g\sin \alpha } \over 3} \times {h \ov... | integer | jee-main-2022-online-27th-june-evening-shift | 12,499 |
1l58hgpl0 | physics | rotational-motion | combined-translational-and-rotational-motion | <p>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</p> | [{"identifier": "A", "content": "$${2 \\over 5}$$"}, {"identifier": "B", "content": "$${2 \\over 7}$$"}, {"identifier": "C", "content": "$${1 \\over 5}$$"}, {"identifier": "D", "content": "$${7 \\over 10}$$"}] | ["B"] | null | <p>$$K{E_R} = {1 \over 2}l{w^2}$$</p>
<p>$$ = {1 \over 2} \times {2 \over 5} \times {\omega ^2} \times (m{R^2})$$</p>
<p>$$K{E_{total}} = {1 \over 2} \times {7 \over 5} \times m{R^2} \times {\omega ^2}$$</p>
<p>$$\therefore$$ $${{K{E_R}} \over {K{E_{total}}}} = {2 \over 7}$$</p> | mcq | jee-main-2022-online-26th-june-evening-shift | 12,500 |
1l6dxzpz1 | physics | rotational-motion | combined-translational-and-rotational-motion | <p>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 :</p> | [{"identifier": "A", "content": "$$\\sqrt{\\frac{5}{3}}$$"}, {"identifier": "B", "content": "$$\\sqrt{\\frac{4}{5}}$$"}, {"identifier": "C", "content": "$$\\sqrt{\\frac{3}{5}}$$"}, {"identifier": "D", "content": "$$\\sqrt{\\frac{14}{15}}$$"}] | ["D"] | null | <p>$$a = {{g\sin \theta } \over {1 + {{{K^2}} \over {{R^2}}}}}$$</p>
<p>$$v = \sqrt {{{2Sg\sin \theta } \over {1 + {{{K^2}} \over {{R^2}}}}}} $$</p>
<p>$$ \Rightarrow {{{v_c}} \over {{v_{ss}}}}\sqrt {{{1 + {{K_{ss}^2} \over {{R^2}}}} \over {1 + {{K_c^2} \over {{R^2}}}}}} = \sqrt {{{1 + {2 \over 5}} \over {1 + {1 \over... | mcq | jee-main-2022-online-25th-july-morning-shift | 12,501 |
1l6gnhwwx | physics | rotational-motion | combined-translational-and-rotational-motion | <p>A disc of mass $$1 \mathrm{~kg}$$ and radius $$\mathrm{R}$$ is free to rotate about a horizontal axis passing through its centre and perpendicular to the plane of disc. A body of same mass as that of disc is fixed at the highest point of the disc. Now the system is released, when the body comes to the lowest positio... | [] | null | 5 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l6v6jjtt/73602aaf-2134-4ea9-9e13-7b2e0cdbd91c/ccd41910-1cd4-11ed-843d-81ad9f680592/file-1l6v6jjtu.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l6v6jjtt/73602aaf-2134-4ea9-9e13-7b2e0cdbd91c/ccd41910-1cd4-11ed-843d-81ad9f680592... | integer | jee-main-2022-online-26th-july-morning-shift | 12,502 |
1ldohfflh | physics | rotational-motion | combined-translational-and-rotational-motion | <p>A solid cylinder is released from rest from the top of an inclined plane of inclination $$30^{\circ}$$ and length $$60 \mathrm{~cm}$$. If the cylinder rolls without slipping, its speed upon reaching the bottom of the inclined plane is __________ $$\mathrm{ms}^{-1}$$. (Given $$\mathrm{g}=10 \mathrm{~ms}^{-2}$$)</p>
<... | [] | null | 2 | Loss in potential energy $=m g h=m g\left[60 \sin 30^{\circ} \mathrm{cm}\right]$
<br/><br/>$\Rightarrow m g\left[\frac{30}{100}\right]=\frac{1}{2} m v^{2}+\frac{1}{2} \frac{m v^{2}}{2}$
<br/><br/>$\Rightarrow 0.3 \times 10=\frac{3}{4} v^{2}$
<br/><br/>$\Rightarrow v^{2}=4$
<br/><br/>$\Rightarrow v=2 \mathrm{~m} / \... | integer | jee-main-2023-online-1st-february-morning-shift | 12,503 |
ldqw6q13 | physics | rotational-motion | combined-translational-and-rotational-motion | <p>A uniform disc of mass $0.5 \mathrm{~kg}$ and radius $r$ is projected with velocity $18 \mathrm{~m} / \mathrm{s}$ at $\mathrm{t}=0$ s on a rough horizontal surface. It starts off with a purely sliding motion at $\mathrm{t}=0 \mathrm{~s}$. After $2 \mathrm{~s}$ it acquires a purely rolling motion (see figure). The to... | [] | null | 54 | <p>$$v = {v_0} - \mu gt$$</p>
<p>$$ \Rightarrow v = 18 - 0.3 \times 10 \times 2 = 12$$ m/s</p>
<p>$$\Rightarrow$$ Kinetic energy $$ = {1 \over 2}m{v^2} + {1 \over 2}{{m{v^2}} \over 2}$$</p>
<p>$$ = {3 \over 4}m{v^2} = {3 \over 4} \times 0.5 \times 144\,\mathrm{J} = 54\,\mathrm{J}$$</p> | integer | jee-main-2023-online-30th-january-evening-shift | 12,505 |
1ldsq2v76 | physics | rotational-motion | combined-translational-and-rotational-motion | <p>A solid sphere of mass 2 kg is making pure rolling on a horizontal surface with kinetic energy 2240 J. The velocity of centre of mass of the sphere will be _______ ms$$^{-1}$$.</p> | [] | null | 40 | $\frac{1}{2} m v_{\mathrm{cm}}^{2}+\frac{1}{2} \times \frac{2}{5} m R^{2} \times \frac{v_{\mathrm{cm}}^{2}}{R^{2}}=2240 \mathrm{~J}$
<br/><br/>
$$
\begin{aligned}
& \frac{7}{10} m v_{\mathrm{cm}}^{2}=2240 \\\\
& v_{\mathrm{cm}}=\sqrt{\frac{2240 \times 10}{7 \times 2}}=40 \mathrm{~m} / \mathrm{sec}
\end{aligned}
$$ | integer | jee-main-2023-online-29th-january-morning-shift | 12,506 |
1lgq29lea | physics | rotational-motion | combined-translational-and-rotational-motion | <p>A disc is rolling without slipping on a surface. The radius of the disc is $$R$$. At $$t=0$$, the top most point on the disc is $$\mathrm{A}$$ as shown in figure. When the disc completes half of its rotation, the displacement of point A from its initial position is</p>
<p><img src="data:image/png;base64,UklGRngHAABX... | [{"identifier": "A", "content": "$$R\\sqrt {({\\pi ^2} + 1)} $$"}, {"identifier": "B", "content": "$$2R$$"}, {"identifier": "C", "content": "$$R\\sqrt {({\\pi ^2} + 4)} $$"}, {"identifier": "D", "content": "$$2R\\sqrt {(1 + 4{\\pi ^2})} $$"}] | ["C"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lh1p3ha0/17d0ed07-a860-4691-9bd7-2e5d828f8a80/390d6290-e664-11ed-be92-49b0900ca5bd/file-1lh1p3ha1.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lh1p3ha0/17d0ed07-a860-4691-9bd7-2e5d828f8a80/390d6290-e664-11ed-be92-49b0900ca5bd/fi... | mcq | jee-main-2023-online-13th-april-morning-shift | 12,507 |
1lgq3lykk | physics | rotational-motion | combined-translational-and-rotational-motion | <p>A solid sphere is rolling on a horizontal plane without slipping. If the ratio of angular momentum about axis of rotation of the sphere to the total energy of moving sphere is $$\pi: 22$$ then, the value of its angular speed will be ____________ $$\mathrm{rad} / \mathrm{s}$$.</p> | [] | null | 4 | Given that the solid sphere is rolling without slipping, we have:
<br/><br/>
Angular momentum $$L = \left(I_{\text{com}}\right)(\omega)$$
<br/><br/>
Kinetic energy $$K = \frac{1}{2}(I_{\text{com}})(\omega^2) + \frac{1}{2}MV_{\text{com}}^2$$
<br/><br/>
For a solid sphere, the moment of inertia is $$I_{\text{com}} = \fra... | integer | jee-main-2023-online-13th-april-morning-shift | 12,508 |
1lgrjp53g | physics | rotational-motion | combined-translational-and-rotational-motion | <p>For a rolling spherical shell, the ratio of rotational kinetic energy and total kinetic energy is $$\frac{x}{5}$$. The value of $$x$$ is ___________.</p> | [] | null | 2 | For a rolling spherical shell, we must consider the fact that it has both translational and rotational kinetic energy. The total kinetic energy ($K_{total}$) can be expressed as the sum of the translational kinetic energy ($K_{trans}$) and the rotational kinetic energy ($K_{rot}$):
<br/><br/>
$$K_{total} = K_{trans} + ... | integer | jee-main-2023-online-12th-april-morning-shift | 12,509 |
1lgyrblmi | physics | rotational-motion | combined-translational-and-rotational-motion | <p>A hollow spherical ball of uniform density rolls up a curved surface with an initial velocity $$3 \mathrm{~m} / \mathrm{s}$$ (as shown in figure). Maximum height with respect to the initial position covered by it will be __________ cm.</p>
<p><img src="data:image/png;base64,UklGRqgLAABXRUJQVlA4IJwLAADQzwCdASoAA1MCP4... | [] | null | 75 | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1ljho089c/307f0af6-f13f-4042-862d-d71614f48756/eead25f0-16c4-11ee-84dd-7526dde12945/file-6y3zli1ljho089d.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1ljho089c/307f0af6-f13f-4042-862d-d71614f48756/eead25f0-16c4-11ee-84... | integer | jee-main-2023-online-8th-april-evening-shift | 12,510 |
lsan8z5e | physics | rotational-motion | combined-translational-and-rotational-motion | A disc of radius $\mathrm{R}$ and mass $\mathrm{M}$ is rolling horizontally without slipping with speed $v$. It then moves up an inclined smooth surface as shown in figure. The maximum height that the disc can go up the incline is :
<br/><br/>
<img src="data:image/png;base64,UklGRqwHAABXRUJQVlA4IKAHAADwbACdASoAAwcBP4HA... | [{"identifier": "A", "content": "$\\frac{3}{4} \\frac{v^2}{\\mathrm{~g}}$"}, {"identifier": "B", "content": "$\\frac{v^2}{g}$"}, {"identifier": "C", "content": "$\\frac{2}{3} \\frac{v^2}{\\mathrm{~g}}$"}, {"identifier": "D", "content": "$\\frac{1}{2} \\frac{v^2}{\\mathrm{~g}}$"}] | ["D"] | null | Only the translational kinetic energy of disc changes into gravitational potential energy. And rotational $\mathrm{KE}$ remains unchanged as there is no friction.
<br/><br/>$$
\begin{aligned}
& \frac{1}{2} \mathrm{mv}^2=\mathrm{mgh} \\\\
& \Rightarrow \mathrm{h}=\frac{\mathrm{v}^2}{2 \mathrm{~g}}
\end{aligned}
$$ | mcq | jee-main-2024-online-1st-february-evening-shift | 12,511 |
jaoe38c1lscpwe1j | physics | rotational-motion | combined-translational-and-rotational-motion | <p>A ring and a solid sphere roll down the same inclined plane without slipping. They start from rest. The radii of both bodies are identical and the ratio of their kinetic energies is $$\frac{7}{x}$$, where $$x$$ is _________.</p> | [] | null | 7 | In pure rolling work done by friction is zero. Hence potential energy is converted into kinetic energy. Since initially the ring and the sphere have same potential energy, finally they will have same kinetic energy too.
<br/><br/>$\therefore$ Ratio of kinetic energies $=1$
<br/><br/>$$
\Rightarrow \frac{7}{x}=1 \Righta... | integer | jee-main-2024-online-27th-january-evening-shift | 12,512 |
jaoe38c1lsf22kej | physics | rotational-motion | combined-translational-and-rotational-motion | <p>A cylinder is rolling down on an inclined plane of inclination $$60^{\circ}$$. It's acceleration during rolling down will be $$\frac{x}{\sqrt{3}} m / s^2$$, where $$x=$$ ________ (use $$\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$$).</p> | [] | null | 10 | <p>To determine the acceleration of a cylinder rolling down an inclined plane without slipping, we can use Newton's second law and the concept of rolling motion. For an inclined plane at an angle $$ \theta $$, the component of gravitational acceleration along the plane is $$ g \sin \theta $$. However, because the cylin... | integer | jee-main-2024-online-29th-january-morning-shift | 12,513 |
luxwdn7o | physics | rotational-motion | combined-translational-and-rotational-motion | <p>A circular disc reaches from top to bottom of an inclined plane of length $$l$$. When it slips down the plane, if takes $$t \mathrm{~s}$$. When it rolls down the plane then it takes $$\left(\frac{\alpha}{2}\right)^{1 / 2} t \mathrm{~s}$$, where $$\alpha$$ is _________.</p> | [] | null | 3 | <p>To find the value of $ \alpha $ from the given problem, we need to analyze the motion of a circular disc moving down an inclined plane in two different modes: slipping and rolling.</p>
<p><b>Slipping:</b></p>
<p>When the disc slips without rolling, it is primarily subjected to kinetic friction and gravity, without... | integer | jee-main-2024-online-9th-april-evening-shift | 12,514 |
lv0vy02i | physics | rotational-motion | combined-translational-and-rotational-motion | <p>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 __________.</p> | [] | null | 7 | <p>To solve this problem, we first note that for both the solid sphere and the hollow cylinder, the total mechanical energy is conserved as they roll up the inclined plane without slipping. The initial kinetic energy (comprised of both translational and rotational kinetic energy) is converted into potential energy at t... | integer | jee-main-2024-online-4th-april-morning-shift | 12,515 |
lv9s25k9 | physics | rotational-motion | combined-translational-and-rotational-motion | <p>A hollow sphere is rolling on a plane surface about its axis of symmetry. The ratio of rotational kinetic energy to its total kinetic energy is $$\frac{x}{5}$$. The value of $$x$$ is _________.</p> | [] | null | 2 | <p>For a hollow sphere rolling on a plane surface without slipping, its total kinetic energy (K.E.) is the sum of its translational kinetic energy and rotational kinetic energy. The translational kinetic energy results from the motion of the center of mass of the sphere, and the rotational kinetic energy is due to its ... | integer | jee-main-2024-online-5th-april-evening-shift | 12,517 |
JtkUq44cynQYPh0c | physics | rotational-motion | moment-of-inertia | Moment of inertia of a circular wire of mass $$M$$ and radius $$R$$ about its diameter is | [{"identifier": "A", "content": "$${{M{R^2}} \\over 2}$$"}, {"identifier": "B", "content": "$$M{R^2}$$ "}, {"identifier": "C", "content": "$$2M{R^2}$$ "}, {"identifier": "D", "content": "$${{M{R^2}} \\over 4}$$"}] | ["A"] | null | Moment of Inertia of a circular wire about an axis $$nn'$$ passing through the centre of the circle and perpendicular to the plane of the circle $$ = M{R^2}$$
<br><img class="question-image" src="https://imagex.cdn.examgoal.net/0QPu6RP5eFIx9MHJ2/HzAB9CBLLSomupsNdLEDkG200sNCI/N6x1o7AreqzJCm2DEZV6zo/image.svg" loading="l... | mcq | aieee-2002 | 12,518 |
vnOj5CKUn7CTtjDi | physics | rotational-motion | moment-of-inertia | One solid sphere $$A$$ and another hollow sphere $$B$$ are of same mass and same outer radii. Their moment of inertia about their diameters are respectively $${I_A}$$ and $${I_B}$$ such that | [{"identifier": "A", "content": "$${I_A} < {I_B}$$ "}, {"identifier": "B", "content": "$${I_A} > {I_B}$$ "}, {"identifier": "C", "content": "$${I_A} = {I_B}$$ "}, {"identifier": "D", "content": "$${{{I_A}} \\over {{I_B}}} = {{{d_A}} \\over {{d_B}}}$$\nwhere $${d_A}$$ and $${d_B}$$ are their densities."}] | ["A"] | null | For solid sphere the moment of inertia of $$A$$ about its diameter
<br><br>$${I_A} = {2 \over 5}M{R^2}.$$
<br><br>The moment of inertia of a hollow sphere $$B$$ about its diameter
<br><br>$${I_B} = {2 \over 3}M{R^2}.$$
<br><br>$$\therefore$$ $${I_A} < {I_B}$$ | mcq | aieee-2004 | 12,520 |
TwPowX0a7PgaEvAY | physics | rotational-motion | moment-of-inertia | The moment of inertia of a uniform semicircular disc of mass $$M$$ and radius $$r$$ about a line perpendicular to the plane of the disc through the center is | [{"identifier": "A", "content": "$${2 \\over 5}M{r^2}$$ "}, {"identifier": "B", "content": "$${1 \\over 4}Mr$$ "}, {"identifier": "C", "content": "$${1 \\over 2}M{r^2}$$ "}, {"identifier": "D", "content": "$$M{r^2}$$ "}] | ["C"] | null | Let mass of the semi circular disc = M
<br><br>Now assume a disc which is combination of two semi circular parts. Let $$I$$ be the moment of inertia of the uniform semicircular disc. So $$2I$$ will be the moment of inertia of the full circular disc and 2M will be the mass.
<br><br>$$ \Rightarrow 2I = {{2M{r^2}} \over ... | mcq | aieee-2005 | 12,521 |
WH2jIqhtfbspNMF1 | physics | rotational-motion | moment-of-inertia | Four point masses, each of value $$m,$$ are placed at the corners of a square $$ABCD$$ of side $$l$$. The moment of inertia of this system about an axis passing through $$A$$ and parallel to $$BD$$ is | [{"identifier": "A", "content": "$$2m{l^2}$$ "}, {"identifier": "B", "content": "$$\\sqrt 3 m{l^2}$$ "}, {"identifier": "C", "content": "$$3m{l^2}$$ "}, {"identifier": "D", "content": "$$m{l^2}$$ "}] | ["C"] | null | <img class="question-image" src="https://imagex.cdn.examgoal.net/rJrxDJDcPXsZ9UyA4/xKnXvCnpOzApdDdfwLnrUIrUGiNAR/W2S6G4kcS8tQsm762ANSeq/image.svg" loading="lazy" alt="AIEEE 2006 Physics - Rotational Motion Question 200 English Explanation">
Let $${I_{A}}$$ is the moment of inertia about an axis passing through A and p... | mcq | aieee-2006 | 12,522 |
Jixn0ultbE5ecu62 | physics | rotational-motion | moment-of-inertia | For the given uniform square lamina $$ABCD$$, whose center is $$O,$$
<img src="data:image/png;base64,UklGRuoIAABXRUJQVlA4IN4IAABQVgCdASraAVoBP4G61mY2LawnIhBpYsAwCWlu4W2k/mNwvj5N7KP9Bj0QD20mPFdtPAB3U8zJWF3Sfum9vE4NsvTpw0Xk6LydF5Oi8ZEZnl/8YM2FQ/4ynnl/2R7SaMzy/+ioi/CQs6Yi2e7c7L3PNY4AWudTy+rDp6ZQPue6KnmJyOcb6/VJ8y2oWYe0GW... | [{"identifier": "A", "content": "$${I_{AC}} = \\sqrt 2 \\,\\,{I_{EF}}$$ "}, {"identifier": "B", "content": "$$\\sqrt 2 {I_{AC}} = {I_{EF}}$$ "}, {"identifier": "C", "content": "$${I_{AD}} = 3{I_{EF}}$$ "}, {"identifier": "D", "content": "$${I_{AC}} = {I_{EF}}$$ "}] | ["D"] | null | <br>By perpendicular axes theorem,
<br><br>$${I_x} = {I_x} + {I_y}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$or,\,\,\,\,\,\,\,\,\,\,\,\,\,{I_z} = 2{I_y}$$
<br><br>( as $${I_x} = {I_y}$$ by symmetry of the figure)
<br><br>$$\therefore$$ $${I_{EF}} = {{{I_z}} \over 2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,....\left( i \right... | mcq | aieee-2007 | 12,523 |
ZTrPc4vdtmcrBxoR | physics | rotational-motion | moment-of-inertia | Consider a uniform square plate of side $$' a '$$ and mass $$'m'$$. The moment of inertia of this plate about an axis perpendicular to its plane and passing through one of its corners is | [{"identifier": "A", "content": "$${5 \\over 6}m{a^2}$$"}, {"identifier": "B", "content": "$${1 \\over 12}m{a^2}$$"}, {"identifier": "C", "content": "$${7 \\over 12}m{a^2}$$"}, {"identifier": "D", "content": "$${2 \\over 3}m{a^2}$$"}] | ["D"] | null | <img class="question-image" src="https://imagex.cdn.examgoal.net/bENXrcH9dAS9IfwU7/aI1XhjAdJDBKlfQ6EGTTIfl3oOv0A/YXl1QQo6q8LjBqzDH0xSAw/image.svg" loading="lazy" alt="AIEEE 2008 Physics - Rotational Motion Question 196 English Explanation">
<br>Moment of inertia for the square plate through O, perpendicular to the plat... | mcq | aieee-2008 | 12,524 |
22tooMeZBtDgsgfY | physics | rotational-motion | moment-of-inertia | A thin uniform rod of length $$l$$ and mass $$m$$ is swinging freely about a horizontal axis passing through its end. Its maximum angular speed is $$\omega $$. Its center of mass rises to a maximum height of: | [{"identifier": "A", "content": "$${1 \\over 6}\\,\\,{{l\\omega } \\over g}$$ "}, {"identifier": "B", "content": "$${1 \\over 2}\\,\\,{{{l^2}{\\omega ^2}} \\over g}$$ "}, {"identifier": "C", "content": "$${1 \\over 6}\\,\\,{{{l^2}{\\omega ^2}} \\over g}$$ "}, {"identifier": "D", "content": "$${1 \\over 3}\\,\\,{{{l^2}{... | ["C"] | null | <img class="question-image" src="https://imagex.cdn.examgoal.net/S55cDI0cWbihwfXQP/3bP4RBCPguUStDWQzDIQEzNLjUXS6/oxFDLm2pNQIrtqZOFEeVzt/image.svg" loading="lazy" alt="AIEEE 2009 Physics - Rotational Motion Question 195 English Explanation">
<br>The moment of inertia of the rod about $$O$$ is $${1 \over 2}m{\ell ^2}.$$... | mcq | aieee-2009 | 12,525 |
a6lKQHUSUq7mYL0t | physics | rotational-motion | moment-of-inertia | From a solid sphere of mass $$M$$ and radius $$R$$ a cube of maximum possible volume is cut. Moment of inertia of cube about an axis passing through its center and perpendicular to one of its face is: | [{"identifier": "A", "content": "$${{4M{R^2}} \\over {9\\sqrt {3\\pi } }}$$ "}, {"identifier": "B", "content": "$${{4M{R^2}} \\over {3\\sqrt {3\\pi } }}$$ "}, {"identifier": "C", "content": "$${{M{R^2}} \\over {32\\sqrt {2\\pi } }}$$ "}, {"identifier": "D", "content": "$${{M{R^2}} \\over {16\\sqrt {2\\pi } }}$$ "}] | ["A"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l91oebzm/4546a40f-3e1e-43ec-be20-b0bdd9fd76ef/c2b4e120-47ff-11ed-8757-0f869593f41f/file-1l91oebzn.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l91oebzm/4546a40f-3e1e-43ec-be20-b0bdd9fd76ef/c2b4e120-47ff-11ed-8757-0f869593f41f/fi... | mcq | jee-main-2015-offline | 12,526 |
vZk3cBXR23HJs2Pg | physics | rotational-motion | moment-of-inertia | The moment of inertia of a uniform cylinder of length $$l$$ and radius R about its perpendicular bisector is $$I$$.
What is the ratio $${l \over R}$$ such that the moment of inertia is minimum? | [{"identifier": "A", "content": "$${3 \\over {\\sqrt 2 }}$$ "}, {"identifier": "B", "content": "$$\\sqrt {{3 \\over 2}} $$ "}, {"identifier": "C", "content": "$${{\\sqrt 3 } \\over 2}$$"}, {"identifier": "D", "content": "1"}] | ["B"] | null | The volume of the cylinder V = $$\pi {R^2}l$$
<br><br>$$\therefore$$ $${R^2} = {V \over {\pi l}}$$
<br><br>We know, moment of inertia of a uniform cylinder of length $$l$$
and radius R about its perpendicular bisector is,
<br><br>$$I = {{M{l^2}} \over {12}} + {{M{R^2}} \over 4}$$
<br><br>[ Putting $${R^2} = {V \over {... | mcq | jee-main-2017-offline | 12,527 |
DTYVpqLOj6guqgSjz4IBq | physics | rotational-motion | moment-of-inertia | A circular hole of radius $${R \over 4}$$ is made in a thin uniform disc having mass M and radius R, as shown in figure. The moment of inertia of the remaining portion of the disc about an axis passing through the point O and perpendicular to the plane of the disc is :
<br/><br/><img src="data:image/png;base64,UklGRrI... | [{"identifier": "A", "content": "$${{219\\,M{R^2}} \\over {256}}$$"}, {"identifier": "B", "content": "$${{237\\,M{R^2}} \\over {512}}$$"}, {"identifier": "C", "content": "$${{19\\,M{R^2}} \\over {512}}$$"}, {"identifier": "D", "content": "$${{197\\,M{R^2}} \\over {256}}$$"}] | ["B"] | null | Mass of removed disc = $${M \over {16}}$$ Radius of removed disc = $${R \over 4}$$
<br><br>Moment of inertia of removed disc about it's own axis (O')
<br><br>= $${1 \over 2}$$ $$ \times $$ $${M \over {16}}$$ $$ \times $$ $${\left( {{R \over 4}} \right)^2}$$ = $${{M{R^2}} \over {512}}$$
<br><br>Moment of inertia of remo... | mcq | jee-main-2017-online-9th-april-morning-slot | 12,528 |
ibSw3Bw4FVm9EVoQZVcjS | physics | rotational-motion | moment-of-inertia | Moment of inertia of an equilateral triangular lamina ABC, about the axis
passing through its centre O and perpendicular to its plane is I<sub>o</sub> as shown in the figure. A cavity DEF is cut out from the lamina, where D, E, F are the mid points of the sides. Moment of inertia of the remaining part of lamina about ... | [{"identifier": "A", "content": "$${7 \\over 8}$$ I<sub>o</sub>"}, {"identifier": "B", "content": "$${15 \\over 16}$$ I<sub>o</sub>"}, {"identifier": "C", "content": "$${{3\\,{{\\rm I}_o}} \\over 4}$$"}, {"identifier": "D", "content": "$${{31\\,{{\\rm I}_o}} \\over 32}$$"}] | ["B"] | null | Let, side of triangle ABC = $$\ell $$
<br><br>According to perpendicular axes theorem, moment of inertia of triangle about it center and perpendicular to its plane,
<br><br>I<sub>O</sub> = $${1 \over {12}}$$ m$$\ell $$<sup>2</sup>
<br><br>In, triangle DEF,
<br><br>DE = DF = EF = $${1 \over 2}$$ AB = $${1 \over 2}$$ $... | mcq | jee-main-2017-online-8th-april-morning-slot | 12,529 |
aTsPG3xs23Pa7hEj10OKH | physics | rotational-motion | moment-of-inertia | A thin circular disk is in the xy plane as shown in the figure. The ratio of its moment of inertia about z and z' axes will be :
<br/><br/><img src="data:image/png;base64,UklGRgIMAABXRUJQVlA4IPYLAACQ0ACdASoAA5oCP4HA3GW2MK2nIXV46sAwCWlu4W5S9mNwvx6Sxl3Dn912vE7fan5/9Ltv/yeNVP187Z5ttqbU2ptTam1NqbU2ptTam1NqbU2psBP0E2B4wZSE... | [{"identifier": "A", "content": "1 : 3"}, {"identifier": "B", "content": "1 : 4"}, {"identifier": "C", "content": "1 : 5"}, {"identifier": "D", "content": "1 : 2"}] | ["A"] | null | As we know,
<br><br>moment of inertia about z axis
<br><br>$${{\rm I}_z} = {{m{R^2}} \over 2}$$
<br><br>and moment of inertia about z'
<br><br>$${\rm I}_z^1 = {3 \over 2}m{R^2}$$
<br><br>$$\therefore\,\,\,\,$$ $${{{{\rm I}_z}} \over {{\rm I}{'_z}}}$$ = $${{{{m{R^2}} \over 2}} \over {{3 \over 2}m{R^2}}}$$ = $${1 \over ... | mcq | jee-main-2018-online-16th-april-morning-slot | 12,530 |
1jKxr2kiGFuwVxRZ | physics | rotational-motion | moment-of-inertia | From a uniform circular disc of radius R and mass 9M, a small disc
of radius R/3 is removed as shown in the figure. The moment of
inertia of the remaining disc about an axis perpendicular to the plane
of the disc and passing through centre of disc is :
<img src="data:image/png;base64,UklGRoISAABXRUJQVlA4IHYSAADwagCdASq... | [{"identifier": "A", "content": "$${{37} \\over 9}M{R^2}$$"}, {"identifier": "B", "content": "$$4M{R^2}$$"}, {"identifier": "C", "content": "$${{40} \\over 9}M{R^2}$$"}, {"identifier": "D", "content": "$$10M{R^2}$$"}] | ["B"] | null | Given that for a uniform circular disc the radius is R and mass 9M
<br><br>$$\therefore$$ The area of uniform circular disc = $$\pi {R^2}$$
<br><br>The radius of removed portion = $${R \over 3}$$
<br><br>$$\therefore$$ The area of removed portion = $${{\pi {R^2}} \over 9}$$
<br><br>So the mass of the removed portion = ... | mcq | jee-main-2018-offline | 12,531 |
F7lnlRb4qhAX6Utl | physics | rotational-motion | moment-of-inertia | Seven identical circular planar disks, each of mass M and radius R are
welded symmetrically as shown. The moment of inertia of the arrangement
about the axis normal to the plane and passing through the point P is :
<img src="data:image/png;base64,UklGRmIVAABXRUJQVlA4IFYVAACwZgCdASoVAQYBPm00lkgkIqIhJLF6uIANiWlu/HyYTMO+... | [{"identifier": "A", "content": "$${{181} \\over 2}M{R^2}$$ "}, {"identifier": "B", "content": "$${{55} \\over 2}M{R^2}$$ "}, {"identifier": "C", "content": "$${{19} \\over 2}M{R^2}$$ "}, {"identifier": "D", "content": "$${{73} \\over 2}M{R^2}$$ "}] | ["A"] | null | Moment of inertia of any disc form its center perpendicular to the plane of disc = $${1 \over 2}M{R^2}$$
<br><br>Moment of inertia of any one of the outer disc about an axis passing through point O and perpendicular to the plane
<br><br>$${I_1} = {1 \over 2}M{R^2} + M{\left( {2R} \right)^2}$$ = $${9 \over 2}M{R^2}$$
<... | mcq | jee-main-2018-offline | 12,532 |
Z4MHHZFbApzENkdU3nzvi | physics | rotational-motion | moment-of-inertia | An equilateral triangle ABC is cut from a thin solid sheet of wood. (see figure) D, E and F are the mid-points of its sides as shown and G is the centre of the triangle. The moment of inertia of the triangle about an axis passing through G and perpendicular to the plane of the triangle is I0. If the smaller triangle DE... | [{"identifier": "A", "content": "$${\\rm I} = {{{{\\rm I}_0}} \\over 4}$$"}, {"identifier": "B", "content": "$${\\rm I} = {{15} \\over {16}}{{\\rm I}_0}$$"}, {"identifier": "C", "content": "$${\\rm I} = {9 \\over {16}}{{\\rm I}_0}$$"}, {"identifier": "D", "content": "$${\\rm I} = {3 \\over 4}{{\\rm I}_0}$$"}] | ["B"] | null | Suppose M is mass and a is side of larger triangle, then $${M \over 4}$$ and $${a \over 2}$$ will be mass and side length of smaller triangle.
<br><br>$${{{{\rm I}_{removed}}} \over {{{\rm I}_{original}}}} = {{{M \over 4}{{\left( {{a \over 2}} \right)}^2}} \over {M{{\left( a \right)}^2}}}$$
<br><br>$${{\rm I}_{removed... | mcq | jee-main-2019-online-11th-january-morning-slot | 12,533 |
Car9pLGRIDoZfpmHFy18hoxe66ijvzn5oc0 | physics | rotational-motion | moment-of-inertia | Two coaxial discs, having moments of inertia
I<sub>1</sub> and I<sub>1</sub>/2, are rotating with respective angular
velocities $$\omega $$<sub>1 </sub> and
$$\omega $$<sub>1</sub>/2
, about their common axis.
They are brought in contact with each other and
thereafter they rotate with a common angular
velocity. If E<su... | [{"identifier": "A", "content": "$${{{I_1}\\omega _1^2} \\over {24}}$$"}, {"identifier": "B", "content": "$${{{I_1}\\omega _1^2} \\over {12}}$$"}, {"identifier": "C", "content": "$${3 \\over 8}{I_1}\\omega _1^2$$"}, {"identifier": "D", "content": "$${{{I_1}\\omega _1^2} \\over {6}}$$"}] | ["A"] | null | $${E_i} = {1 \over 2}{I_I} \times \omega _1^2 + {1 \over 2}{I \over 2} \times {{\omega _1^2} \over 4}$$<br><br>
$$ = {{{I_1}\omega _1^2} \over 2}\left( {{9 \over 8}} \right) = {9 \over {16}}{I_1}\omega _1^2$$<br><br>
$${I_1}{\omega _1} + {{{I_1}{\omega _1}} \over 4} = {{3{I_1}} \over 2}\omega ;{5 \over 4}{I_1}{\omega _... | mcq | jee-main-2019-online-10th-april-morning-slot | 12,535 |
byPYPbwFgB0CXsY4IZ18hoxe66ijvzn0x28 | physics | rotational-motion | moment-of-inertia | A thin disc of mass M and radius R has mass
per unit area $$\sigma $$(r) = kr<sup>2</sup> where r is the distance
from its centre. Its moment of inertia about an
axis going through its centre of mass and
perpendicular to its plane is : | [{"identifier": "A", "content": "$${{M{R^2}} \\over 3}$$"}, {"identifier": "B", "content": "$${{M{R^2}} \\over 6}$$"}, {"identifier": "C", "content": "$${{2M{R^2}} \\over 3}$$"}, {"identifier": "D", "content": "$${{M{R^2}} \\over 2}$$"}] | ["C"] | null | $${I_{Disc}} = \int\limits_0^R {\left( {dm} \right)} {r^2} \Rightarrow {I_{Disc}} = \int\limits_0^R {\left( {\sigma 2\pi rdr} \right)} {r^2}$$<br><br>
$${I_{Disc}} = \int\limits_0^R {\left( {k{r^2}2\pi rdr} \right)} {r^2}$$ Mass of Disc<br><br>
$${I_{Disc}} = 2\pi k\int\limits_0^R {{r^2}dr} \,\,\,\,M ... | mcq | jee-main-2019-online-10th-april-morning-slot | 12,536 |
ANiozDvSSiYK3aYNBxrC5 | physics | rotational-motion | moment-of-inertia | Moment of inertia of a body about a given axis
is 1.5 kg m<sup>2</sup>. Initially the body is at rest. In order
to produce a rotational kinetic energy of
1200 J, the angular accleration of 20 rad/s<sup>2</sup>
must be applied about the axis for a
duration of :- | [{"identifier": "A", "content": "2.5 s"}, {"identifier": "B", "content": "3 s"}, {"identifier": "C", "content": "5s"}, {"identifier": "D", "content": "2 s"}] | ["D"] | null | KE = $${1 \over 2}I{\omega ^2} = 1200$$ (given)<br><br>
$$ \Rightarrow \omega = 40\,rad/s$$<br><br>
$$ \Rightarrow \omega = {\omega _0} + \alpha t$$<br><br>
$$ \Rightarrow 40 = 0 + (20)t$$<br><br>
$$ \Rightarrow t = 2\,\sec $$ | mcq | jee-main-2019-online-9th-april-evening-slot | 12,537 |
oZMeF2P5AgHl84kp3cyQY | physics | rotational-motion | moment-of-inertia | A stationary horizontal disc is free to rotate
about its axis. When a torque is applied on it,
its kinetic energy as a function of $$\theta $$, where $$\theta $$
is the angle by which it has rotated, is given as
k$$\theta $$<sup>2</sup>. If its moment of inertia is I then the
angular acceleration of the disc is : | [{"identifier": "A", "content": "$${k \\over {4I}}\\theta $$"}, {"identifier": "B", "content": "$${k \\over {I}}\\theta $$"}, {"identifier": "C", "content": "$${k \\over {2I}}\\theta $$"}, {"identifier": "D", "content": "$${2k \\over {I}}\\theta $$"}] | ["D"] | null | Kinetic energy KE = $${1 \over 2}l{\omega ^2} = k{\theta ^2}$$<br><br>
$$ \Rightarrow {\omega ^2} = {{2k{\theta ^2}} \over l} \Rightarrow \omega = \sqrt {{{2k} \over l}} \theta $$ .... (A)<br><br>
Differentiate (A) wrt time $$ \to $$<br><br>
$${{d\omega } \over {dt}} = \alpha = \sqrt {{{2k} \over l}} \left( {{{d\thet... | mcq | jee-main-2019-online-9th-april-morning-slot | 12,538 |
ObonSAX7wcIhQ9FqwgIim | physics | rotational-motion | moment-of-inertia | The moment of inertia of a solid sphere, about an axis parallel to its diameter and at a distance of x from it, is 'I(x)'. Which one of the graphs represents the variation of I(x) with x correctly ? | [{"identifier": "A", "content": "<img src=\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264913/exam_images/ezxxvuuq6jgvtiya2nvw.webp\" style=\"max-width: 100%; height: auto;display: block;margin: 0 auto;\" loading=\"lazy\" alt=\"JEE Main 2019 (Online) 12th January Evening Slot Physics - Rotational Motion Qu... | ["C"] | null | <p>The correct answer is <strong>Option C</strong>. Here's why :</p>
<p>The moment of inertia of a solid sphere about an axis passing through its center is given by :</p>
<p>$$I_0 = \frac{2}{5}MR^2$$</p>
<p>Where :</p>
<ul>
<li> $I_0$ is the moment of inertia about the center</li>
<li> $M$ is the mass of the sp... | mcq | jee-main-2019-online-12th-january-evening-slot | 12,540 |
ysXBZyKjbUp7UJb3f0DUM | physics | rotational-motion | moment-of-inertia | Let the moment of inertia of a hollow cylinder of length 30 cm (inner radius 10 cm and outer radius 20 cm), about its axis be I. The radius of a thin cylinder of the same mass such that its moment of inertia about its axis is also I, is : | [{"identifier": "A", "content": "16 cm"}, {"identifier": "B", "content": "12 cm"}, {"identifier": "C", "content": "14 cm"}, {"identifier": "D", "content": "18 cm"}] | ["A"] | null | Consider an element of radius x and thickness dx
<img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266244/exam_images/za03hwqutc2rzcpd2itu.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 12th January Morning Slot Physics - Rotational Mo... | mcq | jee-main-2019-online-12th-january-morning-slot | 12,541 |
E6I8tJQ5uMvamSrh93jgy2xukfrn8vv8 | physics | rotational-motion | moment-of-inertia | Shown in the figure is a hollow icecream cone (it is open at the top). If its mass is M, radius of its
top, R and height, H, then its moment of inertia about its axis is :
<img src="data:image/png;base64,UklGRvQMAABXRUJQVlA4IOgMAAAwUwCdASrrAFwBPm02mEgkIyKhJTE5yIANiWlu/HyYo8OjOzrt/Rn+Rfjb5jfzr+zf1T8bPIb8s/Xvyw/sPoq7iXzX... | [{"identifier": "A", "content": "$${{M\\left( {{R^2} + {H^2}} \\right)} \\over 3}$$"}, {"identifier": "B", "content": "$${{M{R^2}} \\over 2}$$"}, {"identifier": "C", "content": "$${{M{R^2}} \\over 3}$$"}, {"identifier": "D", "content": "$${{M{H^2}} \\over 3}$$"}] | ["B"] | null | Moment of inertia of this cone will same as
circular disk of mass (M) and radius R.
<br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264152/exam_images/nopfaaqzlfjtyk4fbwiz.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 6th Septem... | mcq | jee-main-2020-online-6th-september-morning-slot | 12,544 |
VNPKX8Peju0Jhnpzopjgy2xukg0bmoa7 | physics | rotational-motion | moment-of-inertia | The linear mass density of a thin rod AB of length L varies from A to B as
<br/>$$\lambda \left( x \right) = {\lambda _0}\left( {1 + {x \over L}} \right)$$, where
x is the distance from A. If M is the mass of the rod then its moment of inertia about an axis passing
through A and perpendicular to the rod is : | [{"identifier": "A", "content": "$${2 \\over 5}M{L^2}$$"}, {"identifier": "B", "content": "$${5 \\over {12}}M{L^2}$$"}, {"identifier": "C", "content": "$${7 \\over {18}}M{L^2}$$"}, {"identifier": "D", "content": "$${3 \\over 7}M{L^2}$$"}] | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266741/exam_images/obgkte28otijc0dlh9qi.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 6th September Evening Slot Physics - Rotational Motion Question 105 English Explanation">
<br><br>... | mcq | jee-main-2020-online-6th-september-evening-slot | 12,545 |
zT4AYz1QoP2zM4t8cfjgy2xukfl1siq0 | physics | rotational-motion | moment-of-inertia | A ring is hung on a nail. It can oscillate, without
slipping or sliding <br/>(i) in its plane with a time
period T<sub>1</sub> and, <br/>(ii) back and forth in a direction
perpendicular to its plane, <br/>with a period T<sub>2</sub>. The
ratio $${{{T_1}} \over {{T_2}}}$$ will be : | [{"identifier": "A", "content": "$${{\\sqrt 2 } \\over 3}$$"}, {"identifier": "B", "content": "$${2 \\over {\\sqrt 3 }}$$"}, {"identifier": "C", "content": "$${2 \\over 3}$$"}, {"identifier": "D", "content": "$${3 \\over {\\sqrt 2 }}$$"}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264484/exam_images/rtbzcdo7iyiiwoleekld.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 5th September Evening Slot Physics - Rotational Motion Question 109 English Explanation">
<br><br>... | mcq | jee-main-2020-online-5th-september-evening-slot | 12,546 |
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