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JEE Main · Physics

Rotational Motion Practice Questions

Generate JEE Main level questions on Rotational Motion. Focus on Moment of Inertia, Torque, and Rolling motion.

200 questions · 20 PYQs · 0 AI practice · JEE Main 2027

Q101JEE Main 2020MCQ

Moment of inertia of a cylinder of mass M, length $\ {L}$ and radius $\ {R}$ about an axis passing through its centre and perpendicular to the axis of the cylinder is $\ {I}=\ {M}(\frac{{R}^{2}}{4}+\frac{{L}^{2}}{12}\ ).$ If such a cylinder is to be made for a given mass of material, the ratio L/R for it to have minimum possible I is :-

🚀 Solve in Practice Mode

📖 Explanation

$\ {I}=\ {M}(\frac{{R}^{2}}{4}+\frac{{L}^{2}}{12}\ )$ ......(1) as mass is constant $\Rightarrow {m}=\rho {V}={\text{constant}}$ $\ {V}=$ constant $\pi ^{2} \ {R} l=$ constant $\Rightarrow {R}^{2} \ {L}=$ constant $2 {RL}+ {R}^{2} \frac{{dL}}{\ {dR}}=0$ .....(2) From equation (1) $\frac{{dI}}{ {dR}}= {M}(\frac{2 {R}}{4}+ \frac{2 {L}}{12} \times \frac{{dL}}{ {dr}} )=0$ $\frac{{R}}{2}+\frac{{L}}{6} \frac{{dL}}{ {dR}}=0$ Substituting value of $\frac{{d} {L}}{ {d} {R}}$ from eqution (2) $\frac{{R}}{2}+\frac{{L}}{6}(\frac{-2 {L}}{\ {R}} )=0$ $\frac{{R}}{2}=\frac{{L}^{2}}{3 {R}}$ $\Rightarrow \frac{{L}}{ {R}}=\\sqrt{ \frac{3}{2}}$

Explanation Figure 1

Q102

JEE Main 2020

NAT

A force ${\vec{F}}=( \hat{i}+2 \hat{j}+3 \hat{k}) {N}$ acts at a point $(4 \hat{i}+3\hat{j}-\hat{k}) {m}.$ Then the magnitude of torque about the point $( \hat{i}+2 \hat{j}+ \hat{k}) { m}$ will $b e$ $\sqrt{{x}} {N}- {m} .$ The value of ${x}$ is _______.

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📖 Explanation

${\tau }^{\to }=(\vec{r}_{2}-\vec{r}_{1} ) \times \vec{F}$ $=[(4 \hat{i}+3\hat{j}-\hat{k})-( \hat{i}+2 \hat{j}+3 \hat{k}) ]$ $\times { {F}}^{\to }$ $=(3\hat{i}+\hat{j}-2\hat{k}) \times (\hat{i}+2\hat{j}+3 \hat{k})$ $\tau =\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & -2 \\ 1 & 2 & 3\end{vmatrix}$ $=7 \hat{i}-11 \hat{j}+5 \hat{k}$ $| \vec{\tau }|=\\sqrt{195}$

Q103JEE Main 2020MCQ

Consider two uniform discs of the samethickness and different radii $\ {R}_{1}=\ {R}$ and $\ {R}_{2}=\alpha {R}$ made of the same material. If the ratio of their moments of inertia $\ {I}_{1}$ and $\ {I}_{2}$ , respectively, about their axes is $\ {I}_{1}: {I}_{2}=1: 16$ then the value of $\alpha$ is :

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📖 Explanation

${I}_{1}=\frac{{MR}^{2}}{2}= \frac{\ \rho (\pi {R}^{2} ) {t} .{R}^{2}}{2}$ ${I} \ \propto {R}^{4}$ $\frac{{I}_{1}}{ {I}_{2}}=\frac{{R}_{1}^{4}}{ {R}_{2}^{4}}= \frac{1}{16}$ $\therefore \frac{ {R}_{1}}{ {R}_{2}}= \frac{1}{2}$

Q104JEE Main 2020MCQ

The radius of gyration of a uniform rod of length $l$ , about an axis passing through a point $l/4$ away from the centre of the rod, and perpendicular to it, is : [7-Jan-2020 Shift 1]

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📖 Explanation

$ml^2/12+ml^2/16=mk^2$ ${7l^2}/48=k^2$

Q105JEE Main 2020MCQ

The linear mass density of a thin rod ${\text{AB}}$ of length ${\text{L}}$ varies from ${\text{A}}$ to ${\text{B}}$ as $\lambda({x})=\lambda_{0}(1+{{x}} /{{\text{L}}}),$ where ${x}$ is the distance from A. If ${\text{M}}$ is the mass of the rod then its moment of inertia about an axis passing through A and perpendicular to the rod is:

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📖 Explanation

${I}=∫{r}^{2} {dm}=∫{x}^{2} \lambda{d} {x}$ ${I}= ∫_{0}^{\text{L}}{x}^{2} \lambda_{0}(1+\frac{{x}}{{L}}) {d} {x}$ ${I}=\lambda_{0} ∫_{0}^{\text{L}}({x}^{2}+{{x}^{3}}/{{\text{L}}}) {d} {x}$ ${I}={7 {\text{L}}^{3} \lambda_{0}}/{12}$ ....(i) ${M}= ∫_{0}^{\text{L}} \lambda {d} {x}= ∫_{0}^{\text{L}} \lambda_{0}(1+\frac{{x}}{{L}}) {d} {x}$ ${M}=\lambda_{0}({\text{L}}+{{\text{L}}}/{2})=\lambda_{0} {3 {\text{L}}}/{2}$ $\frac{2}{3} {\text{M}}=(\lambda_{0} {\text{L}})$ ....(ii) From (i) & (ii) ${I}=\frac{7}{12}(\frac{2}{3} {M}) {L}^{2}={7 {\text{ML}}^{2}}/{18}$

Explanation Figure 1

Q106JEE Main 2020MCQ

For a uniform rectangular sheet shown in the figure, the ratio of moments of inertia about the axes perpendicular to the sheet and passing through $\ {O}$ (the centre of mass) and O' (corner point) is :

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📖 Explanation

${I}_{ {O}}=\frac{{M}}{12}[{L}^{2}+ {B}^{2} ]$ $=\frac{{M}}{12} [80^{2}+60^{2} {J} ]$ ${I}_{ {O}'}= {I}_{0}+ {Md}^{2}\{$ parallel axis theorem $\}$ $=\frac{{M}}{12} \$ [80^{2}+60^{2} ]$+\ {M}[50]^{2}{{I}{ {O}}}/{ {I}{ {O}'}}={{M} / 12 $[80^{2}+60^{2} ]$}/{\frac{{M}}{12} $[80^{2}+60^{2} ]$+ {M}[50]^{2}}= \frac{1}{4}$

Explanation Figure 2

Q107JEE Main 2020NAT

Consider a uniform cubical box of side a on a rough floor that is to be moved by applying minimum possible force F at a point b above its centre of mass (see figure). If the coefficient of friction is $\mu = 0.4$ , the maximum possible value of $100 \times b/a$ for a box not to topple before moving is ______ .

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📖 Explanation

F = µmg ...(1) $F(b+a/2)=mg a/2$ ...(2) $\mu mg(b+a/2)=mg\times a/2$ $(b+a/2)\mu =a/2$ $0.4=\mu =a/{2b+a}$ $0.8b+0.4a=a$ $0.8b=0.6a$ $b/a=3/4$

Explanation Figure 2

Q108JEE Main 2020MCQ

Mass per unit area of a circular disc of radius a depends on the distance r from its centre as σ (r) = A + Br. The moment of inertia of the disc about the the axis, perpendicular to the plane and passing through its centre is :

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📖 Explanation

$dI = dmr^2$ $dI = \sigma \;2 \pi r \;dr \;r^2$ $dI = 2 \pi (A + Br) r^3 dr$ $∫dI=2\pi ∫^{a}_{0}(Ar^{3+}Br^4)dr$ $I=2\pi a^4(A/4+{B9}/5)$

Explanation Figure 1

Q109JEE Main 2020NAT

A body of mass m = 10 kg is attached to one end of a wire of length 0.3 m. The maximum angular speed (in rad s $^{-1}$ ) with which it can be rotated about its other end in space station is (Breaking stress of wire = $4.8 \times 10^7 Nm^{-2}$ and area of cross-section of the wire = $10^{-2} cm^2$ ) is

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📖 Explanation

$T=m\omega ^2l$ Breaking stress $=T/A={m\omega ^2l}/A$ $\omega ^2=\frac{4.8\times 10^7\times (10^{-2}\times 10^{-4})}{10\times 0.3}=16$ $\Rightarrow \omega =4$

Q110JEE Main 2020MCQ

Three solid spheres each of m ass m and diameter d are stuck together such that the lines connecting the centres form an equilateral triangle of side of length d. The ratio $I_0/I_A$ of moment of inertia $I_0$ of the system about an axis passing the centroid and about center of any of the spheres $I_A$ and perpendicular to the plane of the triangle is :

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📖 Explanation

From parallel axis theorem $I_0=3\times [2/5M(d/2)^{2+}M(d/\sqrt3)^2]$ $=13/10Md^2$ $I_A=I_0+3M(d/\sqrt3)^2$ $=13/10Md^{2+}Md^2$ $=23/10Md^2$ $I_0/I_A=13/23$

Q111JEE Main 2020MCQ

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:

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📖 Explanation

Area $=\pi {R} l=\pi {R}(\sqrt{{H}^{2}+{R}^{2}})$ Area of element ${d} {A}=2 \pi {rd} l= 2 \pi r \frac{{dh}}{\cos \theta }$ mass of element ${dm}={{M}}/{\pi {R} \sqrt{{H}^{2}+{R}^{2}}} \times \frac{2 \pi {rdh}}{\cos \theta }$ ${dm}={2 {Mh} \tan \theta {dh}}/{{R} \sqrt{{H}^{2}+{R}^{2}} \cos \theta }$ (here ${r}={h} \tan \theta )$ ${I}=∫({dm}) {r}^{2}$ $=∫\frac{{h}^{2} \tan ^{2} \theta }{\cos \theta }(\frac{2 {m}}{{R}} {{h} \tan \theta }/{\sqrt{{R}^{2}+{H}^{2}}}) {dh}$ $=\frac{2 {M}}{\cos \theta {R}} {\tan ^{3} \theta }/{\sqrt{{R}^{2}+{H}^{2}}}$ $∫_{0} ^{\text{H}} {h}^{3} {dh}={{MR}^{2} {H}^{4}}/{2 {RH}^{3} \sqrt{{R}^{2}+{H}^{2}} \cos \theta }$ $={\text{MR}^2}/2$

Explanation Figure 2

Q112JEE Main 2020MCQ

A uniform rod of length 'l' is pivoted at one ofits ends on a vertical shaft of negligible radius.When the shaft rotates at angular speed $\omega$ therod makes an angle $\theta$ with it (see figure). Tofind $\theta$ equate the rate of change of angularmomentum (direction going into the paper ) $\frac{ {m} l^{2}}{12} \omega ^{2} \sin \theta \cos \theta$ about the centre of mass(CM) to the torque provided by the horizontaland vertical forces $\ {F}_{ {H}}$ and ${F}_{ {V}}$ about the ${CM}$ . Thevalue of $\theta$ is then such that:

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📖 Explanation

${F}_{ {V}}= {mg}$ ${F}_{ {H}}= {m} \omega ^{2} \frac{l}{2} \sin \theta$ ${mg} \frac{l}{2} \sin \theta - {m\omega }^{2}\frac{l}{2} \sin \theta \frac{l}{2} \cos \theta$ $=\frac{{m} l^{2}}{12} \omega ^{2} \sin \theta \cos \theta$ $\cos \theta = \frac{3}{2}\frac{{g}}{\omega ^{2} l}$

Explanation Figure 2

Q113JEE Main 2020MCQ

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 $m_1$ and $m_2$ $(m_1 > m_2)$ are attached to the ends of the string. The system is released from rest. The angular speed of the wheel when $m_1$ descents by a distance h is :

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📖 Explanation

by using work energy theorem Wg = ΔKE $(m_1-m_2)gh$ $=1/2(m_1+m_2)V^{2+}1/2I\omega ^2$ $(m_1-m_2)gh$ $=1/2(m_1+m_2) (\omega R)^{2+}1/2I\omega ^2$ $(m_1-m_2)gh$ $= {\omega ^2}/2[(m_1+m_2)R^{2+} I ]$ $\omega =\$ [\frac{2(m_1-m_2)gh}{(m_1+m_2)R^{2+}I}]^{1/2}$

Explanation Figure 2

Q114JEE Main 2020MCQ

Two uniform circular discs are rotating independently in the same direction around their common axis passing through their centres. The moment of inertia and angularvelocity of the first disc are $0.1 \ {kg}-\ {m}^{2}$ and 10 rad $\ {s}^{-1}$ respectively while those for the second one are $0.2 \ {kg}-\ {m}^{2}$ and $5 \ {\text{rad}} \ {s}^{-1}$ respectively. At some instant they get stuck together and start rotating as a single system about their common axis with some angular speed. The Kinetic energy of the combined system is :

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📖 Explanation

◆ Both discs are rotating in same sense ◆ Angular momentum conserved for the system i.e. ${L}_{1}+ {L}_{2}= {L}_{\text{final}}$ ${I}_{1} \omega _{1}+ {I}_{2} \omega _{2}= ( {I}_{1}+ {I}_{2} ) \omega _{ {f}}$ $0.1 \\times 10+0.2 \\times 5$ $=(0.1+0.2) \\times \omega _{ {f}}$ $\omega _{ {f}}= \frac{20}{3}$ ◆ Kinetic energy of combined disc system $\Rightarrow \frac{1}{2}({I}_{1}+ {I}_{2} ) \omega _{ {f}}^{2}$ $= \frac{1}{2}(0.1+0.2) .( \frac{20}{3} )^{2}$ $= \frac{0.3}{2} \\times \frac{400}{9}= \frac{120}{18}= \frac{20}{3} \ {J}$

Q115JEE Main 2020MCQ

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 :

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📖 Explanation

m = 0.5 kg, v = 5 cm/sKE in rolling $=1/2mv^{2+}1/2I\omega ^2$ $=1/2mv^2(1+K^2/R^2)$ $= 8.75 \times 10^{-4}J$

Q116JEE Main 2020MCQ

A ring is hung on a nail. It can oscillate, without slipping or sliding (i) in its plane with a time period ${T}_{1}$ and, (ii) back and forth in a direction perpendicular to its plane, with a period ${T}_{2}$ . the ratio $\frac{ {T}_{1}}{ {T}_{2}}$ will be :

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📖 Explanation

Moment of inertia in case (i) is I $_{1}$ Moment of inertia in case (ii) is ${I}_{2}$ ${I}_{1}=2 {MR}^{2}$ ${I}_{2}= \frac{3}{2} {MR}^{2}$ ${T}_{1}=2 \pi \sqrt{\frac{{I}_{1}}{ {Mgd}}} ;$ ${T}_{2}=2 \pi \sqrt{\frac{{I}_{2}}{ {Mgd}}}$ ${{T}_{1}}{ {T}_{2}}=\sqrt{\frac{{I}_{1}}{ {I}_{2}}}$ $=\sqrt{\frac{2 {MR}^{2}}{ \frac{3}{2} {MR}^{2}}}= \frac{2}{\\sqrt{3}}$

Explanation Figure 1

Q117JEE Main 2020MCQ

A wheel is rotaing freely with an angular speed $\omega$ on a shaft. The moment of inertia of the wheel is I and the moment of inertia of the shaft is negligible. Another wheel of momet of inertia 3I initially at rest is suddenly coupled to the same shaft. The resultant fractional loss in the kinetic energy of the system is :

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📖 Explanation

By anglar momentum conservation $\omega {I}+3 {I} \\times 0=4 {I} \omega '$ $\Rightarrow \omega '= \frac{\omega }{4}$ $({KE})_{ {i}}=\frac{1}{2} {I} \omega ^{2}$ $({KE})_{{f}}={1}{2} \\times (4 {I})\times (\frac{\omega }{4})^{2}$ $=\frac{{I} \omega ^{2}}{8}$ $\Delta{KE}=\frac{3}{8} {I} \omega ^{2}$ fractional loss $=\frac{\Delta{KE}}{{KE}_{1}}={\frac{3}{8} {I} \omega ^{2}}/{{1/2}{ {I} \omega ^{2}}}= \frac{3}{4}$

Explanation Figure 1

Q118JEE Main 2020NAT

A person of $80 \ {kg}$ mass is standing on the rim of a circular platform of mass $200 \ {kg}$ rotating about its axis as 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 whenthe person reaches its centre ________.

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📖 Explanation

${L}_{ {i}}= {L}_{ {f}}$ $\ (80 {R}^{2}+ \frac{200 {R}^{2}}{2} ) \omega$ $= (0+ \frac{200 {R}^{2}}{2} ) \omega _{1}$ $180 \omega _{0}=100 \omega _{1}$ $\omega _{1}=1.8 \omega _{0}=1.8 \\times 5$ $=9 \ {\text{rpm}}$

Q119JEE Main 2020MCQ

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:

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📖 Explanation

${I}={m}(0)^{2}$ $+{m}(\frac{l}{\sqrt{2}})^{2} \times 2$ $+{m}(\sqrt{2} l)^{2}$ $=\frac{2 {m} l^{2}}{2}+2 {m} l^{2}=3 {m} l^{2}$ Angular momentum ${L}={I} \omega$ $=3 {m} l ^2\omega$

Explanation Figure 2

Q120JEE Main 2020NAT

One end of a straight uniform 1m long bar is pivoted on horizontal table. It is released from rest when it makes an angle 30° from the horizontal (see figure). Its angular speed when it hits the table is given as $\text{sqrtn} s^{-1}$ , where n is an integer. The value of n is

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📖 Explanation

From mechanical energy conservation, $U_i+K_i=U_f+K_f$ $\Rightarrow mg l/2\sin30°+0=0+1/2I\omega$ $\Rightarrow mg\times 1/2\times 1/2+0$ $=0+1/2\times {m (1)^2}/3\omega ^2$ $\Rightarrow \omega ^2={3g}/2\Rightarrow \omega =\sqrt15$ $\therefore n=15$

Explanation Figure 2

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