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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

Q121JEE Main 2020NAT

An massless equilateral triangle EFG of side 'a' (As shown in figure) has three particles of mass m situated at its vertices. The moment of intertia of the system about the line EX perpendicularto EG in the plane of EFG is $\frac{{N}}{20} {ma}^{2}$ where $\ {N}$ is an integer. The value of $\ {N}$ is _______.

🚀 Solve in Practice Mode

📖 Explanation

${I}=0+ {m}(\frac{{a}}{2} )^{2}+ {ma}^{2}$ $= \frac{5}{4} {ma}^{2}$

Explanation Figure 2

Q122JEE Main 2020MCQ

A uniform cylinder of mass $\ {M}$ and radius $\ {R}$ isto be pulled over a step of height a $(a< R)$ byapplying a force $\ {F}$ at its centre 'O' perpendicularto the plane through the axes of the cylinderon the edge of the step (see figure). Theminimum value of F required is :

🚀 Solve in Practice Mode

📖 Explanation

$(\tau )_{\ {P}}=0$ F.R. $-\ {mgx}=0$ $x=\\sqrt{R^{2}-(R-a)^{2}}$ ${F}= {mg}\frac{{x}}{ {R}}$ ${F}= {mg} \\sqrt{1-$ \frac{{R}- {a}}{ {R}} )^{2}}=$ minimum value of force to pull

Explanation Figure 2

Explanation Figure 3

Q123JEE Main 2019MCQ

A circular disc of radius b has a hole of radius a at its centre (see figure). If the mass per unit area of the disc varies as $( \frac{\sigma _{0}}{r} )$ then the radius of gyration of the disc about its axis passing through the centre is.

🚀 Solve in Practice Mode

📖 Explanation

$m k^{2}=∫_{a} ^{b}(2 \pi r d r) \frac{\sigma _{0}}{r} r^{2}$ $k^{2} =∫_{a} ^{b} 2 \pi r d r \frac{\sigma _{0}}{r}=∫(2 \pi r d r) \frac{\sigma _{0}}{r} r^{2}$ $k^{2}= 2 \pi \sigma .(b-a)= \frac{2 \pi \sigma _{0}}{3} (b^{3}-a^{3} )$ $k=\sqrt{ \frac{ {a}^{2}+ {b}^{2}+ {ab}}{3}}$

Q124JEE Main 2019MCQ

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 center of mass at the bottom of the incline is same, theratio of the maximum heights they climb is:

🚀 Solve in Practice Mode

📖 Explanation

$M g h=\ \frac{1}{2} {{m V}^{2}}_{c m}(1+K)$ Ratio $=1+1 : 1+\ \frac{1}{2} : 1+ \frac{2}{5}$ $=20 : 15 : 14$

Q125JEE Main 2019MCQ

Two identical spherical balls of mass M and radius R each are stuck on two ends of a rod of length 2R and mass M (see figure). The moment of inertia of the system about the axis passing perpendicularly through the centre of the rod is :

🚀 Solve in Practice Mode

📖 Explanation

For Ball using parallel axis theorem $I_{\text{ball}}$ = $2/5 MR^2 + M(2R)^2$ = ${22}/5 MR^2$ 2 balls or ${44}/5 MR^2$ Irod = for rod ${M(2R)^2}/R$ = ${MR^2}/3$ $I_{\text{system}}$ = $I_{\text{ball}}+I_{\text{rod}}$ = ${44}/5 MR^{2+}{MR^2}/3$ = $\frac{137}{15} MR^2$

Q126JEE Main 2019MCQ

To mop-clean a floor, a cleaning machinepresses a circular mop of radius R verticallydown with a total force F and rotates it with aconstant angular speed about its axis. If theforce F is distributed uniformly over the mopand if coefficient of friction between the mopand the floor is μ, the torque, applied by themachine on the mop is :

🚀 Solve in Practice Mode

📖 Explanation

Consider a strip of radius x & thickness dx,Torque due to friction on this strip. λ dr = $∫^{R}_{0}$ $\frac{x\mu F.2\pi xdx}{\pi R^2}$ τ = ${2\mu F}/R^2$ . $R^3/3$ τ = ${2\mu FR}/3$ ∴ correct answer is (1)

Explanation Figure 1

Q127JEE Main 2019MCQ

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:

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Q128JEE Main 2019MCQ

If the angular momentum of a planet of massm, moving around the Sun in a circular orbitis L, about the center of the Sun, its arealvelocity is : [9-Jan-2019 Shift 1]

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📖 Explanation

$\frac{dA}{dt}$ = $L/{2m}$

Q129JEE Main 2019MCQ

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 : (Take g= 10 m/ $s^2$ )

🚀 Solve in Practice Mode

📖 Explanation

Work Energy Theorem from A to B mgh = $1/2 mv_B^2$ - $1/2 mv_A^2$ 2gh = $v_B^{2-}v_A^2$ 2 × 10 × 10 = $v_B^{2-}5^2$ $v_B$ = 15 m/s Angular momentum about 0 $L_0$ = mvr = 20 × $10^{-3}$ × 20 $L_0$ = 6 kg. $m^2$ /s

Q130JEE Main 2019MCQ

A rod of length 50cm is pivoted at one end. It is raised such that if makes an angle of 30° from the horizontal as shown and released from rest. Its angular speed when it passes through the horizontal (in rad s-1) will be (g = 10 $ms^{-2}$ )

🚀 Solve in Practice Mode

📖 Explanation

Work done by gravity from initial to finalposition is, W = mg $l/2$ sin 30° = ${mgl}/4$ According to work energy theorem W = $1/2 I\omega ^2$ \Rightarrow $1/2 {ml^3}/3 \omega ^2$ = ${mgl}/4$ ω = $\sqrt{\frac{3g}{2l}}$ = $\sqrt{\frac{3\times 0}{2\times 0.5}}$ ω = $\sqrt{30}$ rad/sec ∴ correct answer is (1)

Explanation Figure 2

Q131JEE Main 2019MCQ

A thin smooth rod of length L and mass M is rotating freely with angular speed $\omega _0$ about an axis perpendicular to the rod and passing through its centre. Two beads of mass m and negligible size are at the centre of the rod initially. The beads are free to slide along the rod. The angular speed of the system, when the beads reach the opposite ends of the rod, will be:

🚀 Solve in Practice Mode

📖 Explanation

Angular momentum is conserved $I_{i n i} \omega _{0}=I_{A n} \omega$ $\Rightarrow\frac{m L^{2}}{T^{2}} \omega _{0}$ $=\ (\ \frac{M L^{2}}{12}+\ \frac{m L^{2}}{4}+\ \frac{M L^{2}}{4}\ ) \omega$ $\Rightarrow\omega =\ \frac{M \omega _{0}}{M+6 m}$

Q132JEE Main 2019MCQ

A uniform rod of length l is being rotated in a horizontal plane with a constant angular speed about an axis passing through one of its ends. If the tension generated in the rod due to rotation is T(x) at a distance x from the axis, then which of the following graphs depicts it most closely?

🚀 Solve in Practice Mode

📖 Explanation

$\ {dT} =\ {dmx} \omega ^{2}$ $𝜕{T} =\lambda{dx} \ {x} \omega ^{2}$ $\ {T} =\lambda\omega ^{2} ∫__ {\ {x}}^^ {\ {L}-\ {x}} \ {x} \ {d} \ {x}$ $\ {T} = \frac{\lambda\omega ^{2}}{2}\ [(\ {L}-\ {x})^{2}-\ {x}^{2}\ ]$ $\ {T} =\ \frac{\lambda \omega ^{2}}{2} \$ [\ {L}^{2}-2 \ {L} \ {x}\ ]$ $

Explanation Figure 1

Explanation Figure 2

Q133JEE Main 2019MCQ

A person of mass M is, sitting on a swing of length L and swinging with an angular amplitude $\theta _0$ .If the person stands up when the swing passes through its lowest point, the work done by him, assuming that his centre of mass moves by a distance l(l << L),is close to.

🚀 Solve in Practice Mode

📖 Explanation

Applying angular momentum conservation ${Mv}'( {L}-1)= {mvL}$ $\Rightarrow {v}'= \frac{ {vL}}{ {L}-1}$ 1 Using Work-energy theorem ${W}_{ {m}}- {\text{Mgl}}= \frac{1}{2} {mv'}^{ 2}- \frac{1}{2} {m}$ $\Rightarrow {W}_{ {m}}= {\text{mgl}} [1+\theta _{0}^{2} ]$

Q134JEE Main 2019MCQ

The time dependence of the position of a particle of mass m=2 is given by $\ \vec{r}(t)=2 t \hat{i}-3 t^{2} \hat{j}$ . Its angular momentum, with respect to the origin, at time t=2 is:

🚀 Solve in Practice Mode

📖 Explanation

$\ \vec{L}=\ \vec{r} \\times \vec{p}=\ {r} ^{\to }\times (m \vec{v})$ Where $\vec{r}=2 t \ \hat{i}-3 t^{2} \ \hat{j}$ $\ \vec{v}=\ \frac{d \ \vec{r}}{d t}=2 \ \hat{i}-6 t \ \hat{j}$ $\ \vec{L}=m(\ \vec{r} \\times \ \vec{v})=-48 \ \hat{k}$

Q135JEE Main 2019MCQ

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 perpendicularto the plane of the triangle is $I_0$ . It the smaller triangle DEF is removed from ABC, the moment of inertia of the remaining figure about the same axis is I. Then

🚀 Solve in Practice Mode

📖 Explanation

Suppose M is mass and a is side of largertriangle, then $M/4$ and $a/2$ will be mass and sidelength of smaller triangle. $I_{\text{removed}}/I_{\text{original}}$ = ${M/4}/M . (a/2)^2/(a)^2$ $I_{\text{removed}}$ = $I_0/{16}$ So, I = $I_0-I_0/{16}$ = $\frac{15}{16} I_0$

Q136JEE Main 2019MCQ

The magnitude of torque on a particle of mass1kg is 2.5 Nm about the origin. If the force actingon it is 1 N, and the distance of the particle fromthe origin is 5m, the angle between the force andthe position vector is (in radians) :-

🚀 Solve in Practice Mode

📖 Explanation

2.5 = 1 × 5 sin θ sin θ = 0.5 = $1/2$ θ = $\pi /6$

Q137JEE Main 2019MCQ

A solid sphere and solid cylinder of identical radii approach an incline with the same linear velocity (see figure). Both roll without slipping all throughout. The two climb maximum heights $\text{h}_{\text{sph}}$ and $\text{h}_{\text{cyl}}$ on the incline. The ratio $\text{h}_{\text{sph}}/{\text{h}_{\text{cyl}}}$ is given by:

🚀 Solve in Practice Mode

📖 Explanation

Toal K.E. $= \frac{1}{2} (I_{c m}+M_{R}^{2} ) \frac{V^{2}}{R^{2}}$ $K . E_{ { \text{sph} }}= \frac{1}{2} ( \frac{2}{5}+1 ) m V^{2}$ $K . E_{ c y}= \frac{1}{2} ( \frac{1}{2}+1 ) m V^{2}$ $\text{h}_{\text{sph}}/{\text{h}_{\text{cyl}}}$ $= \frac{K E_{s p h}}{K E_{c y}}=$ $\frac{7 / 5}{3 / 2}= \frac{14}{15}$

Explanation Figure 1

Q138JEE Main 2019MCQ

A rigid massless rod of length 3l has twomasses attached at each end as shown in thefigure. The rod is pivoted at point P on thehorizontal axis (see figure). When released frominitial horizontal position, its instantaneousangular acceleration will be :

🚀 Solve in Practice Mode

📖 Explanation

Applying torque equation about point P. $2M_0$ (2l) - 5 $M_0$ gl = Iα I = $2M_0(2l)^{2+}5M_0I^2$ = 13 $M_0I^2d$ ∴ α = - $\frac{M_0gl}{13M_0l^2}$ \Rightarrow α - $g/{13l}$ ∴ α = $g/{13l}$ anticlockwise

Explanation Figure 2

Q139JEE Main 2019MCQ

A homogeneous solid cylindrical roller of radius R and mass M is pulled on a cricket pitch by a horizontal force. Assuming rolling with out slipping, angular acceleration of the cylinder is:

🚀 Solve in Practice Mode

📖 Explanation

FR = $3/2 MR^2\alpha$ α = $\frac{2F}{3mR}$

Explanation Figure 1

Q140JEE Main 2019MCQ

A rectangular solid box of length 0.3 m is held horizontally, with one of its sides on the edge of a platform of height 5 m. When released, it slips off the table in a very short time $\tau =0.01$ s, remaining essentially horizontal. The angle by which it would rotate when it hits the ground will be (in radians) close to

🚀 Solve in Practice Mode

📖 Explanation

$m g \frac{l}{2}= \frac{m l^{2}}{3} \alpha$ $\Rightarrow \alpha = \frac{3 g}{2 l}= \frac{3 \times 10}{2 \times 0.31}=50$ $rad / {sec}^{2}$ $\omega =50 \times 0.01=0.5 {rad} / {sec}$ $t=\sqrt{ \frac{2 h}{g}}=\sqrt{\ \frac{2 \times 5}{10}}=1 {sec}$ $\theta =\omega t=0.5 {rad}$

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