A force of 49~N acts tangentially at the highest point of a sphere (solid) of mass 20~kg, kept on a rough horizontal…

Q42 jee_main_2026_23_january_morning Angular Momentum

Two small balls with masses m and 2m are attached to both ends of a rigid rod of length d and negligible mass. If angular momentum of this system is L about an axis (A) passing through its centre of mass and perpendicular to the rod then angular velocity of the system about A is:

A. $(3)/(2) Lmd²$ $\frac{3}{2}\frac{L}{md^{2}}$
B. $2Lmd²$ $\frac{2L}{md^{2}}$
C. $(4)/(3) Lmd²$ $\frac{4}{3}\frac{L}{md^{2}}$
D. $2L5md²$ $\frac{2L}{5md^{2}}$

Solution

Related Formula

rcm = m₁r₁ + m₂r₂m₁ + m₂

Icm = μ d² = ( m₁m₂m₁ + m₂)d²

$L = Iω$ $L = I\omega$

Step 1: Calculate Moment of Inertia about COM

Let mass $m$ $m$ be at the origin. Position of $2m$ $2m$ is $d$ $d$.

Xcm = (m(0) + 2m(d))/(m + 2m) = (2d)/(3)

Distance of mass $m$ $m$ from COM is $(2d)/(3)$ $\frac{2d}{3}$. Distance of mass $2m$ $2m$ from COM is $d - (2d)/(3) = (d)/(3)$ $d - \frac{2d}{3} = \frac{d}{3}$.

I = m((2d)/(3))² + 2m((d)/(3))²

I = 4md²9 + 2md²9 = 6md²9 = 2md²3

Step 2: Calculate Angular Velocity

$L = Iω$ $L = I\omega$

ω = (L)/(I)

ω = L 2md²3 = 3L2md²

Pattern Recognition

Sees: "two point masses" + "rotation about COM" → Quickly use reduced mass $μ$ $\mu$ moment of inertia shortcut: $Icm = μ d² = ((m \cdot 2m)/(3m))d² = (2)/(3)md²$ to save time.

Chapter Mix

Class 11 Physics: Systems of Particles and Rotational Motion

Q27 jee_main_2026_23_january_evening Conservation of Momentum

A body of mass 14 kg initially at rest explodes and breaks into three fragments of masses in the ratio 2 : 2 : 3. The two pieces of equal masses fly off perpendicular to each other with a speed of 18 m/s each. The velocity of the heavier fragment is ____m/s.

Conservation of Momentum diagram for Q27 - JEE Main 2026 Evening — Vector diagram illustrating the trajectories of the explosive fragments.

A. $10\sqrt(2)$ $10\sqrt{2}$
B. $12\sqrt(2)$ $12\sqrt{2}$
C. $12$ $12$
D. $24\sqrt(2)$ $24\sqrt{2}$

Solution

Related Formula

pinitial = pfinal

M₁ V₁ + M₂ V₂ + M₃ V₃ = 0

Core Logic

The total mass is $14 kg$ $14 \, \mathrm{kg}$ and breaks in ratio $2:2:3$ $2:2:3$. Let the masses be $M₁$ $M_1$, $M₂$ $M_2$, and $M₃$ $M_3$.

M₁ = (2)/(7) \times 14 = 4 kg

M₂ = (2)/(7) \times 14 = 4 kg

M₃ = (3)/(7) \times 14 = 6 kg

Step 1: Setup Momentum Equations

Since $M₁$ $M_1$ and $M₂$ $M_2$ fly off perpendicular to each other, let their velocity vectors be along the x and y axes. Because the total momentum must be zero, we align the fragments opposite to the final 3rd fragment's direction for simplicity, or just use standard axes.

Let $V₁ = -18 i$ $\vec{V}_1 = -18\hat{i}$ and $V₂ = -18 j$ $\vec{V}_2 = -18\hat{j}$. Note that the solution simplifies the mass ratio directly to $2, 2, 3$ $2, 2, 3$ as relative masses for the momentum equation:

2(-18 i) + 2(-18 j) + 3 V₃ = 0

Step 2: Solve for Velocity

V₃ = 36 i + 36 j3 = 12 i + 12 j

Magnitude of $V₃$ $\vec{V}_3$:

| V₃| = \sqrt(12² + 12²) = 12\sqrt(2) m/s

Pattern Recognition

In a 3-part explosion from rest, the momentum of the third piece must be equal and opposite to the vector sum of the other two pieces. You can use the ratio of masses directly in the momentum conservation equation instead of absolute masses to save time.

Chapter Mix

Class 11 Physics: Center of Mass and Collisions

Q49 jee_main_2026_23_january_evening Moment of Inertia

Suppose there is a uniform circular disc of mass M kg and radius r m shown in figure. The shaded regions are cut out from the disc. The moment of inertia of the remainder about the axis A of the disc is given by

(x)/(256)

Mr

²

. The value of x is ____.

Moment of Inertia diagram for Q49 - JEE Main 2026 Evening — Diagram showing a large circular disc with two smaller circular sections removed.

Numerical Answer. Answer: 109 to 109

Solution

Related Formula

Iremaining = Itotal - Σ Icut

Parallel axis theorem: $I = Icm + md²$ $I = I_{\text{cm}} + md^2$

Core Logic

Total mass of original disc is $M = σ π r²$ $M = \sigma \pi r^2$. Two identical smaller discs are removed. From the diagram, each small cut-out disc has radius $r' = (r)/(4)$ $r' = \frac{r}{4}$. The center of each cut-out disc is at distance $d = (3)/(4)r$ $d = \frac{3}{4}r$ from the main axis A. Mass of each cut-out disc:

m = σ π ((r)/(4))² = (σ π r²)/(16) = (M)/(16)

Step 1: Moment of Inertia of Total Disc

For the main solid disc about its central axis A:

Itotal = (1)/(2) M r²

Step 2: Moment of Inertia of Cut-out Discs

Using parallel axis theorem for one cut-out disc about axis A:

Icut = (1)/(2) m (r')² + m d²

Icut = (1)/(2) m ((r)/(4))² + m ((3)/(4)r)²

Icut = (mr²)/(32) + (9mr²)/(16) = m r² ( (1 + 18)/(32) ) = (19)/(32) m r²

Since there are 2 cut-out discs, the total subtracted inertia is:

Iremoved = 2 \times (19)/(32) m r² = (19)/(16) m r²

Step 3: Calculate Final Inertia in Terms of M

Substitute $m = (M)/(16)$ $m = \frac{M}{16}$:

Iremoved = (19)/(16) ( (M)/(16) ) r² = (19)/(256) M r²

Now, subtract from total:

Iremaining = (1)/(2) M r² - (19)/(256) M r²

Iremaining = (128 - 19)/(256) M r² = (109)/(256) M r²

Thus, $x = 109$ $x = 109$.

Pattern Recognition

In cavity problems, mass is strictly proportional to area ( $R²$ $R^2$). Use parallel axis theorem perfectly on the 'negative mass' segments and subtract.

Chapter Mix

Class 11 Physics: System of Particles and Rotational Motion

Q41 jee_main_2026_24_january_morning Dynamics of Rotational Motion

Two masses 400 g and 350 g are suspended from the ends of a light string passing over a heavy pulley of radius 2 cm. When released from rest the heavier mass is observed to fall 81 cm in 9 s. The rotational inertia of the pulley is ____

kg ²

. (g =

9.8 m/s²

)

A. $9.5 \times 10⁻³$ $9.5 \times 10^{-3}$
B. $4.75 \times 10⁻³$ $4.75 \times 10^{-3}$
C. $1.86 \times 10⁻²$ $1.86 \times 10^{-2}$
D. $8.3 \times 10⁻³$ $8.3 \times 10^{-3}$

Solution

Related Formula

s = ut + (1)/(2)at²

a = ((m₁ - m₂)g)/(m₁ + m₂ + (I)/(R²))

Core Logic

First, calculate the acceleration of the system using kinematics:

s = ut + (1)/(2)at²

0.81 = 0 + (1)/(2) a (9)²

a = (2 \times 0.81)/(81) = 0.02 m/s²

Applying Newton's second law for masses and rotation:

m₁ g - T₁ = m₁ a

T₂ - m₂ g = m₂ a

(T₁ - T₂)R = I \cdot α = I ((a)/(R))

This leads to the standard Atwood machine acceleration with massive pulley:

a = ((m₁ - m₂)g)/(m₁ + m₂ + (I)/(R²))

Step 1: Calculate Moment of Inertia

Substitute known values ( $m₁=0.4 kg$ $m_1=0.4\text{ kg}$, $m₂=0.35 kg$ $m_2=0.35\text{ kg}$, $R=0.02 m$ $R=0.02\text{ m}$):

0.02 = ((0.4 - 0.35) \times 9.8)/((0.4 + 0.35) + (I)/(R²))

0.02 = (0.05 \times 9.8)/(0.75 + (I)/(R²))

0.75 + (I)/(R²) = (0.49)/(0.02) = 24.5

(I)/(R²) = 24.5 - 0.75 = 23.75

I = 23.75 \times (0.02)² = 23.75 \times 4 \times 10⁻⁴

I = 95 \times 10⁻⁴ = 9.5 \times 10⁻³ kg ²

Pattern Recognition

For massive pulley problems, the "effective mass" of the system increases by the pulley's equivalent translating mass $I/R²$ $I/R^2$. Simply use $a = Fₙₑₜ / Meffective$ .

Chapter Mix

Class 11 Physics: System of Particles and Rotational Motion Class 11 Physics: Laws of Motion

Q31 jee_main_2026_24_january_evening Rigid Body Rotation and Energy Conservation

A thin uniform rod (X) of mass M and length L is pivoted at a height

((L)/(3))

as shown in the figure. The rod is allowed to fall from a vertical position and lie horizontally on the table. The angular velocity of this rod when it hits the table top, is ____. (g = gravitational acceleration)

Rigid Body Rotation and Energy Conservation diagram for Q31 - JEE Main 2026 Evening — A uniform rod is pivoted at a distance of L/3 from the table surface and falls from a vertical position to horizontal.

A. $(3)/(2) gL$
B. $3\sqrt(2) gL$
C. $1\sqrt(2) gL$
D. $3gL$ $\sqrt{\frac{3\mathrm{g}}{\mathrm{L}}}$

Solution

Related Formula

Δ K.E. = Δ P.E.

mg Δ hcom = (1)/(2) I ω²

Core Logic

The rod falls such that its center of mass lowers by a distance. The rod is pivoted at $(L)/(3)$ $\frac{L}{3}$ from the bottom, meaning the distance from the pivot to the center of mass (which is at $(L)/(2)$ $\frac{L}{2}$ from either end) is:

hcom = (L)/(2) - (L)/(3) = (L)/(6)

Step 1: Moment of Inertia

Using the parallel axis theorem, the moment of inertia about the pivot is:

I = Icom + m d²

I = (mL²)/(12) + m((L)/(6))² = (mL²)/(12) + (mL²)/(36) = (mL²)/(9)

Step 2: Energy Conservation

Equating the loss in potential energy to the gain in rotational kinetic energy:

mg (L)/(6) = (1)/(2) ((mL²)/(9)) ω²

mg (L)/(6) = (mL²)/(18) ω²

ω² = (3g)/(L)

ω = \sqrt((3g)/(L))

Pattern Recognition

For a hinged rod falling from a vertical orientation to horizontal, always track the displacement of the center of mass and calculate rotational inertia strictly about the hinge using $Ipivot = Icm + md²$ .

Chapter Mix

Class 11 Physics: System of Particles and Rotational Motion

Solution & Explanation

Related Formula

Core Logic

Step 1: Solving for Linear Acceleration

Step 2: Analysis of Friction Force Direction

Pattern Recognition

Chapter Mix

Reference Study Guides

More Rotational Motion Previous-Year Questions — Page 3

Solution

Related Formula

Step 1: Calculate Moment of Inertia about COM

Step 2: Calculate Angular Velocity

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Setup Momentum Equations

Step 2: Solve for Velocity

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Moment of Inertia of Total Disc

Step 2: Moment of Inertia of Cut-out Discs

Step 3: Calculate Final Inertia in Terms of M

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Moment of Inertia

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Moment of Inertia

Step 2: Energy Conservation

Pattern Recognition

Chapter Mix

More Rotational Motion Questions — jee_main_2025_03_april_morning

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

Welcome back