The centre of mass of a thin rectangular plate (fig - x) with sides of length a and b, whose mass per unit area (sigma)…

Year	2026	2025	2024	Total
Questions	19	27	11	57

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

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

Q49 jee_main_2026_24_january_evening Moment of Inertia of Continuous Bodies

A uniform solid cylinder of length L and radius R has moment of inertia about its axis equal to

I₁

. A small co-centric cylinder of length L/2 and radius R/3 carved from this cylinder has moment of inertia about its axis equals to

I₂

. The ratio

I₁/I₂

is

Numerical Answer. Answer: 162 to 162

Solution

Related Formula

I = (1)/(2) M R²

M = ρ \cdot V = ρ \cdot π R² L

Core Logic

Moment of Inertia of Continuous Bodies diagram for Q49 - JEE Main 2026 Evening

For the original cylinder (mass $M$ $M$):

I₁ = (1)/(2) M R²

For the carved cylinder, its mass $m$ $m$ is:

m = ρ \times π ((R)/(3))² \times (L)/(2)

Step 1: Calculate Mass of Carved Cylinder

m = (ρ π R² L)/(18) = (M)/(18)

Step 2: Calculate Inertia of Carved Cylinder

I₂ = (1)/(2) m ((R)/(3))²

I₂ = (1)/(2) ( (M)/(18) ) ( (R²)/(9) )

I₂ = (1)/(324) M R²

Step 3: Finding the Ratio

(I₁)/(I₂) = ((1)/(2) M R²)/((1)/(324) M R²) = (324)/(2) = 162

Pattern Recognition

For similar geometries, mass scales as $R² L$ $R^2 L$. Inertia scales as $M R²$ $M R^2$ which ultimately means $I \propto R⁴ L$ $I \propto R^4 L$. Here $R arrow R/3$ $R \rightarrow R/3$ (factor of $1/81$ $1/81$) and $L arrow L/2$ $L \rightarrow L/2$ (factor of $1/2$ $1/2$), so $I₂$ $I_2$ is $1/162$ $1/162$ of $I₁$ $I_1$.

Chapter Mix

Class 11 Physics: System of Particles and Rotational Motion

Solution & Explanation

Core Logic

Step 1: Final Position Coordinates

Pattern Recognition

Chapter Mix

Reference Study Guides

More System of Particles and 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: 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

Solution

Related Formula

Core Logic

Step 1: Calculate Mass of Carved Cylinder

Step 2: Calculate Inertia of Carved Cylinder

Step 3: Finding the Ratio

Pattern Recognition

Chapter Mix

More System of Particles and Rotational Motion Questions — jee_main_2025_28_jan_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