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

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Q49 jee_main_2026_22_january_morning Moment of Inertia of Circular Discs

A circular disc has radius

R₁

and thickness

T₁

. Another circular disc made of the same material has radius

R₂

and thickness

T₂

. If the moment of inertia of both discs are same and

R₁R₂ = 2

then

T₁T₂ = (1)/(α)

. The value of

α

is \_\_\_\_.

Numerical Answer. Answer: 16 to 16

Solution

Related Formula

I = (mR²)/(2), m = π R² T ρ

Core Logic

Solution disc moment of inertia diagram for Q49 - JEE Main 2026 Morning

Mass of discs:

m₁ = π R₁² T₁ ρ, m₂ = π R₂² T₂ ρ

Moments of inertia:

I₁ = (m₁ R₁²)/(2), I₂ = (m₂ R₂²)/(2)

Equating $I₁ = I₂$ $I_1 = I_2$:

(π R₁² T₁ ρ R₁²)/(2) = (π R₂² T₂ ρ R₂²)/(2) (T₁)/(T₂) = ((R₂)/(R₁))⁴ = ((1)/(2))⁴ = (1)/(16)

Therefore, $α = 16$ $\alpha = 16$.

Pattern Recognition

Sees: Equal moment of inertia for two discs of different radii and thicknesses. Shortcut: Equate $mR²$ $mR^2$ expressions and substitute radius ratio $(R₁)/(R₂) = 2$ $\frac{R_1}{R_2} = 2$. Check: Numerical answer is 16. ✓

Chapter Mix

Class 11 Physics: Rotational Motion

Q29 jee_main_2026_22_january_evening Angular Momentum Conservation

A uniform bar of length 12 cm and mass 20 m lies on a smooth horizontal table. Two point masses m and 2 m are moving in opposite directions with same speed of v and in the same plane as the bar, as shown in figure. These masses strike the bar simultaneously and get stuck to it. After collision the entire system is rotating with angular frequency

ω

. The ratio of v and

ω

is :

Angular momentum collision diagram for Q29 - JEE Main 2026 Evening — The figure illustrates a uniform bar of length 12 cm with two point masses m and 2m approaching it perpendicularly from opposite directions.

A. $33$ $33$
B. $2\sqrt(88)$ $2\sqrt{88}$
C. $66$ $66$
D. $32$ $32$

Solution

Related Formula

$Lᵢ = Lf$ $L_i = L_f$

I = M Lbar²12 + Σ mᵢ rᵢ²

Core Logic

Applying angular momentum conservation about the center of mass of the rod:

Lᵢ = m \cdot v \cdot 4 + 2m \cdot v \cdot 2

Calculating total moment of inertia $Ifinal$ $I_{final}$ after collision:

Ifinal = ( (20m(12)²)/(12) + m(4)² + 2m(2)² )

Ifinal = (240m + 16m + 8m) = 264m

Equating initial and final angular momentum:

4mv + 4mv = 264m \cdot ω

8v = 264 ω (v)/(ω) = (264)/(8) = 33

Rotational dynamics post-collision diagram for Q29 - JEE Main 2026 Evening — The figure illustrates a uniform bar of length 12 cm with two point masses m and 2m approaching it perpendicularly from opposite directions.

Step 1: Final Conclusion

The ratio $(v)/(ω)$ $\frac{v}{\omega}$ is 33.

Pattern Recognition

Collision on rotating bar: Conserve angular momentum about COM of rod since net external torque about COM is zero during collision.

Chapter Mix

Class 11 Physics: System of Particles and Rotational Motion

Q45 jee_main_2026_22_january_evening Kinetic Energy of Multi-Particle Systems

Given below are two statements : Statement I : For a mechanical system of many particles total kinetic energy is the sum of kinetic energies of all the particles. Statement II : The total kinetic energy can be the sum of kinetic energy of the center of mass w.r.t. to the origin and the kinetic energy of all the particles w.r.t. the center of mass as the reference. In the light of the above statements, choose the correct answer from the options given below :

A. Both Statement I and Statement II are true
B. Statement I is true but Statement II is false
C. Statement I is false but Statement II is true
D. Both Statement I and Statement II are false

Solution

Related Formula

Ktotal = Σ (1)/(2) mᵢ vᵢ²

Ktotal = (1)/(2) Mtotal vcm² + Krel, cm

Core Logic

Evaluating Statement I: By definition, total kinetic energy of a discrete system of particles is scalar sum of kinetic energies of individual particles: $KE = Σ (1)/(2)mᵢ vᵢ²$ $KE = \sum \frac{1}{2}m_i v_i^2$. Statement I is true.

Evaluating Statement II: According to König's theorem, total kinetic energy breaks down into kinetic energy of COM translation plus internal kinetic energy relative to COM:

KE = (1)/(2)(m₁+m₂) vcm² + (1)/(2) (m₁ m₂)/(m₁+m₂) | v₁ - v₂|²

Statement II is also true.

Step 1: Final Conclusion

Both Statement I and Statement II are true.

Pattern Recognition

König's Theorem for system of particles: $Ktotal = Kcm + Kw.r.t cm$ . Both definitions represent valid expressions for mechanical energy.

Chapter Mix

Class 11 Physics: System of Particles and Rotational Motion

Q47 jee_main_2026_22_january_evening Pulley-Mass System Dynamics

Two masses m and 2m are connected by a light string going over a pulley (disc) of mass 30m with radius r = 0.1 m. The pulley is mounted in a vertical plane and it is free to rotate about its axis. The 2m mass is released from rest and its speed when it has descended through a height of 3.6 m is ____ m/s. (Assume string does not slip and g = 10 m/s

²

)

Numerical Answer. Answer: 2 to 2

Solution

Related Formula

Δ U + Δ K = 0

Ipulley = (1)/(2) M R² = (1)/(2) (30m) R² = 15m R²

ω = (v)/(R)

Core Logic

Using law of conservation of mechanical energy:

Loss in potential energy = Gain in kinetic energy

2m g h - m g h = (1)/(2) m v² + (1)/(2) (2m) v² + (1)/(2) I ω²

m g h = (3)/(2) m v² + (1)/(2) (15 m R²) ((v²)/(R²))

m g h = (3)/(2) m v² + (15)/(2) m v² = 9 m v²

v = \sqrt((g h)/(9))

Substituting $g = 10 ~m/s²$ $g = 10 \mathrm{~m/s}^2$ and $h = 3.6 ~m$ $h = 3.6 \mathrm{~m}$:

v = \sqrt((10 \times 3.6)/(9)) = \sqrt((36)/(9)) = \sqrt(4) = 2 ~m/s

Pulley mass system energy conservation diagram for Q47 - JEE Main 2026 Evening

Step 1: Final Conclusion

The speed of the mass is $2 ~m/s$ $2 \mathrm{~m/s}$.

Pattern Recognition

Energy conservation on pulley system: Net potential loss $mgh$ $mgh$ = sum of linear and rotational kinetic energies $9mv² v = \sqrt(gh/9)$ $9mv^2 \implies v = \sqrt{gh/9}$.

Chapter Mix

Class 11 Physics: System of Particles and Rotational Motion

Q33 jee_main_2026_23_january_morning Moment of Inertia

The moment of inertia of a square loop made of four uniform solid cylinders, each having radius R and length L (R < L) about an axis passing through the mid points of opposite sides, is (Take the mass of the entire loop as M):

A. $(3)/(8)MR² +(7)/(12)ML²$ $\frac{3}{8}MR^2 +\frac{7}{12}ML^2$
B. $(3)/(4)MR² +(1)/(6)ML²$ $\frac{3}{4}MR^2 +\frac{1}{6}ML^2$
C. $(3)/(4)MR² +(7)/(12)ML²$ $\frac{3}{4}MR^2 +\frac{7}{12}ML^2$
D. $(3)/(8)MR² +(1)/(6)ML²$ $\frac{3}{8}MR^2 +\frac{1}{6}ML^2$

Solution

Related Formula

For a solid cylinder: Parallel to its length passing through center: $I = M'R²2$ $I = \frac{M'R^{2}}{2}$ Perpendicular to length through center: $I = M'R²4 + M'L²12$ Parallel Axis Theorem: $I = Icm + M'd²$ $I = I_{\text{cm}} + M'd^{2}$

Core Logic

The square loop is formed by four identical solid cylinders, each of mass $M' = (M)/(4)$ $M' = \frac{M}{4}$. The given axis passes through the mid-points of two opposite cylinders. This means two cylinders have the axis passing perpendicularly through their centers, and the other two cylinders are parallel to the axis at a distance of $L/2$ $L/2$.

Step 1: Moment of Inertia for cylinders bisected perpendicularly

For the two cylinders perpendicular to the axis of rotation:

I₁ = 2 \times ( M'R²4 + M'L²12)

Step 2: Moment of Inertia for cylinders parallel to axis

For the two cylinders parallel to the axis, distance $d = L/2$ $d = L/2$. Apply the parallel axis theorem:

I₂ = 2 \times [ M'R²2 + M'((L)/(2))²]

Step 3: Total Moment of Inertia

Iₙₑₜ = I₁ + I₂ = 2( M'R²4 + M'L²12) + 2( M'R²2 + M'L²4)

Iₙₑₜ = M'R²2 + M'L²6 + M'R² + M'L²2

Iₙₑₜ = 3M'R²2 + 4M'L²6 = 3M'R²2 + 2M'L²3

Step 4: Substitute Total Mass

Substitute $M' = M/4$ $M' = M/4$:

I = (3)/(2)((M)/(4))R² + (2)/(3)((M)/(4))L²

I = (3)/(8)MR² + (1)/(6)ML²

Pattern Recognition

Sees: "square loop of cylinders" + "mass M of entire loop" → Always remember $Mᵢ = M/4$ $M_{i} = M/4$. Calculate individual moment of inertia carefully considering whether the cylinder is oriented parallel or perpendicular.

Chapter Mix

Class 11 Physics: Systems 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 2

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Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Moment of Inertia for cylinders bisected perpendicularly

Step 2: Moment of Inertia for cylinders parallel to axis

Step 3: Total Moment of Inertia

Step 4: Substitute Total Mass

Pattern Recognition

Chapter Mix

More System of Particles and Rotational Motion Questions — jee_main_2025_28_jan_morning

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