A rod of length 5 ~L is bent right angle keeping one side length as 2 ~L . The position of the centre of mass of the…

Year	2026	2025	2024	Total
Questions	20	27	11	58

Q15 jee_main_2025_29_jan_evening Angular Momentum of a System of Particles

Three equal masses

m

are kept at vertices (A, B, C) of an equilateral triangle of side

a

in free space. At

t = 0

, they are given an initial velocity

VA = V₀ AC

,

VB = V₀ BA

and

VC = V₀ CB

. Here,

AC, CB

and

BA

are unit vectors along the edges of the triangle. If the three masses interact gravitationally, then the magnitude of the net angular momentum of the system at the point of collision is:

Angular Momentum of a System of Particles diagram for Q15 - JEE Main 2025 Evening — The graphic exhibits three mass particles at the vertices of an equilateral triangle with velocity vectors pointed along the cyclic boundary directions.

A. $(1)/(2) ~a ~mV₀$
B. $3amV₀$ $3\mathrm{amV}_0$
C. $\sqrt(3)2 ~a ~mV₀$
D. $(3)/(2) ~a ~mV₀$

Solution

Related Formula

L = Σ ( rᵢ \times m vᵢ)

τₑₓₜ = d Ldt

Core Logic

Since the three masses interact purely through mutual internal gravitational forces, the net external torque acting on the system about any central reference point is zero:

τₑₓₜ = 0 Linitial = Lfinal

Let us compute the total angular momentum about the centroid of the equilateral triangle:

Angular Momentum Calculation Geometry diagram for Q15 - JEE Main 2025 Evening — The graphic exhibits three mass particles at the vertices of an equilateral triangle with velocity vectors pointed along the cyclic boundary directions.

From trigonometry, the perpendicular distance from the centroid to the velocity vector along any edge is:

r⊥ = a2\sqrt(3)

The initial angular momentum for one mass about the centroid is $L₁ = m V₀ r⊥$ $L_1 = m V_0 r_{\perp}$. Since all three particles move cyclically in the same direction, their angular momenta reinforce cleanly:

Lₙₑₜ = 3 \cdot (m V₀ a2\sqrt(3)) = 32\sqrt(3) m V₀ a = \sqrt(3)2 a m V₀

By conservation of angular momentum, this configuration value remains unchanged up to the point of collision.

Pattern Recognition

Mutual internal central forces can never alter the angular momentum of a system. Hence, the solution completely reduces to measuring the static configuration values at $t=0$ $t=0$ about the center of mass symmetry.

Chapter Mix

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

Q19 jee_main_2025_28_jan_morning Centre of Mass of Continuous Mass Distribution

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

(σ)

varies as

σ = (σ₀x)/(ab)

(where

σ₀

is a constant), would be

Centre of Mass diagram for Q19 - JEE Main 2025 Morning — A thin plate with variable linear density coordinates mapped across an XY grid system.

A. $((2)/(3) a, (b)/(2))$
B. $((2)/(3) a, (2)/(3) b)$
C. $((a)/(2),(b)/(2))$ $\left(\frac{a}{2},\frac{b}{2}\right)$
D. $((1)/(3) a, (b)/(2))$

Solution

Core Logic

Since density $σ$ $\sigma$ is independent of the y-coordinate, the vertical center of mass resolves directly by symmetry:

ycm = b2

Integration element tracking for continuous mass distribution on Q19 — A thin plate with variable linear density coordinates mapped across an XY grid system.

To find the horizontal center of mass, evaluate the continuous mass integral along the x-axis:

xcm = \int₀a x dm\int₀a dm = \int₀a x ( σ₀ xab) b dx\int₀a ( σ₀ xab) b dx

xcm = \int₀a x² dx\int₀a x dx = [ x³3 ]₀a[ x²2 ]₀a = a³ / 3a² / 2 = 2a3

Step 1: Final Position Coordinates

The center of mass coordinates are $((2)/(3) a, b2)$ , which matches option (1).

Pattern Recognition

When density varies linearly with position ( $σ \propto x$ $\sigma \propto \mathrm{x}$), the mass distribution shifts outward, moving the center of mass from the geometric midpoint $a2$ $\frac{\mathrm{a}}{2}$ to the $(2)/(3)a$ $\frac{2}{3}\mathrm{a}$ mark.

Chapter Mix

Class 11 Physics: System of Particles and Rotational Motion

Q22 jee_main_2025_28_jan_morning Moment of Inertia

The moment of inertia of a solid disc rotating along its diameter is 2.5 times higher than the moment of inertia of a ring rotating in similar way. The moment of inertia of a solid sphere which has same radius as the disc and rotating in similar way, is n times higher than the moment of inertia of the given ring. Here,

n =

_____. Consider all the bodies have equal masses.

Numerical Answer. Answer: 4 to 4

Solution

Related Formula

Idisc = MR₁²4, Iring = MR₂²2, Isphere = 2MR₁²5

Core Logic

Let's list the relevant moment of inertia formulas based on their rotation axes:

Moment of inertia axial comparison steps for Q22

From the given problem statements:

IdiscIring = 2.5 MR₁²4 MR₂²2 = (5)/(2) R₁²R₂² = 5

Now, evaluating the second geometric layout ratio:

IsphereIring = n 2MR₁²5 MR₂²2 = n 4R₁²5R₂² = n

Substituting our radius parameter ( $R₁²R₂² = 5$ ):

n = (4)/(5) \cdot 5 = 4

Step 1: Final Value Conclusion

The scale value parameter is found to be:

$n = 4$ $\mathrm{n} = 4$

Pattern Recognition

Be careful with rotation axis descriptions. Disc and ring components rotating along their structural diameter axes use values that are half of their standard perpendicular planar formulas.

Chapter Mix

Class 11 Physics: System of Particles and Rotational Motion

Q24 jee_main_2025_28_jan_morning Moment of Inertia

Two iron solid discs of negligible thickness have radii

R₁

and

R₂

and moment of inertia

I₁

and

I₂

, respectively. For

R₂ = 2R₁

, the ratio of

I₁

and

I₂

would be

1 / x

, where

x =

Numerical Answer. Answer: 16 to 16

Solution

Core Logic

Since mass scales with the face surface area for discs of identical thickness and material composition:

M = σ \cdot π R² M \propto R²

Disk mass allocation scaling profile diagram for Q24

M₁ = M₀, M₂ = σ π (2R₁)² = 4M₀

The moment of inertia formula for a disc is:

mathrmI = (1)/(2)MR² I \propto MR² \propto R⁴

Calculating the ratio for the given radii configuration:

I₁I₂ = M₁ R₁²M₂ R₂² = M₀ \cdot R₁²4M₀ \cdot (2R₁)² = (1)/(16)

Step 1: Value Convergence

Comparing this fraction to $1/x$ $1/\mathrm{x}$ yields:

$x = 16$ $\mathrm{x} = 16$

Pattern Recognition

For 2D uniform laminar objects, scaling the radius changes both the mass factor (by $R²$ $\mathrm{R}^2$) and the distribution distance (by $R²$ $\mathrm{R}^2$), resulting in an overall $R⁴$ $\mathrm{R}^4$ dependency rule.

Chapter Mix

Class 11 Physics: System of Particles and Rotational Motion

Q10 jee_main_2025_03_april_morning Rolling Without Slipping

A force of

49~N

acts tangentially at the highest point of a sphere (solid) of mass

20~kg

, kept on a rough horizontal plane. If the sphere rolls without slipping, then the acceleration of the center of the sphere is:

Solid sphere with tangential force at top point for Q10 — A schematic of a solid sphere of mass m resting on a horizontal plane with a force F pointing horizontally to the right at the highest point.

A. $3.5~m/s²$ $3.5\mathrm{~m/s}^2$
B. $0.35~m/s²$ $0.35\mathrm{~m/s}^2$
C. $2.5~m/s²$ $2.5\mathrm{~m/s}^2$
D. $0.25~m/s²$ $0.25\mathrm{~m/s}^2$

Solution

Related Formula

Torque equation about the instantaneous center of zero velocity (bottom contact point $P$ $P$):

τP = IP α

For a solid sphere, the moment of inertia about the center is $Ic = (2)/(5)MR²$ $I_c = \frac{2}{5}MR^2$. By the parallel axis theorem:

IP = Ic + MR² = (7)/(5)MR²

Core Logic

Since the sphere rolls without slipping, we can conveniently write the torque equation about the lowest point of contact $P$ $P$ because static friction passes through this point and exerts zero torque.

Distance from point $P$ $P$ to the top highest point is $2R$ $2R$.
Tangential force $F = 49~N$ $F = 49\mathrm{~N}$.
Mass of solid sphere, $M = 20~kg$ $M = 20\mathrm{~kg}$.

τP = F \times 2R

Substitute $τP$ $\tau_P$ and $IP$ $I_P$ into the torque equation:

F \times 2R = ((7)/(5)MR²) α

Step 1: Solving for Linear Acceleration

For pure rolling, the acceleration of the center of mass $a$ $a$ is related to angular acceleration $α$ $\alpha$ by $a = Rα$ $a = R\alpha$:

2F R = (7)/(5)MR² ((a)/(R))

2F = (7)/(5) M a a = (10F)/(7M)

Substitute the numerical values ( $F = 49~N$ $F = 49\mathrm{~N}$ and $M = 20~kg$ $M = 20\mathrm{~kg}$):

a = (10 \times 49)/(7 \times 20) = (490)/(140) = 3.5~m/s²

Step 2: Analysis of Friction Force Direction

Let's write force equations to verify consistency: $F + f = M a$ $F + f = M a$

49 + f = 20 \times 3.5 = 70 f = 21~N

Since $f$ $f$ is positive, static friction acts in the forward direction. Rolling without slipping is fully maintained since the required static friction coefficient is well within realistic limits.

Pattern Recognition

Calculating torque about the bottom contact point is a powerful shortcut for rolling-without-slipping questions! It completely bypasses having to guess or set up equations for the friction direction.

Chapter Mix

Class 11 Physics: System of Particles and Rotational Motion

A rod of length 5 ~L$5 \mathrm{~L}$ is bent right angle keeping one side length as 2 ~L$2 \mathrm{~L}$ . Diagram of an L-shaped rod aligned with the x and y axes, with lengths 2L and 3L respectively. The position of the centre of mass of the system : (Consider L = 10cm$\mathrm{L} = 10\mathrm{cm}$ )

Solution & Explanation

Related Formula

Core Logic

Step 1: Calculate COM Coordinates

Step 2: Vector Form

Pattern Recognition

Chapter Mix

Reference Study Guides

More System of Particles and Rotational Motion Previous-Year Questions — Page 7

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Final Position Coordinates

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Final Value Conclusion

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Value Convergence

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Solving for Linear Acceleration

Step 2: Analysis of Friction Force Direction

Pattern Recognition

Chapter Mix

More System of Particles and Rotational Motion Questions — jee_main_2025_07_april_morning

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