System of Particles and Rotational Motion Previous Year Questions — JEE Main Physics
58 past-year System of Particles and Rotational Motion questions from JEE Main (Physics) — page 2.
Q6 (2025)
A rod of length $5 \mathrm{~L}$ is bent right angle keeping one side length as $2 \mathrm{~L}$ . {{IMG1}} The position of the centre of mass of the system : (Consider $\mathrm{L} = 10\mathrm{cm}$ )
- $2\hat{\mathbf{i}} + 3\hat{\mathbf{j}}$
- $3\hat{\mathbf{i}} + 7\hat{\mathbf{j}}$
- $5\hat{\mathbf{i}} + 8\hat{\mathbf{j}}$
- $4\hat{\mathbf{i}} + 9\hat{\mathbf{j}}$
### Related Formula
For a continuous system modeled as discrete point masses located at their respective centers of mass:
$$x_{\text{com}} = \frac{m_1x_1 + m_2x_2}{m_1 + m_2}$$
$$y_{\text{com}} = \frac{m_1y_1 + m_2y_2}{m_1 + m_2}$$
### Core Logic
Let the uniform linear mass density of the rod be $\lambda$.
- Total length is $5L$.
- One segment of length $2L$ lies on the x-axis. Its mass is $2m = \lambda(2L)$ and its center of mass is at $(L, 0)$.
- The remaining segment of length $3L$ lies on the y-axis. Its mass is $3m = \lambda(3L)$ and its center of mass is at $(0, 1.5L)$.
### Step 1: Calculate COM Coordinates
Find the coordinates of the system's center of mass:
$$x_{\text{com}} = \frac{2m(L) + 3m(0)}{2m + 3m} = \frac{2L}{5} = 0.4L$$
$$y_{\text{com}} = \frac{2m(0) + 3m(1.5L)}{2m + 3m} = \frac{4.5L}{5} = 0.9L$$
Given $L = 10 \mathrm{~cm}$:
$$x_{\text{com}} = 0.4 \times 10 = 4 \mathrm{~cm}$$
$$y_{\text{com}} = 0.9 \times 10 = 9 \mathrm{~cm}$$
### Step 2: Vector Form
Expressing in vector notation:
$$\vec{r}_{\text{com}} = 4\hat{\mathbf{i}} + 9\hat{\mathbf{j}}$$
### Pattern Recognition
Sees: L-shaped rod formed by bending a total length $L_{\text{total}}$.
Shortcut: Treat each arm as a point mass at its geometric midpoint. For segments of ratio $2:3$, the COM divides the distance between their midpoints in the inverse ratio $3:2$ closer to the heavier segment on the y-axis.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: System of Particles and Rotational Motion
Q21 (2025)
A, B and C are disc, solid sphere and spherical shell respectively with same radii and masses. These masses are placed as shown in figure. {{IMG1}} The moment of inertia of the given system about PQ is $\frac{\mathrm{x}}{15}\mathrm{I},$ where I is the moment of inertia of the disc about its diameter. The value of x is
### Related Formula
Parallel Axis Theorem:
$$I_{\text{axis}} = I_{\text{com}} + M R^2$$
Standard Moments of Inertia about center of mass:
- Disc about diameter: $I_{\text{disc,dia}} = \frac{MR^2}{4}$
- Solid sphere: $I_{\text{sphere}} = \frac{2}{5}MR^2$
- Spherical shell: $I_{\text{shell}} = \frac{2}{3}MR^2$
### Core Logic
The axis of rotation PQ passes through the center of the top disc (A) along its diameter.
- Top disc (A):
$$I_A = \frac{MR^2}{4}$$
- Bottom-left solid sphere (B): Center lies at distance $R$ from the axis PQ.
$$I_B = I_{\text{com}} + M R^2 = \frac{2}{5}MR^2 + MR^2 = \frac{7}{5}MR^2$$
- Bottom-right spherical shell (C): Center lies at distance $R$ from the axis PQ.
$$I_C = I_{\text{com}} + M R^2 = \frac{2}{3}MR^2 + MR^2 = \frac{5}{3}MR^2$$
### Step 1: Calculate Total System Moment of Inertia
Sum the contributions:
$$I_{\text{PQ}} = I_A + I_B + I_C$$
$$I_{\text{PQ}} = \frac{MR^2}{4} + \frac{7}{5}MR^2 + \frac{5}{3}MR^2$$
To add the fractions, find a common denominator ($60$):
$$I_{\text{PQ}} = \left( \frac{15 + 84 + 100}{60} \right) MR^2 = \frac{199}{60} MR^2$$
### Step 2: Express in terms of standard Disc Moment
We are given $I = \frac{MR^2}{4} \implies MR^2 = 4I$. Substitute this in the expression:
$$I_{\text{PQ}} = \frac{199}{60} (4I) = \frac{199}{15} I$$
Comparing with $I_{\text{PQ}} = \frac{x}{15} I$ yields $x = 199$.
### Pattern Recognition
Sees: Composite body consisting of three standard symmetric shapes about a tangent/offset axis.
Shortcut: Sum the central inertia terms and the offset terms separately. Offset masses are only B and C, so the offset sum is $2MR^2$. The central sum is $(1/4 + 2/5 + 2/3)MR^2$. Adding these directly yields the combined fractional factor of $199/60$.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: System of Particles and Rotational Motion
Q2 (2025)
A rod of linear mass density $\lambda^{\prime}$ and length $L$ is bent to form a ring of radius $R$. Moment of inertia of ring about any of its diameter is:
- $\frac{\lambda L^{3}}{16\pi^{2}}$
- $\frac{\lambda L^{3}}{12}$
- $\frac{\lambda L^{3}}{4\pi^{2}}$
- $\frac{\lambda L^{3}}{8\pi^{2}}$
### Related Formula
$$I_{\text{dia}} = \frac{1}{2} M R^2$$
where,
$I_{\text{dia}}$ = moment of inertia of a ring about its diameter
$M$ = total mass of the ring
$R$ = radius of the ring
### Core Logic
Since the linear mass density is $\lambda^{\prime}$ (or $\lambda$ as per the options), the total mass $M$ of the rod of length $L$ is:
$M = \lambda L$
When this rod is bent into a ring of radius $R$, its circumference equals the length of the rod:
$$2\pi R = L \implies R = \frac{L}{2\pi}$$
Substituting $M$ and $R$ into the formula for the moment of inertia about the diameter:
$$I_{\text{dia}} = \frac{1}{2} M R^2 = \frac{1}{2} (\lambda L) \left(\frac{L}{2\pi}\right)^2 = \frac{\lambda L^3}{8\pi^2}$$
### Pattern Recognition
Sees: "Rod of length $L$ bent to form a ring" → $R = \frac{L}{2\pi}$.
Trap: Moment of inertia about the central axis perpendicular to the plane is $MR^2$, but about its diameter, it is half, i.e., $\frac{1}{2}MR^2$.
Shortcut: $I = \frac{1}{2} (\lambda L) \left(\frac{L}{2\pi}\right)^2 = \frac{\lambda L^3}{8\pi^2}$. ✓
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Rotational Motion
Q24 (2025)
A thin solid disk of $1\mathrm{~kg}$ is rotating along its diameter axis at the speed of $1800\mathrm{~rpm}$. By applying an external torque of $25\pi\mathrm{~N\cdot m}$ for $40\mathrm{~s}$, the speed increases to $2100\mathrm{~rpm}$. The diameter of the disk is ________ $\mathrm{~m}$.
### Related Formula
$$\omega = 2\pi \frac{N}{60}$$
$$\omega_f = \omega_i + \alpha t$$
$\tau = I \alpha$
$$I_{\text{dia}} = \frac{1}{4} m R^2$$
where,
$N$ = rotational speed in rpm
$\alpha$ = angular acceleration
$\tau$ = torque applied
$I_{\text{dia}}$ = moment of inertia of a solid disk about its diameter axis
### Core Logic
Given parameters:
- Mass, $m = 1\mathrm{~kg}$
- Initial speed, $N_i = 1800\mathrm{~rpm} \implies \omega_i = \frac{1800 \times 2\pi}{60} = 60\pi\mathrm{~rad/s}$
- Final speed, $N_f = 2100\mathrm{~rpm} \implies \omega_f = \frac{2100 \times 2\pi}{60} = 70\pi\mathrm{~rad/s}$
- Time, $t = 40\mathrm{~s}$
- Torque, $\tau = 25\pi\mathrm{~N\cdot m}$
First, calculate the angular acceleration $\alpha$:
$$\omega_f = \omega_i + \alpha t \implies 70\pi = 60\pi + \alpha (40)$$
$$\alpha = \frac{10\pi}{40} = \frac{\pi}{4}\mathrm{~rad/s}^{2}$$
Now relate torque to moment of inertia:
$$\tau = I \alpha \implies 25\pi = \left( \frac{1}{4} m R^2 \right) \left( \frac{\pi}{4} \right)$$
$$25\pi = \frac{1}{4} (1) R^2 \frac{\pi}{4} \implies 25\pi = R^2 \frac{\pi}{16}$$
$$R^2 = 400 \implies R = 20\mathrm{~m}$$
### Step 1: Compute Diameter
The question asks for the diameter of the disk ($D$):
$$D = 2R = 2 \times 20 = 40\mathrm{~m}$$
### Pattern Recognition
Sees: Torque acting on a rotating disk increasing its speed.
Trap: The axis of rotation is the *diameter* axis, not the normal geometric center axis. This means the moment of inertia is $I = \frac{1}{4} m R^2$, not $\frac{1}{2} m R^2$! Check this detail carefully. ✓
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Rotational Motion
Q25 (2025)
A cube having a side of $10\mathrm{~cm}$ with unknown mass and $200\mathrm{~gm}$ mass were hung at two ends of an uniform rigid rod of $27\mathrm{~cm}$ long. The rod along with masses was placed on a wedge keeping the distance between wedge point and $200\mathrm{~gm}$ weight as $25\mathrm{~cm}$. Initially the masses were not at balance. A beaker is placed beneath the unknown mass and water is added slowly to it. At given point the masses were in balance and half volume of the unknown mass was inside the water.
(Take the density of unknown mass is more than that of the water, the mass did not absorb water and water density is $1\mathrm{~g/cm}^{3}$.)
The unknown mass is ________ $\mathrm{kg}$.
### Related Formula
$$\sum \tau = 0 \quad \text{(Rotational Equilibrium)}$$
$$F_{\text{net}} = m g - F_B$$
$$F_B = \rho_{\text{water}} V_{\text{submerged}} g$$
where,
$\tau$ = torque about the wedge point
$F_B$ = buoyancy force
### Core Logic
Let's list the geometric parameters from the setup:
- Length of uniform rigid rod, $L = 27\mathrm{~cm}$.
- Wedge is positioned such that the distance to the $200\mathrm{~g}$ ($0.2\mathrm{~kg}$) weight is $d_1 = 25\mathrm{~cm}$.
- Therefore, the distance to the unknown mass $M$ at the other end is $d_2 = 27 - 25 = 2\mathrm{~cm}$.
Calculate the volume of the cube:
- Side of the cube, $a = 10\mathrm{~cm} = 0.1\mathrm{~m}$.
- Volume, $V = a^3 = 1000\mathrm{~cm}^3 = 10^{-3}\mathrm{~m}^3$.
When the system is balanced, half the volume of the cube is submerged in water:
- Submerged volume, $V_{\text{sub}} = \frac{V}{2} = 500\mathrm{~cm}^3 = 5 \times 10^{-4}\mathrm{~m}^3$.
- Buoyancy force:
$$F_B = \rho_w V_{\text{sub}} g = 1000\mathrm{~kg/m}^3 \times \left(5 \times 10^{-4}\mathrm{~m}^3\right) \times g = 0.5 g\mathrm{~N}$$
### Step 1: Torque Balance Equation
For rotational equilibrium, balance the torques about the wedge point $O$:
- Torque on the left (unknown mass branch): $\tau_{\text{left}} = \left(M g - F_B\right) \times d_2 = (M g - 0.5 g) \times 2$
- Torque on the right ($200\mathrm{~g}$ mass branch): $\tau_{\text{right}} = 0.2 g \times d_1 = 0.2 g \times 25$
$$\tau_{\text{left}} = \tau_{\text{right}}$$
$$(M g - 0.5 g) \times 2 = 0.2 g \times 25$$
$2 (M - 0.5) = 5$
$$M - 0.5 = 2.5 \implies M = 3\mathrm{~kg}$$
Thus, the unknown mass is $3\mathrm{~kg}$.
### Pattern Recognition
Sees: Rod torque balance + buoyancy force on one end.
Trap: Ensure you measure the distances from the pivot point (the wedge). The unknown mass is at $27 - 25 = 2\mathrm{~cm}$ from the wedge.
Shortcut: Since $g$ appears in both gravity and buoyancy terms, it cancels out immediately. Balancing torque simplifies directly to resolving mass differences: $2(M - 0.5) = 0.2 \times 25 = 5$. This yields $M = 3$ instantly! ✓
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Rotational Motion
Class 11 Physics: Mechanical Properties of Fluids
