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).
Q21 (2025)
A wheel of radius $0.2 \mathrm{~m}$ rotates freely about its center when a string that is wrapped over its rim is pulled by force of $10 \mathrm{~N}$ as shown in figure. The established torque produces an angular acceleration of $2 \mathrm{rad} / \mathrm{s}^{2}$ . Moment of inertia of the wheel is ________ kg $m^2$ .
(Acceleration due to gravity $= 10\mathrm{m} / \mathrm{s}^2$)
{{IMG1}}
### Related Formula
1. Torque ($\tau$) produced by a tangential pulling force:
$$\tau = F \cdot R$$
2. Newton's second law for rotation:
$$\tau = I \cdot \alpha$$
### Core Logic
We are given:
- Radius of the wheel $R = 0.2 \ \mathrm{m}$
- Applied force $F = 10 \ \mathrm{N}$
- Angular acceleration $\alpha = 2 \ \mathrm{rad/s^2}$
### Step 1: Calculate torque and moment of inertia
First, find the torque $\tau$:
$$\tau = F \cdot R = 10 \ \mathrm{N} \times 0.2 \ \mathrm{m} = 2 \ \mathrm{N \cdot m}$$
Next, calculate the moment of inertia $I$ using $\tau = I\alpha$:
$$I = \frac{\tau}{\alpha} = \frac{2 \ \mathrm{N \cdot m}}{2 \ \mathrm{rad/s^2}} = 1 \ \mathrm{kg \cdot m^2}$$
Thus, the moment of inertia is $1 \mathrm{~kg\cdot m^2}$.
### Pattern Recognition
Sees: Pulley/wheel torque with basic rotational dynamics.
Trap: Attempting to integrate gravity ($g = 10\text{ m/s}^2$) into mass equations. Gravity is a redundant distractor here because the pulling tension force is explicitly defined!
Shortcut: Directly compute torque as $\tau = F R$ and divide by the angular acceleration $\alpha$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: System of Particles and Rotational Motion
Q3 (2025)
The moment of inertia of a circular ring of mass $M$ and diameter $r$ about a tangential axis lying in the plane of the ring is:
- $\frac{1}{2} M r^2$
- $\frac{3}{8} M r^2$
- $\frac{3}{2} M r^2$
- $2 M r^2$
### Related Formula
1. Moment of inertia of a circular ring of mass $M$ and radius $R$ about a diametrical axis in its plane:
$$I_{\text{dia}} = \frac{1}{2} M R^2$$
2. Parallel Axis Theorem:
$$I = I_{\text{cm}} + M d^2$$
where $d$ is the perpendicular distance between the center of mass axis and the parallel axis.
### Core Logic
The axis of rotation is tangential and lies in the plane of the ring. Thus, the distance from the center of mass axis (which is also diametrical and in-plane) is equal to the radius $R$.
Using the parallel axis theorem:
$$I_{\text{tangent}} = I_{\text{dia}} + M R^2 = \frac{1}{2} M R^2 + M R^2 = \frac{3}{2} M R^2$$
### Step 1: Express in terms of diameter $r$
The question specifies the diameter of the ring is $r$.
Therefore, the radius $R$ is:
$$R = \frac{r}{2}$$
Substitute $R = \frac{r}{2}$ into the moment of inertia formula:
$$I_{\text{tangent}} = \frac{3}{2} M \left(\frac{r}{2}\right)^2 = \frac{3}{8} M r^2$$
Thus, the moment of inertia is $\frac{3}{8} M r^2$.
### Pattern Recognition
Sees: Circular ring tangential axis in-plane.
Trap: Directly using radius $r$ instead of diameter $r$. Read variables carefully: often $r$ is radius, but here it is explicitly given as diameter!
Shortcut: Standard in-plane tangent is $\frac{3}{2}MR^2$. Substitute $R = \frac{r}{2}$ to get $\frac{3}{8}Mr^2$.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: System of Particles and Rotational Motion
Q88 (2025)
A square Lamina OABC of length $10\mathrm{~cm}$ is pivoted at 'O'. Forces act on the Lamina as shown in the figure. If the Lamina remains stationary, then the magnitude of $F$ is:
{{IMG1}}
- $20\mathrm{~N}$
- 0 (zero)
- $10\mathrm{~N}$
- $10\sqrt{2}\mathrm{~N}$
### Related Formula
$$\tau_O = F \cdot r_{\perp}$$
$$\sum \tau_O = 0 \quad \text{(for rotational equilibrium)}$$
### Core Logic
Let the side length of the square lamina be $l = 10\mathrm{~cm}$.
The lamina is pivoted at point $O(0,0)$ and remains stationary under rotational equilibrium. Therefore, the net torque about $O$ must be zero.
Evaluating torque contributions about point $O$:
- Forces acting directly at pivot $O$ produce zero torque.
- Forces whose lines of action pass through $O$ produce zero torque.
- The $10\mathrm{~N}$ force perpendicular to side $OA$ produces torque:
$$\tau_1 = 10 \times l \quad \text{(Counter-Clockwise)}$$
- The unknown force $F$ acting perpendicular to side $OC$ produces torque:
$$\tau_2 = F \times l \quad \text{(Clockwise)}$$
Setting $\sum \tau_O = 0$:
$$10 \cdot l - F \cdot l = 0 \implies F = 10\mathrm{~N}$$
### Step 1: Final Conclusion
The magnitude of the force $F$ is $10\mathrm{~N}$.
### Pattern Recognition
In pivoted laminas, focus on the pivot and disregard any force vector whose line of action passes through the pivot. For symmetric placements, equate Clockwise torque = Counter-Clockwise torque directly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Rotational Motion
Q3 (2025)
A cord of negligible mass is wound around the rim of a wheel supported by spokes with negligible mass. The mass of wheel is $10\mathrm{~kg}$ and radius is $10\mathrm{~cm}$ and it can freely rotate without any friction. Initially the wheel is at rest. If a steady pull of $20\mathrm{~N}$ is applied on the cord, the angular velocity of the wheel, after the cord is unwound by $1\mathrm{~m}$, would be:
{{IMG1}}
- $20\mathrm{~rad/s}$
- $30\mathrm{~rad/s}$
- $10\mathrm{~rad/s}$
- $0\mathrm{~rad/s}$
### Related Formula
$W_F = F \cdot s$
$$K_R = \frac{1}{2} I \omega^2$$
$$I = M R^2 \quad \text{(Moment of inertia of a ring/rim)}$$
### Core Logic
The work done by the constant pulling force $F = 20\mathrm{~N}$ through distance $s = 1\mathrm{~m}$ is completely converted into the rotational kinetic energy of the wheel.
Work done by force:
$$W_F = F \cdot s = 20 \times 1 = 20\mathrm{~J}$$
Since the spokes are of negligible mass, all mass $M = 10\mathrm{~kg}$ is distributed on the outer rim of radius $R = 10\mathrm{~cm} = 0.1\mathrm{~m}$. The wheel acts as a thin ring:
$$I = M R^2 = 10 \times (0.1)^2 = 0.1\mathrm{~kg\cdot m^2}$$
Using the work-energy theorem:
$$W_F = \Delta K_R = \frac{1}{2} I \omega^2$$
$$20 = \frac{1}{2} \times 0.1 \times \omega^2$$
$$40 = 0.1 \omega^2 \implies \omega^2 = 400 \implies \omega = 20\mathrm{~rad/s}$$
### Step 1: Final Conclusion
The angular velocity of the wheel after the cord is unwound by $1\mathrm{~m}$ is $20\mathrm{~rad/s}$.
### Pattern Recognition
Work done in unwinding a string is $F \cdot s$. Under pure rotation with zero friction, this is exactly $\frac{1}{2} I \omega^2$. Always identify the mass distribution (here, rim with massless spokes acts as a thin cylinder/ring $I = MR^2$).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Rotational Motion
Class 11 Physics: Work, Energy and Power
Q16 (2025)
Moment of inertia of a rod of mass 'M' and length 'L' about an axis passing through its center and normal to its length is 'α'. Now the rod is cut into two equal parts and these parts are joined symmetrically to form a cross shape. Moment of inertia of cross about an axis passing through its center and normal to plane containing cross is:
- $\alpha$
- $\alpha / 4$
- $\alpha / 8$
- $\alpha / 2$
### Related Formula
$$I = \frac{1}{12} M L^2$$
### Core Logic
Initially, the moment of inertia is:
$$\alpha = \frac{M L^2}{12}$$
When cut into two equal parts, each smaller rod has:
- Mass, $m = \frac{M}{2}$
- Length, $l = \frac{L}{2}$
When joined symmetrically as a cross, the target axis passes through their joint intersection perpendicular to their plane.
For each rod, this axis passes through its individual center of mass and is perpendicular to its length.
Thus, the total moment of inertia of the cross is the sum of the moments of inertia of the two rods:
$$I_{\text{cross}} = I_1 + I_2 = 2 \times \left(\frac{1}{12} m l^2\right) = \frac{1}{6} m l^2$$
Substituting $m = \frac{M}{2}$ and $l = \frac{L}{2}$:
$$I_{\text{cross}} = \frac{1}{6} \times \left(\frac{M}{2}\right) \times \left(\frac{L}{2}\right)^2 = \frac{1}{6} \times \frac{M}{2} \times \frac{L^2}{4} = \frac{M L^2}{48}$$
Comparing with $\alpha$:
$$I_{\text{cross}} = \frac{1}{4} \left(\frac{M L^2}{12}\right) = \frac{\alpha}{4}$$$
### Step 1: Final Conclusion
The moment of inertia of the cross is $\alpha / 4$.
### Pattern Recognition
Since mass scales linearly ($M \propto L$), cutting a rod into $n$ equal segments scales the length by $1/n$ and mass by $1/n$. The moment of inertia of each segment scales as $1/n^3$. Reassembling $n$ segments linearly sums their contributions, so the final moment of inertia scales as $n \times \frac{1}{n^3} = \frac{1}{n^2}$.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Rotational Motion
