Two iron solid discs of negligible thickness have radii R₁$\mathbf{R}_1$ and R₂$\mathbf{R}_2$ and moment of inertia I₁$\mathrm{I}_1$ and I₂$\mathrm{I}_2$ , respectively. For R₂ = 2R₁$\mathrm{R}_2 = 2\mathrm{R}_1$ , the ratio of I₁$\mathrm{I}_1$ and I₂$\mathrm{I}_2$ would be 1 / x$1 / \mathrm{x}$ , where x =$\mathrm{x} =$
Numerical Answer Type:
Enter a numerical valueAnswer: 16 to 16+4 marks
Solution & Explanation
Core Logic
Since mass scales with the face surface area for discs of identical thickness and material composition:
M = σ · π R² M ∝ R²$$\mathrm{M} = \sigma \cdot \pi \mathrm{R}^2 \implies \mathrm{M} \propto \mathrm{R}^2$$
Disk mass allocation scaling profile diagram for Q24
Disk mass allocation scaling profile diagram for Q24
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
Keywords:#ratio of moment of inertia for iron discs#JEE Main 2025 Morning Q24#System of Particles and Rotational Motion JEE Main 2025#Moment of Inertia JEE Main 2025
More System of Particles and Rotational Motion Previous-Year Questions — Page 7
Q19jee_main_2025_28_jan_morningCentre 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)$(\sigma)$ varies as σ = (σ₀x)/(ab)$\sigma = \frac{\sigma_0x}{ab}$ (where σ₀$\sigma_0$ is a constant), would be
A thin plate with variable linear density coordinates mapped across an XY grid system.
A.((2)/(3) a, (b)/(2))$\left(\frac{2}{3} a, \frac{b}{2}\right)$
B.((2)/(3) a, (2)/(3) b)$\left(\frac{2}{3} a, \frac{2}{3} b\right)$
The center of mass coordinates are ((2)/(3) a, b2)$\left(\frac{2}{3} \mathrm{a}, \frac{\mathrm{b}}{2}\right)$, which matches option (1).
Pattern Recognition
When density varies linearly with position (σ ∝ 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
Q22jee_main_2025_28_jan_morningMoment 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 =$\mathrm{n} =$ _____. Consider all the bodies have equal masses.
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
Q10jee_main_2025_03_april_morningRolling Without Slipping
A force of 49~N$49\mathrm{~N}$ acts tangentially at the highest point of a sphere (solid) of mass 20~kg$20\mathrm{~kg}$, kept on a rough horizontal plane. If the sphere rolls without slipping, then the acceleration of the center of the sphere is:
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$$\tau_P = I_P \alpha$$
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$$I_P = I_c + MR^2 = \frac{7}{5}MR^2$$
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 × 2R$$\tau_P = F \times 2R$$
Substitute τP$\tau_P$ and IP$I_P$ into the torque equation:
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))$$2F R = \frac{7}{5}MR^2 \left(\frac{a}{R}\right)$$2F = (7)/(5) M a a = (10F)/(7M)$$2F = \frac{7}{5} M a \implies a = \frac{10F}{7M}$$
Substitute the numerical values (F = 49~N$F = 49\mathrm{~N}$ and M = 20~kg$M = 20\mathrm{~kg}$):
Let's write force equations to verify consistency:
F + f = M a$F + f = M a$
49 + f = 20 × 3.5 = 70 f = 21~N$$49 + f = 20 \times 3.5 = 70 \implies f = 21\mathrm{~N}$$
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
Q12jee_main_2025_04_april_eveningRolling Motion
A wheel is rolling on a plane surface. The speed of a particle on the highest point of the rim is 8 m/s. The speed of the particle on the rim of the wheel at the same level as the centre of wheel, will be:
A.4√(2) m/s$4\sqrt{2}\text{ m/s}$
B.8 m/s$8\text{ m/s}$
C.4 m/s$4\text{ m/s}$
D.8√(2) m/s$8\sqrt{2}\text{ m/s}$
Solution
Related Formula
Velocity of a point on a rolling wheel at angular position θ$\theta$ from the lowest point:
v = 2 vcm ((θ)/(2))$$v = 2 v_{\text{cm}} \sin\left(\frac{\theta}{2}\right)$$
Core Logic
At the highest point, θ = 180^°$\theta = 180^\circ$, so:
For a particle on the rim at the same horizontal level as the center, the angle from the lowest point is θ = 90^°$\theta = 90^\circ$.
Step 1: Compute Speed at Mid-Height
Substituting θ = 90^°$\theta = 90^\circ$ into our velocity relation:
vmid = 2vcm (45^°) = 2 × 4 × 1√(2) = 4√(2) m/s$$v_{\text{mid}} = 2v_{\text{cm}} \sin(45^\circ) = 2 \times 4 \times \frac{1}{\sqrt{2}} = 4\sqrt{2}\text{ m/s}$$Instantaneous center of rotation on rolling wheel
Pattern Recognition
The contact point with the ground is the Instantaneous Center of Rotation (ICR). Distance to top point is 2R$2R$, distance to mid-level point is √(R²+R²) = √(2)R$\sqrt{R^2+R^2} = \sqrt{2}R$. Velocity scales linearly with distance from ICR.
Chapter Mix
Class 11 Physics: Rotational Motion
More System of Particles and Rotational Motion Questions — jee_main_2025_28_jan_morning
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
