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 =$\mathrm{x} =$

Related Formula

Conservation of Angular Momentum (since no external torque acts):

For a solid sphere, moment of inertia is:

Mass scales with volume: M ∝ R³$M \propto R^3$

Core Logic

When the radius reduces to R₂ = (R)/(2)$R_2 = \frac{R}{2}$, the mass scales cubically:

Now, compute the new moment of inertia I₂$I_2$:

Step 1: Compute Final Angular Velocity

Using conservation of angular momentum:

Hence, the value of x$x$ is 32.

Pattern Recognition

Since inertia of a solid sphere scales with M R²$M R^2$ and M ∝ R³$M \propto R^3$, the net moment of inertia scales with R⁵$R^5$. Shrinking the radius by half (1/2$1/2$) cuts down inertia by a factor of (1/2)⁵ = 1/32$(1/2)^5 = 1/32$. Velocity must scale up by 32 to conserve momentum.

Chapter Mix

Class 11 Physics: Rotational Motion

Related Formula

Fundamental mathematical definition of torque:

Rotational analogue of Newton's second law:

Linear momentum relations:

Core Logic

Let's check each expression sequentially:

• A. τ = I × L$\vec{\tau} = \vec{I} \times \vec{L}$ is dimensionally incorrect (Moment of inertia I$I$ is primarily treated as a tensor or scalar placeholder, not crossed directly like this).
• B. τ = d Ldt = (d)/(dt)( r × p)$\vec{\tau} = \frac{d\vec{L}}{dt} = \frac{d}{dt}(\vec{r} \times \vec{p})$ is fundamentally correct.
• C. τ = r × F = r × d pdt$\vec{\tau} = \vec{r} \times \vec{F} = \vec{r} \times \frac{d\vec{p}}{dt}$ is correct since F = d pdt$\vec{F} = \frac{d\vec{p}}{dt}$.
• D. τ = I α$\vec{\tau} = I\vec{\alpha}$ is the standard scalar component/fixed axis formulation.
• E. τ = r × F$\vec{\tau} = \vec{r} \times \vec{F}$ is the true physical vector definition.

Thus, statements B, C, D, and E are universally correct representations.

Pattern Recognition

Torque can be represented either through geometric structural parameters (position and force cross products) or via kinematic response properties (rate of change of angular momentum or product of rotational inertia and acceleration).

Chapter Mix

Class 11 Physics: System of Particles and Rotational Motion

Related Formula

Acceleration vector equation via secondary derivation:

Force equation:

Core Logic

Given position tracking trace:

Velocity vector v$\vec{v}$:

Step 1: Differentiate to find Acceleration

Step 2: Establish Force Direction

The minus sign indicates the net centripetal pulling force aligns explicitly opposite to the direction of r$\vec{r}$.

Pattern Recognition

The expression describes a standard uniform circular motion profile. In circular configurations, acceleration and centripetal forces point radially inward, directly opposing the outbound position tracker vector.

Chapter Mix

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

Related Formula

Velocity of a rolling body from mechanical energy conservation:

Core Logic

Evaluate radius of gyration factor coefficients:

• For a circular ring: (k²)/(R²) = 1$\frac{k^2}{R^2} = 1$
• For a solid sphere: (k²)/(R²) = (2)/(5)$\frac{k^2}{R^2} = \frac{2}{5}$

Step 1: Compute Velocity Expressions

Step 2: Calculate the Velocity Ratio

Matching with the prompt expression format √((x)/(5))$\sqrt{\frac{x}{5}}$ reveals: x = 3.5$x = 3.5$

Rounding to the nearest integer yields 4.

Pattern Recognition

Objects with lower mass concentration near the center (lower (k²)/(R²)$\frac{k^2}{R^2}$ like the sphere) convert gravitational potential energy into translational kinetic energy more efficiently, rolling faster than hollow equivalents.

Chapter Mix

Class 11 Physics: System of Particles and Rotational Motion

Related Formula

Idisc = (1)/(2) M R²$$I_{\text{disc}} = \frac{1}{2} M R^2$$ [cite: 833]

I = Icm + M d² (Parallel Axis Theorem)$$I = I_{\text{cm}} + M d^2 \quad \text{(Parallel Axis Theorem)}$$ [cite: 848]

Core Logic

Let the original mass density per unit area be σ$\sigma$. Without any cavity, the moment of inertia is: [cite: 198, 833]

I₁ = (MR²)/(2)$$I_1 = \frac{MR^2}{2}$$ [cite: 833]

The mass of the removed small section scales directly with its cut-out area profile: [cite: 198, 834]

m = (M)/(π R²) × π ((R)/(3))² = (M)/(9)$$m = \frac{M}{\pi R^2} \times \pi \left(\frac{R}{3}\right)^2 = \frac{M}{9}$$ [cite: 198, 834, 844]

The center of mass of the removed disk sits at a distance d = R - (R)/(3) = (2R)/(3)$d = R - \frac{R}{3} = \frac{2R}{3}$ away from the primary center O$O$[cite: 198, 205, 848]. Calculating its partial moment of inertia about O$O$ via the parallel axis theorem: [cite: 199, 848]

I₂ = (m r²)/(2) + m d² = ((M)/(9)((R)/(3))²)/(2) + (M)/(9)((2R)/(3))²$$I_2 = \frac{m r^2}{2} + m d^2 = \frac{\frac{M}{9}\left(\frac{R}{3}\right)^2}{2} + \frac{M}{9}\left(\frac{2R}{3}\right)^2$$ [cite: 848]

I₂ = (MR²)/(162) + (4MR²)/(81) = (MR² + 8MR²)/(162) = (9MR²)/(162) = (MR²)/(18)$$I_2 = \frac{MR^2}{162} + \frac{4MR^2}{81} = \frac{MR^2 + 8MR^2}{162} = \frac{9MR^2}{162} = \frac{MR^2}{18}$$ [cite: 848, 850]

Subtracting the removed component from the original configuration: [cite: 851]

I = I₁ - I₂ = (MR²)/(2) - (MR²)/(18) = (9MR² - MR²)/(18) = (8MR²)/(18) = (4)/(9)MR²$$I = I_1 - I_2 = \frac{MR^2}{2} - \frac{MR^2}{18} = \frac{9MR^2 - MR^2}{18} = \frac{8MR^2}{18} = \frac{4}{9}MR^2$$ [cite: 851]

Matching this with (4)/(x)MR²$\frac{4}{x}MR^2$, we find x = 9$x = 9$[cite: 200, 208, 851, 853].

Pattern Recognition

For uniform planar surfaces, mass always scales squarely with linear dimension changes (r arrow (r)/(3) m arrow (m)/(9)$r \rightarrow \frac{r}{3} \implies m \rightarrow \frac{m}{9}$)[cite: 198, 834, 844]. Always apply the parallel axis theorem to bring the component values to a unified reference point before executing addition or subtraction[cite: 848, 851].

Chapter Mix

Class 11 Physics: System of Particles and Rotational Motion