A cube having a side of 10~cm$10\mathrm{~cm}$ with unknown mass and 200~gm$200\mathrm{~gm}$ mass were hung at two ends of an uniform rigid rod of 27~cm$27\mathrm{~cm}$ long. The rod along with masses was placed on a wedge keeping the distance between wedge point and 200~gm$200\mathrm{~gm}$ weight as 25~cm$25\mathrm{~cm}$. 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~g/cm³$1\mathrm{~g/cm}^{3}$.) The unknown mass is ________ kg$\mathrm{kg}$.

Related Formula

Velocity of a point on a rolling wheel at angular position θ$\theta$ from the lowest point:

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$$v_{\text{mid}} = 2v_{\text{cm}} \sin(45^\circ) = 2 \times 4 \times \frac{1}{\sqrt{2}} = 4\sqrt{2}\text{ m/s}$$

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

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