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

τ = Iα$\tau = I\alpha$

Core Logic

Immediately after one string is cut, the rod starts rotating about the point where the remaining string is attached.

Taking torque about the end where the string is attached (this point has instantaneous acceleration but initially zero vertical velocity):

Gravity provides the torque: τ = mg ((l)/(2))$\tau = mg \left(\frac{l}{2}\right)$.

Step 1: Calculate Angular Acceleration

Moment of inertia about the end is I = (ml²)/(3)$I = \frac{ml^2}{3}$.

Step 2: Calculate Force and Tension

The acceleration of the center of mass (CM) is downwards:

Applying Newton's second law for translational motion of the CM in vertical direction:

Pattern Recognition

Classic 'cut string' rigid body problem. Always take torque about the pivot/hinge point to find α$\alpha$, then relate the center of mass linear acceleration a = rcm α$a = r_{cm} \alpha$ to find the unknown tension using Fₙₑₜ = ma$F_{\text{net}} = ma$.

Chapter Mix

Class 11 Physics: System of Particles and Rotational Motion

Related Formula

Core Logic

Let the rods be Rod 1 (vertical, passing through P at its end) and Rod 2 (horizontal, attached at the other end of Rod 1).

For Rod 2 (length L$L$, mass M$M$): The axis passes parallel to Rod 2's center of mass axis, at a distance L$L$ from it (since it's attached to the bottom end of Rod 1). I₂ = Icm + M d² = (ML²)/(12) + M(L)²$I_2 = I_{\text{cm}} + M d^2 = \frac{ML^2}{12} + M(L)^2$

Step 1: Total Moment of Inertia

We are given that I = (x)/(12) ML²$I = \frac{x}{12} ML^2$ (Correction from source PDF text: the source question text says (x)/(2) ML²$\frac{x}{2} ML^2$, but the solution uses (x)/(12) ML²$\frac{x}{12} ML^2$. Following the solution steps: x=17$x=17$ is consistent if the denominator is 12. Let's assume the question asked for (x)/(12)$\frac{x}{12}$ or x=17/6$x=17/6$, but the official answer gives 17. Our output will state 17).

Pattern Recognition

For composite shapes, calculate I$I$ for each simple shape separately about the desired axis using Parallel Axis Theorem, then sum them up.

Chapter Mix

Class 11 Physics: System of Particles and Rotational Motion

Related Formula

Core Logic

The relative angular momentum of a particle translating uniformly with respect to another moving reference frame can be found using their relative velocity and the perpendicular distance between their lines of motion.

Relative velocity of A with respect to B:

Step 1: Conversion to SI units

Step 2: Final Conclusion

Using the perpendicular distance (r⊥ = 10 m$r_{\perp} = 10 \text{ m}$) and mass (m = 10³ kg$m = 10^3 \text{ kg}$):

Pattern Recognition

For parallel tracks, the angular momentum of one translating body relative to another is simply m × vrelative × track separation$m \times v_{\text{relative}} \times \text{track separation}$.

Chapter Mix

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

Related Formula

Mg - T₂ = Ma$Mg - T_2 = Ma$ T₁ - mg = ma$T_1 - mg = ma$

Core Logic

First, evaluate the moment of inertia of the pulley. The pulley consists of:

• A thin rim of mass M$M$ and radius r$r$.
• Two rods, each of mass M$M$ and length 2r$2r$ (diameter).

Moment of inertia of the rim: Irim = Mr²$I_{\text{rim}} = Mr^2$ Moment of inertia of two rods about the center: Irods = 2 × ((M(2r)²)/(12)) = 2 × (4Mr²)/(12) = (2)/(3)Mr²$I_{\text{rods}} = 2 \times \left(\frac{M(2r)^2}{12}\right) = 2 \times \frac{4Mr^2}{12} = \frac{2}{3}Mr^2$

Total I = Mr² + (2)/(3)Mr² = (5)/(3)Mr²$I = Mr^2 + \frac{2}{3}Mr^2 = \frac{5}{3}Mr^2$

Step 1: Force Equations

From the free body diagrams of the descending block (mass M$M$) and ascending block (mass m$m$):

Torque on the pulley:

Step 2: Final Conclusion

Adding equations (1), (2), and (3):

Substitute I = (5)/(3)Mr²$I = \frac{5}{3}Mr^2$:

Pattern Recognition

For a real pulley system, the effective mass acting against acceleration incorporates an inertia term: meff = m₁ + m₂ + (I)/(R²)$m_{\text{eff}} = m_1 + m_2 + \frac{I}{R^2}$. Breaking the complex pulley into fundamental shapes (ring + rods) solves the inertia safely.

Chapter Mix

Class 11 Physics: Systems of Particles and Rotational Motion Class 11 Physics: Laws of Motion

Related Formula

Core Logic

Substitute the given mass and radius values into the standard moment of inertia formula for spheres about their common tangent at the contact point:

Pattern Recognition

Sees: Two touching solid spheres + moment of inertia about tangent at contact point. Shortcut: Apply parallel/perpendicular axis theorem adjustments directly via standard formula summation. Check: Calculations yield 0.63 kg m²$0.63 \text{ kg m}^2$, matching option (4). ✓

Chapter Mix

Class 11 Physics: Rotational Motion