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.

Related Formula

Parallel Axis Theorem:

Standard Moments of Inertia about center of mass:

• Disc about diameter: Idisc,dia = (MR²)/(4)$I_{\text{disc,dia}} = \frac{MR^2}{4}$
• Solid sphere: Isphere = (2)/(5)MR²$I_{\text{sphere}} = \frac{2}{5}MR^2$
• Spherical shell: Ishell = (2)/(3)MR²$I_{\text{shell}} = \frac{2}{3}MR^2$

Core Logic

The axis of rotation PQ passes through the center of the top disc (A) along its diameter.

• Top disc (A):

• Bottom-left solid sphere (B): Center lies at distance R$R$ from the axis PQ.

• Bottom-right spherical shell (C): Center lies at distance R$R$ from the axis PQ.

Step 1: Calculate Total System Moment of Inertia

Sum the contributions:

To add the fractions, find a common denominator (60$60$):

Step 2: Express in terms of standard Disc Moment

We are given I = (MR²)/(4) MR² = 4I$I = \frac{MR^2}{4} \implies MR^2 = 4I$. Substitute this in the expression:

Comparing with IPQ = (x)/(15) I$I_{\text{PQ}} = \frac{x}{15} I$ yields x = 199$x = 199$.

Pattern Recognition

Sees: Composite body consisting of three standard symmetric shapes about a tangent/offset axis. Shortcut: Sum the central inertia terms and the offset terms separately. Offset masses are only B and C, so the offset sum is 2MR²$2MR^2$. The central sum is (1/4 + 2/5 + 2/3)MR²$(1/4 + 2/5 + 2/3)MR^2$. Adding these directly yields the combined fractional factor of 199/60$199/60$.

Chapter Mix

Class 11 Physics: System of Particles and Rotational Motion

Related Formula

where, Idia$I_{\text{dia}}$ = moment of inertia of a ring about its diameter M$M$ = total mass of the ring R$R$ = radius of the ring

Core Logic

Since the linear mass density is λ$\lambda^{\prime}$ (or λ$\lambda$ as per the options), the total mass M$M$ of the rod of length L$L$ is:

M = λ L$M = \lambda L$

When this rod is bent into a ring of radius R$R$, its circumference equals the length of the rod:

Substituting M$M$ and R$R$ into the formula for the moment of inertia about the diameter:

Pattern Recognition

Sees: "Rod of length L$L$ bent to form a ring" → R = (L)/(2π)$R = \frac{L}{2\pi}$. Trap: Moment of inertia about the central axis perpendicular to the plane is MR²$MR^2$, but about its diameter, it is half, i.e., (1)/(2)MR²$\frac{1}{2}MR^2$. Shortcut: I = (1)/(2) (λ L) ((L)/(2π))² = (λ L³)/(8π²)$I = \frac{1}{2} (\lambda L) \left(\frac{L}{2\pi}\right)^2 = \frac{\lambda L^3}{8\pi^2}$. ✓

Chapter Mix

Class 11 Physics: Rotational Motion

Related Formula

τ = I α$\tau = I \alpha$

where, N$N$ = rotational speed in rpm α$\alpha$ = angular acceleration τ$\tau$ = torque applied Idia$I_{\text{dia}}$ = moment of inertia of a solid disk about its diameter axis

Core Logic

Given parameters:

• Mass, m = 1~kg$m = 1\mathrm{~kg}$
• Initial speed, Nᵢ = 1800~rpm ωᵢ = (1800 × 2π)/(60) = 60π~rad/s$N_i = 1800\mathrm{~rpm} \implies \omega_i = \frac{1800 \times 2\pi}{60} = 60\pi\mathrm{~rad/s}$
• Final speed, Nf = 2100~rpm ωf = (2100 × 2π)/(60) = 70π~rad/s$N_f = 2100\mathrm{~rpm} \implies \omega_f = \frac{2100 \times 2\pi}{60} = 70\pi\mathrm{~rad/s}$
• Time, t = 40~s$t = 40\mathrm{~s}$
• Torque, τ = 25π~N· m$\tau = 25\pi\mathrm{~N\cdot m}$

First, calculate the angular acceleration α$\alpha$:

Now relate torque to moment of inertia:

Step 1: Compute Diameter

The question asks for the diameter of the disk (D$D$):

Pattern Recognition

Sees: Torque acting on a rotating disk increasing its speed. Trap: The axis of rotation is the diameter axis, not the normal geometric center axis. This means the moment of inertia is I = (1)/(4) m R²$I = \frac{1}{4} m R^2$, not (1)/(2) m R²$\frac{1}{2} m R^2$! Check this detail carefully. ✓

Chapter Mix

Class 11 Physics: Rotational Motion

Related Formula

where, τ$\tau$ = torque about the wedge point FB$F_B$ = buoyancy force

Core Logic

Let's list the geometric parameters from the setup:

• Length of uniform rigid rod, L = 27~cm$L = 27\mathrm{~cm}$.
• Wedge is positioned such that the distance to the 200~g$200\mathrm{~g}$ (0.2~kg$0.2\mathrm{~kg}$) weight is d₁ = 25~cm$d_1 = 25\mathrm{~cm}$.
• Therefore, the distance to the unknown mass M$M$ at the other end is d₂ = 27 - 25 = 2~cm$d_2 = 27 - 25 = 2\mathrm{~cm}$.

Calculate the volume of the cube:

• Side of the cube, a = 10~cm = 0.1~m$a = 10\mathrm{~cm} = 0.1\mathrm{~m}$.
• Volume, V = a³ = 1000~cm³ = 10⁻³~m³$V = a^3 = 1000\mathrm{~cm}^3 = 10^{-3}\mathrm{~m}^3$.

When the system is balanced, half the volume of the cube is submerged in water:

• Submerged volume, Vsub = (V)/(2) = 500~cm³ = 5 × 10⁻⁴~m³$V_{\text{sub}} = \frac{V}{2} = 500\mathrm{~cm}^3 = 5 \times 10^{-4}\mathrm{~m}^3$.
• Buoyancy force:

Step 1: Torque Balance Equation

For rotational equilibrium, balance the torques about the wedge point O$O$:

• Torque on the left (unknown mass branch): τleft = (M g - FB) × d₂ = (M g - 0.5 g) × 2$\tau_{\text{left}} = \left(M g - F_B\right) \times d_2 = (M g - 0.5 g) \times 2$
• Torque on the right (200~g$200\mathrm{~g}$ mass branch): τright = 0.2 g × d₁ = 0.2 g × 25$\tau_{\text{right}} = 0.2 g \times d_1 = 0.2 g \times 25$

2 (M - 0.5) = 5$2 (M - 0.5) = 5$

Thus, the unknown mass is 3~kg$3\mathrm{~kg}$.

Pattern Recognition

Sees: Rod torque balance + buoyancy force on one end. Trap: Ensure you measure the distances from the pivot point (the wedge). The unknown mass is at 27 - 25 = 2~cm$27 - 25 = 2\mathrm{~cm}$ from the wedge. Shortcut: Since g$g$ appears in both gravity and buoyancy terms, it cancels out immediately. Balancing torque simplifies directly to resolving mass differences: 2(M - 0.5) = 0.2 × 25 = 5$2(M - 0.5) = 0.2 \times 25 = 5$. This yields M = 3$M = 3$ instantly! ✓

Chapter Mix

Class 11 Physics: Rotational Motion Class 11 Physics: Mechanical Properties of Fluids

Related Formula

Core Logic

Since the three masses interact purely through mutual internal gravitational forces, the net external torque acting on the system about any central reference point is zero:

Let us compute the total angular momentum about the centroid of the equilateral triangle:

The initial angular momentum for one mass about the centroid is L₁ = m V₀ r⊥$L_1 = m V_0 r_{\perp}$. Since all three particles move cyclically in the same direction, their angular momenta reinforce cleanly:

By conservation of angular momentum, this configuration value remains unchanged up to the point of collision.

Pattern Recognition

Mutual internal central forces can never alter the angular momentum of a system. Hence, the solution completely reduces to measuring the static configuration values at t=0$t=0$ about the center of mass symmetry.

Chapter Mix

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