JEE Mains 2026 (24 Jan Shift 1) Physics Solution: Fluid Oscillations & SHM скачать в хорошем качестве

JEE Mains 2026 (24 Jan Shift 1) Physics Solution: Fluid Oscillations & SHM

Welcome to Physics Unlocked ⚛🔓! In this video, we solve a problem combining Oscillations and Properties of Fluids from the JEE Mains 2026 exam (24 January, Shift 1). 📌 Question: A cylindrical block of mass M and area of cross section A is floating in a liquid of density ρ and with its axis vertical. When depressed a little and released the block starts oscillating. The period of oscillation is ________. Options: 1. 2π√(ρA / Mg) 2. π√(ρA / Mg) 3. 2π√(M / ρAg) 4. 2π√(2M / ρAg) ⏱️ Timestamps: 0:00 - Introduction & Problem Statement 0:32 - Equilibrium Condition & Buoyant Force 1:11 - Displacing the Block by 'x' 1:47 - Applying Newton's Second Law 2:16 - Expanding the Equation & Canceling Terms 2:35 - Comparing with Standard SHM Equation 2:49 - Calculating the Final Time Period 3:04 - Outro 📝 Detailed Step-by-Step Solution: Step 1: Understand the equilibrium condition. Initially, the block is floating, so the downward force of gravity (Mg) is perfectly balanced by the upward buoyant force (Fb). Let 'L' be the length of the cylinder submerged in equilibrium. F_gravity - F_buoyancy = 0 Mg - (ρ * A * L * g) = 0 Therefore, the mass can be expressed as: M = ρ * A * L Step 2: Analyze the forces when the block is depressed by a small distance 'x'. When the block is depressed by 'x', the new submerged length becomes (L + x). The force of gravity pulling it down remains the same (Mg), but the upward buoyant force increases because more volume is now displaced. We apply Newton's second law (F_net = ma). Taking the downward direction as positive: ma = F_gravity - F_buoyancy Ma = Mg - [ρ * A * (L + x) * g] Step 3: Expand and simplify the equation. Ma = Mg - (ρ * A * L * g) - (ρ * A * x * g) From Step 1, we established that Mg = ρ * A * L * g. Substituting this into our equation, the first two terms cancel out: Ma = (ρ * A * L * g) - (ρ * A * L * g) - (ρ * A * x * g) Ma = - ρ * A * x * g Step 4: Find the acceleration and compare it to the standard Simple Harmonic Motion (SHM) equation. Isolating 'a', we get: a = - (ρ * A * g / M) * x The standard defining equation for SHM is a = - ω² * x, where ω is the angular frequency. Comparing our derived equation with the standard equation, we get the square of the angular frequency: ω² = (ρ * A * g) / M Taking the square root: ω = √((ρ * A * g) / M) Step 5: Calculate the final Time Period (T). The time period of oscillation is given by the formula T = 2π / ω. Substituting our value of ω: T = 2π / √((ρ * A * g) / M) T = 2π * √(M / (ρ * A * g)) This perfectly matches option 3. The correct option is 3. Don't forget to like, share, and subscribe for more comprehensive JEE physics solutions!