A wooden stick of length L, radius R and density ρ has a small metal piece of mass m (of negligible volume) attached to its one end. The minimum value for the mass m (in terms of given parameters) that would make the stick float vertically in equilibrium in a liquid of density σ is πKR²Lρ[√(σ/ρ)−1]. Find the value of K.
Physics · Mixed Chapter
JEE Mains

Question

A wooden stick of length L, radius R and density ρ has a small metal piece of mass m (of negligible volume) attached to its one end. The minimum value for the mass m (in terms of given parameters) that would make the stick float vertically in equilibrium in a liquid of density σ is πKR²Lρ[√(σ/ρ)−1]. Find the value of K.

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1: Let the stick's cross-sectional area be A=πR² and the length submerged be h. Buoyancy balance gives σAh=ρAL+m, hence h=(ρAL+m)/(σA). Step 2: With the metal piece at the lower end, measure height upwards from that end. The combined centre of mass is z at G=(ρAL)(L/2)/(ρAL+m), while the centre of buoyancy is z at B=h/2. Step 3: For the limiting case of upright equilibrium (slender stick, neglecting metacentric correction), set z at G=z at B. Thus (ρAL+m)²=ρσ A²L². Step 4: mₘᵢₙ=AL(√(ρσ)−ρ)…