Project ForgeA uniform wooden rod (specific gravity = 0.6, diameter = 4 cm and length = 8 m) is immersed in the water and is hinged without friction at point A on the waterline as shown in the figure. A solid spherical ball made of lead (specific gravity = 11.4) is attached to the free end of the rod to keep the assembly in static equilibrium inside the water. For simplicity, assume that the radius of the ball is much smaller than the length of the rod. Assume density of water = 10³ kg/m³ and π = 3.14. Radius of the ball is _______ cm (round off to 2 decimal places).
Explanation
Since all forces (weight, buoyancy) are vertical and the rod is straight, every moment arm about hinge A scales with the same sinθ factor, which cancels out of the moment balance regardless of the rod's actual angle. Taking moments about A: (net upward force on rod)×(L/2) + (net upward force on ball)×L = 0. Net rod force = ρw·g·Vrod·(1−0.6) = 0.4ρw·g·Vrod (upward, wood floats). Net ball force = ρw·g·Vball·(1−11.4) = −10.4ρw·g·Vball (downward, lead sinks). So 0.4Vrod×0.5 = 10.4Vball, giving Vball=Vrod/52. Vrod=(π/4)(0.04)²(8)=0.010048 m³, so Vball=0.00019323 m³=193.23 cm³. Solving (4/3)πr³=193.23 for r: r³=46.15 cm³, r≈3.59 cm.
Source: GATE ME 2022 Session 2 (ME-2) (2022)
This question is part of a 15-day revision cycle — sign in and Project Forge will bring it back the day before you'd forget it.