Physics

Physics

Problem

A solid steel cylinder of radius 2.0 cm2.0\,\text{cm} and height 10 cm10\,\text{cm} has to be made to float in water with its top level with the water surface by drilling a coaxial cylindrical hole through its full height. What should be the radius of the hole? (Density of steel =7.8×103 kg m−3=7.8\times10^{3}\,\text{kg m}^{-3}, density of water =1.0×103 kg m−3=1.0\times10^{3}\,\text{kg m}^{-3}) a) 1.7 cm1.7\,\text{cm} b) 1.9 cm1.9\,\text{cm} c) 2.1 cm2.1\,\text{cm} d) 1.5 cm1.5\,\text{cm}
Intuitive explanation

The key is balancing weight and buoyancy: the hollowed cylinder must displace enough water to support its own remaining steel mass. Set the buoyant force (weight of water displaced by the submerged volume) equal to the gravitational force on the steel that remains after drilling. Since the top floats level with the surface, the entire outer cylinder is submerged, so you're finding what inner radius leaves just enough steel to match the weight of water in that volume.

Intuitive explanation

The key is balancing weight and buoyancy: the hollowed cylinder must displace enough water to support its own remaining steel mass. Set the buoyant force (weight of water displaced by the submerged volume) equal to the gravitational force on the steel that remains after drilling. Since the top floats level with the surface, the entire outer cylinder is submerged, so you're finding what inner radius leaves just enough steel to match the weight of water in that volume.

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