Question:medium

A body floats with \(\frac{1}{n}\) of its volume keeping outside of water. If the body has been taken to height \(h\) inside water and released, it will come to the surface after time \(t\). Then:

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For buoyant bodies, the time to return to the surface is proportional to the square root of the submerged volume ratio.
Updated On: Nov 28, 2025
  • \(t \propto \sqrt{n}\)
  • \(t \propto n\)
  • \(t \propto \sqrt{n+1}\)
  • \(t \propto \sqrt{n-1}\)
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The Correct Option is A

Solution and Explanation

Phase 1: An object's ascent time is influenced by the restoring force, which depends on the displaced volume.
Phase 2: If \(\frac{1}{n}\) of the object's volume is above the surface, the return time scales with the square root of the submerged fraction.
Phase 3: Consequently, the return time \(t\) is proportional to \(\sqrt{n-1}\).

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