Question

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:

• A. \(t \propto \sqrt{n}\)
• B. \(t \propto n\)
• C. \(t \propto \sqrt{n+1}\)
• D. \(t \propto \sqrt{n-1}\)

Hint:

For buoyant bodies, the time to return to the surface is proportional to the square root of the submerged volume ratio.

Answer 1

The Correct Option is A

Step 1: The time taken for a floating object to rise to the surface depends on the restoring force, which is related to the displaced volume of the body.
Step 2: For a body floating with \(\frac{1}{n}\) of its volume above the surface, the time to return to the surface will scale with the square root of the volume fraction submerged.
Step 3: Therefore, the time \(t\) to return to the surface is proportional to \(\sqrt{n-1}\).

Answer 2

F_B = mg ; V (1 - 1/n) ρg = V_dg ρg , ⇒ d = (n - 1)/n σ

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a = (F_B - mg) / m = (Vρ_σg - Vρg) / m = (Vρg / m) (σ/ρ - 1) = g (σ/ρ - 1)

a = g (1/n - 1) = g (1 - n) / n ⇒ a = g (n - 1) / (1 - n) g = -g

a = (Vρ_σg / Vρ(n-1)/n) - g ⇒ a = (g σ / ρ (n - 1) / n) - g ⇒ a = (ng / (n - 1)) - g ⇒ a = (ng - g(n - 1)) / (n - 1) ⇒ a = (ng - ng + g) / (n - 1) ⇒ a = g / (n - 1)

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Now for time, h = ½ at² , h = ½ (g / (n - 1)) t² , ⇒ t = √(2h / g (n - 1)) ⇒ t = √(2h(n - 1) / g) ⇒ t = √(2h / g) √(n - 1) ⇒ t α √(n - 1)