JEE Main · 2024 · 09 Apr 2024 · Shift 1 · Fluid Mechanics

A sphere of relative density

σ

and diameter D has concentric cavity of diameter d. The ratio of

\frac{D}{d}

, if it just floats on water in a tank is:

(A) $(\frac{σ}{σ - 1})^{\frac{1}{3}}$ ✓ Correct
(B) $(\frac{σ + 1}{σ - 1})^{\frac{1}{3}}$
(C) $(\frac{σ - 1}{σ})^{\frac{1}{3}}$
(D) $(\frac{σ - 2}{σ + 2})^{\frac{1}{3}}$

Answer: (A)

Solution

Just floating: weight = buoyant force.

\frac{4}{3} π (\frac{D ^{3} - d ^{3}}{8}) σ g = \frac{4}{3} π (\frac{D ^{3}}{8}) g

, so

(D^{3} - d^{3}) σ = D^{3} \Rightarrow 1 - \frac{d ^{3}}{D ^{3}} = \frac{1}{σ} \Rightarrow (\frac{D}{d}) = (\frac{σ}{σ - 1})^{1/3}

. Option (1).

JEE Main · 2024 · 09 Apr 2024 · Shift 1 · Single correct option

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