A uniform solid sphere of radius R is in equilibrium inside a liquid whose density varies with depth from free surface as $ρ = ρ_{0} (1 + \frac{h}{h _{0}})$ . Where h is depth from free surface. Density of sphere $σ$ will be :

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TestHub · Accepted Answer

Buoyancy = weight of displaced liquid; pair symmetric elements at depths $d - x$ and $d + x$ (sphere's centre at depth d):

$σV g = \int [ρ_{0} (1 + \frac{d - x}{h _{0}}) g d V + ρ_{0} (1 + \frac{d + x}{h _{0}}) g d V]$

$σV = 2 ρ_{0} (1 + \frac{d}{h _{0}}) \int_{0}^{V /2} d V = ρ_{0} V (1 + \frac{d}{h _{0}})$

$σ = ρ_{0} (1 + \frac{d}{h _{0}})$ - (D)

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