First slide

Pressure in a fluid-mechanical properties of fluids

Question

Consider a solid sphere of radius R and mass density $ρ (r) = ρ_{0} (1 - \frac{r^{2}}{R^{2}})$ , $0 < r \leq R$ . The minimum density of a liquid in which it will float is:

Difficult

A

$\frac{2 ρ_{0}}{5}$
B

$\frac{2 ρ_{0}}{3}$
C

$\frac{ρ_{0}}{5}$
D

$\frac{ρ_{0}}{3}$

Solution

$G i v e n : ρ = ρ_{0} (1 - \frac{r^{2}}{R^{2}}) f o r 0 < r \leq R C o n s i d e r a a n e l e m e n t o f t h i c k n e s s d r, a t a d i s \tan c e r f r o m c e n t e r . I t s v o l u m e i s d V = (4 π r^{2}) d r I t s m a s s i s d m = ρ d V = ρ_{0} (1 - \frac{r^{2}}{R^{2}}) (4 π r^{2}) d r I n t e g r a t i n g, m = \int d m = \int_{0}^{R} ρ_{0} 4 π (r^{2} - \frac{r^{4}}{R^{2}}) d r = ρ_{0} 4 π {({\frac{r}{3}}^{3} - \frac{r^{5}}{5 R^{2}})}_{0}^{R} = \frac{8}{15} ρ_{0} π R^{3} T o f l o a t i n l i q u i d, m g = B \Rightarrow \frac{8}{15} ρ_{0} π R^{3} g = ρ_{L} \frac{4}{3} π R^{3} g \Rightarrow \frac{2}{5} ρ_{0} = ρ_{L}$

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