A sphere of relative density sigma and diamet | vedclass

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Question

(A) The weight of the sphere is given by $W = V_{material} \cdot \rho_{sphere} \cdot g = \frac{4}{3} \pi \left( \frac{D^3 - d^3}{8} \right) \sigma \rh

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$\left(\frac{\sigma}{\sigma-1}\right)^{\frac{1}{3}}$

Vedclass · Answer

(A) The weight of the sphere is given by $W = V_{material} \cdot \rho_{sphere} \cdot g = \frac{4}{3} \pi \left( \frac{D^3 - d^3}{8} \right) \sigma \rho_w g$.The buoyant force is equal to the weight of the water displaced by the total volume of the sphere: $F_b = V_{total} \cdot \rho_w \cdot g = \frac{4}{3} \pi \left( \frac{D^3}{8} \right) \rho_w g$.For the sphere to just float,the weight must equal the buoyant force: $W = F_b$.Substituting the expressions: $\frac{4}{3} \pi \left( \frac{D^3 - d^3}{8} \right) \sigma \rho_w g = \frac{4}{3} \pi \left( \frac{D^3}{8} \right) \rho_w g$.Simplifying the equation: $(D^3 - d^3) \sigma = D^3$.Dividing by $D^3 \sigma$: $1 - \frac{d^3}{D^3} = \frac{1}{\sigma}$.Rearranging for the ratio: $\frac{d^3}{D^3} = 1 - \frac{1}{\sigma} = \frac{\sigma - 1}{\sigma}$.Taking the reciprocal and cube root: $\frac{D}{d} = \left( \frac{\sigma}{\sigma - 1} \right)^{\frac{1}{3}}$.

$A$ sphere of relative density $\sigma$ and diameter $D$ has a concentric cavity of diameter $d$. The ratio of $\frac{D}{d}$,if it just floats on water in a tank,is:

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