Question

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:

• A. $\left( \frac{\sigma}{\sigma - 1} \right)^{\frac{1}{3}}$
• B. $\left( \frac{\sigma + 1}{\sigma - 1} \right)^{\frac{1}{3}}$
• C. $\left( \frac{\sigma - 1}{\sigma} \right)^{\frac{1}{3}}$
• D. $\left( \frac{\sigma - 2}{\sigma + 2} \right)^{\frac{1}{3}}$

Answer 1

Answer: (A)

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

Answer 2

Step 1: Weight of the sphere The weight of the sphere w is given by:

$w = \frac{4}{3} \pi \left( D^3 - \frac{d^3}{8} \right) \sigma g,$

where:

• $D^3 - d^3$ accounts for the volume of the sphere minus the cavity,
• $\sigma$ is the relative density,
• $g$ is the acceleration due to gravity.

Step 2: Buoyant force The buoyant force $F_b$ is given by:

$F_b = \frac{4}{3} \pi \left( \frac{D^3}{8} \right) g,$

where $\frac{D^3}{8}$ is the volume of displaced water.

Step 3: Equilibrium condition For the sphere to just float, the weight equals the buoyant force:

$w = F_b.$

Substitute expressions for $w$ and $F_b$:

$\frac{4}{3} \pi \left( D^3 - \frac{d^3}{8} \right) \sigma g = \frac{4}{3} \pi \left( \frac{D^3}{8} \right) g.$

Cancel common terms:

$\left( D^3 - d^3 \right) \sigma = D^3.$

Simplify:

$D^3 - d^3 = \frac{D^3}{\sigma}.$

Step 4: Solve for $\frac{d}{D}$ Divide through by $D^3$:

$1 - \frac{d^3}{D^3} = \frac{1}{\sigma}.$

Rearrange:

$\frac{d^3}{D^3} = 1 - \frac{1}{\sigma}.$

Take the cube root:

$\frac{d}{D} = \left( 1 - \frac{1}{\sigma} \right)^{\frac{1}{3}}.$

Invert to find $\frac{D}{d}$:

$\frac{D}{d} = \left( \frac{\sigma}{\sigma - 1} \right)^{\frac{1}{3}}.$

Final Answer: $\frac{D}{d} = \left( \frac{\sigma}{\sigma - 1} \right)^{\frac{1}{3}}.$