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

The Correct Option is A

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