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:
- \( \left( \frac{\sigma}{\sigma - 1} \right)^{\frac{1}{3}} \)
- \( \left( \frac{\sigma + 1}{\sigma - 1} \right)^{\frac{1}{3}} \)
- \( \left( \frac{\sigma - 1}{\sigma} \right)^{\frac{1}{3}} \)
- \( \left( \frac{\sigma - 2}{\sigma + 2} \right)^{\frac{1}{3}} \)
The Correct Option is A
Approach Solution - 1
To determine the ratio \( \frac{D}{d} \) for the sphere to just float on water, we need to apply the principle of buoyancy. The sphere will float when the weight of the sphere equals the weight of the water displaced by it.
The volume of the whole sphere (including the cavity) is given by:
\(V_{\text{total}} = \frac{4}{3} \pi \left(\frac{D}{2}\right)^3 = \frac{\pi D^3}{6}\)
The volume of the cavity is:
\(V_{\text{cavity}} = \frac{4}{3} \pi \left(\frac{d}{2}\right)^3 = \frac{\pi d^3}{6}\)
Thus, the volume of the material of the sphere is the total volume minus the cavity volume:
\(V_{\text{material}} = \frac{\pi D^3}{6} - \frac{\pi d^3}{6} = \frac{\pi (D^3 - d^3)}{6}\)
Let the density of the sphere's material be \(\rho_{\text{material}} = \sigma \times \rho_{\text{water}}\) where \(\rho_{\text{water}}\) is the density of water.
The weight of the sphere is:
\(W = \rho_{\text{s}} \cdot V_{\text{material}} \cdot g = \sigma \cdot \rho_{\text{water}} \cdot \frac{\pi (D^3 - d^3)}{6} \cdot g\)
For the sphere to just float, the buoyant force equals the weight:
\(\rho_{\text{water}} \cdot V_{\text{total}} \cdot g = \rho_{\text{material}} \cdot \frac{\pi (D^3 - d^3)}{6} \cdot g\)
Substituting known values, we get:
\(\rho_{\text{water}} \cdot \frac{\pi D^3}{6} = \sigma \cdot \rho_{\text{water}} \cdot \frac{\pi (D^3 - d^3)}{6}\)
Dividing through by \(\rho_{\text{water}} \cdot \frac{\pi}{6}\):
\(D^3 = \sigma (D^3 - d^3)\)
Simplifying, we get:
\(D^3 = \sigma D^3 - \sigma d^3\)
Rearranging gives:
\(\sigma d^3 = \sigma D^3 - D^3 = (\sigma - 1) D^3\)
Thus, the ratio \(\frac{D^3}{d^3}\) is:
\(\frac{D^3}{d^3} = \frac{\sigma}{\sigma - 1}\)
Taking the cube root on both sides, we find the ratio:
\(\frac{D}{d} = \left( \frac{\sigma}{\sigma - 1} \right)^{\frac{1}{3}}\)
Therefore, the correct option is:
\( \left( \frac{\sigma}{\sigma - 1} \right)^{\frac{1}{3}} \)
Approach Solution -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: \( \frac{D}{d} = \left( \frac{\sigma}{\sigma - 1} \right)^{\frac{1}{3}}. \)
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