Question:
A hollow spherical shell at outer radius \(R\) floats just submerged under the water surface. The inner radius of the shell is \(r\). If the specific gravity of the shell material is \(\frac{27}{8}\) w.r.t water, the value of \(r\) is :
A hollow spherical shell at outer radius \(R\) floats just submerged under the water surface. The inner radius of the shell is \(r\). If the specific gravity of the shell material is \(\frac{27}{8}\) w.r.t water, the value of \(r\) is :
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Weight of the shell equals the buoyant force of the displaced water (volume \(\frac{4}{3}\pi R^{3}\)); express the shell volume as \(\frac{4}{3}\pi(R^{3}-r^{3})\).
Updated On: Nov 5, 2026
- \(\frac{2}{3}R\)
- \(\frac{4}{9}R\)
- \(\frac{1}{3}R\)
- \(\frac{8}{9}R\)
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The Correct Option is D
Solution and Explanation
Floating just submerged: weight of shell \(=\) weight of water of volume \(\frac{4}{3}\pi R^{3}\).
Shell material density \(=\frac{27}{8}\rho_w\), shell volume \(=\frac{4}{3}\pi\left(R^{3}-r^{3}\right)\).
\(\frac{27}{8}\rho_w\cdot\frac{4}{3}\pi(R^{3}-r^{3})\,g=\rho_w\cdot\frac{4}{3}\pi R^{3}\,g\)
\(\Rightarrow R^{3}-r^{3}=\frac{8}{27}R^{3}\Rightarrow r^{3}=\frac{19}{27}R^{3}\Rightarrow r=\frac{19^{1/3}}{3}R\approx0.89R\approx\frac{8}{9}R\).
The closest option is \(\frac{8}{9}R\).
Answer: (4) \(\frac{8}{9}R\)
Shell material density \(=\frac{27}{8}\rho_w\), shell volume \(=\frac{4}{3}\pi\left(R^{3}-r^{3}\right)\).
\(\frac{27}{8}\rho_w\cdot\frac{4}{3}\pi(R^{3}-r^{3})\,g=\rho_w\cdot\frac{4}{3}\pi R^{3}\,g\)
\(\Rightarrow R^{3}-r^{3}=\frac{8}{27}R^{3}\Rightarrow r^{3}=\frac{19}{27}R^{3}\Rightarrow r=\frac{19^{1/3}}{3}R\approx0.89R\approx\frac{8}{9}R\).
The closest option is \(\frac{8}{9}R\).
Answer: (4) \(\frac{8}{9}R\)
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