JEE Mains · Physics · STD 11 - 9.1 fluid mechanics
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\(......R\)
- A \(0.44\)
- B \(0.88\)
- C \(0.33\)
- D \(0.66\)
Answer & Solution
Correct Answer
(B) \(0.88\)
Step-by-step Solution
Detailed explanation
\(\frac{4}{3} \pi\left( R ^{3}- r ^{3}\right) \rho_{ m } g =\frac{4}{3} \pi R ^{3} \rho_{ w } g\) \(1-\left(\frac{ r }{ R }\right)^{3}=\frac{8}{27}\) \(\Rightarrow \frac{ r }{ R }=\left(\frac{19}{27}\right)^{1 / 3}=\frac{19^{1 / 3}}{3}\) \(=0.88\simeq \frac{8}{ g }\)
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