A sphere of relative density σ and diameter D has concentric cavity of diameter d. The ratio of D/d, if it just floats on water in a tank is: ​

Question

A sphere of relative density \(\sigma\) and diameter D has concentric cavity of diameter d. The ratio of \(\frac{D}{d}\), if it just floats on water in a tank is:

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

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

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

(4) \(\left(\frac{\sigma-2}{\sigma+2}\right)^{\frac{1}{3}}\)   

Answer 1

Correct option is : (1) \(\left(\frac{\sigma}{\sigma-1}\right)^{\frac{1}{3}}\)  

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

Buoyant force \(\left(\mathrm{F}_{\mathrm{b}}\right)=1 \times \frac{4}{3} \pi\left(\frac{\mathrm{D}^{3}}{8}\right) \cdot \mathrm{g}\)

For Just Float \(\Rightarrow \mathrm{w}=\mathrm{F}_{\mathrm{b}}\)

\(\Rightarrow\left(\mathrm{D}^{3}-\mathrm{d}^{3}\right) \sigma=\mathrm{D}^{3}\)

\(\Rightarrow 1-\frac{\mathrm{d}^{3}}{\mathrm{D}^{3}}=\frac{1}{\sigma}\)

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

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