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A sphere of relative density \mathrm{\sigma} and diameter D has concentric cavity of diameter d. It will just float on water in a tank if the ratio \mathrm{D / d}  is

Option: 1

\mathrm{\frac{\sigma}{(\sigma-1)}}


Option: 2

\mathrm{\frac{(\sigma+1)}{\sigma}}


Option: 3

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


Option: 4

\mathrm{\left(\frac{\sigma+1}{\sigma}\right)^{1 / 3}}


Answers (1)

best_answer

Let \mathrm{\rho} be the density of the sphere and \mathrm{\rho_0 } that of water. Volume of metal \mathrm{=\frac{4}{3} \pi\left(R^3-r^3\right)} and volume of water displaced \mathrm{=\frac{4}{3} \pi R^3}. From the principle of floatation, we have

                   \mathrm{ \frac{4}{3} \pi\left(R^3-r^3\right) \rho g=\frac{4}{3} \pi R^3 \rho 0 g }
or
                      \mathrm{ \left(R^3-r^3\right) \frac{\rho}{\rho_0}=R^3 }
or
                        \mathrm{ \left(R^3-r^3\right) \sigma=R^3 }
which gives
                              \mathrm{ \frac{R}{r}=\left(\frac{\sigma}{\sigma-1}\right)^{1 / 3} }
Hence the correct choice is (c).

Posted by

Anam Khan

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