Multiple choice

A sphere of radius R and made of material of relative density $\sigma$ has a concentric cavity of radius r.It just floats when placed in a tank full water.The value of the ratio R/r will be

  1. $\left( \dfrac { \sigma }{ \sigma -1 } \right) ^{ 1/3 }$
  2. $\left( \dfrac { \sigma -1 }{ \sigma } \right) ^{ 1/3 }$
  3. $\left( \dfrac { \sigma \div 1 }{ \sigma } \right) ^{ 1/3 }$
  4. $\left( \frac { \sigma -1 }{ \sigma \div 1 } \right) ^{ 1/3 }$
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A Correct answer
Explanation

For floating equilibrium, the weight of the material equals the weight of the displaced water. Thus, sigma(R^3 - r^3) = R^3, which gives R^3/r^3 = sigma/(sigma - 1). Taking cube roots gives the stated ratio.

AI explanation

For the hollow sphere to just float in water, its overall density must equal the density of water, meaning its weight equals the weight of the displaced water. Setting the mass of the spherical shell equal to the mass of the displaced water gives (4/3)pi(R^3 - r^3) * sigma = (4/3)pi(R^3) * 1. Dividing by (4/3)pi and rearranging the terms yields R^3 * (sigma - 1) = r^3 * sigma, which simplifies to the ratio (R/r)^3 = sigma / (sigma - 1). Taking the cube root of both sides provides the ratio R/r = ( sigma / (sigma - 1) )^(1/3).

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