Question

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

Answer 1

Answer:

$\frac{d_1}{d_2} =(\frac{\sigma-1}{\sigma})^{1/3}$

Step-by-step explanation:

Let the diameter of sphere d_1 and diameter of d_2.

then volume of the sphere V=  $\frac{\pi}{6} (d_1^3-d_2^3)$

According to principal of flotation

Weight of the sphere = weight of water displaced

$\frac{\pi}{6}(d_1^3-d_2^3)\sigma = \frac{\pi}{6}(d_1^3)\times1$

⇒$\sigma = \frac{d_1^3}{d_1^3-d_2^3}$

further solving we get

$\frac{d_1}{d_2} =(\frac{\sigma-1}{\sigma})^{1/3}$