Question 55: Properties of Matter

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 :

Accepted Answer

To solve this problem, we consider the buoyancy and weight force acting on the sphere. For the sphere to just float on water, the weight of the water displaced by the sphere must be equal to the weight of the sphere. The volume of water displaced by the sphere is equivalent to the outer volume of the sphere minus the volume of the cavity inside it.
The volume of a sphere is given by $$V = \frac{4}{3}\pi r^3$$, where $r$ is the radius of the sphere. For the given sphere, its outer radius is $$R = \frac{D}{2}$$, and the radius of the cavity is $$r = \frac{d}{2}$$. Therefore, the volume of the sphere excluding the cavity is:
$$V_{	ext{solid part}} = \frac{4}{3}\pi R^3 - \frac{4}{3}\pi r^3$$
$$V_{	ext{solid part}} = \frac{4}{3}\pi \left(\frac{D}{2}ight)^3 - \frac{4}{3}\pi \left(\frac{d}{2}ight)^3$$
Relative density ($\sigma$) is defined as the ratio of the density of an object to the density of water. This means the actual density of the sphere is $\sigma 	imes 	ext{density of water}$. Since the object just floats, the weight of the displaced water is equal to the weight of the solid part of the sphere (ignoring the cavity), which can be mathematically represented as:
$$	ext{Weight of solid part} = 	ext{Weight of displaced water}$$
$$\sigma \cdot ho_{	ext{water}} \cdot V_{	ext{solid part}} \cdot g = ho_{	ext{water}} \cdot V_{	ext{displaced water}} \cdot g$$
Since the sphere is floating, $V_{	ext{displaced water}} = \frac{4}{3}\pi R^3$, the equation simplifies to:
$$\sigma \left(\frac{4}{3}\pi R^3 - \frac{4}{3}\pi r^3ight) = \frac{4}{3}\pi R^3$$
Cancelling out common terms gives:
$$\sigma (R^3 - r^3) = R^3$$
Given that $\sigma$ is the relative density, we can rearrange the equation to solve for the ratio of $D/d$ or equivalently $R/r$:
$$\sigma = \frac{R^3}{R^3 - r^3}$$
Solving for $R/r$:
$$R^3 (1 - \sigma) + \sigma r^3 = 0$$
$$\frac{R^3}{r^3} = \frac{\sigma}{\sigma - 1}$$
Since $R = \frac{D}{2}$ and $r = \frac{d}{2}$, the ratio of $D/d$ is equal to the ratio of $R/r$, thus:
$$\frac{D}{d} = \left(\frac{\sigma}{\sigma - 1}ight)^{1/3}$$
This matches Option D:
$$\left(\frac{\sigma}{\sigma-1}ight)^{1 / 3}$$

Answer

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

Answer

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

Answer

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

Answer

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

Properties of Matter

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 :

Choose the correct answer: