Question
A hollow spherical body of inner and outer radii 6cm and 8cm respectively floats half submerged in water. Find the density of the material of the sphere.
✓
Answer
Let the density of the spherical body is Pg/cm
By the Principle of the Flotation,
Weight of the spherical body = Weight of the fluid displaced.
$\therefore$ Volume of the spherical body × density of spherical body × g = Volume of body displaced in water × density of water × g
Volume of the spherical body = $\frac{4}{\pi}\big(\text{R}^3-\text{r}^3\big)$
$=\frac{4}{3}\times\frac{22}{7}\times\big(8^3-6^3\big) $
$=\frac{4}{3\pi}\big(8^3-6^3\big)$
Volume of the spherical body displaced in water = Outer volume $=\frac{\text{outer}\ \text{volume}}{2}=\frac{4}{3\pi}\big(8^3\big)$
Also, density of water $=1\text{g}/\text{cm}^3.$
$\therefore\frac{4}{3\pi}\big(\text{R}^3-\text{r}^3\big)\times\rho=\frac{\Big[\frac{4}{3\pi}\big(\text{R}^3\big)\Big]}{2}\times1$
$\therefore\frac{4}{3\pi}\big(8^3-6^3\big)\times\rho=\frac{2}{3\pi}\big(8^3\big)$
$2\big(8^3-6^3\big)\times\rho=(8^3)$
$2\times296\rho=-512$
$\text{p}=0.8648\text{g}/\text{cm}^3$
$\therefore\text{p}=864.8\text{kg}/\text{m}^3.$
Hence, the density of the material of the spherical body = $864.8\text{kg}/\text{m}^3.$
Hope it helps.
By the Principle of the Flotation,
Weight of the spherical body = Weight of the fluid displaced.
$\therefore$ Volume of the spherical body × density of spherical body × g = Volume of body displaced in water × density of water × g
Volume of the spherical body = $\frac{4}{\pi}\big(\text{R}^3-\text{r}^3\big)$
$=\frac{4}{3}\times\frac{22}{7}\times\big(8^3-6^3\big) $
$=\frac{4}{3\pi}\big(8^3-6^3\big)$
Volume of the spherical body displaced in water = Outer volume $=\frac{\text{outer}\ \text{volume}}{2}=\frac{4}{3\pi}\big(8^3\big)$
Also, density of water $=1\text{g}/\text{cm}^3.$
$\therefore\frac{4}{3\pi}\big(\text{R}^3-\text{r}^3\big)\times\rho=\frac{\Big[\frac{4}{3\pi}\big(\text{R}^3\big)\Big]}{2}\times1$
$\therefore\frac{4}{3\pi}\big(8^3-6^3\big)\times\rho=\frac{2}{3\pi}\big(8^3\big)$
$2\big(8^3-6^3\big)\times\rho=(8^3)$
$2\times296\rho=-512$
$\text{p}=0.8648\text{g}/\text{cm}^3$
$\therefore\text{p}=864.8\text{kg}/\text{m}^3.$
Hence, the density of the material of the spherical body = $864.8\text{kg}/\text{m}^3.$
Hope it helps.
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