Question 147: Properties of Matter

Question

Consider a solid sphere of radius R and mass
density $$\rho \left( r \right) = {\rho _0}\left( {1 - {{{r^2}} \over {{R^2}}}} \right)$$ , $$0 < r \le R$$ The
minimum density of a liquid in which it will
float is :

Accepted Answer

Mass of solid = $$\int\limits_0^R {\rho dV} $$
= $$\int\limits_0^R {{\rho _0}\left( {1 - {{{r^2}} \over {{R^2}}}} \right)4\pi {r^2}dr} $$
= $${\rho _0}4\pi \left[ {\int\limits_0^R {{r^2}dr}  - {1 \over {{R^2}}}\int\limits_0^R {{r^4}dr} } \right]$$
= $${\rho _0}4\pi \left[ {{{{R^3}} \over 3} - {{{R^5}} \over 5}} \right]$$
= $${\rho _0}{{8\pi {R^3}} \over {15}}$$
For minimum density of liquid, solid sphere has
to float (completely immersed) in the liquid.
So Weight of the sphere = Buoyant force
mg = F
b
$$ \Rightarrow $$ $${\rho _0}{{8\pi {R^3}} \over {15}}$$g = $${\rho _l}{4 \over 3}\pi {R^3}g$$
$$ \Rightarrow $$ $${\rho _l} = {{2{\rho _0}} \over 5}$$

Answer

$${{2{\rho _0}} \over 3}$$

Answer

$${{2{\rho _0}} \over 5}$$

Answer

$${{{\rho _0}} \over 5}$$

Answer

$${{{\rho _0}} \over 3}$$

Properties of Matter

Consider a solid sphere of radius R and mass density $\rho \left( r \right) = {\rho _0}\left( {1 - {{{r^2}} \over {{R^2}}}} \right) , 0 < r \le R$ The minimum density of a liquid in which it will float is :

Choose the correct answer: