JEE MAIN - Physics (2020 - 8th January Morning Slot - No. 7)
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 :
$$\rho \left( r \right) = {\rho _0}\left( {1 - {{{r^2}} \over {{R^2}}}} \right)$$ , $$0 < r \le R$$
The minimum density of a liquid in which it will float is :
$${{2{\rho _0}} \over 3}$$
$${{2{\rho _0}} \over 5}$$
$${{{\rho _0}} \over 5}$$
$${{{\rho _0}} \over 3}$$
Explanation
_8th_January_Morning_Slot_en_7_1.png)
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 = Fb
$$ \Rightarrow $$ $${\rho _0}{{8\pi {R^3}} \over {15}}$$g = $${\rho _l}{4 \over 3}\pi {R^3}g$$
$$ \Rightarrow $$ $${\rho _l} = {{2{\rho _0}} \over 5}$$
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