Question

A cube and solid sphere both made of the same material are completely submerged in water but to different depths. The sphere and the cube have the same surface area. The buoyant force is-
(A) Greater for the cube than the sphere
(B) Greater for the sphere than the cube
(C) Same for the sphere and the cube
(D) Greater for the object that is submerged deeper

Answer 1

Hint
Firstly find the surface area of both bodies and find the relation between radius of sphere and side of the cube .Now calculate buoyant force in both bodies. Finally compare the buoyant force and calculate the answer. Buoyant force is a kind of force which is applied on the surface of liquid perpendicularly.

Complete step by step solution
Let’s assume the side of the cube is $a$ unit and the radius of the sphere is $r$ unit.
So, surface area of cube will be :${A_1} = 6{a^2}$
And surface of sphere will be: ${A_2} = 4\pi {r^2}$
On equating both the areas:
$\Rightarrow 6{a^2} = 4\pi {r^2}$
$\Rightarrow \dfrac{r}{a} = \sqrt {\dfrac{6}{{4\pi }}}$
$\Rightarrow \dfrac{r}{a} = \sqrt {\dfrac{3}{{2\pi }}}$
$\therefore$ we all know that buoyant force on a body is equal to volume of body submerged in liquid multiplied by specific density of liquid and multiplied by acceleration due to gravity.
so buoyant force on cube is ${B_c} = \dfrac{4}{3}\pi {r^3}\rho g$
and buoyant force on sphere is ${B_s} = {a^3}\rho g$
now
$\Rightarrow \dfrac{{{B_s}}}{{{B_c}}} = \dfrac{{4\pi }}{3}\dfrac{{{r^3}}}{{{a^3}}}$
On putting the value of $\dfrac{r}{a}$in above equation
We get-
$\Rightarrow \dfrac{{{B_s}}}{{{B_c}}} = \dfrac{{4\pi }}{3} \times \dfrac{3}{{2\pi }} \times \sqrt {\dfrac{3}{{2\pi }}}$
$\Rightarrow \dfrac{{{B_s}}}{{{B_c}}} = 2\sqrt {\dfrac{3}{{2\pi }}} > 1$
Hence buoyant force at the sphere will be more than cube.
Correct answer will be option (B).

Note
Buoyant force does not depend upon the depth of the submerged body. It totally depends on the volume of immersed liquid, material of the body and acceleration due to gravity. Hence, mathematically we apply:
$B = v\rho g$
Where, $v$ is volume of body immersed in liquid,
$\rho$ is a specific density of material of the body,
$g$ is acceleration due to gravity.