- Home
- Standard 11
- Physics
A solid cube and a solid sphere both made of same material are completely submerged in water but to different depths. The sphere and the cube have same surface area. The buoyant force is
greater for the cube than the sphere
greater for the sphere than the cube
same for the sphere and the cube
greater for the object that is submerged deeper
Solution
(b)
Given,
Surface area of cube $=$ Surface area of sphere
$\Rightarrow 6 a^2 =4 \pi r^2 \dots(i)$
$\Rightarrow \frac{a}{r}=\sqrt{\left(\frac{4 \pi}{6}\right)}=\sqrt{\frac{2 \pi}{3}}$
Now, buoyant force is $F_B=V_{\text {in }} \cdot \rho_f . g$ So, ratio of buoyant force on cube and sphere is
$\frac{\left(F_B\right)_{\text {cube }}}{\left(F_B\right)_{\text {sphere }}}=\frac{V_{\text {cube }}}{V_{\text {sphere }}}$
$=\frac{a^3}{\frac{4}{3} \pi r^3}=\frac{3}{4 \pi} \times\left(\frac{2 \pi}{3}\right)^{\frac{3}{2}}$
$=\sqrt{\frac{9}{16 \times \pi^2} \times 27}=\sqrt{\frac{\pi}{6}}$
$\therefore \left(F_B\right)_{\text {cube }} < \left(F_B\right)_{\text {sphere }}$
