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
KVPY 2012, Diffcult
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(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
$\frac{\left(F_B\right)_{\text {cube }}}{\left(F_B\right)_{\text {sphere }}}=\frac{V_{\text {cube }}}{V_{\text {sphere }}}$
$=\frac{a^3}{\frac{4}{3} \pi r^3}=\frac{3}{4 \pi} \times\left(\frac{2 \pi}{3}\right)^{\frac{3}{2}}$
$=\sqrt{\frac{9}{16 \times \pi^2} \times 27}=\sqrt{\frac{\pi}{6}}$
$\therefore \left(F_B\right)_{\text {cube }} < \left(F_B\right)_{\text {sphere }}$
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