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(B) Given,surface area of cube $=$ surface area of sphere.$6a^2 = 4\pi r^2 \Rightarrow \frac{a}{r} = \sqrt{\frac{4\pi}{6}} = \sqrt{\frac{2\pi}{3}}$.Bu

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greater for the sphere than the cube

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(B) Given,surface area of cube $=$ surface area of sphere.$6a^2 = 4\pi r^2 \Rightarrow \frac{a}{r} = \sqrt{\frac{4\pi}{6}} = \sqrt{\frac{2\pi}{3}}$.Buoyant force $F_B = V \cdot ho_f \cdot g$. Since both are completely submerged in the same fluid,the ratio of buoyant forces is equal to the ratio of their volumes:$\frac{(F_B)_{	ext{cube}}}{(F_B)_{	ext{sphere}}} = \frac{V_{	ext{cube}}}{V_{	ext{sphere}}} = \frac{a^3}{\frac{4}{3}\pi r^3} = \frac{3}{4\pi} \left( \frac{a}{r} ight)^3$.Substituting $\frac{a}{r} = \sqrt{\frac{2\pi}{3}}$:$\frac{(F_B)_{	ext{cube}}}{(F_B)_{	ext{sphere}}} = \frac{3}{4\pi} \left( \sqrt{\frac{2\pi}{3}} ight)^3 = \frac{3}{4\pi} \cdot \frac{2\pi}{3} \cdot \sqrt{\frac{2\pi}{3}} = \frac{1}{2} \sqrt{\frac{2\pi}{3}} = \sqrt{\frac{2\pi}{3 \cdot 4}} = \sqrt{\frac{\pi}{6}}$.Since $\pi \approx 3.14$,$\frac{\pi}{6} This implies $(F_B)_{	ext{cube}} Thus,the buoyant force is greater for the sphere.

$A$ solid cube and a solid sphere,both made of the same material,are completely submerged in water but at different depths. The sphere and the cube have the same surface area. The buoyant force is

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