AP TET Paper 1 Model Test 2

Concept:Equal surface areas of a cube and a sphere give a relation between their side and radius. The ratio of their volumes is then found using this relation.Explanation:Let the side of cube be $a$ and radius of sphere be $r$.Given: $6a^{2} = 4\pi r^{2}$.Divide both sides by $6r^{2}$: $\frac{a^{2}}{r^{2}} = \frac{4\pi}{6} = \frac{2\pi}{3}$.Take square root: $\frac{a}{r} = \frac{\sqrt{2\pi}}{\sqrt{3}}$.Volume of cube $V_c = a^{3}$; volume of sphere $V_s = \frac{4}{3}\pi r^{3}$.Ratio $\frac{V_c}{V_s} = \frac{a^{3}}{\frac{4}{3}\pi r^{3}} = \frac{3}{4\pi} \left(\frac{a}{r}\right)^{3}$.Substitute $\frac{a}{r} = \frac{\sqrt{2\pi}}{\sqrt{3}}$: $\left(\frac{a}{r}\right)^{3} = \frac{(2\pi)^{3/2}}{3^{3/2}} = \frac{2\sqrt{2\pi}}{3\sqrt{3}}$.Then $\frac{V_c}{V_s} = \frac{3}{4\pi} \cdot \frac{2\sqrt{2\pi}}{3\sqrt{3}} = \frac{\sqrt{2\pi}}{2\pi\sqrt{3}} \cdot 2?$ Wait, simplify carefully:$\frac{3}{4\pi} \times \frac{2\sqrt{2\pi}}{3\sqrt{3}} = \frac{2\sqrt{2\pi}}{4\pi\sqrt{3}} = \frac{\sqrt{2\pi}}{2\pi\sqrt{3}}$.But this seems off. Let's do directly from given existing solution: We had $6a^2 = 4\pi r^2$ → $a^2/r^2 = 4\pi/6 = 2\pi/3$ → $a/r = \sqrt{2\pi}/\sqrt{3}$. Then $V_c/V_s = a^3 / (4/3 \pi r^3) = (3/(4\pi)) (a/r)^3 = (3/(4\pi)) * ( (2\pi)^{3/2} / 3^{3/2} ) = (3/(4\pi)) * (2\pi\sqrt{2\pi} / (3\sqrt{3})) = (3 * 2\pi\sqrt{2\pi}) / (4\pi * 3\sqrt{3}) = (2\pi\sqrt{2\pi})/(12\pi\sqrt{3}) = (\sqrt{2\pi})/(6\sqrt{3}) = \sqrt{2\pi}/\sqrt{108}? This doesn't match options. Better to use the existing solution's approach: Assume$a = 2\sqrt{\pi}$,$r = \sqrt{6}$from$a/r = 2\sqrt{\pi}/\sqrt{6}$(since$a^2/r^2 = 4\pi/6$). Then$V_c = (2\sqrt{\pi})^3 = 8\pi\sqrt{\pi}$,$V_s = \frac{4}{3}\pi (\sqrt{6})^3 = \frac{4}{3}\pi * 6\sqrt{6} = 8\pi\sqrt{6}$. Ratio$V_c:V_s = (8\pi\sqrt{\pi}) : (8\pi\sqrt{6}) = \sqrt{\pi} : \sqrt{6}$. So the correct ratio is$\sqrt{\pi}:\sqrt{6}$, which matches option B. Thus final answer is option B. Now structure correctly: < b>Concept:</b> < br> Equal surface area of a cube and a sphere allows us to express one dimension in terms of the other. The volumes ratio then simplifies to a simple surd. < br> < b>Explanation:</b> < br> Let cube side =$a$, sphere radius =$r$. < br> Given:$6a^2 = 4\pi r^2$. So$a^2/r^2 = 4\pi/6 = 2\pi/3$. < br> Take square root:$a/r = \sqrt{2\pi}/\sqrt{3}$. < br> To avoid fractions, set$a = 2\sqrt{\pi}$,$r = \sqrt{6}$(since$a^2/r^2 = 4\pi/6$). < br> Volume of cube =$a^3 = (2\sqrt{\pi})^3 = 8\pi\sqrt{\pi}$. < br> Volume of sphere =$\frac{4}{3}\pi r^3 = \frac{4}{3}\pi (\sqrt{6})^3 = \frac{4}{3}\pi \cdot 6\sqrt{6} = 8\pi\sqrt{6}$. < br> Ratio of volumes =$(8\pi\sqrt{\pi}) : (8\pi\sqrt{6}) = \sqrt{\pi} : \sqrt{6}$. < br> < b>Answer:</b> < br> Option B:$\sqrt{\pi} : \sqrt{6}$