IBPS PO Mains 5 Nov 2023 Paper

Concept:A cube inscribed in a sphere has its body diagonal equal to the diameter of the sphere.Explanation:Let the side of the cube be $P$ and the radius of the sphere be $R$.Since all vertices of the cube lie on the sphere, $2R = \sqrt{3}P$.Squaring both sides: $4R^2 = 3P^2$, so $\frac{R^2}{P^2} = \frac{3}{4}$.Total surface area of the sphere is $4\pi R^2$.Lateral surface area of the cube is $4P^2$.Required ratio is $\frac{4\pi R^2}{4P^2} = \pi \times \frac{R^2}{P^2}$.Using $\pi = \frac{22}{7}$: $\frac{22}{7} \times \frac{3}{4} = \frac{66}{28} = \frac{33}{14}$.Thus the required ratio is $\frac{33}{14}$.Answer:$\frac{33}{14}$, so the correct option is B.