IBPS PO Mains 5 Nov 2023 Paper

Concept:When a cube is inscribed in a sphere, the body diagonal of the cube equals the diameter of the sphere. This relation, combined with the cone volume formula, gives the required equation.Explanation:Let the side of the cube be $P$, the radius of the sphere be $R$, the radius of the cone be $\sqrt{3}R$, and the height of the cone be $H$.Given: Volume of cone $= \frac{165}{14}$ cubic units.For the cube inside the sphere:$\sqrt{3}P = 2R$Squaring both sides: $3P^{2} = 4R^{2}$Hence, $R^{2} = \frac{3P^{2}}{4}$.Radius of cone squared:$(\sqrt{3}R)^{2} = 3R^{2} = 3 \times \frac{3P^{2}}{4} = \frac{9P^{2}}{4}$.Volume of cone:$\frac{1}{3}\pi r^{2}H = \frac{165}{14}$$\frac{1}{3} \times \frac{22}{7} \times \frac{9P^{2}}{4} \times H = \frac{165}{14}$$\frac{33P^{2}H}{14} = \frac{165}{14}$$33P^{2}H = 165$$P^{2}H = 5$$H = \frac{5}{P^{2}}$Squaring both sides:$H^{2} = \frac{25}{P^{4}}$Therefore, $\frac{25}{P^{4}} = H^{2}$.Answer:The correct relation is $\frac{25}{P^{4}} = H^{2}$, which corresponds to option D.