UPSC CDS 2 2021 Math Paper

Concept:The volume of water left equals the cylinder's volume minus the volume of the solid (a cone with a hemisphere placed on it).Explanation:The cylinder has radius $x$ and height $3x$.Its volume is $\pi x^2 \cdot 3x = 3\pi x^3$.The solid consists of a cone and a hemisphere, each with radius $x$.The cone has height $2x$.Volume of cone = $\frac{1}{3}\pi x^2 \cdot 2x = \frac{2}{3}\pi x^3$.Volume of hemisphere = $\frac{2}{3}\pi x^3$.Total solid volume = $\frac{2}{3}\pi x^3 + \frac{2}{3}\pi x^3 = \frac{4}{3}\pi x^3$.Water left = $3\pi x^3 - \frac{4}{3}\pi x^3 = \frac{5}{3}\pi x^3$.Using $\pi \approx \frac{22}{7}$, we get $\frac{5}{3} \times \frac{22}{7} x^3 = \frac{110}{21} x^3 \approx 5.238 x^3$, which rounds to $5.24 x^3$.Answer:$5.24 x^{3}$ (Option D).