Q1. Volume = Volume of hemisphere + Volume of cone = $\frac{2}{3}\pi r^3 + \frac{1}{3}\pi r^2 h = \frac{2}{3}\pi(2^3) + \frac{1}{3}\pi(2^2)(2) = 8\pi = 25.12\text{ cm}^3$.
Q2. Volume is constant: $\frac{4}{3}\pi (4.2)^3 = \pi (6)^2 h \rightarrow h = \frac{4 \times 74.088}{3 \times 36} = 2.744\text{ cm}$.
Q3. Volume left = Vol of cylinder - (Vol of cone + Vol of hemisphere) = $1.131\text{ m}^3$.
Q4. Volume of frustum = $\frac{1}{3}\pi h (R^2 + r^2 + Rr) = 10449.92\text{ cm}^3 = 10.45\text{ litres}. Cost = $10.45 \times 40 = \text{Rs. } 418$.
Q5. Total volume = Vol of sphere + Vol of cylinder = $\frac{4}{3}\pi R^3 + \pi r^2 h = 321.39 + 25.13 = 346.51\text{ cm}^3. She is incorrect (actual volume is $346.51\text{ cm}^3$).
