(a) Find the volume of a right solid cone of base radius 4cm and perpendicular height 6cm. [\(\pi = 3.142\)] (b) A hemispherical tank of diameter which is 1...
Question 1 Report
(a) Find the volume of a right solid cone of base radius 4cm and perpendicular height 6cm. [\(\pi = 3.142\)]
(b) A hemispherical tank of diameter which is 10m is filled by water issuing from a pipe of radius 20cm at 2m per second. Calculate, correct to three significant figures, the time, in minutes, it takes to fill the tank.
Answer Details
(a) Volume of a right circular cone \(= \dfrac{1}{3}\pi r^2 h\), with \(r = 4\) cm, \(h = 6\) cm, \(\pi = 3.142\):
\[ V = \frac{1}{3} \times 3.142 \times 4^2 \times 6 = \frac{1}{3} \times 3.142 \times 16 \times 6 = 3.142 \times 32 = 100.544\ \text{cm}^3. \]\(V \approx 100.5\ \text{cm}^3\).
(b) The tank is a hemisphere of diameter 10 m, so radius \(R = 5\) m.
\[ \text{Volume of tank} = \frac{2}{3}\pi R^3 = \frac{2}{3} \times 3.142 \times 5^3 = \frac{2}{3} \times 3.142 \times 125 = 261.83\ \text{m}^3. \]The pipe has radius \(20\ \text{cm} = 0.2\ \text{m}\) and water flows at 2 m/s. Volume delivered per second:
\[ \text{Flow rate} = \pi (0.2)^2 \times 2 = 3.142 \times 0.04 \times 2 = 0.25136\ \text{m}^3/\text{s}. \]Time to fill:
\[ t = \frac{261.83}{0.25136} = 1041.7\ \text{s} = \frac{1041.7}{60} = 17.4\ \text{minutes (to 3 s.f.).} \]Download The App On Google Playstore
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