Question.
A conical pit of top diameter $3.5 \mathrm{~m}$ is $12 \mathrm{~m}$ deep. What is its capacity in kilolitres? $\left[\right.$ Assume $\left.\pi=\frac{22}{7}\right]$
A conical pit of top diameter $3.5 \mathrm{~m}$ is $12 \mathrm{~m}$ deep. What is its capacity in kilolitres? $\left[\right.$ Assume $\left.\pi=\frac{22}{7}\right]$
Solution:
Radius $(r)$ of pit $=\left(\frac{3.5}{2}\right) \mathrm{m}=1.75 \mathrm{~m}$
Height $(h)$ of pit = Depth of pit $=12 \mathrm{~m}$
Volume of pit $=\frac{1}{3} \pi r^{2} h$
$=\left[\frac{1}{3} \times \frac{22}{7} \times(1.75)^{2} \times 12\right] \mathrm{cm}^{3}$
$=38.5 \mathrm{~m}^{3}$
Thus, capacity of the pit $=(38.5 \times 1)$ kilolitres $=38.5$ kilolitres
Radius $(r)$ of pit $=\left(\frac{3.5}{2}\right) \mathrm{m}=1.75 \mathrm{~m}$
Height $(h)$ of pit = Depth of pit $=12 \mathrm{~m}$
Volume of pit $=\frac{1}{3} \pi r^{2} h$
$=\left[\frac{1}{3} \times \frac{22}{7} \times(1.75)^{2} \times 12\right] \mathrm{cm}^{3}$
$=38.5 \mathrm{~m}^{3}$
Thus, capacity of the pit $=(38.5 \times 1)$ kilolitres $=38.5$ kilolitres
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