# An inverted cone (base above the vertex) is 2 m high and has a base radius of 1/2 m. If the tank is full, how much work is required to pump the water to a level 1 m above the top of the tank?

An inverted cone (base above the vertex) is 2 m high and has a base radius of $\frac{1}{2}$ m. If the tank is full, how much work is required to pump the water to a level 1 m above the top of the tank?
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cyhuddwyr9
Step 1
Given, an inverted cone is 2 m high and has a base radius of $\frac{1}{2}$ m. If the tank is full, then how much work is required to pump the water to a level 1 m above the top of the tank.
Step 2: Concept Used
If force is a function F(x) of position x then in moving from x = a to x = b work done is
$W={\int }_{a}^{b}F\left(x\right)dx$
Step 3: Calculation
Consider the water tank conical in shape. We will make a small horizontal section of the water at depth h and thickness dh and also assume radius at depth h is w.
From similarity, we will get
$\frac{w}{\frac{1}{2}}=\frac{2-h}{2}$
$2w=\frac{2-h}{2}$
$w=\frac{2-h}{4}$
The weight of the slice under consideration is
$Weight=volume×density×gravitationalcons\mathrm{tan}t$
$Weight=\pi ×{w}^{2}×dh×62.4\left(Densityofwater=62.4\frac{lb}{f{t}^{3}}\right)$
$Weight=62.4\pi {w}^{2}dh$
Now by definition the work done is
$W={\int }_{1}^{2}dW$
$={\int }_{1}^{2}\left(62.4\pi {w}^{2}dh\right)h$
$=62.4\pi ×\frac{1}{16}{\int }_{1}^{2}{\left(2-h\right)}^{2}hdh$
$=62.4\pi ×\frac{1}{16}{\int }_{1}^{2}\left(4+{h}^{2}-4h\right)hdh$
$=62.4\pi ×\frac{1}{16}{\int }_{1}^{2}\left(4h+{h}^{3}-4{h}^{2}\right)dh$
$=62.4\pi ×\frac{1}{16}{\left\{\frac{4{h}^{2}}{2}+\frac{{h}^{4}}{4}-\frac{4{h}^{3}}{3}\right\}}_{1}^{2}$
$=62.4\pi ×\frac{1}{16}\left\{\frac{16}{2}+\frac{16}{4}-\frac{32}{3}-\frac{4}{2}-\frac{1}{4}+\frac{4}{3}\right\}$
$=62.4\pi ×\frac{1}{16}\left\{8+4-\frac{32}{3}-2-\frac{1}{4}+\frac{4}{3}\right\}$
$=1.625\pi lb$
Hence work done is $1.635\pi lb$.