An inverted cone (base above the vertex) is 2 m high and has a base radius of 12 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?

Question

cyhuddwyr9 · Accepted Answer

Step 1  Given, an inverted cone is 2 m high and has a base radius of 12 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=∫abF(x)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  w12=2−h2  2w=2−h2  w=2−h4  The weight of the slice under consideration is  Weight=volume×density×gravitationalconstan⁡t  Weight=π×w2×dh×62.4(Densityofwater=62.4lbft3)  Weight=62.4πw2dh  Now by definition the work done is  W=∫12dW  =∫12(62.4πw2dh)h  =62.4π×116∫12(2−h)2hdh  =62.4π×116∫12(4+h2−4h)hdh  =62.4π×116∫12(4h+h3−4h2)dh  =62.4π×116{4h22+h44−4h33}12  =62.4π×116{162+164−323−42−14+43}  =62.4π×116{8+4−323−2−14+43}  =1.625πlb  Hence work done is 1.635πlb.

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?

Answered question

Answer & Explanation

New Questions in High school geometry