# A conical tank is of height 12 m and surface diameter 8m. Water is pumped into the tank at the rate of 50 m^3/(min). How fast is the water level increasing when the depth of the water is 6 m?

A conical tank is of height 12 m and surface diameter 8m. Water is pumped into the tank at the rate of $50\frac{{m}^{3}}{min}$. How fast is the water level increasing when the depth of the water is 6 m?
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Arnold Odonnell
Step 1
27 (a)Given- height(H)=12 m, diameter(D)=8m and $\frac{dV}{dt}=50\frac{{m}^{3}}{min}$.
To find- the level of the water increasing when the depth of the water 6m.
Formula Used- the volume of the conical tank $\left(V\right)=\frac{1}{3}\pi {r}^{2}h$, when , r= radius and h=height.
Step 2
Explanation- The volume of the tank $\left(V\right)=\frac{1}{3}\pi {r}^{2}h$ ⋯⋯(1)
As we have to find the level of the water increasing when the depth of the water 6m, so we have to eliminate the radius (r) in terms of h, which can be written as follws,
$\frac{H}{R}=\frac{h}{r}$ (using the triangle similarity)
Further, we can write as,
$r=\frac{R}{H}·h$
Now, substituting the value of r in equation (1), we get,
$V=\frac{1}{3}\pi {\left(\frac{R}{H}\cdot h\right)}^{2}\cdot h$
$V=\frac{1}{3}\pi \frac{{R}^{2}}{{H}^{2}}{h}^{3}$
Step 3
Now, diffferentiating the above expression, w.r.t. h, we get,
$\frac{dV}{dt}=\frac{1}{3}\pi \cdot \frac{{R}^{2}}{{H}^{2}}\cdot 3{h}^{2}\cdot \frac{dh}{dt}$
$50=\pi {\left(\frac{4}{12}\right)}^{2}\cdot {\left(6\right)}^{2}\cdot \frac{dh}{dt}$
Further, we can write as,
$\frac{dh}{dt}=\frac{50}{4\pi }$
$=3.978\approx 4\frac{m}{min}$
So, the level of the water increasing when the depth of the water 6m at $3.978\approx \frac{4m}{min}$.
Result: Hence, the level of the water increasing when the depth of the water 6m at $3.978\approx \frac{4m}{min}$.