Question

A water tank is in the shape of a cone.Its diameter is 50 meter and slant edge is also 50 meter.How much water it can store In it ? it releases water to a gated apartment building at the rate of 400 liters per second. In how many minutes tank will be empty?    

Answer 1

To get the most accurate answer possible, we're going to have to go into some unsightly calculation, but bear with me here:

Assessing the situation:

Let's get a feel for the shape of the problem here: what step should we be aiming to get to by the end? We want to find out how long it will take, in minutes, for the tank to drain completely, given a drainage rate of 400 L/s. Let's name a few key variables we'll need to keep track of here:

$V$ - the storage volume of our tank (in liters)
$t$ - the amount of time it will take for the tank to drain (in minutes)

We're about ready to set up an expression using those variables, but first, we should address a subtlety: the question provides us with the drainage rate in liters per second. We want the answer expressed in liters per minute, so we'll have to make that conversion beforehand. Since one second is 1/60 of a minute, a drainage rate of 400 L/s becomes 400 · 60 = 24,000 L/min.

From here, we can set up our expression. We want to find out when the tank is completely drained - when the water volume is equal to 0. If we assume that it starts full with a water volume of $V$ L, and we know that 24,000 L is drained - or subtracted - from that volume every minute, we can model our problem with the equation

$V-24000t=0$

To isolate $t$, we can take the following steps:

$V-24000t=0\\ V=24000t\\ \frac{V}{24000}=t$

So, all we need to do now to find $t$ is find $V$. As it turns out, this is a pretty tall order. Let's begin:

Solving for $V$:

About units: all of our measurements for the cone-shaped tank have been provided for us in meters, which means that our calculations will produce a value for the volume in cubic meters. This is a problem, since our drainage rate is given to us in liters per second. To account for this, we should find the conversion rate between cubic meters and liters so we can use it to convert at the end.

It turns out that 1 cubic meter is equal to 1000 liters, which means that we'll need to multiply our result by 1000 to switch them to the correct units.

Down to business: We begin with the formula for the area of a cone,

$V= \frac{1}{3}\pi r^2h$

which is to say, 1/3 multiplied by the area of the circular base and the height of the cone. We don't know $h$ yet, but we are given the diameter of the base: 50 m. To find the radius $r$, we divide that diameter in half to obtain r = 50/2 = 25 m. All that's left now is to find the height.

To find that, we'll use another piece of information we've been given: a slant edge of 50 m. Together with the height and the radius of the cone, we have a right triangle, with the slant edge as the hypotenuse and the height and radius as legs. Since we've been given the slant edge (50 m) and the radius (25 m), we can use the Pythagorean Theorem to solve for the height $h$:

$h^2+25^2=50^2\\ h^2+625=2500\\ h^2=1875\\ h=\sqrt{1875}=\sqrt{625\cdot3}=25\sqrt{3}$

With $h=25\sqrt{3}$ and $r=25$, we're ready to solve for $V$:

$V= \frac{1}{3} \pi(25)^2\cdot25\sqrt{3}\\ V= \frac{1}{3} \pi\cdot625\cdot25\sqrt{3}\\ V= \frac{1}{3} \pi\cdot15625\sqrt{3}\\\\ V= \frac{15625\sqrt{3}\pi}{3}$

This gives us our volume in cubic meters. To convert it to liters, we multiply this monstrosity by 1000 to obtain:

$\frac{15625\sqrt{3}\pi}{3}\cdot1000= \frac{15625000\sqrt{3}\pi}{3}$

We're almost there.

Bringing it home:

Remember that formula for $t$ we derived at the beginning? Let's revisit that. The number of minutes $t$ that it will take for this tank to drain completely is:

$t= \frac{V}{24000}$

We have our $V$ now, so let's do this:

$t= \frac{\frac{15625000\sqrt{3}\pi}{3}}{24000} \\ t= \frac{15625000\sqrt{3}\pi}{3}\cdot \frac{1}{24000} \\ t=\frac{15625000\sqrt{3}\pi}{3\cdot24000}\\ t=\frac{15625\sqrt{3}\pi}{3\cdot24}\\ t=\frac{15625\sqrt{3}\pi}{72}\\ t\approx1180.86$

So, it will take approximately 1180.86 minutes to completely drain the tank, which can hold approximately $V= \frac{15625000\sqrt{3}\pi}{3}\approx 28340615.06$ L of fluid.