Question

A conical paper cup is 10 cm tall with a radius of 10 cm. The cup is being filled with water so that the water level rises at a rate of 2 cm/sec. At what rate is water being poured into the cup when the water level is 8 cm?

Answer 1

Answer:

Water is being pored in the cone with the rate of 401.92 cm³ per second.

Step-by-step explanation:

There is a cone having height h = 10 cm and radius r = 10 cm

That means h = r

Rate of change of water level $\frac{dh}{dt}=2$ cm per second

Since volume of a cone is represented by the formula,

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

Now we have to find the value of $\frac{dV}{dt}$.

$\frac{dV}{dt}=\frac{d}{dt}(\frac{1}{3}\pi r^{2}h)$

Since r = h

$\frac{dV}{dt}=\frac{d}{dt}(\frac{1}{3}\pi h^{3} )$

$\frac{dV}{dt}=\frac{3}{3}\pi h^{2}.\frac{dh}{dt}$

For h = 8 cm and $\frac{dh}{dt}=2$ cm per second.

$\frac{dV}{dt}=\frac{3}{3}\pi (8)^{2}\times (2)$

$\frac{dV}{dt}=128\pi$ = 401.92 cubic cm per second

Therefore, water is being pored with the rate of 401.92 cm³ per sec.