Question

There is a conical flask of semi vertex angle of 37° in which water is filled at a constant rate 20cm³/s. The rate of height of water, when height of water becomes 4 cm is (Use pi = 3)​

Answer 1

Given:

There is a conical flask of semi vertex angle of 37° in which water is filled at a constant rate 20cm³/s.

To find:

Rate fight of water, when the height of water becomes force.

Calculation:

Volume of a cone is :

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

Now , in ∆AOB ,

$\tan( {37}^{ \circ} ) = \dfrac{r}{h} \\ = > \dfrac{3}{4} = \dfrac{r}{h} \\ = > r = \dfrac{3h}{4}$

Replacing "r" in 1st Equation;

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

$= > V = \dfrac{1}{3} \pi {( \dfrac{3h}{4}) }^{2} h$

$= > V = \dfrac{1}{3} \pi \times \dfrac{9 {h}^{3} }{16}$

$= > V = \dfrac{3\pi {h}^{3} }{16}$

Differentiating w.r.t time ;

$= > \dfrac{d(V)}{dt} = \dfrac{d (\dfrac{3\pi {h}^{3} }{16}) }{dt}$

$= > \dfrac{d(V)}{dt} = \dfrac{3\pi}{16} \times \dfrac{d {h}^{3} }{dt}$

$= > \dfrac{d(V)}{dt} = \dfrac{3\pi}{16} \times 3 {h}^{2} \times \dfrac{dh }{dt}$

Putting available values ;

$= > 20 = \dfrac{3\pi}{16} \times 3 \times {(4)}^{2} \times \dfrac{dh }{dt}$

$= > 20 = \dfrac{3\pi}{16} \times 3 \times 16 \times \dfrac{dh }{dt}$

$= > 20 =9\pi \times \dfrac{dh }{dt}$

$= > \dfrac{dh }{dt} = \dfrac{20}{9\pi}$

Taking π = 3 , we get ;

$= > \dfrac{dh }{dt} = \dfrac{20}{9\times3}$

$= > \dfrac{dh }{dt} = \dfrac{20}{27}$

$= > \dfrac{dh }{dt} = 0.74 \:cm {s}^{-1}$

So, final answer is:

$\boxed{ \bold{ \red{ \dfrac{dh }{dt} = \dfrac{20}{9\pi} \: cm \: {s}^{ - 1} =0.74 \: cm\: {s}^{-1}}}}$