Question

Water is dripping out from a conical funnel at a uniform rate of 4 cm/sec., through a tiny hole at the
wex in the bottom. When the slant height of the water is 3 cm, find the rate of decrease of the slant
Light of the water, given that the vertical angle of the funnel is 120°.​

Answer 1

Answer:

the answer of the is 1cm

Answer 2

Answer:

$\frac{32}{27\pi}\ \frac{cm}{sec}$

Step-by-step explanation:

In this question

Given that,

Rate of dripping out from a conical funnel = $\frac{d_v}{d_t}$ = $4\frac{cm}{sec}$

Volume of water = $\frac{1}{3}\times \pi\ r{^2}\times h$

Also, sin60 = $\frac{r}{l}$

 $r$ = $\frac{\sqrt3}{2l}$

        cos60 = $\frac{h}{l}$

 $h$ = $\frac{l}{2}$

Volume  $v$ = $\frac{1}{3}\times \pi\times \frac{3}{4}\times l^2\times \frac{l}{2}$

$v [t/ex] = [tex]\frac{1}{8}\times \pi\times l^3$

Diffraction with recpect to t,

$\frac{d_v}{d_t}$ = $\frac{1}{8}\times \pi\times 3(l^2)\times \frac{d_l}{d_t}$

At l = 3, then $\frac{d_lv}{d_t}$ = 4

So, $\frac{d_l}{d_t}$ = $\frac{4\times 8}{27\times \pi}$

     $\frac{d_l}{d_t}$ = $\frac{32}{27\pi}\ \frac{cm}{sec}$

Hence, value of rate of decrease of the slant  Light of the water is $\frac{32}{27\pi}\ \frac{cm}{sec}$