It is about finding expression for volume?

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It is about finding expression for volume?

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Answer 1 · 2018-05-09T05:05:51

Let half of the base of isoscles triangle of height $h$ $h$ cm after $t$ $t$ sec of start of filling the tank be $x$ $x$ cm.

So $\frac{h}{x} = {tan 30}^{\circ} \Rightarrow x = \sqrt{3} h$ $h/x=tan30^@=>x=sqrt3h$ .

So the area of triangular surface of water of height will be

$A = x \cdot h = \sqrt{3} h^{2} c m^{2}$ $A=x*h=sqrt3h^2" "cm^2$

(1) Volume of water at t th sec will be $V = A \cdot 40 = 40 \sqrt{3} h^{2} c m^{3}$ $V=A*40=40sqrt3h^2cm^3$

(2) The rate of pouring water in the tank is $200 c m^{3} / s$ $200cm^3"/"s$ .

So the amount of water poured in t sec will be $200 t c m^{3}$ $200tcm^3$

Hence

$40 \sqrt{3} h^{2} c m^{3} = 200 t$ $40sqrt3h^2cm^3=200t$

$\Rightarrow h^{2} = \frac{5}{\sqrt{3}} t$ $=>h^2=5/sqrt3t$

Differentiating w .r to t we get

$2 h \frac{d h}{d t} = \frac{5}{\sqrt{3}}$ $2h(dh)/(dt)=5/sqrt3$

$\Rightarrow \frac{d h}{d t} = \frac{5}{2 \sqrt{3} h}$ $=>(dh)/(dt)=5/(2sqrt3h)$

So rate of increasing height when $h = 5 c m$ $h=5cm$ will be
$\Rightarrow {[\frac{d h}{d t}]}_{h = 5} = \frac{5}{2 \sqrt{3} \cdot 5} = \frac{\sqrt{3}}{6} c m s^{- 1}$ $=>[(dh)/(dt)]_(h=5)=5/(2sqrt3*5)=sqrt3/6cms^-1$

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It is about finding expression for volume?

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