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 $h$ cm after $t$ sec of start of filling the tank be $x$ cm.

So $h/x=tan30^@=>x=sqrt3h$ .

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

$A=x*h=sqrt3h^2" "cm^2$

(1) Volume of water at t th sec will be $V=A*40=40sqrt3h^2cm^3$

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

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

Hence

$40sqrt3h^2cm^3=200t$

$=>h^2=5/sqrt3t$

Differentiating w .r to t we get

$2h(dh)/(dt)=5/sqrt3$

$=>(dh)/(dt)=5/(2sqrt3h)$

So rate of increasing height when $h=5cm$ will be
$=>[(dh)/(dt)]_(h=5)=5/(2sqrt3*5)=sqrt3/6cms^-1$