Question

It is about finding expression for volume?

Answer 1

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`