If `h ,\ S\ a n d\ V` denote respectively the height, curved surface area and volume of a right circular cone, then `3piV h^3-\ S^2h^2+9V^2` is equal to (a) 8 (b)0(c)`4pi`(d)`32pi^2`

Explanation

As we know that the surface area of cone = `pirl`

Here, l=slant height , r=radius and h=height

`l=sqrt(r^2+h^2)`

`therefore` S.A =`pirsqrt(r^2+h^2)`

Volume of a cone(V) = `1/3pir^2h`

Now, `3piVh^3-S^2h^2+9V^2`

`3pi(1/3pir^2h)h^3-pi^2r^2(r^2+h^2)h^2+9(1/9pi^2r^4h^2)

`0`

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