If h,C,V are respectively the height, the curved surface and the volume of a cone, then $$3\pi$$ $$Vh^{3}-C^{2}h^{2}+9V^{2}=$$ ?

Let slant height of cone = $$l$$ units and radius = $$r$$ units

Thus, $$l=\sqrt{h^2+r^2}$$ , $$V=\frac{1}{3}\pi r^2h$$ and $$C=\pi rl$$

To find : $$3\pi$$ $$Vh^{3}-C^{2}h^{2}+9V^{2}$$

= $$[3\pi\times(\frac{1}{3}\pi r^2h)\times h^3]-[(\pi rl)^2\times h^2]+[9\times(\frac{1}{3}\pi r^2h)^2]$$

= $$[\pi^2 r^2h^4]-[\pi^2r^2h^2(r^2+h^2)]+[\pi^2r^4h^2]$$

= $$(\pi^2 r^2h^4)-(\pi^2r^4h^2)-(\pi^2r^2h^4)+(\pi^2r^4h^2)$$

= $$0$$

=> Ans - (A)