$A$ solid right circular cone of height $120 \,cm$ and radius $60 \,cm$ is placed in a right circular cylinder full of water of height $180 \,cm$ such that it touches the bottom. Find the volume of water left in the cylinder,if the radius of the cylinder is equal to the radius of the cone.

(N/A) $(i)$ When a solid object is submerged in a container full of water,the volume of water displaced is equal to the volume of the submerged part of the object.
$(ii)$ The total volume of water initially in the cylinder is equal to the volume of the cylinder.
$(iii)$ The volume of water left in the cylinder $=$ (Total volume of the cylinder) $-$ (Volume of the submerged cone).
Given:
Height of the cone $(h_1)$ $= 120 \,cm$
Radius of the cone $(r)$ $= 60 \,cm$
Height of the cylinder $(h_2)$ $= 180 \,cm$
Radius of the cylinder $(r)$ $= 60 \,cm$
Volume of the cone $= \frac{1}{3} \pi r^2 h_1 = \frac{1}{3} \times \pi \times (60)^2 \times 120 = 144000 \pi \,cm^3$.
Volume of the cylinder $= \pi r^2 h_2 = \pi \times (60)^2 \times 180 = 648000 \pi \,cm^3$.
Since the cone is fully submerged,the volume of water displaced is equal to the volume of the cone,which is $144000 \pi \,cm^3$.
Volume of water left in the cylinder $= 648000 \pi - 144000 \pi = 504000 \pi \,cm^3$.
Using $\pi \approx \frac{22}{7}$:
Volume $= 504000 \times \frac{22}{7} = 72000 \times 22 = 1584000 \,cm^3$.
Converting to cubic meters $(1 \,m^3 = 10^6 \,cm^3)$:
Volume $= \frac{1584000}{1000000} \,m^3 = 1.584 \,m^3$.
Thus,the volume of water left in the cylinder is $1.584 \,m^3$.