A right circular cone of diameter r cm and height 12cm rests on the base of a right circular cylinder of radius r cm. Their bases are in the same plane and the cylinder is filled with water upto a height of 12cm. If the cone is then removed, find the to which the water level will fall.
adius of base of cone=r/2
radius of base ofcylinder=r
height of cone=12cm
h of water before cone ws tkn out=12cm
thus, vol of water left in cylinder aftr cone is tken out=vol of water - vol of cone
=pi x r^2 x 12 - 1/3x pi x(r/2)^2 x 12
=pi x r^2x11
u cn observe that 11 is present height (or "h") of the water(refer cylinder 's volume formula)
Let r and be the radius of base of the cylinder and cone respectively.
Height of the water in the cylinder = 12 cm
Height of the cone, h = 12 cm
Volume of water
= (12 π r 2 – π r 2) cm3
= 11 π r 2 cm3
Let the height of water in the cylinder be h' when the cone is removed.
Volume of water = 11 π r 2 cm3
∴ πr 2 h' = 11 π r 2
⇒ h' = 11
Thus, the level of water in the cylinder is 11 cm.
not even expert can say it wrong
base radius of cylinder be r/2
height of cone & cylinder is 12 cm
volume of water left = volume of water - volume of cone
= pi r^2*h - 1/3 pi r^2*h
= pi r^2*12 - 1/3 pi (r/2)^2 *12
= 12 pi r^2 - pi r^2
=11 pi r^2 cm^3
height of water in cylinder be h when the cone is removed
volume of water = 11 pi r^2 cm^3
thus, the level of water in the cylinder = 11 cm