A solid cone of base radius 9 cm and height 10 cm is lowered into a cylindrical jar of radius 10 cm, which contains water sufficient to submerge the cone completely. Find the rise in water level in the jar.

Radius of cone (r) = 9 cm.

Height of cone (h) = 10 cm.

Volume of water filled in cone = $\dfrac{1}{3}πr^2h$

Let h₁ be the rise in height of water in the jar.

Radius of jar (r₁) = 10.

Since, cone is submerged completely hence, volume of water level rise = volume of cone.

$\therefore πr$1^2h1 = \dfrac{1}{3}πr^2h \\[1em] \Rightarrow π(10)^2 \times h1 = \dfrac{1}{3}π(9)^2 \times 10 \\[1em] \Rightarrow h1 = \dfrac{\dfrac{1}{3}π \times 81 \times 10}{π \times (10)^2} \\[1em] \Rightarrow h1 = \dfrac{\cancel{π} \times \overset{27}{\cancel{81}} \times \cancel{10}}{\cancel{3} \times \cancel{π} \times \cancel{10} \times 10} \\[1em] \Rightarrow h1 = \dfrac{27}{10} \\[1em] \Rightarrow h_1 = 2.7 \text{ cm}.

Hence, the rise in height of water in jar is 2.7 cm.