A conical vessel whose internal base radius is 18 cm and height 60 cm is full of a liquid. The contents ofthe vessel are emptied into a cylindrical vessel with base of internal radius 15 cm. The height (in cm) to which the liquid will rise in the cylindrical vessel is: 

Correct option is B

Given :

Conical vessel radius $$r_1 =$$​18 cm

Height of cone $$h_1$$​ = 60 cm

Cylindrical vessel radius $$r_2 =$$​ 15 cm

Formula Used :

Volume of cone = $$\dfrac{1}{3}\pi r^2 h$$​​

Volume of cylinder =$$\pi r^2 h$$​

Solution :

Let height of liquid in cylinder = h cm

Since the liquid is transferred completely,
Volume of cone = Volume of cylinder

​$$\dfrac{1}{3}\pi (18)^2 \times 60 = \pi (15)^2 \times h$$​​

​$$\dfrac{1}{3} \times 324 \times 60 = 225h$$​​

​$$\frac{19440}{3} = 225$$​h

6480 = 225h

h =$$\frac{6480}{225}$$​= 28.8 cm