Solid cylinder of brass 8 m high and 4 m diameter is melted and recast into a cone of diameter 3 m . Then, the height of cone is

Correct option is A

Given:

Height of cylinder = 8 m

Diameter of cylinder = 4 m => Radius = 2 m

Diameter of cone = 3 m => Radius = 1.5 m

Volume is conserved during melting and recasting

Formula Used:

Volume of cylinder = $$\pi r^2 h$$​​

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

Solution:

Volume of cylinder $$= \pi \times 2^2 \times 8 = \pi \times 4 \times 8 = 32\pi$$​​

Let height of cone be h: 

Volume of cone = $$\frac{1}{3} \pi \times (1.5)^2 \times h$$​​

​$$= \pi \times \frac{9}{4} \times \frac{1}{3} \times h$$​​

​$$= \frac{3\pi}{4} h$$​​

Now equate volumes:

​$$32\pi = \frac{3\pi}{4} h\\ \ \\ 32 = \frac{3}{4} h\\ \ \\ \Rightarrow h = \frac{32 \times 4}{3} = \frac{128}{3}= 42.66 \ m$$​​