Question

A cylindrical container of height 14 m and base diameter 12 m contains oil. This oil is to be transferred to one cylindrical can, one conical can and a spherical can. The base radius of all the containers is same. The height of the conical can is 6 m. While pouring some oil is dropped and hence only $\frac{3}{4}^{th}$ of cylindrical can could be filled. How much oil is dropped?

• A. 54 $\pi m^3$
• B. 36 $\pi m^3$
• C. 46 $\pi m^3$
• D. 50 $\pi m^3$

Answer 1

Answer: (B)

36 $\pi m^3$

Answer 2

Volume of oil = $\pi$ × 62 ×14 = 504$\pi$m3
Volume of conical In = $$$\frac{1}{3}$×$\pi$× (6)2 × 6 = 72$\pi$m3 Volume of spherical can =$\frac{4}{3}$×$\pi$(6)2 = 288π m³ Remaining oil = 504$\pi$(228$\pi$\+ 72$\pi$) = 144$\pi$m3 Volume of cylindrical can =$\pi$× (6)²x h According to question 144$\pi$=$\pi$× 36 × h h = 4m Now$\frac{3}{4}^{th}$of cylindrical can be filled. Oil dropped =$\frac{1}{4}$×$\pi$× (6)2 × 4 = 36$\pi$m3
So the correct option is (B)