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

The Correct Option is B

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)