Question

A cylindrical container of height 14 m and base 12 m contains oil. The 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 6m. While pouring some oil is dropped and hence only \(\frac34\) th of cylindrical can could be filled. How much oil is dropped ? 

(a) 54 p m³ 

(b) 36 p m³

(c) 46 p m³ 

(d) 50 p m³

Answer 1

(b) 36 π m³

Volume of oil = π × (6)² × 14 = 504 πm³ 

Volume of conical can = \(\frac13\) x π x (6)² x 6 72 πm³

Volume of spherical can = \(\frac43\) x π x (6)³ = 288π m³ 

Remaining oil = Vol. of oil – Vol. of oil in (conical can + spherical can) 

= 504 π – (72π + 288π) = 144 πm³ 

Volume of the cylindrical can = π × (6)² × h = 144 π 

⇒ h = 4 m. 

As only \(\frac34\) th  of the cylindrical can could be filled, Vol. of oil dropped = \(\frac14\) x π (6)² x 4

= 36 π m³.