Home » Aptitude » Volume and Surface Area

A cylindrical tub of radius 12 cm contains water to a depth of 20 cm. A spherical iron ball is dropped into the tub and thus the level of water is raised by 6.75 cm. What is the radius of the ball?

Correct Answer: 9 cm

Explanation:

Volume of the spherical ball = Volume of the water displaced
⇒ 4/3 πr3 = π(12)2 x 6.75
⇒ r3 = 144 x 6.75 x 3/4 = 729
∴ r = 9 cm


← Previous Question Next Question→

More Questions from Volume and Surface Area

Discussion & Comments

No comments yet. Be the first to comment!
Join Discussion