Question

A cylindrical tub of radius 12 cm contains water up 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. The radius of the ball is

• A. 6 cm
• B. 4.5 cm
• C. 7.25 cm
• D. 9 cm

Answer 1

The Correct Option is D

Given radius of the cylinder, r = 12 cm
It is also given that a spherical iron ball is dropped into the cylinder and the water level raised by 6.75 cm
Hence volume of water displaced = volume of the iron ball
Height of the raised water level, h = 6.75 m
Volume of water displaced \(=\pi r^{2h}\)

\(=\pi\times12\times12\times6.75cm^3\)

So, Volume of iron ball \(=\pi\times12\times12\times6.75cm^3\) ... (1)

But, volume of iron ball \(=\frac{4}{3\pi r^3}\) ...(2)
From (1) and (2) we get

\(\frac{4}{3\pi r^3}=\pi\times12\times12\times\times6.75\)

On solving, we get

\(r^3=729\)

\(r^3=3^6\) Therefore, r' = 9

Radius of the iron ball is 9 cm.
 
So the correct option is (D)